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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158 | Rectangle $ABCD$ and a semicircle with diameter $AB$ are coplanar and have nonoverlapping interiors. Let $\mathcal{R}$ denote the region enclosed by the semicircle and the rectangle. Line $\ell$ meets the semicircle, segment $AB$, and segment $CD$ at distinct points $N$, $U$, and $T$, respectively. Line $\ell$ divides region $\mathcal{R}$ into two regions with areas in the ratio $1: 2$. Suppose that $AU = 84$, $AN = 126$, and $UB = 168$. Then $DA$ can be represented as $m\sqrt {n}$, where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. Find $m + n$. | 2010 AIME I Problem 13 | Diagram
Solution 1
The center of the semicircle is also the midpoint of $AB$. Let this point be O. Let $h$ be the length of $AD$.
Rescale everything by 42, so $AU = 2, AN = 3, UB = 4$. Then $AB = 6$ so $OA = OB = 3$.
Since $ON$ is a radius of the semicircle, $ON = 3$. Thus $OAN$ is an equilateral triangle.
Let $X$, $Y$, and $Z$ be the areas of triangle $OUN$, sector $ONB$, and trapezoid $UBCT$ respectively.
$X = \frac {1}{2}(UO)(NO)\sin{O} = \frac {1}{2}(1)(3)\sin{60^\circ} = \frac {3}{4}\sqrt {3}$
$Y = \frac {1}{3}\pi(3)^2 = 3\pi$
To find $Z$ we have to find the length of $TC$. Project $T$ and $N$ onto $AB$ to get points $T'$ and $N'$. Notice that $UNN'$ and $TUT'$ are similar. Thus:
$\frac {TT'}{UT'} = \frac {UN'}{NN'} \implies \frac {TT'}{h} = \frac {1/2}{3\sqrt {3}/2} \implies TT' = \frac {\sqrt {3}}{9}h$.
Then $TC = T'C - T'T = UB - TT' = 4 - \frac {\sqrt {3}}{9}h$. So:
$Z = \frac {1}{2}(BU + TC)(CB) = \frac {1}{2}\left(8 - \frac {\sqrt {3}}{9}h\right)h = 4h - \frac {\sqrt {3}}{18}h^2$
Let $L$ be the area of the side of line $l$ containing regions $X, Y, Z$. Then
$L = X + Y + Z = \frac {3}{4}\sqrt {3} + 3\pi + 4h - \frac {\sqrt {3}}{18}h^2$
Obviously, the $L$ is greater than the area on the other side of line $l$. This other area is equal to the total area minus $L$. Thus:
$\frac {2}{1} = \frac {L}{6h + \frac {9}{2}{\pi} - L} \implies 12h + 9\pi = 3L$.
Now just solve for $h$.
\begin{align*} 12h + 9\pi & = \frac {9}{4}\sqrt {3} + 9\pi + 12h - \frac {\sqrt {3}}{6}h^2 \\ 0 & = \frac {9}{4}\sqrt {3} - \frac {\sqrt {3}}{6}h^2 \\ h^2 & = \frac {9}{4}(6) \\ h & = \frac {3}{2}\sqrt {6} \end{align*}
Don't forget to un-rescale at the end to get $AD = \frac {3}{2}\sqrt {6} \cdot 42 = 63\sqrt {6}$.
Finally, the answer is $63 + 6 = \boxed{069}$. | /* geogebra conversion, see azjps userscripts.org/scripts/show/72997 */ import graph; defaultpen(linewidth(0.7)+fontsize(10)); size(500); pen zzttqq = rgb(0.6,0.2,0); pen xdxdff = rgb(0.4902,0.4902,1); /* segments and figures */ draw((0,-154.31785)--(0,0)); draw((0,0)--(252,0)); draw((0,0)--(126,0),zzttqq); draw((126,0)--(63,109.1192),zzttqq); draw((63,109.1192)--(0,0),zzttqq); draw((-71.4052,(+9166.01287-109.1192*-71.4052)/21)--(504.60925,(+9166.01287-109.1192*504.60925)/21)); draw((0,-154.31785)--(252,-154.31785)); draw((252,-154.31785)--(252,0)); draw((0,0)--(84,0)); draw((84,0)--(252,0)); draw((63,109.1192)--(63,0)); draw((84,0)--(84,-154.31785)); draw(arc((126,0),126,0,180)); /* points and labels */ dot((0,0)); label("$A$",(-16.43287,-9.3374),NE/2); dot((252,0)); label("$B$",(255.242,5.00321),NE/2); dot((0,-154.31785)); label("$D$",(3.48464,-149.55669),NE/2); dot((252,-154.31785)); label("$C$",(255.242,-149.55669),NE/2); dot((126,0)); label("$O$",(129.36332,5.00321),NE/2); dot((63,109.1192)); label("$N$",(44.91307,108.57427),NE/2); label("$126$",(28.18236,40.85473),NE/2); dot((84,0)); label("$U$",(87.13819,5.00321),NE/2); dot((113.69848,-154.31785)); label("$T$",(116.61611,-149.55669),NE/2); dot((63,0)); label("$84$",(41.72627,-12.5242),NE/2); label("$168$",(167.60494,-12.5242),NE/2); dot((84,-154.31785)); dot((252,0)); label("$I$",(255.242,5.00321),NE/2); clip((-71.4052,-225.24323)--(-71.4052,171.51361)--(504.60925,171.51361)--(504.60925,-225.24323)--cycle); | [] |
159 | In $\triangle{ABC}$ with $AB = 12$, $BC = 13$, and $AC = 15$, let $M$ be a point on $\overline{AC}$ such that the incircles of $\triangle{ABM}$ and $\triangle{BCM}$ have equal radii. Then $\frac{AM}{CM} = \frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p + q$. | 2010 AIME I Problem 15 | Solution 1
Let $AM = x$, then $CM = 15 - x$. Also let $BM = d$ Clearly, $\frac {[ABM]}{[CBM]} = \frac {x}{15 - x}$. We can also express each area by the rs formula. Then $\frac {[ABM]}{[CBM]} = \frac {p(ABM)}{p(CBM)} = \frac {12 + d + x}{28 + d - x}$. Equating and cross-multiplying yields $25x + 2dx = 15d + 180$ or $d = \frac {25x - 180}{15 - 2x}.$ Note that for $d$ to be positive, we must have $7.2 < x < 7.5$.
By Stewart's Theorem, we have $12^2(15 - x) + 13^2x = d^215 + 15x(15 - x)$ or $432 = 3d^2 + 40x - 3x^2.$ Brute forcing by plugging in our previous result for $d$, we have $432 = \frac {3(25x - 180)^2}{(15 - 2x)^2} + 40x - 3x^2.$ Clearing the fraction and gathering like terms, we get $0 = 12x^4 - 340x^3 + 2928x^2 - 7920x.$
Aside: Since $x$ must be rational in order for our answer to be in the desired form, we can use the Rational Root Theorem to reveal that $6x$ is an integer because we can divide the polynomial by $2$. The only such $x$ in the above-stated range is $\frac {22}3$.
Legitimately solving that quartic, note that $x = 0$ and $x = 15$ should clearly be solutions, corresponding to the sides of the triangle and thus degenerate cevians. Factoring those out, we get $0 = 4x(x - 15)(3x^2 - 40x + 132) = x(x - 15)(x - 6)(3x - 22).$ The only solution in the desired range is thus $\frac {22}3$. Then $CM = \frac {23}3$, and our desired ratio $\frac {AM}{CM} = \frac {22}{23}$, giving us an answer of $\boxed{045}$.
Solution 2
Let $AM = 2x$ and $BM = 2y$ so $CM = 15 - 2x$. Let the incenters of $\triangle ABM$ and $\triangle BCM$ be $I_1$ and $I_2$ respectively, and their equal inradii be $r$. From $r = \sqrt {(s - a)(s - b)(s - c)}/s$, we find that
\begin{align*}r^2 & = \frac {(x + y - 6)( - x + y + 6)(x - y + 6)}{x + y + 6} & (1) \\ & = \frac {( - x + y + 1)(x + y - 1)( - x - y + 14)}{ - x + y + 14}. & (2) \end{align*}
Let the incircle of $\triangle ABM$ meet $AM$ at $P$ and the incircle of $\triangle BCM$ meet $CM$ at $Q$. Then note that $I_1 P Q I_2$ is a rectangle. Also, $\angle I_1 M I_2$ is right because $MI_1$ and $MI_2$ are the angle bisectors of $\angle AMB$ and $\angle CMB$ respectively and $\angle AMB + \angle CMB = 180^\circ$. By properties of tangents to circles $MP = (MA + MB - AB)/2 = x + y - 6$ and $MQ = (MB + MC - BC)/2 = - x + y + 1$. Now notice that the altitude of $M$ to $I_1 I_2$ is of length $r$, so by similar triangles we find that $r^2 = MP \cdot MQ = (x + y - 6)( - x + y + 1)$ (3). Equating (3) with (1) and (2) separately yields
\begin{align*} 2y^2 - 30 = 2xy + 5x - 7y \\ 2y^2 - 70 = - 2xy - 5x + 7y, \end{align*}
and adding these we have
\[4y^2 - 100 = 0\implies y = 5\implies x = 11/3 \\ \implies AM/MC = (22/3)/(15 - 22/3) = 22/23 \implies \boxed{045}.\]
Solution 3
Let the incircle of $ABM$ hit $AM$, $AB$, $BM$ at $X_{1},Y_{1},Z_{1}$, and let the incircle of $CBM$ hit $MC$, $BC$, $BM$ at $X_{2},Y_{2},Z_{2}$. Draw the incircle of $ABC$, and let it be tangent to $AC$ at $X$. Observe that we have a homothety centered at A sending the incircle of $ABM$ to that of $ABC$, and one centered at $C$ taking the incircle of $BCM$ to that of $ABC$. These have the same power, since they take incircles of the same size to the same circle. Also, the power of the homothety is $AX_{1}/AX=CX_{2}/CX$.
By standard computations, $AX=\dfrac{AB+AC-BC}{2}=7$ and $CX=\dfrac{BC+AC-AB}{2}=8$. Now, let $AX_{1}=7x$ and $CX_{2}=8x$. We will now go around and chase lengths. Observe that $BY_{1}=BA-AY_{1}=BA-AX_{1}=12-7x$. Then, $BZ_{1}=12-7x$. We also have $CY_{2}=CX_{2}=8x$, so $BY_{2}=13-8x$ and $BZ_{2}=13-8x$.
Observe now that $X_{1}M+MX_{2}=AC-15x=15(1-x)$. Also,$X_{1}M-MX_{2}=MZ_{1}-MZ_{2}=BZ_{2}-BZ_{1}=BY_{2}-BY_{1}=(1-x)$. Solving, we get $X_{1}M=8-8x$ and $MX_{2}=7-7x$ (as a side note, note that $AX_{1}+MX_{2}=X_{1}M+X_{2}C$, a result that I actually believe appears in Mandelbrot 1995-2003, or some book in that time-range).
Now, we get $BM=BZ_{2}+Z_{2}M=BZ_{2}+MX_{2}=20-15x$. To finish, we will compute area ratios. $\dfrac{[ABM]}{[CBM]}=\dfrac{AM}{MC}=\dfrac{8-x}{7+x}$. Also, since their inradii are equal, we get $\dfrac{[ABM]}{[CBM]}=\dfrac{40-16x}{40-14x}$. Equating and cross multiplying yields the quadratic $3x^{2}-8x+4=0$, so $x=2/3,2$. However, observe that $AX_{1}+CX_{2}=15x<15$, so we take $x=2/3$. Our ratio is therefore $\dfrac{8-2/3}{7+2/3}=\dfrac{22}{23}$, giving the answer $\boxed{045}$.
Note: Once we have $MX_1=8-8x$ and $MX_2=7-7x$, it's bit easier to use use the right triangle of $O_1MO_2$ than chasing the area ratio. The inradius of $\triangle{ABC}$ can be calculated to be $r=\sqrt{14}$, and the inradius of $ABM$ and $ACM$ are $r_1=r_2= xr$, so,
\[O_1O_2^2 = O_1M^2 + O_2M^2 = r_1^2+X_1M^2 + r_2^2 + X_2M^2\]
or,
\[(15(1-x))^2 = 2(\sqrt{14}x)^2 + (7(1-x))^2 + (8(1-x))^2\]
\[112(1-x)^2 = 28x^2\]
\[4(1-x)^2 = x^2\]
We get $x=\frac{2}{3}$ or $x=2$.
Solution 4
Suppose the incircle of $ABM$ touches $AM$ at $X$, and the incircle of $CBM$ touches $CM$ at $Y$. Then
\[r = AX \tan(A/2) = CY \tan(C/2)\]
We have $\cos A = \frac{12^2+15^2-13^2}{2\cdot 12\cdot 15} = \frac{200}{30\cdot 12}=\frac{5}{9}$, $\tan(A/2) = \sqrt{\frac{1-\cos A}{1+\cos A}} = \sqrt{\frac{9-5}{9+5}} = \frac{2}{\sqrt{14}}$
$\cos C = \frac{13^2+15^2-12^2}{2\cdot 13\cdot 15} = \frac{250}{30\cdot 13} = \frac{25}{39}$, $\tan(C/2) = \sqrt{\frac{39-25}{39+25}}=\frac{\sqrt{14}}{8}$,
Therefore $AX/CY = \tan(C/2)/\tan(A/2) = \frac{14}{2\cdot 8}= \frac{7}{8}.$
And since $AX=\frac{1}{2}(12+AM-BM)$, $CY = \frac{1}{2}(13+CM-BM)$,
\[\frac{12+AM-BM}{13+CM-BM} = \frac{7}{8}\]
\[96+8AM-8BM = 91 +7CM-7BM\]
\[BM= 5 + 8AM-7CM = 5 + 15AM - 7(CM+AM) = 5+15(AM-7) \dots\dots (1)\]
Now,
$\frac{AM}{CM} = \frac{[ABM]}{[CBM]} = \frac{\frac{1}{2}(12+AM+BM)r}{\frac{1}{2}(13+CM+BM)r}=\frac{12+AM+BM}{13+CM+BM}= \frac{12+BM}{13+BM} = \frac{17+15(AM-7)}{18+15(AM-7)}$
\[\frac{AM}{15} = \frac{17+15(AM-7)}{35+30(AM-7)} = \frac{15AM-88}{30AM-175}\]
\[6AM^2 - 35AM = 45AM-264\]
\[3AM^2 -40AM+132=0\]
\[(3AM-22)(AM-6)=0\]
So $AM=22/3$ or $6$. But from (1) we know that $5+15(AM-7)>0$, or $AM>7-1/3>6$, so $AM=22/3$, $CM=15-22/3=23/3$, $AM/CM=22/23$.
Solution 5
Let the common inradius equal r, $BM = x$, $AM = y$, $MC = z$
From the prespective of $\triangle{ABM}$ and $\triangle{BMC}$ we get:
$S_{ABM} = rs = r \cdot (\frac{12+x+y}{2})$ $S_{BMC} = rs = r \cdot (\frac{13+x+z}{2})$
Add two triangles up, we get $\triangle{ABC}$ :
$S_{ABC} = S_{ABM} + S_{BMC} = r \cdot \frac{25+2x+y+z}{2}$
Since $y + z = 15$, we get:
$r = \frac{S_{ABC}}{20 + x}$
By drawing an altitude from $I_1$ down to a point $H_1$ and from $I_2$ to $H_2$, we can get:
$r \cdot cot(\frac{\angle A}{2}) =r \cdot A H_1 = r \cdot \frac{12+y-x}{2}$ and
$r \cdot cot(\frac{\angle C}{2}) = r \cdot H_2 C = r \cdot \frac{13+z-x}{2}$
Adding these up, we get:
$r \cdot (cot(\frac{\angle A}{2})+cot(\frac{\angle C}{2})) = \frac{25+y+z-2x}{2} = \frac{40-2x}{2} = 20-x$
$r = \frac{20-x}{cot(\frac{\angle A}{2})+cot(\frac{\angle C}{2})}$
Now, we have 2 values equal to r, we can set them equal to each other:
$\frac{S_{ABC}}{20 + x} = \frac{20-x}{cot(\frac{\angle A}{2})+cot(\frac{\angle C}{2})}$
If we let R denote the incircle of ABC, note:
AC = $(cot(\frac{\angle A}{2})+cot(\frac{\angle C}{2})) \cdot R = 15$ and
$S_{ABC} = \frac{12+13+15}{2} \cdot R = 20 \cdot R$.
By cross multiplying the equation above, we get:
$400 - x^2 = (cot(\frac{\angle A}{2})+cot(\frac{\angle C}{2})) \cdot S_{ABC} = (cot(\frac{\angle A}{2})+cot(\frac{\angle C}{2})) \cdot R \cdot 20 = 15 \cdot 20 = 300$
We can find out x:
$x = 10$.
Now, we can find ratio of y and z:
$\frac{AM}{CM} = \frac{y}{z} = \frac{S_{ABM}}{S_{BCM}} = \frac{r \cdot \frac{22+y}{2} }{r \cdot \frac{23+z}{2}} = \frac{22+y}{23+z} = \frac{22}{23}$
The answer is $\boxed{045}$.
-Alexlikemath
Solution 6 (Similar to Solution 1 with easier computation)
Let $CM=x, AM=rx, BM=d$. $x+rx=15\Rightarrow x=\frac{15}{1+r}$.
Similar to Solution 1, we have
\[r=\frac{[AMB]}{[CMB]}=\frac{12+rx+d}{13+x+d} \Rightarrow d=\frac{13r-12}{1-r}\]
as well as
\[12^2\cdot x + 13^2 rx=15x\cdot rx+15d^2 (\text{via Stewart's Theorem})\]
\[\frac{(12^2 + 13^2r) \cdot 15}{1+r} - \frac{15r\cdot 15^2}{(1+r)^2}=\frac{15(13r-12)^2}{(1-r)^2}\]
\[\frac{169r^2+88r+144}{(1+r)^2}=\frac{(13r-12)^2}{(1-r)^2} =\frac{169r^2-312r+144}{(1-r)^2} =\frac{400r}{4r}=100\]
(here we used the fact that if $\frac{a}{b} = \frac{c}{d} = k,$ then $\frac{a-c}{b-d}=k$ as well.)
