problem_id int64 1 978 | question stringlengths 86 2.11k | source stringlengths 19 76 | solution stringlengths 94 14.7k | asymptote_code stringlengths 44 17.8k | solution_image_url stringclasses 16
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80 | Three of the edges of a cube are $\overline{AB}, \overline{BC},$ and $\overline{CD},$ and $\overline{AD}$ is an interior diagonal. Points $P, Q,$ and $R$ are on $\overline{AB}, \overline{BC},$ and $\overline{CD},$ respectively, so that $AP = 5, PB = 15, BQ = 15,$ and $CR = 10.$ What is the area of the polygon that is the intersection of plane $PQR$ and the cube?
Contents
1 Problem
2 Solution 1
3 Solution 2
4 See also | 1998 AIME Problem 11 | "For non-asymptote version of image, see Image:1998_AIME-11.png"
This approach uses analytic geometry. Let $A$ be at the origin, $B$ at $(20,0,0)$, $C$ at $(20,0,20)$, and $D$ at $(20,20,20)$. Thus, $P$ is at $(5,0,0)$, $Q$ is at $(20,0,15)$, and $R$ is at $(20,10,20)$.
Let the plane $PQR$ have the equation $ax + by + cz = d$. Using point $P$, we get that $5a = d$. Using point $Q$, we get \[20a + 15c = d \Longrightarrow 4d + 15c = d \Longrightarrow d = -5c\]Using point $R$, we get \[20a + 10b + 20c = d \Longrightarrow 4d + 10b - 4d = d \Longrightarrow d = 10b\]Thus plane $PQR$’s equation reduces to \[\frac{d}{5}x + \frac{d}{10}y - \frac{d}{5}z = d \Longrightarrow 2x + y - 2z = 10\]
We know need to find the intersection of this plane with that of $z = 0$, $z = 20$, $x = 0$, and $y = 20$. After doing a little bit of algebra, the intersections are the lines \[y = -2x + 10\] \[y = -2x + 50\] \[y = 2z + 10\] \[z = x + 5\]Thus, there are three more vertices on the polygon, which are at $(0,10,0)(0,20,5)(15,20,20)$.
We can find the lengths of the sides of the polygons now. There are 4 right triangles with legs of length 5 and 10, so their hypotenuses are $5\sqrt{5}$. The other two are of $45-45-90 \triangle$s with legs of length 15, so their hypotenuses are $15\sqrt{2}$. So we have a hexagon with sides $15\sqrt{2},5\sqrt{5}, 5\sqrt{5},15\sqrt{2}, 5\sqrt{5},5\sqrt{5}$ By symmetry, we know that opposite angles of the polygon are congruent. We can also calculate the length of the long diagonal by noting that it is of the same length of a face diagonal, making it $20\sqrt{2}$.
The height of the triangles at the top/bottom is $\frac{20\sqrt{2} - 15\sqrt{2}}{2} = \frac{5}{2}\sqrt{2}$. The Pythagorean Theorem gives that half of the base of the triangles is $\frac{15}{\sqrt{2}}$. We find that the middle rectangle is actually a square, so the total area is $(15\sqrt{2})^2 + 4\left(\frac 12\right)\left(\frac 52\sqrt{2}\right)\left(\frac{15}{\sqrt{2}}\right) = 525$. | // Block 1
import three;
size(280); defaultpen(linewidth(0.6)+fontsize(9));
currentprojection=perspective(30,-60,40);
triple A=(0,0,0),B=(20,0,0),C=(20,0,20),D=(20,20,20);
triple P=(5,0,0),Q=(20,0,15),R=(20,10,20),Pa=(15,20,20),Qa=(0,20,5),Ra=(0,10,0);
draw(box((0,0,0),(20,20,20)));
draw(P--Q--R--Pa--Qa--Ra--cycle,linewidth(0.7));
label("\(A\,(0,0,0)\)",A,SW);
label("\(B\,(20,0,0)\)",B,S);
label("\(C\,(20,0,20)\)",C,SW);
label("\(D\,(20,20,20)\)",D,E);
label("\(P\,(5,0,0)\)",P,SW);
label("\(Q\,(20,0,15)\)",Q,E);
label("\(R\,(20,10,20)\)",R,E);
label("\((15,20,20)\)",Pa,N);
label("\((0,20,5)\)",Qa,W);
label("\((0,10,0)\)",Ra,W);
// Block 2
size(190);
pointpen=black;pathpen=black;
real s=2^.5;
pair P=(0,0),Q=(7.5*s,2.5*s),R=Q+(0,15*s),Pa=(0,20*s),Qa=(-Q.x,Q.y),Ra=(-R.x,R.y);
D(P--Q--R--Pa--Ra--Qa--cycle);D(R--Ra);D(Q--Qa);D(P--Pa);
MP("15\sqrt{2}",(Q+R)/2,E);
MP("5\sqrt{5}",(P+Q)/2,SE);
MP("5\sqrt{5}",(R+Pa)/2,NE);
MP("20\sqrt{2}",(P+Pa)/2,W);
// Block 3
import three; size(280); defaultpen(linewidth(0.6)+fontsize(9)); currentprojection=perspective(30,-60,40); triple A=(0,0,0),B=(20,0,0),C=(20,0,20),D=(20,20,20); triple P=(5,0,0),Q=(20,0,15),R=(20,10,20),Pa=(15,20,20),Qa=(0,20,5),Ra=(0,10,0); draw(box((0,0,0),(20,20,20))); draw(P--Q--R--Pa--Qa--Ra--cycle,linewidth(0.7)); label("\(A\,(0,0,0)\)",A,SW); label("\(B\,(20,0,0)\)",B,S); label("\(C\,(20,0,20)\)",C,SW); label("\(D\,(20,20,20)\)",D,E); label("\(P\,(5,0,0)\)",P,SW); label("\(Q\,(20,0,15)\)",Q,E); label("\(R\,(20,10,20)\)",R,E); label("\((15,20,20)\)",Pa,N); label("\((0,20,5)\)",Qa,W); label("\((0,10,0)\)",Ra,W);
// Block 4
size(190); pointpen=black;pathpen=black; real s=2^.5; pair P=(0,0),Q=(7.5*s,2.5*s),R=Q+(0,15*s),Pa=(0,20*s),Qa=(-Q.x,Q.y),Ra=(-R.x,R.y); D(P--Q--R--Pa--Ra--Qa--cycle);D(R--Ra);D(Q--Qa);D(P--Pa); MP("15\sqrt{2}",(Q+R)/2,E); MP("5\sqrt{5}",(P+Q)/2,SE); MP("5\sqrt{5}",(R+Pa)/2,NE); MP("20\sqrt{2}",(P+Pa)/2,W); | [] |
81 | Given that $\sum_{k=1}^{35}\sin 5k=\tan \frac mn,$ where angles are measured in degrees, and $m_{}$ and $n_{}$ are relatively prime positive integers that satisfy $\frac mn<90,$ find $m+n.$ | 1999 AIME Problem 11 | This solution will work entirely in degrees. Let
\[z = e^{5^\circ i}.\]
For each integer $k$, consider the unit vector in the complex plane
\[v_k = z^k = e^{5k^\circ i}.\]
Geometrically, $v_k$ is a unit vector making an angle of $5k ^\circ$ with the positive real axis. Now draw each vector $v_k$ for $1 \leq k \leq 35$ tip-to-tail on the complex plane. Begin at $O$, the origin, and draw the vector $v_1$. Then starting at where $v_1$ ended, draw $v_2$. Continue until you draw $v_{35}$, and the sum of these $35$ vectors will end at $B$, as seen in the diagram below. Because each successive vector is obtained by rotating the previous vector by $5^\circ$, we have a portion of a regular polygon with an exterior angle of $5$ degrees, so it has $\frac{360^\circ}{5^\circ} = 72$ sides.
Additionally, notice that $B$ is directly above $A$. This is because the vector sum will have its horizontal components completely cancel out. We realize that $\cos{5} + \cos{175} = 0$, and we can form a similar pair for each of the other complex numbers, except for $v_18 = i$.
The above diagram starts at the origin $O$ and draws the $35$ vectors.
Note that
\[\text{Im}(v_k) = \sin{5k}.\] Therefore, $\sum_{k=1}^{35}\sin 5k$ is simply the y-coordinate of the vector sum $v_1 + v_2 + \dots v_{35}$. In the diagram, this is the distance of the vertical blue line connecting $O$ to the topmost point.
Let $A$ be the vertex of the $72$-gon that is directly to the left of $O$ and $B$ be the resulting vertex of summing the $35$ vectors. Looking at right triangle $AOB$, we notice that $AO=1$. Remember that our desired sum equals $BO = \tan{\angle BAO}.$ We now just need to find $\angle BAO$.
Notice that $BA$ bisects $A$'s interior angle, so $\angle BAO$ is half of the interior angle of the $72$-gon. So $\angle BAO = \frac{175}{2}$.
Recall that our sum equals $\tan{\frac{m}{n}} = \tan{\frac{175}{2}}$, so our answer is $175 + 2 = \boxed{177}$.
~lprado | // Block 1
pathpen = black + linewidth(0.5); pointpen = black;
unitsize(0.4cm);
pair O=(0,0);
real s=0.9;
pair[] P;
P[0]=O;
for(int k=1;k<=35;++k) P[k]=P[k-1]+s*dir(5*k);
for(int k=1;k<=35;++k) draw(P[k-1]--P[k],pathpen,Arrow(TeXHead));
int it=0;
for(int k=1;k<=35;++k) if(P[k].y>P[it].y) it=k;
pair B=P[it];
pair A=O+s*dir(180);
draw(O--B,blue+linewidth(0.8));
draw(A--O,red+linewidth(0.8));
draw(A--B,red+linewidth(0.8));
dot(O,pointpen); dot(A,pointpen); dot(B,pointpen);
label("$O$",O,S);
label("$A$",A,W);
label("$B$",B,N);
// Block 2
pathpen = black + linewidth(0.5); pointpen = black; unitsize(0.4cm); pair O=(0,0); real s=0.9; pair[] P; P[0]=O; for(int k=1;k<=35;++k) P[k]=P[k-1]+s*dir(5*k); for(int k=1;k<=35;++k) draw(P[k-1]--P[k],pathpen,Arrow(TeXHead)); int it=0; for(int k=1;k<=35;++k) if(P[k].y>P[it].y) it=k; pair B=P[it]; pair A=O+s*dir(180); draw(O--B,blue+linewidth(0.8)); draw(A--O,red+linewidth(0.8)); draw(A--B,red+linewidth(0.8)); dot(O,pointpen); dot(A,pointpen); dot(B,pointpen); label("$O$",O,S); label("$A$",A,W); label("$B$",B,N); | [] |
82 | The inscribed circle of triangle $ABC$ is tangent to $\overline{AB}$ at $P_{},$ and its radius is $21$. Given that $AP=23$ and $PB=27,$ find the perimeter of the triangle.
Contents
1 Problem
2 Solution
2.1 Solution 1
2.2 Solution 2
3 Solution 3
4 Solution 4 (Trig Bash)
5 See also | 1999 AIME Problem 12 | Solution 1
Let $Q$ be the tangency point on $\overline{AC}$, and $R$ on $\overline{BC}$. By the Two Tangent Theorem, $AP = AQ = 23$, $BP = BR = 27$, and $CQ = CR = x$. Using $rs = A$, where $s = \frac{27 \cdot 2 + 23 \cdot 2 + x \cdot 2}{2} = 50 + x$, we get $(21)(50 + x) = A$. By Heron's formula, $A = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{(50+x)(x)(23)(27)}$. Equating and squaring both sides,
\begin{eqnarray*} [21(50+x)]^2 &=& (50+x)(x)(621)\\ 441(50+x) &=& 621x\\ 180x = 441 \cdot 50 &\Longrightarrow & x = \frac{245}{2} \end{eqnarray*}
We want the perimeter, which is $2s = 2\left(50 + \frac{245}{2}\right) = \boxed{345}$.
Solution 2
Let the incenter be denoted $I$. It is commonly known that the incenter is the intersection of the angle bisectors of a triangle. So let $\angle ABI = \angle CBI = \alpha, \angle BAI = \angle CAI = \beta,$ and $\angle BCI = \angle ACI = \gamma.$
We have that
\begin{eqnarray*} \tan \alpha & = & \frac {21}{27} \\ \tan \beta & = & \frac {21}{23} \\ \tan \gamma & = & \frac {21}x. \end{eqnarray*}
So naturally we look at $\tan \gamma.$ But since $\gamma = \frac \pi2 - (\beta + \alpha)$ we have
\begin{eqnarray*} \tan \gamma & = & \tan\left(\frac \pi2 - (\beta + \alpha)\right) \\ & = & \frac 1{\tan(\alpha + \beta)} \\ \Rightarrow \frac {21}x & = & \frac {1 - \frac {21\cdot 21}{23\cdot 27}}{\frac {21}{27} + \frac {21}{23}} \end{eqnarray*}
Doing the algebra, we get $x = \frac {245}2.$
The perimeter is therefore $2\cdot\frac {245}2 + 2\cdot 23 + 2\cdot 27 = \boxed{345}.$ | // Block 1
pathpen = black + linewidth(0.65); pointpen = black;
pair A=(0,0),B=(50,0),C=IP(circle(A,23+245/2),circle(B,27+245/2)), I=incenter(A,B,C);
path P = incircle(A,B,C);
D(MP("A",A)--MP("B",B)--MP("C",C,N)--cycle);D(P);
D(MP("P",IP(A--B,P)));
pair Q=IP(C--A,P),R=IP(B--C,P);
D(MP("R",R,NE));D(MP("Q",Q,NW));
MP("23",(A+Q)/2,W);MP("27",(B+R)/2,E);
// Block 2
pathpen = black + linewidth(0.65); pointpen = black; pair A=(0,0),B=(50,0),C=IP(circle(A,23+245/2),circle(B,27+245/2)), I=incenter(A,B,C); path P = incircle(A,B,C); D(MP("A",A)--MP("B",B)--MP("C",C,N)--cycle);D(P); D(MP("P",IP(A--B,P))); pair Q=IP(C--A,P),R=IP(B--C,P); D(MP("R",R,NE));D(MP("Q",Q,NW)); MP("23",(A+Q)/2,W);MP("27",(B+R)/2,E); | [] |
82 | The inscribed circle of triangle $ABC$ is tangent to $\overline{AB}$ at $P_{},$ and its radius is $21$. Given that $AP=23$ and $PB=27,$ find the perimeter of the triangle.
Contents
1 Problem
2 Solution
2.1 Solution 1
2.2 Solution 2
3 Solution 3
4 Solution 4 (Trig Bash)
5 See also | 1999 AIME Problem 12 | Let \(CR = CQ = x\). Using the inradius formula for the area of a triangle, we have
\[[ABC] = 1050 + 21x,\]
where \([ABC]\) denotes the area of \(\triangle ABC\).
Next, we find \(\sin \angle CBA\). Note that
\[\angle CBA = \angle RBP = 2\angle OBP.\]
First, we find the hypotenuse \(OB\) using the Pythagorean Theorem:
\[OB^2 = 27^2 + 21^2 = 1170 \quad \Rightarrow \quad OB = 3\sqrt{130}.\]
Let \(\angle OBP = \theta\). Then
\[\sin \theta = \frac{21}{3\sqrt{130}}, \qquad \cos \theta = \frac{27}{3\sqrt{130}}.\]
Using the double-angle identity,
\[\sin(2\theta) = 2\sin \theta \cos \theta= 2\left(\frac{21}{3\sqrt{130}}\right)\left(\frac{27}{3\sqrt{130}}\right)= \frac{1134}{1170}= \frac{63}{65}.\]
Now we compute the area of \(\triangle ABC\) using the formula \(\tfrac{1}{2}ab\sin C\):
\[[ABC] = \frac{1}{2}(50)(27+x)\left(\frac{63}{65}\right)= \frac{315}{13}(27+x).\]
Equating the two expressions for the area,
\[1050 + 21x = \frac{315}{13}(27+x).\]
Solving,
\[\frac{13}{15}x + \frac{130}{3} = 27 + x,\]\[\frac{49}{3} = \frac{2}{15}x,\]
so
\[x = \frac{245}{2}.\]
Finally, we compute the perimeter of \(\triangle ABC\):
\[2(23+27) + 2\left(\frac{245}{2}\right) = \boxed{345}.\]
~Voidling | // Block 1
import olympiad;
pathpen = black + linewidth(0.65); pointpen = black;
pair A=(0,0), B=(50,0);
pair C=IP(circle(A,23+245/2),circle(B,27+245/2));
pair O = incenter(A,B,C);
path P_circle = incircle(A,B,C);
pair P = foot(O, A, B);
pair Q = foot(O, A, C);
pair R = foot(O, B, C);
D(MP("A",A, SW)--MP("B",B, SE)--MP("C",C,N)--cycle);
D(P_circle);
D(O--B, dashed + blue);
D(O--P, blue);
dot(MP("O", O, NW));
D(MP("P", P, S));
D(MP("R", R, NE));
D(MP("Q", Q, NW));
MP("23", (A+P)/2, S);
MP("27", (B+P)/2, S);
MP("23", (A+Q)/2, W);
MP("27", (B+R)/2, E);
MP("x", (C+R)/2, NE);
MP("x", (C+Q)/2, NW);
D(rightanglemark(O,P,B, 1.5));
// Block 2
import olympiad; pathpen = black + linewidth(0.65); pointpen = black; pair A=(0,0), B=(50,0); pair C=IP(circle(A,23+245/2),circle(B,27+245/2)); pair O = incenter(A,B,C); path P_circle = incircle(A,B,C); pair P = foot(O, A, B); pair Q = foot(O, A, C); pair R = foot(O, B, C); D(MP("A",A, SW)--MP("B",B, SE)--MP("C",C,N)--cycle); D(P_circle); D(O--B, dashed + blue); D(O--P, blue); dot(MP("O", O, NW)); D(MP("P", P, S)); D(MP("R", R, NE)); D(MP("Q", Q, NW)); MP("23", (A+P)/2, S); MP("27", (B+P)/2, S); MP("23", (A+Q)/2, W); MP("27", (B+R)/2, E); MP("x", (C+R)/2, NE); MP("x", (C+Q)/2, NW); D(rightanglemark(O,P,B, 1.5)); | [] |
83 | Point $P_{}$ is located inside triangle $ABC$ so that angles $PAB, PBC,$ and $PCA$ are all congruent. The sides of the triangle have lengths $AB=13, BC=14,$ and $CA=15,$ and the tangent of angle $PAB$ is $m/n,$ where $m_{}$ and $n_{}$ are relatively prime positive integers. Find $m+n.$
Contents
1 Problem
2 Solution
2.1 Solution 1
2.2 Solution 2
2.3 Solution 3
2.4 Solution 4 (Law of sines)
3 Solution 5 (Coordinate Bash)
4 See also | 1999 AIME Problem 14 | Solution 1
Drop perpendiculars from $P$ to the three sides of $\triangle ABC$ and let them meet $\overline{AB}, \overline{BC},$ and $\overline{CA}$ at $D, E,$ and $F$ respectively.
Let $BE = x, CF = y,$ and $AD = z$. We have that
\begin{align*}DP&=z\tan\theta\\ EP&=x\tan\theta\\ FP&=y\tan\theta\end{align*}
We can then use the tool of calculating area in two ways
\begin{align*}[ABC]&=[PAB]+[PBC]+[PCA]\\ &=\frac{1}{2}(13)(z\tan\theta)+\frac{1}{2}(14)(x\tan\theta)+\frac{1}{2}(15)(y\tan\theta)\\ &=\frac{1}{2}\tan\theta(13z+14x+15y)\end{align*}
On the other hand,
\begin{align*}[ABC]&=\sqrt{s(s-a)(s-b)(s-c)}\\ &=\sqrt{21\cdot6\cdot7\cdot8}\\ &=84\end{align*}
We still need $13z+14x+15y$ though. We have all these right triangles and we haven't even touched Pythagoras. So we give it a shot:
\begin{align}x^2+x^2\tan^2\theta&=z^2\tan^2\theta+(13-z)^2\\ z^2+z^2\tan^2\theta&=y^2\tan^2\theta+(15-y)^2\\ y^2+y^2\tan^2\theta&=x^2\tan^2\theta+(14-x)^2\end{align}
Adding $(1) + (2) + (3)$ gives
\begin{align*}x^2+y^2+z^2&=(14-x)^2+(15-y)^2+(13-z)^2\\ \Rightarrow13z+14x+15y&=295\end{align*}
Recall that we found that $[ABC]=\frac{1}{2}\tan\theta(13z+14x+15y)=84$. Plugging in $13z+14x+15y=295$, we get $\tan\theta=\frac{168}{295}$, giving us $\boxed{463}$ for an answer.
Solution 2
Let $AB=c$, $BC=a$, $AC=b$, $PA=x$, $PB=y$, and $PC=z$.
So by the Law of Cosines, we have:
\begin{align*}x^2 &= z^2 + b^2 - 2bz\cos{\theta}\\ y^2 &= x^2 + c^2 - 2cx\cos{\theta}\\ z^2 &= y^2 + a^2 - 2ay\cos{\theta}\end{align*}
Adding these equations and rearranging, we have:
\[a^2 + b^2 + c^2 = (2bz + 2cx + 2ay)\cos{\theta}\qquad(1)\]
Now $[CAP] + [ABP] + [BCP] = [ABC] = \sqrt {(21)(8)(7)(6)} = 84$, by Heron's formula.
Now the area of a triangle, $[A] = \frac {mn\sin{\beta}}{2}$, where $m$ and $n$ are sides on either side of an angle, $\beta$. So,
\begin{align*}[CAP] &= \frac {bz\sin{\theta}}{2}\\ [ABP] &= \frac {cx\sin{\theta}}{2}\\ [BCP] &= \frac {ay\sin{\theta}}{2}\end{align*}
Adding these equations yields:
\begin{align*}[ABC]= 84 &= \frac {(bz + cx + ay)\sin{\theta}}{2}\\ \Rightarrow 168&= (bz + cx + ay)\sin{\theta}\qquad (2)\end{align*}
Dividing $(2)$ by $(1)$, we have:
\begin{align*}\frac {168}{a^2 + b^2 + c^2} &= \frac {(bz + cx + ay)\sin{\theta}}{(2bz + 2cx + 2ay)\cos{\theta}}\\ \Rightarrow \tan{\theta} = \frac {336}{a^2 + b^2 + c^2} &= \frac {336}{14^2 + 15^2 + 13^2} = \frac {336}{590} = \frac {168}{295}\end{align*}
Thus, $m + n = 168 + 295 = \boxed{463}$.
Note: In fact, this problem is unfairly easy to those who happen to have learned about Brocard point. The Brocard Angle is given by
\[cot(\theta)=\frac{a^2+b^2+c^2}{4\Delta}\]
Solution 3
Let $\angle{PAB} = \angle{PBC} = \angle{PCA} = x.$ Then, using Law of Cosines on the three triangles containing vertex $P,$ we have
\begin{align*} b^2 &= a^2 + 169 - 26a \cos x \\ c^2 &= b^2 + 196 - 28b \cos x \\ a^2 &= c^2 + 225 - 30c \cos x. \end{align*}
Add the three equations up and rearrange to obtain \[(13a + 14b + 15c) \cos x = 295.\] Also, using $[ABC] = \frac{1}{2}ab \sin \angle C$ we have \[[ABC] = [APB] + [BPC] + [CPA] = \dfrac{\sin x}{2}(13a + 14b + 15c) = 84 \iff (13a + 14b + 15c) \sin x = 168.\] Divide the two equations to obtain $\tan x = \frac{168}{295} \iff \boxed{463}.~\square$
Solution 4 (Law of sines)
Firstly, denote angles $ABC$, $BCA$, and $CAB$ as $B$, $A$, and $C$ respectively. Let $\angle{PAB}=x$.
Notice that by angle chasing that $\angle{BPC}=180-C$ and $\angle{BPA}=180-B$.
Using the nice properties of the 13-14-15 triangle, we have $\sin B = \frac{12}{13}$ and $\sin C = \frac{4}{5}$. $\cos C$ is easily computed, so we have $\cos C=\frac{3}{5}$.
Using Law of Sines,
\begin{align*} \frac{BP}{\sin x} &= \frac{13}{\sin (180 - B)} \\ \frac{BP}{\sin (C - x)} &= \frac{14}{\sin (180 - C)} \end{align*}
hence,
\begin{align*} \frac{BP}{\sin x} &= \frac{13}{\sin B} \\ \frac{BP}{\sin (C - x)} &= \frac{14}{\sin C} \end{align*}
Now, computation carries the rest.
\begin{align*} \frac{13 \sin x}{\sin B} &= \frac{14 \sin (C-x)}{\sin C} \\ \frac{169 \sin x}{12} &= \frac{210 \sin (C-x)}{12} \\ 169 \sin x &= 210 (\sin C \cos x - \cos C \sin x) \\ 169 \sin x &= 210 (\frac{4}{5} \cos x - \frac{3}{5} \sin x) \\ 169 \sin x &= 168 \cos x - 126 \sin x \\ 295 \sin x &= 168 \cos x \\ \tan x &= \frac{168}{295} \end{align*}
Extracting yields $168 + 295 = \boxed{463}$. | // Block 1
real theta = 29.66115; /* arctan(168/295) to five decimal places .. don't know other ways to construct Brocard */
pathpen = black +linewidth(0.65); pointpen = black;
pair A=(0,0),B=(13,0),C=IP(circle(A,15),circle(B,14));
D(MP("A",A)--MP("B",B)--MP("C",C,N)--cycle);
/* constructing P, C is there as check */
pair Aa=A+(B-A)*dir(theta),Ba=B+(C-B)*dir(theta),Ca=C+(A-C)*dir(theta), P=IP(A--Aa,B--Ba);
D(A--MP("P",P,NW)--B);D(P--C);
D(anglemark(B,A,P,30));D(anglemark(C,B,P,30));D(anglemark(A,C,P,30));
MP("13",(A+B)/2,S);MP("15",(A+C)/2,NW);MP("14",(C+B)/2,NE);
// Block 2
real theta = 29.66115; /* arctan(168/295) to five decimal places .. don't know other ways to construct Brocard */ pathpen = black +linewidth(0.65); pointpen = black; pair A=(0,0),B=(13,0),C=IP(circle(A,15),circle(B,14)); D(MP("A",A)--MP("B",B)--MP("C",C,N)--cycle); /* constructing P, C is there as check */ pair Aa=A+(B-A)*dir(theta),Ba=B+(C-B)*dir(theta),Ca=C+(A-C)*dir(theta), P=IP(A--Aa,B--Ba); D(A--MP("P",P,NW)--B);D(P--C); D(anglemark(B,A,P,30));D(anglemark(C,B,P,30));D(anglemark(A,C,P,30)); MP("13",(A+B)/2,S);MP("15",(A+C)/2,NW);MP("14",(C+B)/2,NE);
// Block 3
import olympiad; real theta = 29.66115; /* arctan(168/295) to five decimal places .. don't know other ways to construct Brocard */ pathpen = black +linewidth(0.65); pointpen = black; pair A=(0,0),B=(13,0),C=IP(circle(A,15),circle(B,14)); D(MP("A",A)--MP("B",B)--MP("C",C,N)--cycle); /* constructing P, C is there as check */ pair Aa=A+(B-A)*dir(theta),Ba=B+(C-B)*dir(theta),Ca=C+(A-C)*dir(theta), P=IP(A--Aa,B--Ba); D(A--MP("P",P,SSW)--B);D(P--C); D(anglemark(B,A,P,30));D(anglemark(C,B,P,30));D(anglemark(A,C,P,30)); MP("13",(A+B)/2,S);MP("15",(A+C)/2,NW);MP("14",(C+B)/2,NE); /* constructing D,E,F as foot of perps from P */ pair D=foot(P,A,B),E=foot(P,B,C),F=foot(P,C,A); D(MP("D",D,NE)--P--MP("E",E,SSW),dashed);D(P--MP("F",F),dashed); D(rightanglemark(P,E,C,15));D(rightanglemark(P,F,C,15));D(rightanglemark(P,D,A,15)); | [] |
83 | Point $P_{}$ is located inside triangle $ABC$ so that angles $PAB, PBC,$ and $PCA$ are all congruent. The sides of the triangle have lengths $AB=13, BC=14,$ and $CA=15,$ and the tangent of angle $PAB$ is $m/n,$ where $m_{}$ and $n_{}$ are relatively prime positive integers. Find $m+n.$
Contents
1 Problem
2 Solution
2.1 Solution 1
2.2 Solution 2
2.3 Solution 3
2.4 Solution 4 (Law of sines)
3 Solution 5 (Coordinate Bash)
4 See also | 1999 AIME Problem 14 | Plot $\triangle ABC$ on the coordinate plane with \(B=(0,0), C=(14,0), A=(5,12).\)
Let $\angle PAB=\angle PBC=\angle PCA=a$, and let $P=(x,y)$. For convenience, define $\tan a = k$.
Since $\angle PBC=a$, the slope of $PB$ is $k$, so \(PB:\ y=kx.\)
Let $\angle ADB$ be the angle that line $AB$ makes with the $x$-axis. Since $\tan(\angle ADB)=\frac{12}{5}$, the slope of $PA$ is
\[\tan(\angle ADB+a)=\frac{\frac{12}{5}+k}{1-\frac{12}{5}k}=\frac{12+5k}{5-12k}.\]
Thus the equation of $PA$ is
\[y-12=\frac{12+5k}{5-12k}(x-5),\]
which simplifies to
\[PA:\ y=\frac{12x+5kx-169k}{5-12k}.\]
Similarly, since $\tan(\angle ECA)=-\frac{4}{3}$, the slope of $PC$ is
\[\tan(\angle ECA+a)=\frac{-\frac{4}{3}+k}{1+\frac{4}{3}k}=\frac{3k-4}{4k+3}.\]
Thus the equation of $PC$ is
\[PC:\ y=\frac{3k-4}{4k+3}(x-14),\]
or equivalently,
\[PC:\ y=\frac{3kx-4x-42k+56}{4k+3}.\]
Setting the equations of $PB$ and $PA$ equal gives
\[\frac{12x+5kx-169k}{5-12k}=kx.\]
Clearing denominators yields
\[12x+5kx-169k=5kx-12k^2x,\]
which simplifies to
\[12x-169k=-12k^2x.\]
Next, setting the equations of $PB$ and $PC$ equal gives
\[\frac{3kx-4x-42k+56}{4k+3}=kx.\]
Clearing denominators yields
\[3kx-4x-42k+56=4k^2x+3kx,\]
or
\[-4x-42k+56=4k^2x.\]
Multiplying by $-3$ gives
\[12x+126k-168=-12k^2x.\]
Equating the right-hand sides of the two derived equations gives
\[12x+126k-168=12x-169k,\]
so \(295k=168,\) and hence \(k=\frac{168}{295}.\)
Because the problem asks for $m+n$, where $k=\frac{m}{n}$, the final answer is \(168+295=\boxed{463}.\)
~Voidling | // Block 1
import olympiad;
real theta = 29.66115; /* Brocard angle */
pathpen = black + linewidth(0.65);
pointpen = black;
pair B=(0,0), C=(14,0), A=(5,12);
pair R1=(0,12), R2=(14,12); // Corners of the bounding rectangle
D(B--R1--R2--C, dashed);
D(A--B--C--cycle);
pair Bb = B + (C-B)*dir(theta);
pair Cc = C + (A-C)*dir(theta);
pair P = IP(B--Bb, C--Cc);
D(A--P--B);
D(P--C);
D(anglemark(B,A,P,20));
D(anglemark(C,B,P,20));
D(anglemark(A,C,P,20));
label("$A$", A, N);
label("$B$", B, W);
label("$C$", C, E);
label("$P$", P, NW);
label("$D$", R1, NW);
label("$E$", R2, NW);
label("$(5,12)$", A, 5*N, fontsize(8pt));
label("$(0,0)$", B, 3.5*S, fontsize(8pt));
label("$(14,0)$", C, 3.5*S, fontsize(8pt));
label("$(x,y)$", P, 3.5*S, fontsize(8pt));
label("$13$", (A+B)/2, NW);
label("$15$", (A+C)/2, NE);
label("$14$", (C+B)/2, S);
// Block 2
import olympiad; real theta = 29.66115; /* Brocard angle */ pathpen = black + linewidth(0.65); pointpen = black; pair B=(0,0), C=(14,0), A=(5,12); pair R1=(0,12), R2=(14,12); // Corners of the bounding rectangle D(B--R1--R2--C, dashed); D(A--B--C--cycle); pair Bb = B + (C-B)*dir(theta); pair Cc = C + (A-C)*dir(theta); pair P = IP(B--Bb, C--Cc); D(A--P--B); D(P--C); D(anglemark(B,A,P,20)); D(anglemark(C,B,P,20)); D(anglemark(A,C,P,20)); label("$A$", A, N); label("$B$", B, W); label("$C$", C, E); label("$P$", P, NW); label("$D$", R1, NW); label("$E$", R2, NW); label("$(5,12)$", A, 5*N, fontsize(8pt)); label("$(0,0)$", B, 3.5*S, fontsize(8pt)); label("$(14,0)$", C, 3.5*S, fontsize(8pt)); label("$(x,y)$", P, 3.5*S, fontsize(8pt)); label("$13$", (A+B)/2, NW); label("$15$", (A+C)/2, NE); label("$14$", (C+B)/2, S); | [] |
84 | Consider the paper triangle whose vertices are $(0,0), (34,0),$ and $(16,24).$ The vertices of its midpoint triangle are the midpoints of its sides. A triangular pyramid is formed by folding the triangle along the sides of its midpoint triangle. What is the volume of this pyramid? | 1999 AIME Problem 15 | As shown in the image above, let $D$, $E$, and $F$ be the midpoints of $\overline{BC}$, $\overline{CA}$, and $\overline{AB}$, respectively. Suppose $P$ is the apex of the tetrahedron, and let $O$ be the foot of the altitude from $P$ to $\triangle ABC$. The crux of this problem is the following lemma.
Lemma: The point $O$ is the orthocenter of $\triangle ABC$.
Proof. Observe that \[OF^2 - OE^2 = PF^2 - PE^2 = AF^2 - AE^2;\] the first equality follows by the Pythagorean Theorem, while the second follows from $AF = FP$ and $AE = EP$. Thus, by the Perpendicularity Lemma, $AO$ is perpendicular to $FE$ and hence $BC$. Analogously, $O$ lies on the $B$-altitude and $C$-altitude of $\triangle ABC$, and so $O$ is, indeed, the orthocenter of $\triangle ABC$.
To find the coordinates of $O$, we need to find the intersection point of altitudes $BE$ and $AD$. The equation of $BE$ is simply $x=16$. $AD$ is perpendicular to line $BC$, so the slope of $AD$ is equal to the negative reciprocal of the slope of $BC$. $BC$ has slope $\frac{24-0}{16-34}=-\frac{4}{3}$, therefore $y=\frac{3}{4} x$. These two lines intersect at $(16,12)$, so that's the base of the height of the tetrahedron.
Let $S$ be the foot of altitude $BS$ in $\triangle BPQ$. From the Pythagorean Theorem, $h=\sqrt{BS^2-SO^2}$. However, since $S$ and $O$ are, by coincidence, the same point, $SO=0$ and $h=12$.
The area of the base is $102$, so the volume is $\frac{102*12}{3}=\boxed{408}$.~Shen Kislay Kai | // Block 1
defaultpen(fontsize(9)+linewidth(0.63)); pair A=(0,0), B=(16,24), C=(34,0), P=(8,12), Q=(25,12), R=(17,0); draw(A--B--C--A);draw(P--Q--R--P); draw(A--foot(A,B,C));draw(B--foot(B,A,C));draw(C--foot(C,A,B)); label("\(A\)",A,SW);label("\(B\)",B,NW);label("\(C\)",C,SE); label("\(D\)",foot(A,B,C),NE);label("\(E\)",foot(B,A,C),SW);label("\(F\)",foot(C,A,B),NW);label("\(P\)",P,NW);label("\(Q\)",Q,NE);label("\(R\)",R,SE);
// Block 2
import three; defaultpen(linewidth(0.6));
currentprojection=orthographic(1/2,-1,1/2); triple A=(0,0,0), B=(16,24,0), C=(34,0,0), P=(8,12,0), Q=(25,12,0), R=(17,0,0), S=(16,12,12); draw(A--B--C--A); draw(P--Q--R--P); draw(S--P..S--Q..S--R); draw(S--(16,12,0));
// Block 3
import three; defaultpen(linewidth(0.6)); currentprojection=orthographic(1/2,-1,1/2); triple A=(0,0,0), B=(16,24,0), C=(34,0,0), P=(8,12,0), Q=(25,12,0), R=(17,0,0), S=(16,12,12); draw(A--B--C--A); draw(P--Q--R--P); draw(S--P..S--Q..S--R); draw(S--(16,12,0)); | [] |
85 | Let $u$ and $v$ be integers satisfying $0 < v < u$. Let $A = (u,v)$, let $B$ be the reflection of $A$ across the line $y = x$, let $C$ be the reflection of $B$ across the y-axis, let $D$ be the reflection of $C$ across the x-axis, and let $E$ be the reflection of $D$ across the y-axis. The area of pentagon $ABCDE$ is $451$. Find $u + v$. | 2000 AIME I Problem 2 | Solution 1
Since $A = (u,v)$, we can find the coordinates of the other points: $B = (v,u)$, $C = (-v,u)$, $D = (-v,-u)$, $E = (v,-u)$. If we graph those points, we notice that since the latter four points are all reflected across the x/y-axis, they form a rectangle, and $ABE$ is a triangle. The area of $BCDE$ is $(2u)(2v) = 4uv$ and the area of $ABE$ is $\frac{1}{2}(2u)(u-v) = u^2 - uv$. Adding these together, we get $u^2 + 3uv = u(u+3v) = 451 = 11 \cdot 41$. Since $u,v$ are positive, $u+3v>u$, and by matching factors we get either $(u,v) = (1,150)$ or $(11,10)$. Since $v < u$ the latter case is the answer, and $u+v = \boxed{021}$.
Solution 2
We find the coordinates like in the solution above: $A = (u,v)$, $B = (v,u)$, $C = (-v,u)$, $D = (-v,-u)$, $E = (v,-u)$. Then we notice pentagon $ABCDE$ fits into a rectangle of side lengths $(u+v)$ and $(2u)$, giving us two triangles, each with hypotenuse $AB$ and $BE$. First, we can solve for the first triangle. Using the coordinates of $A$ and $B$, we discover the side lengths are both $(u-v)$, so the area of the triangle of hypotenuse $AB$ is $\frac{1}{2}(u-v)^2$. Next, we can solve for the second triangle. Using the coordinates of $A$ and $E$, we discover the side lengths are $(u-v)$ and $(u+v)$, so the area of the triangle of hypotenuse $AE$ is $\frac{1}{2}(u-v)(u+v) = \frac{1}{2}(u^2-v^2)$. Now, let’s subtract the area of these 2 triangles from the rectangle giving us $(u+v)(2u)-\frac{1}{2}(u-v)^2-\frac{1}{2}(u^2-v^2)=451 —> u^2+3uv=451 -> u(u+3v)=451$. Next, we take note of the fact that $u$ and $u+3v$ are both factors of 451, and since both $u$ and $v$ are positive integers, $u+3v$ must be greater than $u$, thus giving us two cases, where either $u=1$ or $u=11$. After trying both, the only working pair of $(u,v)$ where both $u$ and $v$ are integers are $u=11$ and $v=10$, thus meaning $u + v =$ $\boxed{021}$
~Aeioujyot
Solution 3
We find the coordinates like in the solution above: $A = (u,v)$, $B = (v,u)$, $C = (-v,u)$, $D = (-v,-u)$, $E = (v,-u)$. Then we apply the Shoelace Theorem. \[A = \frac{1}{2}[(u^2 + vu + vu + vu + v^2) - (v^2 - uv - uv - uv -u^2)] = 451\] \[\frac{1}{2}(2u^2 + 6uv) = 451\] \[u(u + 3v) = 451\]
This means that $(u,v) = (11, 10)$ or $(1,150)$, but since $v < u$, then the answer is $\boxed{021}$ | pointpen = black; pathpen = linewidth(0.7) + black; size(180); pair A=(11,10), B=(10,11), C=(-10, 11), D=(-10, -11), E=(10, -11); D(D(MP("A\ (u,v)",A,(1,0)))--D(MP("B",B,N))--D(MP("C",C,N))--D(MP("D",D))--D(MP("E",E))--cycle); D((-15,0)--(15,0),linewidth(0.6),Arrows(5)); D((0,-15)--(0,15),linewidth(0.6),Arrows(5)); D((-15,-15)--(15,15),linewidth(0.6),Arrows(5)); | [] |
86 | In the middle of a vast prairie, a firetruck is stationed at the intersection of two perpendicular straight highways. The truck travels at $50$ miles per hour along the highways and at $14$ miles per hour across the prairie. Consider the set of points that can be reached by the firetruck within six minutes. The area of this region is $m/n$ square miles, where $m$ and $n$ are relatively prime positive integers. Find $m + n$. | 2000 AIME I Problem 13 | Let the intersection of the highways be at the origin $O$, and let the highways be the x and y axes. We consider the case where the truck moves in the positive x direction.
After going $x$ miles, $t=\frac{d}{r}=\frac{x}{50}$ hours has passed. If the truck leaves the highway it can travel for at most $t=\frac{1}{10}-\frac{x}{50}$ hours, or $d=rt=14t=1.4-\frac{7x}{25}$ miles. It can end up anywhere off the highway in a circle with this radius centered at $(x,0)$. All these circles are homothetic with respect to a center at $(5,0)$.
Now consider the circle at $(0,0)$. Draw a line tangent to it at $A$ and passing through $B (5,0)$. By the Pythagorean Theorem $AB^2+AO^2=OB^2 \Longrightarrow AB=\sqrt{OB^2-AO^2}=\sqrt{5^2-1.4^2}=\frac{24}{5}$. Then $\tan(\angle ABO)=\frac{OA}{AB}=\frac{7}{24}$, so the slope of line $AB$ is $\frac{-7}{24}$. Since it passes through $(5,0)$ its equation is $y=\frac{-7}{24}(x-5)$.
This line and the x and y axis bound the region the truck can go if it moves in the positive x direction. Similarly, the line $y=5-\frac{24}{7}x$ bounds the region the truck can go if it moves in positive y direction. The intersection of these two lines is $\left(\frac{35}{31},\frac{35}{31}\right)$. The bounded region in Quadrant I is made up of a square and two triangles. $A=x^2+x(5-x)=5x$. By symmetry, the regions in the other quadrants are the same, so the area of the whole region is $20x=\frac{700}{31}$ so the answer is $700+31=\boxed{731}$. | // Block 1
pair truck(pair P){
pair Q = IP(P--P+(7/10,24/10),(35/31,35/31)--(5,0));
D(P--Q,EndArrow(5)); D(CP(P,Q),linewidth(0.5));
return Q;
}
pointpen = black; pathpen = black+linewidth(0.7); size(250);
pair B=(5,0), C=(35/31,35/31);
D(D(B)--D(C)--D(B*dir(90))--D(C*dir(90))--D(B*dir(180))--D(C*dir(180))--D(B*dir(270))--D(C*dir(270))--cycle);
D((-6,0)--(6,0),Arrows(4)); D((0,-6)--(0,6),Arrows(4)); truck((1,0)); truck((2,0)); truck((3,0)); truck((4,0));
// Block 2
pointpen = black; pathpen = black+linewidth(0.7); size(250);
pair O=(0,0), B=(5,0), A=1.4*expi(atan(24/7)), C=1.4*expi(atan(7/24));
D(D(B)--D(A)--D(O)); D(O--D(C)--D(B*dir(90))--D(A*dir(90))--O--D(C*dir(90))--D(B*dir(180))--D(A*dir(180))--O--D(C*dir(180))--D(B*dir(270))--D(A*dir(270))--O--D(C*dir(270))--B,linewidth(0.5)); D(CR(O,1.4));
D((-6,0)--(6,0),Arrows(4)); D((0,-6)--(0,6),Arrows(4));
MP("A",A,N); MP("B",B); MP("(5,0)",B,N); D(MP("\left(\frac{35}{31},\frac{35}{31}\right)",(35/31,35/31),NE));
D(rightanglemark(O,A,B));
// Block 3
pair truck(pair P){ pair Q = IP(P--P+(7/10,24/10),(35/31,35/31)--(5,0)); D(P--Q,EndArrow(5)); D(CP(P,Q),linewidth(0.5)); return Q; } pointpen = black; pathpen = black+linewidth(0.7); size(250); pair B=(5,0), C=(35/31,35/31); D(D(B)--D(C)--D(B*dir(90))--D(C*dir(90))--D(B*dir(180))--D(C*dir(180))--D(B*dir(270))--D(C*dir(270))--cycle); D((-6,0)--(6,0),Arrows(4)); D((0,-6)--(0,6),Arrows(4)); truck((1,0)); truck((2,0)); truck((3,0)); truck((4,0));
// Block 4
pointpen = black; pathpen = black+linewidth(0.7); size(250); pair O=(0,0), B=(5,0), A=1.4*expi(atan(24/7)), C=1.4*expi(atan(7/24)); D(D(B)--D(A)--D(O)); D(O--D(C)--D(B*dir(90))--D(A*dir(90))--O--D(C*dir(90))--D(B*dir(180))--D(A*dir(180))--O--D(C*dir(180))--D(B*dir(270))--D(A*dir(270))--O--D(C*dir(270))--B,linewidth(0.5)); D(CR(O,1.4)); D((-6,0)--(6,0),Arrows(4)); D((0,-6)--(0,6),Arrows(4)); MP("A",A,N); MP("B",B); MP("(5,0)",B,N); D(MP("\left(\frac{35}{31},\frac{35}{31}\right)",(35/31,35/31),NE)); D(rightanglemark(O,A,B)); | [] |
87 | In triangle $ABC,$ it is given that angles $B$ and $C$ are congruent. Points $P$ and $Q$ lie on $\overline{AC}$ and $\overline{AB},$ respectively, so that $AP = PQ = QB = BC.$ Angle $ACB$ is $r$ times as large as angle $APQ,$ where $r$ is a positive real number. Find $\lfloor 1000r \rfloor$.
Contents
1 Problem
2 Official Solution (MAA)
3 Solution 1
4 Solution 2 (Law of sines)
5 Solution 3
6 Solution 4 (Trig identities)
7 Solution 5
8 See also | 2000 AIME I Problem 14 | Let point $R$ be in $\triangle ABC$ such that $QB = BR = RP$. Then $PQBR$ is a rhombus, so $AB \parallel PR$ and $APRB$ is an isosceles trapezoid. Since $\overline{PB}$ bisects $\angle QBR$, it follows by symmetry in trapezoid $APRB$ that $\overline{RA}$ bisects $\angle BAC$. Thus $R$ lies on the perpendicular bisector of $\overline{BC}$, and $BC = BR = RC$. Hence $\triangle BCR$ is an equilateral triangle.
Now $\angle ABR = \angle BAC = \angle ACR$, and the sum of the angles in $\triangle ABC$ is $\angle ABR + 60^{\circ} + \angle BAC + \angle ACR + 60^{\circ} = 3\angle BAC + 120^{\circ} = 180^{\circ} \Longrightarrow \angle BAC = 20^{\circ}$. Then $\angle APQ = 140^{\circ}$ and $\angle ACB = 80^{\circ}$, so the answer is $\left\lfloor 1000 \cdot \frac{80}{140} \right\rfloor = \left\lfloor \frac{4000}{7} \right\rfloor = \boxed{571}$. | defaultpen(fontsize(8)); size(200); pair A=20*dir(80)+20*dir(60)+20*dir(100), B=(0,0), C=20*dir(0), P=20*dir(80)+20*dir(60), Q=20*dir(80), R=20*dir(60); draw(A--B--C--A);draw(P--Q);draw(A--R--B);draw(P--R);D(R--C,dashed); label("\(A\)",A,(0,1));label("\(B\)",B,(-1,-1));label("\(C\)",C,(1,-1));label("\(P\)",P,(1,1)); label("\(Q\)",Q,(-1,1));label("\(R\)",R,(1,0)); | [] |
87 | In triangle $ABC,$ it is given that angles $B$ and $C$ are congruent. Points $P$ and $Q$ lie on $\overline{AC}$ and $\overline{AB},$ respectively, so that $AP = PQ = QB = BC.$ Angle $ACB$ is $r$ times as large as angle $APQ,$ where $r$ is a positive real number. Find $\lfloor 1000r \rfloor$.
Contents
1 Problem
2 Official Solution (MAA)
3 Solution 1
4 Solution 2 (Law of sines)
5 Solution 3
6 Solution 4 (Trig identities)
7 Solution 5
8 See also | 2000 AIME I Problem 14 | Again, construct $R$ as above.
Let $\angle BAC = \angle QBR = \angle QPR = 2x$ and $\angle ABC = \angle ACB = y$, which means $x + y = 90$.
$\triangle QBC$ is isosceles with $QB = BC$, so $\angle BCQ = 90 - \frac {y}{2}$.
Let $S$ be the intersection of $QC$ and $BP$. Since $\angle BCQ = \angle BQC = \angle BRS$, $BCRS$ is cyclic, which means $\angle RBS = \angle RCS = x$.
Since $APRB$ is an isosceles trapezoid, $BP = AR$, but since $AR$ bisects $\angle BAC$, $\angle ABR = \angle ACR = 2x$.
Therefore we have that $\angle ACB = \angle ACR + \angle RCS + \angle QCB = 2x + x + 90 - \frac {y}{2} = y$.
We solve the simultaneous equations $x + y = 90$ and $2x + x + 90 - \frac {y}{2} = y$ to get $x = 10$ and $y = 80$.
$\angle APQ = 180 - 4x = 140$, $\angle ACB = 80$, so $r = \frac {80}{140} = \frac {4}{7}$.
$\left\lfloor 1000\left(\frac {4}{7}\right)\right\rfloor = \boxed{571}$. | defaultpen(fontsize(8)); size(200); pair A=20*dir(80)+20*dir(60)+20*dir(100), B=(0,0), C=20*dir(0), P=20*dir(80)+20*dir(60), Q=20*dir(80), R=20*dir(60), S; S=intersectionpoint(Q--C,P--B); draw(A--B--C--A);draw(B--P--Q--C--R--Q);draw(A--R--B);draw(P--R--S); label("\(A\)",A,(0,1));label("\(B\)",B,(-1,-1));label("\(C\)",C,(1,-1));label("\(P\)",P,(1,1)); label("\(Q\)",Q,(-1,1));label("\(R\)",R,(1,0));label("\(S\)",S,(-1,0)); | [] |
87 | In triangle $ABC,$ it is given that angles $B$ and $C$ are congruent. Points $P$ and $Q$ lie on $\overline{AC}$ and $\overline{AB},$ respectively, so that $AP = PQ = QB = BC.$ Angle $ACB$ is $r$ times as large as angle $APQ,$ where $r$ is a positive real number. Find $\lfloor 1000r \rfloor$.
Contents
1 Problem
2 Official Solution (MAA)
3 Solution 1
4 Solution 2 (Law of sines)
5 Solution 3
6 Solution 4 (Trig identities)
7 Solution 5
8 See also | 2000 AIME I Problem 14 | Reflect $\triangle ABC$ over $BC$ and translate it to attach side $AB$ onto $AC$, mapping $\triangle ABC$ to $\triangle CAD$. Point $P$ maps to $R$, and $Q$ maps to $P$. Then we have that $BC=BQ=QP=PA=AD=PR=RC$. Notice how $BQ=RC$ and $BQ\parallel RC$, so $BQRC$ is a parallelogram and $QR=BC$. But $BC=QP=PR$, so $\triangle QPR$ is actually equilateral. Set $\angle BAC=\angle ACD=x$. Then notice that $\angle QPC=\angle PQA+\angle PAQ=2x$, but $\angle RPC=\angle PQA=x$. Thus $\angle QPR=3x=60$, so $x=20$. Thus $\angle QPA=140^{\circ}$ and $\angle BCA=80^{\circ}$, so $r=\frac{80}{140}=\frac{4}{7}$. The answer is $\left \lfloor \frac{4000}{7}\right \rfloor =\boxed{571}$.
~ethanzhang1001 | defaultpen(fontsize(8)); size(200); pair A=20*dir(80)+20*dir(60)+20*dir(100), B=(0,0), C=20*dir(0), P=20*dir(80)+20*dir(60), Q=20*dir(80), R=20*dir(0)+20*dir(80), D=20*dir(0)+20*dir(80)+20*dir(60)+20*dir(100); draw(Q--A--D--C--B--Q--P--R--Q); draw(A--C); label("A",A,NW); label("B",B,SW); label("C",C,SE); label("D",D,NE); label("P",P,W); label("Q",Q,W); label("R",R,E); | [] |
88 | One base of a trapezoid is $100$ units longer than the other base. The segment that joins the midpoints of the legs divides the trapezoid into two regions whose areas are in the ratio $2: 3$. Let $x$ be the length of the segment joining the legs of the trapezoid that is parallel to the bases and that divides the trapezoid into two regions of equal area. Find the greatest integer that does not exceed $x^2/100$. | 2000 AIME II Problem 6 | Let the shorter base have length $b$ (so the longer has length $b+100$), and let the height be $h$. The length of the midline of the trapezoid is the average of its bases, which is $\frac{b+b+100}{2} = b+50$. The two regions which the midline divides the trapezoid into are two smaller trapezoids, both with height $h/2$. Then,
\[\frac{\frac 12 (h/2) (b + b+50)}{\frac 12 (h/2) (b + 50 + b + 100)} = \frac{2}{3} \Longrightarrow \frac{b + 75}{b + 25} = \frac 32 \Longrightarrow b = 75\]
Construct the perpendiculars from the vertices of the shorter base to the longer base. This splits the trapezoid into a rectangle and two triangles; it also splits the desired line segment into three partitions with lengths $x_1, 75, x_2$. By similar triangles, we easily find that $\frac{x - 75}{100} = \frac{x_1+x_2}{100} = \frac{h_1}{h}$.
The area of the region including the shorter base must be half of the area of the entire trapezoid, so
\[2 \cdot \frac 12 h_1 (75 + x) = \frac 12 h (75 + 175) \Longrightarrow x = 125 \cdot \frac{h}{h_1} - 75\]
Substituting our expression for $\frac h{h_1}$ from above, we find that
\[x = \frac{12500}{x-75} - 75 \Longrightarrow x^2 - 75x = 5625 + 12500 - 75x \Longrightarrow x^2 = 18125\]
The answer is $\left\lfloor\frac{x^2}{100}\right\rfloor = \boxed{181}$. | // Block 1
pathpen = linewidth(0.7); pen d = linetype("4 4") + linewidth(0.7);
pair A=(0,0),B=(175,0),C=(105,100),D=(30,100);
D(A--B--C--D--cycle); D((A+D)/2 -- (B+C)/2, d); MP("b",(C+D)/2,N);MP("b+100",(A+B)/2);
// Block 2
pathpen = linewidth(0.7); pen d = linetype("4 4") + linewidth(0.7);
pair A=(0,0),B=(175,0),C=(105,100),D=(30,100),E=D*(1.75-(18125)^.5/100),F=IP(B--C,E--(175,E.y));
D(A--B--C--D--cycle); MP("75",(C+D)/2,N);MP("175",(A+B)/2); D(C--(C.x,0),d);D(D--(D.x,0),d); D(E--F,d); D((-20,100)--(-20,0)); MP("h",(-20,50),(-1,0));MP("h_1",(C.x,(C.y+E.y)/2),(-1,0)); MP("x_1",((E.x+D.x)/2,E.y));MP("x_2",((F.x+C.x)/2,E.y));
// Block 3
pathpen = linewidth(0.7); pen d = linetype("4 4") + linewidth(0.7); pair A=(0,0),B=(175,0),C=(105,100),D=(30,100); D(A--B--C--D--cycle); D((A+D)/2 -- (B+C)/2, d); MP("b",(C+D)/2,N);MP("b+100",(A+B)/2);
// Block 4
pathpen = linewidth(0.7); pen d = linetype("4 4") + linewidth(0.7); pair A=(0,0),B=(175,0),C=(105,100),D=(30,100),E=D*(1.75-(18125)^.5/100),F=IP(B--C,E--(175,E.y)); D(A--B--C--D--cycle); MP("75",(C+D)/2,N);MP("175",(A+B)/2); D(C--(C.x,0),d);D(D--(D.x,0),d); D(E--F,d); D((-20,100)--(-20,0)); MP("h",(-20,50),(-1,0));MP("h_1",(C.x,(C.y+E.y)/2),(-1,0)); MP("x_1",((E.x+D.x)/2,E.y));MP("x_2",((F.x+C.x)/2,E.y)); | [] |
89 | In trapezoid $ABCD$, leg $\overline{BC}$ is perpendicular to bases $\overline{AB}$ and $\overline{CD}$, and diagonals $\overline{AC}$ and $\overline{BD}$ are perpendicular. Given that $AB=\sqrt{11}$ and $AD=\sqrt{1001}$, find $BC^2$. | 2000 AIME II Problem 8 | Solution 1
Let $x = BC$ be the height of the trapezoid, and let $y = CD$. Since $AC \perp BD$, it follows that $\triangle BAC \sim \triangle CBD$, so $\frac{x}{\sqrt{11}} = \frac{y}{x} \Longrightarrow x^2 = y\sqrt{11}$.
Let $E$ be the foot of the altitude from $A$ to $\overline{CD}$. Then $AE = x$, and $ADE$ is a right triangle. By the Pythagorean Theorem,
\[x^2 + \left(y-\sqrt{11}\right)^2 = 1001 \Longrightarrow x^4 - 11x^2 - 11^2 \cdot 9 \cdot 10 = 0\]
The positive solution to this quadratic equation is $x^2 = \boxed{110}$.
Solution 2
Let $BC=x$. Dropping the altitude from $A$ and using the Pythagorean Theorem tells us that $CD=\sqrt{11}+\sqrt{1001-x^2}$. Therefore, we know that vector $\vec{BD}=\langle \sqrt{11}+\sqrt{1001-x^2},-x\rangle$ and vector $\vec{AC}=\langle-\sqrt{11},-x\rangle$. Now we know that these vectors are perpendicular, so their dot product is 0.\[\vec{BD}\cdot \vec{AC}=-11-\sqrt{11(1001-x^2)}+x^2=0\]
\[(x^2-11)^2=11(1001-x^2)\]
\[x^4-11x^2-11\cdot 990=0.\]
As above, we can solve this quadratic to get the positve solution $BC^2=x^2=\boxed{110}$.
Solution 3
Let $BC=x$ and $CD=y+\sqrt{11}$. From Pythagoras with $AD$, we obtain $x^2+y^2=1001$. Since $AC$ and $BD$ are perpendicular diagonals of a quadrilateral, then $AB^2+CD^2=BC^2+AD^2$, so we have \[\left(y+\sqrt{11}\right)^2+11=x^2+1001.\] Substituting $x^2=1001-y^2$ and simplifying yields \[y^2+\sqrt{11}y-990=0,\] and the quadratic formula gives $y=9\sqrt{11}$. Then from $x^2+y^2=1001$, we plug in $y$ to find $x^2=\boxed{110}$.