Notice $\frac{12}{13} < r < 1$, so $\frac{13r-12}{1-r} = 10$ and $\boxed{r = \frac{22}{23}}.$
~ asops
Solution 7 (No Stewart's or trig, fast + clever)
Let $BM = d, AM = x, CM = 15 - x$. Observe that we have the equation by the incircle formula:
\[\frac{[ABM]}{12 + AM + MB} = \frac{[CBM]}{13 + CM + MB} \implies \frac{AM}{CM} = \frac{12 + MB}{13 + MB} \implies \frac{x}{15 - x} = \frac{12 + d}{13 + d}.\]
Now let $X$ be the point of tangency between the incircle of $\triangle ABC$ and $AC$. Additionally, let $P$ and $Q$ be the points of tangency between the incircles of $\triangle ABM$ and $\triangle CBM$ with $AC$ respectively. Some easy calculation yields $AX = 7, CX = 8$. By homothety we have
\[\frac{AP}{7} = \frac{CQ}{8} \implies 8(AP) = 7(CQ) \implies 8(12 + x - d) = 7(13 + 15 - x - d) \implies d = 15x - 100.\]
Substituting into the first equation derived earlier it is left to solve
\[\frac{x}{15 - x} = \frac{15x - 88}{15x - 87} \implies 3x^2 - 40x + 132 \implies (x - 6)(3x - 22) = 0.\]
Now $x = 6$ yields $d = -10$ which is invalid, hence $x = \frac{22}{3}$ so $\frac{AM}{CM} = \frac{\frac{22}{3}}{15 - \frac{22}{3}} = \frac{22}{23}.$ The requested sum is $22 + 23 = \boxed{45}$. ~blueprimes | // Block 1
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160 | A point $P$ is chosen at random in the interior of a unit square $S$. Let $d(P)$ denote the distance from $P$ to the closest side of $S$. The probability that $\frac{1}{5}\le d(P)\le\frac{1}{3}$ is equal to $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. | 2010 AIME II Problem 2 | Any point outside the square with side length $\frac{1}{3}$ that has the same center and orientation as the unit square and inside the square with side length $\frac{3}{5}$ that has the same center and orientation as the unit square has $\frac{1}{5}\le d(P)\le\frac{1}{3}$.
Since the area of the unit square is $1$, the probability of a point $P$ with $\frac{1}{5}\le d(P)\le\frac{1}{3}$ is the area of the shaded region, which is the difference of the area of two squares.
$\left(\frac{3}{5}\right)^2-\left(\frac{1}{3}\right)^2=\frac{9}{25}-\frac{1}{9}=\frac{56}{225}$
Thus, the answer is $56 + 225 = \boxed{281}.$ | // Block 1
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161 | Triangle $ABC$ with right angle at $C$, $\angle BAC < 45^\circ$ and $AB = 4$. Point $P$ on $\overline{AB}$ is chosen such that $\angle APC = 2\angle ACP$ and $CP = 1$. The ratio $\frac{AP}{BP}$ can be represented in the form $p + q\sqrt{r}$, where $p$, $q$, $r$ are positive integers and $r$ is not divisible by the square of any prime. Find $p+q+r$. | 2010 AIME II Problem 14 | Let $O$ be the circumcenter of $ABC$ and let the intersection of $CP$ with the circumcircle be $D$. It now follows that $\angle{DOA} = 2\angle ACP = \angle{APC} = \angle{DPB}$. Hence $ODP$ is isosceles and $OD = DP = 2$.
Denote $E$ the projection of $O$ onto $CD$. Now $CD = CP + DP = 3$. By the Pythagorean Theorem, $OE = \sqrt {2^2 - \frac {3^2}{2^2}} = \sqrt {\frac {7}{4}}$. Now note that $EP = \frac {1}{2}$. By the Pythagorean Theorem, $OP = \sqrt {\frac {7}{4} + \frac {1^2}{2^2}} = \sqrt {2}$. Hence it now follows that,
\[\frac {AP}{BP} = \frac {AO + OP}{BO - OP} = \frac {2 + \sqrt {2}}{2 - \sqrt {2}} = 3 + 2\sqrt {2}\]
This gives that the answer is $\boxed{007}$.
An alternate finish for this problem would be to use Power of a Point on $BA$ and $CD$. By Power of a Point Theorem, $CP\cdot PD=1\cdot 2=BP\cdot PA$. Since $BP+PA=4$, we can solve for $BP$ and $PA$, giving the same values and answers as above. | // Block 1
/* geogebra conversion, see azjps userscripts.org/scripts/show/72997 */
import graph; defaultpen(linewidth(0.7)+fontsize(10)); size(250); real lsf = 0.5; /* changes label-to-point distance */
pen xdxdff = rgb(0.49,0.49,1); pen qqwuqq = rgb(0,0.39,0); pen fftttt = rgb(1,0.2,0.2);
/* segments and figures */
draw((0.2,0.81)--(0.33,0.78)--(0.36,0.9)--(0.23,0.94)--cycle,qqwuqq); draw((0.81,-0.59)--(0.93,-0.54)--(0.89,-0.42)--(0.76,-0.47)--cycle,qqwuqq); draw(circle((2,0),2)); draw((0,0)--(0.23,0.94),linewidth(1.6pt)); draw((0.23,0.94)--(4,0),linewidth(1.6pt)); draw((0,0)--(4,0),linewidth(1.6pt)); draw((0.23,(+0.55-0.94*0.23)/0.35)--(4.67,(+0.55-0.94*4.67)/0.35));
/* points and labels */
label("$1$", (0.26,0.42), SE*lsf); draw((1.29,-1.87)--(2,0)); label("$2$", (2.91,-0.11), SE*lsf); label("$2$", (1.78,-0.82), SE*lsf); pair parametricplot0_cus(real t){
return (0.28*cos(t)+0.23,0.28*sin(t)+0.94);
}
draw(graph(parametricplot0_cus,-1.209429202888189,-0.24334747753738661)--(0.23,0.94)--cycle,fftttt); pair parametricplot1_cus(real t){
return (0.28*cos(t)+0.59,0.28*sin(t)+0);
}
draw(graph(parametricplot1_cus,0.0,1.9321634507016043)--(0.59,0)--cycle,fftttt); label("$\theta$", (0.42,0.77), SE*lsf); label("$2\theta$", (0.88,0.38), SE*lsf); draw((2,0)--(0.76,-0.47)); pair parametricplot2_cus(real t){
return (0.28*cos(t)+2,0.28*sin(t)+0);
}
draw(graph(parametricplot2_cus,-1.9321634507016048,0.0)--(2,0)--cycle,fftttt); label("$2\theta$", (2.18,-0.3), SE*lsf); dot((0,0)); label("$B$", (-0.21,-0.2),NE*lsf); dot((4,0)); label("$A$", (4.03,0.06),NE*lsf); dot((2,0)); label("$O$", (2.04,0.06),NE*lsf); dot((0.59,0)); label("$P$", (0.28,-0.27),NE*lsf); dot((0.23,0.94)); label("$C$", (0.07,1.02),NE*lsf); dot((1.29,-1.87)); label("$D$", (1.03,-2.12),NE*lsf); dot((0.76,-0.47)); label("$E$", (0.56,-0.79),NE*lsf); clip((-0.92,-2.46)--(-0.92,2.26)--(4.67,2.26)--(4.67,-2.46)--cycle);
// Block 2
/* geogebra conversion, see azjps userscripts.org/scripts/show/72997 */ import graph; defaultpen(linewidth(0.7)+fontsize(10)); size(250); real lsf = 0.5; /* changes label-to-point distance */ pen xdxdff = rgb(0.49,0.49,1); pen qqwuqq = rgb(0,0.39,0); pen fftttt = rgb(1,0.2,0.2); /* segments and figures */ draw((0.2,0.81)--(0.33,0.78)--(0.36,0.9)--(0.23,0.94)--cycle,qqwuqq); draw((0.81,-0.59)--(0.93,-0.54)--(0.89,-0.42)--(0.76,-0.47)--cycle,qqwuqq); draw(circle((2,0),2)); draw((0,0)--(0.23,0.94),linewidth(1.6pt)); draw((0.23,0.94)--(4,0),linewidth(1.6pt)); draw((0,0)--(4,0),linewidth(1.6pt)); draw((0.23,(+0.55-0.94*0.23)/0.35)--(4.67,(+0.55-0.94*4.67)/0.35)); /* points and labels */ label("$1$", (0.26,0.42), SE*lsf); draw((1.29,-1.87)--(2,0)); label("$2$", (2.91,-0.11), SE*lsf); label("$2$", (1.78,-0.82), SE*lsf); pair parametricplot0_cus(real t){ return (0.28*cos(t)+0.23,0.28*sin(t)+0.94); } draw(graph(parametricplot0_cus,-1.209429202888189,-0.24334747753738661)--(0.23,0.94)--cycle,fftttt); pair parametricplot1_cus(real t){ return (0.28*cos(t)+0.59,0.28*sin(t)+0); } draw(graph(parametricplot1_cus,0.0,1.9321634507016043)--(0.59,0)--cycle,fftttt); label("$\theta$", (0.42,0.77), SE*lsf); label("$2\theta$", (0.88,0.38), SE*lsf); draw((2,0)--(0.76,-0.47)); pair parametricplot2_cus(real t){ return (0.28*cos(t)+2,0.28*sin(t)+0); } draw(graph(parametricplot2_cus,-1.9321634507016048,0.0)--(2,0)--cycle,fftttt); label("$2\theta$", (2.18,-0.3), SE*lsf); dot((0,0)); label("$B$", (-0.21,-0.2),NE*lsf); dot((4,0)); label("$A$", (4.03,0.06),NE*lsf); dot((2,0)); label("$O$", (2.04,0.06),NE*lsf); dot((0.59,0)); label("$P$", (0.28,-0.27),NE*lsf); dot((0.23,0.94)); label("$C$", (0.07,1.02),NE*lsf); dot((1.29,-1.87)); label("$D$", (1.03,-2.12),NE*lsf); dot((0.76,-0.47)); label("$E$", (0.56,-0.79),NE*lsf); clip((-0.92,-2.46)--(-0.92,2.26)--(4.67,2.26)--(4.67,-2.46)--cycle); | [] |
162 | In rectangle $ABCD$, $AB = 12$ and $BC = 10$. Points $E$ and $F$ lie inside rectangle $ABCD$ so that $BE = 9$, $DF = 8$, $\overline{BE} \parallel \overline{DF}$, $\overline{EF} \parallel \overline{AB}$, and line $BE$ intersects segment $\overline{AD}$. The length $EF$ can be expressed in the form $m \sqrt{n} - p$, where $m$, $n$, and $p$ are positive integers and $n$ is not divisible by the square of any prime. Find $m + n + p$. | 2011 AIME I Problem 2 | Let us call the point where $\overline{EF}$ intersects $\overline{AD}$ point $G$, and the point where $\overline{EF}$ intersects $\overline{BC}$ point $H$. Since angles $FHB$ and $EGA$ are both right angles, and angles $BEF$ and $DFE$ are congruent due to parallelism, right triangles $BHE$ and $DGF$ are similar. This implies that $\frac{BH}{GD} = \frac{9}{8}$. Since $BC=10$, $BH+GD=BH+HC=BC=10$. ($HC$ is the same as $GD$ because they are opposite sides of a rectangle.) Now, we have a system:
$\frac{BH}{GD}=\frac{9}8$
$BH+GD=10$
Solving this system (easiest by substitution), we get that:
$BH=\frac{90}{17}$
$GD=\frac{80}{17}$
Using the Pythagorean Theorem, we can solve for the remaining sides of the two right triangles:
$\sqrt{9^2-\left(\frac{90}{17}\right)^2}$ and $\sqrt{8^2-\left(\frac{80}{17}\right)^2}$
Notice that adding these two sides would give us twelve plus the overlap $EF$. This means that:
$EF= \sqrt{9^2-\left(\frac{90}{17}\right)^2}+\sqrt{8^2-\left(\frac{80}{17}\right)^2}-12=3\sqrt{21}-12$
Since $21$ isn't divisible by any perfect square, our answer is:
$3+21+12=\boxed{36}$ | unitsize(0.5cm);
defaultpen(0.8);
pair A=(0,0), B=(12,0), C=(12,10), D=(0,10);
draw(A--B--C--D--cycle);
label("$A$",A,SW);
label("$B$",B,SE);
label("$C$",C,NE);
label("$D$",D,NW);
pair E=(3,5), F=(9,5);
dot("$E$",E,N);
dot("$F$",F,N);
pair G = extension(A, D, E, F);
pair H = extension(B, C, E, F);
draw(G--H);
dot("$G$",G,W);
dot("$H$", H,E); | [] |
163 | In triangle $ABC$, $AB=125$, $AC=117$ and $BC=120$. The angle bisector of angle $A$ intersects $\overline{BC}$ at point $L$, and the angle bisector of angle $B$ intersects $\overline{AC}$ at point $K$. Let $M$ and $N$ be the feet of the perpendiculars from $C$ to $\overline{BK}$ and $\overline{AL}$, respectively. Find $MN$. | 2011 AIME I Problem 4 | Extend ${CM}$ and ${CN}$ such that they intersect line ${AB}$ at points $P$ and $Q$, respectively.
Since ${BM}$ is the angle bisector of angle $B$ and ${CM}$ is perpendicular to ${BM}$, $\triangle BCP$ must be an isoceles triangle, so $BP=BC=120$, and $M$ is the midpoint of ${CP}$. For the same reason, $AQ=AC=117$, and $N$ is the midpoint of ${CQ}$.
Hence $MN=\tfrac 12 PQ$. Since \[PQ=BP+AQ-AB=120+117-125=112,\] so $MN=\boxed{056}$. | // Block 1
defaultpen(fontsize(10)+0.8); size(200);
pen p=fontsize(9)+linewidth(3);
pair A,B,C,D,K,L,M,N,P,Q;
A=origin; B=(125,0); C=IP(CR(A,117),CR(B,120)); L=extension(B,C,A,bisectorpoint(B,A,C)); K=extension(A,C,B,bisectorpoint(C,B,A)); M=foot(C,B,K); N=foot(C,A,L);
draw(A--B--C--A); draw(A--L^^B--K, gray+dashed+0.5); draw(M--C--N^^N--extension(A,B,C,N)^^M--extension(A,B,C,M), gray+0.5);
dot("$A$",A,dir(200),p); dot("$B$",B,right,p); dot("$C$",C,up,p); dot("$L$",L,2*dir(70),p); dot("$N$",N,2*dir(-90),p); dot("$M$",M,2*dir(-90),p); dot("$P$",extension(A,B,C,M),2*down,p); dot("$Q$",extension(A,B,C,N),2*down,p);
label("$125$",A--B,down,fontsize(10)); label("$117$",A--C,2*dir(130),fontsize(10)); label("$120$",B--C,1.5*dir(30),fontsize(10));
// Block 2
defaultpen(fontsize(10)+0.8); size(200); pen p=fontsize(9)+linewidth(3); pair A,B,C,D,K,L,M,N,P,Q; A=origin; B=(125,0); C=IP(CR(A,117),CR(B,120)); L=extension(B,C,A,bisectorpoint(B,A,C)); K=extension(A,C,B,bisectorpoint(C,B,A)); M=foot(C,B,K); N=foot(C,A,L); draw(A--B--C--A); draw(A--L^^B--K, gray+dashed+0.5); draw(M--C--N^^N--extension(A,B,C,N)^^M--extension(A,B,C,M), gray+0.5); dot("$A$",A,dir(200),p); dot("$B$",B,right,p); dot("$C$",C,up,p); dot("$L$",L,2*dir(70),p); dot("$N$",N,2*dir(-90),p); dot("$M$",M,2*dir(-90),p); dot("$P$",extension(A,B,C,M),2*down,p); dot("$Q$",extension(A,B,C,N),2*down,p); label("$125$",A--B,down,fontsize(10)); label("$117$",A--C,2*dir(130),fontsize(10)); label("$120$",B--C,1.5*dir(30),fontsize(10)); | [] |
164 | The probability that a set of three distinct vertices chosen at random from among the vertices of a regular n-gon determine an obtuse triangle is $\frac{93}{125}$ . Find the sum of all possible values of $n$. | 2011 AIME I Problem 10 | Let the regular $n$-gon be inscribed in a circle, allowing the use of the inscribed angle theorem.
We wish to determine the number of three distinct vertices such that they form an obtuse triangle.
WLOG fix vertex $A$
Let $B$ be $k$ edges away from $A$ clockwise
$C$ be $j$ edges away from $A$ in the other direction
To ensure $\triangle ABC$ has no conciding vertices the edges between them $k, j, n - (k+j) > 0$
\begin{cases}
k > 0\\
j > 0\\
k + j < n\\
\end{cases}
The angle of $\triangle ABC$ are $\frac{180^\circ}{n} \cdot (n - k - j), \frac{180^\circ}{n} \cdot j, \frac{180^\circ}{n} \cdot k$ respectively. Solving for the angles $\frac{180^\circ}{n} \cdot x > 90$ for an obtuse,
\begin{cases}
k > \frac{n}{2}\\
j > \frac{n}{2}\\
k + j < \frac{n}{2}
\end{cases}
These are the conditions for each angle of $\triangle ABC$ to be obtuse.
Hence we wish to find lattice point $(k, j)$ that satisfy the inequalities.
As expected these are disjoint, as a triangle contain at most one obtuse angle.
We wish to take the union of lattice points within the pink shaded regions. The blue line represents the conditions that make $k$ and $j$ form a non degenerate triangle.
Case $n = 2c$
Focus on of the pink regions.
The number of lattice points is $1 + 2 + 3 + \cdots + (c - 2) = \frac{1}{2}(c - 2)(c-1)$
Multiply by $3$ for each of the regions resulting in $\frac{3}{2}(c - 2)(c-1)$
Case $n = 2c - 1$
Proceed similarly and get $\frac{3}{2}(c - 2)(c-1)$ too.
At first the WLOG fixing a single vertex, hence $\times n$
Eac $\triangle ABC$ is counted $3$ times, once for each vertex, hence $\div 3$
\[\frac{3}{2}(c - 2)(c-1) \cdot n \div 3 = \frac{n(c-2)(c-1)}{2}\]
Where $c =\left \lfloor \frac{n}{2} \right \rfloor$
The total number of ways to choose 3 distinct vertices on the $n$-gon is $\binom{n}{3}$
\[\frac{n(c-2)(c-1)}{2 \dbinom{n}{3} } = \frac{93}{125}\]
\begin{align*}
125 \cdot \frac{n(c-2)(c-1)}{2} &= 93 \cdot \dbinom{n}{3}\\
125 \cdot \frac{n(c-2)(c-1)}{2} &= 93 \cdot \frac{n(n-1)(n-2)}{6}\\
125 (c-2)(c-1) &= 31 (n-1)(n-2)\\
125 (\left \lfloor \frac{n}{2} \right \rfloor-2)(\left \lfloor \frac{n}{2} \right \rfloor-1) &= 31 (n-1)(n-2)
\end{align*}
Case $n = 2c$
\begin{align*}
125(c-2)(c-1) &= 31(2c - 1)(2c - 2) \\
&= 31 \cdot 2 \cdot (2c - 1)(c-1)\\
125(c-2) &= 62(2c-1)\\
c &= 188\\
n &= 2c = 376
\end{align*}
Case $n = 2c - 1$
\begin{align*}
125(c-2)(c-1) &= 31(2c - 2)(2c - 3) \\
&= 31 \cdot 2 \cdot (c - 1)(2c-3)\\
125(c-2) &= 62(2c-3)\\
c &= 64\\
n &= 2c - 1 = 127
\end{align*}
The final is thus $376 + 127 = \boxed{503}$
~clod | // Block 1
/* Generated by Cloud's Excalidraw to Asymptote */
draw(shift(-950.5, -920.5) * rotate(0) * ellipse((0, 0), 295.5, 295.5), rgb(0.11, 0.11, 0.11)+linewidth(1)+solid);
draw((-1073, -654)--(-817, -654), rgb(0.87, 0.19, 0.19)+linewidth(2)+solid);
draw((-817, -658)--(-662, -855), rgb(0.87, 0.19, 0.19)+linewidth(2)+solid);
draw((-664, -858)--(-722, -1108), rgb(0.4, 0.25, 0.85)+linewidth(2)+solid);
draw((-723, -1109)--(-957, -1213), rgb(0.4, 0.25, 0.85)+linewidth(2)+solid);
draw((-957, -1214)--(-1185, -1096), rgb(0.09, 0.44, 0.76)+linewidth(2)+solid);
draw((-1186, -1097)--(-1234, -844), rgb(0.09, 0.44, 0.76)+linewidth(2)+solid);
draw((-1234, -841)--(-1077, -654), rgb(0.09, 0.44, 0.76)+linewidth(2)+solid);
filldraw(shift(-1077.5, -655.5) * rotate(0) * ellipse((0, 0), 4.5, 4.5), rgb(0.11, 0.11, 0.11), rgb(0.11, 0.11, 0.11)+linewidth(2)+solid);
filldraw(shift(-819.5, -654.5) * rotate(0) * ellipse((0, 0), 4.5, 4.5), rgb(0.11, 0.11, 0.11), rgb(0.11, 0.11, 0.11)+linewidth(2)+solid);
filldraw(shift(-1234.5, -840.5) * rotate(0) * ellipse((0, 0), 4.5, 4.5), rgb(0.11, 0.11, 0.11), rgb(0.11, 0.11, 0.11)+linewidth(2)+solid);
filldraw(shift(-1186.5, -1097.5) * rotate(0) * ellipse((0, 0), 4.5, 4.5), rgb(0.11, 0.11, 0.11), rgb(0.11, 0.11, 0.11)+linewidth(2)+solid);
filldraw(shift(-958.5, -1214.5) * rotate(0) * ellipse((0, 0), 4.5, 4.5), rgb(0.11, 0.11, 0.11), rgb(0.11, 0.11, 0.11)+linewidth(2)+solid);
filldraw(shift(-721.5, -1107.5) * rotate(0) * ellipse((0, 0), 4.5, 4.5), rgb(0.11, 0.11, 0.11), rgb(0.11, 0.11, 0.11)+linewidth(2)+solid);
filldraw(shift(-661.5, -855.5) * rotate(0) * ellipse((0, 0), 4.5, 4.5), rgb(0.11, 0.11, 0.11), rgb(0.11, 0.11, 0.11)+linewidth(2)+solid);
label("A", (-1088, -631.5), fontsize(20pt) + rgb(0.11, 0.11, 0.11)+linewidth(1)+solid);
label("B", (-631, -855.5), fontsize(20pt) + rgb(0.11, 0.11, 0.11)+linewidth(1)+solid);
label("C", (-966, -1258.5), fontsize(20pt) + rgb(0.11, 0.11, 0.11)+linewidth(1)+solid);
draw((-1077, -659)--(-960, -1212), rgb(0.11, 0.11, 0.11)+linewidth(2)+solid);
draw((-1072, -657)--(-662, -852), rgb(0.11, 0.11, 0.11)+linewidth(2)+solid);
draw((-957, -1210)--(-663, -854), rgb(0.11, 0.11, 0.11)+linewidth(2)+solid);
label("$k$ edges", (-749, -624.5), fontsize(20pt) + rgb(0.87, 0.19, 0.19)+linewidth(1)+solid);
label("$j$ edges", (-1294.5, -1026.5), fontsize(20pt) + rgb(0.09, 0.44, 0.76)+linewidth(1)+solid);
label("$n - (k + j)$ edges", (-655, -1174), fontsize(15pt) + rgb(0.4, 0.25, 0.85)+linewidth(1)+solid);
label("$\frac{180^\circ}{n} \cdot j$", (-724, -862), fontsize(15pt) + rgb(0.09, 0.44, 0.76)+linewidth(4)+solid);
label("$\frac{180^\circ}{n} \cdot k$", (-935, -1112), fontsize(15pt) + rgb(0.87, 0.19, 0.19)+linewidth(4)+solid);
label("$\frac{180^\circ}{n} \cdot (n - k - j)$", (-946, -791), fontsize(15pt) + rgb(0.4, 0.25, 0.85)+linewidth(4)+solid);
// Block 2
/* Generated by Cloud's Excalidraw to Asymptote */
filldraw((-1115, -1529)--(-1115, -1729)--(-915, -1729)--(-1115, -1529)--cycle, rgb(1, 0.78, 0.78), rgb(0, 0, 0)+linewidth(0)+solid);
filldraw((-915, -1729)--(-915, -1929)--(-715, -1929)--(-915, -1729)--cycle, rgb(1, 0.78, 0.78), rgb(0, 0, 0)+linewidth(0)+solid);
filldraw((-1115, -1729)--(-1115, -1929)--(-915, -1929)--(-1115, -1729)--cycle, rgb(1, 0.78, 0.78), rgb(0, 0, 0)+linewidth(0)+solid);
label("$k$", (-651, -1927.5), fontsize(32pt) + rgb(0.11, 0.11, 0.11)+linewidth(4)+solid);
label("$j$", (-1117, -1462), fontsize(25pt) + rgb(0.11, 0.11, 0.11)+linewidth(4)+solid);
label("$\frac{n}{2}$", (-916, -1959.5), fontsize(23pt) + rgb(0.11, 0.11, 0.11)+linewidth(4)+solid);
label("$n$", (-714, -1971), fontsize(28pt) + rgb(0.11, 0.11, 0.11)+linewidth(4)+solid);
draw((-1115, -1728)--(-915, -1928), rgb(0.87, 0.19, 0.19)+linewidth(4)+solid);
draw((-1115, -1528)--(-715, -1928), rgb(0.09, 0.44, 0.76)+linewidth(4)+solid);
draw((-1115, -1718)--(-1115, -1488), rgb(0.11, 0.11, 0.11)+linewidth(4)+solid, arrow=ArcArrow(size=20));
draw((-1115, -1718)--(-1115, -1948), rgb(0.11, 0.11, 0.11)+linewidth(4)+solid);
draw((-905, -1928)--(-675, -1928), rgb(0.11, 0.11, 0.11)+linewidth(4)+solid, arrow=ArcArrow(size=20));
draw((-905, -1928)--(-1135, -1928), rgb(0.11, 0.11, 0.11)+linewidth(4)+solid);
draw((-915, -1528)--(-915, -1926), rgb(0.87, 0.19, 0.19)+linewidth(4)+solid);
draw((-715, -1728)--(-1115, -1728), rgb(0.87, 0.19, 0.19)+linewidth(4)+solid);
filldraw((-715, -1908)--(-715, -1948)--cycle, rgb(0.87, 0.19, 0.19), rgb(0.11, 0.11, 0.11)+linewidth(4)+solid);
filldraw((-1135, -1528)--(-1095, -1528)--cycle, rgb(0.87, 0.19, 0.19), rgb(0.11, 0.11, 0.11)+linewidth(4)+solid);
label("$n$", (-1159, -1529), fontsize(28pt) + rgb(0.11, 0.11, 0.11)+linewidth(4)+solid);
label("$\frac{n}{2}$", (-1142, -1727.5), fontsize(23pt) + rgb(0.11, 0.11, 0.11)+linewidth(4)+solid);
label("$k + j < n$", (-776, -1538.5), fontsize(17pt) + rgb(0.09, 0.44, 0.76)+linewidth(4)+solid);
label("$k, j > \frac{n}{2}$", (-867.5, -1636.5), fontsize(14pt) + rgb(0.87, 0.19, 0.19)+linewidth(4)+solid);
label("$k + j < \frac{n}{2}$", (-857.5, -1678.5), fontsize(15pt) + rgb(0.87, 0.19, 0.19)+linewidth(4)+solid);
label("$0 < k, j$", (-784.5, -1578.5), fontsize(17pt) + rgb(0.09, 0.44, 0.76)+linewidth(4)+solid);
filldraw(shift(-1090, -1904) * rotate(0) * ellipse((0, 0), 10, 10), rgb(0.87, 0.19, 0.19), rgb(0, 0, 0)+linewidth(0)+solid);
filldraw(shift(-1090, -1864) * rotate(0) * ellipse((0, 0), 10, 10), rgb(0.87, 0.19, 0.19), rgb(0, 0, 0)+linewidth(0)+solid);
filldraw(shift(-1050, -1904) * rotate(0) * ellipse((0, 0), 10, 10), rgb(0.87, 0.19, 0.19), rgb(0, 0, 0)+linewidth(0)+solid);
filldraw(shift(-1010, -1904) * rotate(0) * ellipse((0, 0), 10, 10), rgb(0.87, 0.19, 0.19), rgb(0, 0, 0)+linewidth(0)+solid);
filldraw(shift(-970, -1904) * rotate(0) * ellipse((0, 0), 10, 10), rgb(0.87, 0.19, 0.19), rgb(0, 0, 0)+linewidth(0)+solid);
filldraw(shift(-1050, -1864) * rotate(0) * ellipse((0, 0), 10, 10), rgb(0.87, 0.19, 0.19), rgb(0, 0, 0)+linewidth(0)+solid);
filldraw(shift(-1010, -1864) * rotate(0) * ellipse((0, 0), 10, 10), rgb(0.87, 0.19, 0.19), rgb(0, 0, 0)+linewidth(0)+solid);
filldraw(shift(-1090, -1824) * rotate(0) * ellipse((0, 0), 10, 10), rgb(0.87, 0.19, 0.19), rgb(0, 0, 0)+linewidth(0)+solid);
filldraw(shift(-1050, -1824) * rotate(0) * ellipse((0, 0), 10, 10), rgb(0.87, 0.19, 0.19), rgb(0, 0, 0)+linewidth(0)+solid);
filldraw(shift(-1090, -1784) * rotate(0) * ellipse((0, 0), 10, 10), rgb(0.87, 0.19, 0.19), rgb(0, 0, 0)+linewidth(0)+solid);
filldraw(shift(-891, -1904) * rotate(0) * ellipse((0, 0), 10, 10), rgb(0.87, 0.19, 0.19), rgb(0, 0, 0)+linewidth(0)+solid);
filldraw(shift(-891, -1864) * rotate(0) * ellipse((0, 0), 10, 10), rgb(0.87, 0.19, 0.19), rgb(0, 0, 0)+linewidth(0)+solid);
filldraw(shift(-851, -1904) * rotate(0) * ellipse((0, 0), 10, 10), rgb(0.87, 0.19, 0.19), rgb(0, 0, 0)+linewidth(0)+solid);
filldraw(shift(-811, -1904) * rotate(0) * ellipse((0, 0), 10, 10), rgb(0.87, 0.19, 0.19), rgb(0, 0, 0)+linewidth(0)+solid);
filldraw(shift(-771, -1904) * rotate(0) * ellipse((0, 0), 10, 10), rgb(0.87, 0.19, 0.19), rgb(0, 0, 0)+linewidth(0)+solid);
filldraw(shift(-851, -1864) * rotate(0) * ellipse((0, 0), 10, 10), rgb(0.87, 0.19, 0.19), rgb(0, 0, 0)+linewidth(0)+solid);
filldraw(shift(-811, -1864) * rotate(0) * ellipse((0, 0), 10, 10), rgb(0.87, 0.19, 0.19), rgb(0, 0, 0)+linewidth(0)+solid);
filldraw(shift(-891, -1824) * rotate(0) * ellipse((0, 0), 10, 10), rgb(0.87, 0.19, 0.19), rgb(0, 0, 0)+linewidth(0)+solid);
filldraw(shift(-851, -1824) * rotate(0) * ellipse((0, 0), 10, 10), rgb(0.87, 0.19, 0.19), rgb(0, 0, 0)+linewidth(0)+solid);
filldraw(shift(-891, -1784) * rotate(0) * ellipse((0, 0), 10, 10), rgb(0.87, 0.19, 0.19), rgb(0, 0, 0)+linewidth(0)+solid);
filldraw(shift(-1090, -1704) * rotate(0) * ellipse((0, 0), 10, 10), rgb(0.87, 0.19, 0.19), rgb(0, 0, 0)+linewidth(0)+solid);
filldraw(shift(-1090, -1664) * rotate(0) * ellipse((0, 0), 10, 10), rgb(0.87, 0.19, 0.19), rgb(0, 0, 0)+linewidth(0)+solid);
filldraw(shift(-1050, -1704) * rotate(0) * ellipse((0, 0), 10, 10), rgb(0.87, 0.19, 0.19), rgb(0, 0, 0)+linewidth(0)+solid);
filldraw(shift(-1010, -1704) * rotate(0) * ellipse((0, 0), 10, 10), rgb(0.87, 0.19, 0.19), rgb(0, 0, 0)+linewidth(0)+solid);
filldraw(shift(-970, -1704) * rotate(0) * ellipse((0, 0), 10, 10), rgb(0.87, 0.19, 0.19), rgb(0, 0, 0)+linewidth(0)+solid);
filldraw(shift(-1050, -1664) * rotate(0) * ellipse((0, 0), 10, 10), rgb(0.87, 0.19, 0.19), rgb(0, 0, 0)+linewidth(0)+solid);
filldraw(shift(-1010, -1664) * rotate(0) * ellipse((0, 0), 10, 10), rgb(0.87, 0.19, 0.19), rgb(0, 0, 0)+linewidth(0)+solid);
filldraw(shift(-1090, -1624) * rotate(0) * ellipse((0, 0), 10, 10), rgb(0.87, 0.19, 0.19), rgb(0, 0, 0)+linewidth(0)+solid);
filldraw(shift(-1050, -1624) * rotate(0) * ellipse((0, 0), 10, 10), rgb(0.87, 0.19, 0.19), rgb(0, 0, 0)+linewidth(0)+solid);
filldraw(shift(-1090, -1584) * rotate(0) * ellipse((0, 0), 10, 10), rgb(0.87, 0.19, 0.19), rgb(0, 0, 0)+linewidth(0)+solid);
// Block 3
/* Generated by Cloud's Excalidraw to Asymptote */ draw(shift(-950.5, -920.5) * rotate(0) * ellipse((0, 0), 295.5, 295.5), rgb(0.11, 0.11, 0.11)+linewidth(1)+solid); draw((-1073, -654)--(-817, -654), rgb(0.87, 0.19, 0.19)+linewidth(2)+solid); draw((-817, -658)--(-662, -855), rgb(0.87, 0.19, 0.19)+linewidth(2)+solid); draw((-664, -858)--(-722, -1108), rgb(0.4, 0.25, 0.85)+linewidth(2)+solid); draw((-723, -1109)--(-957, -1213), rgb(0.4, 0.25, 0.85)+linewidth(2)+solid); draw((-957, -1214)--(-1185, -1096), rgb(0.09, 0.44, 0.76)+linewidth(2)+solid); draw((-1186, -1097)--(-1234, -844), rgb(0.09, 0.44, 0.76)+linewidth(2)+solid); draw((-1234, -841)--(-1077, -654), rgb(0.09, 0.44, 0.76)+linewidth(2)+solid); filldraw(shift(-1077.5, -655.5) * rotate(0) * ellipse((0, 0), 4.5, 4.5), rgb(0.11, 0.11, 0.11), rgb(0.11, 0.11, 0.11)+linewidth(2)+solid); filldraw(shift(-819.5, -654.5) * rotate(0) * ellipse((0, 0), 4.5, 4.5), rgb(0.11, 0.11, 0.11), rgb(0.11, 0.11, 0.11)+linewidth(2)+solid); filldraw(shift(-1234.5, -840.5) * rotate(0) * ellipse((0, 0), 4.5, 4.5), rgb(0.11, 0.11, 0.11), rgb(0.11, 0.11, 0.11)+linewidth(2)+solid); filldraw(shift(-1186.5, -1097.5) * rotate(0) * ellipse((0, 0), 4.5, 4.5), rgb(0.11, 0.11, 0.11), rgb(0.11, 0.11, 0.11)+linewidth(2)+solid); filldraw(shift(-958.5, -1214.5) * rotate(0) * ellipse((0, 0), 4.5, 4.5), rgb(0.11, 0.11, 0.11), rgb(0.11, 0.11, 0.11)+linewidth(2)+solid); filldraw(shift(-721.5, -1107.5) * rotate(0) * ellipse((0, 0), 4.5, 4.5), rgb(0.11, 0.11, 0.11), rgb(0.11, 0.11, 0.11)+linewidth(2)+solid); filldraw(shift(-661.5, -855.5) * rotate(0) * ellipse((0, 0), 4.5, 4.5), rgb(0.11, 0.11, 0.11), rgb(0.11, 0.11, 0.11)+linewidth(2)+solid); label("A", (-1088, -631.5), fontsize(20pt) + rgb(0.11, 0.11, 0.11)+linewidth(1)+solid); label("B", (-631, -855.5), fontsize(20pt) + rgb(0.11, 0.11, 0.11)+linewidth(1)+solid); label("C", (-966, -1258.5), fontsize(20pt) + rgb(0.11, 0.11, 0.11)+linewidth(1)+solid); draw((-1077, -659)--(-960, -1212), rgb(0.11, 0.11, 0.11)+linewidth(2)+solid); draw((-1072, -657)--(-662, -852), rgb(0.11, 0.11, 0.11)+linewidth(2)+solid); draw((-957, -1210)--(-663, -854), rgb(0.11, 0.11, 0.11)+linewidth(2)+solid); label("$k$ edges", (-749, -624.5), fontsize(20pt) + rgb(0.87, 0.19, 0.19)+linewidth(1)+solid); label("$j$ edges", (-1294.5, -1026.5), fontsize(20pt) + rgb(0.09, 0.44, 0.76)+linewidth(1)+solid); label("$n - (k + j)$ edges", (-655, -1174), fontsize(15pt) + rgb(0.4, 0.25, 0.85)+linewidth(1)+solid); label("$\frac{180^\circ}{n} \cdot j$", (-724, -862), fontsize(15pt) + rgb(0.09, 0.44, 0.76)+linewidth(4)+solid); label("$\frac{180^\circ}{n} \cdot k$", (-935, -1112), fontsize(15pt) + rgb(0.87, 0.19, 0.19)+linewidth(4)+solid); label("$\frac{180^\circ}{n} \cdot (n - k - j)$", (-946, -791), fontsize(15pt) + rgb(0.4, 0.25, 0.85)+linewidth(4)+solid);
// Block 4