Solution 4
Let $E$ be the intersection of the diagonals. Since the diagonals are perpendicular, applying the Pythagorean Theorem multiple times we have
\begin{align*} BC^2&=BE^2+CE^2 \\ &=(AB^2-AE^2)+(CD^2-DE^2) \\ &=CD^2+\sqrt{11}^2-(AE^2+DE^2) \\ &=CD^2+11-AD^2 \\ &=CD^2-990 \end{align*}
Followed by dropping the perpendicular like in solution 1, we obtain system of equation
\[BC^2=CD^2-990\]
\[BC^2+CD^2-2\sqrt{11}CD=990\]
Rearrange the first equation yields
\[CD^2-BC^2=990\]
Equating it with the second equation we have
\[CD^2-BC^2=BC^2+CD^2-2\sqrt{11}CD\]
Which gives $CD=\frac{BC^2}{\sqrt{11}}$.
Substituting into equation 1 obtains the quadratic in terms of $BC^2$
\[(BC^2)^2-11BC^2-11\cdot990=0\]
Solving the quadratic to obtain $BC^2=\boxed{110}$.
~Nafer ~edits by fermat_sLastAMC | size(200); pathpen = linewidth(0.7); pair C=(0,0),B=(0,110^.5),A=(11^.5,B.y),D=(10*11^.5,0),E=foot(A,C,D); D(MP("A",A,(2,.5))--MP("B",B,W)--MP("C",C)--MP("D",D)--cycle); D(A--C);D(B--D);D(A--E,linetype("4 4") + linewidth(0.7)); MP("\sqrt{11}",(A+B)/2,N);MP("\sqrt{1001}",(A+D)/2,NE);MP("\sqrt{1001}",(A+D)/2,NE);MP("x",(B+C)/2,W);MP("y",(D+C)/2);D(rightanglemark(B,IP(A--C,B--D),C,20)); | [] |
90 | The coordinates of the vertices of isosceles trapezoid $ABCD$ are all integers, with $A=(20,100)$ and $D=(21,107)$. The trapezoid has no horizontal or vertical sides, and $\overline{AB}$ and $\overline{CD}$ are the only parallel sides. The sum of the absolute values of all possible slopes for $\overline{AB}$ is $m/n$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. | 2000 AIME II Problem 11 | For simplicity, we translate the points so that $A$ is on the origin and $D = (1,7)$. Suppose $B$ has integer coordinates; then $\overrightarrow{AB}$ is a vector with integer parameters (vector knowledge is not necessary for this solution). We construct the perpendicular from $A$ to $\overline{CD}$, and let $D' = (a,b)$ be the reflection of $D$ across that perpendicular. Then $ABCD'$ is a parallelogram, and $\overrightarrow{AB} = \overrightarrow{D'C}$. Thus, for $C$ to have integer coordinates, it suffices to let $D'$ have integer coordinates.[1]
Let the slope of the perpendicular be $m$. Then the midpoint of $\overline{DD'}$ lies on the line $y=mx$, so $\frac{b+7}{2} = m \cdot \frac{a+1}{2}$. Also, $AD = AD'$ implies that $a^2 + b^2 = 1^2 + 7^2 = 50$. Combining these two equations yields
\[a^2 + \left(7 - (a+1)m\right)^2 = 50\]
Since $a$ is an integer, then $7-(a+1)m$ must be an integer. There are $12$ pairs of integers whose squares sum up to $50,$ namely $( \pm 1, \pm 7), (\pm 7, \pm 1), (\pm 5, \pm 5)$. We exclude the cases $(\pm 1, \pm 7)$ because they lead to degenerate parallelograms (rectangle, line segment, vertical and horizontal sides). Thus we have
\[7 - 8m = \pm 1, \quad 7 + 6m = \pm 1, \quad 7 - 6m = \pm 5, 7 + 4m = \pm 5\]
These yield $m = 1, \frac 34, -1, -\frac 43, 2, \frac 13, -3, - \frac 12$. Therefore, the corresponding slopes of $\overline{AB}$ are $-1, -\frac 43, 1, \frac 34, -\frac 12, -3, \frac 13$, and $2$. The sum of their absolute values is $\frac{119}{12}$. The answer is $m+n= \boxed{131}$
^ In other words, since $ABCD'$ is a parallelogram, the difference between the x-coordinates and the y-coordinates of $C$ and $D'$ are, respectively, the difference between the x-coordinates and the y-coordinates of $A$ and $B$. But since the latter are integers, then the former are integers also, so $C$ has integer coordinates iff $D'$ has integer coordinates.
Note: It may not seem like we considered that the coordinates of $B$ have to be integers, but since the slopes of $\overline{AB}$ are all rational, we can just extend $\overline{AB}$ by some arbitrary amount so that $B$ becomes a lattice point, which won't affect the position of $D'$. ~inaccessibles | // Block 1
pathpen = linewidth(0.7);
pair A=(0,0), D=(1,7), Da = MP("D'",D((-7,1)),N), B=(-8,-6), C=B+Da, F=foot(A,C,D);
D(MP("A",A)--MP("B",B)--MP("C",C,N)--MP("D",D,N)--cycle); D(F--A--Da,linetype("4 4"));
// Block 2
pathpen = linewidth(0.7); pair A=(0,0), D=(1,7), Da = MP("D'",D((-7,1)),N), B=(-8,-6), C=B+Da, F=foot(A,C,D); D(MP("A",A)--MP("B",B)--MP("C",C,N)--MP("D",D,N)--cycle); D(F--A--Da,linetype("4 4")); | [] |
91 | In triangle $ABC$, angles $A$ and $B$ measure $60$ degrees and $45$ degrees, respectively. The bisector of angle $A$ intersects $\overline{BC}$ at $T$, and $AT=24$. The area of triangle $ABC$ can be written in the form $a+b\sqrt{c}$, where $a$, $b$, and $c$ are positive integers, and $c$ is not divisible by the square of any prime. Find $a+b+c$. | 2001 AIME I Problem 4 | First, draw a good diagram.
We realize that $\angle C = 75^\circ$, and $\angle CAT = 30^\circ$. Therefore, $\angle CTA = 75^\circ$ as well, making $\triangle CAT$ an isosceles triangle. $AT$ and $AC$ are congruent, so $AC=24$. We now drop an altitude from $C$, and call the foot this altitude point $D$.
By 30-60-90 triangles, $AD=12$ and $CD=12\sqrt{3}$.
We also notice that $\triangle CDB$ is an isosceles right triangle. $CD$ is congruent to $BD$, which makes $BD=12\sqrt{3}$. The base $AB$ is $12+12\sqrt{3}$, and the altitude $CD=12\sqrt{3}$. We can easily find that the area of triangle $ABC$ is $216+72\sqrt{3}$, so $a+b+c=\boxed{291}$.
-youyanli | // Block 1
size(200);
defaultpen(linewidth(0.4)+fontsize(8));
pair A,B,C,D,T,F;
A = origin;
T = scale(24)*dir(30);
C = scale(24)*dir(60);
B = extension(C,T,A,(1,0));
F = foot(T,A,B);
D = foot(C,A,B);
draw(A--B--C--A--T, black+0.8);
draw(C--D, dashed);
label(rotate(degrees(T-A))*"$24$", A--T, N);
label(rotate(degrees(C-A))*"$24$", A--C, 2*NW);
label("$12\sqrt 3$", C--D, E);
label("$12\sqrt 3$", D--B, S);
label("$12$", A--D, S);
pen p = fontsize(8)+red;
MA("45^\circ", C,B,A,2);
MA("30^\circ", B,A,T,2.5);
MA("30^\circ", T,A,C,3.5);
dot("$A$", A, SW);
dot("$B$", B, SE);
dot("$C$", C, N);
dot("$T$", T, NE);
dot("$D$", D, S);
// Block 2
size(200); defaultpen(linewidth(0.4)+fontsize(8)); pair A,B,C,D,T,F; A = origin; T = scale(24)*dir(30); C = scale(24)*dir(60); B = extension(C,T,A,(1,0)); F = foot(T,A,B); D = foot(C,A,B); draw(A--B--C--A--T, black+0.8); draw(C--D, dashed); label(rotate(degrees(T-A))*"$24$", A--T, N); label(rotate(degrees(C-A))*"$24$", A--C, 2*NW); label("$12\sqrt 3$", C--D, E); label("$12\sqrt 3$", D--B, S); label("$12$", A--D, S); pen p = fontsize(8)+red; MA("45^\circ", C,B,A,2); MA("30^\circ", B,A,T,2.5); MA("30^\circ", T,A,C,3.5); dot("$A$", A, SW); dot("$B$", B, SE); dot("$C$", C, N); dot("$T$", T, NE); dot("$D$", D, S); | [] |
92 | An equilateral triangle is inscribed in the ellipse whose equation is $x^2+4y^2=4$. One vertex of the triangle is $(0,1)$, one altitude is contained in the y-axis, and the square of the length of each side is $\sqrt{\frac{m}{n}}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.
Contents
1 Problem
2 Solution
2.1 Solution 1
2.2 Solution 2
2.3 Solution 3
2.4 Solution 4
2.5 Solution 5
2.6 Solution 6
3 See also | 2001 AIME I Problem 5 | Solution 1
Denote the vertices of the triangle $A,B,$ and $C,$ where $B$ is in quadrant 4 and $C$ is in quadrant $3.$
Note that the slope of $\overline{AC}$ is $\tan 60^\circ = \sqrt {3}.$ Hence, the equation of the line containing $\overline{AC}$ is
\[y = x\sqrt {3} + 1.\]
This will intersect the ellipse when
\begin{eqnarray*}4 = x^{2} + 4y^{2} & = & x^{2} + 4(x\sqrt {3} + 1)^{2} \\ & = & x^{2} + 4(3x^{2} + 2x\sqrt {3} + 1) \implies x(13x+8\sqrt 3)=0\implies x = \frac { - 8\sqrt {3}}{13}. \end{eqnarray*}
We ignore the $x=0$ solution because it is not in quadrant 3.
Since the triangle is symmetric with respect to the y-axis, the coordinates of $B$ and $C$ are now $\left(\frac {8\sqrt {3}}{13},y_{0}\right)$ and $\left(\frac { - 8\sqrt {3}}{13},y_{0}\right),$ respectively, for some value of $y_{0}.$
It is clear that the value of $y_{0}$ is irrelevant to the length of $BC$. Our answer is
\[BC = 2 \cdot \frac {8\sqrt {3}}{13}=\sqrt {4\left(\frac {8\sqrt {3}}{13}\right)^{2}} = \sqrt {\frac {768}{169}}\implies m + n = \boxed{937}.\]
Solution 2
Solving for $y$ in terms of $x$ gives $y= \frac {\sqrt{4-x^2}}{2}$, so the two other points of the triangle are $(x,\frac {\sqrt{4-x^2}}{2})$ and $(-x,\frac {\sqrt{4-x^2}}{2})$, which are a distance of $2x$ apart. Thus $2x$ equals the distance between $(x,\frac {\sqrt{4-x^2}}{2})$ and $(0,1)$, so by the distance formula we have
\[2x=\sqrt{x^2+(\frac {1-\sqrt{4-x^2}}{2})^2}.\]
Squaring both sides and simplifying through algebra yields $x^2= \frac {192}{169}$, so $2x=\sqrt{\frac{768}{169}}$ and the answer is $\boxed{937}$.
Solution 3
Since the altitude goes along the $y$ axis, this means that the base is a horizontal line, which means that the endpoints of the base are $(x,y)$ and $(-x,y)$, and WLOG, we can say that $x$ is positive.
Now, since all sides of an equilateral triangle are the same, we can do this (distance from one of the endpoints of the base to the vertex and the length of the base):
$\sqrt{x^2 + (y-1)^2} = 2x$
Square both sides,
$x^2 + (y-1)^2 = 4x^2\implies (y-1)^2 = 3x^2$
Now, with the equation of the ellipse:
$x^2 + 4y^2 = 4$
$x^2 = 4-4y^2$
$3x^2 = 12-12y^2$
Substituting,
$12-12y^2 = y^2 - 2y +1$
Moving stuff around and solving:
$y = \frac{-11}{13}, 1$
The second is found to be extraneous, so, when we go back and figure out $x$ and then $2x$ (which is the side length), we find it to be:
$\sqrt{\frac{768}{169}}$
and so we get the desired answer of $\boxed{937}$.
Solution 4
Denote $(0,1)$ as vertex $A,$ $B$ as the vertex to the left of the $y$-axis and $C$ as the vertex to the right of the $y$-axis. Let $D$ be the intersection of $BC$ and the $y$-axis.
Let $x_0$ be the $x$-coordinate of $C.$ This implies
\[C=\left(x_0 , \sqrt{\frac{4-x_0^2}{4}}\right)\]
and
\[B=\left(-x_0 , \sqrt{\frac{4-x_0^2}{4}}\right).\]
Note that $BC=2x_0$ and
\[\frac{BC}{\sqrt3}=AD=1-\sqrt{\frac{4-x_0^2}{4}}.\]
This yields
\[\frac{2x_0}{\sqrt3}=1-\sqrt{\frac{4-x_0^2}{4}}.\]
Re-arranging and squaring, we have
\[\frac{4-x_0^2}{4}=\frac{4x_0^2}{3}-\frac{4x_0}{\sqrt3} +1.\]
Simplifying and solving for $x_0$, we have
\[x_0=\frac{48}{13\sqrt 3}.\]
As the length of each side is $2x_0,$ our desired length is
\[4x_0^2=\frac{768}{169}\]
which means our desired answer is
\[768+169=\boxed{937}\]
~ASAB
Solution 5
Notice that $x^2+4y^2=4$ can be rewritten as $(x)^2+(2y)^2=2^2$. The points of the triangle are $(0, 1)$, $(-x, 1-x\sqrt{3})$, and $(x, 1-x\sqrt{3})$. When plugging the second coordinate into the equation, we get $x^2+4-8x\sqrt{3}+12x^2=4$, which equals $13x^2-8x\sqrt{3}=0$.
This yields $x(13x-8\sqrt{3})=0$. Obviously x can't be 0, so $x=\frac{8\sqrt{3}}{13}$. The side length of the equilateral triangle is twice of this, so $\frac{16\sqrt{3}}{13}$.
This can be rewritten as $\sqrt{\frac{256\cdot3}{169}}=\sqrt{\frac{768}{169}}$.
$768+169=\boxed{937}$.
~ MC413551
Solution 6
Consider the transformation $(x,y)$ to $(x/2, y).$ This sends the ellipse to the unit circle. If we let $n$ be one-fourth of the side length of the triangle, the equilateral triangle is sent to an isosceles triangle with side lengths $2n, n\sqrt{13}, n\sqrt{13}.$ Let the triangle be $ABC$ such that $AB=AC.$ Let the foot of the altitude from A be $X.$ Then $BX=n,$ and $AX=2n\sqrt{3}.$ Let $C$ be a point such that $AC$ is a diameter of the unit circle. Then $XC=2-2n\sqrt{3}.$ Using power of a point on X,
\[n^2=2n\sqrt{3}(2-2n\sqrt{3})\]
Simplifying gets us to $13n^2=4n\sqrt{3}.$ Then, $n=\dfrac{4\sqrt{3}}{13}$ which means the side length is $\dfrac{16\sqrt{3}}{13}=\sqrt{\dfrac{768}{169}}.$
Thus, the answer is $768+169=\boxed{937}.$ | // Block 1
pointpen = black; pathpen = black + linewidth(0.7);
path e = xscale(2)*unitcircle; real x = -8/13*3^.5;
D((-3,0)--(3,0)); D((0,-2)--(0,2)); /* axes */
D(e); D(D((0,1))--(x,x*3^.5+1)--(-x,x*3^.5+1)--cycle);
// Block 2
pointpen = black; pathpen = black + linewidth(0.7); path e = xscale(2)*unitcircle; real x = -8/13*3^.5; D((-3,0)--(3,0)); D((0,-2)--(0,2)); /* axes */ D(e); D(D((0,1))--(x,x*3^.5+1)--(-x,x*3^.5+1)--cycle); | [] |
93 | Triangle $ABC$ has $AB=21$, $AC=22$ and $BC=20$. Points $D$ and $E$ are located on $\overline{AB}$ and $\overline{AC}$, respectively, such that $\overline{DE}$ is parallel to $\overline{BC}$ and contains the center of the inscribed circle of triangle $ABC$. Then $DE=m/n$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.
Contents
1 Problem
2 Solution 1
3 Solution 2
4 Solution 3 (mass points)
5 Solution 4 (Faster)
6 Solution 5
7 Solution 6
8 Solution 7
9 Solution 8 (vectors)
10 See also | 2001 AIME I Problem 7 | Let $I$ be the incenter of $\triangle ABC$, so that $BI$ and $CI$ are angle bisectors of $\angle ABC$ and $\angle ACB$ respectively. Then, $\angle BID = \angle CBI = \angle DBI,$ so $\triangle BDI$ is isosceles, and similarly $\triangle CEI$ is isosceles. It follows that $DE = DB + EC$, so the perimeter of $\triangle ADE$ is $AD + AE + DE = AB + AC = 43$. Hence, the ratio of the perimeters of $\triangle ADE$ and $\triangle ABC$ is $\frac{43}{63}$, which is the scale factor between the two similar triangles, and thus $DE = \frac{43}{63} \times 20 = \frac{860}{63}$. Thus, $m + n = \boxed{923}$. | // Block 1
pointpen = black; pathpen = black+linewidth(0.7);
pair B=(0,0), C=(20,0), A=IP(CR(B,21),CR(C,22)), I=incenter(A,B,C), D=IP((0,I.y)--(20,I.y),A--B), E=IP((0,I.y)--(20,I.y),A--C);
D(MP("A",A,N)--MP("B",B)--MP("C",C)--cycle); D(MP("I",I,NE)); D(MP("E",E,NE)--MP("D",D,NW));
// D((A.x,0)--A,linetype("4 4")+linewidth(0.7)); D((I.x,0)--I,linetype("4 4")+linewidth(0.7)); D(rightanglemark(B,(A.x,0),A,30));
D(B--I--C);
MP("20",(B+C)/2); MP("21",(A+B)/2,NW); MP("22",(A+C)/2,NE);
// Block 2
pointpen = black; pathpen = black+linewidth(0.7); pair B=(0,0), C=(20,0), A=IP(CR(B,21),CR(C,22)), I=incenter(A,B,C), D=IP((0,I.y)--(20,I.y),A--B), E=IP((0,I.y)--(20,I.y),A--C); D(MP("A",A,N)--MP("B",B)--MP("C",C)--cycle); D(MP("I",I,NE)); D(MP("E",E,NE)--MP("D",D,NW)); // D((A.x,0)--A,linetype("4 4")+linewidth(0.7)); D((I.x,0)--I,linetype("4 4")+linewidth(0.7)); D(rightanglemark(B,(A.x,0),A,30)); D(B--I--C); MP("20",(B+C)/2); MP("21",(A+B)/2,NW); MP("22",(A+C)/2,NE); | [] |
93 | Triangle $ABC$ has $AB=21$, $AC=22$ and $BC=20$. Points $D$ and $E$ are located on $\overline{AB}$ and $\overline{AC}$, respectively, such that $\overline{DE}$ is parallel to $\overline{BC}$ and contains the center of the inscribed circle of triangle $ABC$. Then $DE=m/n$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.
Contents
1 Problem
2 Solution 1
3 Solution 2
4 Solution 3 (mass points)
5 Solution 4 (Faster)
6 Solution 5
7 Solution 6
8 Solution 7
9 Solution 8 (vectors)
10 See also | 2001 AIME I Problem 7 | The semiperimeter of $ABC$ is $s = \frac{20 + 21 + 22}{2} = \frac{63}{2}$. By Heron's formula, the area of the whole triangle is $A = \sqrt{s(s-a)(s-b)(s-c)} = \frac{21\sqrt{1311}}{4}$. Using the formula $A = rs$, we find that the inradius is $r = \frac{A}{s} = \frac{\sqrt{1311}}6$. Since $\triangle ADE \sim \triangle ABC$, the ratio of the heights of triangles $ADE$ and $ABC$ is equal to the ratio between sides $DE$ and $BC$. From $A=\frac{1}{2}bh$, we find $h_{ABC} = \frac{21\sqrt{1311}}{40}$. Thus, we have
$\frac{h_{ADE}}{h_{ABC}} = \frac{h_{ABC}-r}{h_{ABC}} = \frac{21\sqrt{1311}/40-\sqrt{1311}/6}{21\sqrt{1311}/40}=\frac{DE}{20}.$
Solving for $DE$ gives $DE=\frac{860}{63},$ so the answer is $m+n=\boxed{923}$.
Or we have the area of the triangle as $S$.
Using the ratio of heights to ratio of bases of $ADE$ and $ABC$
$\frac {\frac {2S}{20}-\frac {2S}{63}}{\frac {2S}{20}}= \frac {DE}{BC(20)}$
from that it is easy to deduce that $DE=\frac{860}{63}$. | // Block 1
pointpen = black; pathpen = black+linewidth(0.7);
pair B=(0,0), C=(20,0), A=IP(CR(B,21),CR(C,22)), I=incenter(A,B,C), D=IP((0,I.y)--(20,I.y),A--B), E=IP((0,I.y)--(20,I.y),A--C);
D(MP("A",A,N)--MP("B",B)--MP("C",C)--cycle); D(incircle(A,B,C)); D(MP("I",I,NE)); D(MP("E",E,NE)--MP("D",D,NW));
D((A.x,0)--A,linetype("4 4")+linewidth(0.7)); D((I.x,0)--I,linetype("4 4")+linewidth(0.7)); D(rightanglemark(B,(A.x,0),A,30));
MP("20",(B+C)/2); MP("21",(A+B)/2,NW); MP("22",(A+C)/2,NE);
// Block 2
pointpen = black; pathpen = black+linewidth(0.7); pair B=(0,0), C=(20,0), A=IP(CR(B,21),CR(C,22)), I=incenter(A,B,C), D=IP((0,I.y)--(20,I.y),A--B), E=IP((0,I.y)--(20,I.y),A--C); D(MP("A",A,N)--MP("B",B)--MP("C",C)--cycle); D(incircle(A,B,C)); D(MP("I",I,NE)); D(MP("E",E,NE)--MP("D",D,NW)); D((A.x,0)--A,linetype("4 4")+linewidth(0.7)); D((I.x,0)--I,linetype("4 4")+linewidth(0.7)); D(rightanglemark(B,(A.x,0),A,30)); MP("20",(B+C)/2); MP("21",(A+B)/2,NW); MP("22",(A+C)/2,NE); | [] |
93 | Triangle $ABC$ has $AB=21$, $AC=22$ and $BC=20$. Points $D$ and $E$ are located on $\overline{AB}$ and $\overline{AC}$, respectively, such that $\overline{DE}$ is parallel to $\overline{BC}$ and contains the center of the inscribed circle of triangle $ABC$. Then $DE=m/n$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.
Contents
1 Problem
2 Solution 1
3 Solution 2
4 Solution 3 (mass points)
5 Solution 4 (Faster)
6 Solution 5
7 Solution 6
8 Solution 7
9 Solution 8 (vectors)
10 See also | 2001 AIME I Problem 7 | Let $P$ be the incenter; then it is be the intersection of all three angle bisectors. Draw the bisector $AP$ to where it intersects $BC$, and name the intersection $F$.
Using the angle bisector theorem, we know the ratio $BF:CF$ is $21:22$, thus we shall assign a weight of $22$ to point $B$ and a weight of $21$ to point $C$, giving $F$ a weight of $43$. In the same manner, using another bisector, we find that $A$ has a weight of $20$. So, now we know $P$ has a weight of $63$, and the ratio of $FP:PA$ is $20:43$. Therefore, the smaller similar triangle $ADE$ is $43/63$ the height of the original triangle $ABC$. So, $DE$ is $43/63$ the size of $BC$. Multiplying this ratio by the length of $BC$, we find $DE$ is $860/63 = m/n$. Therefore, $m+n=\boxed{923}$. | pointpen = black; pathpen = black+linewidth(0.7); pen d = linewidth(0.7) + linetype("4 4"); pair B=(0,0), C=(20,0), A=IP(CR(B,21),CR(C,22)), I=incenter(A,B,C), D=IP((0,I.y)--(20,I.y),A--B), E=IP((0,I.y)--(20,I.y),A--C); D(MP("A",A,N)--MP("B",B)--MP("C",C)--cycle); D(incircle(A,B,C)); D(MP("P",I,(1,2))); D(MP("E",E,NE)--MP("D",D,NW)); MP("20",(B+C)/2); MP("21",(A+B)/2,NW); MP("22",(A+C)/2,NE); /* construct angle bisectors */ path anglebisector (pair X, pair Y, pair Z, path K) { return Y -- IP(Y -- Y + 30 * (bisectorpoint(X,Y,Z)-Y) , K); } D(anglebisector(C,A,B,B--C), d); D(anglebisector(B,C,A,A--B),d); D(anglebisector(C,B,A,A--C),d); | [] |
93 | Triangle $ABC$ has $AB=21$, $AC=22$ and $BC=20$. Points $D$ and $E$ are located on $\overline{AB}$ and $\overline{AC}$, respectively, such that $\overline{DE}$ is parallel to $\overline{BC}$ and contains the center of the inscribed circle of triangle $ABC$. Then $DE=m/n$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.
Contents
1 Problem
2 Solution 1
3 Solution 2
4 Solution 3 (mass points)
5 Solution 4 (Faster)
6 Solution 5
7 Solution 6
8 Solution 7
9 Solution 8 (vectors)
10 See also | 2001 AIME I Problem 7 | Diagram borrowed from Solution 3.
Let the angle bisector of $\angle{A}$ intersects $BC$ at $F$.
Applying the Angle Bisector Theorem on $\angle{A}$ we have
\[\frac{AB}{BF}=\frac{AC}{CF}\]
\[BF=BC\cdot(\frac{AB}{AB+AC})\]
\[BF=\frac{420}{43}\]
Since $BP$ is the angle bisector of $\angle{B}$, we can once again apply the Angle Bisector Theorem on $\angle{B}$ which gives
\[\frac{BA}{AP}=\frac{BF}{FP}\]
\[\frac{AP}{PF}=\frac{AB}{BF}=\frac{41}{20}\]
Since $\bigtriangleup ADE\sim\bigtriangleup ABC$ we have
\[\frac{DE}{BC}=\frac{AP}{AF}\]
\[DE=BC\cdot(\frac{AP}{(\frac{61}{41})\cdot AP})\]
Solving gets $DE=\frac{860}{63}$. Thus $m+n=860+63=\boxed{923}$.
~ Nafer | // Block 1
pointpen = black; pathpen = black+linewidth(0.7); pen d = linewidth(0.7) + linetype("4 4");
pair B=(0,0), C=(20,0), A=IP(CR(B,21),CR(C,22)), I=incenter(A,B,C), D=IP((0,I.y)--(20,I.y),A--B), E=IP((0,I.y)--(20,I.y),A--C);
D(MP("A",A,N)--MP("B",B)--MP("C",C)--cycle); D(incircle(A,B,C)); D(MP("P",I,(1,2))); D(MP("E",E,NE)--MP("D",D,NW));
MP("20",(B+C)/2); MP("21",(A+B)/2,NW); MP("22",(A+C)/2,NE);
/* construct angle bisectors */
path anglebisector (pair X, pair Y, pair Z, path K) { return Y -- IP(Y -- Y + 30 * (bisectorpoint(X,Y,Z)-Y) , K); }
D(anglebisector(C,A,B,B--C), d); D(anglebisector(C,B,A,A--C),d);
// Block 2
pointpen = black; pathpen = black+linewidth(0.7); pen d = linewidth(0.7) + linetype("4 4"); pair B=(0,0), C=(20,0), A=IP(CR(B,21),CR(C,22)), I=incenter(A,B,C), D=IP((0,I.y)--(20,I.y),A--B), E=IP((0,I.y)--(20,I.y),A--C); D(MP("A",A,N)--MP("B",B)--MP("C",C)--cycle); D(incircle(A,B,C)); D(MP("P",I,(1,2))); D(MP("E",E,NE)--MP("D",D,NW)); MP("20",(B+C)/2); MP("21",(A+B)/2,NW); MP("22",(A+C)/2,NE); /* construct angle bisectors */ path anglebisector (pair X, pair Y, pair Z, path K) { return Y -- IP(Y -- Y + 30 * (bisectorpoint(X,Y,Z)-Y) , K); } D(anglebisector(C,A,B,B--C), d); D(anglebisector(C,B,A,A--C),d); | [] |
94 | In triangle $ABC$, $AB=13$, $BC=15$ and $CA=17$. Point $D$ is on $\overline{AB}$, $E$ is on $\overline{BC}$, and $F$ is on $\overline{CA}$. Let $AD=p\cdot AB$, $BE=q\cdot BC$, and $CF=r\cdot CA$, where $p$, $q$, and $r$ are positive and satisfy $p+q+r=2/3$ and $p^2+q^2+r^2=2/5$. The ratio of the area of triangle $DEF$ to the area of triangle $ABC$ can be written in the form $m/n$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. | 2001 AIME I Problem 9 | Solution 1
We let $[\ldots]$ denote area; then the desired value is
$\frac mn = \frac{[DEF]}{[ABC]} = \frac{[ABC] - [ADF] - [BDE] - [CEF]}{[ABC]}$
Using the formula for the area of a triangle $\frac{1}{2}ab\sin C$, we find that
$\frac{[ADF]}{[ABC]} = \frac{\frac 12 \cdot p \cdot AB \cdot (1-r) \cdot AC \cdot \sin \angle CAB}{\frac 12 \cdot AB \cdot AC \cdot \sin \angle CAB} = p(1-r)$
and similarly that $\frac{[BDE]}{[ABC]} = q(1-p)$ and $\frac{[CEF]}{[ABC]} = r(1-q)$. Thus, we wish to find
\begin{align*}\frac{[DEF]}{[ABC]} &= 1 - \frac{[ADF]}{[ABC]} - \frac{[BDE]}{[ABC]} - \frac{[CEF]}{[ABC]} \\ &= 1 - p(1-r) - q(1-p) - r(1-q)\\ &= (pq + qr + rp) - (p + q + r) + 1 \end{align*}
We know that $p + q + r = \frac 23$, and also that $(p+q+r)^2 = p^2 + q^2 + r^2 + 2(pq + qr + rp) \Longleftrightarrow pq + qr + rp = \frac{\left(\frac 23\right)^2 - \frac 25}{2} = \frac{1}{45}$. Substituting, the answer is $\frac 1{45} - \frac 23 + 1 = \frac{16}{45}$, and $m+n = \boxed{061}$.
Solution 2
By the barycentric area formula, our desired ratio is equal to
\begin{align*} \begin{vmatrix} 1-p & p & 0 \\ 0 & 1-q & q \\ r & 0 & 1-r \notag \end{vmatrix} &=1-p-q-r+pq+qr+pr\\ &=1-(p+q+r)+\frac{(p+q+r)^2-(p^2+q^2+r^2)}{2}\\ &=1-\frac{2}{3}+\frac{\frac{4}{9}-\frac{2}{5}}{2}\\ &=\frac{16}{45}, \end{align*} so the answer is $\boxed{061}$
Solution 3 (Informal)
Since the only conditions are that $p + q + r = \frac{2}{3}$ and $p^2 + q^2 + r^2 = \frac{2}{5}$, we can simply let one of the variables be equal to 0. In this case, let $p = 0$. Then, $q + r = \frac{2}{3}$ and $q^2 + r^2$ = $\frac{2}{5}$. Note that the ratio between the area of $DEF$ and $ABC$ is equivalent to $(1-q)(1-r)$. Solving this system of equations, we get $q = \frac{1}{3} \pm \sqrt{\frac{4}{45}}$, and $r = \frac{1}{3} \mp \sqrt{\frac{4}{45}}$. Plugging back into $(1-q)(1-r)$, we get $\frac{16}{45}$, so the answer is $\boxed{061}$
Note
Because the givens in the problem statement are all regarding the ratios of the sides, the side lengths of triangle $ABC$, namely $13, 15, 17$, are actually not necessary to solve the problem. This is clearly demonstrated in all of the above solutions, as the side lengths are not used at all. | /* -- arbitrary values, I couldn't find nice values for pqr please replace if possible -- */ real p = 0.5, q = 0.1, r = 0.05; /* -- arbitrary values, I couldn't find nice values for pqr please replace if possible -- */ pointpen = black; pathpen = linewidth(0.7) + black; pair A=(0,0),B=(13,0),C=IP(CR(A,17),CR(B,15)), D=A+p*(B-A), E=B+q*(C-B), F=C+r*(A-C); D(D(MP("A",A))--D(MP("B",B))--D(MP("C",C,N))--cycle); D(D(MP("D",D))--D(MP("E",E,NE))--D(MP("F",F,NW))--cycle); | [] |
95 | A sphere is inscribed in the tetrahedron whose vertices are $A = (6,0,0), B = (0,4,0), C = (0,0,2),$ and $D = (0,0,0).$ The radius of the sphere is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$ | 2001 AIME I Problem 12 | The center $I$ of the insphere must be located at $(r,r,r)$ where $r$ is the sphere's radius.
$I$ must also be a distance $r$ from the plane $ABC$
The signed distance between a plane and a point $I$ can be calculated as $\frac{(I-G) \cdot P}{|P|}$, where G is any point on the plane, and P is a vector perpendicular to ABC.
A vector $P$ perpendicular to plane $ABC$ can be found as $V=(A-C)\times(B-C)=\langle 8, 12, 24 \rangle$
Thus $\frac{(I-C) \cdot P}{|P|}=-r$ where the negative comes from the fact that we want $I$ to be in the opposite direction of $P$
\begin{align*}\frac{(I-C) \cdot P}{|P|}&=-r\\ \frac{(\langle r, r, r \rangle-\langle 0, 0, 2 \rangle) \cdot P}{|P|}&=-r\\ \frac{\langle r, r, r-2 \rangle \cdot \langle 8, 12, 24 \rangle}{\langle 8, 12, 24 \rangle}&=-r\\ \frac{44r -48}{28}&=-r\\ 44r-48&=-28r\\ 72r&=48\\ r&=\frac{2}{3} \end{align*}
Finally $2+3=\boxed{005}$ | // Block 1
import three;
currentprojection = perspective(-2,9,4);
triple A = (6,0,0), B = (0,4,0), C = (0,0,2), D = (0,0,0);
triple E = (2/3,0,0), F = (0,2/3,0), G = (0,0,2/3), L = (0,2/3,2/3), M = (2/3,0,2/3), N = (2/3,2/3,0);
triple I = (2/3,2/3,2/3);
triple J = (6/7,20/21,26/21);
draw(C--A--D--C--B--D--B--A--C);
draw(L--F--N--E--M--G--L--I--M--I--N--I--J);
label("$I$",I,W);
label("$A$",A,S);
label("$B$",B,S);
label("$C$",C,W*-1);
label("$D$",D,W*-1);
// Block 2
import three; currentprojection = perspective(-2,9,4); triple A = (6,0,0), B = (0,4,0), C = (0,0,2), D = (0,0,0); triple E = (2/3,0,0), F = (0,2/3,0), G = (0,0,2/3), L = (0,2/3,2/3), M = (2/3,0,2/3), N = (2/3,2/3,0); triple I = (2/3,2/3,2/3); triple J = (6/7,20/21,26/21); draw(C--A--D--C--B--D--B--A--C); draw(L--F--N--E--M--G--L--I--M--I--N--I--J); label("$I$",I,W); label("$A$",A,S); label("$B$",B,S); label("$C$",C,W*-1); label("$D$",D,W*-1); | [] |
96 | The numbers $1, 2, 3, 4, 5, 6, 7,$ and $8$ are randomly written on the faces of a regular octahedron so that each face contains a different number. The probability that no two consecutive numbers, where $8$ and $1$ are considered to be consecutive, are written on faces that share an edge is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$ | 2001 AIME I Problem 15 | The probability is equivalent to counting the number of Hamiltonian cycles in this 3D graph over $7!.$ This is because each Hamiltonian cycle corresponds to eight unique ways to label the faces. Label the vertices $AR,BR,CR,DR,AX,BX,CX,DX$ where vertices $ab$ and $cd$ are connected if $a=c$ or $b=d.$
Case 1: Four of the vertical edges are used. $6\cdot 2=12.$
Case 2: Two of the vertical edges are used. $4\cdot 3 \cdot 2\cdot 2=48.$
So, the probability is $\frac{60}{5040}=\frac{1}{84}.$ Therefore, our answer is $\boxed{085}$ | // Block 1
import three;
draw((0,0,0)--(0,0,1)--(0,1,1)--(1,1,1)--(1,0,1)--(1,0,0)--(1,1,0)--(0,1,0)--(0,0,0));
draw((1,1,0)--(1,1,1));
draw((0,1,0)--(0,1,1));
draw((0,0,0)--(1,0,0));
draw((0,0,1)--(1,0,1));
for(int i = 0; i < 2; ++i) {
for(int j = 0; j < 2; ++j) {
for(int k = 0; k < 2; ++k) {
dot((i,j,k));
}
}
}
// dot((0,0,1),blue);
// dot((0,1,0),green);
// dot((1,0,0),red);
draw((0,0,0)--(1,1,0));
draw((0,1,0)--(1,0,0));
draw((0,0,1)--(1,1,1));
draw((0,1,1)--(1,0,1));
// Block 2
import three; draw((0,0,0)--(0,0,1)--(0,1,1)--(1,1,1)--(1,0,1)--(1,0,0)--(1,1,0)--(0,1,0)--(0,0,0)); draw((1,1,0)--(1,1,1)); draw((0,1,0)--(0,1,1)); draw((0,0,0)--(1,0,0)); draw((0,0,1)--(1,0,1)); for(int i = 0; i < 2; ++i) { for(int j = 0; j < 2; ++j) { for(int k = 0; k < 2; ++k) { dot((i,j,k)); } } } // dot((0,0,1),blue); // dot((0,1,0),green); // dot((1,0,0),red); draw((0,0,0)--(1,1,0)); draw((0,1,0)--(1,0,0)); draw((0,0,1)--(1,1,1)); draw((0,1,1)--(1,0,1)); | [] |
97 | Let $R = (8,6)$. The lines whose equations are $8y = 15x$ and $10y = 3x$ contain points $P$ and $Q$, respectively, such that $R$ is the midpoint of $\overline{PQ}$. The length of $PQ$ equals $\frac {m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$. | 2001 AIME II Problem 4 | The coordinates of $P$ can be written as $\left(8a, 15a\right)$ and the coordinates of point $Q$ can be written as $\left(10b, 3b\right)$. By the midpoint formula, we have $8a+10b=16$ and $15a+3b=12$. Substituting $b=4-5a$ we derive $a = \frac{4}{7}$ and $b = \frac{8}{7}$. Thus $P$ is $\left(\frac{32}7, \frac{60}7\right)$, Q is $\left(\frac{80}7, \frac{24}7\right)$, and the coordinates form a 3-4-5 triangle dilated by $\frac{12}7$. Finally the distance $PQ$ must be $\frac{60}{7}$ so the answer is $\boxed{067}$. | // Block 1
pointpen = black; pathpen = black+linewidth(0.7);
pair R = (8,6), P = (32,60)/7, Q= (80,24)/7;
D((0,0)--MP("x",(13,0),E),EndArrow(6)); D((0,0)--MP("y",(0,10),N),EndArrow(6));
D((0,0)--(10/(15/8),10),EndArrow(6)); D((0,0)--(13,13 * 3/10),EndArrow(6));
D(D(MP("P",P,NW))--D(MP("Q",Q),SE),linetype("4 4")); D(MP("R",R,NE));
// Block 2
pointpen = black; pathpen = black+linewidth(0.7); pair R = (8,6), P = (32,60)/7, Q= (80,24)/7; D((0,0)--MP("x",(13,0),E),EndArrow(6)); D((0,0)--MP("y",(0,10),N),EndArrow(6)); D((0,0)--(10/(15/8),10),EndArrow(6)); D((0,0)--(13,13 * 3/10),EndArrow(6)); D(D(MP("P",P,NW))--D(MP("Q",Q),SE),linetype("4 4")); D(MP("R",R,NE)); | [] |
98 | Square $ABCD$ is inscribed in a circle. Square $EFGH$ has vertices $E$ and $F$ on $\overline{CD}$ and vertices $G$ and $H$ on the circle. If the area of square $ABCD$ is $1$, then the area of square $EFGH$ can be expressed as $\frac {m}{n}$ where $m$ and $n$ are relatively prime positive integers and $m < n$. Find $10n + m$. | 2001 AIME II Problem 6 | Begin by drawing another square congruent to $EFGH$ but on side $AB$ instead of side $CD$. Call this square $IJKL$, such that $I$ lies on $\overline{AB}$ and $\overline{IJ}$ lies on line $FG$. Let the side length of $EFGH$ and $IJKL$ be $s$ and the side length of $ABCD$ be $1$.
By Power of a Point on point $F$, we find that
\[FC \cdot FD = FG \cdot FJ\]
\[\frac{1-s}{2} \cdot \frac{1+s}{2} = s(1+s)\]
Expanding and simplifying we get the quadratic $5s^2 +4s-1 = 0$. Factoring, we get $(5s-1)(s+1) = 0$. Therefore, $s= \frac{1}{5}$ and the ratio of the squares' areas is $\frac{1}{25}$. Our answer is therefore $10 \cdot 25 + 1 = \boxed{251}$.
~lprado | // Block 1
size(150); pointpen=black; pathpen=black+linewidth(0.7);
pair A=(-1,-1), B=(-1,1), C=(1,1), D=(1,-1), O=(0,0);
pair E=(1,-0.2), F=(1,0.2), G=(1.4,0.2), H=(1.4,-0.2);
pair I=(-1,0.2), J=(-1.4,0.2), K=(-1.4,-0.2), L=(-1,-0.2);
draw(A--B--C--D--cycle);
draw(circle(O,sqrt(2)));
draw(E--F--G--H--cycle);
draw(I--J--K--L--cycle);
label("$A$",A,SW); label("$B$",B,NW); label("$C$",C,NE); label("$D$",D,SE);
label("$E$",E,SW); label("$F$",F,NW); label("$G$",G,NE); label("$H$",H,SE);
label("$I$",I,NE); label("$J$",J,NW); label("$K$",K,SW); label("$L$",L,SE);
// Block 2
size(150); pointpen=black; pathpen=black+linewidth(0.7); pair A=(-1,-1), B=(-1,1), C=(1,1), D=(1,-1), O=(0,0); pair E=(1,-0.2), F=(1,0.2), G=(1.4,0.2), H=(1.4,-0.2); pair I=(-1,0.2), J=(-1.4,0.2), K=(-1.4,-0.2), L=(-1,-0.2); draw(A--B--C--D--cycle); draw(circle(O,sqrt(2))); draw(E--F--G--H--cycle); draw(I--J--K--L--cycle); label("$A$",A,SW); label("$B$",B,NW); label("$C$",C,NE); label("$D$",D,SE); label("$E$",E,SW); label("$F$",F,NW); label("$G$",G,NE); label("$H$",H,SE); label("$I$",I,NE); label("$J$",J,NW); label("$K$",K,SW); label("$L$",L,SE); | [] |
98 | Square $ABCD$ is inscribed in a circle. Square $EFGH$ has vertices $E$ and $F$ on $\overline{CD}$ and vertices $G$ and $H$ on the circle. If the area of square $ABCD$ is $1$, then the area of square $EFGH$ can be expressed as $\frac {m}{n}$ where $m$ and $n$ are relatively prime positive integers and $m < n$. Find $10n + m$. | 2001 AIME II Problem 6 | Let $O$ be the center of the circle, $2a$ be the side length of square $ABCD$, and $2b$ be the side length of square $EFGH$. By symmetry, the horizontal and vertical displacements of $C$ from $O$ are both $\frac{2a}{2} = a$, so by the Pythagorean theorem, the radius of the circle is $OC = \sqrt{a^2+a^2} = a\sqrt{2}$.
Now let $I$ be the midpoint of $\overline{GH}$, giving $\overline{GI} = \frac{2b}{2} = b$, and let $J$ be the point where $\overline{OI}$ intersects $\overline{CD}$. We observe that since a diameter bisects a chord perpendicular to it, $\overline{GH}$ must be perpendicular to the diameter passing through $I$. This means that triangle $OGI$ has a right angle at $I$, and that $\overline{OJ}$ and $\overline{JI}$ are both parallel to $\overline{BC}$ and $\overline{FG}$. As the horizontal displacement of $C$ from $O$ is $a$ (from above), it follows that $\overline{OJ} = a$, and hence \[\overline{OI} = \overline{OJ}+\overline{JI} = a+\overline{FG} = a+2b,\] so by the Pythagorean Theorem again,
\begin{align*} \overline{OG}^2 = \overline{OI}^2 + \overline{GI}^2 &\iff \left(a\sqrt{2}\right)^2 = (a+2b)^2 + b^2 \\ &\iff 2a^2 = a^2+4ab+4b^2+b^2 \\ &\iff a^2-4ab-5b^2 = 0 \\ &\iff (a-5b)(a+b) = 0 \end{align*}
Since lengths must be positive, we clearly cannot have $a+b = 0$, so the solution must instead be $a = 5b$. Since the ratio of the areas of two similar figures is the square of the ratio of their corresponding side lengths, we deduce that the required ratio is $\left(\frac{1}{5}\right)^2 = \frac{1}{25}$, and so the answer is $10 \cdot 25 + 1 = \boxed{251}$.
Alternative Finish for Solution 1:
After establishing that $a^2-4ab-5b^2 = 0$, another way to proceed is to observe that what we actually need to calculate is $\frac{b}{a}$. Accordingly, we divide both sides of this equation by $a^2$ (or equivalently, choose units such that the area of square $ABCD$ is $1$, without loss of generality), giving
\[1-4\left(\frac{b}{a}\right)-5\left(\frac{b}{a}\right)^2 = 0,\]
which is a quadratic in precisely the variable $\frac{b}{a}$. Thus, by solving it, we immediately obtain $\frac{b}{a} = \frac{1}{5}$ or $\frac{b}{a} = -1$, and as in Solution 1a, the negative root is obviously extraneous. Hence the required ratio of areas is $\left(\frac{1}{5}\right)^2 = \frac{1}{25}$, and so the answer is $10 \cdot 25 + 1 = \boxed{251}$. | // Block 1
size(150); pointpen = black; pathpen = black+linewidth(0.7); pen d = linetype("4 4") + blue + linewidth(0.7);
pair C=(1,1), D=(1,-1), B=(-1,1), A=(-1,-1), E= (1, -0.2), F=(1, 0.2), G=(1.4, 0.2), H=(1.4, -0.2);
D(MP("A",A)--MP("B",B,N)--MP("C",C,N)--MP("D",D)--cycle); D(MP("E",E,SW)--MP("F",F,NW)--MP("G",G,NE)--MP("H",H,SE)--cycle); D(CP(D(MP("O",(0,0))), A));
D((0,0) -- (2^.5, 0), d); D((0,0) -- G -- (G.x,0), d);
// Block 2
size(150); pointpen = black; pathpen = black+linewidth(0.7); pen d = linetype("4 4") + blue + linewidth(0.7); pair C=(1,1), D=(1,-1), B=(-1,1), A=(-1,-1), E= (1, -0.2), F=(1, 0.2), G=(1.4, 0.2), H=(1.4, -0.2); D(MP("A",A)--MP("B",B,N)--MP("C",C,N)--MP("D",D)--cycle); D(MP("E",E,SW)--MP("F",F,NW)--MP("G",G,NE)--MP("H",H,SE)--cycle); D(CP(D(MP("O",(0,0))), A)); D((0,0) -- (2^.5, 0), d); D((0,0) -- G -- (G.x,0), d); | [] |
99 | Let $\triangle{PQR}$ be a right triangle with $PQ = 90$, $PR = 120$, and $QR = 150$. Let $C_{1}$ be the inscribed circle. Construct $\overline{ST}$ with $S$ on $\overline{PR}$ and $T$ on $\overline{QR}$, such that $\overline{ST}$ is perpendicular to $\overline{PR}$ and tangent to $C_{1}$. Construct $\overline{UV}$ with $U$ on $\overline{PQ}$ and $V$ on $\overline{QR}$ such that $\overline{UV}$ is perpendicular to $\overline{PQ}$ and tangent to $C_{1}$. Let $C_{2}$ be the inscribed circle of $\triangle{RST}$ and $C_{3}$ the inscribed circle of $\triangle{QUV}$. The distance between the centers of $C_{2}$ and $C_{3}$ can be written as $\sqrt {10n}$. What is $n$?
Contents
1 Problem
2 Solution
2.1 Solution 1 (analytic)
2.2 Solution 2 (synthetic)
2.3 Solution 3
2.4 Note
2.5 Solution 4 (easy but hard to see)
3 See also | 2001 AIME II Problem 7 | Solution 1 (analytic)
Let $P = (0,0)$ be at the origin. Using the formula $A = rs$ on $\triangle PQR$, where $r_{1}$ is the inradius (similarly define $r_2, r_3$ to be the radii of $C_2, C_3$), $s = \frac{PQ + QR + RP}{2} = 180$ is the semiperimeter, and $A = \frac 12 bh = 5400$ is the area, we find $r_{1} = \frac As = 30$. Or, the inradius could be directly by using the formula $\frac{a+b-c}{2}$, where $a$ and $b$ are the legs of the right triangle and $c$ is the hypotenuse. (This formula should be used only for right triangles.) Thus $ST, UV$ lie respectively on the lines $y = 60, x = 60$, and so $RS = 60, UQ = 30$.
Note that $\triangle PQR \sim \triangle STR \sim \triangle UQV$. Since the ratio of corresponding lengths of similar figures are the same, we have
\[\frac{r_{1}}{PR} = \frac{r_{2}}{RS} \Longrightarrow r_{2} = 15\ \text{and} \ \frac{r_{1}}{PQ} = \frac{r_{3}}{UQ} \Longrightarrow r_{3} = 10.\]
Let the centers of $\odot C_2, C_3$ be $O_2 = (0 + r_{2}, 60 + r_{2}) = (15, 75), O_3 = (60 + r_{3}, 0 + r_{3}) = (70,10)$, respectively; then by the distance formula we have $O_2O_3 = \sqrt{55^2 + 65^2} = \sqrt{10 \cdot 725}$. Therefore, the answer is $n = \boxed{725}$.
Solution 2 (synthetic)
We compute $r_1 = 30, r_2 = 15, r_3 = 10$ as above. Let $A_1, A_2, A_3$ respectively the points of tangency of $C_1, C_2, C_3$ with $QR$.
By the Two Tangent Theorem, we find that $A_{1}Q = 60$, $A_{1}R = 90$. Using the similar triangles, $RA_{2} = 45$, $QA_{3} = 20$, so $A_{2}A_{3} = QR - RA_2 - QA_3 = 85$. Thus $(O_{2}O_{3})^{2} = (15 - 10)^{2} + (85)^{2} = 7250\implies n=\boxed{725}$.
Solution 3
The radius of an incircle is $r=A_t/\text{semiperimeter}$. The area of the triangle is equal to $\frac{90\times120}{2} = 5400$ and the semiperimeter is equal to $\frac{90+120+150}{2} = 180$. The radius, therefore, is equal to $\frac{5400}{180} = 30$. Thus using similar triangles the dimensions of the triangle circumscribing the circle with center $C_2$ are equal to $120-2(30) = 60$, $\frac{1}{2}(90) = 45$, and $\frac{1}{2}\times150 = 75$. The radius of the circle inscribed in this triangle with dimensions $45\times60\times75$ is found using the formula mentioned at the very beginning. The radius of the incircle is equal to $15$.
Defining $P$ as $(0,0)$, $C_2$ is equal to $(60+15,15)$ or $(75,15)$. Also using similar triangles, the dimensions of the triangle circumscribing the circle with center $C_3$ are equal to $90-2(30)$, $\frac{1}{3}\times120$, $\frac{1}{3}\times150$ or $30,40,50$. The radius of $C_3$ by using the formula mentioned at the beginning is $10$. Using $P$ as $(0,0)$, $C_3$ is equal to $(10, 60+10)$ or $(10,70)$. Using the distance formula, the distance between $C_2$ and $C_3$: $\sqrt{(75-10)^2 +(15-70)^2}$ this equals $\sqrt{7250}$ or $\sqrt{725\times10}$, thus $n$ is $\boxed{725}$.
Note
The problem can be reduced to a $3-4-5$ triangle for the initial calculations, where $C_2$ is calculated to be ($\frac{5}{2}, \frac{1}{2})$, and $C_3$ is calculated to be ($\frac{1}{3}, \frac{7}{3})$. After we find the incenters the points can be scaled up by a factor of $30$ for the final distance calculation.
Solution 4 (easy but hard to see)
We can calculate the inradius of each triangle as with the previous solutions. Now, notice that the pentagon $PSTVU$ is a square with its corner cut off. We literally complete the square and mark the last corner as point X. Now, construct right triangle with $C_3C_2$ as its hypotenuse. The right angle will be at point $Y$. We will now find the length of each leg and use the Pythagorean Theorem to compute the desired length. We see that the length of $C_3Y$ is $50 + 15 = 65$, as seen by the inradius of $C_2$ and $10$ less than the square's side length. $C_2Y$ is $45 + 10 = 55$, which is $15$ less than the square plus the inradius of $C_3$. Our final answer is $\sqrt{65^2 + 55^2} = \sqrt{7250} = \sqrt{(10)(725)} \rightarrow \boxed{725}$. | // Block 1
pointpen = black; pathpen = black + linewidth(0.7); pair P = (0,0), Q = (90, 0), R = (0, 120), S=(0, 60), T=(45, 60), U = (60,0), V=(60, 40), O1 = (30,30), O2 = (15, 75), O3 = (70, 10); D(MP("P",P)--MP("Q",Q)--MP("R",R,W)--cycle); D(MP("S",S,W) -- MP("T",T,NE)); D(MP("U",U) -- MP("V",V,NE)); D(O2 -- O3, rgb(0.2,0.5,0.2)+ linewidth(0.7) + linetype("4 4")); D(CR(D(O1), 30)); D(CR(D(O2), 15)); D(CR(D(O3), 10));
// Block 2
pointpen = black; pathpen = black + linewidth(0.7); pair P = (0,0), Q = (90, 0), R = (0, 120), S=(0, 60), T=(45, 60), U = (60,0), V=(60, 40), O1 = (30,30), O2 = (15, 75), O3 = (70, 10); D(MP("P",P)--MP("Q",Q)--MP("R",R,W)--cycle); D(MP("S",S,W) -- MP("T",T,NE)); D(MP("U",U) -- MP("V",V,NE)); D(O2 -- O3, rgb(0.2,0.5,0.2)+ linewidth(0.7) + linetype("4 4")); D(CR(D(O1), 30)); D(CR(D(O2), 15)); D(CR(D(O3), 10)); pair A2 = IP(incircle(R,S,T), Q--R), A3 = IP(incircle(Q,U,V), Q--R); D(D(MP("A_2",A2,NE)) -- O2, linetype("4 4")+linewidth(0.6)); D(D(MP("A_3",A3,NE)) -- O3 -- foot(O3, A2, O2), linetype("4 4")+linewidth(0.6)); | [] |
100 | In quadrilateral $ABCD$, $\angle{BAD}\cong\angle{ADC}$ and $\angle{ABD}\cong\angle{BCD}$, $AB = 8$, $BD = 10$, and $BC = 6$. The length $CD$ may be written in the form $\frac {m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$. | 2001 AIME II Problem 13 | Notice that $\angle BAD = \angle ADC$ and $\angle ABD = \angle DCB$, which motivates us to find similar triangles. We extend $CB$ and $DA$ to point $E$. We know that $\triangle ABD \sim \triangle DCE$. Hence $\angle ADB = \angle DEC$, and $\triangle BDE$ is isosceles. Then $BD = BE = 10$.
Using the similarity, we have:
\[\frac{AB}{BD} = \frac 8{10} = \frac{CD}{CE} = \frac{CD}{16} \Longrightarrow CD = \frac{64}5\]
The answer is $m+n = \boxed{069}$.
~lprado
Extension: To Find $AD$, use Law of Cosines on $\triangle BCD$ to get $\cos(\angle BCD)=\frac{13}{20}$
Then since $\angle BCD=\angle ABD$ use Law of Cosines on $\triangle ABD$ to find $AD=2\sqrt{15}$ | // Block 1
/* We arbitrarily set AD = x */
real x = 60^.5, anglesize = 28;
pointpen = black; pathpen = black+linewidth(0.7); pen d = linetype("6 6")+linewidth(0.7);
pair A=(0,0), D=(x,0), B=IP(CR(A,8),CR(D,10)), E=(-3x/5,0), C=IP(CR(E,16),CR(D,64/5));
D(MP("A",A)--MP("B",B,NW)--MP("C",C,NW)--MP("D",D)--cycle); D(B--D); D(A--MP("E",E)--B,d);
D(anglemark(D,A,B,anglesize));D(anglemark(C,D,A,anglesize));D(anglemark(A,B,D,anglesize));D(anglemark(E,C,D,anglesize));D(anglemark(A,B,D,5/4*anglesize));D(anglemark(E,C,D,5/4*anglesize));
MP("10",(B+D)/2,SW);MP("8",(A+B)/2,W);MP("6",(B+C)/2,NW);
// Block 2
/* We arbitrarily set AD = x */ real x = 60^.5, anglesize = 28; pointpen = black; pathpen = black+linewidth(0.7); pen d = linetype("6 6")+linewidth(0.7); pair A=(0,0), D=(x,0), B=IP(CR(A,8),CR(D,10)), E=(-3x/5,0), C=IP(CR(E,16),CR(D,64/5)); D(MP("A",A)--MP("B",B,NW)--MP("C",C,NW)--MP("D",D)--cycle); D(B--D); D(A--MP("E",E)--B,d); D(anglemark(D,A,B,anglesize));D(anglemark(C,D,A,anglesize));D(anglemark(A,B,D,anglesize));D(anglemark(E,C,D,anglesize));D(anglemark(A,B,D,5/4*anglesize));D(anglemark(E,C,D,5/4*anglesize)); MP("10",(B+D)/2,SW);MP("8",(A+B)/2,W);MP("6",(B+C)/2,NW); | [] |
101 | Let $EFGH$, $EFDC$, and $EHBC$ be three adjacent square faces of a cube, for which $EC = 8$, and let $A$ be the eighth vertex of the cube. Let $I$, $J$, and $K$ be the points on $\overline{EF}$, $\overline{EH}$, and $\overline{EC}$, respectively, so that $EI = EJ = EK = 2$. A solid $S$ is obtained by drilling a tunnel through the cube. The sides of the tunnel are planes parallel to $\overline{AE}$, and containing the edges, $\overline{IJ}$, $\overline{JK}$, and $\overline{KI}$. The surface area of $S$, including the walls of the tunnel, is $m + n\sqrt {p}$, where $m$, $n$, and $p$ are positive integers and $p$ is not divisible by the square of any prime. Find $m + n + p$. | 2001 AIME II Problem 15 | Set the coordinate system so that vertex $E$, where the drilling starts, is at $(8,8,8)$. Using a little visualization (involving some similar triangles, because we have parallel lines) shows that the tunnel meets the bottom face (the xy plane one) in the line segments joining $(1,0,0)$ to $(2,2,0)$, and $(0,1,0)$ to $(2,2,0)$, and similarly for the other three faces meeting at the origin (by symmetry). So one face of the tunnel is the polygon with vertices (in that order), $S(1,0,0), T(2,0,2), U(8,6,8), V(8,8,6), W(2,2,0)$, and the other two faces of the tunnel are congruent to this shape.