/* Generated by Cloud's Excalidraw to Asymptote */ filldraw((-1115, -1529)--(-1115, -1729)--(-915, -1729)--(-1115, -1529)--cycle, rgb(1, 0.78, 0.78), rgb(0, 0, 0)+linewidth(0)+solid); filldraw((-915, -1729)--(-915, -1929)--(-715, -1929)--(-915, -1729)--cycle, rgb(1, 0.78, 0.78), rgb(0, 0, 0)+linewidth(0)+solid); filldraw((-1115, -1729)--(-1115, -1929)--(-915, -1929)--(-1115, -1729)--cycle, rgb(1, 0.78, 0.78), rgb(0, 0, 0)+linewidth(0)+solid); label("$k$", (-651, -1927.5), fontsize(32pt) + rgb(0.11, 0.11, 0.11)+linewidth(4)+solid); label("$j$", (-1117, -1462), fontsize(25pt) + rgb(0.11, 0.11, 0.11)+linewidth(4)+solid); label("$\frac{n}{2}$", (-916, -1959.5), fontsize(23pt) + rgb(0.11, 0.11, 0.11)+linewidth(4)+solid); label("$n$", (-714, -1971), fontsize(28pt) + rgb(0.11, 0.11, 0.11)+linewidth(4)+solid); draw((-1115, -1728)--(-915, -1928), rgb(0.87, 0.19, 0.19)+linewidth(4)+solid); draw((-1115, -1528)--(-715, -1928), rgb(0.09, 0.44, 0.76)+linewidth(4)+solid); draw((-1115, -1718)--(-1115, -1488), rgb(0.11, 0.11, 0.11)+linewidth(4)+solid, arrow=ArcArrow(size=20)); draw((-1115, -1718)--(-1115, -1948), rgb(0.11, 0.11, 0.11)+linewidth(4)+solid); draw((-905, -1928)--(-675, -1928), rgb(0.11, 0.11, 0.11)+linewidth(4)+solid, arrow=ArcArrow(size=20)); draw((-905, -1928)--(-1135, -1928), rgb(0.11, 0.11, 0.11)+linewidth(4)+solid); draw((-915, -1528)--(-915, -1926), rgb(0.87, 0.19, 0.19)+linewidth(4)+solid); draw((-715, -1728)--(-1115, -1728), rgb(0.87, 0.19, 0.19)+linewidth(4)+solid); filldraw((-715, -1908)--(-715, -1948)--cycle, rgb(0.87, 0.19, 0.19), rgb(0.11, 0.11, 0.11)+linewidth(4)+solid); filldraw((-1135, -1528)--(-1095, -1528)--cycle, rgb(0.87, 0.19, 0.19), rgb(0.11, 0.11, 0.11)+linewidth(4)+solid); label("$n$", (-1159, -1529), fontsize(28pt) + rgb(0.11, 0.11, 0.11)+linewidth(4)+solid); label("$\frac{n}{2}$", (-1142, -1727.5), fontsize(23pt) + rgb(0.11, 0.11, 0.11)+linewidth(4)+solid); label("$k + j < n$", (-776, -1538.5), fontsize(17pt) + rgb(0.09, 0.44, 0.76)+linewidth(4)+solid); label("$k, j > \frac{n}{2}$", (-867.5, -1636.5), fontsize(14pt) + rgb(0.87, 0.19, 0.19)+linewidth(4)+solid); label("$k + j < \frac{n}{2}$", (-857.5, -1678.5), fontsize(15pt) + rgb(0.87, 0.19, 0.19)+linewidth(4)+solid); label("$0 < k, j$", (-784.5, -1578.5), fontsize(17pt) + rgb(0.09, 0.44, 0.76)+linewidth(4)+solid); filldraw(shift(-1090, -1904) * rotate(0) * ellipse((0, 0), 10, 10), rgb(0.87, 0.19, 0.19), rgb(0, 0, 0)+linewidth(0)+solid); filldraw(shift(-1090, -1864) * rotate(0) * ellipse((0, 0), 10, 10), rgb(0.87, 0.19, 0.19), rgb(0, 0, 0)+linewidth(0)+solid); filldraw(shift(-1050, -1904) * rotate(0) * ellipse((0, 0), 10, 10), rgb(0.87, 0.19, 0.19), rgb(0, 0, 0)+linewidth(0)+solid); filldraw(shift(-1010, -1904) * rotate(0) * ellipse((0, 0), 10, 10), rgb(0.87, 0.19, 0.19), rgb(0, 0, 0)+linewidth(0)+solid); filldraw(shift(-970, -1904) * rotate(0) * ellipse((0, 0), 10, 10), rgb(0.87, 0.19, 0.19), rgb(0, 0, 0)+linewidth(0)+solid); filldraw(shift(-1050, -1864) * rotate(0) * ellipse((0, 0), 10, 10), rgb(0.87, 0.19, 0.19), rgb(0, 0, 0)+linewidth(0)+solid); filldraw(shift(-1010, -1864) * rotate(0) * ellipse((0, 0), 10, 10), rgb(0.87, 0.19, 0.19), rgb(0, 0, 0)+linewidth(0)+solid); filldraw(shift(-1090, -1824) * rotate(0) * ellipse((0, 0), 10, 10), rgb(0.87, 0.19, 0.19), rgb(0, 0, 0)+linewidth(0)+solid); filldraw(shift(-1050, -1824) * rotate(0) * ellipse((0, 0), 10, 10), rgb(0.87, 0.19, 0.19), rgb(0, 0, 0)+linewidth(0)+solid); filldraw(shift(-1090, -1784) * rotate(0) * ellipse((0, 0), 10, 10), rgb(0.87, 0.19, 0.19), rgb(0, 0, 0)+linewidth(0)+solid); filldraw(shift(-891, -1904) * rotate(0) * ellipse((0, 0), 10, 10), rgb(0.87, 0.19, 0.19), rgb(0, 0, 0)+linewidth(0)+solid); filldraw(shift(-891, -1864) * rotate(0) * ellipse((0, 0), 10, 10), rgb(0.87, 0.19, 0.19), rgb(0, 0, 0)+linewidth(0)+solid); filldraw(shift(-851, -1904) * rotate(0) * ellipse((0, 0), 10, 10), rgb(0.87, 0.19, 0.19), rgb(0, 0, 0)+linewidth(0)+solid); filldraw(shift(-811, -1904) * rotate(0) * ellipse((0, 0), 10, 10), rgb(0.87, 0.19, 0.19), rgb(0, 0, 0)+linewidth(0)+solid); filldraw(shift(-771, -1904) * rotate(0) * ellipse((0, 0), 10, 10), rgb(0.87, 0.19, 0.19), rgb(0, 0, 0)+linewidth(0)+solid); filldraw(shift(-851, -1864) * rotate(0) * ellipse((0, 0), 10, 10), rgb(0.87, 0.19, 0.19), rgb(0, 0, 0)+linewidth(0)+solid); filldraw(shift(-811, -1864) * rotate(0) * ellipse((0, 0), 10, 10), rgb(0.87, 0.19, 0.19), rgb(0, 0, 0)+linewidth(0)+solid); filldraw(shift(-891, -1824) * rotate(0) * ellipse((0, 0), 10, 10), rgb(0.87, 0.19, 0.19), rgb(0, 0, 0)+linewidth(0)+solid); filldraw(shift(-851, -1824) * rotate(0) * ellipse((0, 0), 10, 10), rgb(0.87, 0.19, 0.19), rgb(0, 0, 0)+linewidth(0)+solid); filldraw(shift(-891, -1784) * rotate(0) * ellipse((0, 0), 10, 10), rgb(0.87, 0.19, 0.19), rgb(0, 0, 0)+linewidth(0)+solid); filldraw(shift(-1090, -1704) * rotate(0) * ellipse((0, 0), 10, 10), rgb(0.87, 0.19, 0.19), rgb(0, 0, 0)+linewidth(0)+solid); filldraw(shift(-1090, -1664) * rotate(0) * ellipse((0, 0), 10, 10), rgb(0.87, 0.19, 0.19), rgb(0, 0, 0)+linewidth(0)+solid); filldraw(shift(-1050, -1704) * rotate(0) * ellipse((0, 0), 10, 10), rgb(0.87, 0.19, 0.19), rgb(0, 0, 0)+linewidth(0)+solid); filldraw(shift(-1010, -1704) * rotate(0) * ellipse((0, 0), 10, 10), rgb(0.87, 0.19, 0.19), rgb(0, 0, 0)+linewidth(0)+solid); filldraw(shift(-970, -1704) * rotate(0) * ellipse((0, 0), 10, 10), rgb(0.87, 0.19, 0.19), rgb(0, 0, 0)+linewidth(0)+solid); filldraw(shift(-1050, -1664) * rotate(0) * ellipse((0, 0), 10, 10), rgb(0.87, 0.19, 0.19), rgb(0, 0, 0)+linewidth(0)+solid); filldraw(shift(-1010, -1664) * rotate(0) * ellipse((0, 0), 10, 10), rgb(0.87, 0.19, 0.19), rgb(0, 0, 0)+linewidth(0)+solid); filldraw(shift(-1090, -1624) * rotate(0) * ellipse((0, 0), 10, 10), rgb(0.87, 0.19, 0.19), rgb(0, 0, 0)+linewidth(0)+solid); filldraw(shift(-1050, -1624) * rotate(0) * ellipse((0, 0), 10, 10), rgb(0.87, 0.19, 0.19), rgb(0, 0, 0)+linewidth(0)+solid); filldraw(shift(-1090, -1584) * rotate(0) * ellipse((0, 0), 10, 10), rgb(0.87, 0.19, 0.19), rgb(0, 0, 0)+linewidth(0)+solid); | [] |
165 | A cube with side length 10 is suspended above a plane. The vertex closest to the plane is labeled $A$. The three vertices adjacent to vertex $A$ are at heights 10, 11, and 12 above the plane. The distance from vertex $A$ to the plane can be expressed as $\frac{r-\sqrt{s}}{t}$, where $r$, $s$, and $t$ are positive integers. Find $r+s+t$. | 2011 AIME I Problem 13 | First, shift the cube down by 11 so the vertices adjacent to $A$ are $-1$, $0$, and $1$ above the plane.
Consider the two points 1 above and 1 below the plane:
Now we rotate the cube so that all 3 vertices adjacent to $A$ are on the plane. We can calculate $A$ to be $\frac{10\sqrt3}{3}$ below the plane.
Notice that when $A$ is rotated, it gets slightly closer to the plane.
The triangle shown in the diagram has legs of ration $1:7$, because it is similar to the triangle in the first diagram, since we are rotating. (kind of bad explanation)
So, we can compute $h$, where $h^2 + \left(\dfrac{h}{7}\right)^2 = \left(\dfrac{10\sqrt3}{3}\right)^2$.
Solving yields $h=\frac{7\sqrt{6}}{3}$.
Remember, this is the distance that $A$ is below the plane, so when we shift up by 11, $A$ is $11-\frac{7\sqrt{6}}{3}$ above the plane. Answer extraction yields $\boxed{330}$. | // Block 1
defaultpen(fontsize(10)+0.8); size(300);
pen p=fontsize(9)+linewidth(3);
// Define points
pair A=(-7,0), B=(0,0), C=(7,0);
pair D=(-7,-1), E=(7,1);
// Draw segments
draw(A--B--C, linewidth(1.2));
draw(A--D, linewidth(1.2));
draw(C--E, linewidth(1.2));
draw(D--E, linewidth(1.2));
// Label lengths
label("$7$", A--B, dir(90));
label("$7$", B--C, dir(270));
label("$1$", A--D, dir(180));
label("$1$", C--E, dir(0));
label("$5\sqrt{2}$", D--B, dir(225));
label("$5\sqrt{2}$", B--E, dir(45));
// Block 2
defaultpen(fontsize(10)+0.8); size(250);
pen p=fontsize(9)+linewidth(1.2);
// Define points
pair A=(-5*sqrt(2),0), B=(0,0), C=(5*sqrt(2),0);
pair D=(0,-10*sqrt(3)/3);
pair E=(7*sqrt(3)/15, -49*sqrt(3)/15);
// Draw triangle ABC base
draw(A--B--C, linewidth(1));
// Draw vertical line from B to D
draw(B--D, linewidth(1));
// Draw right triangle BDE
draw(B--E--D--B, linewidth(1));
// Label points
dot(D, p); label("$A$", D, S, p);
// Label lengths (approximate positions)
label("$\frac{10\sqrt{3}}{3}$", B--D, W);
label("$h$", B--E, dir(0));
// Block 3
defaultpen(fontsize(10)+0.8); size(300); pen p=fontsize(9)+linewidth(3); // Define points pair A=(-7,0), B=(0,0), C=(7,0); pair D=(-7,-1), E=(7,1); // Draw segments draw(A--B--C, linewidth(1.2)); draw(A--D, linewidth(1.2)); draw(C--E, linewidth(1.2)); draw(D--E, linewidth(1.2)); // Label lengths label("$7$", A--B, dir(90)); label("$7$", B--C, dir(270)); label("$1$", A--D, dir(180)); label("$1$", C--E, dir(0)); label("$5\sqrt{2}$", D--B, dir(225)); label("$5\sqrt{2}$", B--E, dir(45));
// Block 4
defaultpen(fontsize(10)+0.8); size(250); pen p=fontsize(9)+linewidth(1.2); // Define points pair A=(-5*sqrt(2),0), B=(0,0), C=(5*sqrt(2),0); pair D=(0,-10*sqrt(3)/3); pair E=(7*sqrt(3)/15, -49*sqrt(3)/15); // Draw triangle ABC base draw(A--B--C, linewidth(1)); // Draw vertical line from B to D draw(B--D, linewidth(1)); // Draw right triangle BDE draw(B--E--D--B, linewidth(1)); // Label points dot(D, p); label("$A$", D, S, p); // Label lengths (approximate positions) label("$\frac{10\sqrt{3}}{3}$", B--D, W); label("$h$", B--E, dir(0)); | [] |
166 | On square $ABCD$, point $E$ lies on side $AD$ and point $F$ lies on side $BC$, so that $BE=EF=FD=30$. Find the area of the square $ABCD$. | 2011 AIME II Problem 2 | Drawing the square and examining the given lengths,
you find that the three segments cut the square into three equal horizontal sections. Therefore, ($x$ being the side length), $\sqrt{x^2+(x/3)^2}=30$, or $x^2+(x/3)^2=900$. Multiplying both sides by $9$ and simplifying, we find that $10x^2=8100$. Dividing by ten gives $x^2=810.$
Area of the square is $\fbox{810}$. | // Block 1
size(2inch, 2inch);
currentpen = fontsize(8pt);
pair A = (0, 0); dot(A); label("$A$", A, plain.SW);
pair B = (3, 0); dot(B); label("$B$", B, plain.SE);
pair C = (3, 3); dot(C); label("$C$", C, plain.NE);
pair D = (0, 3); dot(D); label("$D$", D, plain.NW);
pair E = (0, 1); dot(E); label("$E$", E, plain.W);
pair F = (3, 2); dot(F); label("$F$", F, plain.E);
label("$\frac x3$", E--A);
label("$\frac x3$", F--C);
label("$x$", A--B);
label("$x$", C--D);
label("$\frac {2x}3$", B--F);
label("$\frac {2x}3$", D--E);
label("$30$", B--E);
label("$30$", F--E);
label("$30$", F--D);
draw(B--C--D--F--E--B--A--D);
// Block 2
size(2inch, 2inch); currentpen = fontsize(8pt); pair A = (0, 0); dot(A); label("$A$", A, plain.SW); pair B = (3, 0); dot(B); label("$B$", B, plain.SE); pair C = (3, 3); dot(C); label("$C$", C, plain.NE); pair D = (0, 3); dot(D); label("$D$", D, plain.NW); pair E = (0, 1); dot(E); label("$E$", E, plain.W); pair F = (3, 2); dot(F); label("$F$", F, plain.E); label("$\frac x3$", E--A); label("$\frac x3$", F--C); label("$x$", A--B); label("$x$", C--D); label("$\frac {2x}3$", B--F); label("$\frac {2x}3$", D--E); label("$30$", B--E); label("$30$", F--E); label("$30$", F--D); draw(B--C--D--F--E--B--A--D); | [] |
167 | In triangle $ABC$, $AB=20$ and $AC=11$. The angle bisector of $\angle A$ intersects $BC$ at point $D$, and point $M$ is the midpoint of $AD$. Let $P$ be the point of the intersection of $AC$ and $BM$. The ratio of $CP$ to $PA$ can be expressed in the form $\dfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. | 2011 AIME II Problem 4 | Let $D'$ be on $\overline{AC}$ such that $BP \parallel DD'$. It follows that $\triangle BPC \sim \triangle DD'C$, so \[\frac{PC}{D'C} = 1 + \frac{BD}{DC} = 1 + \frac{AB}{AC} = \frac{31}{11}\] by the Angle Bisector Theorem. Similarly, we see by the Midline Theorem that $AP = PD'$. Thus, \[\frac{CP}{PA} = \frac{1}{\frac{PD'}{PC}} = \frac{1}{1 - \frac{D'C}{PC}} = \frac{31}{20},\] and $m+n = \boxed{51}$. | // Block 1
pointpen = black; pathpen = linewidth(0.7);
pair A = (0,0), C= (11,0), B=IP(CR(A,20),CR(C,18)), D = IP(B--C,CR(B,20/31*abs(B-C))), M = (A+D)/2, P = IP(M--2*M-B, A--C), D2 = IP(D--D+P-B, A--C);
D(MP("A",D(A))--MP("B",D(B),N)--MP("C",D(C))--cycle); D(A--MP("D",D(D),NE)--MP("D'",D(D2))); D(B--MP("P",D(P))); D(MP("M",M,NW)); MP("20",(B+D)/2,ENE); MP("11",(C+D)/2,ENE);
// Block 2
pointpen = black; pathpen = linewidth(0.7); pair A = (0,0), C= (11,0), B=IP(CR(A,20),CR(C,18)), D = IP(B--C,CR(B,20/31*abs(B-C))), M = (A+D)/2, P = IP(M--2*M-B, A--C), D2 = IP(D--D+P-B, A--C); D(MP("A",D(A))--MP("B",D(B),N)--MP("C",D(C))--cycle); D(A--MP("D",D(D),NE)--MP("D'",D(D2))); D(B--MP("P",D(P))); D(MP("M",M,NW)); MP("20",(B+D)/2,ENE); MP("11",(C+D)/2,ENE); | [] |
168 | Let $\triangle ABC$ be a right triangle with right angle at $C.$ Let $D$ and $E$ be points on $\overline{AB}$ with $D$ between $A$ and $E$ such that $\overline{CD}$ and $\overline{CE}$ trisect $\angle C.$ If $\frac{DE}{BE} = \frac{8}{15},$ then $\tan B$ can be written as $\frac{m \sqrt{p}}{n},$ where $m$ and $n$ are relatively prime positive integers, and $p$ is a positive integer not divisible by the square of any prime. Find $m+n+p.$ | 2012 AIME I Problem 12 | We have $\angle BCE = \angle ECD = \angle DCA = \tfrac 13 \cdot 90^\circ = 30^\circ$. Drop the altitude from $D$ to $CB$ and call the foot $F$.
Let $CD = 8a$. Using angle bisector theorem on $\triangle CDB$, we get $CB = 15a$. Now $CDF$ is a $30$-$60$-$90$ triangle, so $CF = 4a$, $FD = 4a\sqrt{3}$, and $FB = 11a$. Finally, $\tan{B} = \tfrac{DF}{FB}=\tfrac{4\sqrt{3}a}{11a} = \tfrac{4\sqrt{3}}{11}$. Our final answer is $4 + 3 + 11 = \boxed{018}$. | // Block 1
import cse5;size(200);
defaultpen(linewidth(0.4)+fontsize(8));
pair A,B,C,D,E0,F;
C=origin;
B=(10,0);
A=(0,5);
E0=extension(C,dir(30),A,B);
D=extension(C,dir(60),A,B);
F=foot(D,C,B);
draw(A--B--C--A, black+0.8);
draw(C--D--F^^C--E0, gray);
dot("$A$",A,N);
dot("$B$",B,SE);
dot("$C$",C,SW);
dot("$D$",D,NE);
dot("$E$",E0,2*NE);
dot("$F$",F,S);
label("$8$",D--E0,2*NE);
label("$15$",E0--B,2*NE);
label("$11a$",C--B,2*S);
label(rotate(60)*"$8a$",C--D,2*NW);
label(rotate(-90)*"$4a\sqrt{3}$",D--F, E);
label("$4a$",C--F,2*S);
// Block 2
import cse5;size(200); defaultpen(linewidth(0.4)+fontsize(8)); pair A,B,C,D,E0,F; C=origin; B=(10,0); A=(0,5); E0=extension(C,dir(30),A,B); D=extension(C,dir(60),A,B); F=foot(D,C,B); draw(A--B--C--A, black+0.8); draw(C--D--F^^C--E0, gray); dot("$A$",A,N); dot("$B$",B,SE); dot("$C$",C,SW); dot("$D$",D,NE); dot("$E$",E0,2*NE); dot("$F$",F,S); label("$8$",D--E0,2*NE); label("$15$",E0--B,2*NE); label("$11a$",C--B,2*S); label(rotate(60)*"$8a$",C--D,2*NW); label(rotate(-90)*"$4a\sqrt{3}$",D--F, E); label("$4a$",C--F,2*S); | [] |
169 | Three concentric circles have radii $3,$ $4,$ and $5.$ An equilateral triangle with one vertex on each circle has side length $s.$ The largest possible area of the triangle can be written as $a + \tfrac{b}{c} \sqrt{d},$ where $a,$ $b,$ $c,$ and $d$ are positive integers, $b$ and $c$ are relatively prime, and $d$ is not divisible by the square of any prime. Find $a+b+c+d.$ | 2012 AIME I Problem 13 | Reinterpret the problem in the following manner. Equilateral triangle $ABC$ has a point $X$ on the interior such that $AX = 5,$ $BX = 4,$ and $CX = 3.$ A $60^\circ$ counter-clockwise rotation about vertex $A$ maps $X$ to $X'$ and $B$ to $C.$
Note that angle $XAX'$ is $60$ and $XA = X'A = 5$ which tells us that triangle $XAX'$ is equilateral and that $XX' = 5.$ We now notice that $XC = 3$ and $X'C = 4$ (recall that $C$ is the image of $B$ after rotation) which tells us that angle $XCX'$ is $90$ because there is a $3$-$4$-$5$ Pythagorean triple. Now note that $\angle ABC + \angle ACB = 120^\circ$ and $\angle XCA + \angle XBA = 90^\circ,$ so $\angle XCB+\angle XBC = 30^\circ$ and $\angle BXC = 150^\circ.$ Applying the law of cosines on triangle $BXC$ yields
\[BC^2 = BX^2+CX^2 - 2 \cdot BX \cdot CX \cdot \cos150^\circ = 4^2+3^2-24 \cdot \frac{-\sqrt{3}}{2} = 25+12\sqrt{3}\]
and thus the area of $ABC$ equals \[\frac{\sqrt{3}}{4}\cdot BC^2 = 25\frac{\sqrt{3}}{4}+9.\]
so our final answer is $3+4+25+9 = \boxed{041}.$ | // Block 1
import cse5;
size(200);
defaultpen(linewidth(0.4)+fontsize(8));
pair O = origin;
pair A,B,C,Op,Bp,Cp;
path c3,c4,c5;
c3 = CR(O,3);
c4 = CR(O,4);
c5 = CR(O,5);
draw(c3^^c4^^c5, gray+0.25);
A = 5*dir(96.25);
Op = rotate(60,A)*O;
B = OP(CR(Op,4),c3);
Bp = IP(CR(Op,4),c3);
C = rotate(-60,A)*B;
Cp = rotate(-60,A)*Bp;
draw(A--B--C--A, black+0.8);
draw(A--Bp--Cp--A, royalblue+0.8);
draw(CR(Op,4), heavygreen+0.25);
dot("$A$",A,N);
dot("$X$",O,E);
dot("$X'$",Op,E);
dot("$C$",B,SE);
dot("$C'$",Bp,NE);
dot("$B$",C,2*SW);
dot("$B'$",Cp,2*S);
// Block 2
import cse5; size(200); defaultpen(linewidth(0.4)+fontsize(8)); pair O = origin; pair A,B,C,Op,Bp,Cp; path c3,c4,c5; c3 = CR(O,3); c4 = CR(O,4); c5 = CR(O,5); draw(c3^^c4^^c5, gray+0.25); A = 5*dir(96.25); Op = rotate(60,A)*O; B = OP(CR(Op,4),c3); Bp = IP(CR(Op,4),c3); C = rotate(-60,A)*B; Cp = rotate(-60,A)*Bp; draw(A--B--C--A, black+0.8); draw(A--Bp--Cp--A, royalblue+0.8); draw(CR(Op,4), heavygreen+0.25); dot("$A$",A,N); dot("$X$",O,E); dot("$X'$",Op,E); dot("$C$",B,SE); dot("$C'$",Bp,NE); dot("$B$",C,2*SW); dot("$B'$",Cp,2*S); | [] |
169 | Three concentric circles have radii $3,$ $4,$ and $5.$ An equilateral triangle with one vertex on each circle has side length $s.$ The largest possible area of the triangle can be written as $a + \tfrac{b}{c} \sqrt{d},$ where $a,$ $b,$ $c,$ and $d$ are positive integers, $b$ and $c$ are relatively prime, and $d$ is not divisible by the square of any prime. Find $a+b+c+d.$ | 2012 AIME I Problem 13 | Let's call the circle center $X$. It has a distance of 3, 4, 5 to an equilateral triangle $LMN$. Consider $X$’s pedal triangle $ABC$. Since $X$’s antipedal triangle is equilateral, $X$ must be the one of the isogonic centers of $\triangle{ABC}$. We’ll take the one inside $ABC$, i.e., the Fermat point, because it leads to larger $\triangle LMN$. Now we construct the three equilateral triangles $ABD$, $ACE$, and $BCF$, the same way the Fermat point is constructed. Then we have $\angle DXE = \angle EXF = \angle FXE = 120$. Since $AEMCX$ is concyclic with $XM$=4 as diameter, we have $AC=4\sin(60)$. Similarly, $AB=3\sin(60)$, and $BC=5\sin(60)$. So $\triangle ABC$ is a 3-4-5 right triangle with $\angle BAC=90$. With some more angle chasing we get
\[\angle MXC+\angle LXB = \angle MAC + \angle LAB = 180 – \angle BAC = 90\]
\[\angle LXM = 360 – (\angle MXC + \angle LXB + \angle BXC) = 360 –(90+120)=150\]
By Law of Cosines, we have
\[LM^2 = 3^2+4^2-2*3*4\cos(150)=25+12\sqrt 3\]
And the area follows.
\[[LMN] = \frac{25}{4}\sqrt{3} + 9; \boxed{041}.\]
By Mathdummy | // Block 1
import olympiad;
import cse5;
import graph;
dotfactor = 2;
unitsize(0.3inch);
pair B = (0,0), C= (5,0), A = (sqrt(9-2.4*2.4),2.4);
pair D = rotate(60,B)*A, E=rotate(60,A)*C, F=rotate(60,C)*B;
pair X = extension(A,F,D,C);
pair L = (-1.5,2), M = (6.2,3), N = rotate(-60,L)*M;
dot("$C$", C, dir(0)); dot("$A$", A, dir(90));dot("$B$", B, dir(180));
dot("$D$", D, NE); dot("$E$", E, dir(90));dot("$F$", F, dir(270));
dot("$M$", M, NE); dot("$N$", N, dir(270));dot("$L$", L, NW);
dot("$X$", X, dir(250));
draw(L--X); draw(M--X); draw(N--X);
draw(A--B--C--cycle);
draw(A--D--B); draw(B--F--C); draw(A--E--C);
draw(A--F,dashed); draw(D--C,dashed); draw(B--E,dashed);
draw(L--M--N--cycle);
// Block 2
import olympiad; import cse5; import graph; dotfactor = 2; unitsize(0.3inch); pair B = (0,0), C= (5,0), A = (sqrt(9-2.4*2.4),2.4); pair D = rotate(60,B)*A, E=rotate(60,A)*C, F=rotate(60,C)*B; pair X = extension(A,F,D,C); pair L = (-1.5,2), M = (6.2,3), N = rotate(-60,L)*M; dot("$C$", C, dir(0)); dot("$A$", A, dir(90));dot("$B$", B, dir(180)); dot("$D$", D, NE); dot("$E$", E, dir(90));dot("$F$", F, dir(270)); dot("$M$", M, NE); dot("$N$", N, dir(270));dot("$L$", L, NW); dot("$X$", X, dir(250)); draw(L--X); draw(M--X); draw(N--X); draw(A--B--C--cycle); draw(A--D--B); draw(B--F--C); draw(A--E--C); draw(A--F,dashed); draw(D--C,dashed); draw(B--E,dashed); draw(L--M--N--cycle); | [] |
169 | Three concentric circles have radii $3,$ $4,$ and $5.$ An equilateral triangle with one vertex on each circle has side length $s.$ The largest possible area of the triangle can be written as $a + \tfrac{b}{c} \sqrt{d},$ where $a,$ $b,$ $c,$ and $d$ are positive integers, $b$ and $c$ are relatively prime, and $d$ is not divisible by the square of any prime. Find $a+b+c+d.$ | 2012 AIME I Problem 13 | We have $x=3$, $y=4$, and $z=5$. Because $AD=m$ is the median of $\triangle AOB$, by Stewart's Theorem we have \[m^2=\frac 12 (x^2+y^2)-\frac 14 \cdot z^2\quad \Rightarrow \quad m = \frac 52.\] Because $CD$ is the altitude of equilateral triangle $OBC$, we have $CD=\frac{\sqrt{3}}2\cdot z$. Then in $\triangle ADC$, we have $\angle ADC=\varphi+90^\circ$, so $\cos(\angle ADC)=-\sin\varphi$, and the Law of Cosines gives\[s^2=m^2+\frac 34\cdot z^2 + mz\sqrt{3}\sin\varphi = 25 \left(1+\frac {\sqrt{3}}{2}\cdot\sin\varphi \right)\]To calculate $\sin\phi$ we apply the Law of Cosines to $\triangle ADO$ to get\[mz\cos\varphi = m^2+\frac 14\cdot z^2 - x^2 \quad \Rightarrow \quad \cos\varphi = \frac 7{25}\quad \Rightarrow \quad \sin\varphi = \frac {24}{25}.\]
Finally, we get $s^2=25+12\sqrt{3}$ and and thus the area of $\triangle ABC$ equals\[\frac{\sqrt{3}}{4}\cdot s^2 = 25\frac{\sqrt{3}}{4}+9.\]
so our final answer is $3+4+25+9 = \boxed{041}.$
vladimir.shelomovskii@gmail.com, vvsss | // Block 1
import cse5; size(350); defaultpen(linewidth(0.6)+fontsize(12));
pair O=origin; pair Op,Bp,Cp;
path c3 = CR(O,3), c4 = CR(O,4), c5 = CR(O,5);
var theta=55.75;
pair A = 3*dir(theta), Ap = rotate(150,O)*A, F=IP(c4,O--2*Ap), C=rotate(60,A)*F, E=rotate(60,A)*C, B=IP(c5,O--E), D=foot(C,O,E);
filldraw(A--O--B--cycle, rgb(255,255,200));
draw(c3, cyan);
draw(c4, green);
draw(c5, purple);
draw(A--F--C--A--E--C, red+0.8);
draw(D--B--C--D--O--C, purple+0.6);
draw(A--O^^E--B, cyan+0.6);
draw(A--B^^O--F, heavygreen+0.6);
draw(A--D);
dot("$A$",A,N); dot("$O$",O,dir(A-B)); dot("$F$",F,dir(F-E)); dot("$C$",C,SE); dot("$B$",B,dir(-45)); dot("$D$",D,dir(B-C)); dot("$E$",E,dir(E-F));
label("$s$",A--F,dir(A-E),red);
label("$s$",F--C,dir(C-B),red);
label("$s$",A--C,dir(C-E),red);
label("$s$",A--E,dir(E-C),red);
label("$s$",C--E,dir(C),red);
label("$x$",B--E,down,royalblue);
label("$x$",O--A,dir(B-A),royalblue);
label("$y$",A--B,dir(F-A),heavygreen);
label("$y$",O--F,dir(C),heavygreen);
label("$z$",C--B,dir(E-A),purple);
label("$z$",O--C,dir(F),purple);
label("$m$",A--D,dir(0),blue+fontsize(9));
markscalefactor=0.03;
draw(rightanglemark(C,D,B), gray);
MA("\varphi",A,D,O,0.25,blue);
// Block 2
import cse5; size(350); defaultpen(linewidth(0.6)+fontsize(12)); pair O=origin; pair Op,Bp,Cp; path c3 = CR(O,3), c4 = CR(O,4), c5 = CR(O,5); var theta=55.75; pair A = 3*dir(theta), Ap = rotate(150,O)*A, F=IP(c4,O--2*Ap), C=rotate(60,A)*F, E=rotate(60,A)*C, B=IP(c5,O--E), D=foot(C,O,E); filldraw(A--O--B--cycle, rgb(255,255,200)); draw(c3, cyan); draw(c4, green); draw(c5, purple); draw(A--F--C--A--E--C, red+0.8); draw(D--B--C--D--O--C, purple+0.6); draw(A--O^^E--B, cyan+0.6); draw(A--B^^O--F, heavygreen+0.6); draw(A--D); dot("$A$",A,N); dot("$O$",O,dir(A-B)); dot("$F$",F,dir(F-E)); dot("$C$",C,SE); dot("$B$",B,dir(-45)); dot("$D$",D,dir(B-C)); dot("$E$",E,dir(E-F)); label("$s$",A--F,dir(A-E),red); label("$s$",F--C,dir(C-B),red); label("$s$",A--C,dir(C-E),red); label("$s$",A--E,dir(E-C),red); label("$s$",C--E,dir(C),red); label("$x$",B--E,down,royalblue); label("$x$",O--A,dir(B-A),royalblue); label("$y$",A--B,dir(F-A),heavygreen); label("$y$",O--F,dir(C),heavygreen); label("$z$",C--B,dir(E-A),purple); label("$z$",O--C,dir(F),purple); label("$m$",A--D,dir(0),blue+fontsize(9)); markscalefactor=0.03; draw(rightanglemark(C,D,B), gray); MA("\varphi",A,D,O,0.25,blue); | [] |
170 | Ana, Bob, and Cao bike at constant rates of $8.6$ meters per second, $6.2$ meters per second, and $5$ meters per second, respectively. They all begin biking at the same time from the northeast corner of a rectangular field whose longer side runs due west. Ana starts biking along the edge of the field, initially heading west, Bob starts biking along the edge of the field, initially heading south, and Cao bikes in a straight line across the field to a point $D$ on the south edge of the field. Cao arrives at point $D$ at the same time that Ana and Bob arrive at $D$ for the first time. The ratio of the field's length to the field's width to the distance from point $D$ to the southeast corner of the field can be represented as $p : q : r$, where $p$, $q$, and $r$ are positive integers with $p$ and $q$ relatively prime. Find $p+q+r$. | 2012 AIME II Problem 4 | Let $a,b,c$ be the labeled lengths as shown in the diagram. Also, assume WLOG the time taken is $1$ second.
Observe that $\dfrac{2a+b+c}{8.6}=1$ or $2a+b+c=8.6$, and $\dfrac{b+c}{6.2}=1$ or $b+c=6.2$. Subtracting the second equation from the first gives $2a=2.4$, or $a=1.2$.
Now, let us solve $b$ and $c$. Note that $\dfrac{\sqrt{b^2+c^2}}{5}=1$, or $b^2+c^2=25$. We also have $b+c=6.2$.
We have a system of equations: \[\left\{\begin{array}{l}b+c=6.2\\ b^2+c^2=25\end{array}\right.\]
Squaring the first equation gives $b^2+2bc+c^2=38.44$, and subtracting the second from this gives $2bc=13.44$. Now subtracting this from $b^2+c^2=25$ gives $b^2-2bc+c^2=(b-c)^2=11.56$, or $b-c=3.4$. Now we have the following two equations:
\[\left\{\begin{array}{l}b+c=6.2\\ b-c=3.4\end{array}\right.\]
Adding the equations and dividing by two gives $b=4.8$, and it follows that $c=1.4$.
The ratios we desire are therefore $1.4:6:4.8=7:30:24$, and our answer is $7+30+24=\boxed{061}$.