Observe that this shape is made up of two congruent trapezoids each with height $\sqrt {2}$ and bases $7\sqrt {3}$ and $6\sqrt {3}$. Together they make up an area of $\sqrt {2}(7\sqrt {3} + 6\sqrt {3}) = 13\sqrt {6}$. The total area of the tunnel is then $3\cdot13\sqrt {6} = 39\sqrt {6}$. Around the corner $E$ we're missing an area of $6$, the same goes for the corner opposite $E$ . So the outside area is $6\cdot 64 - 2\cdot 6 = 372$. Thus the the total surface area is $372 + 39\sqrt {6}$, and the answer is $372 + 39 + 6 = \boxed{417}$. | // Block 1
import three; currentprojection = orthographic(camera=(1/4,2,3/4)); defaultpen(linewidth(0.7)); pen l = linewidth(0.5) + linetype("10 2");
triple S=(1,0,0), T=(2,0,2), U=(8,6,8), V=(8,8,6), W=(2,2,0), X=(6,8,8);
draw((1,0,0)--(8,0,0)--(8,0,8)--(0,0,8)--(0,0,1)); draw((1,0,0)--(8,0,0)--(8,8,0)--(0,8,0)--(0,1,0),l); draw((0,8,0)--(0,8,8)); draw((0,8,8)--(0,0,8)--(0,0,1)); draw((8,8,0)--(8,8,6),l); draw((8,0,8)--(8,6,8)); draw((0,8,8)--(6,8,8));
draw(S--T--U--V--W--cycle); draw((0,0,1)--T--U--X--(0,2,2)--cycle); draw((0,1,0)--W--V--X--(0,2,2)--cycle);
// Block 2
import three; currentprojection = orthographic(camera=(1/2,1/3,-1/4)); defaultpen(linewidth(0.7)); pen l = linewidth(0.5) + linetype("10 2");
triple S=(1,0,0), T=(2,0,2), U=(8,6,8), V=(8,8,6), W=(2,2,0), X=(6,8,8);
draw((1,0,0)--(8,0,0)--(8,0,8),l); draw((8,0,8)--(0,0,8)); draw((0,0,8)--(0,0,1),l); draw((8,0,0)--(8,8,0)); draw((8,8,0)--(0,8,0)); draw((0,8,0)--(0,1,0),l); draw((0,8,0)--(0,8,8)); draw((0,0,8)--(0,0,1),l); draw((8,8,0)--(8,8,6)); draw((8,0,8)--(8,6,8)); draw((0,0,8)--(0,8,8)--(6,8,8));
draw(S--T--U--V--W--cycle); draw((0,0,1)--T--U--X--(0,2,2)--cycle); draw((0,1,0)--W--V--X--(0,2,2)--cycle);
// Block 3
import three; currentprojection = orthographic(camera=(1/4,2,3/4)); defaultpen(linewidth(0.7)); pen l = linewidth(0.5) + linetype("10 2"); triple S=(1,0,0), T=(2,0,2), U=(8,6,8), V=(8,8,6), W=(2,2,0), X=(6,8,8); draw((1,0,0)--(8,0,0)--(8,0,8)--(0,0,8)--(0,0,1)); draw((1,0,0)--(8,0,0)--(8,8,0)--(0,8,0)--(0,1,0),l); draw((0,8,0)--(0,8,8)); draw((0,8,8)--(0,0,8)--(0,0,1)); draw((8,8,0)--(8,8,6),l); draw((8,0,8)--(8,6,8)); draw((0,8,8)--(6,8,8)); draw(S--T--U--V--W--cycle); draw((0,0,1)--T--U--X--(0,2,2)--cycle); draw((0,1,0)--W--V--X--(0,2,2)--cycle);
// Block 4
import three; currentprojection = orthographic(camera=(1/2,1/3,-1/4)); defaultpen(linewidth(0.7)); pen l = linewidth(0.5) + linetype("10 2"); triple S=(1,0,0), T=(2,0,2), U=(8,6,8), V=(8,8,6), W=(2,2,0), X=(6,8,8); draw((1,0,0)--(8,0,0)--(8,0,8),l); draw((8,0,8)--(0,0,8)); draw((0,0,8)--(0,0,1),l); draw((8,0,0)--(8,8,0)); draw((8,8,0)--(0,8,0)); draw((0,8,0)--(0,1,0),l); draw((0,8,0)--(0,8,8)); draw((0,0,8)--(0,0,1),l); draw((8,8,0)--(8,8,6)); draw((8,0,8)--(8,6,8)); draw((0,0,8)--(0,8,8)--(6,8,8)); draw(S--T--U--V--W--cycle); draw((0,0,1)--T--U--X--(0,2,2)--cycle); draw((0,1,0)--W--V--X--(0,2,2)--cycle); | [] |
102 | In triangle $ABC$ the medians $\overline{AD}$ and $\overline{CE}$ have lengths $18$ and $27$, respectively, and $AB=24$. Extend $\overline{CE}$ to intersect the circumcircle of $ABC$ at $F$. The area of triangle $AFB$ is $m\sqrt{n}$, where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. Find $m+n$. | 2002 AIME I Problem 13 | Applying Stewart's Theorem to medians $AD, CE$, we have:
\begin{align*} BC^2 + 4 \cdot 18^2 &= 2\left(24^2 + AC^2\right) \\ 24^2 + 4 \cdot 27^2 &= 2\left(AC^2 + BC^2\right) \end{align*}
Substituting the first equation into the second and simplification yields $24^2 = 2\left(3AC^2 + 2 \cdot 24^2 - 4 \cdot 18^2\right)- 4 \cdot 27^2$ $\Longrightarrow AC = \sqrt{2^5 \cdot 3 + 2 \cdot 3^5 + 2^4 \cdot 3^3 - 2^7 \cdot 3} = 3\sqrt{70}$.
By the Power of a Point Theorem on $E$, we get $EF = \frac{12^2}{27} = \frac{16}{3}$. The Law of Cosines on $\triangle ACE$ gives
\begin{align*} \cos \angle AEC = \left(\frac{12^2 + 27^2 - 9 \cdot 70}{2 \cdot 12 \cdot 27}\right) = \frac{3}{8} \end{align*}
Hence $\sin \angle AEC = \sqrt{1 - \cos^2 \angle AEC} = \frac{\sqrt{55}}{8}$ (taking the positive square root since $0^{\circ} < \angle AEC < 180^{\circ}$, so $\sin\angle AEC > 0$). Using the identity $\sin x^{\circ} \equiv \sin\left(180-x\right)^{\circ}$, we now deduce $\sin \angle AEF = \frac{\sqrt{55}}{8}$. Finally, because $\triangle AEF, BEF$ have the same height and equal bases, they have the same area, yielding \[[ABF] = 2[AEF] = 2 \cdot \frac 12 \cdot AE \cdot EF \sin \angle AEF = 12 \cdot \frac{16}{3} \cdot \frac{\sqrt{55}}{8} = 8\sqrt{55},\] and so the answer is $8 + 55 = \boxed{063}$. | // Block 1
size(150); pathpen = linewidth(0.7); pointpen = black; pen f = fontsize(8); pair A=(0,0), B=(24,0), E=(A+B)/2, C=IP(CR(A,3*70^.5),CR(E,27)), D=(B+C)/2, F=IP(circumcircle(A,B,C),E--C+2*(E-C));
D(D(MP("A",A))--D(MP("B",B))--D(MP("C",C,NW))--cycle); D(circumcircle(A,B,C)); D(MP("F",F)); D(A--D); D(C--F); D(A--F--B); D(MP("E",E,NE)); D(MP("D",D,NE)); MP("12",(A+E)/2,SE,f);MP("12",(B+E)/2,f); MP("27",(C+E)/2,SW,f); MP("18",(A+D)/2,SE,f);
// Block 2
size(150); pathpen = linewidth(0.7); pointpen = black; pen f = fontsize(8); pair A=(0,0), B=(24,0), E=(A+B)/2, C=IP(CR(A,3*70^.5),CR(E,27)), D=(B+C)/2, F=IP(circumcircle(A,B,C),E--C+2*(E-C)); D(D(MP("A",A))--D(MP("B",B))--D(MP("C",C,NW))--cycle); D(circumcircle(A,B,C)); D(MP("F",F)); D(A--D); D(C--F); D(A--F--B); D(MP("E",E,NE)); D(MP("D",D,NE)); MP("12",(A+E)/2,SE,f);MP("12",(B+E)/2,f); MP("27",(C+E)/2,SW,f); MP("18",(A+D)/2,SE,f); | [] |
103 | Polyhedron $ABCDEFG$ has six faces. Face $ABCD$ is a square with $AB = 12;$ face $ABFG$ is a trapezoid with $\overline{AB}$ parallel to $\overline{GF},$ $BF = AG = 8,$ and $GF = 6;$ and face $CDE$ has $CE = DE = 14.$ The other three faces are $ADEG, BCEF,$ and $EFG.$ The distance from $E$ to face $ABCD$ is 12. Given that $EG^2 = p - q\sqrt {r},$ where $p, q,$ and $r$ are positive integers and $r$ is not divisible by the square of any prime, find $p + q + r.$ | 2002 AIME I Problem 15 | Let's put the polyhedron onto a coordinate plane. For simplicity, let the origin be the center of the square: $A(-6,6,0)$, $B(-6,-6,0)$, $C(6,-6,0)$ and $D(6,6,0)$. Since $ABFG$ is an isosceles trapezoid and $CDE$ is an isosceles triangle, we have symmetry about the $xz$-plane.
Therefore, the $y$-component of $E$ is 0. We are given that the $z$ component is 12, and it lies over the square, so we must have $E(2,0,12)$ so $CE=\sqrt{4^2+6^2+12^2}=\sqrt{196}=14$ (the other solution, $E(10,0,12)$ does not lie over the square). Now let $F(a,-3,b)$ and $G(a,3,b)$, so $FG=6$ is parallel to $\overline{AB}$. We must have $BF=8$, so $(a+6)^2+b^2=8^2-3^2=55$.
The last piece of information we have is that $ADEG$ (and its reflection, $BCEF$) are faces of the polyhedron, so they must all lie in the same plane. Since we have $A$, $D$, and $E$, we can derive this plane.* Let $H$ be the extension of the intersection of the lines containing $\overline{AG}, \overline{BF}$. It follows that the projection of $\triangle AHB$ onto the plane $x = 6$ must coincide with the $\triangle CDE'$, where $E'$ is the projection of $E$ onto the plane $x = 6$. $\triangle GHF \sim \triangle AHB$ by a ratio of $1/2$, so the distance from $H$ to the plane $x = -6$ is \[\sqrt{\left(\sqrt{(2 \times 8)^2 - 6^2}\right)^2 - 12^2} = 2\sqrt{19};\] and by the similarity, the distance from $G$ to the plane $x = -6$ is $\sqrt{19}$. The altitude from $G$ to $ABCD$ has height $12/2 = 6$. By similarity, the x-coordinate of $G$ is $-6/2 = -3$. Then $G = (-6 \pm \sqrt{19}, -3, 6)$.
Now that we have located $G$, we can calculate $EG^2$:
\[EG^2=(8\pm\sqrt{19})^2+3^2+6^2=64\pm16\sqrt{19}+19+9+36=128\pm16\sqrt{19}.\] Taking the negative root because the answer form asks for it, we get $128-16\sqrt{19}$, and $128+16+19=\fbox{163}$.
One may also do this by vectors; $\overrightarrow{AD}\times\overrightarrow{DE}=(12,0,0)\times(-4,-6,12)=(0,-144,-72)=-72(0,2,1)$, so the plane is $2y+z=2\cdot6=12$. Since $G$ lies on this plane, we must have $2\cdot3+b=12$, so $b=6$. Therefore, $a=-6\pm\sqrt{55-6^2}=-6\pm\sqrt{19}$. So $G(-6\pm\sqrt{19},-3,6)$. | // Block 1
size(200);
import three;
import graph;
defaultpen(linewidth(0.7)+fontsize(8));
currentprojection=orthographic(-30,50,40);
triple A=(-6,-6,0), B = (-6,6,0), C = (6,6,0), D = (6,-6,0), E = (2,0,12), H=(-6+2*sqrt(19),0,12), H1=(-6-2*sqrt(19),0,12), F, G, E1 = (6,0,12);
F = 1/2*H+1/2*B;
G = 1/2*H+1/2*A;
draw((A--B--C--D--A)^^(D--E--C)^^(A--G--F--B)^^(G--E--F));draw((G--H--F)^^(H--E1),gray(0.6));
dot(H1^^H,linewidth(2));
label("$A$",A,( 0,-1, 0));
label("$B$",B,( 0, 1, 0));
label("$C$",C,( 0, 1, 0));
label("$D$",D,( 0,-1, 0));
label("$E$",E,(-1,-1, 1));
label("$F$",F,( 0, 1, 0));
label("$G$",G,(-1,-1, 1));
label("$H$",H,( 1,-1, 1));
label("$H'$",H1,(-1,-1, 1));
// Block 2
size(200); import three; import graph; defaultpen(linewidth(0.7)+fontsize(8)); currentprojection=orthographic(-30,50,40); triple A=(-6,-6,0), B = (-6,6,0), C = (6,6,0), D = (6,-6,0), E = (2,0,12), H=(-6+2*sqrt(19),0,12), H1=(-6-2*sqrt(19),0,12), F, G, E1 = (6,0,12); F = 1/2*H+1/2*B; G = 1/2*H+1/2*A; draw((A--B--C--D--A)^^(D--E--C)^^(A--G--F--B)^^(G--E--F));draw((G--H--F)^^(H--E1),gray(0.6)); dot(H1^^H,linewidth(2)); label("$A$",A,( 0,-1, 0)); label("$B$",B,( 0, 1, 0)); label("$C$",C,( 0, 1, 0)); label("$D$",D,( 0,-1, 0)); label("$E$",E,(-1,-1, 1)); label("$F$",F,( 0, 1, 0)); label("$G$",G,(-1,-1, 1)); label("$H$",H,( 1,-1, 1)); label("$H'$",H1,(-1,-1, 1)); | [] |
104 | A basketball player has a constant probability of $.4$ of making any given shot, independent of previous shots. Let $a_n$ be the ratio of shots made to shots attempted after $n$ shots. The probability that $a_{10} = .4$ and $a_n\le.4$ for all $n$ such that $1\le n\le9$ is given to be $p^aq^br/\left(s^c\right)$ where $p$, $q$, $r$, and $s$ are primes, and $a$, $b$, and $c$ are positive integers. Find $\left(p+q+r+s\right)\left(a+b+c\right)$. | 2002 AIME II Problem 12 | Solution 1
We graph the $10$ shots on a grid. Suppose that a made shot is represented by a step of $(0,1)$, and a missed shot is represented by $(1,0)$. Then the basketball player's shots can be represented by the number of paths from $(0,0)$ to $(6,4)$ that always stay below the line $y=\frac{2x}{3}$. We can find the number of such paths using a Pascal's Triangle type method below, computing the number of paths to each point that only move right and up.
Therefore, there are $23$ ways to shoot $4$ makes and $6$ misses under the given conditions. The probability of each possible sequence occurring is $(.4)^4(.6)^6$. Hence the desired probability is
\[\frac{23\cdot 2^4\cdot 3^6}{5^{10}},\]
and the answer is $(23+2+3+5)(4+6+10)=\boxed{660}$.
Solution 2
The first restriction is that $a_{10} = .4$, meaning that the player gets exactly 4 out of 10 baskets. The second restriction is $a_n\le.4$. This means that the player may never have a shooting average over 40%. Thus, the first and second shots must fail, since $\frac{1}{1}$ and $\frac{1}{2}$ are both over $.4$, but the player may make the third basket, since $\frac{1}{3} \le .4$ In other words, the earliest the first basket may be made is attempt 3. Using similar reasoning, the earliest the second basket may be made is attempt 5, the earliest the third basket may be made is attempt 8, and the earliest the fourth basket may be made is attempt 10.
Using X to represent a basket and O to represent a failure, this 'earliest' solution may be represented as:
OOXOXOOXOX
To simplify counting, note that the first, second, and tenth shots are predetermined. The first two shots must fail, and the last shot must succeed. Thus, only slots 3-9 need to be counted, and can be abbreviated as follows:
XOXOOXO
The problem may be separated into five cases, since the first shot may be made on attempt 3, 4, 5, 6, or 7. The easiest way to count the problem is to remember that each X may slide to the right, but NOT to the left.
First shot made on attempt 3:
XOXOOXO
XOXOOOX
XOOXOXO
XOOXOOX
XOOOXXO
XOOOXOX
XOOOOXX
Total - 7
First shot made on attempt 4:
Note that all that needs to be done is change each line in the prior case from starting with "XO....." to "OX.....".
Total - 7
First shot made on attempt 5:
OOXXOXO
OOXXOOX
OOXOXXO
OOXOXOX
OOXOOXX
Total - 5
First shot made on attempt 6:
OOOXXXO
OOOXXOX
OOOXOXX
Total - 3
First shot made on attempt 7:
OOOOXXX
Total - 1
The total number of ways the player may satisfy the requirements is $7+7+5+3+1=23$.
The chance of hitting any individual combination (say, for example, OOOOOOXXXX) is $\left(\frac{3}{5}\right)^6\left(\frac{2}{5}\right)^4$
Thus, the chance of hitting any of these 23 combinations is $23\left(\frac{3}{5}\right)^6\left(\frac{2}{5}\right)^4 = \frac{23\cdot3^6\cdot2^4}{5^{10}}$
Thus, the final answer is $(23+3+2+5)(6+4+10)=\boxed{660}$
Solution 3
Note $a_{10}=.4$. Therefore the player made 4 shots out of 10. He must make the 10th shot, because if he doesn't, then $a_9=\frac{4}{9}>.4$. Since $a_n\leq .4$ for all $n$ less than 11, we know that $a_1=a_2=0$. Now we must look at the 3rd through 9th shot.
Now let's take a look at those un-determined shots. Let's put them into groups: the 3rd, 4th, and 5th shots in group A, and the 6th, 7th, 8th, and 9th shots in group B. The total number of shots made in groups A and B must be 3, since the player makes the 10th shot. We cannot have all three shots made in group A, since $a_5\leq .4$. Therefore we can have two shots made, one shot made, or no shots made in group A.
Case 1: Group A contains no made shots.
The number of ways this can happen in group A is 1. Now we must arrange the shots in group B accordingly. There are four ways to arrange them total, and all of them work. There are $\textbf{4}$ possibilities here.
Case 2: Group A contains one made shot.
The number of ways this could happen in group A is 3. Now we must arrange the shots in group B accordingly. There are six ways to arrange them total, but the arrangement "hit hit miss miss" fails, because that would mean $a_7=\frac{3}{7}>.4$. All the rest work. Therefore there are $3\cdot5=\textbf{15}$ possibilities here.
Case 3: Group A contains two made shots.
The number of ways this could happen in group A is 2 (hit hit miss doesn't work but the rest do). Now we must arrange the shots in group B accordingly. Note hit miss miss miss and miss hit miss miss fail. Therefore there are only 2 ways to do this, and there are $2\cdot 2=\textbf{4}$ total possibilities for this case.
Taking all these cases into account, we find that there are $4+15+4=23$ ways to have $a_{10} = .4$ and $a_n\leq .4$. Each of these has a probability of $.4^4\cdot.6^6=\frac{2^4\cdot 3^6}{5^{10}}$. Therefore the probability that we have $a_{10} = .4$ and $a_n\leq .4$ is $\frac{23\cdot2^4\cdot3^6}{5^{10}}$. Now we are asked to find the product of the sum of the primes and the sum of the exponents, which is $(23+2+3+5)(4+6+10)=33\cdot20=\boxed{660}$.
Solution 4
To satisfy $a_{10} = 0.4$, the player must make exactly 4 shots out of 10. Let $k_n$ be the total number of shots made after $n$ attempts. The condition $a_n \le 0.4$ means $k_n \le 0.4n$. The maximum allowable values of $k_n$ are:
\[\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|} \hline \text{Shot } (n) & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ \hline \text{Max Makes } (k_n) & 0 & 0 & 1 & 1 & 2 & 2 & 2 & 3 & 3 & 4 \\ \hline \end{array}\]
At $n=9$, the player can have at most 3 makes. Since 4 makes are required by $n=10$, the tenth shot must be a make. Split the sequence into Group A (shots 1–7) and Group B (shots 8–10). The remaining three makes must be distributed between these two groups.
Case 1: 1 shot made in Group A, 3 shots made in Group B
Group A: The make cannot be on shots 1 or 2, since the maximum allowed number of makes there is 0. Thus the make may occur on any of shots 3 through 7. There are 5 ways.
Group B: All three shots must be makes. At $n=9$, $k_9 = 3$, and $3/9 \le 0.4$, so this case is valid. There is 1 way.
Subtotal: $5 \times 1 = 5$ ways.
Case 2: 2 shots made in Group A, 2 shots made in Group B
Group A: Two makes may be placed as follows.
If both occur among shots 5, 6, or 7, there are $\binom{3}{2} = 3$ ways.
If one occurs among shots 3 or 4 and one among shots 5, 6, or 7, there are $2 \times 3 = 6$ ways.
Thus Group A contributes $3 + 6 = 9$ ways.
Group B: Since shot 10 must be a make, the remaining make must occur on shot 8 or 9. There are 2 ways.
Subtotal: $9 \times 2 = 18$ ways.
Final Calculation
The total number of valid sequences is $5 + 18 = 23$. Each sequence has probability
\[(0.4)^4(0.6)^6 = \left(\frac{2}{5}\right)^4\left(\frac{3}{5}\right)^6 = \frac{2^4 \cdot 3^6}{5^{10}}.\]
Thus the desired probability is
\[\frac{23 \cdot 2^4 \cdot 3^6}{5^{10}}.\]
Writing this in the form $\frac{p^a q^b r}{s^c}$ gives $p=2$, $q=3$, $r=23$, $s=5$, with $a=4$, $b=6$, and $c=10$. Therefore,
\[(p+q+r+s)(a+b+c) = 33 \cdot 20 = \boxed{660}.\] | size(150); for (int i=0;i<7;++i) {draw((i,0)--(i,4));} for (int i=0;i<5;++i) {draw((0,i)--(6,i));} draw((0,0)--(6,4),dashed); label("$1$",(0,0),SE,fontsize(8)); label("$1$",(1,0),SE,fontsize(8)); label("$1$",(2,0),SE,fontsize(8)); label("$1$",(2,1),SE,fontsize(8)); label("$1$",(3,0),SE,fontsize(8)); label("$2$",(3,1),SE,fontsize(8)); label("$2$",(3,2),SE,fontsize(8)); label("$1$",(4,0),SE,fontsize(8)); label("$3$",(4,1),SE,fontsize(8)); label("$5$",(4,2),SE,fontsize(8)); label("$1$",(5,0),SE,fontsize(8)); label("$4$",(5,1),SE,fontsize(8)); label("$9$",(5,2),SE,fontsize(8)); label("$9$",(5,3),SE,fontsize(8)); label("$1$",(6,0),SE,fontsize(8)); label("$5$",(6,1),SE,fontsize(8)); label("$14$",(6,2),SE,fontsize(8)); label("$23$",(6,3),SE,fontsize(8)); label("$23$",(6,4),SE,fontsize(8)); | [] |
105 | In triangle $ABC,$ point $D$ is on $\overline{BC}$ with $CD = 2$ and $DB = 5,$ point $E$ is on $\overline{AC}$ with $CE = 1$ and $EA = 3,$ $AB = 8,$ and $\overline{AD}$ and $\overline{BE}$ intersect at $P.$ Points $Q$ and $R$ lie on $\overline{AB}$ so that $\overline{PQ}$ is parallel to $\overline{CA}$ and $\overline{PR}$ is parallel to $\overline{CB}.$ It is given that the ratio of the area of triangle $PQR$ to the area of triangle $ABC$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m + n$. | 2002 AIME II Problem 13 | Let $X$ be the intersection of $\overline{CP}$ and $\overline{AB}$.
Since $\overline{PQ} \parallel \overline{CA}$ and $\overline{PR} \parallel \overline{CB}$, $\angle CAB = \angle PQR$ and $\angle CBA = \angle PRQ$. So $\Delta ABC \sim \Delta QRP$, and thus, $\frac{[\Delta PQR]}{[\Delta ABC]} = \left(\frac{PX}{CX}\right)^2$.
Using mass points:
WLOG, let $W_C=15$.
Then:
$W_A = \left(\frac{CE}{AE}\right)W_C = \frac{1}{3}\cdot15=5$.
$W_B = \left(\frac{CD}{BD}\right)W_C = \frac{2}{5}\cdot15=6$.
$W_X=W_A+W_B=5+6=11$.
$W_P=W_C+W_X=15+11=26$.
Thus, $\frac{PX}{CX}=\frac{W_C}{W_P}=\frac{15}{26}$. Therefore, $\frac{[\Delta PQR]}{[\Delta ABC]} = \left( \frac{15}{26} \right)^2 = \frac{225}{676}$, and $m+n=\boxed{901}$. Note we can just use mass points to get $\left( \frac{15}{26} \right)^2= \frac{225}{676}$ which is $\boxed{901}$. | // Block 1
size(10cm);
pair A,B,C,D,E,X,P,Q,R;
A=(0,0);
B=(8,0);
C=(1.9375,3.4994);
D=(3.6696,2.4996);
E=(1.4531,2.6246);
X=(4.3636,0);
P=(2.9639,2.0189);
Q=(1.8462,0);
R=(6.4615,0);
dot(A);
dot(B);
dot(C);
dot(D);
dot(E);
dot(X);
dot(P);
dot(Q);
dot(R);
label("$A$",A,WSW);
label("$B$",B,ESE);
label("$C$",C,NNW);
label("$D$",D,NE);
label("$E$",E,WNW);
label("$X$",X,SSE);
label("$P$",P,NNE);
label("$Q$",Q,SSW);
label("$R$",R,SE);
draw(A--B--C--cycle);
draw(P--Q--R--cycle);
draw(A--D);
draw(B--E);
draw(C--X);
// Block 2
size(10cm); pair A,B,C,D,E,X,P,Q,R; A=(0,0); B=(8,0); C=(1.9375,3.4994); D=(3.6696,2.4996); E=(1.4531,2.6246); X=(4.3636,0); P=(2.9639,2.0189); Q=(1.8462,0); R=(6.4615,0); dot(A); dot(B); dot(C); dot(D); dot(E); dot(X); dot(P); dot(Q); dot(R); label("$A$",A,WSW); label("$B$",B,ESE); label("$C$",C,NNW); label("$D$",D,NE); label("$E$",E,WNW); label("$X$",X,SSE); label("$P$",P,NNE); label("$Q$",Q,SSW); label("$R$",R,SE); draw(A--B--C--cycle); draw(P--Q--R--cycle); draw(A--D); draw(B--E); draw(C--X); | [] |
105 | In triangle $ABC,$ point $D$ is on $\overline{BC}$ with $CD = 2$ and $DB = 5,$ point $E$ is on $\overline{AC}$ with $CE = 1$ and $EA = 3,$ $AB = 8,$ and $\overline{AD}$ and $\overline{BE}$ intersect at $P.$ Points $Q$ and $R$ lie on $\overline{AB}$ so that $\overline{PQ}$ is parallel to $\overline{CA}$ and $\overline{PR}$ is parallel to $\overline{CB}.$ It is given that the ratio of the area of triangle $PQR$ to the area of triangle $ABC$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m + n$. | 2002 AIME II Problem 13 | First draw $\overline{CP}$ and extend it so that it meets with $\overline{AB}$ at point $X$.
We have that $[ABC]=\frac{1}{2}\cdot AC \cdot BC\sin{C}=\frac{1}{2}\cdot 4\cdot {7}\sin{C}=14\sin{C}$
By Ceva's, \[3\cdot{\frac{2}{5}}\cdot{\frac{BX}{AX}}=1\implies BX=\frac{5\cdot AX}{6}\] That means that \[\frac{11\cdot {AX}}{6}=8\implies AX=\frac{48}{11} \ \text{and} \ BX=\frac{40}{11}\]
Now we apply mass points. Assume WLOG that $W_{A}=1$. That means that
\[W_{C}=3, W_{B}=\frac{6}{5}, W_{X}=\frac{11}{5}, W_{D}=\frac{21}{5}, W_{E}=4, W_{P}=\frac{26}{5}\]
Notice now that $\triangle{PBQ}$ is similar to $\triangle{EBA}$. Therefore,
\[\frac{PQ}{EA}=\frac{PB}{EB}\implies \frac{PQ}{3}=\frac{10}{13}\implies PQ=\frac{30}{13}\]
Also, $\triangle{PRA}$ is similar to $\triangle{DBA}$. Therefore,
\[\frac{PA}{DA}=\frac{PR}{DB}\implies \frac{21}{26}=\frac{PR}{5}\implies PR=\frac{105}{26}\]
Because $\triangle{PQR}$ is similar to $\triangle{CAB}$, $\angle{C}=\angle{P}$.
As a result, $[PQR]=\frac{1}{2}\cdot PQ \cdot PR \sin{C}=\frac{1}{2}\cdot \frac{30}{13}\cdot \frac{105}{26}\sin{P}=\frac{1575}{338}\sin{C}$.
Therefore, \[\frac{[PQR]}{[ABC]}=\frac{\frac{1575}{338}\sin{C}}{14\sin{C}}=\frac{225}{676}\implies 225+676=\boxed{901}\]
Not the author writing here, but a note is that Ceva's Theorem was actually not necessary to solve this problem. The information was just nice to know :) | // Block 1
size(10cm);
pair A,B,C,D,E,X,P,Q,R;
A=(0,0);
B=(8,0);
C=(1.9375,3.4994);
D=(3.6696,2.4996);
E=(1.4531,2.6246);
X=(4.3636,0);
P=(2.9639,2.0189);
Q=(1.8462,0);
R=(6.4615,0);
dot(A);
dot(B);
dot(C);
dot(D);
dot(E);
dot(X);
dot(P);
dot(Q);
dot(R);
label("$A$",A,WSW);
label("$B$",B,ESE);
label("$C$",C,NNW);
label("$D$",D,NE);
label("$E$",E,WNW);
label("$X$",X,SSE);
label("$P$",P,NNE);
label("$Q$",Q,SSW);
label("$R$",R,SE);
draw(A--B--C--cycle);
draw(P--Q--R--cycle);
draw(A--D);
draw(B--E);
draw(C--X);
// Block 2
size(10cm); pair A,B,C,D,E,X,P,Q,R; A=(0,0); B=(8,0); C=(1.9375,3.4994); D=(3.6696,2.4996); E=(1.4531,2.6246); X=(4.3636,0); P=(2.9639,2.0189); Q=(1.8462,0); R=(6.4615,0); dot(A); dot(B); dot(C); dot(D); dot(E); dot(X); dot(P); dot(Q); dot(R); label("$A$",A,WSW); label("$B$",B,ESE); label("$C$",C,NNW); label("$D$",D,NE); label("$E$",E,WNW); label("$X$",X,SSE); label("$P$",P,NNE); label("$Q$",Q,SSW); label("$R$",R,SE); draw(A--B--C--cycle); draw(P--Q--R--cycle); draw(A--D); draw(B--E); draw(C--X); | [] |
105 | In triangle $ABC,$ point $D$ is on $\overline{BC}$ with $CD = 2$ and $DB = 5,$ point $E$ is on $\overline{AC}$ with $CE = 1$ and $EA = 3,$ $AB = 8,$ and $\overline{AD}$ and $\overline{BE}$ intersect at $P.$ Points $Q$ and $R$ lie on $\overline{AB}$ so that $\overline{PQ}$ is parallel to $\overline{CA}$ and $\overline{PR}$ is parallel to $\overline{CB}.$ It is given that the ratio of the area of triangle $PQR$ to the area of triangle $ABC$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m + n$. | 2002 AIME II Problem 13 | Use the mass of point. Denoting the mass of $C=15,B=6,A=5,D=21,E=20$, we can see that the mass of $P$ is $26$, hence we know that $\frac{BP}{PE}=\frac{10}{3}$, now we can find that $\frac{PQ}{AE}=\frac{10}{13}$ which implies $PQ=\frac{30}{13}$, it is obvious that $\triangle{PQR}$ is similar to $\triangle{ACB}$ so we need to find the ration between PQ and AC, which is easy, it is $\frac{15}{26}$, so our final answer is $\left( \frac{15}{26} \right)^2= \frac{225}{676}$ which is $\boxed{901}$. ~bluesoul | // Block 1
size(10cm);
pair A,B,C,D,E,P,Q,R;
A=(0,0);
B=(8,0);
C=(1.9375,3.4994);
D=(3.6696,2.4996);
E=(1.4531,2.6246);
P=(2.9639,2.0189);
Q=(1.8462,0);
R=(6.4615,0);
dot(A);
dot(B);
dot(C);
dot(D);
dot(E);
dot(P);
dot(Q);
dot(R);
label("$A$",A,WSW);
label("$B$",B,ESE);
label("$C$",C,NNW);
label("$D$",D,NE);
label("$E$",E,WNW);
label("$P$",P,NNE);
label("$Q$",Q,SSW);
label("$R$",R,SE);
draw(A--B--C--cycle);
draw(P--Q--R--cycle);
draw(A--D);
draw(B--E);
// Block 2
size(10cm); pair A,B,C,D,E,P,Q,R; A=(0,0); B=(8,0); C=(1.9375,3.4994); D=(3.6696,2.4996); E=(1.4531,2.6246); P=(2.9639,2.0189); Q=(1.8462,0); R=(6.4615,0); dot(A); dot(B); dot(C); dot(D); dot(E); dot(P); dot(Q); dot(R); label("$A$",A,WSW); label("$B$",B,ESE); label("$C$",C,NNW); label("$D$",D,NE); label("$E$",E,WNW); label("$P$",P,NNE); label("$Q$",Q,SSW); label("$R$",R,SE); draw(A--B--C--cycle); draw(P--Q--R--cycle); draw(A--D); draw(B--E); | [] |
105 | In triangle $ABC,$ point $D$ is on $\overline{BC}$ with $CD = 2$ and $DB = 5,$ point $E$ is on $\overline{AC}$ with $CE = 1$ and $EA = 3,$ $AB = 8,$ and $\overline{AD}$ and $\overline{BE}$ intersect at $P.$ Points $Q$ and $R$ lie on $\overline{AB}$ so that $\overline{PQ}$ is parallel to $\overline{CA}$ and $\overline{PR}$ is parallel to $\overline{CB}.$ It is given that the ratio of the area of triangle $PQR$ to the area of triangle $ABC$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m + n$. | 2002 AIME II Problem 13 | Construct $\overline{CP}$ and extend it to line $\overline{AB}$ at point $F$.
Using Ceva's Theorem on triangle $ABC$ and point $P$, we get \[\frac{AF}{BF} \times \frac{BD}{CD} \times \frac{CE}{AE}=1\]
Thus, $\frac{AF}{BF}=\frac{6}{5}$.
Then, using this info, we apply Menelaus on triangle $ACF$ and line $BE$, obtaining \[\frac{AE}{BE} \times \frac{CP}{FP} \times \frac{FB}{AB}=1\]
Simplifying and substituting, we find that $\frac{CP}{FP}=\frac{11}{15}$.
Alternatively, $\frac{FP}{CF}=\frac{15}{26}$, which is also the ratio between the heights of the desired triangles.
Finishing, $\left( \frac{15}{26} \right)^2= \frac{225}{676}$ achieving the final answer of $\boxed{901}$.
~faliure167 | // Block 1
size(10cm);
pair A,B,C,D,E,F,P,Q,R;
A=(0,0);
B=(8,0);
C=(1.9375,3.4994);
D=(3.6696,2.4996);
E=(1.4531,2.6246);
F=(4.3636,0);
P=(2.9639,2.0189);
Q=(1.8462,0);
R=(6.4615,0);
dot(A);
dot(B);
dot(C);
dot(D);
dot(E);
dot(F);
dot(P);
dot(Q);
dot(R);
label("$A$",A,WSW);
label("$B$",B,ESE);
label("$C$",C,NNW);
label("$D$",D,NE);
label("$E$",E,WNW);
label("$F$",F,SSE);
label("$P$",P,NNE);
label("$Q$",Q,SSW);
label("$R$",R,SE);
draw(A--B--C--cycle);
draw(P--Q--R--cycle);
draw(A--D);
draw(B--E);
draw(C--F);
// Block 2
size(10cm); pair A,B,C,D,E,F,P,Q,R; A=(0,0); B=(8,0); C=(1.9375,3.4994); D=(3.6696,2.4996); E=(1.4531,2.6246); F=(4.3636,0); P=(2.9639,2.0189); Q=(1.8462,0); R=(6.4615,0); dot(A); dot(B); dot(C); dot(D); dot(E); dot(F); dot(P); dot(Q); dot(R); label("$A$",A,WSW); label("$B$",B,ESE); label("$C$",C,NNW); label("$D$",D,NE); label("$E$",E,WNW); label("$F$",F,SSE); label("$P$",P,NNE); label("$Q$",Q,SSW); label("$R$",R,SE); draw(A--B--C--cycle); draw(P--Q--R--cycle); draw(A--D); draw(B--E); draw(C--F); | [] |
106 | Circles $\mathcal{C}_{1}$ and $\mathcal{C}_{2}$ intersect at two points, one of which is $(9,6)$, and the product of the radii is $68$. The x-axis and the line $y = mx$, where $m > 0$, are tangent to both circles. It is given that $m$ can be written in the form $a\sqrt {b}/c$, where $a$, $b$, and $c$ are positive integers, $b$ is not divisible by the square of any prime, and $a$ and $c$ are relatively prime. Find $a + b + c$. | 2002 AIME II Problem 15 | Let $r_1$ and $r_2$ be the radii of the circles. Then the centers of the circles are of the form $(kr_1,r_1)$ and $(kr_2,r_2)$ for the same constant $k,$ since the two centers are collinear with the origin. Since $(9,6)$ lies on both circles,
\[(kr_i - 9)^2 + (r_i - 6)^2 = r^2,\]where $r_i$ represents either radius. Expanding, we get
\[k^2 r^2 - (18k + 12) r + 117 = 0.\]We are told the product of the circles is 68, so by Vieta's formulas, $\frac{117}{k^2} = 68.$ Hence, $k^2 = \frac{117}{68},$ and $k = \sqrt{\frac{117}{68}}.$
Since the circle is tangent to the line $y = mx,$ the distance from the center $(kr,r)$ to the line is $r.$ We can write $y = mx$ as $y - mx = 0,$ so from the distance formula,
\[\frac{|r - krm|}{\sqrt{1 + m^2}} = r.\]Squaring both sides, we get
\[\frac{(r - krm)^2}{1 + m^2} = r^2,\]so $(r - krm)^2 = r^2 (1 + m^2).$ Since $r \neq 0,$ we can divide both sides by r, to get
\[(1 - km)^2 = 1 + m^2.\]Then $1 - 2km + k^2 m^2 = 1 + m^2,$ so $m^2 (1 - k^2) + 2km = 0.$ Since $m \neq 0,$
\[m(1 - k^2) + 2k = 0.\]Hence,
\[m = \frac{2k}{k^2 - 1} = \frac{2 \sqrt{\frac{117}{68}}}{\frac{117}{68} - 1} = \frac{12 \sqrt{221}}{49}.\]
Our answer is thus $12 + 221 + 49 = \boxed{282}$ | // Block 1
unitsize(0.25 cm);
pair[] O;
real[] r;
pair P;
r[1] = 4.096;
r[2] = 16.6;
O[1] = (r[1]/(2/3*sqrt(17/13)),r[1]);
O[2] = (r[2]/(2/3*sqrt(17/13)),r[2]);
P = reflect(O[1],O[2])*(9,6);
draw(Circle(O[1],r[1]));
//draw(Circle(O[2],r[2]));
draw(arc(O[2],r[2],130,300));
draw((0,0)--(8,12*sqrt(221)/49*8));
draw((0,0)--(30,0));
draw((0,0)--O[1]--(O[1].x,0));
draw(O[1]--(O[1] + reflect((0,0),(10,12*sqrt(221)/49*10))*(O[1]))/2);
label("$y = mx$", (8,12*sqrt(221)/49*8), N);
dot("$(9,6)$", (9,6), NE);
dot("$(kr,r)$", O[1], N);
dot(P,red);
// Block 2
unitsize(0.25 cm); pair[] O; real[] r; pair P; r[1] = 4.096; r[2] = 16.6; O[1] = (r[1]/(2/3*sqrt(17/13)),r[1]); O[2] = (r[2]/(2/3*sqrt(17/13)),r[2]); P = reflect(O[1],O[2])*(9,6); draw(Circle(O[1],r[1])); //draw(Circle(O[2],r[2])); draw(arc(O[2],r[2],130,300)); draw((0,0)--(8,12*sqrt(221)/49*8)); draw((0,0)--(30,0)); draw((0,0)--O[1]--(O[1].x,0)); draw(O[1]--(O[1] + reflect((0,0),(10,12*sqrt(221)/49*10))*(O[1]))/2); label("$y = mx$", (8,12*sqrt(221)/49*8), N); dot("$(9,6)$", (9,6), NE); dot("$(kr,r)$", O[1], N); dot(P,red); | [] |
107 | Consider the set of points that are inside or within one unit of a rectangular parallelepiped (box) that measures $3$ by $4$ by $5$ units. Given that the volume of this set is $\frac{m + n\pi}{p},$ where $m, n,$ and $p$ are positive integers, and $n$ and $p$ are relatively prime, find $m + n + p.$ | 2003 AIME I Problem 5 | The set can be broken into several parts: the big $3\times 4 \times 5$ parallelepiped, $6$ external parallelepipeds that each share a face with the large parallelepiped and have a height of $1$, the $1/8$ spheres (one centered at each vertex of the large parallelepiped), and the $1/4$ cylinders connecting each adjacent pair of spheres.
The volume of the parallelepiped is $3 \times 4 \times 5 = 60$ cubic units.
The volume of the external parallelepipeds is $2(3 \times 4 \times 1)+2(3 \times 5 \times 1 )+2(4 \times 5 \times 1)=94$.
There are $8$ of the $1/8$ spheres, each of radius $1$. Together, their volume is $\frac{4}{3}\pi$.
There are $12$ of the $1/4$ cylinders, so $3$ complete cylinders can be formed. Their volumes are $3\pi$, $4\pi$, and $5\pi$, adding up to $12\pi$.
The combined volume of these parts is $60+94+\frac{4}{3}\pi+12\pi = \frac{462+40\pi}{3}$. Thus, the answer is $m+n+p = 462+40+3 = \boxed{505}$. | // Block 1
size(220);
import three; currentprojection = perspective(5,4,3); defaultpen(linetype("8 8")+linewidth(0.6));
draw(box((0,-.1,0),(0.4,0.6,0.3)));
draw(box((-.1,0,0),(0.5,0.5,0.3)));
draw(box((0,0,-.1),(0.4,0.5,0.4)));
draw(box((0,0,0),(0.4,0.5,0.3)),linewidth(1.2)+linetype("1"));
// Block 2
size(220); import three; currentprojection = perspective(5,4,3); defaultpen(linetype("8 8")+linewidth(0.6)); draw(box((0,-.1,0),(0.4,0.6,0.3))); draw(box((-.1,0,0),(0.5,0.5,0.3))); draw(box((0,0,-.1),(0.4,0.5,0.4))); draw(box((0,0,0),(0.4,0.5,0.3)),linewidth(1.2)+linetype("1")); | [] |
108 | The sum of the areas of all triangles whose vertices are also vertices of a $1$ by $1$ by $1$ cube is $m + \sqrt{n} + \sqrt{p},$ where $m, n,$ and $p$ are integers. Find $m + n + p.$ | 2003 AIME I Problem 6 | Since there are $8$ vertices of a cube, there are ${8 \choose 3} = 56$ total triangles to consider. They fall into three categories: there are those which are entirely contained within a single face of the cube (whose sides are two edges and one face diagonal), those which lie in a plane perpendicular to one face of the cube (whose sides are one edge, one face diagonal and one space diagonal of the cube) and those which lie in a plane oblique to the edges of the cube, whose sides are three face diagonals of the cube.
Each face of the cube contains ${4\choose 3} = 4$ triangles of the first type, and there are $6$ faces, so there are $24$ triangles of the first type. Each of these is a right triangle with legs of length $1$, so each triangle of the first type has area $\frac 12$.
Each edge of the cube is a side of exactly $2$ of the triangles of the second type, and there are $12$ edges, so there are $24$ triangles of the second type. Each of these is a right triangle with legs of length $1$ and $\sqrt 2$, so each triangle of the second type has area $\frac{\sqrt{2}}{2}$.
Each vertex of the cube is associated with exactly one triangle of the third type (whose vertices are its three neighbors), and there are $8$ vertices of the cube, so there are $8$ triangles of the third type. Each of the these is an equilateral triangle with sides of length $\sqrt 2$, so each triangle of the third type has area $\frac{\sqrt 3}2$.
Thus the total area of all these triangles is $24 \cdot \frac12 + 24\cdot\frac{\sqrt2}2 + 8\cdot\frac{\sqrt3}2 = 12 + 12\sqrt2 + 4\sqrt3 = 12 + \sqrt{288} + \sqrt{48}$ and the answer is $12 + 288 + 48 = \boxed{348}$. | // Block 1
size(120);
import three;
unitsize(1cm);
draw((0,0,0)--(1,0,0)--(1,1,0)--(0,1,0)--(0,0,0)--(0,0,1)--(0,1,1)--(1,1,1)--(1,0,1)--(0,0,1));
draw((0,1,1)--(0,1,0)); draw((1,1,1)--(1,1,0)); draw((1,0,1)--(1,0,0));
draw((1,1,1)--(1,0,0)--(1,1,0)--(1,1,1), blue);
// Block 2
size(120);
import three;
unitsize(1cm);
draw((0,0,0)--(1,0,0)--(1,1,0)--(0,1,0)--(0,0,0)--(0,0,1)--(0,1,1)--(1,1,1)--(1,0,1)--(0,0,1));
draw((0,1,1)--(0,1,0)); draw((1,1,1)--(1,1,0)); draw((1,0,1)--(1,0,0));
draw((0,1,1)--(1,0,0)--(1,1,0)--(0,1,1), blue);
// Block 3
size(120);
import three;
unitsize(1cm);
draw((0,0,0)--(1,0,0)--(1,1,0)--(0,1,0)--(0,0,0)--(0,0,1)--(0,1,1)--(1,1,1)--(1,0,1)--(0,0,1));
draw((0,1,1)--(0,1,0)); draw((1,1,1)--(1,1,0)); draw((1,0,1)--(1,0,0));
draw((1,0,0)--(0,1,0)--(1,1,1)--(1,0,0), blue);
// Block 4
size(120); import three; unitsize(1cm); draw((0,0,0)--(1,0,0)--(1,1,0)--(0,1,0)--(0,0,0)--(0,0,1)--(0,1,1)--(1,1,1)--(1,0,1)--(0,0,1)); draw((0,1,1)--(0,1,0)); draw((1,1,1)--(1,1,0)); draw((1,0,1)--(1,0,0)); draw((1,1,1)--(1,0,0)--(1,1,0)--(1,1,1), blue);
// Block 5
size(120); import three; unitsize(1cm); draw((0,0,0)--(1,0,0)--(1,1,0)--(0,1,0)--(0,0,0)--(0,0,1)--(0,1,1)--(1,1,1)--(1,0,1)--(0,0,1)); draw((0,1,1)--(0,1,0)); draw((1,1,1)--(1,1,0)); draw((1,0,1)--(1,0,0)); draw((0,1,1)--(1,0,0)--(1,1,0)--(0,1,1), blue);
// Block 6
size(120); import three; unitsize(1cm); draw((0,0,0)--(1,0,0)--(1,1,0)--(0,1,0)--(0,0,0)--(0,0,1)--(0,1,1)--(1,1,1)--(1,0,1)--(0,0,1)); draw((0,1,1)--(0,1,0)); draw((1,1,1)--(1,1,0)); draw((1,0,1)--(1,0,0)); draw((1,0,0)--(0,1,0)--(1,1,1)--(1,0,0), blue); | [] |
109 | Point $B$ is on $\overline{AC}$ with $AB = 9$ and $BC = 21.$ Point $D$ is not on $\overline{AC}$ so that $AD = CD,$ and $AD$ and $BD$ are integers. Let $s$ be the sum of all possible perimeters of $\triangle ACD$. Find $s.$ | 2003 AIME I Problem 7 | Denote the height of $\triangle ACD$ as $h$, $x = AD = CD$, and $y = BD$. Using the Pythagorean theorem, we find that $h^2 = y^2 - 6^2$ and $h^2 = x^2 - 15^2$. Thus, $y^2 - 36 = x^2 - 225 \Longrightarrow x^2 - y^2 = 189$. The LHS is difference of squares, so $(x + y)(x - y) = 189$. As both $x,\ y$ are integers, $x+y,\ x-y$ must be integral divisors of $189$.
The pairs of divisors of $189$ are $(1,189)\ (3,63)\ (7,27)\ (9,21)$. This yields the four potential sets for $(x,y)$ as $(95,94)\ (33,30)\ (17,10)\ (15,6)$. The last is not a possibility since it simply degenerates into a line. The sum of the three possible perimeters of $\triangle ACD$ is equal to $3(AC) + 2(x_1 + x_2 + x_3) = 90 + 2(95 + 33 + 17) = \boxed{380}$. | // Block 1
size(220);
pointpen = black; pathpen = black + linewidth(0.7);
pair O=(0,0),A=(-15,0),B=(-6,0),C=(15,0),D=(0,8);
D(D(MP("A",A))--D(MP("C",C))--D(MP("D",D,NE))--cycle);
D(D(MP("B",B))--D); D((0,-4)--(0,12),linetype("4 4")+linewidth(0.7));
MP("6",B/2); MP("15",C/2); MP("9",(A+B)/2);
// Block 2
size(220); pointpen = black; pathpen = black + linewidth(0.7); pair O=(0,0),A=(-15,0),B=(-6,0),C=(15,0),D=(0,8); D(D(MP("A",A))--D(MP("C",C))--D(MP("D",D,NE))--cycle); D(D(MP("B",B))--D); D((0,-4)--(0,12),linetype("4 4")+linewidth(0.7)); MP("6",B/2); MP("15",C/2); MP("9",(A+B)/2); | [] |
110 | Triangle $ABC$ is isosceles with $AC = BC$ and $\angle ACB = 106^\circ.$ Point $M$ is in the interior of the triangle so that $\angle MAC = 7^\circ$ and $\angle MCA = 23^\circ.$ Find the number of degrees in $\angle CMB.$
Contents
1 Problem
2 Solutions
2.1 Solution 1
2.2 Solution 2
2.3 Solution 3
2.4 Solution 4
2.5 Solution 5 (Ceva)
2.6 Solution 6
3 See also | 2003 AIME I Problem 10 | Solution 1
Take point $N$ inside $\triangle ABC$ such that $\angle CBN = 7^\circ$ and $\angle BCN = 23^\circ$.
$\angle MCN = 106^\circ - 2\cdot 23^\circ = 60^\circ$. Also, since $\triangle AMC$ and $\triangle BNC$ are congruent (by ASA), $CM = CN$. Hence $\triangle CMN$ is an equilateral triangle, so $\angle CNM = 60^\circ$.
Then $\angle MNB = 360^\circ - \angle CNM - \angle CNB = 360^\circ - 60^\circ - 150^\circ = 150^\circ$. We now see that $\triangle MNB$ and $\triangle CNB$ are congruent. Therefore, $CB = MB$, so $\angle CMB = \angle MCB = \boxed{83^\circ}$.
Solution 2
From the givens, we have the following angle measures: $m\angle AMC = 150^\circ$, $m\angle MCB = 83^\circ$. If we define $m\angle CMB = \theta$ then we also have $m\angle CBM = 97^\circ - \theta$. Then apply the Law of Sines to triangles $\triangle AMC$ and $\triangle BMC$ to get
\[\frac{\sin 150^\circ}{\sin 7^\circ} = \frac{AC}{CM} = \frac{BC}{CM} = \frac{\sin \theta}{\sin (97^\circ - \theta)}\]
Clearing denominators, evaluating $\sin 150^\circ = \frac 12$ and applying one of our trigonometric identities to the result gives
\[\frac{1}{2} \cos (7^\circ - \theta )= \sin 7^\circ \sin \theta\]
and multiplying through by 2 and applying the double angle formulas gives
\[\cos 7^\circ\cos\theta + \sin7^\circ\sin\theta = 2 \sin7^\circ \sin\theta\]
and so $\cos 7^\circ \cos \theta = \sin 7^\circ \sin\theta \Longleftrightarrow \tan 7^{\circ} = \cot \theta$; since $0^\circ < \theta < 180^\circ$, we must have $\theta = 83^\circ$, so the answer is $\boxed{83}$.
Solution 3
Without loss of generality, let $AC = BC = 1$. Then, using the Law of Sines in triangle $AMC$, we get $\frac {1}{\sin 150} = \frac {MC}{\sin 7}$, and using the sine addition formula to evaluate $\sin 150 = \sin (90 + 60)$, we get $MC = 2 \sin 7$.
Then, using the Law of Cosines in triangle $MCB$, we get $MB^2 = 4\sin^2 7 + 1 - 4\sin 7(\cos 83) = 1$, since $\cos 83 = \sin 7$. So triangle $MCB$ is isosceles, and $\angle CMB = \boxed{83}$.
Solution 4
Note: A diagram would be much appreciated; I cannot make one since I'm bad at asymptote. Also, please make this less cluttered :) ~tauros
First, take point $E$ outside of $\triangle{ABC}$ so that $\triangle{CEB}$ is equilateral. Then, connect $A$, $C$, and $M$ to $E$. Also, let $ME$ intersect $AB$ at $F$. $\angle{MCE} = 83^\circ - 60^\circ = 23^\circ$, $CE = AB$, and (trivially) $CM = CM$, so $\triangle{MCE} \cong \triangle{MCA}$ by SAS congruence. Also, $\angle{CMA} = \angle{CME} = 150^\circ$, so $\angle{AME} = 60^\circ$, and $AM = ME$,
making $\triangle{AME}$ also equilateral. (it is isosceles with a $60^\circ$ angle) $\triangle{MAF} \cong \triangle{EAF}$ by SAS ($MA = AE$,
$AF = AF$, and $m\angle{MAF} = m\angle{EAF} = 30^\circ$), and $\triangle{MAB} \cong \triangle{EAB}$ by SAS ($MA = AE$, $AB = AB$, and
$m\angle{MAB} = m\angle{EAB} = 30^\circ$). Thus, $\triangle{BME}$ is isosceles, with $m\angle{BME} = m\angle{BEM} = 60^\circ + 7^\circ = 67^\circ$. Also, $\angle{EMB} + \angle {CMB} = \angle{CME} = 150^\circ$, so $\angle{CME} = 150^\circ - 67^\circ = \boxed{83^\circ}$.
Solution 5 (Ceva)
Noticing that we have three concurrent cevians, we apply Ceva's theorem:
\[(\sin \angle ACM)(\sin \angle BAM)(\sin \angle CBM) = (\sin \angle CAM)(\sin \angle ABM)(\sin \angle BCM)\]
\[(\sin 23)(\sin 30)(\sin x) = (\sin 7)(\sin 37-x)(\sin 83)\]
using the fact that $\sin 83 = \cos 7$ and $(\sin 7)(\cos 7) = 1/2 (\sin 14)$ we have:
\[(\sin 23)(\sin x) = (\sin 14)(\sin 37-x)\]
By inspection, $x=14^\circ$ works, so the answer is $180-83-14= \boxed{083}$
Solution 6
Let $\angle{APC} = \theta^{\circ}$
Using sine rule on $\triangle{APB}, \triangle{APC}$, letting $AP=d$ we get :
$\frac{d}{1} = \frac{\sin{7^{\circ}}}{\sin{150^{\circ}}} = 2\sin{7^{\circ}}= \frac{\sin{14^{\circ}}}{\cos{7^{\circ}}}= \frac{\sin{14^{\circ}}}{\sin{83^{\circ}}}= \frac{\sin{(97-\theta)^{\circ}}}{\sin{\theta^{\circ}}}$
Simplifying, we get that
$\cos{(14-\theta)^{\circ}}-\cos{(14+\theta)^{\circ}}=\cos{(14-\theta)^{\circ}}-\cos{(180-\theta)^{\circ}},$ from where $\cos{(14-\theta)^{\circ}}=\cos{(180-\theta)^{\circ}}$
Simplifying more, we get that $\sin{97^{\circ}} \cdot \sin{(\theta-83)^{\circ}} = 0$, so $\theta = 83^{\circ}$
NOTE: The simplifications were carried out by the product-to-sum and sum-to-product identities
~Prabh1512 | // Block 1
pointpen = black; pathpen = black+linewidth(0.7); size(220);
/* We will WLOG AB = 2 to draw following */
pair A=(0,0), B=(2,0), C=(1,Tan(37)), M=IP(A--(2Cos(30),2Sin(30)),B--B+(-2,2Tan(23)));
D(MP("A",A)--MP("B",B)--MP("C",C,N)--cycle); D(A--D(MP("M",M))--B); D(C--M);
// Block 2
pointpen = black; pathpen = black+linewidth(0.7); size(220); /* We will WLOG AB = 2 to draw following */ pair A=(0,0), B=(2,0), C=(1,Tan(37)), M=IP(A--(2Cos(30),2Sin(30)),B--B+(-2,2Tan(23))); D(MP("A",A)--MP("B",B)--MP("C",C,N)--cycle); D(A--D(MP("M",M))--B); D(C--M);
// Block 3
pointpen = black; pathpen = black+linewidth(0.7); size(220); /* We will WLOG AB = 2 to draw following */ pair A=(0,0), B=(2,0), C=(1,Tan(37)), M=IP(A--(2Cos(30),2Sin(30)),B--B+(-2,2Tan(23))), N=(2-M.x,M.y); D(MP("A",A)--MP("B",B)--MP("C",C,N)--cycle); D(A--D(MP("M",M))--B); D(C--M); D(C--D(MP("N",N))--B--N--M,linetype("6 6")+linewidth(0.7)); | [] |
111 | In convex quadrilateral $ABCD, \angle A \cong \angle C, AB = CD = 180,$ and $AD \neq BC.$ The perimeter of $ABCD$ is $640$. Find $\lfloor 1000 \cos A \rfloor.$ (The notation $\lfloor x \rfloor$ means the greatest integer that is less than or equal to $x.$) | 2003 AIME I Problem 12 | Solution 1
By the Law of Cosines on $\triangle ABD$ at angle $A$ and on $\triangle BCD$ at angle $C$ (note $\angle C = \angle A$),
\[180^2 + AD^2 - 360 \cdot AD \cos A = 180^2 + BC^2 - 360 \cdot BC \cos A\]
\[(AD^2 - BC^2) = 360(AD - BC) \cos A\]
\[(AD - BC)(AD + BC) = 360(AD - BC) \cos A\]
\[(AD + BC) = 360 \cos A\]
We know that $AD + BC = 640 - 360 = 280$. $\cos A = \dfrac{280}{360} = \dfrac{7}{9} = 0.777 \ldots$
$\lfloor 1000 \cos A \rfloor = \boxed{777}$.
Solution 2
Notice that $AB = CD$, and $BD = DB$, and $\angle{DAB} \cong \angle{BCD}$, so we have side-side-angle matching on triangles $ABD$ and $CDB$. Since the problem does not allow $\triangle{ABD} \cong \triangle{CDB}$, we know that $\angle{ADB}$ is not a right angle, and there is a unique other triangle with the matching side-side-angle.