Note that in our diagram, we labeled the part of the bottom $b$ and the side $c$. However, these labels are interchangeable. We can cancel out the case where the side is $4.8$ and the part of the bottom is $1.4$ by noting a restriction of the problem: "...a rectangular field whose longer side runs due west." If we had the side be $4.8$, then the entire bottom would be $1.2+1.4=2.6$, clearly less than $4.8$ and therefore violating our restriction. | // Block 1
draw((1.2,0)--(0,0)--(0,1.4)--(6,1.4)--(6,0)--(1.2,0)--(6,1.4));
label("$D$", (1.2,0),dir(-90));
dot((6,1.4));
dot((1.2,0));
label("$a$", (0.6,0),dir(-90));
label("$b$", (3.6,0),dir(-90));
label("$c$", (6,0.7),dir(0));
// Block 2
draw((1.2,0)--(0,0)--(0,1.4)--(6,1.4)--(6,0)--(1.2,0)--(6,1.4)); label("$D$", (1.2,0),dir(-90)); dot((6,1.4)); dot((1.2,0)); label("$a$", (0.6,0),dir(-90)); label("$b$", (3.6,0),dir(-90)); label("$c$", (6,0.7),dir(0)); | [] |
171 | Triangle $ABC$ is inscribed in circle $\omega$ with $AB=5$, $BC=7$, and $AC=3$. The bisector of angle $A$ meets side $\overline{BC}$ at $D$ and circle $\omega$ at a second point $E$. Let $\gamma$ be the circle with diameter $\overline{DE}$. Circles $\omega$ and $\gamma$ meet at $E$ and a second point $F$. Then $AF^2 = \frac mn$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. | 2012 AIME II Problem 15 | Use the angle bisector theorem to find $CD=\tfrac{21}{8}$, $BD=\tfrac{35}{8}$, and use Stewart's Theorem to find $AD=\tfrac{15}{8}$. Use Power of Point $D$ to find $DE=\tfrac{49}{8}$, and so $AE=8$. Use law of cosines to find $\angle CAD = \tfrac{\pi} {3}$, hence $\angle BAD = \tfrac{\pi}{3}$ as well, and $\triangle BCE$ is equilateral, so $BC=CE=BE=7$.
In triangle $AEF$, let $X$ be the foot of the altitude from $A$; then $EF=EX+XF$, where we use signed lengths. Writing $EX=AE \cdot \cos \angle AEF$ and $XF=AF \cdot \cos \angle AFE$, we get
\begin{align}\tag{1} EF = AE \cdot \cos \angle AEF + AF \cdot \cos \angle AFE. \end{align}
Note $\angle AFE = \angle ACE$, and the Law of Cosines in $\triangle ACE$ gives $\cos \angle ACE = -\tfrac 17$.
Also, $\angle AEF = \angle DEF$, and $\angle DFE = \tfrac{\pi}{2}$ ($DE$ is a diameter), so $\cos \angle AEF = \tfrac{EF}{DE} = \tfrac{8}{49}\cdot EF$.
Plugging in all our values into equation $(1)$, we get:
\[EF = \tfrac{64}{49} EF -\tfrac{1}{7} AF \quad \Longrightarrow \quad EF = \tfrac{7}{15} AF.\]
The Law of Cosines in $\triangle AEF$, with $EF=\tfrac 7{15}AF$ and $\cos\angle AFE = -\tfrac 17$ gives
\[8^2 = AF^2 + \tfrac{49}{225} AF^2 + \tfrac 2{15} AF^2 = \tfrac{225+49+30}{225}\cdot AF^2\]
Thus $AF^2 = \frac{900}{19}$. The answer is $\boxed{919}$.
~Shen Kislay Kai | // Block 1
size(150);
defaultpen(fontsize(9pt));
picture pic;
pair A,B,C,D,E,F,W;
B=MP("B",origin,dir(180)); C=MP("C",(7,0),dir(0)); A=MP("A",IP(CR(B,5),CR(C,3)),N); D=MP("D",extension(B,C,A,bisectorpoint(C,A,B)),dir(220)); path omega=circumcircle(A,B,C); E=MP("E",OP(omega,A--(A+20*(D-A))),S); path gamma=CR(midpoint(D--E),length(D-E)/2); F=MP("F",OP(omega,gamma),SE);
draw(omega^^A--B--C--cycle^^gamma); draw(pic, A--E--F--cycle, gray); add(pic);
dot("$W$",circumcenter(A,B,C),dir(180)); label("$\gamma$",gamma,dir(180));
// Block 2
size(150); defaultpen(fontsize(9pt)); picture pic; pair A,B,C,D,E,F,W; B=MP("B",origin,dir(180)); C=MP("C",(7,0),dir(0)); A=MP("A",IP(CR(B,5),CR(C,3)),N); D=MP("D",extension(B,C,A,bisectorpoint(C,A,B)),dir(220)); path omega=circumcircle(A,B,C); E=MP("E",OP(omega,A--(A+20*(D-A))),S); path gamma=CR(midpoint(D--E),length(D-E)/2); F=MP("F",OP(omega,gamma),SE); draw(omega^^A--B--C--cycle^^gamma); draw(pic, A--E--F--cycle, gray); add(pic); dot("$W$",circumcenter(A,B,C),dir(180)); label("$\gamma$",gamma,dir(180)); | [] |
171 | Triangle $ABC$ is inscribed in circle $\omega$ with $AB=5$, $BC=7$, and $AC=3$. The bisector of angle $A$ meets side $\overline{BC}$ at $D$ and circle $\omega$ at a second point $E$. Let $\gamma$ be the circle with diameter $\overline{DE}$. Circles $\omega$ and $\gamma$ meet at $E$ and a second point $F$. Then $AF^2 = \frac mn$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. | 2012 AIME II Problem 15 | Let $a = BC$, $b = CA$, $c = AB$ for convenience. Let $M$ be the midpoint of segment $BC$. We claim that $\angle MAD=\angle DAF$.
$\textit{Proof}$. Since $AE$ is the angle bisector, it follows that $EB = EC$ and consequently $EM\perp BC$. Therefore, $M\in \gamma$. Now let $X = FD\cap \omega$. Since $\angle EFX=90^\circ$, $EX$ is a diameter, so $X$ lies on the perpendicular bisector of $BC$; hence $E$, $M$, $X$ are collinear. From $\angle DAG = \angle DMX = 90^\circ$, quadrilateral $ADMX$ is cyclic. Therefore, $\angle MAD = \angle MXD$. But $\angle MXD$ and $\angle EAF$ are both subtended by arc $EF$ in $\omega$, so they are equal. Thus $\angle MAD=\angle DAF$, as claimed.
As a result, $\angle CAM = \angle FAB$. Combined with $\angle BFA=\angle MCA$, we get $\triangle ABF\sim\triangle AMC$ and therefore\[\frac c{AM}=\frac {AF}b\qquad \Longrightarrow \qquad AF^2=\frac{b^2c^2}{AM^2} = \frac{15^2}{AM^2}\]
By Stewart's Theorem on $\triangle ABC$ (with cevian $AM$), we get \[AM^2 = \tfrac 12 (b^2+c^2)-\tfrac 14 a^2 = \tfrac{19}{4},\] so $AF^2 = \tfrac{900}{19}$, so the answer is $900+19=\boxed{919}$.
-Solution by thecmd999 | // Block 1
size(175);
defaultpen(fontsize(10pt));
picture pic;
pair A,B,C,D,E,F,W;
B=MP("B",origin,dir(180)); C=MP("C",(7,0),dir(0)); A=MP("A",IP(CR(B,5),CR(C,3)),N); D=MP("D",extension(B,C,A,bisectorpoint(C,A,B)),dir(220)); path omega=circumcircle(A,B,C); E=MP("E",OP(omega,A--(A+20*(D-A))),S); path gamma=CR(midpoint(D--E),length(D-E)/2); F=MP("F",OP(omega,gamma),SE); pair X=MP("X",IP(omega,F--(F+2*(D-F))),N); pair M=MP("M",midpoint(B--C),dir(220));
draw(omega^^A--B--C--cycle^^gamma); draw(A--E--F--cycle, gray); draw(E--X--F, gray); draw(pic, A--M--C--cycle^^A--B--F--cycle); draw(A--M, royalblue);
dot("$W$",circumcenter(A,B,C),dir(180)); dot(circumcenter(D,E,F)); label("$\gamma$",gamma,dir(180));
draw(A--B--F--cycle, black+1);
// Block 2
size(175); defaultpen(fontsize(10pt)); picture pic; pair A,B,C,D,E,F,W; B=MP("B",origin,dir(180)); C=MP("C",(7,0),dir(0)); A=MP("A",IP(CR(B,5),CR(C,3)),N); D=MP("D",extension(B,C,A,bisectorpoint(C,A,B)),dir(220)); path omega=circumcircle(A,B,C); E=MP("E",OP(omega,A--(A+20*(D-A))),S); path gamma=CR(midpoint(D--E),length(D-E)/2); F=MP("F",OP(omega,gamma),SE); pair X=MP("X",IP(omega,F--(F+2*(D-F))),N); pair M=MP("M",midpoint(B--C),dir(220)); draw(omega^^A--B--C--cycle^^gamma); draw(A--E--F--cycle, gray); draw(E--X--F, gray); draw(pic, A--M--C--cycle^^A--B--F--cycle); draw(A--M, royalblue); dot("$W$",circumcenter(A,B,C),dir(180)); dot(circumcenter(D,E,F)); label("$\gamma$",gamma,dir(180)); draw(A--B--F--cycle, black+1); | [] |
171 | Triangle $ABC$ is inscribed in circle $\omega$ with $AB=5$, $BC=7$, and $AC=3$. The bisector of angle $A$ meets side $\overline{BC}$ at $D$ and circle $\omega$ at a second point $E$. Let $\gamma$ be the circle with diameter $\overline{DE}$. Circles $\omega$ and $\gamma$ meet at $E$ and a second point $F$. Then $AF^2 = \frac mn$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. | 2012 AIME II Problem 15 | Use the angle bisector theorem to find $CD=\tfrac{21}{8}$, $BD=\tfrac{35}{8}$, and use Stewart's Theorem to find $AD=\tfrac{15}{8}$. Use Power of Point $D$ to find $DE=\tfrac{49}{8}$, and so $AE=8$. Then use the Extended Law of Sine to find that the length of the circumradius of $\triangle ABC$ is $\tfrac{7\sqrt{3}}{3}$.
Since $DE$ is the diameter of circle $\gamma$, $\angle DFE$ is $90^\circ$. Extending $DF$ to intersect circle $\omega$ at $X$, we find that $XE$ is the diameter of $\omega$ (since $\angle DFE$ is $90^\circ$). Therefore, $XE=\tfrac{14\sqrt{3}}{3}$.
Let $EF=x$, $XD=u$, and $DF=v$. Then $XE^2-XF^2=EF^2=DE^2-DF^2$, so we get
\[(u+v)^2-v^2=\frac{196}{3}-\frac{2401}{64}\]
which simplifies to
\[u^2+2uv = \frac{5341}{192}.\]
By Power of Point $D$, $uv=BD \cdot DC=735/64$. Combining with above, we get
\[XD^2=u^2=\frac{931}{192}.\]
Note that $\triangle XDE\sim \triangle ADF$ and the ratio of similarity is $\rho = AD : XD = \tfrac{15}{8}:u$. Then $AF=\rho\cdot XE = \tfrac{15}{8u}\cdot R$ and \[AF^2 = \frac{225}{64}\cdot \frac{R^2}{u^2} = \frac{900}{19}.\]
The answer is $900+19=\boxed{919}$.
-Solution by TheBoomBox77 | // Block 1
size(175);
defaultpen(fontsize(9pt));
pair A,B,C,D,E,F,W;
B=MP("B",origin,dir(180)); C=MP("C",(7,0),dir(0)); A=MP("A",IP(CR(B,5),CR(C,3)),N); D=MP("D",extension(B,C,A,bisectorpoint(C,A,B)),dir(220)); path omega=circumcircle(A,B,C); E=MP("E",OP(omega,A--(A+20*(D-A))),S); path gamma=CR(midpoint(D--E),length(D-E)/2); F=MP("F",OP(omega,gamma),SE); pair X=MP("X",IP(omega,F--(F+2*(D-F))),N);
draw(omega^^A--B--C--cycle^^gamma); draw(A--E--F--cycle, gray); draw(E--X--F, royalblue);
dot("$W$",circumcenter(A,B,C),dir(180)); dot(circumcenter(D,E,F)); label("$\gamma$",gamma,dir(180)); label("$u$",X--D,dir(60)); label("$v$",D--F,dir(70));
// Block 2
size(175); defaultpen(fontsize(9pt)); pair A,B,C,D,E,F,W; B=MP("B",origin,dir(180)); C=MP("C",(7,0),dir(0)); A=MP("A",IP(CR(B,5),CR(C,3)),N); D=MP("D",extension(B,C,A,bisectorpoint(C,A,B)),dir(220)); path omega=circumcircle(A,B,C); E=MP("E",OP(omega,A--(A+20*(D-A))),S); path gamma=CR(midpoint(D--E),length(D-E)/2); F=MP("F",OP(omega,gamma),SE); pair X=MP("X",IP(omega,F--(F+2*(D-F))),N); draw(omega^^A--B--C--cycle^^gamma); draw(A--E--F--cycle, gray); draw(E--X--F, royalblue); dot("$W$",circumcenter(A,B,C),dir(180)); dot(circumcenter(D,E,F)); label("$\gamma$",gamma,dir(180)); label("$u$",X--D,dir(60)); label("$v$",D--F,dir(70)); | [] |
172 | A paper equilateral triangle $ABC$ has side length $12$. The paper triangle is folded so that vertex $A$ touches a point on side $\overline{BC}$ a distance $9$ from point $B$. The length of the line segment along which the triangle is folded can be written as $\frac{m\sqrt{p}}{n}$, 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 $m+n+p$. | 2013 AIME I Problem 9 | Let $M$ and $N$ be the points on $\overline{AB}$ and $\overline{AC}$, respectively, where the paper is folded. Let $D$ be the point on $\overline{BC}$ where the folded $A$ touches it.
We have $AF=6\sqrt{3}$ and $FD=3$, so $AD=3\sqrt{13}$. Denote $\angle DAF = \theta$; we get $\cos\theta = 2\sqrt{3}/\sqrt{13}$.
In triangle $AXY$, $AY=\tfrac 12 AD = \tfrac 32 \sqrt{13}$, and $AX=AY\sec\theta =\tfrac{13}{4}\sqrt{3}$.
In triangle $AMX$, we get $\angle AMX=60^\circ-\theta$ and then use sine-law to get $MX=\tfrac 12 AX\csc(60^\circ-\theta)$; similarly, from triangle $ANX$ we get $NX=\tfrac 12 AX\csc(60^\circ+\theta)$. Thus \[MN=\tfrac 12 AX(\csc(60^\circ-\theta) +\csc(60^\circ+\theta)).\] Since $\sin(60^\circ\pm \theta) = \tfrac 12 (\sqrt{3}\cos\theta \pm \sin\theta)$, we get
\begin{align*} \csc(60^\circ-\theta) +\csc(60^\circ+\theta) &= \frac{\sqrt{3}\cos\theta}{\cos^2\theta - \tfrac 14} = \frac{24 \cdot \sqrt{13}}{35} \end{align*}
Then \[MN = \frac 12 AX \cdot \frac{24 \cdot \sqrt{13}}{35} = \frac{39\sqrt{39}}{35}\]
The answer is $39 + 39 + 35 = \boxed{113}$. | // Block 1
import cse5; size(8cm); pen tpen = defaultpen + 1.337;
real a = 39/5.0; real b = 39/7.0;
pair B = MP("B", (0,0), dir(200)); pair A = MP("A", 12*dir(60), dir(90)); pair C = MP("C", (12,0), dir(-20)); pair D = MP("D", (9,0), dir(-80)); pair Y = MP("Y", midpoint(A--D), dir(-50)); pair M = MP("M", extension(A,B,Y,Y+(dir(90)*(D-A))), dir(180)); pair N = MP("N", extension(A,C,M,Y), dir(20)); pair F = MP("F", foot(A,B,C), dir(-90)); pair X = MP("X", extension(A,F,M,N), dir(-120));
draw(B--A--C--cycle, tpen); draw(M--N^^F--A--D); draw(rightanglemark(D,F,A,15)); draw(rightanglemark(A,Y,M,15)); MA("\theta",F,A,D,1.8);
// Block 2
import cse5; size(8cm); pen tpen = defaultpen + 1.337; real a = 39/5.0; real b = 39/7.0; pair B = MP("B", (0,0), dir(200)); pair A = MP("A", 12*dir(60), dir(90)); pair C = MP("C", (12,0), dir(-20)); pair D = MP("D", (9,0), dir(-80)); pair Y = MP("Y", midpoint(A--D), dir(-50)); pair M = MP("M", extension(A,B,Y,Y+(dir(90)*(D-A))), dir(180)); pair N = MP("N", extension(A,C,M,Y), dir(20)); pair F = MP("F", foot(A,B,C), dir(-90)); pair X = MP("X", extension(A,F,M,N), dir(-120)); draw(B--A--C--cycle, tpen); draw(M--N^^F--A--D); draw(rightanglemark(D,F,A,15)); draw(rightanglemark(A,Y,M,15)); MA("\theta",F,A,D,1.8); | [] |
172 | A paper equilateral triangle $ABC$ has side length $12$. The paper triangle is folded so that vertex $A$ touches a point on side $\overline{BC}$ a distance $9$ from point $B$. The length of the line segment along which the triangle is folded can be written as $\frac{m\sqrt{p}}{n}$, 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 $m+n+p$. | 2013 AIME I Problem 9 | We let the original position of $A$ be $A$, and the position of $A$ after folding be $D$. Also, we put the triangle on the coordinate plane such that $A=(0,0)$, $B=(-6,-6\sqrt3)$, $C=(6,-6\sqrt3)$, and $D=(3,-6\sqrt3)$.