Overlay the triangles $\triangle BDA$ and $\triangle BDC$ on each other as in the diagram above (where the red labels correspond to $\triangle BDA$ and the blue labels correspond to $\triangle BDC$). Here we assume without loss of generality that $BC>AD$. Furthermore, let $\theta$ be the angle $A$ referenced in the problem; we need to find $\lfloor 1000\cos\theta\rfloor$.
Since the perimeter of $ABCD$ is $640$, we have $AD+BC=640-360=280$. Thus let $BC=a$ and $AD=280-a$ for some positive real number $a$. But the sides that correspond to $\overline{BD}$ above are congruent, so we can drop a perpendicular from the topmost point to a point $E$, where the base of the isosceles triangle is bisected. Notice that the base of the isosceles triangle has length $a-(280-a)=2a-280$, so in the diagram above, $AE=280-a+\frac{1}{2}(2a-280)=280-a+a-140=140$.
Looking above at the right triangle containing $\theta$, we see that $\cos\theta=\frac{AE}{180}=\frac{140}{180}=\frac{7}{9}$. Hence $\left\lfloor 1000\cdot\frac{7}{9}\right\rfloor=\boxed{777}$ is our answer.
~ eevee9406
Solution 3
Start the same as solution 1. We get \[180^2+a^2-360a \cos A = 180^2+b^2-360b \cos A \Rightarrow a^2-360a \cos A = b^2-360b \cos A,\]where $a$ is the length of $BC$ and $b$ is the length of $AD$. Let the common value of $a^2-360a \cos A$ and $b^2-360b \cos A$ be $c$. Then, the quadratic in $x$ \[x^2-360 \cos A \cdot x - c = 0\]has solutions $a$ and $b$. Therefore, by Vieta's, $a+b = 360 \cos A$. However, we know that the perimeter of $ABCD$ is $640$, so $a+b+180+180=640$, so $a+b=280$. Therefore, \[360 \cos A = 280 \Rightarrow \cos A = \frac{280}{360} \Rightarrow \cos A = \frac{7}{9} \Rightarrow \lfloor 1000 \cos A \rfloor = \boxed{777}.\] | // Block 1
real x = 1.60; /* arbitrary */ pointpen = black; pathpen = black+linewidth(0.7); size(180); real BD = x*x + 1.80*1.80 - 2 * 1.80 * x * 7 / 9; pair A=(0,0),B=(1.8,0),D=IP(CR(A,x),CR(B,BD)),C=OP(CR(D,1.8),CR(B,2.80 - x)); D(MP("A",A)--MP("B",B)--MP("C",C)--MP("D",D,N)--B--A--D); MP("180",(A+B)/2); MP("180",(C+D)/2,NE); D(anglemark(B,A,D)); D(anglemark(D,C,B));
// Block 2
pair A,B,C,D,F; A=(0,0);B=(5,4.04061);C=(6,0);D=(4,0);F=(5,0); draw(B--A--C--B--D); label("A",A,SW,red);label("E",F,S);label("B",B,NW,red);label("D",D,S,red);label("B",C,SE,blue);label("C",A,NW,blue);label("D",B,NE,blue);label("$\theta$",A,(4.5,1.5)); draw(anglemark(D,A,B,20));label("180",(A+B)/2,NW); draw(B--F,dashed);draw(rightanglemark(B,F,A));draw(D--B--C,dashed); add(pathticks(B--C,1,0.5,1,10));add(pathticks(B--D,1,0.5,1,10)); | [] |
112 | In $\triangle ABC, AB = 360, BC = 507,$ and $CA = 780.$ Let $M$ be the midpoint of $\overline{CA},$ and let $D$ be the point on $\overline{CA}$ such that $\overline{BD}$ bisects angle $ABC.$ Let $F$ be the point on $\overline{BC}$ such that $\overline{DF} \perp \overline{BD}.$ Suppose that $\overline{DF}$ meets $\overline{BM}$ at $E.$ The ratio $DE: EF$ can be written in the form $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$ | 2003 AIME I Problem 15 | In the following, let the name of a point represent the mass located there. Since we are looking for a ratio, we assume that $AB=120$, $BC=169$, and $CA=260$ in order to simplify our computations.
First, reflect point $F$ over angle bisector $BD$ to a point $F'$.
As $BD$ is an angle bisector of both triangles $BAC$ and $BF'F$, we know that $F'$ lies on $AB$. We can now balance triangle $BF'C$ at point $D$ using mass points.
By the Angle Bisector Theorem, we can place mass points on $C,D,A$ of $120,\,289,\,169$ respectively. Thus, a mass of $\frac {289}{2}$ belongs at both $F$ and $F'$ because BD is a median of triangle $BF'F$ . Therefore, $CB/FB=\frac{289}{240}$.
Now, we reassign mass points to determine $FE/FD$. This setup involves $\triangle CFD$ and transversal $MEB$. For simplicity, put masses of $240$ and $289$ at $C$ and $F$ respectively. To find the mass we should put at $D$, we compute $CM/MD$. Applying the Angle Bisector Theorem again and using the fact $M$ is a midpoint of $AC$, we find
\[\frac {MD}{CM} = \frac {\frac{169}{289}\cdot 260 - 130}{130} = \frac {49}{289}\]
At this point we could find the mass at $D$ but it's unnecessary.
\[\frac {DE}{EF} = \frac {F}{D} = \frac {F}{C}\cdot\frac {C}{D} = \frac {289}{240}\cdot\frac {49}{289} = \boxed{\frac {49}{240}}\]
and the answer is $49 + 240 = \boxed{289}$. | // Block 1
size(400); pointpen = black; pathpen = black+linewidth(0.7);
pair A=(0,0),C=(7.8,0),B=IP(CR(A,3.6),CR(C,5.07)), M=(A+C)/2, Da = bisectorpoint(A,B,C), D=IP(B--B+(Da-B)*10,A--C), F=IP(D--D+10*(B-D)*dir(270),B--C), E=IP(B--M,D--F);pair Fprime=2*D-F; /* scale down by 100x */
D(MP("A",A,NW)--MP("B",B,N)--MP("C",C)--cycle); D(B--D(MP("D",D))--D(MP("F",F,NE))); D(B--D(MP("M",M)));D(A--MP("F'",Fprime,SW)--D); MP("E",E,NE); D(rightanglemark(F,D,B,4)); MP("390",(M+C)/2); MP("390",(M+C)/2); MP("360",(A+B)/2,NW); MP("507",(B+C)/2,NE);
// Block 2
size(400); pointpen = black; pathpen = black+linewidth(0.7); pair A=(0,0),C=(7.8,0),B=IP(CR(A,3.6),CR(C,5.07)), M=(A+C)/2, Da = bisectorpoint(A,B,C), D=IP(B--B+(Da-B)*10,A--C), F=IP(D--D+10*(B-D)*dir(270),B--C), E=IP(B--M,D--F);pair Fprime=2*D-F; /* scale down by 100x */ D(MP("A",A,NW)--MP("B",B,N)--MP("C",C)--cycle); D(B--D(MP("D",D))--D(MP("F",F,NE))); D(B--D(MP("M",M)));D(A--MP("F'",Fprime,SW)--D); MP("E",E,NE); D(rightanglemark(F,D,B,4)); MP("390",(M+C)/2); MP("390",(M+C)/2); MP("360",(A+B)/2,NW); MP("507",(B+C)/2,NE); | [] |
113 | In triangle $ABC,$ $AB = 13,$ $BC = 14,$ $AC = 15,$ and point $G$ is the intersection of the medians. Points $A',$ $B',$ and $C',$ are the images of $A,$ $B,$ and $C,$ respectively, after a $180^\circ$ rotation about $G.$ What is the area of the union of the two regions enclosed by the triangles $ABC$ and $A'B'C'?$ | 2003 AIME II Problem 6 | Since a $13-14-15$ triangle is a $5-12-13$ triangle and a $9-12-15$ triangle "glued" together on the $12$ side, $[ABC]=\frac{1}{2}\cdot12\cdot14=84$.
There are six points of intersection between $\Delta ABC$ and $\Delta A'B'C'$. Connect each of these points to $G$.
There are $12$ smaller congruent triangles which make up the desired area. Also, $\Delta ABC$ is made up of $9$ of such triangles.
Therefore, $\left[\Delta ABC \bigcup \Delta A'B'C'\right] = \frac{12}{9}[\Delta ABC]= \frac{4}{3}\cdot84=\boxed{112}$. | // Block 1
size(8cm);
pair A,B,C,G,D,E,F,A_1,A_2,B_1,B_2,C_1,C_2;
B=(0,0);
A=(5,12);
C=(14,0);
E=(12.6667,8);
D=(7.6667,-4);
F=(-1.3333,8);
G=(6.3333,4);
B_1=(4.6667,0);
B_2=(1.6667,4);
A_1=(3.3333,8);
A_2=(8,8);
C_1=(11,4);
C_2=(9.3333,0);
dot(A);
dot(B);
dot(C);
dot(G);
dot(D);
dot(E);
dot(F);
dot(A_1);
dot(B_1);
dot(C_1);
dot(A_2);
dot(B_2);
dot(C_2);
draw(B--A--C--cycle);
draw(E--D--F--cycle);
draw(B_1--A_2);
draw(A_1--C_2);
draw(C_1--B_2);
label("$B$",B,WSW);
label("$A$",A,N);
label("$C$",C,ESE);
label("$G$",G,S);
label("$B'$",E,ENE);
label("$A'$",D,S);
label("$C'$",F,WNW);
// Block 2
size(8cm); pair A,B,C,G,D,E,F,A_1,A_2,B_1,B_2,C_1,C_2; B=(0,0); A=(5,12); C=(14,0); E=(12.6667,8); D=(7.6667,-4); F=(-1.3333,8); G=(6.3333,4); B_1=(4.6667,0); B_2=(1.6667,4); A_1=(3.3333,8); A_2=(8,8); C_1=(11,4); C_2=(9.3333,0); dot(A); dot(B); dot(C); dot(G); dot(D); dot(E); dot(F); dot(A_1); dot(B_1); dot(C_1); dot(A_2); dot(B_2); dot(C_2); draw(B--A--C--cycle); draw(E--D--F--cycle); draw(B_1--A_2); draw(A_1--C_2); draw(C_1--B_2); label("$B$",B,WSW); label("$A$",A,N); label("$C$",C,ESE); label("$G$",G,S); label("$B'$",E,ENE); label("$A'$",D,S); label("$C'$",F,WNW); | [] |
114 | Let $A = (0,0)$ and $B = (b,2)$ be points on the coordinate plane. Let $ABCDEF$ be a convex equilateral hexagon such that $\angle FAB = 120^\circ,$ $\overline{AB}\parallel \overline{DE},$ $\overline{BC}\parallel \overline{EF,}$ $\overline{CD}\parallel \overline{FA},$ and the y-coordinates of its vertices are distinct elements of the set $\{0,2,4,6,8,10\}.$ The area of the hexagon can be written in the form $m\sqrt {n},$ where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. Find $m + n.$ | 2003 AIME II Problem 14 | From this image, we can see that the y-coordinate of F is 4, and from this, we can gather that the coordinates of E, D, and C, respectively, are 8, 10, and 6.
Let the angle between the $x$-axis and segment $AB$ be $\theta$, as shown above. Thus, as $\angle FAB=120^\circ$, the angle between the $x$-axis and segment $AF$ is $60-\theta$, so $\sin{(60-\theta)}=2\sin{\theta}$. Expanding, we have
$\sin{60}\cos{\theta}-\cos{60}\sin{\theta}=\frac{\sqrt{3}\cos{\theta}}{2}-\frac{\sin{\theta}}{2}=2\sin{\theta}$
Isolating $\sin{\theta}$ we see that $\frac{\sqrt{3}\cos{\theta}}{2}=\frac{5\sin{\theta}}{2}$, or $\cos{\theta}=\frac{5}{\sqrt{3}}\sin{\theta}$. Using the fact that $\sin^2{\theta}+\cos^2{\theta}=1$, we have $\frac{28}{3}\sin^2{\theta}=1$, or $\sin{\theta}=\sqrt{\frac{3}{28}}$. Letting the side length of the hexagon be $y$, we have $\frac{2}{y}=\sqrt{\frac{3}{28}}$. After simplification we find that that $y=\frac{4\sqrt{21}}{3}$.
In particular, note that by the Pythagorean theorem, $b^2+2^2=y^2$, hence $b=10\sqrt{3}/3$. Also, if $C=(c,6)$, then $y^2=BC^2=4^2+(c-b)^2$, hence $c-b=8\sqrt{3}/3,$ and thus $c=18\sqrt{3}/3$. Using similar methods (or symmetry), we determine that $D=(10\sqrt{3}/3,10)$, $E=(0,8)$, and $F=(-8\sqrt{3}/3,4)$. By the Shoelace theorem,
\[[ABCDEF]=\frac12\left|\begin{array}{cc} 0&0\\ 10\sqrt{3}/3&2\\ 18\sqrt{3}/3&6\\ 10\sqrt{3}/3&10\\ 0&8\\ -8\sqrt{3}/3&4\\ 0&0\\ \end{array}\right|=\frac12|60+180+80-36-60-(-64)|\sqrt{3}/3=48\sqrt{3}.\]
Hence the answer is $\boxed{51}$.
Note
By symmetry the area of $ABCDEF$ is twice the area of $ABCF$. Therefore, you only need to calculate the coordinates of $B$, $C$, and $F$. | // Block 1
size(200);
draw((0,0)--(10/sqrt(3),2)--(18/sqrt(3),6)--(10/sqrt(3),10)--(0,8)--(-8/sqrt(3),4)--cycle);
dot((0,0));dot((10/sqrt(3),2));dot((18/sqrt(3),6));dot((10/sqrt(3),10));dot((0,8));dot((-8/sqrt(3),4));
label("$A (0,0)$",(0,0),S);label("$B (b,2)$",(10/sqrt(3),2),SE);label("$C$",(18/sqrt(3),6),E);label("$D$",(10/sqrt(3),10),N);label("$E$",(0,8),NW);label("$F$",(-8/sqrt(3),4),W);
// Block 2
size(200);
draw((0,0)--(10/sqrt(3),2)--(18/sqrt(3),6)--(10/sqrt(3),10)--(0,8)--(-8/sqrt(3),4)--cycle);
dot((0,0));dot((10/sqrt(3),2));dot((18/sqrt(3),6));dot((10/sqrt(3),10));dot((0,8));dot((-8/sqrt(3),4));
label("$A (0,0)$",(0,0),SE);label("$B (b,2)$",(10/sqrt(3),2),SE);label("$C$",(18/sqrt(3),6),E);label("$D$",(10/sqrt(3),10),N);label("$E$",(0,8),NW);label("$F$",(-8/sqrt(3),4),W);
xaxis("$x$");yaxis("$y$");
pair b=foot((10/sqrt(3),2),(0,0),(10,0));
pair f=foot((-8/sqrt(3),4),(0,0),(-10,0));
draw(b--(10/sqrt(3),2),dotted);
draw(f--(-8/sqrt(3),4),dotted);
label("$\theta$",(0,0),7*dir((0,0)--(10/sqrt(3),2)+(4*sqrt(21)/3,0)));
// Block 3
size(200); draw((0,0)--(10/sqrt(3),2)--(18/sqrt(3),6)--(10/sqrt(3),10)--(0,8)--(-8/sqrt(3),4)--cycle); dot((0,0));dot((10/sqrt(3),2));dot((18/sqrt(3),6));dot((10/sqrt(3),10));dot((0,8));dot((-8/sqrt(3),4)); label("$A (0,0)$",(0,0),S);label("$B (b,2)$",(10/sqrt(3),2),SE);label("$C$",(18/sqrt(3),6),E);label("$D$",(10/sqrt(3),10),N);label("$E$",(0,8),NW);label("$F$",(-8/sqrt(3),4),W);
// Block 4
size(200); draw((0,0)--(10/sqrt(3),2)--(18/sqrt(3),6)--(10/sqrt(3),10)--(0,8)--(-8/sqrt(3),4)--cycle); dot((0,0));dot((10/sqrt(3),2));dot((18/sqrt(3),6));dot((10/sqrt(3),10));dot((0,8));dot((-8/sqrt(3),4)); label("$A (0,0)$",(0,0),SE);label("$B (b,2)$",(10/sqrt(3),2),SE);label("$C$",(18/sqrt(3),6),E);label("$D$",(10/sqrt(3),10),N);label("$E$",(0,8),NW);label("$F$",(-8/sqrt(3),4),W); xaxis("$x$");yaxis("$y$"); pair b=foot((10/sqrt(3),2),(0,0),(10,0)); pair f=foot((-8/sqrt(3),4),(0,0),(-10,0)); draw(b--(10/sqrt(3),2),dotted); draw(f--(-8/sqrt(3),4),dotted); label("$\theta$",(0,0),7*dir((0,0)--(10/sqrt(3),2)+(4*sqrt(21)/3,0))); | [] |
114 | Let $A = (0,0)$ and $B = (b,2)$ be points on the coordinate plane. Let $ABCDEF$ be a convex equilateral hexagon such that $\angle FAB = 120^\circ,$ $\overline{AB}\parallel \overline{DE},$ $\overline{BC}\parallel \overline{EF,}$ $\overline{CD}\parallel \overline{FA},$ and the y-coordinates of its vertices are distinct elements of the set $\{0,2,4,6,8,10\}.$ The area of the hexagon can be written in the form $m\sqrt {n},$ where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. Find $m + n.$ | 2003 AIME II Problem 14 | First, we see that the y-coordinates of F, E, D, and C must be $4$, $8$, and $10$, and $6$, respectively, as in the previous solutions. We can draw a rectangle around the hexagon ABCDEF and use negative space to find the area of the hexagon. If we call the distance from the foot of the perpendiculars of B and F to A $x$ and $z$, respectively, and the distance from the bottom left vertex of the rectangle to the foot of the perpendicular from B $y$. This tells us that the area of the entire rectangle is $10(x+y+z)$, since the opposite sides are parallel and thus the length of the rectangle is $4+4+2=10$. Then,
if we find the area of the extra triangles and subtract, we find that the area of hexagon ABCDEF as $6x+8z+2y$. However, noticing that $x=y$, the area of ABCDEF can also be expressed as $8(x+z)$. Now we just need to find $x+z$. Since $AB=AF$ and $\angle BAF = 120$ degrees, $BF=AB\sqrt{3}$. However, we can find AB by using the Pythagorean Theorem on either of the right triangles formed by dropping perpendiculars from B and F to the x-axis (let's call them ABX and AFY).
From triangle ABX we have that $AB=\sqrt{4+x^2}$, so $BF=\sqrt{3x^2+12}$. Since AB=AF, we can also form the equation $4+x^2=16+z^2$.
We can also find BF by dropping a perpendicular from B to line FY and using the Pythagorean Theorem on the right triangle formed. This gives us $BF=\sqrt{4+(x+z)^2}$. Setting our two values of BF equal and substituting $x^2$ as $12+z^2$ and simplifying, we get the equation $3z^4-16z^2-1024=0$. Now we can use the quadratic formula to get that $z^2=\frac{64}{3}$ or $-18$, so $z^2=\frac{64}{3}$. Plugging this value back into the equation $x^2=12+z^2$, we get that $x^2=\frac{100}{3}$. Now we get that $x+z$ is $6\sqrt{3}$, so the area of the hexagon is $8 \cdot 6\sqrt{3}=48\sqrt{3}$, so the answer is $48+3=\boxed{051}$
~ant08 and sky2025 | size(200); draw((0,0)--(10/sqrt(3),2)--(18/sqrt(3),6)--(10/sqrt(3),10)--(0,8)--(-8/sqrt(3),4)--cycle); dot((0,0));dot((10/sqrt(3),2));dot((18/sqrt(3),6));dot((10/sqrt(3),10));dot((0,8));dot((-8/sqrt(3),4)); label("$A (0,0)$",(0,0),SE);label("$B (b,2)$",(10/sqrt(3),2),SE);label("$C$",(18/sqrt(3),6),E);label("$D$",(10/sqrt(3),10),N);label("$E$",(0,8),NW);label("$F$",(-8/sqrt(3),4),W); xaxis("$x$");yaxis("$y$"); pair b=foot((10/sqrt(3),2),(0,0),(10,0)); pair f=foot((-8/sqrt(3),4),(0,0),(-10,0)); draw(b--(10/sqrt(3),2),dotted); draw(f--(-8/sqrt(3),4),dotted); label("$\theta$",(0,0),7*dir((0,0)--(10/sqrt(3),2)+(4*sqrt(21)/3,0))); | [] |
115 | Square $ABCD$ has sides of length 2. Set $S$ is the set of all line segments that have length 2 and whose endpoints are on adjacent sides of the square. The midpoints of the line segments in set $S$ enclose a region whose area to the nearest hundredth is $k$. Find $100k$. | 2004 AIME I Problem 4 | Without loss of generality, let $(0,0)$, $(2,0)$, $(0,2)$, and $(2,2)$ be the vertices of the square. Suppose the endpoints of the segment lie on the two sides of the square determined by the vertex $(0,0)$. Let the two endpoints of the segment have coordinates $(x,0)$ and $(0,y)$. Because the segment has length 2, $x^2+y^2=4$. Using the midpoint formula, we find that the midpoint of the segment has coordinates $\left(\frac{x}{2},\frac{y}{2}\right)$. Let $d$ be the distance from $(0,0)$ to $\left(\frac{x}{2},\frac{y}{2}\right)$. Using the distance formula we see that $d=\sqrt{\left(\frac{x}{2}\right)^2+\left(\frac{y}{2}\right)^2}= \sqrt{\frac{1}{4}\left(x^2+y^2\right)}=\sqrt{\frac{1}{4}(4)}=1$. Thus the midpoints lying on the sides determined by vertex $(0,0)$ form a quarter-circle with radius 1.
The set of all midpoints forms a quarter circle at each corner of the square. The area enclosed by all of the midpoints is $4-4\cdot \left(\frac{\pi}{4}\right)=4-\pi \approx .86$ to the nearest hundredth. Thus $100\cdot k=\boxed{86}$ | // Block 1
size(100);
pointpen=black;pathpen = black+linewidth(0.7);
pair A=(0,0),B=(2,0),C=(2,2),D=(0,2);
D(A--B--C--D--A);
picture p;
draw(p,CR(A,1));draw(p,CR(B,1));draw(p,CR(C,1));draw(p,CR(D,1));
clip(p,A--B--C--D--cycle);
add(p);
// Block 2
size(100); pointpen=black;pathpen = black+linewidth(0.7); pair A=(0,0),B=(2,0),C=(2,2),D=(0,2); D(A--B--C--D--A); picture p; draw(p,CR(A,1));draw(p,CR(B,1));draw(p,CR(C,1));draw(p,CR(D,1)); clip(p,A--B--C--D--cycle); add(p); | [] |
115 | Square $ABCD$ has sides of length 2. Set $S$ is the set of all line segments that have length 2 and whose endpoints are on adjacent sides of the square. The midpoints of the line segments in set $S$ enclose a region whose area to the nearest hundredth is $k$. Find $100k$. | 2004 AIME I Problem 4 | If we imagine an arbitrary line with length $2$ connecting two sides of the square, we can draw the rectangle formed by drawing a perpendicular from where that line touches the square.
Drawing the other diagonal of the rectangle, it also has length two, and it bisects with the original line. Since their intersection is the midpoint of both lines, the distance from the corner to the midpoint is always $1$, which forms a circle with radius $1$ centered at the corner of the square.
The area of the shape then follows from simple calculations. | // Block 1
size(100);
pointpen=black;pathpen = black+linewidth(0.7);
draw((0,0)--(2,0)--(2,2)--(0,2)--cycle);
draw(arc((0,2),1,270,360));
draw((0,1)--(1.7,2)); draw((0,2)--(1.7,1)); draw((0,1)--(1.7,1)--(1.7,2));
// Block 2
size(100); pointpen=black;pathpen = black+linewidth(0.7); draw((0,0)--(2,0)--(2,2)--(0,2)--cycle); draw(arc((0,2),1,270,360)); draw((0,1)--(1.7,2)); draw((0,2)--(1.7,1)); draw((0,1)--(1.7,1)--(1.7,2)); | [] |
116 | Let $ABC$ be a triangle with sides 3, 4, and 5, and $DEFG$ be a 6-by-7 rectangle. A segment is drawn to divide triangle $ABC$ into a triangle $U_1$ and a trapezoid $V_1$ and another segment is drawn to divide rectangle $DEFG$ into a triangle $U_2$ and a trapezoid $V_2$ such that $U_1$ is similar to $U_2$ and $V_1$ is similar to $V_2.$ The minimum value of the area of $U_1$ can be written in the form $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | 2004 AIME I Problem 9 | We let $AB=3, AC=4, DE=6, DG=7$ for the purpose of labeling. Clearly, the dividing segment in $DEFG$ must go through one of its vertices, without loss of generality $D$. The other endpoint ($D'$) of the segment can either lie on $\overline{EF}$ or $\overline{FG}$. $V_2$ is a trapezoid with a right angle then, from which it follows that $V_1$ contains one of the right angles of $\triangle ABC$, and so $U_1$ is similar to $ABC$. Thus $U_1$, and hence $U_2$, are $3-4-5\,\triangle$s.
Suppose we find the ratio $r$ of the smaller base to the larger base for $V_2$, which consequently is the same ratio for $V_1$. By similar triangles, it follows that $U_1 \sim \triangle ABC$ by the same ratio $r$, and since the ratio of the areas of two similar figures is equal to the square of the ratio of their corresponding lengths, it follows that $[U_1] = r^2 \cdot [ABC] = 6r^2$.
If $D'$ lies on $\overline{EF}$, then $ED' = \frac34 \cdot 6 = \frac{9}{2},\, 8$ from maintaining the $3-4-5$ triangular ratio; the latter ($\frac43 \cdot 6 = 8$) can be discarded as extraneous since we need $ED'<EF$. Therefore, $D'F = \frac{5}{2}$, and the ratio $r = \frac{D'F}{DG} = \frac{5}{14}$. The area of $[U_1] = 6\left(\frac{5}{14}\right)^2$ in this case.
If $D'$ lies on $\overline{FG}$, then $GD' = \frac{21}{4},\, \frac{28}{3}$; the latter can be discarded as extraneous since we need $GD'<GF$. Therefore, $D'F = \frac{3}{4}$, and the ratio $r = \frac{D'F}{DE} = \frac{1}{8}$. The area of $[U_1] = 6\left(\frac{1}{8}\right)^2$ in this case.
Of the two cases, the second is smaller; the answer is $\frac{3}{32}$, and $m+n = \boxed{035}$. | // Block 1
defaultpen(linewidth(0.7));
pair A=(0,0),B=(0,3),C=(4,0);
draw(MP("A",A)--MP("B",B,N)--MP("C",C)--cycle);
draw((9*4/14,0)--(9*4/14,5*3/14),dashed);
label("\(V_1\)",(1,1.2)); label("\(U_1\)",(3,0.3));
// Block 2
defaultpen(linewidth(0.7)); pointpen = black;
pair D=(0,0),E=(0,6),F=(7,6),G=(7,0),H=(4.5,6);
draw(MP("D",D)--MP("E",E,N)--MP("F",F,N)--MP("G",G)--cycle);
draw(D--D(MP("D'",H,N)),dashed);
label("\(U_2\)",(1,3)); label("\(V_2\)",(5,3)); MP("7",(D+G)/2,S); MP("6",(D+E)/2,W); MP("9/2",(E+H)/2,N);
// Block 3
defaultpen(linewidth(0.7));
pair A=(0,0),B=(0,3),C=(4,0);
draw(MP("A",A)--MP("B",B,N)--MP("C",C)--cycle);
draw((3.5,0)--(3.5,3/8),dashed);
label("\(V_1\)",(1.5,1)); label("\(U_1\)",(3.8,0.4));
// Block 4
defaultpen(linewidth(0.7)); pointpen = black;
pair D=(0,0),E=(0,6),F=(7,6),G=(7,0),H=(7,21/4);
draw(MP("D",D)--MP("E",E,N)--MP("F",F,N)--MP("G",G)--cycle);
draw(D--D(MP("D'",H,NW)),dashed);
label("\(V_2\)",(2,3)); label("\(U_2\)",(5,1)); MP("7",(D+G)/2,S); MP("6",(D+E)/2,W); MP("21/4",(G+H)/2,(-1,0));
// Block 5
defaultpen(linewidth(0.7)); pair A=(0,0),B=(0,3),C=(4,0); draw(MP("A",A)--MP("B",B,N)--MP("C",C)--cycle); draw((9*4/14,0)--(9*4/14,5*3/14),dashed); label("\(V_1\)",(1,1.2)); label("\(U_1\)",(3,0.3));
// Block 6
defaultpen(linewidth(0.7)); pointpen = black; pair D=(0,0),E=(0,6),F=(7,6),G=(7,0),H=(4.5,6); draw(MP("D",D)--MP("E",E,N)--MP("F",F,N)--MP("G",G)--cycle); draw(D--D(MP("D'",H,N)),dashed); label("\(U_2\)",(1,3)); label("\(V_2\)",(5,3)); MP("7",(D+G)/2,S); MP("6",(D+E)/2,W); MP("9/2",(E+H)/2,N);
// Block 7
defaultpen(linewidth(0.7)); pair A=(0,0),B=(0,3),C=(4,0); draw(MP("A",A)--MP("B",B,N)--MP("C",C)--cycle); draw((3.5,0)--(3.5,3/8),dashed); label("\(V_1\)",(1.5,1)); label("\(U_1\)",(3.8,0.4));
// Block 8
defaultpen(linewidth(0.7)); pointpen = black; pair D=(0,0),E=(0,6),F=(7,6),G=(7,0),H=(7,21/4); draw(MP("D",D)--MP("E",E,N)--MP("F",F,N)--MP("G",G)--cycle); draw(D--D(MP("D'",H,NW)),dashed); label("\(V_2\)",(2,3)); label("\(U_2\)",(5,1)); MP("7",(D+G)/2,S); MP("6",(D+E)/2,W); MP("21/4",(G+H)/2,(-1,0)); | [] |
117 | A unicorn is tethered by a $20$-foot silver rope to the base of a magician's cylindrical tower whose radius is $8$ feet. The rope is attached to the tower at ground level and to the unicorn at a height of $4$ feet. The unicorn has pulled the rope taut, the end of the rope is $4$ feet from the nearest point on the tower, and the length of the rope that is touching the tower is $\frac{a-\sqrt{b}}c$ feet, where $a, b,$ and $c$ are positive integers, and $c$ is prime. Find $a+b+c.$ | 2004 AIME I Problem 14 | Looking from an overhead view, call the center of the circle $O$, the tether point to the unicorn $A$ and the last point where the rope touches the tower $B$. $\triangle OAB$ is a right triangle because $OB$ is a radius and $BA$ is a tangent line at point $B$. We use the Pythagorean Theorem to find the horizontal component of $AB$ has length $4\sqrt{5}$.
Now look at a side view and "unroll" the cylinder to be a flat surface. Let $C$ be the bottom tether of the rope, let $D$ be the point on the ground below $A$, and let $E$ be the point directly below $B$. Triangles $\triangle CDA$ and $\triangle CEB$ are similar right triangles. By the Pythagorean Theorem $CD=8\cdot\sqrt{6}$.
Let $x$ be the length of $CB$. \[\frac{CA}{CD}=\frac{CB}{CE}\implies \frac{20}{8\sqrt{6}}=\frac{x}{8\sqrt{6}-4\sqrt{5}}\implies x=\frac{60-\sqrt{750}}{3}\]
Therefore $a=60, b=750, c=3, a+b+c=\boxed{813}$. | // Block 1
/* Settings */
import three; defaultpen(fontsize(10)+linewidth(0.62));
currentprojection = perspective(-2,-50,15); size(200);
/* Variables */
real x = 20 - ((750)^.5)/3, CE = 8*(6^.5) - 4*(5^.5), CD = 8*(6^.5), h = 4*CE/CD;
pair Cxy = 8*expi((3*pi)/2-CE/8);
triple Oxy = (0,0,0), A=(4*5^.5,-8,4), B=(0,-8,h), C=(Cxy.x,Cxy.y,0), D=(A.x,A.y,0), E=(B.x,B.y,0), O=(O.x,O.y,h);
pair L = 8*expi(pi+0.05), R = 8*expi(-0.22); /* left and right cylinder lines, numbers from trial/error */
/* Drawing */
draw(B--A--D--E--B--C);
draw(circle(Oxy,8));
draw(circle(O,8));
draw((L.x,L.y,0)--(L.x,L.y,h)); draw((R.x,R.y,0)--(R.x,R.y,h));
draw(O--B--(A.x,A.y,h)--cycle,dashed);
/* Labeling */
label("\(A\)",A,NE); dot(A);
label("\(B\)",B,NW); dot(B);
label("\(C\)",C,W); dot(C);
label("\(D\)",D,E); dot(D);
label("\(E\)",E,S); dot(E);
label("\(O\)",O,NW); dot(O);
pair O1 = (25,4), A1=O1+(4*sqrt(5),-8), B1=O1+(0,-8);
draw(circle(O1,8));
draw(O1--A1--B1--O1);
label("\(A\)",A1,SE);label("\(B\)",B1,SW);label("\(O\)",O1,N);
label("$8$",O1--B1,W);label("$4$",(5*A1+O1)/6,0.8*(1,1));
label("$8$",O1--A1,0.8*(-0.5,2));label("$4\sqrt{5}$",B1/2+A1/2,(0,-1));
// Block 2
defaultpen(fontsize(10)+linewidth(0.62));
pair A=(-4*sqrt(5),4), B=(0,4*(8*sqrt(6)-4*sqrt(5))/(8*sqrt(6))), C=(8*sqrt(6)-4*sqrt(5),0), D=(-4*sqrt(5),0), E=(0,0);
draw(A--C--D--A);draw(B--E);
label("\(A\)",A,(-1,1));label("\(B\)",B,(1,1));label("\(C\)",C,(1,0));label("\(D\)",D,(-1,-1));label("\(E\)",E,(0,-1));
label("$4\sqrt{5}$",D/2+E/2,(0,-1));label("$8\sqrt{6}-4\sqrt{5}$",C/2+E/2,(0,-1));
label("$4$",D/2+A/2,(-1,0));label("$x$",C/2+B/2,(1,0.5));label("$20-x$",0.7*A+0.3*B,(1,0.5));
dot(A^^B^^C^^D^^E);
// Block 3
/* Settings */ import three; defaultpen(fontsize(10)+linewidth(0.62)); currentprojection = perspective(-2,-50,15); size(200); /* Variables */ real x = 20 - ((750)^.5)/3, CE = 8*(6^.5) - 4*(5^.5), CD = 8*(6^.5), h = 4*CE/CD; pair Cxy = 8*expi((3*pi)/2-CE/8); triple Oxy = (0,0,0), A=(4*5^.5,-8,4), B=(0,-8,h), C=(Cxy.x,Cxy.y,0), D=(A.x,A.y,0), E=(B.x,B.y,0), O=(O.x,O.y,h); pair L = 8*expi(pi+0.05), R = 8*expi(-0.22); /* left and right cylinder lines, numbers from trial/error */ /* Drawing */ draw(B--A--D--E--B--C); draw(circle(Oxy,8)); draw(circle(O,8)); draw((L.x,L.y,0)--(L.x,L.y,h)); draw((R.x,R.y,0)--(R.x,R.y,h)); draw(O--B--(A.x,A.y,h)--cycle,dashed); /* Labeling */ label("\(A\)",A,NE); dot(A); label("\(B\)",B,NW); dot(B); label("\(C\)",C,W); dot(C); label("\(D\)",D,E); dot(D); label("\(E\)",E,S); dot(E); label("\(O\)",O,NW); dot(O); pair O1 = (25,4), A1=O1+(4*sqrt(5),-8), B1=O1+(0,-8); draw(circle(O1,8)); draw(O1--A1--B1--O1); label("\(A\)",A1,SE);label("\(B\)",B1,SW);label("\(O\)",O1,N); label("$8$",O1--B1,W);label("$4$",(5*A1+O1)/6,0.8*(1,1)); label("$8$",O1--A1,0.8*(-0.5,2));label("$4\sqrt{5}$",B1/2+A1/2,(0,-1));
// Block 4
defaultpen(fontsize(10)+linewidth(0.62)); pair A=(-4*sqrt(5),4), B=(0,4*(8*sqrt(6)-4*sqrt(5))/(8*sqrt(6))), C=(8*sqrt(6)-4*sqrt(5),0), D=(-4*sqrt(5),0), E=(0,0); draw(A--C--D--A);draw(B--E); label("\(A\)",A,(-1,1));label("\(B\)",B,(1,1));label("\(C\)",C,(1,0));label("\(D\)",D,(-1,-1));label("\(E\)",E,(0,-1)); label("$4\sqrt{5}$",D/2+E/2,(0,-1));label("$8\sqrt{6}-4\sqrt{5}$",C/2+E/2,(0,-1)); label("$4$",D/2+A/2,(-1,0));label("$x$",C/2+B/2,(1,0.5));label("$20-x$",0.7*A+0.3*B,(1,0.5)); dot(A^^B^^C^^D^^E); | [] |
118 | $ABCD$ is a rectangular sheet of paper that has been folded so that corner $B$ is matched with point $B'$ on edge $AD.$ The crease is $EF,$ where $E$ is on $AB$ and $F$ is on $CD.$ The dimensions $AE=8, BE=17,$ and $CF=3$ are given. The perimeter of rectangle $ABCD$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | 2004 AIME II Problem 7 | Solution 1 (Synthetic)
Since $EF$ is the perpendicular bisector of $\overline{BB'}$, it follows that $BE = B'E$ (by SAS). By the Pythagorean Theorem, we have $AB' = 15$. Similarly, from $BF = B'F$, we have
\begin{align*} BC^2 + CF^2 = B'D^2 + DF^2 &\Longrightarrow BC^2 + 9 = (BC - 15)^2 + 484 \\ BC &= \frac{70}{3} \end{align*}
Thus the perimeter of $ABCD$ is $2\left(25 + \frac{70}{3}\right) = \frac{290}{3}$, and our answer is $m+n=\boxed{293}$.
Solution 2 (analytic)
Let $A = (0,0), B=(0,25)$, so $E = (0,8)$ and $F = (l,22)$, and let $l = AD$ be the length of the rectangle. The slope of $EF$ is $\frac{14}{l}$ and so the equation of $EF$ is $y -8 = \frac{14}{l}x$. We know that $EF$ is perpendicular to and bisects $BB'$. The slope of $BB'$ is thus $\frac{-l}{14}$, and so the equation of $BB'$ is $y -25 = \frac{-l}{14}x$. Let the point of intersection of $EF, BB'$ be $G$. Then the y-coordinate of $G$ is $\frac{25}{2}$, so
\begin{align*} \frac{14}{l}x &= y-8 = \frac{9}{2}\\ \frac{-l}{14}x &= y-25 = -\frac{25}{2}\\ \end{align*}
Dividing the two equations yields
$l^2 = \frac{25 \cdot 14^2}{9} \Longrightarrow l = \frac{70}{3}$
The answer is $\boxed{293}$ as above.
Solution 3 (Coordinate Bashing)
Firstly, observe that if we are given that $AE=8$ and $BE=17$, the length of the triangle is given and the height depends solely on the length of $CF$. Let Point $A = (0,0)$. Since $AE=8$, point E is at (8,0). Next, point $B$ is at $(25,0)$ since $BE=17$ and point $B'$ is at $(0,15)$ since $BE=AE$ by symmetry. Draw line segment $BB'$. Notice that this is perpendicular to $EF$ by symmetry. Next, find the slope of EB, which is $\frac{15}{25}=\frac{3}{5}$. Then, the slope of $EF$ is -$\frac{5}{3}$.
Line EF can be written as y=$-\frac{5}{3}x+b$. Plug in the point $(8,0)$, and we get the equation of EF to be y=$_\frac{5}{3}x+\frac{40}{3}$. Since the length of $AB$=25, a point on line $EF$ lies on $DC$ when $x=25-3=22$. Plug in $x=22$ into our equation to get $y=-\frac{70}{3}$. $|y|=BC=\frac{70}{3}$. Therefore, our answer is $2(AB+BC)=2\left(25+\frac{70}{3}\right)=2\left(\frac{145}{3}\right)=\frac{290}{3}= \boxed{293}$.
Solution 4 (Trig)
Firstly, note that $B'E=BE=17$, so $AB'=\sqrt{17^2-8^2}=15$. Then let $\angle BEF=\angle B'EF=\theta$, so $\angle B'EA = \pi-2\theta$. Then $\tan(\pi-2\theta)=\frac{15}{8}$, or
\[\frac{2\tan(\theta)}{\tan^2(\theta)-1}=\frac{15}{8}\] using supplementary and double angle identities. Multiplying though and factoring yields
\[(3\tan(\theta)-5)(5\tan(\theta)+3)=0\]
It is clear from the problem setup that $0<\theta<\frac\pi2$, so the correct value is $\tan(\theta)=\frac53$. Next, extend rays $\overrightarrow{BC}$ and $\overrightarrow{EF}$ to intersect at $C'$. Then $\tan(\theta)=\frac{BC'}{17}=\frac53$, so $BC'=\frac{85}{3}$. By similar triangles, $CC'=\frac{3}{17}BC'=\frac{15}{3}$, so $BC=\frac{70}{3}$. The perimeter is $\frac{140}{3}+50=\frac{290}{3}\Longrightarrow \boxed{293}$
An even faster way to finish is, to draw a line segment $FF'$ where $F'$ is a point on $EB$ such that $FF'$ is perpendicular to $EB$. This makes right triangle $FF'E$, Also, note that $F'B$ has length of $3$ (draw the diagram out, and note the $F'B =FC$). From here, through $\tan \theta = \frac{5}{3}$, we can note that $\frac{FF'}{EF'} = \frac{5}{3} \implies \frac{FF'}{14} = \frac{5}{3} \implies FF' = \frac{70}{3}$. $FF'$ is parallel and congrurent to $CB$ and $AD$, and hence we can use this to calculate the perimeter. The perimeter is simply $\frac{70}{3} + \frac{70}{3} + 25 + 25 = \frac{290}{3} \Longrightarrow \boxed{293}$
Solution 5 (Fast, Pythagorean)
Use the prepared diagram for this solution.
Call the intersection of DF and B'C' G. AB'E is an 8-15-17 right triangle, and so are B'DG and C'FG. Since C'F is 3, then using the properties of similar triangles GF is 51/8. DF is 22, so DG is 125/8. Finally, DB can to calculated to be 25/3.
Add all the sides together to get $\boxed{293}$.
-jackshi2006
Solution 6(fast as wind[rufeng])
Call the intersection of $B'C'$, $BC$, and $EF$ $G$. Since $FCBE$ and $FC'B'E$ are congruent, we know that the three lines intersect.
We already know $AB$ so we just need to find $CB$, call it $x$. Drop an altitude from $F$ to $AB$ and call it $H$. $EH=EB-FC=14$. Using Pythagorean Theorem, we have $EF=\sqrt{x^2+14^2}$. Triangles $EFH$ and $EGB$ are similar (AA), so we get
\[\frac{HF}{BG}=\frac{EH}{EB}\]
\[\frac{x}{x+GC}=\frac{14}{17}\]
Simplify and we get $GC=\frac{3x}{14}$.
We find the area of $FCBE$ by using the fact that it is a trapezoid.
$[FCBE]=\frac{(3+17)x}{2}=10x$
A different way to find the area:
$[FCBE]=\frac{1}{2} EG\cdot($height of $EGB$ with $EG$ as base$)-[FGC]$
Since $GBE$ and $G'B'E$ are congruent(SAS), their height from $EG$ is the same. $B'B=\sqrt{AB'^2+AB^2}=5\sqrt{34}$.
$EG=\sqrt{EB^2+BG^2}=\sqrt{(\frac{17x}{14})^2+17^2}=17\sqrt{\frac{x^2}{196}+1}$
\[[FCBE]=\frac{1}{2} \cdot 17 \cdot \sqrt{\frac{x^2}{196}+1} \cdot \frac{5\sqrt{34}}{2}-\frac{9x}{28}\]
\[280x+9x=7\cdot 5 \cdot \sqrt{34} \cdot 17 \cdot \sqrt{\frac{x^2}{196}+1}\]
\[17^4 x^2=49 \cdot 25 \cdot 34 \cdot 17^2 \cdot (\frac{x^2}{196}+1)\]
\[17x^2=\frac{25}{2}x^2+2450\]
\[x=\frac{70}{3}\]
The perimeter is $\frac{140}{3}+50=\frac{290}{3},$ so our answer is $\boxed{293}$.
Solution 7 (Similar to solution 5, more in depth)
Let the endpoint of the intersection of the fold near $F$ be $G$. Since trapezoid $BCFE$ is folded, it is congruent to trapezoid $B'C'FE$. Therefore, $BE=B'E=17$. Since $\triangle AB'E$ is a right triangle, $AB'=15$ from the pythagorean theorem. From here, we can see that triangles $\triangle AEB \sim \triangle DGB' \sim \triangle C'GF$ by AA similarity. From here, we find $BC$ from a lot of similarities. Let $BC=x$.
Since $\triangle ABE' \sim \triangle DGB'$:
\[\frac {AE}{AB'} = \frac{DB}{DG}\]
\[\frac {8}{15} = \frac {x-15}{DG}\]
\[DG = \frac {15(x-15)}{8}\]
\[GF = DC-DG-FC\]
\[GF = \frac{-15x+401}{8}\]
Since $\triangle ABE' \sim \triangle C'GF'$,
\[\frac {AE}{B'E} = \frac {C'F}{GF}\]
\[\frac {8}{17} = \frac{3}{\frac {-15x+401}{8}}\]
from which we get $x= \frac {70}{3}$.
Finally, our answer is $2(\frac {70}{3}) + 2(25)=\frac {290}{3}$, which is $290+3=\boxed{293}$.
~ Wesserwessey7254 | pointpen = black; pathpen = black +linewidth(0.7); pair A=(0,0),B=(0,25),C=(70/3,25),D=(70/3,0),E=(0,8),F=(70/3,22),G=(15,0); D(MP("A",A)--MP("B",B,N)--MP("C",C,N)--MP("D",D)--cycle); D(MP("E",E,W)--MP("F",F,(1,0))); D(B--G); D(E--MP("B'",G)--F--B,dashed); MP("8",(A+E)/2,W);MP("17",(B+E)/2,W);MP("22",(D+F)/2,(1,0)); | [] |
119 | Let $ABCD$ be an isosceles trapezoid, whose dimensions are $AB = 6, BC=5=DA,$and $CD=4.$ Draw circles of radius 3 centered at $A$ and $B,$ and circles of radius 2 centered at $C$ and $D.$ A circle contained within the trapezoid is tangent to all four of these circles. Its radius is $\frac{-k+m\sqrt{n}}p,$ where $k, m, n,$ and $p$ are positive integers, $n$ is not divisible by the square of any prime, and $k$ and $p$ are relatively prime. Find $k+m+n+p.$ | 2004 AIME II Problem 12 | Let the radius of the center circle be $r$ and its center be denoted as $O$.
Clearly line $AO$ passes through the point of tangency of circle $A$ and circle $O$. Let $y$ be the height from the base of the trapezoid to $O$. From the Pythagorean Theorem,
\[3^2 + y^2 = (r + 3)^2 \Longrightarrow y = \sqrt {r^2 + 6r}.\]
We use a similar argument with the line $DO$, and find the height from the top of the trapezoid to $O$, $z$, to be $z = \sqrt {r^2 + 4r}$.
Now $y + z$ is simply the height of the trapezoid. Let $D'$ be the foot of the perpendicular from $D$ to $AB$; then $AD' = 3 - 2 = 1$. By the Pythagorean Theorem, $(AD')^2 + (DD')^2 = (AD)^2 \Longrightarrow DD' = \sqrt{24}$ so we need to solve the equation $\sqrt {r^2 + 4r} + \sqrt {r^2 + 6r} = \sqrt {24}$. We can solve this by moving one radical to the other side, and squaring the equation twice to end with a quadratic equation.
Solving this, we get $r = \frac { - 60 + 48\sqrt {3}}{23}$, and the answer is $k + m + n + p = 60 + 48 + 3 + 23 = \boxed{134}$. | // Block 1
pointpen = black; pathpen = black+linewidth(0.7); pen d = linewidth(0.7) + linetype("4 4"); pen f = fontsize(8);
real r = (-60 + 48 * 3^.5)/23;
pair A=(0,0), B=(6,0), D=(1, 24^.5), C=(5,D.y), O = (3,(r^2 + 6*r)^.5);
D(MP("A",A)--MP("B",B)--MP("C",C,N)--MP("D",D,N)--cycle); D(CR(A,3));D(CR(B,3));D(CR(C,2));D(CR(D,2));D(CR(O,r));
D(O); D((3,0)--(3,D.y),d); D(A--O--D,d); MP("3",(3/2,0),S,f);MP("2",(2,D.y),N,f);
// Block 2
pointpen = black; pathpen = black+linewidth(0.7); pen d = linewidth(0.7) + linetype("4 4"); pen f = fontsize(8); real r = (-60 + 48 * 3^.5)/23; pair A=(0,0), B=(6,0), D=(1, 24^.5), C=(5,D.y), O = (3,(r^2 + 6*r)^.5); D(MP("A",A)--MP("B",B)--MP("C",C,N)--MP("D",D,N)--cycle); D(CR(A,3));D(CR(B,3));D(CR(C,2));D(CR(D,2));D(CR(O,r)); D(O); D((3,0)--(3,D.y),d); D(A--O--D,d); MP("3",(3/2,0),S,f);MP("2",(2,D.y),N,f); | [] |
120 | Let $ABCDE$ be a convex pentagon with $AB \parallel CE, BC \parallel AD, AC \parallel DE, \angle ABC=120^\circ, AB=3, BC=5,$ and $DE = 15.$ Given that the ratio between the area of triangle $ABC$ and the area of triangle $EBD$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers, find $m+n.$ | 2004 AIME II Problem 13 | Let the intersection of $\overline{AD}$ and $\overline{CE}$ be $F$. Since $AB \parallel CE, BC \parallel AD,$ it follows that $ABCF$ is a parallelogram, and so $\triangle ABC \cong \triangle CFA$. Also, as $AC \parallel DE$, it follows that $\triangle ABC \sim \triangle EFD$.
By the Law of Cosines, $AC^2 = 3^2 + 5^2 - 2 \cdot 3 \cdot 5 \cos 120^{\circ} = 49 \Longrightarrow AC = 7$. Thus the length similarity ratio between $\triangle ABC$ and $\triangle EFD$ is $\frac{AC}{ED} = \frac{7}{15}$.
Let $h_{ABC}$ and $h_{BDE}$ be the lengths of the altitudes in $\triangle ABC, \triangle BDE$ to $AC, DE$ respectively. Then, the ratio of the areas $\frac{[ABC]}{[BDE]} = \frac{\frac 12 \cdot h_{ABC} \cdot AC}{\frac 12 \cdot h_{BDE} \cdot DE} = \frac{7}{15} \cdot \frac{h_{ABC}}{h_{BDE}}$.
However, $h_{BDE} = h_{ABC} + h_{CAF} + h_{EFD}$, with all three heights oriented in the same direction. Since $\triangle ABC \cong \triangle CFA$, it follows that $h_{ABC} = h_{CAF}$, and from the similarity ratio, $h_{EFD} = \frac{15}{7}h_{ABC}$. Hence $\frac{h_{ABC}}{h_{BDE}} = \frac{h_{ABC}}{2h_{ABC} + \frac {15}7h_{ABC}} = \frac{7}{29}$, and the ratio of the areas is $\frac{7}{15} \cdot \frac 7{29} = \frac{49}{435}$. The answer is $m+n = \boxed{484}$. | // Block 1
pointpen = black; pathpen = black+linewidth(0.7);
pair D=(0,0), E=(15,0), F=IP(CR(D, 75/7), CR(E, 45/7)), A=D+ (5+(75/7))/(75/7) * (F-D), C = E+ (3+(45/7))/(45/7) * (F-E), B=IP(CR(A,3), CR(C,5));
D(MP("A",A,(1,0))--MP("B",B,N)--MP("C",C,NW)--MP("D",D)--MP("E",E)--cycle); D(D--A--C--E); D(MP("F",F)); MP("5",(B+C)/2,NW); MP("3",(A+B)/2,NE); MP("15",(D+E)/2);
// Block 2
pointpen = black; pathpen = black+linewidth(0.7); pair D=(0,0), E=(15,0), F=IP(CR(D, 75/7), CR(E, 45/7)), A=D+ (5+(75/7))/(75/7) * (F-D), C = E+ (3+(45/7))/(45/7) * (F-E), B=IP(CR(A,3), CR(C,5)); D(MP("A",A,(1,0))--MP("B",B,N)--MP("C",C,NW)--MP("D",D)--MP("E",E)--cycle); D(D--A--C--E); D(MP("F",F)); MP("5",(B+C)/2,NW); MP("3",(A+B)/2,NE); MP("15",(D+E)/2); | [] |
121 | In quadrilateral $ABCD,\ BC=8,\ CD=12,\ AD=10,$ and $m\angle A= m\angle B = 60^\circ.$ Given that $AB = p + \sqrt{q},$ where $p$ and $q$ are positive integers, find $p+q.$
Contents
1 Problem
2 Solution
2.1 Solution 1
2.2 Solution 2
2.3 Solution 3
2.4 Solution 4
2.5 Solution 5 (EASY EASY VERY TREMENDOUSLY EASY)
3 See also | 2005 AIME I Problem 7 | Solution 1
Draw line segment $DE$ such that line $DE$ is concurrent with line $BC$. Then, $ABED$ is an isosceles trapezoid so $AD=BE=10$, and $BC=8$ and $EC=2$. We are given that $DC=12$. Since $\angle CED = 120^{\circ}$, using Law of Cosines on $\bigtriangleup CED$ gives \[12^2=DE^2+4-2(2)(DE)(\cos 120^{\circ})\] which gives \[144-4=DE^2+2DE\]. Adding $1$ to both sides gives $141=(DE+1)^2$, so $DE=\sqrt{141}-1$. $\bigtriangleup DAP$ and $\bigtriangleup EBQ$ are both $30-60-90$, so $AP=5$ and $BQ=5$. $PQ=DE$, and therefore $AB=AP+PQ+BQ=5+\sqrt{141}-1+5=9+\sqrt{141} \rightarrow (p,q)=(9,141) \rightarrow \boxed{150}$.
Solution 2
[Image: images/aime/2005_AIME_I_Problem_7_0.png]
Draw the perpendiculars from $C$ and $D$ to $AB$, labeling the intersection points as $E$ and $F$. This forms 2 $30-60-90$ right triangles, so $AE = 5$ and $BF = 4$. Also, if we draw the horizontal line extending from $C$ to a point $G$ on the line $DE$, we find another right triangle $\triangle DGC$. $DG = DE - CF = 5\sqrt{3} - 4\sqrt{3} = \sqrt{3}$. The Pythagorean Theorem yields that $GC^2 = 12^2 - \sqrt{3}^2 = 141$, so $EF = GC = \sqrt{141}$. Therefore, $AB = 5 + 4 + \sqrt{141} = 9 + \sqrt{141}$, and $p + q = \boxed{150}$.
Solution 3
[Image: images/aime/2005_AIME_I_Problem_7_1.png]
Extend $AD$ and $BC$ to an intersection at point $E$. We get an equilateral triangle $ABE$. We denote the length of a side of $\triangle ABE$ as $s$ and solve for it using the Law of Cosines: \[12^2 = (s - 10)^2 + (s - 8)^2 - 2(s - 10)(s - 8)\cos{60}\]
\[144 = 2s^2 - 36s + 164 - (s^2 - 18s + 80)\] This simplifies to $s^2 - 18s - 60=0$; the quadratic formula yields the (discard the negative result) same result of $9 + \sqrt{141}$.
Solution 4
Extend $BC$ and $AD$ to meet at point $E$, forming an equilateral triangle $\triangle ABE$. Draw a line from $C$ parallel to $AB$ so that it intersects $AD$ at point $F$. Then, apply Stewart's Theorem on $\triangle CFE$. Let $CE=x$. \[2x(x-2) + 12^2x = 2x^2 + x^2(x-2)\] \[x^3 - 2x^2 - 140x = 0\] By the quadratic formula (discarding the negative result), $x = 1 + \sqrt{141}$, giving $AB = 9 + \sqrt{141}$ for a final answer of $p+q=150$.
Solution 5 (EASY EASY VERY TREMENDOUSLY EASY)
Draw a line from point $C$ to a new point $E$ on $AD$ parallel to $AB$. Draw a line from point $E$ to a new point $F$ on $AB$ parallel to $CD$. This creates parallelogram $CEFB$.
The reasoning for this is to connect the two angles that are congruent.
$\angle CEF = 60^{\circ}$ and triangle $AEF$ is an equilateral triangle with side length 8. Thus, $AF = 8$.
$ED = 10 - 8 = 2$. By the Law of Cosines, $CD^2 = ED^2 + EC^2 - 2 \cdot {ED} \cdot {EC} \cdot \cos{\angle DEC}$. Thus, $144 = 4 + c^2 - 2c$ and $c = 1 + \sqrt{141}$. $EC = FB$.
$AB = AF + FB = 8 + 1 + \sqrt{141} = 9 + \sqrt{141}$ so the answer is $p+q=150$.
-unhappyfarmer | draw((0,0)--(20.87,0)--(15.87,8.66)--(5,8.66)--cycle); draw((5,8.66)--(5,0)); draw((15.87,8.66)--(15.87,0)); draw((5,8.66)--(16.87,6.928)); label("$A$",(0,0),SW); label("$B$",(20.87,0),SE); label("$E$",(15.87,8.66),NE); label("$D$",(5,8.66),NW); label("$P$",(5,0),S); label("$Q$",(15.87,0),S); label("$C$",(16.87,7),E); label("$12$",(10.935,7.794),S); label("$10$",(2.5,4.5),W); label("$10$",(18.37,4.5),E); | ["https://artofproblemsolving.com/wiki/images/4/41/AIME_2005I_Solution_7_1.png", "https://artofproblemsolving.com/wiki/images/2/2f/AIME_2005I_Solution_7_2.png"] |
122 | A particle moves in the Cartesian plane according to the following rules:
From any lattice point $(a,b),$ the particle may only move to $(a+1,b), (a,b+1),$ or $(a+1,b+1).$
There are no right angle turns in the particle's path.
How many different paths can the particle take from $(0,0)$ to $(5,5)$?
Contents
1 Problem
2 Solution
2.1 Solution 1
2.2 Solution 2
3 See also | 2005 AIME I Problem 13 | Solution 1
The length of the path (the number of times the particle moves) can range from $l = 5$ to $9$; notice that $d = 10-l$ gives the number of diagonals. Let $R$ represent a move to the right, $U$ represent a move upwards, and $D$ to be a move that is diagonal. Casework upon the number of diagonal moves:
Case $d = 1$: It is easy to see only $2$ cases.
Case $d = 2$: There are two diagonals. We need to generate a string with $3$ $R$'s, $3$ $U$'s, and $2$ $D$'s such that no two $R$'s or $U$'s are adjacent. The $D$'s split the string into three sections ($-D-D-$): by the Pigeonhole principle all of at least one of the two letters must be all together (i.e., stay in a row).
If both $R$ and $U$ stay together, then there are $3 \cdot 2=6$ ways.
If either $R$ or $U$ splits, then there are $3$ places to put the letter that splits, which has $2$ possibilities. The remaining letter must divide into $2$ in one section and $1$ in the next, giving $2$ ways. This totals $6 + 3\cdot 2\cdot 2 = 18$ ways.
Case $d = 3$: Now $2$ $R$'s, $2$ $U$'s, and $3$ $D$'s, so the string is divided into $4$ partitions ($-D-D-D-$).
If the $R$'s and $U$'s stay together, then there are $4 \cdot 3 = 12$ places to put them.
If one of them splits and the other stays together, then there are $4 \cdot {3\choose 2}$ places to put them, and $2$ ways to pick which splits, giving $4 \cdot 3 \cdot 2 = 24$ ways.
If both groups split, then there are ${4\choose 2}=6$ ways to arrange them. These add up to $12 + 24 + 6 = 42$ ways.
Case $d = 4$: Now $1$ $R$, $1$ $U$, $4$ $D$'s ($-D-D-D-D-$). There are $5$ places to put $R$, $4$ places to put $U$, giving $20$ ways.
Case $d = 5$: It is easy to see only $1$ case.
Together, these add up to $2 + 18 + 42 + 20 + 1 = \boxed{083}$.
Solution 2
Another possibility is to use block-walking and recursion: for each vertex, the number of ways to reach it is $a + b + c$, where $a$ is the number of ways to reach the vertex from the left (without having come to that vertex (the one on the left) from below), $b$ is the number of ways to reach the vertex from the vertex diagonally down and left, and $c$ is the number of ways to reach the vertex from below (without having come to that vertex (the one below) from the left).
Assign to each point $(i,j)$ the triplet $(a_{i,j}, b_{i,j}, c_{i,j})$. Let $s(i,j) = a_{i,j}+ b_{i,j}+ c_{i,j}$. Let all lattice points that contain exactly one negative coordinate be assigned to $(0,0,0)$. This leaves the lattice points of the first quadrant, the positive parts of the $x$ and $y$ axes, and the origin unassigned. As a seed, assign to $(0,1,0)$. (We will see how this correlates with the problem.) Then define for each lattice point $(i,j)$ its triplet thus:
\begin{align*} a_{i,j} &= s(i-1,j) - c_{i-1,j}\\ b_{i,j} &= s(i-1,j-1) \\ c_{i,j} &= s(i,j-1) - a_{i,j-1}. \end{align*}
It is evident that $s(i,j)$ is the number of ways to reach $(i,j)$ from $(0,0)$. Therefore we compute vertex by vertex the triplets $(a_{i,j}, b_{i,j}, c_{i,j})$ with $0 \leq i, j \leq 5$.
Finally, after simple but tedious calculations, we find that $(a_{5,5}, b_{5,5}, c_{5,5}) = (28,27,28)$, so $s(i,j)=28+27+28 = \boxed{083}$. | defaultpen(fontsize(8)+0.8+heavyblue); size(250); for(int i = 0; i<6; ++i) { draw((0,i)--(5,i)^^(i,0)--(i,5), gray+0.25); } label("$(0,0,0)$", (0,0), N); for(int i = 1; i<6; ++i) { label("$(0,0,1)$", (0,i), N); label("$(1,0,0)$", (i,0), N); label("$({"+string(i-1)+"},1,0)$", (i,1), N); label("$(0,1,{"+string(i-1)+"})$", (1,i), N);} real[] val={1,2,4,7}; for(int i = 2; i<6; ++i) { label("$(1,{"+string(i-1)+"}, {"+string(val[i-2])+"})$", (2,i), N); label("$({"+string(val[i-2])+"},{"+string(i-1)+"}, 1)$", (i,2), N);} label("$(3,3,3)$", (3,3), N); label("$(9,9,9)$", (4,4), N); label("$(28,27,28)$", (5,5), N); label("$(4,5,6)$", (3,4), N); label("$(6,5,4)$", (4,3), N); label("$(5,8,11)$", (3,5), N); label("$(11,8,5)$", (5,3), N); label("$(13,15,18)$", (4,5), N); label("$(18,15,13)$", (5,4), N); | [] |
123 | Triangle $ABC$ has $BC=20.$ The incircle of the triangle evenly trisects the median $AD.$ If the area of the triangle is $m \sqrt{n}$ where $m$ and $n$ are integers and $n$ is not divisible by the square of a prime, find $m+n.$ | 2005 AIME I Problem 15 | Let $E$, $F$ and $G$ be the points of tangency of the incircle with $BC$, $AC$ and $AB$, respectively. Without loss of generality, let $AC < AB$, so that $E$ is between $D$ and $C$. Let the length of the median be $3m$. Then by two applications of the Power of a Point Theorem, $DE^2 = 2m \cdot m = AF^2$, so $DE = AF$. Now, $CE$ and $CF$ are two tangents to a circle from the same point, so by the Two Tangent Theorem $CE = CF = c$ and thus $AC = AF + CF = DE + CE = CD = 10$. Then $DE = AF = AG = 10 - c$ so $BG = BE = BD + DE = 20 - c$ and thus $AB = AG + BG = 30 - 2c$.
Now, by Stewart's Theorem in triangle $\triangle ABC$ with cevian $\overline{AD}$, we have
\[(3m)^2\cdot 20 + 20\cdot10\cdot10 = 10^2\cdot10 + (30 - 2c)^2\cdot 10.\]
Our earlier result from Power of a Point was that $2m^2 = (10 - c)^2$, so we combine these two results to solve for $c$ and we get
\[9(10 - c)^2 + 200 = 100 + (30 - 2c)^2 \quad \Longrightarrow \quad c^2 - 12c + 20 = 0.\]
Thus $c = 2$ or $= 10$. We discard the value $c = 10$ as extraneous (it gives us a line) and are left with $c = 2$, so our triangle has area $\sqrt{28 \cdot 18 \cdot 8 \cdot 2} = 24\sqrt{14}$ and so the answer is $24 + 14 = \boxed{038}$. | // Block 1
size(300); pointpen=black;pathpen=black+linewidth(0.65); pen s = fontsize(10);
pair A=(0,0),B=(26,0),C=IP(circle(A,10),circle(B,20)),D=(B+C)/2,I=incenter(A,B,C);
path cir = incircle(A,B,C); pair E1=IP(cir,B--C),F=IP(cir,A--C),G=IP(cir,A--B),P=IP(A--D,cir),Q=OP(A--D,cir);
D(MP("A",A,s)--MP("B",B,s)--MP("C",C,N,s)--cycle); D(cir);
D(A--MP("D",D,NE,s)); D(MP("E",E1,NE,s)); D(MP("F",F,NW,s)); D(MP("G",G,s)); D(MP("P",P,SW,s)); D(MP("Q",Q,SE,s));
MP("10",(B+D)/2,NE); MP("10",(C+D)/2,NE);
// Block 2
size(300); pointpen=black;pathpen=black+linewidth(0.65); pen s = fontsize(10); pair A=(0,0),B=(26,0),C=IP(circle(A,10),circle(B,20)),D=(B+C)/2,I=incenter(A,B,C); path cir = incircle(A,B,C); pair E1=IP(cir,B--C),F=IP(cir,A--C),G=IP(cir,A--B),P=IP(A--D,cir),Q=OP(A--D,cir); D(MP("A",A,s)--MP("B",B,s)--MP("C",C,N,s)--cycle); D(cir); D(A--MP("D",D,NE,s)); D(MP("E",E1,NE,s)); D(MP("F",F,NW,s)); D(MP("G",G,s)); D(MP("P",P,SW,s)); D(MP("Q",Q,SE,s)); MP("10",(B+D)/2,NE); MP("10",(C+D)/2,NE); | [] |
124 | Circles $C_1$ and $C_2$ are externally tangent, and they are both internally tangent to circle $C_3.$ The radii of $C_1$ and $C_2$ are 4 and 10, respectively, and the centers of the three circles are all collinear. A chord of $C_3$ is also a common external tangent of $C_1$ and $C_2.$ Given that the length of the chord is $\frac{m\sqrt{n}}p$ where $m,n,$ and $p$ are positive integers, $m$ and $p$ are relatively prime, and $n$ is not divisible by the square of any prime, find $m+n+p.$ | 2005 AIME II Problem 8 | Let $O_1, O_2, O_3$ be the centers and $r_1 = 4, r_2 = 10,r_3 = 14$ the radii of the circles $C_1, C_2, C_3$. Let $T_1, T_2$ be the points of tangency from the common external tangent of $C_1, C_2$, respectively, and let the extension of $\overline{T_1T_2}$ intersect the extension of $\overline{O_1O_2}$ at a point $H$. Let the endpoints of the chord/tangent be $A,B$, and the foot of the perpendicular from $O_3$ to $\overline{AB}$ be $T$. From the similar right triangles $\triangle HO_1T_1 \sim \triangle HO_2T_2 \sim \triangle HO_3T$,
\[\frac{HO_1}{4} = \frac{HO_1+14}{10} = \frac{HO_1+10}{O_3T}.\]
It follows that $HO_1 = \frac{28}{3}$, and that $O_3T = \frac{58}{7}\dagger$. By the Pythagorean Theorem on $\triangle ATO_3$, we find that
\[AB = 2AT = 2\left(\sqrt{r_3^2 - O_3T^2}\right) = 2\sqrt{14^2 - \frac{58^2}{7^2}} = \frac{8\sqrt{390}}{7}\]
and the answer is $m+n+p=\boxed{405}$.
$\dagger$ Alternatively, drop an altitude from $O_1$ to $O_3T$ at $O_3'$, and to $O_2T_2$ at $O_2'$. Then, $O_2O_2'=10-4=6$, and $O_1O_2=14$. But $O_1O_3O_3'$ is similar to $O_1O_2O_2'$ so $O_3O_3'=\frac{6}{14} \cdot 10=\frac{30}{7}$. From rectangles, $O_3'T=O_1T_1=4$ so $O_3T=4+\frac{30}{7}=\frac{58}{7}$. | // Block 1
pointpen = black;
pathpen = black + linewidth(0.7);
size(200);
pair C1 = (-10,0), C2 = (4,0), C3 = (0,0), H = (-10-28/3,0), T = 58/7*expi(pi-acos(3/7));
path cir1 = CR(C1,4.01), cir2 = CR(C2,10), cir3 = CR(C3,14), t = H--T+2*(T-H);
pair A = OP(cir3, t), B = IP(cir3, t), T1 = IP(cir1, t), T2 = IP(cir2, t);
draw(cir1); draw(cir2); draw(cir3);
draw((14,0)--(-14,0));
draw(A--B);
MP("H",H,W);
draw((-14,0)--H--A, linewidth(0.7) + linetype("4 4"));
draw(MP("O_1",C1));
draw(MP("O_2",C2));
draw(MP("O_3",C3));
draw(MP("T",T,N));
draw(MP("A",A,NW));
draw(MP("B",B,NE));
draw(C1--MP("T_1",T1,N));
draw(C2--MP("T_2",T2,N));
draw(C3--T);
draw(rightanglemark(C3,T,H));
// Block 2
pointpen = black; pathpen = black + linewidth(0.7); size(200); pair C1 = (-10,0), C2 = (4,0), C3 = (0,0), H = (-10-28/3,0), T = 58/7*expi(pi-acos(3/7)); path cir1 = CR(C1,4.01), cir2 = CR(C2,10), cir3 = CR(C3,14), t = H--T+2*(T-H); pair A = OP(cir3, t), B = IP(cir3, t), T1 = IP(cir1, t), T2 = IP(cir2, t); draw(cir1); draw(cir2); draw(cir3); draw((14,0)--(-14,0)); draw(A--B); MP("H",H,W); draw((-14,0)--H--A, linewidth(0.7) + linetype("4 4")); draw(MP("O_1",C1)); draw(MP("O_2",C2)); draw(MP("O_3",C3)); draw(MP("T",T,N)); draw(MP("A",A,NW)); draw(MP("B",B,NE)); draw(C1--MP("T_1",T1,N)); draw(C2--MP("T_2",T2,N)); draw(C3--T); draw(rightanglemark(C3,T,H)); | [] |
125 | Given that $O$ is a regular octahedron, that $C$ is the cube whose vertices are the centers of the faces of $O,$ and that the ratio of the volume of $O$ to that of $C$ is $\frac mn,$ where $m$ and $n$ are relatively prime integers, find $m+n.$
Contents
1 Problem
2 Solutions
2.1 Solution 1
2.2 Solution 2
3 See also | 2005 AIME II Problem 10 | Solution 1
Let the side of the octahedron be of length $s$. Let the vertices of the octahedron be $A, B, C, D, E, F$ so that $A$ and $F$ are opposite each other and $AF = s\sqrt2$. The height of the square pyramid $ABCDE$ is $\frac{AF}2 = \frac s{\sqrt2}$ and so it has volume $\frac 13 s^2 \cdot \frac s{\sqrt2} = \frac {s^3}{3\sqrt2}$ and the whole octahedron has volume $\frac {s^3\sqrt2}3$.
Let $M$ be the midpoint of $BC$, $N$ be the midpoint of $DE$, $G$ be the centroid of $\triangle ABC$ and $H$ be the centroid of $\triangle ADE$. Then $\triangle AMN \sim \triangle AGH$ and the symmetry ratio is $\frac 23$ (because the medians of a triangle are trisected by the centroid), so $GH = \frac{2}{3}MN = \frac{2s}3$. $GH$ is also a diagonal of the cube, so the cube has side-length $\frac{s\sqrt2}3$ and volume $\frac{2s^3\sqrt2}{27}$. The ratio of the volumes is then $\frac{\left(\frac{s^3\sqrt2}{3}\right)}{\left(\frac{2s^3\sqrt2}{27}\right)} = \frac92$ and so the answer is $\boxed{011}$.
Solution 2
Let the octahedron have vertices $(\pm 3, 0, 0), (0, \pm 3, 0), (0, 0, \pm 3)$. Then the vertices of the cube lie at the centroids of the faces, which have coordinates $(\pm 1, \pm 1, \pm 1)$. The cube has volume 8. The region of the octahedron lying in each octant is a tetrahedron with three edges mutually perpendicular and of length 3. Thus the octahedron has volume $8 \cdot \left(\frac 16 \cdot3^3\right) = 36$, so the ratio is $\frac {36}{8} = \frac 92$ and so the answer is $\boxed{011}$. | // Block 1
import three; currentprojection = perspective(4,-15,4); defaultpen(linewidth(0.7));
draw(box((-1,-1,-1),(1,1,1)));
draw((-3,0,0)--(0,0,3)--(0,-3,0)--(-3,0,0)--(0,0,-3)--(0,-3,0)--(3,0,0)--(0,0,-3)--(0,3,0)--(0,0,3)--(3,0,0)--(0,3,0)--(-3,0,0));
// Block 2
import three; currentprojection = perspective(4,-15,4); defaultpen(linewidth(0.7)); draw(box((-1,-1,-1),(1,1,1))); draw((-3,0,0)--(0,0,3)--(0,-3,0)--(-3,0,0)--(0,0,-3)--(0,-3,0)--(3,0,0)--(0,0,-3)--(0,3,0)--(0,0,3)--(3,0,0)--(0,3,0)--(-3,0,0)); | [] |
126 | Square $ABCD$ has center $O,\ AB=900,\ E$ and $F$ are on $AB$ with $AE<BF$ and $E$ between $A$ and $F, m\angle EOF =45^\circ,$ and $EF=400.$ Given that $BF=p+q\sqrt{r},$ where $p,q,$ and $r$ are positive integers and $r$ is not divisible by the square of any prime, find $p+q+r.$
Contents
1 Problem
2 Solutions
2.1 Solution 1 (trigonometry)
2.2 Solution 2 (synthetic)
2.3 Solution 3 (similar triangles)
2.4 Solution 4 (Abusing Stewart)
2.5 Solution 5 (Complex Numbers)
2.6 Solution 6
2.7 Solution 7 (Using a Circle)
2.8 Solution 8 (More Similar Triangles)
2.9 Solution 9
2.10 Solution 10(Similar Triangles)
3 See also | 2005 AIME II Problem 12 | Solution 1 (trigonometry)
Let $G$ be the foot of the perpendicular from $O$ to $AB$. Denote $x = EG$ and $y = FG$, and $x > y$ (since $AE < BF$ and $AG = BG$). Then $\tan \angle EOG = \frac{x}{450}$, and $\tan \angle FOG = \frac{y}{450}$.
By the tangent addition rule $\left( \tan (a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b} \right)$, we see that \[\tan 45 = \tan (EOG + FOG) = \frac{\frac{x}{450} + \frac{y}{450}}{1 - \frac{x}{450} \cdot \frac{y}{450}}.\] Since $\tan 45 = 1$, this simplifies to $1 - \frac{xy}{450^2} = \frac{x + y}{450}$. We know that $x + y = 400$, so we can substitute this to find that $1 - \frac{xy}{450^2} = \frac 89 \Longrightarrow xy = 150^2$.
Substituting $x = 400 - y$ again, we know have $xy = (400 - y)y = 150^2$. This is a quadratic with roots $200 \pm 50\sqrt{7}$. Since $y < x$, use the smaller root, $200 - 50\sqrt{7}$.
Now, $BF = BG - FG = 450 - (200 - 50\sqrt{7}) = 250 + 50\sqrt{7}$. The answer is $250 + 50 + 7 = \boxed{307}$.
Solution 2 (synthetic)
Label $BF=x$, so $EA =$ $500 - x$. Rotate $\triangle{OEF}$ about $O$ until $EF$ lies on $BC$. Now we know that $\angle{EOF}=45^\circ$ therefore $\angle BOF+\angle AOE=45^\circ$ also since $O$ is the center of the square. Label the new triangle that we created $\triangle OGJ$. Now we know that rotation preserves angles and side lengths, so $BG=500-x$ and $JC=x$. Draw $GF$ and $OB$. Notice that $\angle BOG =\angle OAE$ since rotations preserve the same angles so
$\angle{FOG}=45^\circ$ too. By SAS we know that $\triangle FOE\cong \triangle FOG,$ so $FG=400$. Now we have a right $\triangle BFG$ with legs $x$ and $500-x$ and hypotenuse $400$. By the Pythagorean Theorem,
\begin{align*} (500-x)^2+x^2&=400^2 \\ 250000-1000x+2x^2&=16000 \\ 90000-1000x+2x^2&=0 \end{align*}
and applying the quadratic formula we get that
$x=250\pm 50\sqrt{7}$. Since $BF > AE,$ we take the positive root, and our answer is $p+q+r = 250 + 50 + 7 = 307$.
Solution 3 (similar triangles)
Let the midpoint of $\overline{AB}$ be $M$ and let $FB = x$, so then $MF = 450 - x$ and $AF = 900 - x$. Drawing $\overline{AO}$, we have $\triangle OEF\sim\triangle AOF$, so
\[\frac{OF}{EF} = \frac{AF}{OF}\Rightarrow (OF)^2 = 400(900 - x).\]
By the Pythagorean Theorem on $\triangle OMF$,
\[(OF)^2 = 450^2 + (450 - x)^2.\]
Setting these two expressions for $(OF)^2$ equal and solving for $x$ (it is helpful to scale the problem down by a factor of 50 first), we get $x = 250\pm 50\sqrt{7}$. Since $BF > AE$, we want the value $x = 250 + 50\sqrt{7}$, and the answer is $250 + 50 + 7 = \boxed{307}$.
Solution 4 (Abusing Stewart)
Let $x = BF$, so $AE = 500-x$. Let $a = OE$, $b = OF$. Applying Stewart's Theorem on triangles $AOB$ twice, first using $E$ as the base point and then $F$, we arrive at the equations \[(450 \sqrt{2})^2 (900) = 900(500-x)(400+x) + a^2 (900)\] and \[(450 \sqrt{2})^2 (900) = 900x(900-x) + b^2 (900)\] Now applying law of sines and law of cosines on $\triangle EOF$ yields \[\frac{1}{2} ab \sin 45^{\circ} = \frac{4}{9} \times \frac{1}{4} \times 900^2 = 202500\] and \[a^2+b^2- 2 ab \cos 45^{\circ} = 160000\] Solving for $ab$ from the sines equation and plugging into the law of cosines equation yields $a^2+b^2 = 290000$. We now finish by adding the two original stewart equations and obtaining: \[2(450\sqrt{2})^2 = (500-x)(400+x)+x(900-x)+520000\] This is a quadratic which only takes some patience to solve for $x = 250 + 50\sqrt{7}$
Solution 5 (Complex Numbers)
Let lower case letters be the complex numbers correspond to their respective upper case points in the complex plane, with $o = 0, a = -450 + 450i, b = 450 + 450i$, and $f = x + 450i$. Since $EF$ = 400, $e = (x-400) + 450i$. From $\angle{EOF} = 45^{\circ}$, we can deduce that the rotation of point $F$ 45 degrees counterclockwise, $E$, and the origin are collinear. In other words, \[\dfrac{e^{i \frac{\pi}{4}} \cdot (x + 450i)}{(x - 400) + 450i}\] is a real number. Simplyfying using the fact that $e^{i \frac{\pi}{4}} = \dfrac{\sqrt{2}}{2} + i \dfrac{\sqrt{2}}{2}$, clearing the denominator, and setting the imaginary part equal to $0$, we eventually get the quadratic \[x^2 - 400x + 22500 = 0\] which has solutions $x = 200 \pm 50\sqrt{7}$. It is given that $AE < BF$, so $x = 200 - 50\sqrt{7}$ and \[BF = 450 - (200 - 50\sqrt{7}) = 250 + 50\sqrt{7} \Rightarrow \boxed{307}.\]
-MP8148
Solution 6
Let G be a point such that it lies on AB, and GOE is 90 degrees. Let H be foot of the altitude from O to AB.
Since $\triangle GOE \sim \triangle OHE$, $\frac{GO}{OE} = \frac{450}{x}$, and by Angle Bisector Theorem, $\frac{GF}{FE} = \frac{450}{x}$. Thus, $GF = \frac{450 \cdot 400}{x}$. $AF = AH-FH = 50+x$, and $KA = EB$ (90 degree rotation), and now we can bash on 2 similar triangles $\triangle GAK \sim \triangle GHO$.
\[\frac{GA}{AK} = \frac{GH}{OH}\]
\[\frac{\frac{450 \cdot 400}{x}-50-x}{450-x} = \frac{\frac{450 \cdot 400}{x}+400-x}{450}\]
I hope you like expanding
\[x^2 - 850x + \frac{81000000}{x} = -450x - 22500 + \frac{81000000}{x}\]
\[x^2 - 400x + 22500 = 0\]
Quadratic formula gives us
\[x = 200 \pm 50 \sqrt{7}\]
Since AE < BF
\[x = 200 - 50 \sqrt{7}\]
Thus,
\[BF = 250 + 50 \sqrt{7}\]
So, our answer is $\boxed{307}$.
-AlexLikeMath
Solution 7 (Using a Circle)
We know that G is on the perpendicular bisector of $EF$, which means that $EJ=JF=200$, $EG=GF=200\sqrt{2}$ and $GH=250$. Now, let $HO$ be equal to $x$. We can set up an equation with the Pythagorean Theorem:
\begin{align*} \sqrt{x^2+250^2}&=(200\sqrt{2})^2 \\ x^2+62500&=80000 \\ x^2&=17500 \\ x&=50\sqrt{7} \end{align*}
Now, since $IO=450$,
\begin{align*} HI&=450-x \\ &=450-50\sqrt{7} \\ \end{align*} \\
Since $HI=AJ$, we now have:
\begin{align*} BF&=AB-AJ-JF \\ &=900-(450-50\sqrt{7})-200 \\ &=250+50\sqrt{7} \\ \end{align*}
This means that our answer would be $250+50+7=\boxed{307}$
~Jerry_Guo
Solution 8 (More Similar Triangles)
Construct $BO, AO.$ Let $\angle{FOB} = \alpha.$ Also let $FB = x$ then $AE = 500-x.$ We then have from simple angle-chasing:
\begin{align*} \angle{BFO} = 135 - \alpha \\ \angle{OFE} = 45 + \alpha \\ \angle{EOA} = 45 - \alpha \\ \angle{AEO} = 90 + \alpha \\ \angle{OEF} = 90 - \alpha. \end{align*}
From AA similarity we have \[\triangle{EOB} \sim \triangle{EFO}.\] This gives the ratios,
\[\dfrac{400 + x}{EO} = \dfrac{450\sqrt{2}}{FO}.\]
Similarly from AA similarity \[\triangle{FOA} \sim \triangle{FEO}.\] So we get the ratios \[\dfrac{EO}{450\sqrt{2}} = \dfrac{FO}{900-x}.\] We can multiply to get
\[\dfrac{400 + x}{450\sqrt{2}} = \dfrac{450\sqrt{2}}{900 - x}.\] Cross-multiplying reveals
\[360000 + 500x - x^2 = 405000.\] Bringing everything to one side we have \[x^2 - 500x + 45000 = 0.\] By the quadratic formula we get \[x = \dfrac{500 + \sqrt{500^2 - 4\cdot45000}}{2} = \dfrac{500 + \sqrt{70000}}{2} = \dfrac{500 + 100\sqrt{7}}{2} = 250 + 50\sqrt{7}.\]
Therefore
\[p+q+r = 250 + 50 + 7 = \boxed{307}.\]
~aa1024
Solution 9
We use ratio lemma and Stewart's theorem:
Connect $OA, OE, OF, OB$ and let $AE = x$ and $BF = 500 - x.$
Let angle $AOE = y,$ hence $BOF = 45 - y.$
Now, we apply Stewart's theorem in triangles $AOF$ and $BOE$ to get $OE$ and $OF$ in terms of $x$
finally, calculate $x/400$ and $500-x/400$ using ratio lemma to find $x$ and $y$
Solution 10(Similar Triangles)
Draw AO, OB, and extend OB to D. Let $\angle{FOB} = \alpha.$ Then, after angle chasing, we find that \[\angle{AEB} = 90 + \alpha\].
Using this, we draw a line perpendicular to $AB$ at $E$ to meet $BD$ at $M$. Since $\angle{MEO} = \alpha$ and $\angle{EMO} = 45$, we have that \[\triangle{EMO} \sim \triangle{OBF}\]
Let $FB = x$. Then $EM = 400+x$. Since $FB/BO = \frac{x}{450\sqrt{2}}$, and $MO/EM = FB/OB$, we have \[MO = \frac{(400+x)x}{450\sqrt{2}}\]
Since $\triangle{EBM}$ is a $45-45-90$ triangle, \[(400+x)\sqrt{2} = 450 \sqrt{2} + \frac{(400+x)x}{450\sqrt{2}}\]
Solving for $x$, we get that $x=250 +- 50s\sqrt{7}$, but since $FB>AE$, $FB = 250+50\sqrt{7}$, thus \[p+q+r=\boxed{307}\]
-dchang0524 | // Block 1
size(200); defaultpen(linewidth(0.7)+fontsize(10)); pair A=(0,9), B=(9,9), C=(9,0), D=(0,0), E=(2.5-0.5*sqrt(7),9), F=(6.5-0.5*sqrt(7),9), G=(4.5,9), O=(4.5,4.5); draw(A--B--C--D--A);draw(E--O--F);draw(G--O); dot(A^^B^^C^^D^^E^^F^^G^^O); label("\(A\)",A,(-1,1));label("\(B\)",B,(1,1));label("\(C\)",C,(1,-1));label("\(D\)",D,(-1,-1)); label("\(E\)",E,(0,1));label("\(F\)",F,(1,1));label("\(G\)",G,(-1,1));label("\(O\)",O,(1,-1)); label("\(x\)",E/2+G/2,(0,1));label("\(y\)",G/2+F/2,(0,1)); label("\(450\)",(O+G)/2,(-1,1));
// Block 2
size(200); defaultpen(linewidth(0.7)+fontsize(10)); pair A=(0,9), B=(9,9), C=(9,0), D=(0,0), E=(2.5-0.5*sqrt(7),9), F=(6.5-0.5*sqrt(7),9), O=(4.5,4.5), G=O+(E-O)*dir(-90), J=O+(F-O)*dir(-90); draw(A--B--C--D--A);draw(E--O--F);draw(G--O--J);draw(F--G,linetype("4 4")); dot(A^^B^^C^^D^^E^^F^^G^^J^^O); label("\(A\)",A,(-1,1));label("\(B\)",B,(1,1));label("\(C\)",C,(1,-1));label("\(D\)",D,(-1,-1)); label("\(E\)",E,(0,1));label("\(F\)",F,(1,1));label("\(G\)",G,(1,0));label("\(J\)",J,(1,0));label("\(O\)",O,(1,-1)); label("\(x\)",(B+F)/2,(0,1)); label("\(400\)",(E+F)/2,(0,1)); label("\(900\)",(C+D)/2,(0,-1));
// Block 3
size(3inch); pair A, B, C, D, M, O, X, Y; A = (0,900); B = (900,900); C = (900,0); D = (0,0); M = (450,900); O = (450,450); X = (250 - 50*sqrt(7),900); Y = (650 - 50*sqrt(7),900); draw(A--B--C--D--cycle); draw(X--O--Y); draw(M--O--A); label("$A$",A,NW); label("$B$",B,NE); label("$C$",C,SE); label("$D$",D,SW); label("$E$",X,N); label("$F$",Y,NNE); label("$O$",O,S); label("$M$",M,N);
// Block 4
size(250); pair A,B,C,D,O,E,F,G,H,K; A = (0,0); B = (900,0); C = (900,900); D = (0,900); O = (450,450); E = (600,0); F = (150,0); G = (-600,0); H = (450,0); K = (0,270); draw(A--B--C--D--cycle); draw(O--E); draw(O--F); draw(O--G); draw(A--G); draw(O--H); label("O",O,N); label("A",A,S); label("B",B,SE); label("C",C,NE); label("D",D,NW); label("E",E,SE); label("F",F,S); label("H",H,SW); label("G",G,SW); label("x",H--E,S); label("K",K,NW);
// Block 5
size(200); defaultpen(linewidth(0.7)+fontsize(10)); pair A=(0,9), B=(9,9), C=(9,0), D=(0,0), E=(2.5-0.5*sqrt(7),9), F=(6.5-0.5*sqrt(7),9), G=(4.5-0.5*sqrt(7),7), O=(4.5,4.5), H=(4.5-0.5*sqrt(7),4.5), I=(0,4.5), J=(4.5-0.5*sqrt(7),9); draw(A--B--C--D--A); draw(E--O--F); draw(J--G); draw(E--G--F); draw(G--H--O--G); draw(I--H); draw(circle(G,2*sqrt(2))); markscalefactor=0.05; draw(rightanglemark(E,G,F)); dot(A^^B^^C^^D^^E^^F^^G^^H^^I^^J^^O); label("\(A\)",A,(-1,1)); label("\(B\)",B,(1,1)); label("\(C\)",C,(1,-1)); label("\(D\)",D,(-1,-1)); label("\(E\)",E,(0,1)); label("\(F\)",F,(1,1)); label("\(G\)",G,(1,0)); label("\(H\)",H,(-1,1)); label("\(I\)",I,(-1,0)); label("\(J\)",J,(0,1)); label("\(O\)",O,(1,-1)); | [] |
127 | In triangle $ABC, AB=13, BC=15,$ and $CA = 14.$ Point $D$ is on $\overline{BC}$ with $CD=6.$ Point $E$ is on $\overline{BC}$ such that $\angle BAE\cong \angle CAD.$ Given that $BE=\frac pq$ where $p$ and $q$ are relatively prime positive integers, find $q.$ | 2005 AIME II Problem 14 | By the Law of Sines and since $\angle BAE = \angle CAD, \angle BAD = \angle CAE$, we have
\begin{align*} \frac{CD \cdot CE}{AC^2} &= \frac{\sin CAD}{\sin ADC} \cdot \frac{\sin CAE}{\sin AEC} \\ &= \frac{\sin BAE \sin BAD}{\sin ADB \sin AEB} \\ &= \frac{\sin BAE}{\sin AEB} \cdot \frac{\sin BAD}{\sin ADB}\\ &= \frac{BE \cdot BD}{AB^2} \end{align*}
Substituting our knowns, we have $\frac{CE}{BE} = \frac{3 \cdot 14^2}{2 \cdot 13^2} = \frac{BC - BE}{BE} = \frac{15}{BE} - 1 \Longrightarrow BE = \frac{13^2 \cdot 15}{463}$. The answer is $q = \boxed{463}$. | // Block 1
import olympiad; import cse5; import geometry; size(150);
defaultpen(fontsize(10pt));
defaultpen(0.8);
dotfactor = 4;
pair A = origin;
pair C = rotate(15,A)*(A+dir(-50));
pair B = rotate(15,A)*(A+dir(-130));
pair D = extension(A,A+dir(-68),B,C);
pair E = extension(A,A+dir(-82),B,C);
label("$A$",A,N);
label("$B$",B,SW);
label("$D$",D,SE);
label("$E$",E,S);
label("$C$",C,SE);
draw(A--B--C--cycle);
draw(A--E);
draw(A--D);
draw(anglemark(B,A,E,5));
draw(anglemark(D,A,C,5));
// Block 2
import olympiad; import cse5; import geometry; size(150); defaultpen(fontsize(10pt)); defaultpen(0.8); dotfactor = 4; pair A = origin; pair C = rotate(15,A)*(A+dir(-50)); pair B = rotate(15,A)*(A+dir(-130)); pair D = extension(A,A+dir(-68),B,C); pair E = extension(A,A+dir(-82),B,C); label("$A$",A,N); label("$B$",B,SW); label("$D$",D,SE); label("$E$",E,S); label("$C$",C,SE); draw(A--B--C--cycle); draw(A--E); draw(A--D); draw(anglemark(B,A,E,5)); draw(anglemark(D,A,C,5)); | [] |
127 | In triangle $ABC, AB=13, BC=15,$ and $CA = 14.$ Point $D$ is on $\overline{BC}$ with $CD=6.$ Point $E$ is on $\overline{BC}$ such that $\angle BAE\cong \angle CAD.$ Given that $BE=\frac pq$ where $p$ and $q$ are relatively prime positive integers, find $q.$ | 2005 AIME II Problem 14 | Note: We first recall some helpful tips regarding 13, 14, 15 triangles. Drawing an altitude H from B to AC results in AHB being a 5-12-13 right triangle and CHB being a 3-4-5 (9-12-15) right triangle.
Now we start by drawing altitudes from D and E onto AC, labeling them as F and G, and labelling $\angle DAG = \alpha$. Now we know that $\overline{DF} = \frac{24}{5}$ and $\overline{FC} = \frac{18}{5}$. Therefore, $\overline{AF} = \frac{52}{5}$, so $\tan{(\alpha)} = \frac{6}{13}$. Our goal now is to use tangent $\angle EAG$ in triangle $AEG$. We set $\overline{BE}$ to $x$, so $\overline{ED} = 9 - x$ and $\overline{EC} = 15 - x$, so $\overline{EG} = \frac{4}{5}(15-x)$ and $\overline{GC} = \frac{3}{5}(15-x)$ so $\overline{AG} = \frac{3x+25}{5}$. Now we just need tangent of $\angle EAG$.
We find this using $\tan{(EAG)} = \tan{(A - \alpha)} = \frac{\tan{A} - \tan{\alpha}}{1 + \tan{A}\tan{\alpha}}$, which is $\frac{\frac{12}{5} - \frac{6}{13}}{1 + \frac{12}{5} \cdot \frac{6}{13}}$ or $\frac{126}{137}$. Now we solve the equation $\tan{\angle EAG} = \frac{126}{137} = \frac{\frac{60-4x}{5}}{\frac{3x+25}{5}}$, so $x = \frac{2535}{463}$ | // Block 1
import olympiad; import cse5; import geometry; size(300);
defaultpen(fontsize(10pt));
defaultpen(0.8);
dotfactor = 4;
pair A = origin;
pair C = rotate(15,A)*(A+dir(-50));
pair B = rotate(15,A)*(A+dir(-130));
pair D = extension(A,A+dir(-68),B,C);
pair E = extension(A,A+dir(-82),B,C);
label("$A$",A,N);
label("$B$",B,SW);
label("$D$",D,SE);
label("$E$",E,S);
label("$C$",C,SE);
draw(A--B--C--cycle);
draw(A--E);
draw(A--D);
draw(anglemark(B,A,E,5));
draw(anglemark(D,A,C,5));
pair G = foot(E,C,A);
pair F = foot(D,C,A);
draw(D--F);
draw(E--G);
label("$G$",G,N);
label("$F$",F,N);
// Block 2
import olympiad; import cse5; import geometry; size(300); defaultpen(fontsize(10pt)); defaultpen(0.8); dotfactor = 4; pair A = origin; pair C = rotate(15,A)*(A+dir(-50)); pair B = rotate(15,A)*(A+dir(-130)); pair D = extension(A,A+dir(-68),B,C); pair E = extension(A,A+dir(-82),B,C); label("$A$",A,N); label("$B$",B,SW); label("$D$",D,SE); label("$E$",E,S); label("$C$",C,SE); draw(A--B--C--cycle); draw(A--E); draw(A--D); draw(anglemark(B,A,E,5)); draw(anglemark(D,A,C,5)); pair G = foot(E,C,A); pair F = foot(D,C,A); draw(D--F); draw(E--G); label("$G$",G,N); label("$F$",F,N); | [] |
127 | In triangle $ABC, AB=13, BC=15,$ and $CA = 14.$ Point $D$ is on $\overline{BC}$ with $CD=6.$ Point $E$ is on $\overline{BC}$ such that $\angle BAE\cong \angle CAD.$ Given that $BE=\frac pq$ where $p$ and $q$ are relatively prime positive integers, find $q.$ | 2005 AIME II Problem 14 | Diagram borrowed from Solution 1.
Applying Law of Cosines on $\bigtriangleup ABC$ with respect to $\angle C$ we have
\[AB^2=AC^2+BC^2-2(AC)(BC)\cos C\]
Solving gets $\cos C=\frac{3}{5}$, which implies that
\[\sin C=\sqrt{1-\cos C}=\frac{4}{5}\]
Applying Stewart's Theorem with cevian $AD$ we have
\[(BC)(AD)^2+(BC)(BD)(CD)=(CD)(AB)^2+(BD)(AC)^2\]
Solving gets $AD=\frac{4\sqrt{205}}{5}$.
Applying Law of Sines on $\bigtriangleup ACD$ to solve for $\sin CAD$ we have
\[\frac{AD}{\sin C}=\frac{CD}{\sin CAD}\]
Solving gets $\sin CAD=\frac{6\sqrt{205}}{205}$. Thus $\sin BAE=\sin CAD=\frac{6\sqrt{205}}{205}$.
Applying Law of Sines on $\bigtriangleup ABC$ we have
\[\frac{AC}{\sin B}=\frac{AB}{\sin C}\]
Solving gets $\sin B=\frac{56}{65}$.
Applying Stewart's Theorem with cevian $AE$ we have
\[(BC)(AE)^2+(BC)(BE)(CE)=(CE)(AB)^2+(BE)(AC)^2\]
\[(BC)(AE)^2+(BC)(BE)(BC-BE)=(BC-BE)(AB)^2+(BE)(AC)^2\]
Solving gets $AE=\sqrt{\frac{15BE^2-198BE+2535}{15}}$
Finally, applying Law of Sines on $\bigtriangleup BAE$ we have
\[\frac{AE}{\sin B}=\frac{BE}{\sin BAE}\]
\[\frac{\sqrt{\frac{15BE^2-198BE+2535}{15}}}{\frac{56}{65}}=\frac{BE}{\frac{6\sqrt{205}}{205}}\]
\[7605BE^2-32342BE+2535=0\]
Solving the easy quadratic equation gets $BE=\frac{1632}{463}\Longrightarrow q=\boxed{463}$
~ Nafer | // Block 1
import olympiad; import cse5; import geometry; size(150);
defaultpen(fontsize(10pt));
defaultpen(0.8);
dotfactor = 4;
pair A = origin;
pair C = rotate(15,A)*(A+dir(-50));
pair B = rotate(15,A)*(A+dir(-130));
pair D = extension(A,A+dir(-68),B,C);
pair E = extension(A,A+dir(-82),B,C);
label("$A$",A,N);
label("$B$",B,SW);
label("$D$",D,SE);
label("$E$",E,S);
label("$C$",C,SE);
draw(A--B--C--cycle);
draw(A--E);
draw(A--D);
draw(anglemark(B,A,E,5));
draw(anglemark(D,A,C,5));
// Block 2
import olympiad; import cse5; import geometry; size(150); defaultpen(fontsize(10pt)); defaultpen(0.8); dotfactor = 4; pair A = origin; pair C = rotate(15,A)*(A+dir(-50)); pair B = rotate(15,A)*(A+dir(-130)); pair D = extension(A,A+dir(-68),B,C); pair E = extension(A,A+dir(-82),B,C); label("$A$",A,N); label("$B$",B,SW); label("$D$",D,SE); label("$E$",E,S); label("$C$",C,SE); draw(A--B--C--cycle); draw(A--E); draw(A--D); draw(anglemark(B,A,E,5)); draw(anglemark(D,A,C,5)); | [] |
127 | In triangle $ABC, AB=13, BC=15,$ and $CA = 14.$ Point $D$ is on $\overline{BC}$ with $CD=6.$ Point $E$ is on $\overline{BC}$ such that $\angle BAE\cong \angle CAD.$ Given that $BE=\frac pq$ where $p$ and $q$ are relatively prime positive integers, find $q.$ | 2005 AIME II Problem 14 | We begin by drawing a line through point $B$ parallel to side $AC$. Extend Lines $AE$ and $AD$ to meet the new parallel line at points $F$ and $G$, respectively. This will help us create a lot of similar triangles.
The first pair of similar triangles that is easy to spot is triangle $ACD$ and triangle $GBD$ by AA similarity ($\angle{ADC}=\angle{GDB}$ by vertical angles, and $\angle{DAC}=\angle{DGB}$ by alternate interior angles). With the ratios $\frac{AC}{BG}=\frac{CD}{BD}$ $\rightarrow$ $\frac{14}{BG}=\frac{6}{9}$ $\rightarrow$ $BG=21$.
The next pair of similar triangles is triangle $BAG$ and triangle $BFA$ by AA similarity ($\angle{CAD}=\angle{BAF}$ as given, and $\angle{CAD}=\angle{BGA}$ by alternate interior angles, so by transitive property, $\angle{BAF}=\angle{BGA}$. Additionaly, $\angle{ABF}=\angle{GBA}$ as they are the same angle.). With the ratios $\frac{AB}{BG}=\frac{BF}{AB}$ $\rightarrow$ $\frac{13}{21}=\frac{BF}{13}$ $\rightarrow$ $BF=\frac{169}{21}$.
The last pair of similar triangles we need is triangle $BEF$ and triangle $CAE$ by AA similarity (similar by vertical angles and alternate interior angles). With the ratios $\frac{AC}{BF}=\frac{CE}{BE}$ $\rightarrow$ $\frac{14}{\frac{169}{21}}=\frac{15-x}{x}$ $\rightarrow$ $x=\frac{169*15}{463}$. This cannot be simplified further, and the problem asks for the denominator, so the answer is $\boxed{463}$.
~ChaitraliKA | // Block 1
// 13–14–15 triangle with AB = 13, AC = 14, BC = 15;
// BF→G extended; “9-x” hugging ED and nudged right; “6” nudged up;
// B and C inverted (B at top, C on right); “13” label shifted left.
import olympiad;
size(350);
pair A = (0,0);
pair B = (5,12); // now AB = 13 (5-0,12-0)
pair C = (14,0); // now AC = 14, BC = 15
real L = 15, DC = 6, x = 4;
pair E = B + (x/L)*(C - B);
pair D = B + ((L-DC)/L)*(C - B);
pair BB = B + (C - A);
pair F = extension(A, E, B, BB);
pair G = extension(A, D, B, BB);
// Draw
draw(A--B--C--cycle);
draw(B--BB--G, Arrows(5));
draw(A--E, red+1bp);
draw(E--F, red+1bp+dashed);
draw(A--D, red+1bp);
draw(D--G, red+1bp+dashed);
// Dots
dot(A); dot(B); dot(C);
dot(D); dot(E);
dot(F); dot(G);
// Labels
label("$A$", A, SW);
label("$B$", B, N);
label("$C$", C, S);
label("$E$", E, NE);
label("$D$", D, NE);
label("$F$", F, N);
label("$G$", G, NE);
// Side‐length labels
label("$13$", midpoint(A--B)+(-0.5,0), N); // AB = 13, shifted left
label("$14$", midpoint(A--C), S); // AC = 14 below
// ("15" label removed)
// Segment labels
label("$x$", midpoint(B--E), W);
label("$9-x$", midpoint(E--D)+(0.9,0), dir(90)); // hugging ED, shifted right
label("$6$", midpoint(D--C)+(0,0.3), NE); // "6" slightly up
// Block 2
// 13–14–15 triangle with AB = 13, AC = 14, BC = 15; // BF→G extended; “9-x” hugging ED and nudged right; “6” nudged up; // B and C inverted (B at top, C on right); “13” label shifted left. import olympiad; size(350); pair A = (0,0); pair B = (5,12); // now AB = 13 (5-0,12-0) pair C = (14,0); // now AC = 14, BC = 15 real L = 15, DC = 6, x = 4; pair E = B + (x/L)*(C - B); pair D = B + ((L-DC)/L)*(C - B); pair BB = B + (C - A); pair F = extension(A, E, B, BB); pair G = extension(A, D, B, BB); // Draw draw(A--B--C--cycle); draw(B--BB--G, Arrows(5)); draw(A--E, red+1bp); draw(E--F, red+1bp+dashed); draw(A--D, red+1bp); draw(D--G, red+1bp+dashed); // Dots dot(A); dot(B); dot(C); dot(D); dot(E); dot(F); dot(G); // Labels label("$A$", A, SW); label("$B$", B, N); label("$C$", C, S); label("$E$", E, NE); label("$D$", D, NE); label("$F$", F, N); label("$G$", G, NE); // Side‐length labels label("$13$", midpoint(A--B)+(-0.5,0), N); // AB = 13, shifted left label("$14$", midpoint(A--C), S); // AC = 14 below // ("15" label removed) // Segment labels label("$x$", midpoint(B--E), W); label("$9-x$", midpoint(E--D)+(0.9,0), dir(90)); // hugging ED, shifted right label("$6$", midpoint(D--C)+(0,0.3), NE); // "6" slightly up | [] |
128 | Let $w_1$ and $w_2$ denote the circles $x^2+y^2+10x-24y-87=0$ and $x^2 +y^2-10x-24y+153=0,$ respectively. Let $m$ be the smallest positive value of $a$ for which the line $y=ax$ contains the center of a circle that is externally tangent to $w_2$ and internally tangent to $w_1.$ Given that $m^2=\frac pq,$ where $p$ and $q$ are relatively prime integers, find $p+q.$ | 2005 AIME II Problem 15 | Rewrite the given equations as $(x+5)^2 + (y-12)^2 = 256$ and $(x-5)^2 + (y-12)^2 = 16$.
Let $w_3$ have center $(x,y)$ and radius $r$. Now, if two circles with radii $r_1$ and $r_2$ are externally tangent, then the distance between their centers is $r_1 + r_2$, and if they are internally tangent, it is $|r_1 - r_2|$. So we have
\begin{align*} r + 4 &= \sqrt{(x-5)^2 + (y-12)^2} \\ 16 - r &= \sqrt{(x+5)^2 + (y-12)^2} \end{align*}
Solving for $r$ in both equations and setting them equal, then simplifying, yields
\begin{align*} 20 - \sqrt{(x+5)^2 + (y-12)^2} &= \sqrt{(x-5)^2 + (y-12)^2} \\ 20+x &= 2\sqrt{(x+5)^2 + (y-12)^2} \end{align*}
Squaring again and canceling yields $1 = \frac{x^2}{100} + \frac{(y-12)^2}{75}.$
So the locus of points that can be the center of the circle with the desired properties is an ellipse.
Since the center lies on the line $y = ax$, we substitute for $y$ and expand:
\[1 = \frac{x^2}{100} + \frac{(ax-12)^2}{75} \Longrightarrow (3+4a^2)x^2 - 96ax + 276 = 0.\]
We want the value of $a$ that makes the line $y=ax$ tangent to the ellipse, which will mean that for that choice of $a$ there is only one solution to the most recent equation. But a quadratic has one solution iff its discriminant is $0$, so $(-96a)^2 - 4(3+4a^2)(276) = 0$.
Solving yields $a^2 = \frac{69}{100}$, so the answer is $\boxed{169}$. | // Block 1
size(220); pointpen = black; pen d = linewidth(0.7); pathpen = d;
pair A = (-5, 12), B = (5, 12), C = (0, 0);
D(CR(A,16));D(CR(B,4));D(shift((0,12)) * yscale(3^.5 / 2) * CR(C,10), linetype("2 2") + d + red);
D((0,30)--(0,-10),Arrows(4));D((15,0)--(-25,0),Arrows(4));D((0,0)--MP("y=ax",(14,14 * (69/100)^.5),E),EndArrow(4));
void bluecirc (real x) {
pair P = (x, (3 * (25 - x^2 / 4))^.5 + 12); dot(P, blue);
D(CR(P, ((P.x - 5)^2 + (P.y - 12)^2)^.5 - 4) , blue + d + linetype("4 4"));
}
bluecirc(-9.2); bluecirc(-4); bluecirc(3);
// Block 2
size(220); pointpen = black; pen d = linewidth(0.7); pathpen = d; pair A = (-5, 12), B = (5, 12), C = (0, 0); D(CR(A,16));D(CR(B,4));D(shift((0,12)) * yscale(3^.5 / 2) * CR(C,10), linetype("2 2") + d + red); D((0,30)--(0,-10),Arrows(4));D((15,0)--(-25,0),Arrows(4));D((0,0)--MP("y=ax",(14,14 * (69/100)^.5),E),EndArrow(4)); void bluecirc (real x) { pair P = (x, (3 * (25 - x^2 / 4))^.5 + 12); dot(P, blue); D(CR(P, ((P.x - 5)^2 + (P.y - 12)^2)^.5 - 4) , blue + d + linetype("4 4")); } bluecirc(-9.2); bluecirc(-4); bluecirc(3); | [] |
128 | Let $w_1$ and $w_2$ denote the circles $x^2+y^2+10x-24y-87=0$ and $x^2 +y^2-10x-24y+153=0,$ respectively. Let $m$ be the smallest positive value of $a$ for which the line $y=ax$ contains the center of a circle that is externally tangent to $w_2$ and internally tangent to $w_1.$ Given that $m^2=\frac pq,$ where $p$ and $q$ are relatively prime integers, find $p+q.$ | 2005 AIME II Problem 15 | As above, we rewrite the equations as $(x+5)^2 + (y-12)^2 = 256$ and $(x-5)^2 + (y-12)^2 = 16$. Let $F_1=(-5,12)$ and $F_2=(5,12)$. If a circle with center $C=(a,b)$ and radius $r$ is externally tangent to $w_2$ and internally tangent to $w_1$, then $CF_1=16-r$ and $CF_2=4+r$. Therefore, $CF_1+CF_2=20$. In particular, the locus of points $C$ that can be centers of circles must be an ellipse with foci $F_1$ and $F_2$ and major axis $20$.
Clearly, the minimum value of the slope $a$ will occur when the line $y=ax$ is tangent to this ellipse. Suppose that this point of tangency is denoted by $T$, and the line $y=ax$ is denoted by $\ell$. Then we reflect the ellipse over $\ell$ to a new ellipse with foci $F_1'$ and $F_2'$ as shown below.
By the reflection property of ellipses (i.e., the angle of incidence to a tangent line is equal to the angle of reflection for any path that travels between the foci), we know that $F_1$, $T$, and $F_2'$ are collinear, and similarly, $F_2$, $T$ and $F_1'$ are collinear. Therefore, $OF_1F_2F_2'F_1'$ is a pentagon with $OF_1=OF_2=OF_1'=OF_2'=13$, $F_1F_2=F_1'F_2'=10$, and $F_1F_2'=F_1'F_2=20$. Note that $\ell$ bisects $\angle F_1'OF_1$. We can bisect this angle by bisecting $\angle F_1'OF_2$ and $F_2OF_1$ separately.
We proceed using complex numbers. Triangle $F_2OF_1'$ is isosceles with side lengths $13,13,20$. The height of this from the base of $20$ is $\sqrt{69}$. Therefore, the complex number $\sqrt{69}+10i$ represents the bisection of $\angle F_1'OF_2$.
Similarly, using the 5-12-13 triangles, we easily see that $12+5i$ represents the bisection of the angle $F_2OF_1$. Therefore, we can add these two angles together by multiplying the complex numbers, finding
\[\text{arg}\left((\sqrt{69}+10i)(12+5i)\right)=\frac{1}{2}\angle F_1'OF_1.\]
Now the point $F_1$ is given by the complex number $-5+12i$. Therefore, to find a point on line $\ell$, we simply subtract $\frac{1}{2}\angle F_1'OF_1$, which is the same as multiplying $-5+12i$ by the conjugate of $(\sqrt{69}+10i)(12+5i)$. We find
\[(-5+12i)(\sqrt{69}-10i)(12-5i)=169(10+i\sqrt{69}).\]
In particular, note that the tangent of the argument of this complex number is $\sqrt{69}/10$, which must be the slope of the tangent line. Hence $a^2=69/100$, and the answer is $\boxed{169}$. | // Block 1
size(220);
pair F1 = (-5, 12), F2 = (5, 12),C=(0,12);
draw(circle(F1,16));
draw(circle(F2,4));
draw(ellipse(C,10,5*sqrt(3)));
xaxis("$x$",Arrows);
yaxis("$y$",Arrows);
dot(F1^^F2^^C);
real l(real x) {return sqrt(69)*x/10;}
path g=graph(l,-7,14);
draw(g);
draw(reflect((0,0),(10,l(10)))*ellipse(C,10,5*sqrt(3)));
pair T=intersectionpoint(ellipse(C,10,5*sqrt(3)),(0,0)--(10,l(10)));
dot(T);
pair F1P=reflect((0,0),(10,l(10)))*F1;
pair F2P=reflect((0,0),(10,l(10)))*F2;
dot(F1P^^F2P);
dot((0,0));
label("$F_1$",F1,N,fontsize(9));
label("$F_2$",F2,N,fontsize(9));
label("$F_1'$",F1P,SE,fontsize(9));
label("$F_2'$",F2P,SE,fontsize(9));
label("$O$",(0,0),NW,fontsize(9));
label("$\ell$",(13,l(13)),SE,fontsize(9));
label("$T$",T,NW,fontsize(9));
draw((0,0)--F1--F2--F2P--F1P--cycle);
draw(F1--F2P^^F2--F1P);
// Block 2
size(220); pair F1 = (-5, 12), F2 = (5, 12),C=(0,12); draw(circle(F1,16)); draw(circle(F2,4)); draw(ellipse(C,10,5*sqrt(3))); xaxis("$x$",Arrows); yaxis("$y$",Arrows); dot(F1^^F2^^C); real l(real x) {return sqrt(69)*x/10;} path g=graph(l,-7,14); draw(g); draw(reflect((0,0),(10,l(10)))*ellipse(C,10,5*sqrt(3))); pair T=intersectionpoint(ellipse(C,10,5*sqrt(3)),(0,0)--(10,l(10))); dot(T); pair F1P=reflect((0,0),(10,l(10)))*F1; pair F2P=reflect((0,0),(10,l(10)))*F2; dot(F1P^^F2P); dot((0,0)); label("$F_1$",F1,N,fontsize(9)); label("$F_2$",F2,N,fontsize(9)); label("$F_1'$",F1P,SE,fontsize(9)); label("$F_2'$",F2P,SE,fontsize(9)); label("$O$",(0,0),NW,fontsize(9)); label("$\ell$",(13,l(13)),SE,fontsize(9)); label("$T$",T,NW,fontsize(9)); draw((0,0)--F1--F2--F2P--F1P--cycle); draw(F1--F2P^^F2--F1P); | [] |
129 | In quadrilateral $ABCD$, $\angle B$ is a right angle, diagonal $\overline{AC}$ is perpendicular to $\overline{CD}$, $AB=18$, $BC=21$, and $CD=14$. Find the perimeter of $ABCD$. | 2006 AIME I Problem 1 | We construct the following diagram:
Using the Pythagorean Theorem, we get the following two equations:
\[AD^2 = AC^2 + CD^2\]
\[AC^2 = AB^2 + BC^2\]
Substituting $AB^2 + BC^2$ for $AC^2$ gives us $AD^2 = AB^2 + BC^2 + CD^2$. Plugging in the given information, we get $AD^2 = 18^2 + 21^2 + 14^2 = 961 \implies AD = 31$, so the perimeter is $AB+BC+CD+AD = 18+21+14+31 = \boxed{84}$. | // Block 1
pathpen = black;
pair C=(0,0),D=(0,-14),A=(-sqrt(765),0),B=IP(circle(C,21),circle(A,18));
D(MP("A",A,W)--MP("B",B,N)--MP("C",C,E)--MP("D",D,E)--A--C);
D(rightanglemark(A,C,D,40));
D(rightanglemark(A,B,C,40));
// Block 2
pathpen = black; pair C=(0,0),D=(0,-14),A=(-sqrt(765),0),B=IP(circle(C,21),circle(A,18)); D(MP("A",A,W)--MP("B",B,N)--MP("C",C,E)--MP("D",D,E)--A--C); D(rightanglemark(A,C,D,40)); D(rightanglemark(A,B,C,40)); | [] |
130 | Hexagon $ABCDEF$ is divided into five rhombuses, $\mathcal{P, Q, R, S,}$ and $\mathcal{T,}$ as shown. Rhombuses $\mathcal{P, Q, R,}$ and $\mathcal{S}$ are congruent, and each has area $\sqrt{2006}.$ Let $K$ be the area of rhombus $\mathcal{T}$. Given that $K$ is a positive integer, find the number of possible values for $K$. | 2006 AIME I Problem 8 | To determine the possible values of $[GIHJ],$ we must determine the maximum and minimum possible areas.