Note that since $A$ is reflected over the fold line to $D$, the fold line is the perpendicular bisector of $AD$. We know $A=(0,0)$ and $D=(3,-6\sqrt3)$. The midpoint of $AD$ (which is a point on the fold line) is $(\tfrac32, -3\sqrt3)$. Also, the slope of $AD$ is $\frac{-6\sqrt3}{3}=-2\sqrt3$, so the slope of the fold line (which is perpendicular), is the negative of the reciprocal of the slope of $AD$, or $\frac{1}{2\sqrt3}=\frac{\sqrt3}{6}$. Then, using point slope form, the equation of the fold line is
\[y+3\sqrt3=\frac{\sqrt3}{6}\left(x-\frac32\right)\]\[y=\frac{\sqrt3}{6}x-\frac{13\sqrt3}{4}\]
Note that the equations of lines $AB$ and $AC$ are $y=\sqrt3x$ and $y=-\sqrt3x$, respectively. We will first find the intersection of $AB$ and the fold line by substituting for $y$:
\[\sqrt3 x=\frac{\sqrt3}{6}x-\frac{13\sqrt3}{4}\]\[\frac{5\sqrt3}{6}x=-\frac{13\sqrt3}{4} \implies x=-\frac{39}{10}\]
Therefore, the point of intersection is $\left(-\tfrac{39}{10},-\tfrac{39\sqrt3}{10}\right)$. Now, lets find the intersection with $AC$. Substituting for $y$ yields
\[-\sqrt3 x=\frac{\sqrt3}{6}x-\frac{13\sqrt3}{4}\]\[\frac{-7\sqrt3}{6}x=-\frac{13\sqrt3}{4} \implies x=\frac{39}{14}\]
Therefore, the point of intersection is $\left(\tfrac{39}{14},-\tfrac{39\sqrt3}{14}\right)$. Now, we just need to use the distance formula to find the distance between $\left(-\tfrac{39}{10},-\tfrac{39\sqrt3}{10}\right)$ and $\left(\tfrac{39}{14},-\tfrac{39\sqrt3}{14}\right)$.
\[\sqrt{\left(\frac{39}{14}+\frac{39}{10}\right)^2+\left(-\frac{39\sqrt3}{14}+\frac{39\sqrt3}{10}\right)^2}\]
The number 39 is in all of the terms, so let's factor it out:
\[39\sqrt{\left(\frac{1}{14}+\frac{1}{10}\right)^2+\left(-\frac{\sqrt3}{14}+\frac{\sqrt3}{10}\right)^2}=39\sqrt{\left(\frac{6}{35}\right)^2+\left(\frac{\sqrt3}{35}\right)^2}\]\[\frac{39}{35}\sqrt{6^2+\sqrt3^2}=\frac{39\sqrt{39}}{35}\]
Therefore, our answer is $39+39+35=\boxed{113}$, and we are done.
Solution by nosaj. | // Block 1
size(10cm);
pen tpen = defaultpen + 1.337;
real a = 39/5.0;
real b = 39/7.0;
pair B = MP("B", (0,0), dir(200));
pair A = (9,0);
pair C = MP("C", (12,0), dir(-20));
pair K = (6,10.392);
pair M = (a*B+(12-a)*K) / 12;
pair N = (b*C+(12-b)*K) / 12;
draw(B--M--N--C--cycle);
draw(M--A--N--cycle);
label("$D$", A, S);
pair X = (6,6*sqrt(3));
draw(B--X--C);
label("$A$",X,dir(90));
draw(A--X);
// Block 2
size(10cm); pen tpen = defaultpen + 1.337; real a = 39/5.0; real b = 39/7.0; pair B = MP("B", (0,0), dir(200)); pair A = (9,0); pair C = MP("C", (12,0), dir(-20)); pair K = (6,10.392); pair M = (a*B+(12-a)*K) / 12; pair N = (b*C+(12-b)*K) / 12; draw(B--M--N--C--cycle); draw(M--A--N--cycle); label("$D$", A, S); pair X = (6,6*sqrt(3)); draw(B--X--C); label("$A$",X,dir(90)); draw(A--X); | [] |
172 | A paper equilateral triangle $ABC$ has side length $12$. The paper triangle is folded so that vertex $A$ touches a point on side $\overline{BC}$ a distance $9$ from point $B$. The length of the line segment along which the triangle is folded can be written as $\frac{m\sqrt{p}}{n}$, 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 $m+n+p$. | 2013 AIME I Problem 9 | Thanks to Solution 1 for the diagram below:
We will use the notation already on the diagram, but our solution is slightly different.
We will only need M and N.
Let NC be length a, which implies NA be 12-a.
Also, AC = 3 because AB = 9
By the Law of Cosines on NCA,
$(12-a)^2=a^2+3^2-2(a)(3)(cos60)$
which simplifies to:
$a=\frac{45}{7}$
Which means that NC = 45/7 and NA = 39/7.
We can do the same thing for MBA.
This time, MB = b.
$(12-b)^2=b^2+81-2(9)(b)(cos60)$
Which gives:
$b=\frac{21}{5}$
Which implies that MA = $\frac{39}{5}$
Now, since MAN is 60 degrees, we can apply the Law of Cosines again(I know, I don't like bashy things too) to get:
$c^2=(\frac{39}{5})^2+39/7^2-2(39/7)(39/5)(cos60)$
Which leads us to our answer => 113
~MC | // Block 1
import cse5; size(8cm); pen tpen = defaultpen + 1.337;
real a = 39/5.0; real b = 39/7.0;
pair B = MP("B", (0,0), dir(200)); pair A = MP("A", 12*dir(60), dir(90)); pair C = MP("C", (12,0), dir(-20)); pair D = MP("D", (9,0), dir(-80)); pair Y = MP("Y", midpoint(A--D), dir(-50)); pair M = MP("M", extension(A,B,Y,Y+(dir(90)*(D-A))), dir(180)); pair N = MP("N", extension(A,C,M,Y), dir(20)); pair F = MP("F", foot(A,B,C), dir(-90)); pair X = MP("X", extension(A,F,M,N), dir(-120));
draw(B--A--C--cycle, tpen); draw(M--N^^F--A--D); draw(rightanglemark(D,F,A,15)); draw(rightanglemark(A,Y,M,15)); MA("\theta",F,A,D,1.8);
// Block 2
import cse5; size(8cm); pen tpen = defaultpen + 1.337; real a = 39/5.0; real b = 39/7.0; pair B = MP("B", (0,0), dir(200)); pair A = MP("A", 12*dir(60), dir(90)); pair C = MP("C", (12,0), dir(-20)); pair D = MP("D", (9,0), dir(-80)); pair Y = MP("Y", midpoint(A--D), dir(-50)); pair M = MP("M", extension(A,B,Y,Y+(dir(90)*(D-A))), dir(180)); pair N = MP("N", extension(A,C,M,Y), dir(20)); pair F = MP("F", foot(A,B,C), dir(-90)); pair X = MP("X", extension(A,F,M,N), dir(-120)); draw(B--A--C--cycle, tpen); draw(M--N^^F--A--D); draw(rightanglemark(D,F,A,15)); draw(rightanglemark(A,Y,M,15)); MA("\theta",F,A,D,1.8); | [] |
172 | A paper equilateral triangle $ABC$ has side length $12$. The paper triangle is folded so that vertex $A$ touches a point on side $\overline{BC}$ a distance $9$ from point $B$. The length of the line segment along which the triangle is folded can be written as $\frac{m\sqrt{p}}{n}$, 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 $m+n+p$. | 2013 AIME I Problem 9 | As shown in the diagram, let $MA'=x$. Since the triangle is equilateral, we have $BM=12-x$. Similarly, let $NA'=y$, so $NC=12-y$.
Because $\angle MBA=\angle NCA=60^\circ$, we may apply the Law of Cosines in the form $a^2+b^2-ab=c^2$. (Note that we could also find these side lengths through similar triangles as mentioned in the previous solutions)
Solving for $x$, we have
\[(12-x)^2+9^2-9(12-x)=x^2.\]
Simplifying,
\[x^2-24x+144-27+9x=x^2,\]
so
\[15x=117 \quad \Rightarrow \quad x=\frac{39}{5}.\]
Solving for $y$ similarly,
\[(12-y)^2+3^2-3(12-y)=y^2.\]
Simplifying,
\[y^2-24y+144+9-36+3y=y^2,\]
which gives
\[21y=117 \quad \Rightarrow \quad y=\frac{39}{7}.\]
Now apply the Law of Cosines again to find $MN$:
\[MN^2=\left(\frac{39}{5}\right)^2+\left(\frac{39}{7}\right)^2-\left(\frac{39}{5}\right)\left(\frac{39}{7}\right).\]
Factoring,
\[MN^2=39^2\left(\frac{1}{25}+\frac{1}{49}-\frac{1}{35}\right)=\frac{39^3}{35^2}.\]
Thus,
\[MN=\frac{39\sqrt{39}}{35}.\]
Since the problem asks for $m+n+p$ when the length is written as $\frac{m\sqrt{p}}{n}$, we have
\[m+n+p=39+39+35=\boxed{113}.\]
~Voidling | // Block 1
import cse5;
size(8cm);
pen tpen = defaultpen + 1.337;
real a = 39/5.0;
real b = 39/7.0;
pair B = (0,0);
pair Ap = (9,0);
pair C = (12,0);
pair K = (6, 10.392);
pair M = (a*B+(12-a)*K) / 12;
pair N = (b*C+(12-b)*K) / 12;
draw(B--M--N--C--cycle, tpen);
draw(M--Ap--N--cycle, tpen);
fill(M--Ap--N--cycle, mediumgrey);
label("B", B, dir(200));
label("A'", Ap, dir(-90));
label("C", C, dir(-20));
label("M", M, NW);
label("N", N, NE);
label("$x$", (M+Ap)/2, SW*2, fontsize(10pt));
label("$y$", (N+Ap)/2, dir(20), fontsize(10pt));
label("$12-x$", (B+M)/2, W, fontsize(10pt));
label("$12-y$", (C+N)/2, E, fontsize(10pt));
label("9", (B+Ap)/2, S, fontsize(9pt));
label("3", (Ap+C)/2, S, fontsize(9pt));
// Block 2
import cse5; size(8cm); pen tpen = defaultpen + 1.337; real a = 39/5.0; real b = 39/7.0; pair B = (0,0); pair Ap = (9,0); pair C = (12,0); pair K = (6, 10.392); pair M = (a*B+(12-a)*K) / 12; pair N = (b*C+(12-b)*K) / 12; draw(B--M--N--C--cycle, tpen); draw(M--Ap--N--cycle, tpen); fill(M--Ap--N--cycle, mediumgrey); label("B", B, dir(200)); label("A'", Ap, dir(-90)); label("C", C, dir(-20)); label("M", M, NW); label("N", N, NE); label("$x$", (M+Ap)/2, SW*2, fontsize(10pt)); label("$y$", (N+Ap)/2, dir(20), fontsize(10pt)); label("$12-x$", (B+M)/2, W, fontsize(10pt)); label("$12-y$", (C+N)/2, E, fontsize(10pt)); label("9", (B+Ap)/2, S, fontsize(9pt)); label("3", (Ap+C)/2, S, fontsize(9pt)); | [] |
173 | In equilateral $\triangle ABC$ let points $D$ and $E$ trisect $\overline{BC}$. Then $\sin(\angle DAE)$ can be expressed in the form $\frac{a\sqrt{b}}{c}$, where $a$ and $c$ are relatively prime positive integers, and $b$ is an integer that is not divisible by the square of any prime. Find $a+b+c$. | 2013 AIME II Problem 5 | Without loss of generality, assume the triangle sides have length 3. Then the trisected side is partitioned into segments of length 1, making your computation easier.
Let $M$ be the midpoint of $\overline{DE}$. Then $\Delta MCA$ is a 30-60-90 triangle with $MC = \dfrac{3}{2}$, $AC = 3$ and $AM = \dfrac{3\sqrt{3}}{2}$. Since the triangle $\Delta AME$ is right, then we can find the length of $\overline{AE}$ by pythagorean theorem, $AE = \sqrt{7}$. Therefore, since $\Delta AME$ is a right triangle, we can easily find $\sin(\angle EAM) = \dfrac{1}{2\sqrt{7}}$ and $\cos(\angle EAM) = \sqrt{1-\sin(\angle EAM)^2}=\dfrac{3\sqrt{3}}{2\sqrt{7}}$. So we can use the double angle formula for sine, $\sin(\angle EAD) = 2\sin(\angle EAM)\cos(\angle EAM) = \dfrac{3\sqrt{3}}{14}$. Therefore, $a + b + c = \boxed{020}$. | // Block 1
pair A = (1, sqrt(3)), B = (0, 0), C= (2, 0);
pair M = (1, 0);
pair D = (2/3, 0), E = (4/3, 0);
draw(A--B--C--cycle);
label("$A$", A, N);
label("$B$", B, SW);
label("$C$", C, SE);
label("$D$", D, S);
label("$M$", M, S);
label("$E$", E, S);
draw(A--D);
draw(A--M);
draw(A--E);
// Block 2
pair A = (1, sqrt(3)), B = (0, 0), C= (2, 0); pair M = (1, 0); pair D = (2/3, 0), E = (4/3, 0); draw(A--B--C--cycle); label("$A$", A, N); label("$B$", B, SW); label("$C$", C, SE); label("$D$", D, S); label("$M$", M, S); label("$E$", E, S); draw(A--D); draw(A--M); draw(A--E); | [] |
174 | A hexagon that is inscribed in a circle has side lengths $22$, $22$, $20$, $22$, $22$, and $20$ in that order. The radius of the circle can be written as $p+\sqrt{q}$, where $p$ and $q$ are positive integers. Find $p+q$. | 2013 AIME II Problem 8 | Solution 1
Let us call the hexagon $ABCDEF$, where $AB=CD=DE=AF=22$, and $BC=EF=20$.
We can just consider one half of the hexagon, $ABCD$, to make matters simpler.
Draw a line from the center of the circle, $O$, to the midpoint of $BC$, $X$. Now, draw a line from $O$ to the midpoint of $AB$, $Y$. Clearly, $\angle BXO=90^{\circ}$, because $BO=CO$, and $\angle BYO=90^{\circ}$, for similar reasons. Also notice that $\angle AOX=90^{\circ}$.
Let us call $\angle BOY=\theta$. Therefore, $\angle AOB=2\theta$, and so $\angle BOX=90-2\theta$. Let us label the radius of the circle $r$. This means \[\sin{\theta}=\frac{BY}{r}=\frac{11}{r}\] \[\sin{(90-2\theta)}=\frac{BX}{r}=\frac{10}{r}\]
Now we can use simple trigonometry to solve for $r$.
Recall that $\sin{(90-\alpha)}=\cos(\alpha)$: That means $\sin{(90-2\theta)}=\cos{2\theta}=\frac{10}{r}$.
Recall that $\cos{2\alpha}=1-2\sin^2{\alpha}$: That means $\cos{2\theta}=1-2\sin^2{\theta}=\frac{10}{r}$.
Let $\sin{\theta}=x$.
Substitute to get $x=\frac{11}{r}$ and $1-2x^2=\frac{10}{r}$
Now substitute the first equation into the second equation: $1-2\left(\frac{11}{r}\right)^2=\frac{10}{r}$
Multiplying both sides by $r^2$ and reordering gives us the quadratic \[r^2-10r-242=0\]
Using the quadratic equation to solve, we get that $r=5+\sqrt{267}$ (because $5-\sqrt{267}$ gives a negative value), so the answer is $5+267=\boxed{272}$.
Solution 2
Using the trapezoid $ABCD$ mentioned above, draw an altitude of the trapezoid passing through point $B$ onto $AD$ at point $J$. Now, we can use the pythagorean theorem: $(22^2-(r-10)^2)+10^2=r^2$. Expanding and combining like terms gives us the quadratic \[r^2-10r-242=0\] and solving for $r$ gives $r=5+\sqrt{267}$. So the solution is $5+267=\boxed{272}$.
Solution 3
Join the diameter of the circle $AD$ and let the length be $d$. By Ptolemy's Theorem on trapezoid $ADEF$, $(AD)(EF) + (AF)(DE) = (AE)(DF)$. Since it is an isosceles trapezoid, both diagonals are equal. Let them be equal to $x$ each. Then
\[20d + 22^2 = x^2\]
Since $\angle AED$ is subtended by the diameter, it is right. Hence by the Pythagorean Theorem with right $\triangle AED$:
\[(AE)^2 + (ED)^2 = (AD)^2\]
\[x^2 + 22^2 = d^2\]
From the above equations, we have:
\[x^2 = d^2 - 22^2 = 20d + 22^2\]
\[d^2 - 20d = 2\times22^2\]
\[d^2 - 20d + 100 = 968+100 = 1068\]
\[(d-10) = \sqrt{1068}\]
\[d = \sqrt{1068} + 10 = 2\times(\sqrt{267}+5)\]
Since the radius is half the diameter, it is $\sqrt{267}+5$, so the answer is $5+267 \Rightarrow \boxed{272}$.
Solution 4
As we can see this image, it is symmetrical hence the diameter divides the hexagon into two congruent quadrilateral. Now we can apply the Ptolemy's theorem. Denote the radius is r, we can get \[22*2x+440=\sqrt{4x^2-400}\sqrt{4x^2-484}\], after simple factorization, we can get \[x^4-342x^2-2420x=0\], it is easy to see that $x=-10, x=0$ are two solutions for the equation, so we can factorize that into \[x(x+10)(x^2-10x-242)\]so we only need to find the solution for \[x^2-10x-242=0\] and we can get $x=(\sqrt{267}+5)$ is the desired answer for the problem, and our answer is $5+267 \Rightarrow \boxed{272}$.～bluesoul
Solution 6 (Trig Bash)
Let $\angle{AOB} = \theta$. So, we have $\sin \dfrac{\theta}{2} = \dfrac{11}{r}$ and $\cos \dfrac{\theta}{2} = \dfrac{\sqrt{r^{2} - 121}}{r}$. So, $\sin \theta = 2 \sin \dfrac{\theta}{2} \cos \dfrac{\theta}{2} = \dfrac{22 \sqrt{r^{2} - 121}}{r^{2}}$. Let $H$ be the foot of the perpendicular from $B$ to $\overline{AD}$. We have $BF = 2 BH = 2 r \sin \theta = \dfrac{44 \sqrt{r^{2} - 121}}{r}$. Using Pythagorean theorem on $\triangle BCF$, to get $(\dfrac{44 \sqrt{r^{2} - 121}}{r})^{2} + 20^{2} = (2r)^{2}$, or $\dfrac{44^{2}r^{2} - 44^{2} \cdot 121}{r^{2}} + 20^{2} = 4r^{4}$. Multiplying by $r^{2}$, we get $44^{2} r^{2} - 44^{2} \cdot 121 + 20^{2} r^{2} = 4r^{4}$. Rearranging and simplifying, we get a quadratic in $r^{2}$: \[r^{4} - 584r^{2} + 242^{2} = 0 \text{,}\] which gives us $r^{2} = 292 \pm 10\sqrt{267}$. Because $r$ is in the form $p + \sqrt{q}$, we know to choose the larger option, meaning $r^2 = 292 + 10\sqrt{267}$, so $p\sqrt{q} = 5\sqrt{267}$ and $p^2 + q = 292$. By inspection, we get $(p, q) = (5, 267)$, so our answer is $5 + 267 = \boxed{272}$.