In the case where the $4$ rhombi are squares, we have $[GIHJ]=0,$ implying the minimum possible positive-integer-valued area is $1.$
Denote the length $HC=a$ and $KH=b.$ We have
\[KI=\sqrt{a^2-b^2}\]
by the Pythagorean Theorem, which implies
\[[IBCH]=a\sqrt{a^2-b^2}=\sqrt{2006}\]
and
\[[GIHJ]=2b\sqrt{a^2-b^2}.\]
The first equation yields
\[\sqrt{a^2-b^2}=\frac{\sqrt{2006}}{a}.\]
Plugging into the second, we have
\[[GIHJ]=2\sqrt{2006}\frac{b}{a}.\]
The maximal value of $\frac{b}{a}$ occurs when the height of $ABCDEF$ is minimized, which means
\[\frac{b}{a}\leq 1.\]
Plugging back up, we have
\[[GIHJ]\leq 2\sqrt{2006}=\sqrt{8024}.\]
We have
\[\lfloor \sqrt{8024} \rfloor = 89,\]
thus our answer is
\[89-1+1=\boxed{089}.\] | // Block 1
size(8cm);
pair A=(0,0), B=(4.2,0), C=(5.85,-1.6), D=(4.2,-3.2), EE=(0,-3.2), F=(-1.65,-1.6), G=(0.45,-1.6), H=(3.75,-1.6), I=(2.1,0), J=(2.1,-3.2), K=(2.1,-1.6);
draw(A--B--C--D--EE--F--cycle);
label("$A$",A,2*N);
label("$B$",B,2*N);
label("$C$",C,2*E);
label("$D$",D,2*S);
label("$E$",EE,2*S);
label("$F$",F,2*W);
label("$G$",(0.47,-1.55),NW);
label("$H$",(3.73,-1.55),NE);
label("$I$",I,2*N);
label("$J$",J,2*S);
label("$K$",K,2*SW);
draw(F--C);
draw(F--G--(2.1,0));
draw(C--H--(2.1,0));
draw(G--(2.1,-3.2));
draw(H--(2.1,-3.2));
draw((2.1,0)--(2.1,-3.2));
// Block 2
size(8cm); pair A=(0,0), B=(4.2,0), C=(5.85,-1.6), D=(4.2,-3.2), EE=(0,-3.2), F=(-1.65,-1.6), G=(0.45,-1.6), H=(3.75,-1.6), I=(2.1,0), J=(2.1,-3.2), K=(2.1,-1.6); draw(A--B--C--D--EE--F--cycle); label("$A$",A,2*N); label("$B$",B,2*N); label("$C$",C,2*E); label("$D$",D,2*S); label("$E$",EE,2*S); label("$F$",F,2*W); label("$G$",(0.47,-1.55),NW); label("$H$",(3.73,-1.55),NE); label("$I$",I,2*N); label("$J$",J,2*S); label("$K$",K,2*SW); draw(F--C); draw(F--G--(2.1,0)); draw(C--H--(2.1,0)); draw(G--(2.1,-3.2)); draw(H--(2.1,-3.2)); draw((2.1,0)--(2.1,-3.2)); | [] |
131 | Eight circles of diameter 1 are packed in the first quadrant of the coordinate plane as shown. Let region $\mathcal{R}$ be the union of the eight circular regions. Line $l,$ with slope 3, divides $\mathcal{R}$ into two regions of equal area. Line $l$'s equation can be expressed in the form $ax=by+c,$ where $a, b,$ and $c$ are positive integers whose greatest common divisor is 1. Find $a^2+b^2+c^2.$ | 2006 AIME I Problem 10 | Solution 1
The line passing through the tangency point of the bottom left circle and the one to its right and through the tangency of the top circle in the middle column and the one beneath it is the line we are looking for: a line passing through the tangency of two circles cuts congruent areas, so our line cuts through the four aforementioned circles splitting into congruent areas, and there are an additional two circles on each side. The line passes through $\left(1,\frac 12\right)$ and $\left(\frac 32,2\right)$, which can be easily solved to be $6x = 2y + 5$. Thus, $a^2 + b^2 + c^2 = \boxed{065}$.
Solution 2
Assume that if unit squares are drawn circumscribing the circles, then the line will divide the area of the concave hexagonal region of the squares equally (as of yet, there is no substantiation that such would work, and definitely will not work in general). Denote the intersection of the line and the x-axis as $(x, 0)$.
The line divides the region into 2 sections. The left piece is a trapezoid, with its area $\frac{1}{2}((x) + (x+1))(3) = 3x + \frac{3}{2}$. The right piece is the addition of a trapezoid and a rectangle, and the areas are $\frac{1}{2}((1-x) + (2-x))(3)$ and $2 \cdot 1 = 2$, totaling $\frac{13}{2} - 3x$. Since we want the two regions to be equal, we find that $3x + \frac 32 = \frac {13}2 - 3x$, so $x = \frac{5}{6}$.
We have that $\left(\frac 56, 0\right)$ is a point on the line of slope 3, so $y - 0 = 3\left(x - \frac 56\right) \Longrightarrow 6x = 2y + 5$. Our answer is $2^2 + 5^2 + 6^2 = 65$.
We now assess the validity of our starting assumption. We can do that by seeing that our answer passes through the tangency of the two circles, cutting congruent areas, a result explored in solution 1.
Solution 3
This problem looks daunting at a first glance, but we can make geometric inequality inferences by drawing lines that simplify the problem by removing sections of the total area. To begin, we can eliminate the possibility of the line intersecting the circle on the top left (call it circle A), or the circle on the bottom right (call it circle B). This is can be seen visually by drawing a line with slope 3 that is tangent to either of these circles. The area is clearly larger on one side; this can be proven by counting full circles. We can go on with the same mindset and eliminate the circle below circle A and the circle above circle B. By removing pairs of circles and proving the line will never intersect with them, we can safely work with whatever is remaining. By now you should have 4 circles making an L shape (waluigi style). Now the two biggest contenders for this method are the two circles on the bottom row. Using the same strategy, we can see that a line that goes through the tangent point of these two circles also goes through the tangent point of the other two circles. This clearly will cut the 4 circles into two regions of equal area. Using linear algebra we get: $6x = 2y + 5$. The answer is then $6^2 + 2^2 + 5^2 = \boxed{065}$. This solution is an alternate explanation to solution 1.
-jackshi2006 | // Block 1
size(150);defaultpen(linewidth(0.7));
draw((6.5,0)--origin--(0,6.5), Arrows(5));
int[] array={3,3,2};
int i,j;
for(i=0; i<3; i=i+1) {
for(j=0; j<array[i]; j=j+1) {
draw(Circle((1+2*i,1+2*j),1));
}}
label("x", (7,0));
label("y", (0,7));
draw((5/3,0)--(23/6,6.5),blue);
// Block 2
size(150);defaultpen(linewidth(0.7)); draw((6.5,0)--origin--(0,6.5), Arrows(5)); int[] array={3,3,2}; int i,j; for(i=0; i<3; i=i+1) { for(j=0; j<array[i]; j=j+1) { draw(Circle((1+2*i,1+2*j),1)); }} label("x", (7,0)); label("y", (0,7)); draw((5/3,0)--(23/6,6.5),blue); | [] |
132 | A tripod has three legs each of length $5$ feet. When the tripod is set up, the angle between any pair of legs is equal to the angle between any other pair, and the top of the tripod is $4$ feet from the ground. In setting up the tripod, the lower 1 foot of one leg breaks off. Let $h$ be the height in feet of the top of the tripod from the ground when the broken tripod is set up. Then $h$ can be written in the form $\frac m{\sqrt{n}},$ where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. Find $\lfloor m+\sqrt{n}\rfloor.$ (The notation $\lfloor x\rfloor$ denotes the greatest integer that is less than or equal to $x.$) | 2006 AIME I Problem 14 | We will use $[...]$ to denote volume (four letters), area (three letters) or length (two letters).
Let $T$ be the top of the tripod, $A,B,C$ are end points of three legs. Let $S$ be the point on $TA$ such that $[TS] = 4$ and $[SA] = 1$. Let $O$ be the center of the base equilateral triangle $ABC$. Let $M$ be the midpoint of segment $BC$. Let $h$ be the distance from $T$ to the triangle $SBC$ ($h$ is what we want to find).
We have the volume ratio $\frac {[TSBC]}{[TABC]} = \frac {[TS]}{[TA]} = \frac {4}{5}$.
So $\frac {h\cdot [SBC]}{[TO]\cdot [ABC]} = \frac {4}{5}$.
We also have the area ratio $\frac {[SBC]}{[ABC]} = \frac {[SM]}{[AM]}$.
The triangle $TOA$ is a $3-4-5$ right triangle so $[AM] = \frac {3}{2}\cdot[AO] = \frac {9}{2}$ and $\cos{\angle{TAO}} = \frac {3}{5}$.
Applying Law of Cosines to the triangle $SAM$ with $[SA] = 1$, $[AM] = \frac {9}{2}$ and $\cos{\angle{SAM}} = \frac {3}{5}$, we find:
$[SM] = \frac {\sqrt {5\cdot317}}{10}.$
Putting it all together, we find $h = \frac {144}{\sqrt {5\cdot317}}$.
$\lfloor 144+\sqrt{5 \cdot 317}\rfloor =144+ \lfloor \sqrt{5 \cdot 317}\rfloor =144+\lfloor \sqrt{1585} \rfloor =144+39=\boxed{183}$. | // Block 1
size(200);
import three; pointpen=black;pathpen=black+linewidth(0.65);pen ddash = dashed+linewidth(0.65);
currentprojection = perspective(1,-10,3.3);
triple O=(0,0,0),T=(0,0,5),C=(0,3,0),A=(-3*3^.5/2,-3/2,0),B=(3*3^.5/2,-3/2,0);
triple M=(B+C)/2,S=(4*A+T)/5;
draw(T--S--B--T--C--B--S--C);draw(B--A--C--A--S,ddash);draw(T--O--M,ddash);
label("$T$",T,N);label("$A$",A,SW);label("$B$",B,SE);label("$C$",C,NE);label("$S$",S,NW);label("$O$",O,SW);label("$M$",M,NE);
label("$4$",(S+T)/2,NW);label("$1$",(S+A)/2,NW);label("$5$",(B+T)/2,NE);label("$4$",(O+T)/2,W);
dot(S);dot(O);
// Block 2
size(200); import three; pointpen=black;pathpen=black+linewidth(0.65);pen ddash = dashed+linewidth(0.65); currentprojection = perspective(1,-10,3.3); triple O=(0,0,0),T=(0,0,5),C=(0,3,0),A=(-3*3^.5/2,-3/2,0),B=(3*3^.5/2,-3/2,0); triple M=(B+C)/2,S=(4*A+T)/5; draw(T--S--B--T--C--B--S--C);draw(B--A--C--A--S,ddash);draw(T--O--M,ddash); label("$T$",T,N);label("$A$",A,SW);label("$B$",B,SE);label("$C$",C,NE);label("$S$",S,NW);label("$O$",O,SW);label("$M$",M,NE); label("$4$",(S+T)/2,NW);label("$1$",(S+A)/2,NW);label("$5$",(B+T)/2,NE);label("$4$",(O+T)/2,W); dot(S);dot(O); | [] |
132 | A tripod has three legs each of length $5$ feet. When the tripod is set up, the angle between any pair of legs is equal to the angle between any other pair, and the top of the tripod is $4$ feet from the ground. In setting up the tripod, the lower 1 foot of one leg breaks off. Let $h$ be the height in feet of the top of the tripod from the ground when the broken tripod is set up. Then $h$ can be written in the form $\frac m{\sqrt{n}},$ where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. Find $\lfloor m+\sqrt{n}\rfloor.$ (The notation $\lfloor x\rfloor$ denotes the greatest integer that is less than or equal to $x.$) | 2006 AIME I Problem 14 | Diagram borrowed from Solution 1
Apply Pythagorean Theorem on $\bigtriangleup TOB$ yields
\[BO=\sqrt{TB^2-TO^2}=3\]
Since $\bigtriangleup ABC$ is equilateral, we have $\angle MOB=60^{\circ}$ and
\[BC=2BM=2(OB\sin MOB)=3\sqrt{3}\]
Apply Pythagorean Theorem on $\bigtriangleup TMB$ yields
\[TM=\sqrt{TB^2-BM^2}=\sqrt{5^2-(\frac{3\sqrt{3}}{2})^2}=\frac{\sqrt{73}}{2}\]
Apply Law of Cosines on $\bigtriangleup TBC$ we have
\[BC^2=TB^2+TC^2-2(TB)(TC)\cos BTC\]
\[(3\sqrt{3})^2=5^2+5^2-2(5)^2\cos BTC\]
\[\cos BTC=\frac{23}{50}\]
Apply Law of Cosines on $\bigtriangleup STB$ using the fact that $\angle STB=\angle BTC$ we have
\[SB^2=ST^2+BT^2-2(ST)(BT)\cos STB\]
\[SB=\sqrt{4^2+5^2-2(4)(5)\cos BTC}=\frac{\sqrt{565}}{5}\]
Apply Pythagorean Theorem on $\bigtriangleup BSM$ yields
\[SM=\sqrt{SB^2-BM^2}=\frac{\sqrt{1585}}{10}\]
Let the perpendicular from $T$ hits $SBC$ at $P$. Let $SP=x$ and $PM=\frac{\sqrt{1585}}{10}-x$. Apply Pythagorean Theorem on $TSP$ and $TMP$ we have
\[TP^2=TS^2-SP^2=TM^2-PM^2\]
\[4^2-x^2=(\frac{\sqrt{73}}{2})^2-(\frac{\sqrt{1585}}{10}-x)^2\]
Cancelling out the $x^2$ term and solving gets $x=\frac{181}{2\sqrt{1585}}$.
Finally, by Pythagorean Theorem,
\[TP=\sqrt{TS^2-SP^2}=\frac{144}{\sqrt{1585}}\]
so $\lfloor m+\sqrt{n}\rfloor=\boxed{183}$
~ Nafer | // Block 1
size(200);
import three; pointpen=black;pathpen=black+linewidth(0.65);pen ddash = dashed+linewidth(0.65);
currentprojection = perspective(1,-10,3.3);
triple O=(0,0,0),T=(0,0,5),C=(0,3,0),A=(-3*3^.5/2,-3/2,0),B=(3*3^.5/2,-3/2,0);
triple M=(B+C)/2,S=(4*A+T)/5;
draw(T--S--B--T--C--B--S--C);draw(B--A--C--A--S,ddash);draw(T--O--M,ddash);
label("$T$",T,N);label("$A$",A,SW);label("$B$",B,SE);label("$C$",C,NE);label("$S$",S,NW);label("$O$",O,SW);label("$M$",M,NE);
label("$4$",(S+T)/2,NW);label("$1$",(S+A)/2,NW);label("$5$",(B+T)/2,NE);label("$4$",(O+T)/2,W);
dot(S);dot(O);
// Block 2
size(200); import three; pointpen=black;pathpen=black+linewidth(0.65);pen ddash = dashed+linewidth(0.65); currentprojection = perspective(1,-10,3.3); triple O=(0,0,0),T=(0,0,5),C=(0,3,0),A=(-3*3^.5/2,-3/2,0),B=(3*3^.5/2,-3/2,0); triple M=(B+C)/2,S=(4*A+T)/5; draw(T--S--B--T--C--B--S--C);draw(B--A--C--A--S,ddash);draw(T--O--M,ddash); label("$T$",T,N);label("$A$",A,SW);label("$B$",B,SE);label("$C$",C,NE);label("$S$",S,NW);label("$O$",O,SW);label("$M$",M,NE); label("$4$",(S+T)/2,NW);label("$1$",(S+A)/2,NW);label("$5$",(B+T)/2,NE);label("$4$",(O+T)/2,W); dot(S);dot(O); | [] |
133 | In convex hexagon $ABCDEF$, all six sides are congruent, $\angle A$ and $\angle D$ are right angles, and $\angle B, \angle C, \angle E,$ and $\angle F$ are congruent. The area of the hexagonal region is $2116(\sqrt{2}+1).$ Find $AB$. | 2006 AIME II Problem 1 | Because $\angle B$, $\angle C$, $\angle E$, and $\angle F$ are congruent, the degree-measure of each of them is ${{720-2\cdot90}\over4}= 135$. Lines $BF$ and $CE$ divide the hexagonal region into two right triangles and a rectangle. Let $AB=x$. Then $BF=x\sqrt2$. Thus
\begin{align*} 2116(\sqrt2+1)&=[ABCDEF]\\ &=2\cdot {1\over2}x^2+x\cdot x\sqrt2=x^2(1+\sqrt2), \end{align*}so $x^2=2116$, and $x=\boxed{046}$. | // Block 1
pair A,B,C,D,E,F;
A=(0,0);
B=(7,0);
C=(13,6);
E=(6,13);
D=(13,13);
F=(0,7);
dot(A);
dot(B);
dot(C);
dot(D);
dot(E);
dot(F);
draw(A--B--C--D--E--F--cycle,linewidth(0.7));
label("{\tiny $A$}",A,S);
label("{\tiny $B$}",B,S);
label("{\tiny $C$}",C,S);
label("{\tiny $D$}",D,N);
label("{\tiny $E$}",E,N);
label("{\tiny $F$}",F,W);
// Block 2
pair A,B,C,D,E,F; A=(0,0); B=(7,0); C=(13,6); E=(6,13); D=(13,13); F=(0,7); dot(A); dot(B); dot(C); dot(D); dot(E); dot(F); draw(A--B--C--D--E--F--cycle,linewidth(0.7)); label("{\tiny $A$}",A,S); label("{\tiny $B$}",B,S); label("{\tiny $C$}",C,S); label("{\tiny $D$}",D,N); label("{\tiny $E$}",E,N); label("{\tiny $F$}",F,W); | [] |
134 | Square $ABCD$ has sides of length 1. Points $E$ and $F$ are on $\overline{BC}$ and $\overline{CD},$ respectively, so that $\triangle AEF$ is equilateral. A square with vertex $B$ has sides that are parallel to those of $ABCD$ and a vertex on $\overline{AE}.$ The length of a side of this smaller square is $\frac{a-\sqrt{b}}{c},$ where $a, b,$ and $c$ are positive integers and $b$ is not divisible by the square of any prime. Find $a+b+c.$ | 2006 AIME II Problem 6 | Call the vertices of the new square A', B', C', and D', in relation to the vertices of $ABCD$, and define $s$ to be one of the sides of that square. Since the sides are parallel, by corresponding angles and AA~ we know that triangles $AA'D'$ and $D'C'E$ are similar. Thus, the sides are proportional: $\frac{AA'}{A'D'} = \frac{D'C'}{C'E} \Longrightarrow \frac{1 - s}{s} = \frac{s}{1 - s - CE}$. Simplifying, we get that $s^2 = (1 - s)(1 - s - CE)$.
$\angle EAF$ is $60$ degrees, so $\angle BAE = \frac{90 - 60}{2} = 15$. Thus, $\cos 15 = \cos (45 - 30) = \frac{\sqrt{6} + \sqrt{2}}{4} = \frac{1}{AE}$, so $AE = \frac{4}{\sqrt{6} + \sqrt{2}} \cdot \frac{\sqrt{6} - \sqrt{2}}{\sqrt{6} - \sqrt{2}} = \sqrt{6} - \sqrt{2}$. Since $\triangle AEF$ is equilateral, $EF = AE = \sqrt{6} - \sqrt{2}$. $\triangle CEF$ is a $45-45-90 \triangle$, so $CE = \frac{AE}{\sqrt{2}} = \sqrt{3} - 1$. Substituting back into the equation from the beginning, we get $s^2 = (1 - s)(2 - \sqrt{3} - s)$, so $(3 - \sqrt{3})s = 2 - \sqrt{3}$. Therefore, $s = \frac{2 - \sqrt{3}}{3 - \sqrt{3}} \cdot \frac{3 + \sqrt{3}}{3 + \sqrt{3}} = \frac{3 - \sqrt{3}}{6}$, and $a + b + c = 3 + 3 + 6 = \boxed{12}$.
Here's an alternative geometric way to calculate $AE$ (as opposed to trigonometric): The diagonal $\overline{AC}$ is made of the altitude of the equilateral triangle and the altitude of the $45-45-90 \triangle$. The former is $\frac{AE\sqrt{3}}{2}$, and the latter is $\frac{AE}{2}$; thus $\frac{AE\sqrt{3} + AE}{2} = AC = \sqrt{2} \Longrightarrow AE= \sqrt{6}-\sqrt{2}$. The solution continues as above. | // Block 1
unitsize(32mm);
defaultpen(linewidth(.8pt)+fontsize(10pt));
dotfactor=3;
pair B = (0, 0), C = (1, 0), D = (1, 1), A = (0, 1);
pair Ep = (2 - sqrt(3), 0), F = (1, sqrt(3) - 1);
pair Ap = (0, (3 - sqrt(3))/6);
pair Cp = ((3 - sqrt(3))/6, 0);
pair Dp = ((3 - sqrt(3))/6, (3 - sqrt(3))/6);
pair[] dots = {A, B, C, D, Ep, F, Ap, Cp, Dp};
draw(A--B--C--D--cycle);
draw(A--F--Ep--cycle);
draw(Ap--B--Cp--Dp--cycle);
dot(dots);
label("$A$", A, NW);
label("$B$", B, SW);
label("$C$", C, SE);
label("$D$", D, NE);
label("$E$", Ep, SE);
label("$F$", F, E);
label("$A'$", Ap, W);
label("$C'$", Cp, SW);
label("$D'$", Dp, E);
label("$s$", Ap--B, W);
label("$1$", A--D, N);
// Block 2
unitsize(32mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=3; pair B = (0, 0), C = (1, 0), D = (1, 1), A = (0, 1); pair Ep = (2 - sqrt(3), 0), F = (1, sqrt(3) - 1); pair Ap = (0, (3 - sqrt(3))/6); pair Cp = ((3 - sqrt(3))/6, 0); pair Dp = ((3 - sqrt(3))/6, (3 - sqrt(3))/6); pair[] dots = {A, B, C, D, Ep, F, Ap, Cp, Dp}; draw(A--B--C--D--cycle); draw(A--F--Ep--cycle); draw(Ap--B--Cp--Dp--cycle); dot(dots); label("$A$", A, NW); label("$B$", B, SW); label("$C$", C, SE); label("$D$", D, NE); label("$E$", Ep, SE); label("$F$", F, E); label("$A'$", Ap, W); label("$C'$", Cp, SW); label("$D'$", Dp, E); label("$s$", Ap--B, W); label("$1$", A--D, N); | [] |
134 | Square $ABCD$ has sides of length 1. Points $E$ and $F$ are on $\overline{BC}$ and $\overline{CD},$ respectively, so that $\triangle AEF$ is equilateral. A square with vertex $B$ has sides that are parallel to those of $ABCD$ and a vertex on $\overline{AE}.$ The length of a side of this smaller square is $\frac{a-\sqrt{b}}{c},$ where $a, b,$ and $c$ are positive integers and $b$ is not divisible by the square of any prime. Find $a+b+c.$ | 2006 AIME II Problem 6 | Since $AEF$ is equilateral, $AE=EF$. Let $BE=x$. By the Pythagorean theorem, $1+x^2=2(1-x)^2$. Simplifying, we get $x^2-4x+1=0$. By the quadratic formula, the roots are $2 \pm \sqrt{3}$. Since $x<1$, we discard the root with the "+", giving $x=2-\sqrt{3}$.
Let the side length of the square be s. Since $MEK$ is similar to $ABE$, $s=\frac{2-\sqrt{3}-s}{2-\sqrt{3}}$. Solving, we get $s=\frac{3-\sqrt{3}}{6}$ and the final answer is $\boxed{012}$. | // Block 1
real n;
n=0.26794919243;
real m;
m=0.2113248654;
draw((0,0)--(0,n)--(1,0)--(0,0));
draw((0,m)--(m,m)--(m,0));
label((0,0), "$B$",SW);
label((0,n), "$E$",SW);
label((0,m), "$M$",SW);
label((1,0), "$A$",SW);
label((m,0), "$N$",SW);
label((m,m), "$K$",NE);
// Block 2
real n; n=0.26794919243; real m; m=0.2113248654; draw((0,0)--(0,n)--(1,0)--(0,0)); draw((0,m)--(m,m)--(m,0)); label((0,0), "$B$",SW); label((0,n), "$E$",SW); label((0,m), "$M$",SW); label((1,0), "$A$",SW); label((m,0), "$N$",SW); label((m,m), "$K$",NE); | [] |
135 | Equilateral $\triangle ABC$ is inscribed in a circle of radius $2$. Extend $\overline{AB}$ through $B$ to point $D$ so that $AD=13,$ and extend $\overline{AC}$ through $C$ to point $E$ so that $AE = 11.$ Through $D,$ draw a line $l_1$ parallel to $\overline{AE},$ and through $E,$ draw a line $l_2$ parallel to $\overline{AD}.$ Let $F$ be the intersection of $l_1$ and $l_2.$ Let $G$ be the point on the circle that is collinear with $A$ and $F$ and distinct from $A.$ Given that the area of $\triangle CBG$ can be expressed in the form $\frac{p\sqrt{q}}{r},$ where $p, q,$ and $r$ are positive integers, $p$ and $r$ are relatively prime, and $q$ is not divisible by the square of any prime, find $p+q+r.$ | 2006 AIME II Problem 12 | Notice that $\angle{E} = \angle{BGC} = 120^\circ$ because $\angle{A} = 60^\circ$. Also, $\angle{GBC} = \angle{GAC} = \angle{FAE}$ because they both correspond to arc ${GC}$. So $\Delta{GBC} \sim \Delta{EAF}$.
\[[EAF] = \frac12 (AE)(EF)\sin \angle AEF = \frac12\cdot11\cdot13\cdot\sin{120^\circ} = \frac {143\sqrt3}4.\]
Because the ratio of the area of two similar figures is the square of the ratio of the corresponding sides, $[GBC] = \frac {BC^2}{AF^2}\cdot[EAF] = \frac {12}{11^2 + 13^2 - 2\cdot11\cdot13\cdot\cos120^\circ}\cdot\frac {143\sqrt3}4 = \frac {429\sqrt3}{433}$. Therefore, the answer is $429+433+3=\boxed{865}$. | // Block 1
size(250); pointpen = black; pathpen = black + linewidth(0.65); pen s = fontsize(8);
pair A=(0,0),B=(-3^.5,-3),C=(3^.5,-3),D=13*expi(-2*pi/3),E1=11*expi(-pi/3),F=E1+D;
path O = CP((0,-2),A); pair G = OP(A--F,O);
D(MP("A",A,N,s)--MP("B",B,W,s)--MP("C",C,E,s)--cycle);D(O);
D(B--MP("D",D,W,s)--MP("F",F,s)--MP("E",E1,E,s)--C);
D(A--F);D(B--MP("G",G,SW,s)--C);
MP("11",(A+E1)/2,NE);MP("13",(A+D)/2,NW);MP("l_1",(D+F)/2,SW);MP("l_2",(E1+F)/2,SE);
// Block 2
size(250); pointpen = black; pathpen = black + linewidth(0.65); pen s = fontsize(8); pair A=(0,0),B=(-3^.5,-3),C=(3^.5,-3),D=13*expi(-2*pi/3),E1=11*expi(-pi/3),F=E1+D; path O = CP((0,-2),A); pair G = OP(A--F,O); D(MP("A",A,N,s)--MP("B",B,W,s)--MP("C",C,E,s)--cycle);D(O); D(B--MP("D",D,W,s)--MP("F",F,s)--MP("E",E1,E,s)--C); D(A--F);D(B--MP("G",G,SW,s)--C); MP("11",(A+E1)/2,NE);MP("13",(A+D)/2,NW);MP("l_1",(D+F)/2,SW);MP("l_2",(E1+F)/2,SE); | [] |
136 | In isosceles triangle $\triangle ABC$, $A$ is located at the origin and $B$ is located at $(20,0)$. Point $C$ is in the first quadrant with $AC = BC$ and angle $BAC = 75^{\circ}$. If triangle $ABC$ is rotated counterclockwise about point $A$ until the image of $C$ lies on the positive $y$-axis, the area of the region common to the original and the rotated triangle is in the form $p\sqrt{2} + q\sqrt{3} + r\sqrt{6} + s$, where $p,q,r,s$ are integers. Find $\frac{p-q+r-s}2$. | 2007 AIME I Problem 12 | Solution 1
Let the new triangle be $\triangle AB'C'$ ($A$, the origin, is a vertex of both triangles). Let $\overline{B'C'}$ intersect with $\overline{AC}$ at point $D$, $\overline{BC}$ intersect with $\overline{B'C'}$ at $E$, and $\overline{BC}$ intersect with $\overline{AB'}$ at $F$. The region common to both triangles is the quadrilateral $ADEF$. Notice that $[ADEF] = [\triangle ADB'] - [\triangle EFB']$, where we let $[\ldots]$ denote area.
To find $[\triangle ADB']$:
Since $\angle B'AC'$ and $\angle BAC$ both have measures $75^{\circ}$, both of their complements are $15^{\circ}$, and $\angle DAB' = 90 - 2(15) = 60^{\circ}$. We know that $\angle DB'A = 75^{\circ}$, so $\angle ADB' = 180 - 60 - 75 = 45^{\circ}$.
Thus $\triangle ADB'$ is a $45 - 60 - 75 \triangle$. It can be solved by drawing an altitude splitting the $75^{\circ}$ angle into $30^{\circ}$ and $45^{\circ}$ angles, forming a $30-60-90$ right triangle and a $45-45-90$ isosceles right triangle. Since we know that $AB' = 20$, the base of the $30-60-90$ triangle is $10$, the base of the $45-45-90$ is $10\sqrt{3}$, and their common height is $10\sqrt{3}$. Thus, the total area of $[\triangle ADB'] = \frac{1}{2}(10\sqrt{3})(10\sqrt{3} + 10) = \boxed{150 + 50\sqrt{3}}$.
To find $[\triangle EFB']$:
Since $\triangle AFB$ is also a $15-75-90$ triangle,
$AF = 20\sin 75 = 20 \sin (30 + 45) = 20\left(\frac{\sqrt{2} + \sqrt{6}}4\right) = 5\sqrt{2} + 5\sqrt{6}$
and
$FB' = AB' - AF = 20 - 5\sqrt{2} - 5\sqrt{6}$
Since $[\triangle EFB'] = \frac{1}{2} (FB' \cdot EF) = \frac{1}{2} (FB') (FB' \tan 75^{\circ})$. With some horrendous algebra, we can calculate
\begin{align*} [\triangle EFB'] &= \frac{1}{2}\tan (30 + 45) \cdot (20 - 5\sqrt{2} - 5\sqrt{6})^2 \\ &= 25 \left(\frac{\frac{1}{\sqrt{3}} + 1}{1 - \frac{1}{\sqrt{3}}}\right) \left(8 - 2\sqrt{2} - 2\sqrt{6} - 2\sqrt{2} + 1 + \sqrt{3} - 2\sqrt{6} + \sqrt{3} + 3\right) \\ &= 25(2 + \sqrt{3})(12 - 4\sqrt{2} - 4\sqrt{6} + 2\sqrt{3}) \\ [\triangle EFB'] &= \boxed{- 500\sqrt{2} + 400\sqrt{3} - 300\sqrt{6} +750}. \end{align*}
To finish,
\begin{align*} [ADEF] &= [\triangle ADB'] - [\triangle EFB']\\ &= \left(150 + 50\sqrt{3}\right) - \left(-500\sqrt{2} + 400\sqrt{3} - 300\sqrt{6} + 750\right)\\ &=500\sqrt{2} - 350\sqrt{3} + 300\sqrt{6} - 600\\ \end{align*}
Hence, $\frac{p-q+r-s}{2} = \frac{500 + 350 + 300 + 600}2 = \frac{1750}2 = \boxed{\boxed{875}}$.
Solution 2
Redefine the points in the same manner as the last time ($\triangle AB'C'$, intersect at $D$, $E$, and $F$). This time, notice that $[ADEF] = [\triangle AB'C'] - ([\triangle ADC'] + [\triangle EFB'])$.
The area of $[\triangle AB'C'] = [\triangle ABC]$. The altitude of $\triangle ABC$ is clearly $10 \tan 75 = 10 \tan (30 + 45)$. The tangent addition rule yields $10(2 + \sqrt{3})$ (see above). Thus, $[\triangle ABC] = \frac{1}{2} 20 \cdot (20 + 10\sqrt{3}) = 200 + 100\sqrt{3}$.
The area of $[\triangle ADC']$ (with a side on the y-axis) can be found by splitting it into two triangles, $30-60-90$ and $15-75-90$ right triangles. $AC' = AC = \frac{10}{\sin 15}$. The sine subtraction rule shows that $\frac{10}{\sin 15} = \frac{10}{\frac{\sqrt{6} - \sqrt{2}}4} = \frac{40}{\sqrt{6} - \sqrt{2}} = 10(\sqrt{6} + \sqrt{2})$. $AC'$, in terms of the height of $\triangle ADC'$, is equal to $h(\sqrt{3} + \tan 75) = h(\sqrt{3} + 2 + \sqrt{3})$.
\begin{align*} [ADC'] &= \frac 12 AC' \cdot h \\ &= \frac 12 (10\sqrt{6} + 10\sqrt{2})\left(\frac{10\sqrt{6} + 10\sqrt{2}}{2\sqrt{3} + 2}\right) \\ &= \frac{(800 + 400\sqrt{3})}{(2 + \sqrt{3})}\cdot\frac{2 - \sqrt{3}}{2-\sqrt{3}} \\ &= \frac{400\sqrt{3} + 400}8 = 50\sqrt{3} + 50 \end{align*}
The area of $[\triangle EFB']$ was found in the previous solution to be $- 500\sqrt{2} + 400\sqrt{3} - 300\sqrt{6} +750$.
Therefore, $[ADEF]$ $= (200 + 100\sqrt{3}) - \left((50 + 50\sqrt{3}) + (-500\sqrt{2} + 400\sqrt{3} - 300\sqrt{6} +750)\right)$ $= 500\sqrt{2} - 350\sqrt{3} + 300\sqrt{6} - 600$, and our answer is $\boxed{875}$.
Solution 3
[Image: images/aime/2007_AIME_I_Problem_12_0.png]
Call the points of the intersections of the triangles $D$, $E$, and $F$ as noted in the diagram (the points are different from those in the diagram for solution 1). $\overline{AD}$ bisects $\angle EDE'$.
Through HL congruency, we can find that $\triangle AED$ is congruent to $\triangle AE'D$. This divides the region $AEDF$ (which we are trying to solve for) into two congruent triangles and an isosceles right triangle.
$AE = 20 \cos 15 = 20 \cos (45 - 30) = 20 \cdot \frac{\sqrt{6} + \sqrt{2}}{4} = 5\sqrt{6} + 5\sqrt{2}$
Since $FE' = AE' = AE$, we find that $[AE'F] = \frac 12 (5\sqrt{6} + 5\sqrt{2})^2 = 100 + 50\sqrt{3}$.
Now, we need to find $[AED] = [AE'D]$. The acute angles of the triangles are $\frac{15}{2}$ and $90 - \frac{15}{2}$. By repeated application of the half-angle formula, we can find that $\tan \frac{15}{2} = \sqrt{2} - \sqrt{3} + \sqrt{6} - 2$.
The area of $[AED] = \frac 12 \left(20 \cos 15\right)^2 \left(\tan \frac{15}{2}\right)$. Thus, $[AED] + [AE'D] = 2\left(\frac 12((5\sqrt{6} + 5\sqrt{2})^2 \cdot (\sqrt{2} - \sqrt{3} + \sqrt{6} - 2))\right)$, which eventually simplifies to $500\sqrt{2} - 350\sqrt{3} + 300\sqrt{6} - 600$.
Adding them together, we find that the solution is $[AEDF] = [AE'F] + [AED] + [AE'D]$ $= 100 + 50\sqrt{3} + 500\sqrt{2} - 350\sqrt{3} + 300\sqrt{6} - 600=$ $= 500\sqrt{2} - 350\sqrt{3} + 300\sqrt{6} - 600$, and the answer is $\boxed{875}$.
Solution 4
From the given information, calculate the coordinates of all the points (by finding the equations of the lines, equating). Then, use the shoelace method to calculate the area of the intersection.
$\overline{AC}$: $y = (\tan 75) x = (2 + \sqrt{3})x$
$\overline{AB'}$: $y = (\tan 15) x = (2 - \sqrt{3})x$
$\overline{BC}$: It passes thru $(20,0)$, and has a slope of $-\tan75 = -(2 + \sqrt{3})$. The equation of its line is $y = (2+\sqrt{3})(20 - x)$.
$\overline{B'C'}$: $AC' = AC = \frac{10}{\cos 75} = 10\sqrt{6} + 10\sqrt{2}$, so it passes thru point $(0, 10\sqrt{6} + 10\sqrt{2})$. It has a slope of $-\tan 60 = -\sqrt{3}$. So the equation of its line is $y = -\sqrt{3}x + 10(\sqrt{6} + \sqrt{2})$.
Now, we can equate the equations to find the intersections of all the points.
$A (0,0)$
$D$ is the intersection of $\overline{BC},\ \overline{B'C'}$. $(2+\sqrt{3})(20-x) = -\sqrt{3}x + 10(\sqrt{6} + \sqrt{2})$. Therefore, $x = 5(4 + 2\sqrt{3}-\sqrt{6}-\sqrt{2})$, $y = 5(3\sqrt{6} + 5\sqrt{2}-4\sqrt{3}-6)$.
$E$ is the intersection of $\overline{AB'},\ \overline{BC}$. $(2 - \sqrt{3})x =(2+\sqrt{3})(20-x)$. Therefore, $x = 5(2+\sqrt{3})$, $y = 5$.
$F$ is the intersection of $\overline{AC},\ \overline{B'C'}$. $(2+\sqrt{3})x = -\sqrt{3}x + 10(\sqrt{6} + \sqrt{2})$. Therefore, $x = 5\sqrt{2}$, $y = 10\sqrt{2}+ 5\sqrt{6}$.
We take these points and tie them together by shoelace, and the answer should come out to be $\boxed{875}$. | // Block 1
defaultpen(fontsize(12)+0.6); size(300);
var theta=15;
pair A=origin, B=(20,0), C=extension(A,dir(75),B/2,bisectorpoint(A,B)), Cp=rotate(theta,A)*C, Bp=rotate(theta,A)*B, X=extension(A,Bp,B,C), Y=extension(B,C,Bp,Cp);
draw(A--B--C--A); draw(A--Bp--Cp--A, royalblue);
markscalefactor=0.1;
draw(rightanglemark(Y,X,A));
dot("$A$",A,dir(210)); dot("$B$",B,dir(-20)); dot("$C$",C,up); dot("$B'$",Bp,dir(-10)); dot("$C'$",Cp,dir(100));
MA("60^\circ",Bp,A,C,2); MA("15^\circ",B,A,Bp,5); MA("75^\circ",C,B,A,2); MA("30^\circ",A,C,B,4); MA("30^\circ",A,Cp,Bp,4); MA("45^\circ",A,extension(A,C,Bp,Cp),Bp,3); MA("15^\circ",C,Y,Cp,8); MA("15^\circ",C,A,Cp,9);
// Block 2
defaultpen(fontsize(12)+0.6); size(300); var theta=15; pair A=origin, B=(20,0), C=extension(A,dir(75),B/2,bisectorpoint(A,B)), Cp=rotate(theta,A)*C, Bp=rotate(theta,A)*B, X=extension(A,Bp,B,C), Y=extension(B,C,Bp,Cp); draw(A--B--C--A); draw(A--Bp--Cp--A, royalblue); markscalefactor=0.1; draw(rightanglemark(Y,X,A)); dot("$A$",A,dir(210)); dot("$B$",B,dir(-20)); dot("$C$",C,up); dot("$B'$",Bp,dir(-10)); dot("$C'$",Cp,dir(100)); MA("60^\circ",Bp,A,C,2); MA("15^\circ",B,A,Bp,5); MA("75^\circ",C,B,A,2); MA("30^\circ",A,C,B,4); MA("30^\circ",A,Cp,Bp,4); MA("45^\circ",A,extension(A,C,Bp,Cp),Bp,3); MA("15^\circ",C,Y,Cp,8); MA("15^\circ",C,A,Cp,9); | ["https://artofproblemsolving.com/wiki/images/thumb/2/25/I-07-12.png/250px-I-07-12.png"] |
137 | Square $ABCD$ has side length $13$, and points $E$ and $F$ are exterior to the square such that $BE=DF=5$ and $AE=CF=12$. Find $EF^{2}$.
Contents
1 Problem
2 Solution
2.1 Solution 1
2.2 Solution 2
2.3 Solution 3
2.4 Solution 4
2.5 Solution 5 (Ptolemy's Theorem)
2.6 Solution 6 (Coordinate Bash)
2.7 Solution 7 (Trig Bash)
3 See also | 2007 AIME II Problem 3 | Solution 1
Let $\angle FCD = \alpha$, so that $FB = \sqrt{12^2 + 13^2 + 2\cdot12\cdot13\sin(\alpha)} = \sqrt{433}$. By the diagonal, $DB = 13\sqrt{2}, DB^2 = 338$.
The sum of the squares of the sides of a parallelogram is the sum of the squares of the diagonals.
\[EF^2 = 2\cdot(5^2 + 433) - 338 = 578.\]
Solution 2
Extend $\overline{AE}, \overline{DF}$ and $\overline{BE}, \overline{CF}$ to their points of intersection. Since $\triangle ABE \cong \triangle CDF$ and are both $5-12-13$ right triangles, we can come to the conclusion that the two new triangles are also congruent to these two (use ASA, as we know all the sides are $13$ and the angles are mostly complementary). Thus, we create a square with sides $5 + 12 = 17$.
$\overline{EF}$ is the diagonal of the square, with length $17\sqrt{2}$; the answer is $EF^2 = (17\sqrt{2})^2 = 578$.
Solution 3
A slightly more analytic/brute-force approach:
[Image: images/aime/2007_AIME_II_Problem_3_0.png]
Drop perpendiculars from $E$ and $F$ to $I$ and $J$, respectively; construct right triangle $EKF$ with right angle at K and $EK || BC$. Since $2[CDF]=DF*CF=CD*JF$, we have $JF=5\times12/13 = \frac{60}{13}$. Similarly, $EI=\frac{60}{13}$. Since $\triangle DJF \sim \triangle DFC$, we have $DJ=\frac{5JF}{12}=\frac{25}{13}$.
Now, we see that $FK=DC-(DJ+IB)=DC-2DJ=13-\frac{50}{13}=\frac{119}{13}$. Also, $EK=BC+(JF+IE)=BC+2JF=13+\frac{120}{13}=\frac{289}{13}$. By the Pythagorean Theorem, we have $EF=\sqrt{\left(\frac{289}{13}\right)^2+\left(\frac{119}{13} \right)^2}=\frac{\sqrt{(17^2)(17^2+7^2)}}{13}$$=\frac{17\sqrt{338}}{13}=\frac{17(13\sqrt{2})}{13}=17\sqrt{2}$. Therefore, $EF^2=(17\sqrt{2})^2=578$.
Solution 4
Based on the symmetry, we know that $F$ is a reflection of $E$ across the center of the square, which we will denote as $O$. Since $\angle BEA$ and $\angle AOB$ are right, we can conclude that figure $AOBE$ is a cyclic quadrilateral. Pythagorean Theorem yields that $BO=AO=\frac{13\sqrt{2}}{2}$. Now, using Ptolemy's Theorem, we get that
\[AO\cdot BE + BO\cdot AE = AB\cdot AO\]
\[\frac{13\sqrt{2}}{2}\cdot 5+\frac{13\sqrt{2}}{2}\cdot 12 = 13\cdot OE\]
\[OE=\frac{17\sqrt{2}}{2}\]
Now, since we stated in the first step that $F$ is a reflection of $E$ across $O$, we can say that $EF=2EO=17\sqrt{2}$. This gives that \[EF^2=(17\sqrt{2})^2=578\] AWD with this bash solution
Solution 5 (Ptolemy's Theorem)
Drawing $EF$, it clearly passes through the center of $ABCD$. Letting this point be $P$, we note that $AEBP$ and $CFDP$ are congruent cyclic quadrilaterals, and that $AP=BP=CP=DP=\frac{13}{\sqrt{2}}.$ Now, from Ptolemy's, $13\cdot EP=\frac{13}{\sqrt{2}}(12+5)\implies EP=\frac{17\sqrt{2}}{2}$. Since $EF=EP+FP=2\cdot EP$, the answer is $(17\sqrt{2})^2=\boxed{578}.$
Solution 6 (Coordinate Bash)
Place the vertices of the square as follows:
\begin{align}
D &= (0,\;0),\\
C &= (13,\;0),\\
A &= (0,\;13),\\
B &= (13,\;13).
\end{align}
Compute the coordinates of points E and F using the distance formula.
Place $E$ at $(x,y)$ and $F$ at $(a,b)$.
\begin{aligned}
x^2+(y-13)^2 = 144,\\
(x-13)^2+(y-13)^2 = 25
\end{aligned}
\begin{aligned}
a^2+b^2=25,\\
(a-13)^2+b^2=144.
\end{aligned}
Solving and taking appropriate solutions to get:
$E = (144/13,229/13)$, and
$F = (25/13, -60/13)$
Computing distance between $E$ and $F$ = $17\sqrt{2}$
\[EF^2 = 578\]
Solution 7 (Trig Bash)
We first see that the whole figure is symmetrical and reflections across the center that we will denote as $O$ bring each half of the figure to the other half. Thus we consider a single part of the figure, namely $EO.$
First note that $\angle BAO = 45^{\circ}$ since $O$ is the center of square $ABCD.$ Also note that $\angle EAB = \arccos{\left(\frac{12}{13}\right)}$ or $\arcsin{\left(\frac{5}{13}\right)}.$ Finally, we know that $AO =\frac{13\sqrt{2}}{2}.$ Now we apply laws of cosines on $\bigtriangleup AEO.$
We have $EO^2 = 12^2 + (\frac{13\sqrt{2}}{2})^2 - 2 \cdot 12 \cdot \left(\frac{13\sqrt{2}}{2}\right) \cdot \cos{\angle EAO}.$ We know that $\angle EAO = 45^{\circ} + \arccos{\left(\frac{12}{13}\right)}.$ Thus we have $\cos{\angle EAO} = \cos\left(45^{\circ} + \arccos{\left(\frac{12}{13}\right)}\right)$ which applying the cosine sum identity yields $\cos{45^{\circ}}\cos{\arccos\frac{12}{13}} - \sin{45^{\circ}}\sin\arcsin{\frac{5}{12}} =\frac{12\sqrt{2}}{26} -\frac{5\sqrt{2}}{26} =\frac{7\sqrt{2}}{26}.$
Note that we are looking for $4EO^2$ so we multiply $EO^2 = 12^2 + \left(\frac{13\sqrt{2}}{2}\right)^2 - 2 \cdot 12 \cdot \left(\frac{13\sqrt{2}}{2}\right) \cdot \cos{\angle EAO}$ by $4$ obtaining $4EO^2 = 576 + 338 - 8 \cdot\left(\frac{13\sqrt{2}}{2}\right) \cdot 12 \cdot\frac{7\sqrt{2}}{26} = 576 + 338 - 4 \cdot 12 \cdot 7 = \boxed{578}.$ | unitsize(0.25 cm); pair A, B, C, D, E, F, G, H; A = (0,13); B = (13,13); C = (13,0); D = (0,0); E = A + (12*12/13,5*12/13); F = D + (5*5/13,-5*12/13); G = rotate(90,(A + C)/2)*(E); H = rotate(90,(A + C)/2)*(F); draw(A--B--C--D--cycle); draw(E--G--F--H--cycle); dot("$A$", A, N); dot("$B$", B, dir(0)); dot("$C$", C, S); dot("$D$", D, W); dot("$E$", E, N); dot("$F$", F, S); dot("$G$", G, W); dot("$H$", H, dir(0)); | ["https://artofproblemsolving.com/wiki/images/5/59/AIME_II_prob10_bruteforce.PNG"] |
138 | Four circles $\omega,$ $\omega_{A},$ $\omega_{B},$ and $\omega_{C}$ with the same radius are drawn in the interior of triangle $ABC$ such that $\omega_{A}$ is tangent to sides $AB$ and $AC$, $\omega_{B}$ to $BC$ and $BA$, $\omega_{C}$ to $CA$ and $CB$, and $\omega$ is externally tangent to $\omega_{A},$ $\omega_{B},$ and $\omega_{C}$. If the sides of triangle $ABC$ are $13,$ $14,$ and $15,$ the radius of $\omega$ can be represented in the form $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n.$
Contents
1 Problem
2 Diagram
3 Solution 1 (Homothety)
4 Solution 2
5 Solution 3 (elementary)
6 Solution 4
7 Sidenote (Generalization)
8 Solution 5
9 Solution 6
10 Video Solution
11 See also | 2007 AIME II Problem 15 | First, apply Heron's formula to find that $[ABC] = \sqrt{21 \cdot 8 \cdot 7 \cdot 6} = 84$. The semiperimeter is $21$, so the inradius is $\frac{A}{s} = \frac{84}{21} = 4$.
Now consider the incenter $I$ of $\triangle ABC$. Let the radius of one of the small circles be $r$. Let the centers of the three little circles tangent to the sides of $\triangle{ABC}$ be $O_A$, $O_B$, and $O_C$. Let the center of the circle tangent to those three circles be $O$. Note that $\triangle O_A O_B O_C$ and $\triangle ABC$ are similar because they have parallel sides and have the same incenter. Also, since the inradius of $\triangle O_A O_B O_C$ is $4-r$ and the inradius of $\triangle ABC$ is $4,$ it follows that $\triangle O_A O_B O_C$ is the result of a homothety $\mathcal{H}\left(I, \frac{4-r}{4}\right)$ on $\triangle ABC.$ Now, note that $OO_A = OO_B = OO_C = 2r$ are all circumradii of $\triangle O_A O_B O_C$, and the circumradius of $\triangle ABC$ is $R = \frac{abc}{4[ABC]} = \frac{65}{8}.$ From the ratio of the homothety, we have that $\frac{2r}{R}=\frac{4-r}{4}.$ Plugging in $R=\frac{65}{8}$ and solving yields $r = \frac{260}{129}$ and the final answer is $\boxed{389}$.
https://latex.artofproblemsolving.com/9/4/7/947b7f06d947dbf8bc5d8f61cdd193c330377372.png | // Block 1
defaultpen(fontsize(12)+0.8); size(350);
pair A,B,C,X,Y,Z,P,Q,R,Zp;
B=origin; C=15*right; A=IP(CR(B,13),CR(C,14)); P=incenter(A,B,C);
real r=260/129;
Q=r*(rotate(-90)*A/length(A)); X=extension(A,P,Q,Q+A); Y=extension(B,P,Q,Q+A);
R=rotate(90,C)*(C+r*(A-C)/length(A-C)); Z=extension(C,P,R,R+A-C); Zp=circumcenter(X,Y,Z);
draw(A--B--C--A); draw(CR(X,r)^^CR(Y,r)^^CR(Z,r), gray); draw(A--P--B^^P--C^^X--Y--Z--X, royalblue); draw(X--foot(X,A,C)^^Z--foot(Z,A,C),royalblue); draw(CR(Zp,r), gray); draw(incircle(A,B,C)^^incircle(X,Y,Z)^^P--foot(P,A,C), heavygreen+0.6); draw(rightanglemark(X,foot(X,A,C),C,10),linewidth(0.6)); draw(rightanglemark(Z,foot(Z,A,C),A,10),linewidth(0.6));
label("$A$",A,N); label("$B$",B,0.15*(B-P)); label("$C$",C,0.1*(C-P));
pen p=fontsize(10)+linewidth(3);
dot("$O_A$",X,right,p); dot("$O_B$",Y,left+up,p); dot("$O_C$",Z,down,p); dot("$O$",Zp,dir(-45),p+red); dot("$I$",P,left+up,p);
// Block 2
defaultpen(fontsize(12)+0.8); size(350); pair A,B,C,X,Y,Z,P,Q,R,Zp; B=origin; C=15*right; A=IP(CR(B,13),CR(C,14)); P=incenter(A,B,C); real r=260/129; Q=r*(rotate(-90)*A/length(A)); X=extension(A,P,Q,Q+A); Y=extension(B,P,Q,Q+A); R=rotate(90,C)*(C+r*(A-C)/length(A-C)); Z=extension(C,P,R,R+A-C); Zp=circumcenter(X,Y,Z); draw(A--B--C--A); draw(CR(X,r)^^CR(Y,r)^^CR(Z,r), gray); draw(A--P--B^^P--C^^X--Y--Z--X, royalblue); draw(X--foot(X,A,C)^^Z--foot(Z,A,C),royalblue); draw(CR(Zp,r), gray); draw(incircle(A,B,C)^^incircle(X,Y,Z)^^P--foot(P,A,C), heavygreen+0.6); draw(rightanglemark(X,foot(X,A,C),C,10),linewidth(0.6)); draw(rightanglemark(Z,foot(Z,A,C),A,10),linewidth(0.6)); label("$A$",A,N); label("$B$",B,0.15*(B-P)); label("$C$",C,0.1*(C-P)); pen p=fontsize(10)+linewidth(3); dot("$O_A$",X,right,p); dot("$O_B$",Y,left+up,p); dot("$O_C$",Z,down,p); dot("$O$",Zp,dir(-45),p+red); dot("$I$",P,left+up,p); | [] |
138 | Four circles $\omega,$ $\omega_{A},$ $\omega_{B},$ and $\omega_{C}$ with the same radius are drawn in the interior of triangle $ABC$ such that $\omega_{A}$ is tangent to sides $AB$ and $AC$, $\omega_{B}$ to $BC$ and $BA$, $\omega_{C}$ to $CA$ and $CB$, and $\omega$ is externally tangent to $\omega_{A},$ $\omega_{B},$ and $\omega_{C}$. If the sides of triangle $ABC$ are $13,$ $14,$ and $15,$ the radius of $\omega$ can be represented in the form $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n.$
Contents
1 Problem
2 Diagram
3 Solution 1 (Homothety)
4 Solution 2
5 Solution 3 (elementary)
6 Solution 4
7 Sidenote (Generalization)
8 Solution 5
9 Solution 6
10 Video Solution
11 See also | 2007 AIME II Problem 15 | Let $O_A, O_B, O_C, O$ be the centers of $w_A, w_B, w_C, w$, respectively. Also, let $I$ be the incenter of $ABC$ and $r$ be the radius of circle $w$. Since $AB||O_AO_B$, $BC||O_BO_C$, and $CA||O_CO_A$, we know that
\[\angle BAI = \angle O_BO_AI, \angle CBI = \angle O_CO_BI, \angle ACI = \angle O_AO_CI \text{ and }\angle CAI = \angle O_CO_AI, \angle ABI = \angle O_AO_BI, \angle BCI = \angle O_BO_CI.\]
That means $\angle ABC = \angle O_AO_BO_C$, $\angle BAC = \angle O_BO_AO_C$, and $\angle ACB = \angle O_AO_CO_B$. Thus, $\triangle ABC \sim \triangle O_AO_BO_C$. We also know that we are scaling each side of $\triangle ABC$ (from $AB$ to $O_AO_B$ for instance), about $I$ (since A,O_A,I are collinear; same apply with $B$ and $C$).
Now, let the homothety $\mathcal{H} (I, x)$ map $\triangle ABC$ to $\triangle O_AO_BO_C$. To start off, we know the circumradius of $O_AO_BO_C$ is $O$, since $OO_A = OO_B = OO_C = 2r$. Since $O_AO_B = 13x$, $O_BO_C = 14x$, $O_CO_A = 15x$, we can get an relationship involving $x$ and $r$ via another way to find the circumradius:
\[[\triangle O_AO_BO_C ] =\frac{abc}{4R} \Longrightarrow 84x^2 =\frac{13x\cdot 14x\cdot 15x}{4\cdot 2r} \Longrightarrow r=\frac{65x}{16}\]
Take notice of the inradius of $ABC$. We get the inradius to be $[\triangle ABC ] = sr_0 \Longrightarrow r_0=4$. Let the tangency point of the incircle and side $BC$ be $H$. We know $IH = 4$. We also know that we can cut off the part of $IH$ that is outside of $\triangle O_AO_BO_C$ to get the inradius of $\triangle O_AO_BO_C$. To part that is outside $\triangle O_AO_BO_C$ turns out just to be the radius of circle $w_B$ (as seen in the picture). That means the inradius of $\triangle O_AO_BO_C$ is just $4-r$. We can calculate that incradius in another way, though. We know that the inradius of $\triangle ABC$ is $4$, which means the inradius of $\triangle O_AO_BO_C$ is just $4x$ (by our homethety ratio).
Thus, we have $4x = 4-r = 4-\dfrac{65x}{16} \Longrightarrow x = \dfrac{64}{129} \Longrightarrow r = \dfrac{260}{129}$. That gives $\boxed{389}$ as our final answer.
The homethety turned out to be $\mathcal{H} \left(I, \dfrac{64}{129}\right)$
~sml1809 | // Block 1
defaultpen(fontsize(12)+0.8); size(300);
pair A,B,C,X,Y,Z,P,Q,R;
B=origin; C=15*right; A=IP(CR(B,13),CR(C,14)); P=incenter(A,B,C);
real r=260/129;
Q=r*(rotate(-90)*A/length(A)); X=extension(A,P,Q,Q+A); Y=extension(B,P,Q,Q+A);
R=rotate(90,C)*(C+r*(A-C)/length(A-C)); Z=extension(C,P,R,R+A-C);
draw(A--B--C--A); draw(CR(X,r)^^CR(Y,r)^^CR(Z,r)^^A--P--B^^P--C, gray); draw(CR(circumcenter(X,Y,Z),r), gray);
label("$A$",A,N); label("$B$",B,0.15*(B-P)); label("$C$",C,0.1*(C-P));
draw(X--Y--Z--cycle); draw(Y--(3.5,0),blue); draw(P--(7,0), blue);
pen p=fontsize(10)+linewidth(3);
dot("$O_A$",X,right,p); dot("$O_B$",Y,left+up,p); dot("$O_C$",Z,right+up,p); dot("$O$",circumcenter(X,Y,Z),right+down,p); dot("$I$",P,left+up,p); dot("$H$",(7,0),down,p);
// Block 2
defaultpen(fontsize(12)+0.8); size(300); pair A,B,C,X,Y,Z,P,Q,R; B=origin; C=15*right; A=IP(CR(B,13),CR(C,14)); P=incenter(A,B,C); real r=260/129; Q=r*(rotate(-90)*A/length(A)); X=extension(A,P,Q,Q+A); Y=extension(B,P,Q,Q+A); R=rotate(90,C)*(C+r*(A-C)/length(A-C)); Z=extension(C,P,R,R+A-C); draw(A--B--C--A); draw(CR(X,r)^^CR(Y,r)^^CR(Z,r)^^A--P--B^^P--C, gray); draw(CR(circumcenter(X,Y,Z),r), gray); label("$A$",A,N); label("$B$",B,0.15*(B-P)); label("$C$",C,0.1*(C-P)); draw(X--Y--Z--cycle); draw(Y--(3.5,0),blue); draw(P--(7,0), blue); pen p=fontsize(10)+linewidth(3); dot("$O_A$",X,right,p); dot("$O_B$",Y,left+up,p); dot("$O_C$",Z,right+up,p); dot("$O$",circumcenter(X,Y,Z),right+down,p); dot("$I$",P,left+up,p); dot("$H$",(7,0),down,p); | [] |
139 | Square $AIME$ has sides of length $10$ units. Isosceles triangle $GEM$ has base $EM$, and the area common to triangle $GEM$ and square $AIME$ is $80$ square units. Find the length of the altitude to $EM$ in $\triangle GEM$. | 2008 AIME I Problem 2 | Note that if the altitude of the triangle is at most $10$, then the maximum area of the intersection of the triangle and the square is $5\cdot10=50$.
This implies that vertex G must be located outside of square $AIME$.
Let $GE$ meet $AI$ at $X$ and let $GM$ meet $AI$ at $Y$. Clearly, $XY=6$ since the area of trapezoid $XYME$ is $80$. Also, $\triangle GXY \sim \triangle GEM$.
Let the height of $GXY$ be $h$. By the similarity, $\dfrac{h}{6} = \dfrac{h + 10}{10}$, we get $h = 15$. Thus, the height of $GEM$ is $h + 10 = \boxed{025}$. | // Block 1
pair E=(0,0), M=(10,0), I=(10,10), A=(0,10);
draw(A--I--M--E--cycle);
pair G=(5,25);
draw(G--E--M--cycle);
label("\(G\)",G,N);
label("\(A\)",A,NW);
label("\(I\)",I,NE);
label("\(M\)",M,NE);
label("\(E\)",E,NW);
label("\(10\)",(M+E)/2,S);
// Block 2
pair E=(0,0), M=(10,0), I=(10,10), A=(0,10); draw(A--I--M--E--cycle); pair G=(5,25); draw(G--E--M--cycle); label("\(G\)",G,N); label("\(A\)",A,NW); label("\(I\)",I,NE); label("\(M\)",M,NE); label("\(E\)",E,NW); label("\(10\)",(M+E)/2,S); | [] |
140 | Let $ABCD$ be an isosceles trapezoid with $\overline{AD}||\overline{BC}$ whose angle at the longer base $\overline{AD}$ is $\dfrac{\pi}{3}$. The diagonals have length $10\sqrt {21}$, and point $E$ is at distances $10\sqrt {7}$ and $30\sqrt {7}$ from vertices $A$ and $D$, respectively. Let $F$ be the foot of the altitude from $C$ to $\overline{AD}$. The distance $EF$ can be expressed in the form $m\sqrt {n}$, where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. Find $m + n$. | 2008 AIME I Problem 10 | Key observation. $AD = 20\sqrt{7}$.
Proof 1. By the triangle inequality, we can immediately see that $AD \geq 20\sqrt{7}$. However, notice that $10\sqrt{21} = 20\sqrt{7}\cdot\sin\frac{\pi}{3}$, so by the law of sines, when $AD = 20\sqrt{7}$, $\angle ACD$ is right and the circle centered at $A$ with radius $10\sqrt{21}$, which we will call $\omega$, is tangent to $\overline{CD}$. Thus, if $AD$ were increased, $\overline{CD}$ would have to be moved even farther outwards from $A$ to maintain the angle of $\frac{\pi}{3}$ and $\omega$ could not touch it, a contradiction.
Proof 2. Again, use the triangle inequality to obtain $AD \geq 20\sqrt{7}$. Let $x = AD$ and $y = CD$. By the law of cosines on $\triangle ADC$, $2100 = x^2+y^2-xy \iff y^2-xy+(x^2-2100) = 0$. Viewing this as a quadratic in $y$, the discriminant $\Delta$ must satisfy $\Delta = x^2-4(x^2-2100) = 8400-3x^2 \geq 0 \iff x \leq 20\sqrt{7}$. Combining these two inequalities yields the desired conclusion.
This observation tells us that $E$, $A$, and $D$ are collinear, in that order.
Then, $\triangle ADC$ and $\triangle ACF$ are $30-60-90$ triangles. Hence $AF = 15\sqrt {7}$, and
$EF = EA + AF = 10\sqrt {7} + 15\sqrt {7} = 25\sqrt {7}$.
Finally, the answer is $25+7=\boxed{032}$. | // Block 1
size(300);
defaultpen(1);
pair A=(0,0), D=(4,0), B= A+2 expi(1/3*pi), C= D+2expi(2/3*pi), E=(-4/3,0), F=(3,0);
draw(F--C--B--A);
draw(E--A--D--C);
draw(A--C,dashed);
draw(circle(A,abs(C-A)),dotted);
label("\(A\)",A,S);
label("\(B\)",B,NW);
label("\(C\)",C,NE);
label("\(D\)",D,SE);
label("\(E\)",E,N);
label("\(F\)",F,S);
clip(currentpicture,(-1.5,-1)--(5,-1)--(5,3)--(-1.5,3)--cycle);
// Block 2
size(300); defaultpen(1); pair A=(0,0), D=(4,0), B= A+2 expi(1/3*pi), C= D+2expi(2/3*pi), E=(-4/3,0), F=(3,0); draw(F--C--B--A); draw(E--A--D--C); draw(A--C,dashed); draw(circle(A,abs(C-A)),dotted); label("\(A\)",A,S); label("\(B\)",B,NW); label("\(C\)",C,NE); label("\(D\)",D,SE); label("\(E\)",E,N); label("\(F\)",F,S); clip(currentpicture,(-1.5,-1)--(5,-1)--(5,3)--(-1.5,3)--cycle); | [] |
141 | Let
\[p(x,y) = a_0 + a_1x + a_2y + a_3x^2 + a_4xy + a_5y^2 + a_6x^3 + a_7x^2y + a_8xy^2 + a_9y^3.\]
Suppose that
\[p(0,0) = p(1,0) = p( - 1,0) = p(0,1) = p(0, - 1)\\ = p(1,1) = p(1, - 1) = p(2,2) = 0.\]
There is a point $\left(\frac {a}{c},\frac {b}{c}\right)$ for which $p\left(\frac {a}{c},\frac {b}{c}\right) = 0$ for all such polynomials, where $a$, $b$, and $c$ are positive integers, $a$ and $c$ are relatively prime, and $c > 1$. Find $a + b + c$. | 2008 AIME I Problem 13 | Solution 1
\begin{align*} p(0,0) &= a_0 \\ &= 0 \\ p(1,0) &= a_0 + a_1 + a_3 + a_6 \\ &= a_1 + a_3 + a_6 \\ &= 0 \\ p(-1,0) &= -a_1 + a_3 - a_6 \\ &= 0 \end{align*}
Adding the above two equations gives $a_3 = 0$, and so we can deduce that $a_6 = -a_1$.
Similarly, plugging in $(0,1)$ and $(0,-1)$ gives $a_5 = 0$ and $a_9 = -a_2$. Now,
\begin{align*} p(1,1) &= a_0 + a_1 + a_2 + a_3 + a_4 + a_5 + a_6 + a_7 + a_8 + a_9 \\ &= 0 + a_1 + a_2 + 0 + a_4 + 0 - a_1 + a_7 + a_8 - a_2 \\ &= a_4 + a_7 + a_8 \\ &= 0 \\ p(1,-1) &= a_0 + a_1 - a_2 + 0 - a_4 + 0 - a_1 - a_7 + a_8 + a_2 \\ &= -a_4 - a_7 + a_8 \\ &= 0 \end{align*}
Therefore $a_8 = 0$ and $a_7 = -a_4$. Finally,
\begin{align*} p(2,2) &= 0 + 2a_1 + 2a_2 + 0 + 4a_4 + 0 - 8a_1 - 8a_4 +0 - 8a_2 \\ &= -6 a_1 - 6 a_2 - 4 a_4 \\ &= 0 \end{align*}
So, $3a_1 + 3a_2 + 2a_4 = 0$, or equivalently $a_4 = -\frac{3(a_1 + a_2)}{2}$.
Substituting these equations into the original polynomial $p$, we find that at $\left(\frac{a}{c}, \frac{b}{c}\right)$,
\[a_1x + a_2y + a_4xy - a_1x^3 - a_4x^2y - a_2y^3 = 0 \iff\]
\[a_1x + a_2y - \frac{3(a_1 + a_2)}{2}xy - a_1x^3 + \frac{3(a_1 + a_2)}{2}x^2y - a_2y^3 = 0 \iff\]
\[a_1x(x - 1)\left(x + 1 - \frac{3}{2}y\right) + a_2y\left(y^2 - 1 - \frac{3}{2}x(x - 1)\right) = 0\].
The remaining coefficients $a_1$ and $a_2$ are now completely arbitrary because the original equations impose no more restrictions on them. Hence, for the final equation to hold for all possible $p$, we must have $x(x - 1)\left(x + 1 - \frac{3}{2}y\right) = y\left(y^2 - 1 - \frac{3}{2}x(x - 1)\right) = 0$.
As the answer format implies that the $x$-coordinate of the root is non-integral, $x(x - 1)\left(x + 1 - \frac{3}{2}y\right) = 0 \iff x + 1 - \frac{3}{2}y = 0 \iff y = \frac{2}{3}(x + 1)\ (1)$. The format also implies that $y$ is positive, so $y\left(y^2 - 1 - \frac{3}{2}x(x - 1)\right) = 0 \iff y^2 - 1 - \frac{3}{2}x(x - 1) = 0\ (2)$. Substituting $(1)$ into $(2)$ and reducing to a quadratic yields $(19x - 5)(x - 2) = 0$, in which the only non-integral root is $x = \frac{5}{19}$, so $y = \frac{16}{19}$.
The answer is $5 + 16 + 19 = \boxed{040}$.
Solution 2
Consider the cross section of $z = p(x, y)$ on the plane $z = 0$. We realize that we could construct the lines/curves in the cross section such that their equations multiply to match the form of $p(x, y)$ (same degree of $x$ and $y$ in terms) and they include the eight given points. One simple way to do this would be to use the equations $x = 0$, $x = 1$, and $y = \frac{2}{3}x + \frac{2}{3}$, giving us
$p_1(x, y) = x\left(x - 1\right)\left( \frac{2}{3}x - y + \frac{2}{3}\right) = \frac{2}{3}x + xy + \frac{2}{3}x^3-x^2y$.
Another way to do this would to use the line $y = x$ and the ellipse, $x^2 + xy + y^2 = 1$. This would give
$p_2(x, y) = \left(x - y\right)\left(x^2 + xy + y^2 - 1\right) = -x + y + x^3 - y^3$. (But
(Another way would be to use the hyperbola from Solution 1. Interesting that different curves both work.)
At this point, we consider that $p_1$ and $p_2$ both must have $\left(\frac{a}{c}, \frac{b}{c}\right)$ as a zero. A quick graph of the 4 lines and the ellipse used to create $p_1$ and $p_2$ gives nine intersection points. Eight of them are the given ones, and the ninth is $\left(\frac{5}{19}, \frac{16}{19}\right)$. The last intersection point can be found by finding the intersection points of $y = \frac{2}{3}x + \frac{2}{3}$ and $x^2 + xy + y^2 = 1$.
Finally, just add the values of $a$, $b$, and $c$ to get $5 + 16 + 19 = \boxed{040}$ | // Block 1
unitsize(1.2 cm); real upperhyper (real x) { return(sqrt((3*x^2 - 3*x + 2)/2)); } real lowerhyper (real x) { return(-sqrt((3*x^2 - 3*x + 2)/2)); } int i; for (i = -3; i <= 3; ++i) { draw((-3,i)--(3,i),gray(0.7)); draw((i,-3)--(i,3),gray(0.7)); } draw((0,-3)--(0,3),red); draw((1,-3)--(1,3),red); draw((-3,-4/3)--(3,8/3),red); draw((-3,0)--(3,0),blue); draw(graph(upperhyper,-1.863,2.863),blue); draw(graph(lowerhyper,-1.836,2.863),blue); dot("$(0,0)$", (0,0), NE, fontsize(8)); dot("$(1,0)$", (1,0), NE, fontsize(8)); dot("$(-1,0)$", (-1,0), NW, fontsize(8)); dot("$(0,1)$", (0,1), SW, fontsize(8)); dot("$(0,-1)$", (0,-1), NW, fontsize(8)); dot("$(1,1)$", (1,1), SE, fontsize(8)); dot("$(1,-1)$", (1,-1), NE, fontsize(8)); dot("$(2,2)$", (2,2), SE, fontsize(8)); dot((5/19,16/19), green);
// Block 2
unitsize(1.2 cm); real upperellipse(real x) { return((sqrt(4- 3*x^2 )-x)/2); } real lowerellipse(real x) { return((-sqrt(4- 3*x^2 )-x)/2); } int i; for (i = -3; i <= 3; ++i) { draw((-3,i)--(3,i),gray(0.7)); draw((i,-3)--(i,3),gray(0.7)); } draw((0,-3)--(0,3),red); draw((1,-3)--(1,3),red); draw((-3,-4/3)--(3,8/3),red); draw((-3, -3)--(3,3),blue); draw(graph(upperellipse,-1.1547,1.1547),blue); draw(graph(lowerellipse, -1.1547,1.1547),blue); dot("$(0,0)$", (0,0), NE, fontsize(8)); dot("$(1,0)$", (1,0), NE, fontsize(8)); dot("$(-1,0)$", (-1,0), NW, fontsize(8)); dot("$(0,1)$", (0,1), SW, fontsize(8)); dot("$(0,-1)$", (0,-1), NW, fontsize(8)); dot("$(1,1)$", (1,1), SE, fontsize(8)); dot("$(1,-1)$", (1,-1), NE, fontsize(8)); dot("$(2,2)$", (2,2), SE, fontsize(8)); dot((5/19,16/19), green); | [] |
142 | Let $\overline{AB}$ be a diameter of circle $\omega$. Extend $\overline{AB}$ through $A$ to $C$. Point $T$ lies on $\omega$ so that line $CT$ is tangent to $\omega$. Point $P$ is the foot of the perpendicular from $A$ to line $CT$. Suppose $\overline{AB} = 18$, and let $m$ denote the maximum possible length of segment $BP$. Find $m^{2}$. | 2008 AIME I Problem 14 | Solution 1
Let $x = OC$. Since $OT, AP \perp TC$, it follows easily that $\triangle APC \sim \triangle OTC$. Thus $\frac{AP}{OT} = \frac{CA}{CO} \Longrightarrow AP = \frac{9(x-9)}{x}$. By the Law of Cosines on $\triangle BAP$,
\begin{align*}BP^2 = AB^2 + AP^2 - 2 \cdot AB \cdot AP \cdot \cos \angle BAP \end{align*}
where $\cos \angle BAP = \cos (180 - \angle TOA) = - \frac{OT}{OC} = - \frac{9}{x}$, so:
\begin{align*}BP^2 &= 18^2 + \frac{9^2(x-9)^2}{x^2} + 2(18) \cdot \frac{9(x-9)}{x} \cdot \frac 9x = 405 + 729\left(\frac{2x - 27}{x^2}\right)\end{align*}
Let $k = \frac{2x-27}{x^2} \Longrightarrow kx^2 - 2x + 27 = 0$; this is a quadratic, and its discriminant must be nonnegative: $(-2)^2 - 4(k)(27) \ge 0 \Longleftrightarrow k \le \frac{1}{27}$. Thus,
\[BP^2 \le 405 + 729 \cdot \frac{1}{27} = \boxed{432}\]
Equality holds when $x = 27$.~Shen Kislay Kai
Solution 1.1 (Calculus)
Proceed as follows for Solution 1.
Once you approach the function $k=(2x-27)/x^2$, find the maximum value by setting $dk/dx=0$.
Simplifying $k$ to take the derivative, we have $2/x-27/x^2$, so $dk/dx=-2/x^2+54/x^3$. Setting $dk/dx=0$, we have $2/x^2=54/x^3$.
Solving, we obtain $x=27$ as the critical value. Hence, $k$ has the maximum value of $(2*27-27)/27^2=1/27$. Since $BP^2=405+729k$, the maximum value of $\overline {BP}$ occurs at $k=1/27$, so $BP^2$ has a maximum value of $405+729/27=\fbox{432}$.
Note: Please edit this solution if it feels inadequate.
~Shen Kislay Kai
Solution 2
From the diagram, we see that $BQ = OT + BO \sin\theta = 9 + 9\sin\theta = 9(1 + \sin\theta)$, and that $QP = BA\cos\theta = 18\cos\theta$.
\begin{align*}BP^2 &= BQ^2 + QP^2 = 9^2(1 + \sin\theta)^2 + 18^2\cos^2\theta\\ &= 9^2[1 + 2\sin\theta + \sin^2\theta + 4(1 - \sin^2\theta)]\\ BP^2 &= 9^2[5 + 2\sin\theta - 3\sin^2\theta]\end{align*}
This is a quadratic equation, maximized when $\sin\theta = \frac { - 2}{ - 6} = \frac {1}{3}$. Thus, $m^2 = 9^2[5 + \frac {2}{3} - \frac {1}{3}] = \boxed{432}$.
Solution 3 (Calculus Bash)
(Diagram credit goes to Solution 2)
We let $AC=x$. From similar triangles, we have that $PC=\frac{x\sqrt{x^2+18x}}{x+9}$ (Use Pythagorean on $\triangle\omega TC$ and then using $\triangle\omega CT\sim\triangle ACP$). Similarly, $TP=QT=\frac{9\sqrt{x^2+18x}}{x+9}$. Using the Pythagorean Theorem again and $\triangle CAP\sim\triangle CBQ$, $BQ=\sqrt{(x+18)^2-(\frac{(x+18)\sqrt{x^2+18x}}{x+9})^2}$. Using the Pythagorean Theorem $\bold{again}$, $BP=\sqrt{(x+18)^2-(\frac{(x+18)\sqrt{x^2+18x}}{x+9})^2+(\frac{18\sqrt{x^2+18x}}{x+9})^2}$. After a large bashful simplification, $BP=\sqrt{405+\frac{1458x-6561}{x^2+18x+81}}$. The fraction is equivalent to $729\frac{2x-9}{(x+9)^2}$. Taking the derivative of the fraction and solving for x, we get that $x=18$. Plugging $x=18$ back into the expression for $BP$ yields $\sqrt{432}$, so the answer is $(\sqrt{432})^2=\boxed{432}$.
Solution 4
(Diagram credit goes to Solution 2)
Let $AC=x$. The only constraint on $x$ is that it must be greater than $0$. Using similar triangles, we can deduce that $PA=\frac{9x}{x+9}$. Now, apply law of cosines on $\triangle PAB$. \[BP^2=\left(\frac{9x^2}{x+9}\right)^2+18^2-2(18)\left(\frac{9x}{x+9}\right)\cos(\angle PAB).\] We can see that $\cos(\angle PAB)=\cos(180^{\circ}-\angle PAC)=\cos(\angle PAC -90^{\circ})=-\sin(\angle PCA)$. We can find $-\sin(\angle PCA)=-\frac{9}{x+9}$. Plugging this into our equation, we get:
\[BP^2=\left(\frac{9x^2}{x+9}\right)^2+18^2-2(18)\left(\frac{9x}{x+9}\right)\left(-\frac{9}{x+9}\right).\] Eventually, \[BP^2 = 81\left(\frac{x^2+36x}{(x+9)^2}+4\right).\] We want to maximize $\frac{x^2+36x}{(x+9)^2}$. There are many ways to maximize this expression, discussed here: https://artofproblemsolving.com/community/c4h2292700_maximization. The maximum result of that expression is $\frac{4}{3}$. Finally, evaluating $BP^2$ for this value $81\left(\frac{4}{3}+4\right) = \boxed{432}$.
~superagh
Solution 5 (Clean)
Let $h$ be the distance from $A$ to $CT$. Observe that $h$ takes any value from $0$ to $r$, where $r$ is the radius of the circle.
Let $Q$ be the foot of the altitude from $B$ to $CT$. It is clear that $T$ is the midpoint of $PQ$, and so the length $OT$ is the average of $AP$ and $BQ$. It follows thus that $BQ = 2r - h$.
We compute $PT = \sqrt{r^2 - (r - h)^2} = \sqrt{h(2r - h)},$
and so $BP^2 = PQ^2 + BQ^2 = 4PT^2 + BQ^2 = 4h(2r - h) + (2r-h)^2 = (2r-h)(2r + 3h)$.
This is $\frac{1}{3}(6r - 3h)(2r + 3h) \le \frac{1}{3} \cdot \left( \frac{8r}{2} \right)^2$. Equality is attained, so thus we extract the answer of $\frac{16 \cdot 9^2}{3} = 27 \cdot 16 = \boxed{432}.$ | // Block 1
size(250); defaultpen(0.70 + fontsize(10)); import olympiad; pair O = (0,0), B = O - (9,0), A= O + (9,0), C=A+(18,0), T = 9 * expi(-1.2309594), P = foot(A,C,T); draw(Circle(O,9)); draw(B--C--T--O); draw(A--P); dot(A); dot(B); dot(C); dot(O); dot(T); dot(P); draw(rightanglemark(O,T,C,30)); draw(rightanglemark(A,P,C,30)); draw(anglemark(B,A,P,35)); draw(B--P, blue); label("\(A\)",A,NW); label("\(B\)",B,NW); label("\(C\)",C,NW); label("\(O\)",O,NW); label("\(P\)",P,SE); label("\(T\)",T,SE); label("\(9\)",(O+A)/2,N); label("\(9\)",(O+B)/2,N); label("\(x-9\)",(C+A)/2,N);
// Block 2
unitsize(3mm); pair B=(0,13.5), C=(23.383,0); pair O=(7.794, 9), P=(2*7.794,0); pair T=(7.794,0), Q=(0,0); pair A=(2*7.794,4.5); draw(Q--B--C--Q); draw(O--T); draw(A--P); draw(Circle(O,9)); dot(A);dot(B);dot(C);dot(T);dot(P);dot(O);dot(Q); label("\(B\)",B,NW); label("\(A\)",A,NE); label("\(O\)",O,N); label("\(P\)",P,S); label("\(T\)",T,S); label("\(Q\)",Q,S); label("\(C\)",C,E); label("\(\theta\)",C + (-1.7,-0.2), NW); label("\(9\)", (B+O)/2, N); label("\(9\)", (O+A)/2, N); label("\(9\)", (O+T)/2,W);
// Block 3
unitsize(3mm); pair B=(0,13.5), C=(23.383,0); pair O=(7.794, 9), P=(2*7.794,0); pair T=(7.794,0), Q=(0,0); pair A=(2*7.794,4.5); draw(Q--B--C--Q); draw(O--T); draw(A--P); draw(Circle(O,9)); dot(A);dot(B);dot(C);dot(T);dot(P);dot(O);dot(Q); label("\(B\)",B,NW); label("\(A\)",A,NE); label("\(\omega\)",O,N); label("\(P\)",P,S); label("\(T\)",T,S); label("\(Q\)",Q,S); label("\(C\)",C,E); label("\(9\)", (B+O)/2, N); label("\(9\)", (O+A)/2, N); label("\(9\)", (O+T)/2,W);
// Block 4
unitsize(3mm); pair B=(0,13.5), C=(23.383,0); pair O=(7.794, 9), P=(2*7.794,0); pair T=(7.794,0), Q=(0,0); pair A=(2*7.794,4.5); draw(Q--B--C--Q); draw(O--T); draw(A--P); draw(Circle(O,9)); dot(A);dot(B);dot(C);dot(T);dot(P);dot(O);dot(Q); label("\(B\)",B,NW); label("\(A\)",A,NE); label("\(\omega\)",O,N); label("\(P\)",P,S); label("\(T\)",T,S); label("\(Q\)",Q,S); label("\(C\)",C,E); label("\(9\)", (B+O)/2, N); label("\(9\)", (O+A)/2, N); label("\(9\)", (O+T)/2,W); | [] |
142 | Let $\overline{AB}$ be a diameter of circle $\omega$. Extend $\overline{AB}$ through $A$ to $C$. Point $T$ lies on $\omega$ so that line $CT$ is tangent to $\omega$. Point $P$ is the foot of the perpendicular from $A$ to line $CT$. Suppose $\overline{AB} = 18$, and let $m$ denote the maximum possible length of segment $BP$. Find $m^{2}$. | 2008 AIME I Problem 14 | (Diagram Credits to Solution 2)
We can see that $AP = \frac{9x}{x+9}$. From the Law of Cosines, we can now find the length of BP.
$BP^2 = (\frac{9x}{x+9})^2 + 18^2 - 2(18)(\frac{9x}{x+9})(-\frac{9}{x+9})$
$BP^2 = 81(\frac{x^2}{(x+9)^2} + 4 + \frac{36x}{(x+9)^2})$
$BP^2 = 81(\frac{x^2+36x}{(x+9)^2}+4)$
From here, we see that the value we need to maximize is $\frac{x^2+36x}{(x+9)^2}$. We shall now do some calculus. Let $f(x) = \frac{x^2+36x}{(x+9)^2}$. We need to find the first derivative. Using the quotient rule $(\frac{u}{v})' = \frac{u'v - uv'}{v^2}$, we can find the derivative which is:
$f'(x) = \frac{(2x+36)(x+9)^2-(x^2+36x)(2)(x+9)}{((x+9)^2)^2}$
This simplifies to
$f'(x) = \frac{-18x+324}{(x+9)^3}$
Now we find the critical point by setting $f'(x)$ to zero and solve for x.
$-18x + 324 = 0$
$-18x = -324$
$x = 18$
We now plug this value of x back into the function to find the max value of f(x)
$f(18) = \frac{18^2+36(18)}{27^2}$
$f(18) = \frac{4}{3}$
We now go back to the formula for $BP^2$ and plug in $\frac{4}{3}$ for that ugly fraction.
$BP^2 = 81(\frac{4}{3}+4)$
$BP^2 = 81(\frac{16}{3})$
$BP^2 = 432$
Thus, the value of $m^2$ (or $BP^2$) with the maximum value of $m$ is $\boxed{432}$.
~ROGER8432V3 | // Block 1
size(250); defaultpen(0.70 + fontsize(10)); import olympiad;
pair O = (0,0), B = O - (9,0), A= O + (9,0), C=A+(18,0), T = 9 * expi(-1.2309594), P = foot(A,C,T);
draw(Circle(O,9)); draw(B--C--T--O); draw(A--P); dot(A); dot(B); dot(C); dot(O); dot(T); dot(P);
draw(rightanglemark(O,T,C,30)); draw(rightanglemark(A,P,C,30)); draw(anglemark(B,A,P,35));
draw(B--P, blue);
label("\(A\)",A,NW);
label("\(B\)",B,NW);
label("\(C\)",C,NW);
label("\(O\)",O,NW);
label("\(P\)",P,SE);
label("\(T\)",T,SE);
label("\(9\)",(O+A)/2,N);
label("\(9\)",(O+B)/2,N);
label("\(x\)",(C+A)/2,N);
// Block 2
size(250); defaultpen(0.70 + fontsize(10)); import olympiad; pair O = (0,0), B = O - (9,0), A= O + (9,0), C=A+(18,0), T = 9 * expi(-1.2309594), P = foot(A,C,T); draw(Circle(O,9)); draw(B--C--T--O); draw(A--P); dot(A); dot(B); dot(C); dot(O); dot(T); dot(P); draw(rightanglemark(O,T,C,30)); draw(rightanglemark(A,P,C,30)); draw(anglemark(B,A,P,35)); draw(B--P, blue); label("\(A\)",A,NW); label("\(B\)",B,NW); label("\(C\)",C,NW); label("\(O\)",O,NW); label("\(P\)",P,SE); label("\(T\)",T,SE); label("\(9\)",(O+A)/2,N); label("\(9\)",(O+B)/2,N); label("\(x\)",(C+A)/2,N); | [] |
143 | A square piece of paper has sides of length $100$. From each corner a wedge is cut in the following manner: at each corner, the two cuts for the wedge each start at a distance $\sqrt{17}$ from the corner, and they meet on the diagonal at an angle of $60^{\circ}$ (see the figure below). The paper is then folded up along the lines joining the vertices of adjacent cuts. When the two edges of a cut meet, they are taped together. The result is a paper tray whose sides are not at right angles to the base. The height of the tray, that is, the perpendicular distance between the plane of the base and the plane formed by the upper edges, can be written in the form $\sqrt[n]{m}$, where $m$ and $n$ are positive integers, $m<1000$, and $m$ is not divisible by the $n$th power of any prime. Find $m+n$. | 2008 AIME I Problem 15 | In the original picture, let $P$ be the corner, and $M$ and $N$ be the two points whose distance is $\sqrt{17}$ from $P$. Also, let $R$ be the point where the two cuts intersect.
Using $\triangle{MNP}$ (a 45-45-90 triangle), $MN=MP\sqrt{2}\quad\Longrightarrow\quad MN=\sqrt{34}$. $\triangle{MNR}$ is equilateral, so $MR = NR = \sqrt{34}$. (Alternatively, we could find this by the Law of Sines.)
The length of the perpendicular from $P$ to $MN$ in $\triangle{MNP}$ is $\frac{\sqrt{17}}{\sqrt{2}}$, and the length of the perpendicular from $R$ to $MN$ in $\triangle{MNR}$ is $\frac{\sqrt{51}}{\sqrt{2}}$. Adding those two lengths, $PR=\frac{\sqrt{17}+\sqrt{51}}{\sqrt{2}}$. (Alternatively, we could have used that $\sin 75^{\circ} = \sin (30+45) = \frac{\sqrt{6}+\sqrt{2}}{4}$.)
Drop a perpendicular from $R$ to the side of the square containing $M$ and let the intersection be $G$.
\begin{align*}PG&=\frac{PR}{\sqrt{2}}=\frac{\sqrt{17}+\sqrt{51}}{2}\\ MG=PG-PM&=\frac{\sqrt{17}+\sqrt{51}}{2}-\sqrt{17}=\frac{\sqrt{51}-\sqrt{17}}{2}\end{align*}
Let $A'B'C'D'$ be the smaller square base of the tray and let $ABCD$ be the larger square, such that $AA'$, etc, are edges. Let $F$ be the foot of the perpendicular from $A$ to plane $A'B'C'D'$.
We know $AA'=MR=\sqrt{34}$ and $A'F=MG\sqrt{2}=\frac{\sqrt{51}-\sqrt{17}}{\sqrt{2}}$. Now, use the Pythagorean Theorem on triangle $AFA'$ to find $AF$:
\begin{align*}\left(\frac{\sqrt{51}-\sqrt{17}}{\sqrt{2}}\right)^2+AF^2&=\left(\sqrt{34}\right)^2\\ \frac{51-34\sqrt{3}+17}{2}+AF^2&=34\\AF&=\sqrt{34-\frac{68-34\sqrt{3}}{2}}\\AF&=\sqrt{\frac{34\sqrt{3}}{2}}\\AF&=\sqrt[4]{867}\end{align*}
The answer is $867 + 4 = \boxed{871}$. | // Block 1
import cse5;
size(200);
pathpen=black;
real s=sqrt(17), r=(sqrt(51)+s)/(sqrt(2));
pair P=(0,0), N=(0,sqrt(17)), M=(sqrt(17),0), R=r*dir(45), G=((sqrt(51)+sqrt(17))/2,0);
D(2*N--P--2*M); D(N--R--M); D(P--R);
D((R.x,2*N.y)--R--(2*M.x,R.y));
MP("30^\circ",R-(0.25,1),SW);
MP("30^\circ",R-(1,0.5),SW);
MP("\sqrt{17}",N/2,W);
MP("\sqrt{17}",M/2,S);
D(N--M,dashed);
D(G--R,dashed);
MP("P",P,SW); MP("N",N,SW); MP("M",M,SW); MP("R",R,NE);
MP("G",G,SW);
// Block 2
import cse5; size(200); pathpen=black; real s=sqrt(17), r=(sqrt(51)+s)/(sqrt(2)); pair P=(0,0), N=(0,sqrt(17)), M=(sqrt(17),0), R=r*dir(45), G=((sqrt(51)+sqrt(17))/2,0); D(2*N--P--2*M); D(N--R--M); D(P--R); D((R.x,2*N.y)--R--(2*M.x,R.y)); MP("30^\circ",R-(0.25,1),SW); MP("30^\circ",R-(1,0.5),SW); MP("\sqrt{17}",N/2,W); MP("\sqrt{17}",M/2,S); D(N--M,dashed); D(G--R,dashed); MP("P",P,SW); MP("N",N,SW); MP("M",M,SW); MP("R",R,NE); MP("G",G,SW); | [] |
144 | In trapezoid $ABCD$ with $\overline{BC}\parallel\overline{AD}$, let $BC = 1000$ and $AD = 2008$. Let $\angle A = 37^\circ$, $\angle D = 53^\circ$, and $M$ and $N$ be the midpoints of $\overline{BC}$ and $\overline{AD}$, respectively. Find the length $MN$. | 2008 AIME II Problem 5 | Solution 1
Extend $\overline{AB}$ and $\overline{CD}$ to meet at a point $E$. Then $\angle AED = 180 - 53 - 37 = 90^{\circ}$.
As $\angle AED = 90^{\circ}$, note that the midpoint of $\overline{AD}$, $N$, is the center of the circumcircle of $\triangle AED$. We can do the same with the circumcircle about $\triangle BEC$ and $M$ (or we could apply the homothety to find $ME$ in terms of $NE$). It follows that
\[NE = ND = \frac {AD}{2} = 1004, \quad ME = MC = \frac {BC}{2} = 500.\]
Thus $MN = NE - ME = \boxed{504}$.
For purposes of rigor we will show that $E,M,N$ are collinear. Since $\overline{BC} \parallel \overline{AD}$, then $BC$ and $AD$ are homothetic with respect to point $E$ by a ratio of $\frac{BC}{AD} = \frac{125}{251}$. Since the homothety carries the midpoint of $\overline{BC}$, $M$, to the midpoint of $\overline{AD}$, which is $N$, then $E,M,N$ are collinear.
Solution 2
Let $F,G,H$ be the feet of the perpendiculars from $B,C,M$ onto $\overline{AD}$, respectively. Let $x = NH$, so $DG = 1004 - 500 - x = 504 - x$ and $AF = 1004 - (500 - x) = 504 + x$. Also, let $h = BF = CG = HM$.
By AA~, we have that $\triangle AFB \sim \triangle CGD$, and so
\[\frac{BF}{AF} = \frac {DG}{CG} \Longleftrightarrow \frac{h}{504+x} = \frac{504-x}{h} \Longrightarrow x^2 + h^2 = 504^2.\]
By the Pythagorean Theorem on $\triangle MHN$,
\[MN^{2} = x^2 + h^2 = 504^2,\] so $MN = \boxed{504}$.
Solution 3
If you drop perpendiculars from $B$ and $C$ to $AD$, and call the points where they meet $\overline{AD}$, $E$ and $F$ respectively, then $FD = x$ and $EA = 1008-x$ , and so you can solve an equation in tangents. Since $\angle{A} = 37$ and $\angle{D} = 53$, you can solve the equation [by cross-multiplication]:
\begin{align*}\tan{37}\times (1008-x) &= \tan{53} \times x\\ \frac{(1008-x)}{x} &= \frac{\tan{53}}{\tan{37}} = \frac{\sin{53}}{\cos{53}} \times\frac{\cos{37}}{\sin{37}}\end{align*}
However, we know that $\cos{90-x} = \sin{x}$ and $\sin{90-x} = \cos{x}$ are co-functions. Applying this,
\begin{align*}\frac{(1008-x)}{x} &= \frac{\sin^2{53}}{\cos^2{53}} \\ x\sin^2{53} &= 1008\cos^2{53} - x\cos^2{53}\\ x(\sin^2{53} + \cos^2{53}) &= 1008\cos^2{53}\\ x = 1008\cos^2{53} &\Longrightarrow 1008-x = 1008\sin^2{53} \end{align*}
Now, if we can find $1004 - (EA + 500)$, and the height of the trapezoid, we can create a right triangle and use the Pythagorean Theorem to find $MN$.
The leg of the right triangle along the horizontal is:
\[1004 - 1008\sin^2{53} - 500 = 504 - 1008\sin^2{53}.\]
Now to find the other leg of the right triangle (also the height of the trapezoid), we can simplify the following expression:
\begin{align*}\tan{37} \times 1008 \sin^2{53} = \tan{37} \times 1008 \cos^2{37} = 1008\cos{37}\sin{37} = 504\sin74\end{align*}
Now we used Pythagorean Theorem and get that $MN$ is equal to:
\begin{align*}&\sqrt{(1008\sin^2{53} + 500 -1004)^2 + (504\sin{74})^2} = 504\sqrt{1-2\sin^2{53} + \sin^2{74}} \end{align*}
However, $1-2\sin^2{53} = \cos^2{106}$ and $\sin^2{74} = \sin^2{106}$ so now we end up with:
\[504\sqrt{\cos^2{106} + \sin^2{106}} =\fbox{504}.\]
Solution 4
Plot the trapezoid such that $B=\left(1000\cos 37^\circ, 0\right)$, $C=\left(0, 1000\sin 37^\circ\right)$, $A=\left(2008\cos 37^\circ, 0\right)$, and $D=\left(0, 2008\sin 37^\circ\right)$.
The midpoints of the requested sides are $\left(500\cos 37^\circ, 500\sin 37^\circ\right)$ and $\left(1004\cos 37^\circ, 1004\sin 37^\circ\right)$.
To find the distance from $M$ to $N$, we simply apply the distance formula and the Pythagorean identity $\sin^2 x + \cos^2 x = 1$ to get $MN=\boxed{504}$.
Solution 5
Similar to solution 1; Notice that it forms a right triangle. Remembering that the median to the hypotenuse is simply half the length of the hypotenuse, we quickly see that the length is $\frac{2008}{2}-\frac{1000}{2}=504$.
Solution 6
Obviously, these angles are random--the only special thing about them is that they add up to 90. So we might as well let the given angles equal 45 and 45, and now the answer is trivially $\boxed{504}$. (The trapezoid is isosceles, and you see two 45-45-90 triangles;from there you can get the answer.) | // Block 1
size(220); defaultpen(0.7+fontsize(10)); real f=100, r=1004/f; pair A=(0,0), D=(2*r, 0), N=(r,0), E=N+r*expi(74*pi/180); pair B=(126*A+125*E)/251, C=(126*D + 125*E)/251; pair[] M = intersectionpoints(N--E,B--C); draw(A--B--C--D--cycle); draw(B--E--C,dashed); draw(M[0]--N); draw(N--E,dashed); draw(rightanglemark(D,E,A,2)); picture p = new picture; draw(p,Circle(N,r),dashed+linewidth(0.5)); clip(p,A--D--D+(0,20)--A+(0,20)--cycle); add(p); label("\(A\)",A,SW); label("\(B\)",B,NW); label("\(C\)",C,NE); label("\(D\)",D,SE); label("\(E\)",E,NE); label("\(M\)",M[0],SW); label("\(N\)",N,S); label("\(1004\)",(N+D)/2,S); label("\(500\)",(M[0]+C)/2,S);
// Block 2
size(220); defaultpen(0.7+fontsize(10)); real f=100, r=1004/f; pair A=(0,0), D=(2*r, 0), N=(r,0), E=N+r*expi(74*pi/180); pair B=(126*A+125*E)/251, C=(126*D + 125*E)/251; pair[] M = intersectionpoints(N--E,B--C); pair F = foot(B,A,D), G=foot(C,A,D), H=foot(M[0],A,D); draw(A--B--C--D--cycle); draw(M[0]--N); draw(B--F,dashed); draw(C--G,dashed); draw(M[0]--H,dashed); label("\(A\)",A,SW); label("\(B\)",B,NW); label("\(C\)",C,NE); label("\(D\)",D,NE); label("\(F\)",F,S); label("\(G\)",G,SW); label("\(M\)",M[0],SW); label("\(N\)",N,S); label("\(H\)",H,S); label("\(x\)",(N+H)/2+(0,1),S); label("\(h\)",(B+F)/2,W); label("\(h\)",(C+G)/2,W); label("\(1000\)",(B+C)/2,NE); label("\(504-x\)",(G+D)/2,S); label("\(504+x\)",(A+F)/2,S); label("\(h\)",(M[0]+H)/2,(1,0)); | [] |
145 | The diagram below shows a $4\times4$ rectangular array of points, each of which is $1$ unit away from its nearest neighbors.
Define a growing path to be a sequence of distinct points of the array with the property that the distance between consecutive points of the sequence is strictly increasing. Let $m$ be the maximum possible number of points in a growing path, and let $r$ be the number of growing paths consisting of exactly $m$ points. Find $mr$. | 2008 AIME II Problem 10 | We label our points using coordinates $0 \le x,y \le 3$, with the bottom-left point being $(0,0)$. By the Pythagorean Theorem, the distance between two points is $\sqrt{d_x^2 + d_y^2}$ where $0 \le d_x, d_y \le 3$; these yield the possible distances (in decreasing order)
\[\sqrt{18},\ \sqrt{13},\ \sqrt{10},\ \sqrt{9},\ \sqrt{8},\ \sqrt{5},\ \sqrt{4},\ \sqrt{2},\ \sqrt{1}\]
As these define $9$ lengths, so the maximum value of $m$ is $10$. For now, we assume that $m = 10$ is achievable. Because it is difficult to immediately impose restrictions on a path with increasing distances, we consider the paths in shrinking fashion. Note that the shrinking paths and growing paths are equivalent, but there are restrictions upon the locations of the first edges of the former.
The $\sqrt{18}$ length is only possible for one of the long diagonals, so our path must start with one of the $4$ corners of the grid. Without loss of generality (since the grid is rotationally symmetric), we let the vertex be $(0,0)$ and the endpoint be $(3,3)$.
The $\sqrt{13}$ length can now only go to $2$ points; due to reflectional symmetry about the main diagonal, we may WLOG let the next endpoint be $(1,0)$.
From $(1,0)$, there are two possible ways to move $\sqrt{10}$ away, either to $(0,3)$ or $(2,3)$. However, from $(0,3)$, there is no way to move $\sqrt{9}$ away, so we discard it as a possibility.
From $(2,3)$, the lengths of $\sqrt{8},\ \sqrt{5},\ \sqrt{4},\ \sqrt{2}$ fortunately are all determined, with the endpoint sequence being $(2,3)-(2,0)-(0,2)-(2,1)-(0,1)-(1,2)$.
From $(1,2)$, there are $3$ possible lengths of $\sqrt{1}$ (to either $(1,1),(2,2),(1,3)$). Thus, the number of paths is $r = 4 \cdot 2 \cdot 3 = 24$, and the answer is $mr = 10 \cdot 24 = \boxed{240}$. | // Block 1
unitsize(0.25inch);
defaultpen(linewidth(0.7)); dotfactor = 4; pen s = linewidth(4);
int i, j;
for(i = 0; i < 4; ++i)
for(j = 0; j < 4; ++j)
dot(((real)i, (real)j));
dot((0,0)^^(3,3),s); draw((0,0)--(3,3));
// Block 2
unitsize(0.25inch);
defaultpen(linewidth(0.7)); dotfactor = 4; pen s = linewidth(4); pen c = rgb(0.5,0.5,0.5);
int i, j;
for(i = 0; i < 4; ++i)
for(j = 0; j < 4; ++j)
dot(((real)i, (real)j));
dot((0,0)^^(3,3)^^(1,0),s); draw((0,0)--(3,3),c); draw((3,3)--(1,0));
// Block 3
unitsize(0.25inch);
defaultpen(linewidth(0.7)); dotfactor = 4; pen s = linewidth(4); pen c = rgb(0.5,0.5,0.5);
int i, j;
for(i = 0; i < 4; ++i)
for(j = 0; j < 4; ++j)
dot(((real)i, (real)j));
dot((0,0)^^(3,3)^^(1,0)^^(2,3)^^(2,0)^^(0,2)^^(2,1)^^(0,1)^^(1,2),s); draw((0,0)--(3,3)--(1,0)--(2,3)--(2,0)--(0,2)--(2,1)--(0,1)--(1,2));
// Block 4
unitsize(0.25inch); defaultpen(linewidth(0.7)); dotfactor = 4; pen s = linewidth(4); int i, j; for(i = 0; i < 4; ++i) for(j = 0; j < 4; ++j) dot(((real)i, (real)j)); dot((0,0)^^(3,3),s); draw((0,0)--(3,3));
// Block 5
unitsize(0.25inch); defaultpen(linewidth(0.7)); dotfactor = 4; pen s = linewidth(4); pen c = rgb(0.5,0.5,0.5); int i, j; for(i = 0; i < 4; ++i) for(j = 0; j < 4; ++j) dot(((real)i, (real)j)); dot((0,0)^^(3,3)^^(1,0),s); draw((0,0)--(3,3),c); draw((3,3)--(1,0));
// Block 6
unitsize(0.25inch); defaultpen(linewidth(0.7)); dotfactor = 4; pen s = linewidth(4); pen c = rgb(0.5,0.5,0.5); int i, j; for(i = 0; i < 4; ++i) for(j = 0; j < 4; ++j) dot(((real)i, (real)j)); dot((0,0)^^(3,3)^^(1,0)^^(2,3)^^(2,0)^^(0,2)^^(2,1)^^(0,1)^^(1,2),s); draw((0,0)--(3,3)--(1,0)--(2,3)--(2,0)--(0,2)--(2,1)--(0,1)--(1,2)); | [] |
146 | In triangle $ABC$, $AB = AC = 100$, and $BC = 56$. Circle $P$ has radius $16$ and is tangent to $\overline{AC}$ and $\overline{BC}$. Circle $Q$ is externally tangent to $P$ and is tangent to $\overline{AB}$ and $\overline{BC}$. No point of circle $Q$ lies outside of $\triangle ABC$. The radius of circle $Q$ can be expressed in the form $m - n\sqrt {k}$, where $m$, $n$, and $k$ are positive integers and $k$ is the product of distinct primes. Find $m + nk$. | 2008 AIME II Problem 11 | Let $X$ and $Y$ be the feet of the perpendiculars from $P$ and $Q$ to $BC$, respectively. Let the radius of $\odot Q$ be $r$. We know that $PQ = r + 16$. From $Q$ draw segment $\overline{QM} \parallel \overline{BC}$ such that $M$ is on $PX$. Clearly, $QM = XY$ and $PM = 16-r$. Also, we know $QPM$ is a right triangle.
To find $XC$, consider the right triangle $PCX$. Since $\odot P$ is tangent to $\overline{AC},\overline{BC}$, then $PC$ bisects $\angle ACB$. Let $\angle ACB = 2\theta$; then $\angle PCX = \angle QBX = \theta$. Dropping the altitude from $A$ to $BC$, we recognize the $7 - 24 - 25$ right triangle, except scaled by $4$.
So we get that $\tan(2\theta) = 24/7$. From the half-angle identity, we find that $\tan(\theta) = \frac {3}{4}$. Therefore, $XC = \frac {64}{3}$. By similar reasoning in triangle $QBY$, we see that $BY = \frac {4r}{3}$.
We conclude that $XY = 56 - \frac {4r + 64}{3} = \frac {104 - 4r}{3}$.
So our right triangle $QPM$ has sides $r + 16$, $r - 16$, and $\frac {104 - 4r}{3}$.
By the Pythagorean Theorem, simplification, and the quadratic formula, we can get $r = 44 - 6\sqrt {35}$, for a final answer of $\fbox{254}$. | // Block 1
size(200);
pathpen=black;pointpen=black;pen f=fontsize(9);
real r=44-6*35^.5;
pair A=(0,96),B=(-28,0),C=(28,0),X=C-(64/3,0),Y=B+(4*r/3,0),P=X+(0,16),Q=Y+(0,r),M=foot(Q,X,P);
path PC=CR(P,16),QC=CR(Q,r);
D(A--B--C--cycle); D(Y--Q--P--X); D(Q--M); D(P--C,dashed);
D(PC); D(QC);
MP("A",A,N,f);MP("B",B,f);MP("C",C,f);MP("X",X,f);MP("Y",Y,f);D(MP("P",P,NW,f));D(MP("Q",Q,NW,f));
// Block 2
size(200); pathpen=black;pointpen=black;pen f=fontsize(9); real r=44-6*35^.5; pair A=(0,96),B=(-28,0),C=(28,0),X=C-(64/3,0),Y=B+(4*r/3,0),P=X+(0,16),Q=Y+(0,r),M=foot(Q,X,P); path PC=CR(P,16),QC=CR(Q,r); D(A--B--C--cycle); D(Y--Q--P--X); D(Q--M); D(P--C,dashed); D(PC); D(QC); MP("A",A,N,f);MP("B",B,f);MP("C",C,f);MP("X",X,f);MP("Y",Y,f);D(MP("P",P,NW,f));D(MP("Q",Q,NW,f)); | [] |
147 | A regular hexagon with center at the origin in the complex plane has opposite pairs of sides one unit apart. One pair of sides is parallel to the imaginary axis. Let $R$ be the region outside the hexagon, and let $S = \left\lbrace\frac{1}{z}|z \in R\right\rbrace$. Then the area of $S$ has the form $a\pi + \sqrt{b}$, where $a$ and $b$ are positive integers. Find $a + b$. | 2008 AIME II Problem 13 | If a point $z = r\text{cis}\,\theta$ is in $R$, then the point $\frac{1}{z} = \frac{1}{r} \text{cis}\, \left(-\theta\right)$ is in $S$ (where cis denotes $\text{cis}\, \theta = \cos \theta + i \sin \theta$). Since $R$ is symmetric every $60^{\circ}$ about the origin, it suffices to consider the area of the result of the transformation when $-30 \le \theta \le 30$, and then to multiply by $6$ to account for the entire area.
We note that if the region $S_2 = \left\lbrace\frac{1}{z}|z \in R_2\right\rbrace$, where $R_2$ is the region (in green below) outside the circle of radius $1/\sqrt{3}$ centered at the origin, then $S_2$ is simply the region inside a circle of radius $\sqrt{3}$ centered at the origin. It now suffices to find what happens to the mapping of the region $R-R_2$ (in blue below).
The equation of the hexagon side in that region is $x = r \cos \theta = \frac{1}{2}$, which is transformed to $\frac{1}{r} \cos (-\theta) = \frac{1}{r} \cos \theta =$2 . Let $r\text{cis}\,\theta = a+bi$ where $a,b \in \mathbb{R}$; then $r = \sqrt{a^2 + b^2}, \cos \theta = \frac{a}{\sqrt{a^2 + b^2}}$, so the equation becomes $a^2 - 2a + b^2 = 0 \Longrightarrow (a-1)^2 + b^2 = 1$. Hence the side is sent to an arc of the unit circle centered at $(1,0)$, after considering the restriction that the side of the hexagon is a segment of length $1/\sqrt{3}$.
Including $S_2$, we find that $S$ is the union of six unit circles centered at $\text{cis}\, \frac{k\pi}{6}$, $k = 0,1,2,3,4,5$, as shown below.
$\Longrightarrow$
The area of the regular hexagon is $6 \cdot \left( \frac{\left(\sqrt{3}\right)^2 \sqrt{3}}{4} \right) = \frac{9}{2}\sqrt{3}$. The total area of the six $120^{\circ}$ sectors is $6\left(\frac{1}{3}\pi - \frac{1}{2} \cdot \frac{1}{2} \cdot \sqrt{3}\right) = 2\pi - \frac{3}{2}\sqrt{3}$. Their sum is $2\pi + \sqrt{27}$, and $a+b = \boxed{029}$.
- Th3Numb3rThr33 | // Block 1
unitsize(1.5cm);
defaultpen(linewidth(0.7)); picture p; real max = .5 + 1/3^.5; pen d = linetype("4 4"); fill(1.5*expi(-pi/6)--arc((0,0),1,-30,30)--1.5*expi(pi/6)--cycle,rgb(0.5,1,0.5));fill(arc((0,0),1,-30,30)--cycle,rgb(0.5,0.5,1));
draw(p,expi(pi/6)--expi(-pi/6)--(0,0)--cycle);for(int i = 0; i < 6; ++i) add(rotate(i*60)*p); draw((0,max)--(0,-max),d,Arrows(4));draw((max,0)--(-max,0),d,Arrows(4)); draw(Circle((0,0),1),d); draw(expi(pi/6)--1.5*expi(pi/6),EndArrow(4)); draw(expi(-pi/6)--1.5*expi(-pi/6),EndArrow(4)); label("$1/\sqrt{3}$",(0,-0.5),W,fontsize(8));
// Block 2
unitsize(4.5cm);
defaultpen(linewidth(0.7)); picture p; fill((0,0)--arc((0,0),1,-30,30)--cycle,rgb(0.5,1,0.5));fill(arc((0,0),1,-30,30)--arc(1/3^.5,1/3^.5,60,-60)--cycle,rgb(0.5,0.5,1));
draw(p,expi(pi/6)--expi(-pi/6)--(0,0)--cycle);draw(p,arc(1/3^.5,1/3^.5,-60,60)); draw(arc(1/3^.5*expi(pi/3),1/3^.5,120,359.99),linetype("4 4")); draw(expi(pi/2)--1/3^.5*expi(pi/3)--expi(pi/6),linetype("4 4")); draw(Circle((0,0),1),linetype("4 4")); label("$\sqrt{3}$",(0,-0.5),W,fontsize(8));
add(p);add(rotate(60)*p);add(rotate(120)*p);add(rotate(180)*p);add(rotate(240)*p);add(rotate(300)*p);
// Block 3
unitsize(1.5cm); defaultpen(linewidth(0.7)); picture p; real max = .5 + 1/3^.5; pen d = linetype("4 4"); fill(1.5*expi(-pi/6)--arc((0,0),1,-30,30)--1.5*expi(pi/6)--cycle,rgb(0.5,1,0.5));fill(arc((0,0),1,-30,30)--cycle,rgb(0.5,0.5,1)); draw(p,expi(pi/6)--expi(-pi/6)--(0,0)--cycle);for(int i = 0; i < 6; ++i) add(rotate(i*60)*p); draw((0,max)--(0,-max),d,Arrows(4));draw((max,0)--(-max,0),d,Arrows(4)); draw(Circle((0,0),1),d); draw(expi(pi/6)--1.5*expi(pi/6),EndArrow(4)); draw(expi(-pi/6)--1.5*expi(-pi/6),EndArrow(4)); label("$1/\sqrt{3}$",(0,-0.5),W,fontsize(8));
// Block 4
unitsize(4.5cm); defaultpen(linewidth(0.7)); picture p; fill((0,0)--arc((0,0),1,-30,30)--cycle,rgb(0.5,1,0.5));fill(arc((0,0),1,-30,30)--arc(1/3^.5,1/3^.5,60,-60)--cycle,rgb(0.5,0.5,1)); draw(p,expi(pi/6)--expi(-pi/6)--(0,0)--cycle);draw(p,arc(1/3^.5,1/3^.5,-60,60)); draw(arc(1/3^.5*expi(pi/3),1/3^.5,120,359.99),linetype("4 4")); draw(expi(pi/2)--1/3^.5*expi(pi/3)--expi(pi/6),linetype("4 4")); draw(Circle((0,0),1),linetype("4 4")); label("$\sqrt{3}$",(0,-0.5),W,fontsize(8)); add(p);add(rotate(60)*p);add(rotate(120)*p);add(rotate(180)*p);add(rotate(240)*p);add(rotate(300)*p); | [] |
148 | There is a complex number $z$ with imaginary part $164$ and a positive integer $n$ such that
\[\frac {z}{z + n} = 4i.\]
Find $n$. | 2009 AIME I Problem 2 | Below is an image of the complex plane. Let $\operatorname{Im}(z)$ denote the imaginary part of a complex number $z$.
$z$ must lie on the line $\operatorname{Im}(z)=164$. $z+n$ must also lie on the same line, since $n$ is real and does not affect the imaginary part of $z$.
Consider $z$ and $z+n$ in terms of their magnitude (distance from the origin) and phase (angle formed by the point, the origin, and the positive real axis, measured counterclockwise from said axis). When multiplying/dividing two complex numbers, you can multiply/divide their magnitudes and add/subtract their phases to get the magnitude and phase of the product/quotient. Expressed in a formula, we have $z_1z_2 = r_1\angle\theta_1 \cdot r_2\angle\theta_2 = r_1r_2\angle(\theta_1+\theta_2)$ and $\frac{z_1}{z_2} = \frac{r_1\angle\theta_1}{r_2\angle\theta_2} = \frac{r_1}{r_2}\angle(\theta_1-\theta_2)$, where $r$ is the magnitude and $\theta$ is the phase, and $z_n=r_n\angle\theta_n$.
Since $4i$ has magnitude $4$ and phase $90^\circ$ (since the positive imaginary axis points in a direction $90^\circ$ counterclockwise from the positive real axis), $z$ must have a magnitude $4$ times that of $z+n$. We denote the length from the origin to $z+n$ with the value $x$ and the length from the origin to $z$ with the value $4x$. Additionally, $z$, the origin, and $z+n$ must form a right angle, with $z$ counterclockwise from $z+n$.
This means that $z$, the origin, and $z+n$ form a right triangle. The hypotenuse is the length from $z$ to $z+n$ and has length $n$, since $n$ is defined to be a positive integer. The area of the triangle can be expressed using the two legs, as $\frac{x \cdot 4x}{2}$, or using the hypotenuse and its corresponding altitude, as $\frac{164n}{2}$, so $\frac{x \cdot 4x}{2} = \frac{164n}{2} \implies x^2 = 41n$. By Pythagorean Theorem, $x^2+(4x)^2 = n^2 \implies 17x^2 = n^2$. Substituting out $x^2$ using the earlier equation, we get $17\cdot41n = n^2 \implies n = \boxed{697}$. ~emerald_block | // Block 1
unitsize(1cm);
xaxis("Re",Arrows);
yaxis("Im",Arrows);
real f(real x) {return 164;}
pair F(real x) {return (x,f(x));}
draw(graph(f,-700,100),red,Arrows);
label("Im$(z)=164$",F(-330),N);
pair z = (-656,164);
dot(Label("$z$",align=N),z);
dot(Label("$z+n$",align=N),z+(697,0));
draw(Label("$4x$"),z--(0,0));
draw(Label("$x$"),(0,0)--z+(697,0));
markscalefactor=2;
draw(rightanglemark(z,(0,0),z+(697,0)));
// Block 2
unitsize(1cm); xaxis("Re",Arrows); yaxis("Im",Arrows); real f(real x) {return 164;} pair F(real x) {return (x,f(x));} draw(graph(f,-700,100),red,Arrows); label("Im$(z)=164$",F(-330),N); pair z = (-656,164); dot(Label("$z$",align=N),z); dot(Label("$z+n$",align=N),z+(697,0)); draw(Label("$4x$"),z--(0,0)); draw(Label("$x$"),(0,0)--z+(697,0)); markscalefactor=2; draw(rightanglemark(z,(0,0),z+(697,0))); | [] |
149 | In parallelogram $ABCD$, point $M$ is on $\overline{AB}$ so that $\frac {AM}{AB} = \frac {17}{1000}$ and point $N$ is on $\overline{AD}$ so that $\frac {AN}{AD} = \frac {17}{2009}$. Let $P$ be the point of intersection of $\overline{AC}$ and $\overline{MN}$. Find $\frac {AC}{AP}$. | 2009 AIME I Problem 4 | We approach the problem using mass points on triangle $ABD$ as displayed below.
But as $MN$ does not protrude from a vertex, we will have to "split the mass" at point $A$. First, we know that $DO$ is congruent to $BO$ because diagonals of parallelograms bisect each other. Therefore, we can assign equal masses to points $B$ and $D$. In this case, we assign $B$ and $D$ a mass of 17 each. Now we split the mass at $A$, so we balance segments $AB$ and $AD$ separately, and then the mass of $A$ is the sum of those masses. A mass of 983 is required to balance segment $AB$, while a mass of 1992 is required to balance segment $AD$. Therefore, $A$ has a mass of $1992+983=2975$. Also, $O$ has a mass of 34. Therefore, $\frac{AO}{AP}=\frac{2975+34}{34}=\frac{3009}{34}$, so $\frac{AC}{AP}=\frac{2 (3009)}{34}=177$. | // Block 1
pair A=(0,0),B=(50,0),D=(10,40),C=B+D,M=(8,0),NN=(2,8);
draw(A--B--C--D--cycle);
draw(B--D^^A--C^^M--NN);
pair O=extension(A,C,B,D);
pair P=extension(A,C,M,NN);
dot(A);dot(B);dot(C);dot(D);dot(O);dot(M);dot(NN);dot(P);
label("$A$",A,SW);
label("$B$",B,SE);
label("$C$",C,NE);
label("$D$",D,NW);
label("$M$",M,S);
label("$N$",NN,NW);
label("$P$",P,NNE);
label("$O$",O,N);
// Block 2
pair A=(0,0),B=(50,0),D=(10,40),C=B+D,M=(8,0),NN=(2,8); draw(A--B--C--D--cycle); draw(B--D^^A--C^^M--NN); pair O=extension(A,C,B,D); pair P=extension(A,C,M,NN); dot(A);dot(B);dot(C);dot(D);dot(O);dot(M);dot(NN);dot(P); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,NE); label("$D$",D,NW); label("$M$",M,S); label("$N$",NN,NW); label("$P$",P,NNE); label("$O$",O,N); | [] |
150 | Triangle $ABC$ has $AC = 450$ and $BC = 300$. Points $K$ and $L$ are located on $\overline{AC}$ and $\overline{AB}$ respectively so that $AK = CK$, and $\overline{CL}$ is the angle bisector of angle $C$. Let $P$ be the point of intersection of $\overline{BK}$ and $\overline{CL}$, and let $M$ be the point on line $BK$ for which $K$ is the midpoint of $\overline{PM}$. If $AM = 180$, find $LP$. | 2009 AIME I Problem 5 | Since $K$ is the midpoint of $\overline{PM}$ and $\overline{AC}$, quadrilateral $AMCP$ is a parallelogram, which implies $AM||LP$ and $\bigtriangleup{AMB}$ is similar to $\bigtriangleup{LPB}$
Thus,
\[\frac {AM}{LP}=\frac {AB}{LB}=\frac {AL+LB}{LB}=\frac {AL}{LB}+1\]
Now let's apply the angle bisector theorem.
\[\frac {AL}{LB}=\frac {AC}{BC}=\frac {450}{300}=\frac {3}{2}\]
\[\frac {AM}{LP}=\frac {AL}{LB}+1=\frac {5}{2}\]
\[\frac {180}{LP}=\frac {5}{2}\]
\[LP=\boxed {072}\]
Note: Another way to realize that $\bigtriangleup{AMB}$ is similar to $\bigtriangleup{LPB}$ is through angle chasing and congruent triangles. Angle $AKM$ is congruent to angle $CKP$. Segments $MK$ and $PK$ are congruent as stated in the problem. Segments $AK$ and $CK$ are congruent as stated in the problem. Thus, by Side-Angle-Side congruency $\bigtriangleup{AMK}$ is similar to $\bigtriangleup{CPK}$. Then angle $AMK = AMB$ is congruent to angle $CPK = BPL$. Notice triangles $\bigtriangleup{AMB}$ and $\bigtriangleup{LPB}$ both share angle $MBA$. Thus the two triangles are similar because of AA similarity with angles [$AMB$ and $LPB$] and [$MBA = PBL$]. *Solution in Beauty of Math video written out | // Block 1
import markers;
defaultpen(fontsize(8));
size(300);
pair A=(0,0), B=(30*sqrt(331),0), C, K, L, M, P;
C = intersectionpoints(Circle(A,450), Circle(B,300))[0];
K = midpoint(A--C);
L = (3*B+2*A)/5;
P = extension(B,K,C,L);
M = 2*K-P;
draw(A--B--C--cycle);
draw(C--L);draw(B--M--A);
markangle(n=1,radius=15,A,C,L,marker(markinterval(stickframe(n=1),true)));
markangle(n=1,radius=15,L,C,B,marker(markinterval(stickframe(n=1),true)));
dot(A^^B^^C^^K^^L^^M^^P);
label("$A$",A,(-1,-1));label("$B$",B,(1,-1));label("$C$",C,(1,1));
label("$K$",K,(0,2));label("$L$",L,(0,-2));label("$M$",M,(-1,1));
label("$P$",P,(1,1));
label("$180$",(A+M)/2,(-1,0));label("$180$",(P+C)/2,(-1,0));label("$225$",(A+K)/2,(0,2));label("$225$",(K+C)/2,(0,2));
label("$300$",(B+C)/2,(1,1));
// Block 2
import markers; defaultpen(fontsize(8)); size(300); pair A=(0,0), B=(30*sqrt(331),0), C, K, L, M, P; C = intersectionpoints(Circle(A,450), Circle(B,300))[0]; K = midpoint(A--C); L = (3*B+2*A)/5; P = extension(B,K,C,L); M = 2*K-P; draw(A--B--C--cycle); draw(C--L);draw(B--M--A); markangle(n=1,radius=15,A,C,L,marker(markinterval(stickframe(n=1),true))); markangle(n=1,radius=15,L,C,B,marker(markinterval(stickframe(n=1),true))); dot(A^^B^^C^^K^^L^^M^^P); label("$A$",A,(-1,-1));label("$B$",B,(1,-1));label("$C$",C,(1,1)); label("$K$",K,(0,2));label("$L$",L,(0,-2));label("$M$",M,(-1,1)); label("$P$",P,(1,1)); label("$180$",(A+M)/2,(-1,0));label("$180$",(P+C)/2,(-1,0));label("$225$",(A+K)/2,(0,2));label("$225$",(K+C)/2,(0,2)); label("$300$",(B+C)/2,(1,1)); | [] |
151 | In right $\triangle ABC$ with hypotenuse $\overline{AB}$, $AC = 12$, $BC = 35$, and $\overline{CD}$ is the altitude to $\overline{AB}$. Let $\omega$ be the circle having $\overline{CD}$ as a diameter. Let $I$ be a point outside $\triangle ABC$ such that $\overline{AI}$ and $\overline{BI}$ are both tangent to circle $\omega$. The ratio of the perimeter of $\triangle ABI$ to the length $AB$ can be expressed in the form $\frac {m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$. | 2009 AIME I Problem 12 | Let $AP=AD=x$, let $BQ=BD=y$, and let $IP=IQ=z$. Let $OD=r$. We find $AB=37$. Let $\alpha$, $\beta$, and $\theta$ be the angles $OAD$, $OBD$, and $OPI$ respectively. Then $\alpha + \beta + \theta = 90^\circ$, so \[\theta = 90^\circ - (\alpha+\beta).\]
The perimeter of $\triangle ABI$ is $2(x+y+z)=2(37+z)$. The desired ratio is then
\[\rho = 2\left(1+\frac z{37}\right)\]
We need to find $z$. In $\triangle OPI$, $z=r\cot\theta = r\tan (\alpha+\beta)$. We get \[\tan\alpha = \frac{OD}{AD} = \frac 12 \frac{CD}{AD} = \frac 12 \tan A = \frac 12 \frac{BC}{AC} = \frac{35}{24}.\] Similarly, $\tan\beta = \tfrac 6{35}$. Then \[z = r\cdot \tan (\alpha+\beta) = r\cdot \frac{\tan\alpha + \tan\beta}{1-\tan\alpha\tan\beta}= \frac{37^2\cdot r}{18\cdot 35}\]
Computing $[ABC]$ in two ways we get $CD = \tfrac{12\cdot 35}{37}$, so $r=\tfrac{6\cdot 35}{37}$. Using this value of $r$ we get $z=\tfrac {37}3$. Thus \[\rho = 2\left(1+\frac 1{3}\right) = \frac 8{3},\]
and $8+3=\boxed{011}$. | // Block 1
size(300);
defaultpen(linewidth(0.4)+fontsize(10));
pen s = linewidth(0.8)+fontsize(8);
pair A,B,C,D,O,X;
C=origin;
A=(0,12);
B=(18,0);
D=foot(C,A,B);
O = (C+D)/2;
real r = length(D-C)/2;
path c = CR(O, r);
pair OA = (O+A)/2;
real rA = length(A-O)/2;
pair Ap = OP(CR(OA,rA), c);
pair OB = (O+B)/2;
real rB = length(B-O)/2;
pair Bp = OP(CR(OB,rB), c);
X=extension(A,Ap,B,Bp);
draw(A--B--C--A, s);
draw(C--D^^B--O--A^^Ap--O--X, gray+0.25);
draw(c^^A--X--B);
dot("$A$", A, N);
dot("$B$", B, SE);
dot("$C$", C, SW);
dot("$D$", D, 0.2*(D-C));
dot("$I$", X, 0.5*(X-C));
dot("$P$", Ap, 0.3*(Ap-O));
dot("$Q$", Bp, 0.3*(Bp-O));
dot("$O$", O, W);
label("$\beta$",B,10*dir(157));
label("$\alpha$",A,5*dir(-55));
label("$\theta$",X,5*dir(55));
// Block 2
size(300); defaultpen(linewidth(0.4)+fontsize(10)); pen s = linewidth(0.8)+fontsize(8); pair A,B,C,D,O,X; C=origin; A=(0,12); B=(18,0); D=foot(C,A,B); O = (C+D)/2; real r = length(D-C)/2; path c = CR(O, r); pair OA = (O+A)/2; real rA = length(A-O)/2; pair Ap = OP(CR(OA,rA), c); pair OB = (O+B)/2; real rB = length(B-O)/2; pair Bp = OP(CR(OB,rB), c); X=extension(A,Ap,B,Bp); draw(A--B--C--A, s); draw(C--D^^B--O--A^^Ap--O--X, gray+0.25); draw(c^^A--X--B); dot("$A$", A, N); dot("$B$", B, SE); dot("$C$", C, SW); dot("$D$", D, 0.2*(D-C)); dot("$I$", X, 0.5*(X-C)); dot("$P$", Ap, 0.3*(Ap-O)); dot("$Q$", Bp, 0.3*(Bp-O)); dot("$O$", O, W); label("$\beta$",B,10*dir(157)); label("$\alpha$",A,5*dir(-55)); label("$\theta$",X,5*dir(55)); | [] |
152 | In rectangle $ABCD$, $AB=100$. Let $E$ be the midpoint of $\overline{AD}$. Given that line $AC$ and line $BE$ are perpendicular, find the greatest integer less than $AD$. | 2009 AIME II Problem 3 | Solution 1
From the problem, $AB=100$ and triangle $FBA$ is a right triangle. As $ABCD$ is a rectangle, triangles $BCA$, and $ABE$ are also right triangles. By $AA$, $\triangle FBA \sim \triangle BCA$, and $\triangle FBA \sim \triangle ABE$, so $\triangle ABE \sim \triangle BCA$. This gives $\frac {AE}{AB}= \frac {AB}{BC}$. $AE=\frac{AD}{2}$ and $BC=AD$, so $\frac {AD}{2AB}= \frac {AB}{AD}$, or $(AD)^2=2(AB)^2$, so $AD=AB \sqrt{2}$, or $100 \sqrt{2}$, so the answer is $\boxed{141}$.
Solution 2
Let $x$ be the ratio of $BC$ to $AB$. On the coordinate plane, plot $A=(0,0)$, $B=(100,0)$, $C=(100,100x)$, and $D=(0,100x)$. Then $E=(0,50x)$. Furthermore, the slope of $\overline{AC}$ is $x$ and the slope of $\overline{BE}$ is $-x/2$. They are perpendicular, so they multiply to $-1$, that is,
\[x\cdot-\frac{x}{2}=-1,\]
which implies that $-x^2=-2$ or $x=\sqrt 2$. Therefore $AD=100\sqrt 2\approx 141.42$ so $\lfloor AD\rfloor=\boxed{141}$.
Solution 3
Similarly to Solution 2, let the positive x-axis be in the direction of ray $BC$ and let the positive y-axis be in the direction of ray $BA$. Thus, the vector $BE=(x,100)$ and the vector $AC=(2x,-100)$ are perpendicular and thus have a dot product of 0. Thus, calculating the dot product:
\[x\cdot2x+(100)\cdot(-100)=2x^2-10000=0\]
\[2x^2-10000=0\rightarrow x^2=5000\]
Substituting AD/2 for x:
\[(AD/2)^2=5000\rightarrow AD^2=20000\]
\[AD=100\sqrt2\]
Solution 4
Draw $CX$ and $EX$ to form a parallelogram $AEXC$. Since $EX \parallel AC$, $\angle BEX=90^\circ$ by the problem statement, so $\triangle BEX$ is right.
Letting $AE=y$, we have $BE=\sqrt{100^2+y^2}$ and $AC=EX=\sqrt{100^2+(2y)^2}$. Since $CX=EA$, $\sqrt{100^2+y^2}^2+\sqrt{100^2+(2y)^2}=(3y)^2$. Solving this, we have
\[100^2+ 100^2 + y^2 + 4y^2 = 9y^2\]
\[2\cdot 100^2 = 4y^2\]
\[\frac{100^2}{2}=y^2\]
\[\frac{100}{\sqrt{2}}=y\]
\[\frac{100\sqrt{2}}{2}=y\]
\[100\sqrt{2}=2y=AD\], so the answer is $\boxed{141}$. | // Block 1
pair A=(0,10), B=(0,0), C=(14,0), D=(14,10), Q=(0,5); draw (A--B--C--D--cycle); pair E=(7,10); draw (B--E); draw (A--C); pair F=(6.7,6.7); label("\(E\)",E,N); label("\(A\)",A,NW); label("\(B\)",B,SW); label("\(C\)",C,SE); label("\(D\)",D,NE); label("\(F\)",F,W); label("\(100\)",Q,W);
// Block 2
pair A=(0,10), B=(0,0), C=(14,0), D=(14,10), Q=(0,5),X=(21,0); draw (A--B--C--D--cycle); pair E=(7,10); draw (B--E); draw (A--C); draw(C--X--E); label("\(E\)",E,N); label("\(A\)",A,NW); label("\(B\)",B,SW); label("\(C\)",C,SE); label("\(D\)",D,NE); label("\(X\)",X,S); label("\(100\)",Q,W); | [] |
153 | Four lighthouses are located at points $A$, $B$, $C$, and $D$. The lighthouse at $A$ is $5$ kilometers from the lighthouse at $B$, the lighthouse at $B$ is $12$ kilometers from the lighthouse at $C$, and the lighthouse at $A$ is $13$ kilometers from the lighthouse at $C$. To an observer at $A$, the angle determined by the lights at $B$ and $D$ and the angle determined by the lights at $C$ and $D$ are equal. To an observer at $C$, the angle determined by the lights at $A$ and $B$ and the angle determined by the lights at $D$ and $B$ are equal. The number of kilometers from $A$ to $D$ is given by $\frac {p\sqrt{q}}{r}$, where $p$, $q$, and $r$ are relatively prime positive integers, and $r$ is not divisible by the square of any prime. Find $p$ + $q$ + $r$. | 2009 AIME II Problem 10 | Let $CD$ and $AB$ meet at $A'$; then $\overline{AD}$ is an angle bisector of isosceles $\triangle AA'C$. Then by the Angle Bisector Theorem, $A'D=\frac {10}{10+13} \cdot 13=\frac {130}{23}$, and $\cos \angle DA'A=\frac {5}{13}$. By the Law of Cosines on $\triangle AA'D$, we have \[AD^2=10^2+\left(\frac {130}{23}\right)^2-2 \cdot 10 \cdot \frac {130}{23} \cdot \frac {5}{13}=\frac {2^4 \cdot 3^2 \cdot 5^2 \cdot 13}{23^2} \Longrightarrow AD=\frac {60\sqrt {13}}{23}\] and the answer is $60+13+23=\boxed{096}$.
-azjps | size(120); pathpen = linewidth(0.7); pointpen = black; pen f = fontsize(10); pair B=(0,0), A=(5,0), C=(0,13), E=(-5,0), O = incenter(E,C,A), D=IP(A -- A+3*(O-A),E--C); D(A--B--C--cycle); D(A--D--C); D(D--E--B, linetype("4 4")); MP("5", (A+B)/2, f); MP("13", (A+C)/2, NE,f); MP("A",D(A),f); MP("B",D(B),f); MP("C",D(C),N,f); MP("A'",D(E),f); MP("D",D(D),NW,f); D(rightanglemark(C,B,A,20)); D(anglemark(D,A,E,35));D(anglemark(C,A,D,30)); | [] |
153 | Four lighthouses are located at points $A$, $B$, $C$, and $D$. The lighthouse at $A$ is $5$ kilometers from the lighthouse at $B$, the lighthouse at $B$ is $12$ kilometers from the lighthouse at $C$, and the lighthouse at $A$ is $13$ kilometers from the lighthouse at $C$. To an observer at $A$, the angle determined by the lights at $B$ and $D$ and the angle determined by the lights at $C$ and $D$ are equal. To an observer at $C$, the angle determined by the lights at $A$ and $B$ and the angle determined by the lights at $D$ and $B$ are equal. The number of kilometers from $A$ to $D$ is given by $\frac {p\sqrt{q}}{r}$, where $p$, $q$, and $r$ are relatively prime positive integers, and $r$ is not divisible by the square of any prime. Find $p$ + $q$ + $r$. | 2009 AIME II Problem 10 | Using the law of sines on $\bigtriangleup ADE$,
\begin{align*} AD &= AE \cdot \dfrac{\sin(2\alpha)}{\sin(180^\circ-3\alpha)}\\ &= AE \cdot \dfrac{\sin(2\alpha)}{\sin(\alpha+2\alpha)}\\ &= 10 \cdot \dfrac{\sin(2\alpha)}{\sin(\alpha)\cos(2\alpha)+\cos(\alpha)\sin(2\alpha)}\\ &= 10 \cdot \dfrac{\sin(2\alpha)}{\sqrt{\dfrac{1-\cos(2\alpha)}{2}}\cdot\cos(2\alpha)+\sqrt{\dfrac{1+\cos(2\alpha)}{2}}\cdot\sin(2\alpha)}\\ &= 10 \cdot \dfrac{\dfrac{12}{13}}{\sqrt{\dfrac{8}{26}}\cdot\dfrac{5}{13}+\sqrt{\dfrac{18}{26}}\cdot\dfrac{12}{13}}\\ &= 10 \cdot \dfrac{12\sqrt{13}}{10+36} = \dfrac{60\sqrt{13}}{23} \end{align*}
$\therefore$ the answer is $60+13+23=\boxed{096}$.
-m1sterzer0 | // Block 1
pathpen = linewidth(0.7);
pen f = fontsize(10);
size(144);
pair B=(0,0);
pair A=(5,0);
pair C=(0,12);
pair E=(-5,0);
pair abA = A+unit(C-A)+unit(B-A);
pair D = extension(A,abA,C,E);
D(A--C);
D(C--B);
D(A--B);
D(C--D);
D(A--D);
D(D--E,dashed);
D(B--E,dashed);
D(rightanglemark(C,B,A,20),red);
MP("A",D(A),plain.SE,f);
MP("B",D(B),plain.S,f);
MP("C",D(C),plain.N,f);
MP("D",D(D),plain.NW,f);
MP("E",D(E),plain.SW,f);
MP("5",(A+B)/2,plain.S,f);
MP("12",(B+C)/2,plain.E,f);
MP("13",(A+C)/2,plain.NE,f);
MP("\alpha",A,4*(unit(D-A)+unit(C-A))/2,f);
MP("\alpha",A,4*(unit(B-A)+unit(D-A))/2,f);
MP("2\alpha",E,4*(unit(C-E)+unit(B-E))/2,f);
MP("\beta",C,6*(unit(A-C)+unit(B-C))/2,f);
MP("\beta",C,6*(unit(E-C)+unit(B-C))/2,f);
// Block 2
pathpen = linewidth(0.7); pen f = fontsize(10); size(144); pair B=(0,0); pair A=(5,0); pair C=(0,12); pair E=(-5,0); pair abA = A+unit(C-A)+unit(B-A); pair D = extension(A,abA,C,E); D(A--C); D(C--B); D(A--B); D(C--D); D(A--D); D(D--E,dashed); D(B--E,dashed); D(rightanglemark(C,B,A,20),red); MP("A",D(A),plain.SE,f); MP("B",D(B),plain.S,f); MP("C",D(C),plain.N,f); MP("D",D(D),plain.NW,f); MP("E",D(E),plain.SW,f); MP("5",(A+B)/2,plain.S,f); MP("12",(B+C)/2,plain.E,f); MP("13",(A+C)/2,plain.NE,f); MP("\alpha",A,4*(unit(D-A)+unit(C-A))/2,f); MP("\alpha",A,4*(unit(B-A)+unit(D-A))/2,f); MP("2\alpha",E,4*(unit(C-E)+unit(B-E))/2,f); MP("\beta",C,6*(unit(A-C)+unit(B-C))/2,f); MP("\beta",C,6*(unit(E-C)+unit(B-C))/2,f); | [] |
154 | Let $\overline{MN}$ be a diameter of a circle with diameter 1. Let $A$ and $B$ be points on one of the semicircular arcs determined by $\overline{MN}$ such that $A$ is the midpoint of the semicircle and $MB=\frac{3}5$. Point $C$ lies on the other semicircular arc. Let $d$ be the length of the line segment whose endpoints are the intersections of diameter $\overline{MN}$ with chords $\overline{AC}$ and $\overline{BC}$. The largest possible value of $d$ can be written in the form $r-s\sqrt{t}$, where $r, s$ and $t$ are positive integers and $t$ is not divisible by the square of any prime. Find $r+s+t$. | 2009 AIME II Problem 15 | Solution 1 (Quick Calculus)
Let $V = \overline{NM} \cap \overline{AC}$ and $W = \overline{NM} \cap \overline{BC}$. Further more let $\angle NMC = \alpha$ and $\angle MNC = 90^\circ - \alpha$. Angle chasing reveals $\angle NBC = \angle NAC = \alpha$ and $\angle MBC = \angle MAC = 90^\circ - \alpha$. Additionally $NB = \frac{4}{5}$ and $AN = AM$ by the Pythagorean Theorem.
By the Angle Bisector Formula,
\[\frac{NV}{MV} = \frac{\sin (\alpha)}{\sin (90^\circ - \alpha)} = \tan (\alpha)\]
\[\frac{MW}{NW} = \frac{3\sin (90^\circ - \alpha)}{4\sin (\alpha)} = \frac{3}{4} \cot (\alpha)\]
As $NV + MV =MW + NW = 1$ we compute $NW = \frac{1}{1+\frac{3}{4}\cot(\alpha)}$ and $MV = \frac{1}{1+\tan (\alpha)}$, and finally $VW = NW + MV - 1 = \frac{1}{1+\frac{3}{4}\cot(\alpha)} + \frac{1}{1+\tan (\alpha)} - 1$. Taking the derivative of $VW$ with respect to $\alpha$, we arrive at
\[VW' = \frac{7\cos^2 (\alpha) - 4}{(\sin(\alpha) + \cos(\alpha))^2(4\sin(\alpha)+3\cos(\alpha))^2}\]
Clearly the maximum occurs when $\alpha = \cos^{-1}\left(\frac{2}{\sqrt{7}}\right)$. Plugging this back in, using the fact that $\tan(\cos^{-1}(x)) = \frac{\sqrt{1-x^2}}{x}$ and $\cot(\cos^{-1}(x)) = \frac{x}{\sqrt{1-x^2}}$, we get
$VW = 7 - 4\sqrt{3}$
with $7 + 4 + 3 = \boxed{014}$
~always_correct
Solution 2 (Projective)
Since $MA = \frac{\sqrt{2}}{2} \approx 0.707 > \frac{3}{5}$, point $B$ lies between $M$ and $A$ on the semicircular arc. We will first compute the length of $\overline{AB}$. By the law of cosines, $\cos \angle MOB = \frac{-(3/5)^2 + 2(1/2)^2}{2(1/2)^2} = \frac{7}{25}$, so $\cos \angle AOB = \sin \angle MOB = \frac{24}{25}$. Then $AB^2 = 2\left(\frac{1}{2}\right)^2 - 2\left(\frac{1}{2}\right)^2 \cdot \frac{24}{25} = \frac{1}{50}$, so $AB = \frac{1}{5\sqrt{2}}$.
Let $P = AC \cap MN$ and $Q = BC \cap MN$, and let $MQ = x$, $PQ = d$, $PN = y$. Note that\[(M, P; Q, N) \stackrel{C}{=} (M, A; B, N),\]that is,\[\frac{QP}{QM} \div \frac{NP}{NM} = \frac{BA}{BM} \div \frac{NA}{NM}\]or\[\frac{d}{xy} = \frac{1/(5\sqrt{2})}{(3/5) \cdot (\sqrt{2}/2)} = \frac{1}{3}.\]Hence $d = \frac{1}{3}xy$, and we also know $d+x+y=1$. Now AM-GM gives\[\frac{x+y}{2} \ge \sqrt{xy} \implies \frac{1-d}{2} \ge \sqrt{3d}.\]This gives the quadratic inequality $d^2 - 14d + 1 \ge 0$, which solves as\[d \in \left(-\infty, 7-4\sqrt3\right] \cup \left[7+4\sqrt3, \infty\right).\]But $d \le 1$, so the greatest possible value of $d$ is $7-4\sqrt3$. The answer is $7+4+3=\boxed{014}$.
~MSTang
Solution 3 (Calculus)
Let $O$ be the center of the circle. Define $\angle{MOC}=t$, $\angle{BOA}=2a$, and let $BC$ and $AC$ intersect $MN$ at points $X$ and $Y$, respectively. We will express the length of $XY$ as a function of $t$ and maximize that function in the interval $[0, \pi]$.
Let $C'$ be the foot of the perpendicular from $C$ to $MN$. We compute $XY$ as follows.
(a) By the Extended Law of Sines in triangle $ABC$, we have
\[CA\]
\[= \sin\angle{ABC}\]
\[= \sin\left(\frac{\widehat{AN} + \widehat{NC}}{2}\right)\]
\[= \sin\left(\frac{\frac{\pi}{2} + (\pi-t)}{2}\right)\]
\[= \sin\left(\frac{3\pi}{4} - \frac{t}{2}\right)\]
\[= \sin\left(\frac{\pi}{4} + \frac{t}{2}\right)\]
(b) Note that $CC' = CO\sin(t) = \left(\frac{1}{2}\right)\sin(t)$ and $AO = \frac{1}{2}$. Since $CC'Y$ and $AOY$ are similar right triangles, we have $CY/AY = CC'/AO = \sin(t)$, and hence,
\[CY/CA\]
\[= \frac{CY}{CY + AY}\]
\[= \frac{\sin(t)}{1 + \sin(t)}\]
\[= \frac{\sin(t)}{\sin\left(\frac{\pi}{2}\right) + \sin(t)}\]
\[= \frac{\sin(t)}{2\sin\left(\frac{\pi}{4} + \frac{t}{2}\right)\cos\left(\frac{\pi}{4} - \frac{t}{2}\right)}\]
(c) We have $\angle{XCY} = \frac{\widehat{AB}}{2}=a$ and $\angle{CXY} = \frac{\widehat{MB}+\widehat{CN}}{2} = \frac{\left(\frac{\pi}{2} - 2a\right) + (\pi - t)}{2} = \frac{3\pi}{4} - a - \frac{t}{2}$, and hence by the Law of Sines,
\[XY/CY\]
\[= \frac{\sin\angle{XCY}}{\sin\angle{CXY}}\]
\[= \frac{\sin(a)}{\sin\left(\frac{3\pi}{4} - a - \frac{t}{2}\right)}\]
\[= \frac{\sin(a)}{\sin\left(\frac{\pi}{4} + a + \frac{t}{2}\right)}\]
(d) Multiplying (a), (b), and (c), we have
\[XY\]
\[= CA * (CY/CA) * (XY/CY)\]
\[= \frac{\sin(t)\sin(a)}{2\cos\left(\frac{\pi}{4} - \frac{t}{2}\right)\sin\left(\frac{\pi}{4} + a + \frac{t}{2}\right)}\]
\[= \frac{\sin(t)\sin(a)}{\sin\left(\frac{\pi}{2} + a\right) + \sin(a + t)}\]
\[= \sin(a)\times\frac{\sin(t)}{\sin(t + a) + \cos(a)}\],
which is a function of $t$ (and the constant $a$). Differentiating this with respect to $t$ yields
\[\sin(a)\times\frac{\cos(t)(\sin(t + a) + \cos(a)) - \sin(t)\cos(t + a)}{(\sin(t + a) + \cos(a))^2}\],
and the numerator of this is
\[\sin(a) \times(\sin(t + a)\cos(t) - \cos(t + a)\sin(t) + \cos(a)\cos(t))\]
\[= \sin(a) \times (\sin(a) + \cos(a)\cos(t))\],
which vanishes when $\sin(a) + \cos(a)\cos(t) = 0$. Therefore, the length of $XY$ is maximized when $t=t'$, where $t'$ is the value in $[0, \pi]$ that satisfies $\cos(t') = -\tan(a)$.
Note that
\[\frac{1 - \tan(a)}{1 + \tan(a)} = \tan\left(\frac{\pi}{4} - a\right) = \tan((\widehat{MB})/2) = \tan\angle{MNB} = \frac{3}{4}\],
so $\tan(a) = \frac{1}{7}$. We compute
\[\sin(a) = \frac{\sqrt{2}}{10}\]
\[\cos(a) = \frac{7\sqrt{2}}{10}\]
\[\cos(t') = -\tan(a) = -\frac{1}{7}\]
\[\sin(t') = \frac{4\sqrt{3}}{7}\]
\[\sin(t' + a)=\sin(t')\cos(a) + \cos(t')\sin(a) = \frac{28\sqrt{6} - \sqrt{2}}{70}\],
so the maximum length of $XY$ is $\sin(a)\times\frac{\sin(t')}{\sin(t' + a) + \cos(a)} = 7 - 4\sqrt{3}$, and the answer is $7 + 4 + 3 = \boxed{014}$.
Solution 4
Suppose $\overline{AC}$ and $\overline{BC}$ intersect $\overline{MN}$ at $D$ and $E$, respectively, and let $MC = x$ and $NC = y$. Since $A$ is the midpoint of arc $MN$, $\overline{CA}$ bisects $\angle MCN$, and we get
\[\frac{MC}{MD} = \frac{NC}{ND}\Rightarrow MD = \frac{x}{x + y}.\]
To find $ME$, we note that $\triangle BNE\sim\triangle MCE$ and $\triangle BME\sim\triangle NCE$, so
\begin{align*} \frac{BN}{NE} &= \frac{MC}{CE} \\ \frac{ME}{BM} &= \frac{CE}{NC}. \end{align*}
Writing $NE = 1 - ME$, we can substitute known values and multiply the equations to get
\[\frac{4(ME)}{3 - 3(ME)} = \frac{x}{y}\Rightarrow ME = \frac{3x}{3x + 4y}.\]
The value we wish to maximize is
\begin{align*} DE &= MD - ME \\ &= \frac{x}{x + y} - \frac{3x}{3x + 4y} \\ &= \frac{xy}{3x^2 + 7xy + 4y^2} \\ &= \frac{1}{3(x/y) + 4(y/x) + 7}. \end{align*}
By the AM-GM inequality, $3(x/y) + 4(y/x)\geq 2\sqrt{12} = 4\sqrt{3}$, so
\[DE\leq \frac{1}{4\sqrt{3} + 7} = 7 - 4\sqrt{3},\]
giving the answer of $7 + 4 + 3 = \boxed{014}$. Equality is achieved when $3(x/y) = 4(y/x)$ subject to the condition $x^2 + y^2 = 1$, which occurs for $x = \frac{2\sqrt{7}}{7}$ and $y = \frac{\sqrt{21}}{7}$.
Solution 5 (Projective)
By Pythagoras in $\triangle BMN,$ we get $BN=\dfrac{4}{5}.$
Since cross ratios are preserved upon projecting, note that $(M,Y;X,N)\stackrel{C}{=}(M,B;A,N).$ By definition of a cross ratio, this becomes \[\dfrac{XM}{NY}:\dfrac{NM}{NY}=\dfrac{AM}{AB}:\dfrac{MN}{NB}.\] Let $MY=a,YX=b,XN=c$ such that $a+b+c=1.$ We know that $XM=a+b,XY=b,NM=1,NY=b+c,$ so the LHS becomes $\dfrac{(a+b)(b+c)}{b}.$
In the RHS, we are given every value except for $AB.$ However, Ptolemy's Theorem on $MBAN$ gives $AB\cdot MN+AN\cdot BM=AM\cdot BN\implies AB+\dfrac{3}{5\sqrt{2}}=\dfrac{4}{5\sqrt{2}}\implies AB=\dfrac{1}{5\sqrt{2}}.$ Substituting, we get $\dfrac{(a+b)(b+c)}{b}=4\implies b(a+b+c)+ac=4b, b=\dfrac{ac}{3}$ where we use $a+b+c=1.$
Again using $a+b+c=1,$ we have $a+b+c=1\implies a+\dfrac{ac}{3}+c=1\implies a=3\dfrac{1-c}{c+3}.$ Then $b=\dfrac{ac}{3}=\dfrac{c-c^2}{c+3}.$ Since this is a function in $c,$ we differentiate WRT $c$ to find its maximum. By quotient rule, it suffices to solve \[(-2c+1)(c+3)-(c-c^2)=0 \implies c^2+6c-3,c=-3+2\sqrt{3}.\] Substituting back yields $b=7-4\sqrt{3},$ so $7+4+3=\boxed{014}$ is the answer.
~Generic_Username | unitsize(144); pair A, B, C, M, n; A = (0,1); B = (-7/25, 24/25); C=(1/7,-4*sqrt(3)/7); M = (-1,0); n = (1,0); pair [] D = intersectionpoints(A--C,M--n); pair [] e = intersectionpoints(B--C,M--n); draw(circle((0,0),1)); draw(M--n--B--M--A--n--C--A--B--C--cycle); label("$A$",A,N); label("$B$",B,NNW); label("$M$",M,W); label("$C$",C,SSE); label("$N$",n,E); label("$D$",D[0],SE); label("$E$",e[0],SW); label("$x$",(M+C)/2,SW); label("$y$",(n+C)/2,SE); | [] |
155 | Let $P(x)$ be a quadratic polynomial with real coefficients satisfying $x^2 - 2x + 2 \le P(x) \le 2x^2 - 4x + 3$ for all real numbers $x$, and suppose $P(11) = 181$. Find $P(16)$. | 2010 AIME I Problem 6 | Solution 1
Let $Q(x) = x^2 - 2x + 2$, $R(x) = 2x^2 - 4x + 3$. Completing the square, we have $Q(x) = (x-1)^2 + 1$, and $R(x) = 2(x-1)^2 + 1$, so it follows that $P(x) \ge Q(x) \ge 1$ for all $x$ (by the Trivial Inequality).
Also, $1 = Q(1) \le P(1) \le R(1) = 1$, so $P(1) = 1$, and $P$ obtains its minimum at the point $(1,1)$. Then $P(x)$ must be of the form $c(x-1)^2 + 1$ for some constant $c$; substituting $P(11) = 181$ yields $c = \frac 95$. Finally, $P(16) = \frac 95 \cdot (16 - 1)^2 + 1 = \boxed{406}$.
Solution 2
It can be seen that the function $P(x)$ must be in the form $P(x) = ax^2 - 2ax + c$ for some real $a$ and $c$. This is because the derivative of $P(x)$ is $2ax - 2a$, and a global minimum occurs only at $x = 1$ (in addition, because of this derivative, the vertex of any quadratic polynomial occurs at $\frac{-b}{2a}$). Substituting $(1,1)$ and $(11, 181)$ we obtain two equations:
$P(11) = 99a + c = 181$, and $P(1) = -a + c = 1$.
Solving, we get $a = \frac{9}{5}$ and $c = \frac{14}{5}$, so $P(x) = \frac{9}{5}x^2 - \frac{18}{5}x + \frac {14}{5}$. Therefore, $P(16) = \boxed{406}$.
Solution 3
Let $y = x^2 - 2x + 2$; note that $2y - 1 = 2x^2 - 4x + 3$. Setting $y = 2y - 1$, we find that equality holds when $y = 1$ and therefore when $x^2 - 2x + 2 = 1$; this is true iff $x = 1$, so $P(1) = 1$.
Let $Q(x) = P(x) - x$; clearly $Q(1) = 0$, so we can write $Q(x) = (x - 1)Q'(x)$, where $Q'(x)$ is some linear function. Plug $Q(x)$ into the given inequality:
$x^2 - 3x + 2 \le Q(x) \le 2x^2 - 5x + 3$
$(x - 1)(x - 2) \le (x - 1)Q'(x) \le (x - 1)(2x - 3)$, and thus
$x - 2 \le Q'(x) \le 2x - 3$
For all $x > 1$; note that the inequality signs are flipped if $x < 1$, and that the division is invalid for $x = 1$. However,
$\lim_{x \to 1} x - 2 = \lim_{x \to 1} 2x - 3 = -1$,
and thus by the sandwich theorem $\lim_{x \to 1} Q'(x) = -1$; by the definition of a continuous function, $Q'(1) = -1$. Also, $Q(11) = 170$, so $Q'(11) = 170/(11-1) = 17$; plugging in and solving, $Q'(x) = (9/5)(x - 1) - 1$. Thus $Q(16) = 390$, and so $P(16) = \boxed{406}$.
Solution 4
Let $Q(x) = P(x) - (x^2-2x+2)$, then $0\le Q(x) \le (x-1)^2$ (note this is derived from the given inequality chain). Therefore, $0\le Q(x+1) \le x^2 \Rightarrow Q(x+1) = Ax^2$ for some real value A.
$Q(11) = 10^2A \Rightarrow P(11)-(11^2-22+2)=100A \Rightarrow 80=100A \Rightarrow A=\frac{4}{5}$.
$Q(16)=15^2A=180 \Rightarrow P(16)-(16^2-32+2) = 180 \Rightarrow P(16)=180+226= \boxed{406}$
Solution 5
Let $P(x) = ax^2 + bx + c$. Plugging in $x = 1$ to the expressions on both sides of the inequality, we see that $a + b + c = 1$. We see from the problem statement that $121a + 11b + c = 181$. Since we know the vertex of $P(x)$ lies at $x = 1$, by symmetry we get $81a -9b + c = 181$ as well. Since we now have three equations, we can solve this trivial system and get our answer of $\boxed{406}$.
Solution 6
Similar to Solution 5, let $P(x) = ax^2 + bx + c$. Note that $(1,1)$ is a vertex of the polynomial. Additionally, this means that $b = -2a$ (since $\frac{-b}{2a}$ is the minimum $x$ point). Thus, we have $P(x) = ax^2 - 2ax + c$. Therefore $a - 2a + c = 1$. Moreover, $99a + c = 181$. And so our polynomial is $\frac{9}{5}x^2 - \frac{18}{5}x + \frac{14}{5}$. Plug in $x = 16$ to get $\boxed{406}$.
Solution 7
Very similar to Solution 6, start by noticing that $P(x)$ is between two polynomials, we try to set them equal to find a point where the two polynomials intersect, meaning that $P(x)$ would also have to intersect that point (it must be between the two graphs). Setting $x^2 - 2x + 2 = 2x^2 - 4x + 3$, we find that $x = 1$. Note that both of these graphs have the same vertex (at $x = 1$), and so $P(x)$ must also have the same vertex $(1, 1)$. Setting $P(x) = ax^2 - 2ax + a + 1$ (this is where we have a vertex at $(1, 1)$), we plug in $11$ and find that $a = 1.8$. Evaluating $1.8x^2 - 3.6x + 2.8$ when $x = 16$ (our intended goal), we find that $P(16) = \boxed{406}$. | import graph; real min = -0.5, max = 2.5; pen dark = linewidth(1); real P(real x) { return 8*(x-1)^2/5+1; } real Q(real x) { return (x-1)^2+1; } real R(real x) { return 2*(x-1)^2+1; } draw(graph(P,min,max),dark); draw(graph(Q,min,max),linetype("6 2")+linewidth(0.7)); draw(graph(R,min,max),linetype("6 2")+linewidth(0.7)); dot((1,1)); label("$P(x)$",(max,P(max)),E,fontsize(10)); label("$Q(x)$",(max,Q(max)),E,fontsize(10)); label("$R(x)$",(max,R(max)),E,fontsize(10)); /* axes */ Label f; f.p=fontsize(8); xaxis(-2, 3, Ticks(f, 5, 1)); yaxis(-1, 5, Ticks(f, 6, 1)); | [] |
156 | For a real number $a$, let $\lfloor a \rfloor$ denote the greatest integer less than or equal to $a$. Let $\mathcal{R}$ denote the region in the coordinate plane consisting of points $(x,y)$ such that $\lfloor x \rfloor ^2 + \lfloor y \rfloor ^2 = 25$. The region $\mathcal{R}$ is completely contained in a disk of radius $r$ (a disk is the union of a circle and its interior). The minimum value of $r$ can be written as $\frac {\sqrt {m}}{n}$, where $m$ and $n$ are integers and $m$ is not divisible by the square of any prime. Find $m + n$. | 2010 AIME I Problem 8 | The desired region consists of 12 boxes, whose lower-left corners are integers solutions of $x^2 + y^2 = 25$, namely $(\pm5,0), (0,\pm5), (\pm3,\pm4), (\pm4,\pm3).$ Since the points themselves are symmetric about $(0,0)$, the boxes are symmetric about $\left(\frac12,\frac12\right)$. The distance from $\left(\frac12,\frac12\right)$ to the furthest point on a box that lays on an axis, for instance $(6,1)$, is $\sqrt {\frac {11}2^2 + \frac12^2} = \sqrt {\frac {122}4}.$ The distance from $\left(\frac12,\frac12\right)$ to the furthest point on a box in the middle of a quadrant, for instance $(5,4)$, is $\sqrt {\frac92^2 + \frac72^2} = \sqrt {\frac {130}4}.$ The latter is the larger, and is $\frac {\sqrt {130}}2$, giving an answer of $130 + 2 = \boxed{132}$. | // Block 1
import graph; size(10.22cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-5.68,xmax=6.54,ymin=-5.52,ymax=6.5;
pen cqcqcq=rgb(0.75,0.75,0.75), evevff=rgb(0.9,0.9,1), zzttqq=rgb(0.6,0.2,0);
filldraw((-3,4)--(-2,4)--(-2,5)--(-3,5)--cycle,evevff,blue); filldraw((3,4)--(4,4)--(4,5)--(3,5)--cycle,evevff,blue); filldraw((4,3)--(5,3)--(5,4)--(4,4)--cycle,evevff,blue); filldraw((5,0)--(6,0)--(6,1)--(5,1)--cycle,evevff,blue); filldraw((4,-3)--(5,-3)--(5,-2)--(4,-2)--cycle,evevff,blue); filldraw((3,-3)--(3,-4)--(4,-4)--(4,-3)--cycle,evevff,blue); filldraw((0,-5)--(1,-5)--(1,-4)--(0,-4)--cycle,evevff,blue); filldraw((-3,-4)--(-2,-4)--(-2,-3)--(-3,-3)--cycle,evevff,blue); filldraw((-4,-3)--(-3,-3)--(-3,-2)--(-4,-2)--cycle,evevff,blue); filldraw((-4,3)--(-3,3)--(-3,4)--(-4,4)--cycle,evevff,blue); filldraw((-5,0)--(-4,0)--(-4,1)--(-5,1)--cycle,evevff,blue); filldraw((0,6)--(0,5)--(1,5)--(1,6)--cycle,evevff,blue);
/*grid*/ pen gs=linewidth(0.7)+cqcqcq+linetype("2 2"); real gx=1,gy=1;
for(real i=ceil(xmin/gx)*gx;i<=floor(xmax/gx)*gx;i+=gx) draw((i,ymin)--(i,ymax),gs); for(real i=ceil(ymin/gy)*gy;i<=floor(ymax/gy)*gy;i+=gy) draw((xmin,i)--(xmax,i),gs);
Label laxis; laxis.p=fontsize(10);
xaxis(xmin,xmax,defaultpen+black,Ticks(laxis,Step=1.0,Size=2,NoZero),Arrows(6),above=true); yaxis(ymin,ymax,defaultpen+black,Ticks(laxis,Step=1.0,Size=2,NoZero),Arrows(6),above=true); draw(circle((0,0),5),linewidth(1.6)); draw(circle((0.5,0.5),5.7),linetype("2 2")); draw((-3,4)--(-2,4),zzttqq); draw((-2,4)--(-2,5),zzttqq); draw((-2,5)--(-3,5),zzttqq); draw((-3,5)--(-3,4),zzttqq); draw((3,4)--(4,4),zzttqq); draw((4,4)--(4,5),zzttqq); draw((4,5)--(3,5),zzttqq); draw((3,5)--(3,4),zzttqq); draw((4,3)--(5,3),zzttqq); draw((5,3)--(5,4),zzttqq); draw((5,4)--(4,4),zzttqq); draw((4,4)--(4,3),zzttqq); draw((5,0)--(6,0),zzttqq); draw((6,0)--(6,1),zzttqq); draw((6,1)--(5,1),zzttqq); draw((5,1)--(5,0),zzttqq); draw((4,-3)--(5,-3),zzttqq); draw((5,-3)--(5,-2),zzttqq); draw((5,-2)--(4,-2),zzttqq); draw((4,-2)--(4,-3),zzttqq); draw((3,-3)--(3,-4),zzttqq); draw((3,-4)--(4,-4),zzttqq); draw((4,-4)--(4,-3),zzttqq); draw((4,-3)--(3,-3),zzttqq); draw((0,-5)--(1,-5),zzttqq); draw((1,-5)--(1,-4),zzttqq); draw((1,-4)--(0,-4),zzttqq); draw((0,-4)--(0,-5),zzttqq); draw((-3,-4)--(-2,-4),zzttqq); draw((-2,-4)--(-2,-3),zzttqq); draw((-2,-3)--(-3,-3),zzttqq); draw((-3,-3)--(-3,-4),zzttqq); draw((-4,-3)--(-3,-3),zzttqq); draw((-3,-3)--(-3,-2),zzttqq); draw((-3,-2)--(-4,-2),zzttqq); draw((-4,-2)--(-4,-3),zzttqq); draw((-4,3)--(-3,3),zzttqq); draw((-3,3)--(-3,4),zzttqq); draw((-3,4)--(-4,4),zzttqq); draw((-4,4)--(-4,3),zzttqq); draw((-5,0)--(-4,0),zzttqq); draw((-4,0)--(-4,1),zzttqq); draw((-4,1)--(-5,1),zzttqq); draw((-5,1)--(-5,0),zzttqq); draw((0,6)--(0,5),zzttqq); draw((0,5)--(1,5),zzttqq); draw((1,5)--(1,6),zzttqq); draw((1,6)--(0,6),zzttqq);
dot((0,5),ds); dot((3,4),ds); dot((4,3),ds); dot((5,0),ds); dot((4,-3),ds); dot((3,-4),ds); dot((0,-5),ds); dot((-3,-4),ds); dot((-4,-3),ds); dot((-4,3),ds); dot((-3,4),ds); dot((-2,4),ds); dot((-2,5),ds); dot((-3,5),ds); dot((4,4),ds); dot((4,5),ds); dot((3,5),ds); dot((5,3),ds); dot((5,4),ds); dot((4,4),ds); dot((6,0),ds); dot((6,1),ds); dot((5,1),ds); dot((5,-3),ds); dot((5,-2),ds); dot((4,-2),ds); dot((3,-3),ds); dot((4,-4),ds); dot((4,-3),ds); dot((1,-5),ds); dot((1,-4),ds); dot((0,-4),ds); dot((-2,-4),ds); dot((-2,-3),ds); dot((-3,-3),ds); dot((-3,-2),ds); dot((-4,-2),ds); dot((-3,3),ds); dot((-3,4),ds); dot((-4,4),ds); dot((-5,0),ds); dot((-4,0),ds); dot((-4,1),ds); dot((-5,1),ds); dot((0,6),ds); dot((1,5),ds); dot((1,6),ds);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
// Block 2
import graph; size(10.22cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-5.68,xmax=6.54,ymin=-5.52,ymax=6.5; pen cqcqcq=rgb(0.75,0.75,0.75), evevff=rgb(0.9,0.9,1), zzttqq=rgb(0.6,0.2,0); filldraw((-3,4)--(-2,4)--(-2,5)--(-3,5)--cycle,evevff,blue); filldraw((3,4)--(4,4)--(4,5)--(3,5)--cycle,evevff,blue); filldraw((4,3)--(5,3)--(5,4)--(4,4)--cycle,evevff,blue); filldraw((5,0)--(6,0)--(6,1)--(5,1)--cycle,evevff,blue); filldraw((4,-3)--(5,-3)--(5,-2)--(4,-2)--cycle,evevff,blue); filldraw((3,-3)--(3,-4)--(4,-4)--(4,-3)--cycle,evevff,blue); filldraw((0,-5)--(1,-5)--(1,-4)--(0,-4)--cycle,evevff,blue); filldraw((-3,-4)--(-2,-4)--(-2,-3)--(-3,-3)--cycle,evevff,blue); filldraw((-4,-3)--(-3,-3)--(-3,-2)--(-4,-2)--cycle,evevff,blue); filldraw((-4,3)--(-3,3)--(-3,4)--(-4,4)--cycle,evevff,blue); filldraw((-5,0)--(-4,0)--(-4,1)--(-5,1)--cycle,evevff,blue); filldraw((0,6)--(0,5)--(1,5)--(1,6)--cycle,evevff,blue); /*grid*/ pen gs=linewidth(0.7)+cqcqcq+linetype("2 2"); real gx=1,gy=1; for(real i=ceil(xmin/gx)*gx;i<=floor(xmax/gx)*gx;i+=gx) draw((i,ymin)--(i,ymax),gs); for(real i=ceil(ymin/gy)*gy;i<=floor(ymax/gy)*gy;i+=gy) draw((xmin,i)--(xmax,i),gs); Label laxis; laxis.p=fontsize(10); xaxis(xmin,xmax,defaultpen+black,Ticks(laxis,Step=1.0,Size=2,NoZero),Arrows(6),above=true); yaxis(ymin,ymax,defaultpen+black,Ticks(laxis,Step=1.0,Size=2,NoZero),Arrows(6),above=true); draw(circle((0,0),5),linewidth(1.6)); draw(circle((0.5,0.5),5.7),linetype("2 2")); draw((-3,4)--(-2,4),zzttqq); draw((-2,4)--(-2,5),zzttqq); draw((-2,5)--(-3,5),zzttqq); draw((-3,5)--(-3,4),zzttqq); draw((3,4)--(4,4),zzttqq); draw((4,4)--(4,5),zzttqq); draw((4,5)--(3,5),zzttqq); draw((3,5)--(3,4),zzttqq); draw((4,3)--(5,3),zzttqq); draw((5,3)--(5,4),zzttqq); draw((5,4)--(4,4),zzttqq); draw((4,4)--(4,3),zzttqq); draw((5,0)--(6,0),zzttqq); draw((6,0)--(6,1),zzttqq); draw((6,1)--(5,1),zzttqq); draw((5,1)--(5,0),zzttqq); draw((4,-3)--(5,-3),zzttqq); draw((5,-3)--(5,-2),zzttqq); draw((5,-2)--(4,-2),zzttqq); draw((4,-2)--(4,-3),zzttqq); draw((3,-3)--(3,-4),zzttqq); draw((3,-4)--(4,-4),zzttqq); draw((4,-4)--(4,-3),zzttqq); draw((4,-3)--(3,-3),zzttqq); draw((0,-5)--(1,-5),zzttqq); draw((1,-5)--(1,-4),zzttqq); draw((1,-4)--(0,-4),zzttqq); draw((0,-4)--(0,-5),zzttqq); draw((-3,-4)--(-2,-4),zzttqq); draw((-2,-4)--(-2,-3),zzttqq); draw((-2,-3)--(-3,-3),zzttqq); draw((-3,-3)--(-3,-4),zzttqq); draw((-4,-3)--(-3,-3),zzttqq); draw((-3,-3)--(-3,-2),zzttqq); draw((-3,-2)--(-4,-2),zzttqq); draw((-4,-2)--(-4,-3),zzttqq); draw((-4,3)--(-3,3),zzttqq); draw((-3,3)--(-3,4),zzttqq); draw((-3,4)--(-4,4),zzttqq); draw((-4,4)--(-4,3),zzttqq); draw((-5,0)--(-4,0),zzttqq); draw((-4,0)--(-4,1),zzttqq); draw((-4,1)--(-5,1),zzttqq); draw((-5,1)--(-5,0),zzttqq); draw((0,6)--(0,5),zzttqq); draw((0,5)--(1,5),zzttqq); draw((1,5)--(1,6),zzttqq); draw((1,6)--(0,6),zzttqq); dot((0,5),ds); dot((3,4),ds); dot((4,3),ds); dot((5,0),ds); dot((4,-3),ds); dot((3,-4),ds); dot((0,-5),ds); dot((-3,-4),ds); dot((-4,-3),ds); dot((-4,3),ds); dot((-3,4),ds); dot((-2,4),ds); dot((-2,5),ds); dot((-3,5),ds); dot((4,4),ds); dot((4,5),ds); dot((3,5),ds); dot((5,3),ds); dot((5,4),ds); dot((4,4),ds); dot((6,0),ds); dot((6,1),ds); dot((5,1),ds); dot((5,-3),ds); dot((5,-2),ds); dot((4,-2),ds); dot((3,-3),ds); dot((4,-4),ds); dot((4,-3),ds); dot((1,-5),ds); dot((1,-4),ds); dot((0,-4),ds); dot((-2,-4),ds); dot((-2,-3),ds); dot((-3,-3),ds); dot((-3,-2),ds); dot((-4,-2),ds); dot((-3,3),ds); dot((-3,4),ds); dot((-4,4),ds); dot((-5,0),ds); dot((-4,0),ds); dot((-4,1),ds); dot((-5,1),ds); dot((0,6),ds); dot((1,5),ds); dot((1,6),ds); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); | [] |
157 | Let $\mathcal{R}$ be the region consisting of the set of points in the coordinate plane that satisfy both $|8 - x| + y \le 10$ and $3y - x \ge 15$. When $\mathcal{R}$ is revolved around the line whose equation is $3y - x = 15$, the volume of the resulting solid is $\frac {m\pi}{n\sqrt {p}}$, where $m$, $n$, and $p$ are positive integers, $m$ and $n$ are relatively prime, and $p$ is not divisible by the square of any prime. Find $m + n + p$. | 2010 AIME I Problem 11 | The inequalities are equivalent to $y \ge x/3 + 5, y \le 10 - |x - 8|$. We can set them equal to find the two points of intersection, $x/3 + 5 = 10 - |x - 8| \Longrightarrow |x - 8| = 5 - x/3$. This implies that one of $x - 8, 8 - x = 5 - x/3$, from which we find that $(x,y) = \left(\frac 92, \frac {13}2\right), \left(\frac{39}{4}, \frac{33}{4}\right)$. The region $\mathcal{R}$ is a triangle, as shown above. When revolved about the line $y = x/3+5$, the resulting solid is the union of two right cones that share the same base and axis.
Let $h_1,h_2$ denote the height of the left and right cones, respectively (so $h_1 > h_2$), and let $r$ denote their common radius. The volume of a cone is given by $\frac 13 Bh$; since both cones share the same base, then the desired volume is $\frac 13 \cdot \pi r^2 \cdot (h_1 + h_2)$. The distance from the point $(8,10)$ to the line $x - 3y + 15 = 0$ is given by $\left|\frac{(8) - 3(10) + 15}{\sqrt{1^2 + (-3)^2}}\right| = \frac{7}{\sqrt{10}}$. The distance between $\left(\frac 92, \frac {13}2\right)$ and $\left(\frac{39}{4}, \frac{33}{4}\right)$ is given by $h_1 + h_2 = \sqrt{\left(\frac{18}{4} - \frac{39}{4}\right)^2 + \left(\frac{26}{4} - \frac{33}{4}\right)^2} = \frac{7\sqrt{10}}{4}$. Thus, the answer is $\frac{343\sqrt{10}\pi}{120} = \frac{343\pi}{12\sqrt{10}} \Longrightarrow 343 + 12 + 10 = \boxed{365}$. | // Block 1
size(280);
import graph; real min = 2, max = 12; pen dark = linewidth(1);
real P(real x) { return x/3 + 5; }
real Q(real x) { return 10 - abs(x - 8); }
path p = (2,P(2))--(8,P(8))--(12,P(12)), q = (2,Q(2))--(12,Q(12));
pair A = (8,10), B = (4.5,6.5), C= (9.75,8.25), F=foot(A,B,C), G=2*F-A; fill(A--B--C--cycle,rgb(0.9,0.9,0.9));
draw(graph(P,min,max),dark);
draw(graph(Q,min,max),dark);
draw(Arc((8,7.67),A,G,CW),dark,EndArrow(8)); draw(B--C--G--cycle,linetype("4 4"));
label("$y \ge x/3 + 5$",(max,P(max)),E,fontsize(10));
label("$y \le 10 - |x-8|$",(max,Q(max)),E,fontsize(10));
label("$\mathcal{R}$",(6,Q(6)),NW);
/* axes */
Label f; f.p=fontsize(8);
xaxis(0, max, Ticks(f, 6, 1));
yaxis(0, 10, Ticks(f, 5, 1));
// Block 2
size(200);
import three; currentprojection = perspective(0,0,10); defaultpen(linewidth(0.7)); pen dark=linewidth(1.3);
pair Fxy = foot((8,10),(4.5,6.5),(9.75,8.25));
triple A = (8,10,0), B = (4.5,6.5,0), C= (9.75,8.25,0), F=(Fxy.x,Fxy.y,0), G=2*F-A, H=(F.x,F.y,abs(F-A)),I=(F.x,F.y,-abs(F-A));
real theta1 = 1.2, theta2 = -1.7,theta3= abs(F-A),theta4=-2.2;
triple J=F+theta1*unit(A-F)+(0,0,((abs(F-A))^2-(theta1)^2)^.5 ),K=F+theta2*unit(A-F)+(0,0,((abs(F-A))^2-(theta2)^2)^.5 ),L=F+theta3*unit(A-F)+(0,0,((abs(F-A))^2-(theta3)^2)^.5 ),M=F+theta4*unit(A-F)-(0,0,((abs(F-A))^2-(theta4)^2)^.5 );
draw(C--A--B--G--cycle,linetype("4 4")+dark); draw(A..H..G..I..A); draw(C--B^^A--G,linetype("4 4")); draw(J--C--K); draw(L--B--M);
dot(B);dot(C);dot(F);
label("$h_1$",(B+F)/2,SE,fontsize(10));
label("$h_2$",(C+F)/2,S,fontsize(10));
label("$r$",(A+F)/2,E,fontsize(10));
// Block 3
size(280); import graph; real min = 2, max = 12; pen dark = linewidth(1); real P(real x) { return x/3 + 5; } real Q(real x) { return 10 - abs(x - 8); } path p = (2,P(2))--(8,P(8))--(12,P(12)), q = (2,Q(2))--(12,Q(12)); pair A = (8,10), B = (4.5,6.5), C= (9.75,8.25), F=foot(A,B,C), G=2*F-A; fill(A--B--C--cycle,rgb(0.9,0.9,0.9)); draw(graph(P,min,max),dark); draw(graph(Q,min,max),dark); draw(Arc((8,7.67),A,G,CW),dark,EndArrow(8)); draw(B--C--G--cycle,linetype("4 4")); label("$y \ge x/3 + 5$",(max,P(max)),E,fontsize(10)); label("$y \le 10 - |x-8|$",(max,Q(max)),E,fontsize(10)); label("$\mathcal{R}$",(6,Q(6)),NW); /* axes */ Label f; f.p=fontsize(8); xaxis(0, max, Ticks(f, 6, 1)); yaxis(0, 10, Ticks(f, 5, 1));
// Block 4
size(200); import three; currentprojection = perspective(0,0,10); defaultpen(linewidth(0.7)); pen dark=linewidth(1.3); pair Fxy = foot((8,10),(4.5,6.5),(9.75,8.25)); triple A = (8,10,0), B = (4.5,6.5,0), C= (9.75,8.25,0), F=(Fxy.x,Fxy.y,0), G=2*F-A, H=(F.x,F.y,abs(F-A)),I=(F.x,F.y,-abs(F-A)); real theta1 = 1.2, theta2 = -1.7,theta3= abs(F-A),theta4=-2.2; triple J=F+theta1*unit(A-F)+(0,0,((abs(F-A))^2-(theta1)^2)^.5 ),K=F+theta2*unit(A-F)+(0,0,((abs(F-A))^2-(theta2)^2)^.5 ),L=F+theta3*unit(A-F)+(0,0,((abs(F-A))^2-(theta3)^2)^.5 ),M=F+theta4*unit(A-F)-(0,0,((abs(F-A))^2-(theta4)^2)^.5 ); draw(C--A--B--G--cycle,linetype("4 4")+dark); draw(A..H..G..I..A); draw(C--B^^A--G,linetype("4 4")); draw(J--C--K); draw(L--B--M); dot(B);dot(C);dot(F); label("$h_1$",(B+F)/2,SE,fontsize(10)); label("$h_2$",(C+F)/2,S,fontsize(10)); label("$r$",(A+F)/2,E,fontsize(10)); | [] |
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