~Puck_0 | // Block 1
import olympiad;
import math;
real a;
a=2*asin(11/(5+sqrt(267)));
pair A,B,C,D,E,F;
A=expi(pi); B=expi(pi-a); C=expi(a); D=expi(0); E=expi(-a); F=expi(pi+a);
draw(A--B--C--D--E--F--A--D);
dot(A); dot(B); dot(C); dot(D); dot(E); dot(F);
draw(unitcircle);
label("$A$",A,W);label("$B$",B,NW);label("$C$",C,NE);label("$D$",D,dir(0));label("$E$",E,SE);label("$F$",F,SW);
// Block 2
import olympiad; import math; real a; a=2*asin(11/(5+sqrt(267))); pair A,B,C,D,E,F; A=expi(pi); B=expi(pi-a); C=expi(a); D=expi(0); E=expi(-a); F=expi(pi+a); draw(A--B--C--D--E--F--A--D); dot(A); dot(B); dot(C); dot(D); dot(E); dot(F); draw(unitcircle); label("$A$",A,W);label("$B$",B,NW);label("$C$",C,NE);label("$D$",D,dir(0));label("$E$",E,SE);label("$F$",F,SW); | [] |
174 | A hexagon that is inscribed in a circle has side lengths $22$, $22$, $20$, $22$, $22$, and $20$ in that order. The radius of the circle can be written as $p+\sqrt{q}$, where $p$ and $q$ are positive integers. Find $p+q$. | 2013 AIME II Problem 8 | We know that $AD=x$ is a diameter, hence $ABD$ and $ACD$ are right triangles. Let $AB=BC=22$, and $CD=20.$ Hence, $ABD$ is a right triangle with legs $22,\sqrt{x^2-484},$ and hypotenuse, $x,$ and $ACD$ is a right triangle with legs $20, \sqrt{x^2-400},$ with hypotenuse $x$. By Ptolemy's we have \[22(x+20)=\sqrt{x^2-400}\sqrt{x^2-484}\].
We square both sides to get \[484(x+20)^2=(x^2-400)(x^2-484) \implies 484(x+20)=(x-20)(x^2-484) \implies 484x=x^3-20x^2-484x \implies x(x^2-20x-968)=0\]
We solve for $x$ via the Quadratic Formula and receive $x=10+2\sqrt{267}$, but we must divide by $2$ since we want the radius, and hence $267+5=\boxed{272}.$
~SirAppel | // Block 1
import olympiad;
import math;
real a;
a=2*asin(11/(5+sqrt(267)));
pair A,B,C,D,E,F;
A=expi(pi); B=expi(pi-a); C=expi(a); D=expi(0); E=expi(-a); F=expi(pi+a);
draw(A--B--C--D--E--F--A--D);
draw(B--D);
draw(A--C);
dot(A); dot(B); dot(C); dot(D); dot(E); dot(F);
draw(unitcircle);
label("$A$",A,W);label("$B$",B,NW);label("$C$",C,NE);label("$D$",D,dir(0));label("$E$",E,SE);label("$F$",F,SW);
// Block 2
import olympiad; import math; real a; a=2*asin(11/(5+sqrt(267))); pair A,B,C,D,E,F; A=expi(pi); B=expi(pi-a); C=expi(a); D=expi(0); E=expi(-a); F=expi(pi+a); draw(A--B--C--D--E--F--A--D); draw(B--D); draw(A--C); dot(A); dot(B); dot(C); dot(D); dot(E); dot(F); draw(unitcircle); label("$A$",A,W);label("$B$",B,NW);label("$C$",C,NE);label("$D$",D,dir(0));label("$E$",E,SE);label("$F$",F,SW); | [] |
175 | Given a circle of radius $\sqrt{13}$, let $A$ be a point at a distance $4 + \sqrt{13}$ from the center $O$ of the circle. Let $B$ be the point on the circle nearest to point $A$. A line passing through the point $A$ intersects the circle at points $K$ and $L$. The maximum possible area for $\triangle BKL$ can be written in the form $\frac{a - b\sqrt{c}}{d}$, where $a$, $b$, $c$, and $d$ are positive integers, $a$ and $d$ are relatively prime, and $c$ is not divisible by the square of any prime. Find $a+b+c+d$. | 2013 AIME II Problem 10 | Now we put the figure in the Cartesian plane, let the center of the circle $O (0,0)$, then $B (\sqrt{13},0)$, and $A(4+\sqrt{13},0)$
The equation for Circle O is $x^2+y^2=13$, and let the slope of the line $AKL$ be $k$, then the equation for line $AKL$ is $y=k(x-4-\sqrt{13})$.
Then we get $(k^2+1)x^2-2k^2(4+\sqrt{13})x+k^2\cdot (4+\sqrt{13})^2-13=0$. According to Vieta's Formulas, we get
$x_1+x_2=\frac{2k^2(4+\sqrt{13})}{k^2+1}$, and $x_1x_2=\frac{(4+\sqrt{13})^2\cdot k^2-13}{k^2+1}$
So, $LK=\sqrt{1+k^2}\cdot \sqrt{(x_1+x_2)^2-4x_1x_2}$
Also, the distance between $B$ and $LK$ is $\frac{k\times \sqrt{13}-(4+\sqrt{13})\cdot k}{\sqrt{1+k^2}}=\frac{-4k}{\sqrt{1+k^2}}$
So the area $S=0.5ah=\frac{-4k\sqrt{(16+8\sqrt{13})k^2-13}}{k^2+1}$
Then the maximum value of $S$ is $\frac{104-26\sqrt{13}}{3}$
So the answer is $104+26+13+3=\boxed{146}$. | // Block 1
import math;
import olympiad;
import graph;
pair A, B, K, L;
B = (sqrt(13), 0); A=(4+sqrt(13), 0);
dot(B);
dot(A);
draw(Circle((0,0), sqrt(13)));
label("$O$", (0,0), S);label("$B$", B, SW);label("$A$", A, S);
dot((0,0));
// Block 2
import math; import olympiad; import graph; pair A, B, K, L; B = (sqrt(13), 0); A=(4+sqrt(13), 0); dot(B); dot(A); draw(Circle((0,0), sqrt(13))); label("$O$", (0,0), S);label("$B$", B, SW);label("$A$", A, S); dot((0,0)); | [] |
175 | Given a circle of radius $\sqrt{13}$, let $A$ be a point at a distance $4 + \sqrt{13}$ from the center $O$ of the circle. Let $B$ be the point on the circle nearest to point $A$. A line passing through the point $A$ intersects the circle at points $K$ and $L$. The maximum possible area for $\triangle BKL$ can be written in the form $\frac{a - b\sqrt{c}}{d}$, where $a$, $b$, $c$, and $d$ are positive integers, $a$ and $d$ are relatively prime, and $c$ is not divisible by the square of any prime. Find $a+b+c+d$. | 2013 AIME II Problem 10 | Draw $OC$ perpendicular to $KL$ at $C$. Draw $BD$ perpendicular to $KL$ at $D$.
\[\frac{\triangle OKL}{\triangle BKL} = \frac{OC}{BD} = \frac{AO}{AB} = \frac{4+\sqrt{13}}{4}\]
Therefore, to maximize area of $\triangle BKL$, we need to maximize area of $\triangle OKL$.
\[\triangle OKL = \frac12 r^2 \sin{\angle KOL}\]
So when area of $\triangle OKL$ is maximized, $\angle KOL = \frac{\pi}{2}$.
Eventually, we get \[\triangle BKL= \frac12 \cdot (\sqrt{13})^2\cdot(\frac{4}{4+\sqrt{13}})=\frac{104-26\sqrt{13}}{3}\]
So the answer is $104+26+13+3=\boxed{146}$. | // Block 1
import math;
import olympiad;
import graph;
pair A, B, K, L;
B = (sqrt(13), 0); A=(4+sqrt(13), 0);
dot(B);
dot(A);
draw(Circle((0,0), sqrt(13)));
label("$O$", (0,0), S);label("$B$", B, SW);label("$A$", A, S);
dot((0,0));
// Block 2
import math; import olympiad; import graph; pair A, B, K, L; B = (sqrt(13), 0); A=(4+sqrt(13), 0); dot(B); dot(A); draw(Circle((0,0), sqrt(13))); label("$O$", (0,0), S);label("$B$", B, SW);label("$A$", A, S); dot((0,0)); | [] |
176 | In $\triangle ABC$, $AC = BC$, and point $D$ is on $\overline{BC}$ so that $CD = 3\cdot BD$. Let $E$ be the midpoint of $\overline{AD}$. Given that $CE = \sqrt{7}$ and $BE = 3$, the area of $\triangle ABC$ 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 $m+n$. | 2013 AIME II Problem 13 | Let $X$ be the foot of the altitude from $C$ with other points labelled as shown below.
Now we proceed using mass points. To balance along the segment $BC$, we assign $B$ a mass of $3$ and $C$ a mass of $1$. Therefore, $D$ has a mass of $4$. As $E$ is the midpoint of $AD$, we must assign $A$ a mass of $4$ as well. This gives $L$ a mass of $5$ and $M$ a mass of $7$.
Now let $AB=b$ be the base of the triangle, and let $CX=h$ be the height. Then as $AM:MB=3:4$, and as $AX=\frac{b}{2}$, we know that \[MX=\frac{b}{2}-\frac{3b}{7}=\frac{b}{14}.\] Also, as $CE:EM=7:1$, we know that $EM=\frac{1}{\sqrt{7}}$. Therefore, by the Pythagorean Theorem on $\triangle {XCM}$, we know that \[\frac{b^2}{196}+h^2=\left(\sqrt{7}+\frac{1}{\sqrt{7}}\right)^2=\frac{64}{7}.\]
Also, as $LE:BE=5:3$, we know that $BL=\frac{8}{5}\cdot 3=\frac{24}{5}$. Furthermore, as $\triangle YLA\sim \triangle XCA$, and as $AL:LC=1:4$, we know that $LY=\frac{h}{5}$ and $AY=\frac{b}{10}$, so $YB=\frac{9b}{10}$. Therefore, by the Pythagorean Theorem on $\triangle BLY$, we get \[\frac{81b^2}{100}+\frac{h^2}{25}=\frac{576}{25}.\]
Solving this system of equations yields $b=2\sqrt{7}$ and $h=3$. Therefore, the area of the triangle is $3\sqrt{7}$, giving us an answer of $\boxed{010}$. | // Block 1
size(200);
pair A=(0,0),B=(2*sqrt(7),0),C=(sqrt(7),3),D=(3*B+C)/4,L=C/5,M=3*B/7;
draw(A--B--C--cycle);draw(A--D^^B--L^^C--M);
label("$A$",A,SW);label("$B$",B,SE);label("$C$",C,N);label("$D$",D,NE);label("$L$",L,NW);label("$M$",M,S);
pair X=foot(C,A,B), Y=foot(L,A,B);
pair EE=D/2;
label("$X$",X,S);label("$E$",EE,NW);label("$Y$",Y,S);
draw(C--X^^L--Y,dotted);
draw(rightanglemark(B,X,C)^^rightanglemark(B,Y,L));
// Block 2
size(200); pair A=(0,0),B=(2*sqrt(7),0),C=(sqrt(7),3),D=(3*B+C)/4,L=C/5,M=3*B/7; draw(A--B--C--cycle);draw(A--D^^B--L^^C--M); label("$A$",A,SW);label("$B$",B,SE);label("$C$",C,N);label("$D$",D,NE);label("$L$",L,NW);label("$M$",M,S); pair X=foot(C,A,B), Y=foot(L,A,B); pair EE=D/2; label("$X$",X,S);label("$E$",EE,NW);label("$Y$",Y,S); draw(C--X^^L--Y,dotted); draw(rightanglemark(B,X,C)^^rightanglemark(B,Y,L)); | [] |
176 | In $\triangle ABC$, $AC = BC$, and point $D$ is on $\overline{BC}$ so that $CD = 3\cdot BD$. Let $E$ be the midpoint of $\overline{AD}$. Given that $CE = \sqrt{7}$ and $BE = 3$, the area of $\triangle ABC$ 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 $m+n$. | 2013 AIME II Problem 13 | Let $BD = x$. Then $CD = 3x$ and $AC = 4x$. Also, let $AE = ED = l$. Using Stewart's Theorem on $\bigtriangleup CEB$ gives us the equation $(x)(3x)(4x) + (4x)(l^2) = 27x + 7x$ or, after simplifying, $4l^2 = 34 - 12x^2$. We use Stewart's again on $\bigtriangleup CAD$: $(l)(l)(2l) + 7(2l) = (16x^2)(l) + (9x^2)(l)$, which becomes $2l^2 = 25x^2 - 14$. Substituting $2l^2 = 17 - 6x^2$, we see that $31x^2 = 31$, or $x = 1$. Then $l^2 = \frac{11}{2}$.
We now use Law of Cosines on $\bigtriangleup CAD$. $(2l)^2 = (4x)^2 + (3x)^2 - 2(4x)(3x)\cos C$. Plugging in for $x$ and $l$, $22 = 16 + 9 - 2(4)(3)\cos C$, so $\cos C = \frac{1}{8}$. Using the Pythagorean trig identity $\sin^2 + \cos^2 = 1$, $\sin^2 C = 1 - \frac{1}{64}$, so $\sin C = \frac{3\sqrt{7}}{8}$.
$[ABC] = \frac{1}{2} AC \cdot BC \sin C = (\frac{1}{2})(4)(4)(\frac{3\sqrt{7}}{8}) = 3\sqrt{7}$, and our answer is $3 + 7 = \boxed{010}$.
Note to writter: Couldn't we just use Heron's formula for $[CEB]$ after $x$ is solved then noticing that $[ABC] = 2 \times [CEB]$? | // Block 1
size(200);
pair A=(0,0),B=(2*sqrt(7),0),C=(sqrt(7),3),D=(3*B+C)/4,L=C/5,M=3*B/7;
draw(A--B--C--cycle);draw(A--D^^B--L^^C--M);
label("$A$",A,SW);label("$B$",B,SE);label("$C$",C,N);label("$D$",D,NE);label("$L$",L,NW);label("$M$",M,S);
pair EE=D/2;
label("$\sqrt{7}$", C--EE, W); label("$x$", D--B, E); label("$3x$", C--D, E); label("$l$", EE--D, N); label("$3$", EE--B, N);
label("$E$",EE,NW);
// Block 2
size(200); pair A=(0,0),B=(2*sqrt(7),0),C=(sqrt(7),3),D=(3*B+C)/4,L=C/5,M=3*B/7; draw(A--B--C--cycle);draw(A--D^^B--L^^C--M); label("$A$",A,SW);label("$B$",B,SE);label("$C$",C,N);label("$D$",D,NE);label("$L$",L,NW);label("$M$",M,S); pair EE=D/2; label("$\sqrt{7}$", C--EE, W); label("$x$", D--B, E); label("$3x$", C--D, E); label("$l$", EE--D, N); label("$3$", EE--B, N); label("$E$",EE,NW); | [] |
176 | In $\triangle ABC$, $AC = BC$, and point $D$ is on $\overline{BC}$ so that $CD = 3\cdot BD$. Let $E$ be the midpoint of $\overline{AD}$. Given that $CE = \sqrt{7}$ and $BE = 3$, the area of $\triangle ABC$ 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 $m+n$. | 2013 AIME II Problem 13 | Let $AB = 2x$ and let $y = BD.$ Then $CD = 3y$ and $AC = 4y.$
By the Law of Cosines on triangle $ABC,$
\[\cos C = \frac{16y^2 + 16y^2 - 4x^2}{2 \cdot 4y \cdot 4y} = \frac{32y^2 - 4x^2}{32y^2} = \frac{8y^2 - x^2}{8y^2}.\]Then by the Law of Cosines on triangle $ACD,$
\begin{align*}
AD^2 &= 16y^2 + 9y^2 - 2 \cdot 4y \cdot 3y \cdot \cos C \\
&= 25y^2 - 24y^2 \cdot \frac{8y^2 - x^2}{8y^2} \\
&= 3x^2 + y^2.
\end{align*}Applying Stewart's Theorem to median $\overline{BE}$ in triangle $ABD,$ we get
\[BE^2 + AE \cdot DE = \frac{AB^2 + BD^2}{2}.\]Thus,
\[9 + \frac{3x^2 + y^2}{4} = \frac{4x^2 + y^2}{2}.\]This simplifies to $5x^2 + y^2 = 36.$
Applying Stewart's Theorem to median $\overline{CE}$ in triangle $ACD,$ we get
\[CE^2 + AE \cdot DE = \frac{AC^2 + CD^2}{2}.\]Thus,
\[7 + \frac{3x^2 + y^2}{4} = \frac{16y^2 + 9y^2}{2}.\]This simplifies to $3x^2 + 28 = 49y^2.$
Solving the system $5x^2 + y^2 = 36$ and $3x^2 + 28 = 49y^2,$ we find $x^2 = 7$ and $y^2 = 1,$ so $x = \sqrt{7}$ and $y = 1.$
Plugging this back in for our equation for $\cos C$ gives us $\frac{1}{8}$, so $\sin C = \frac{3\sqrt{7}}{8}.$ We can apply the alternative area of a triangle formula, where $AC \cdot BC \cdot \sin C \cdot \frac{1}{2} = 3\sqrt{7}.$ Therefore, our answer is $\boxed{010}$. | // Block 1
unitsize(1.5 cm);
pair A, B, C, D, E;
A = (-sqrt(7),0);
B = (sqrt(7),0);
C = (0,3);
D = interp(B,C,1/4);
E = (A + D)/2;
draw(A--B--C--cycle);
draw(A--D);
draw(B--E--C);
label("$A$", A, SW);
label("$B$", B, SE);
label("$C$", C, N);
label("$D$", D, NE);
label("$E$", E, NW);
label("$2x$", (A + B)/2, S);
label("$y$", (B + D)/2, NE);
label("$3y$", (C + D)/2, NE);
label("$4y$", (A + C)/2, NW);
label("$3$", (B + E)/2, N);
label("$\sqrt{7}$", (C + E)/2, W);
// Block 2
unitsize(1.5 cm); pair A, B, C, D, E; A = (-sqrt(7),0); B = (sqrt(7),0); C = (0,3); D = interp(B,C,1/4); E = (A + D)/2; draw(A--B--C--cycle); draw(A--D); draw(B--E--C); label("$A$", A, SW); label("$B$", B, SE); label("$C$", C, N); label("$D$", D, NE); label("$E$", E, NW); label("$2x$", (A + B)/2, S); label("$y$", (B + D)/2, NE); label("$3y$", (C + D)/2, NE); label("$4y$", (A + C)/2, NW); label("$3$", (B + E)/2, N); label("$\sqrt{7}$", (C + E)/2, W); | [] |
177 | Let $w$ and $z$ be complex numbers such that $|w| = 1$ and $|z| = 10$. Let $\theta = \arg \left(\tfrac{w-z}{z}\right)$. The maximum possible value of $\tan^2 \theta$ can be written as $\tfrac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p+q$. (Note that $\arg(w)$, for $w \neq 0$, denotes the measure of the angle that the ray from $0$ to $w$ makes with the positive real axis in the complex plane) | 2014 AIME I Problem 7 | Without the loss of generality one can let $z$ lie on the positive x axis and since $arg(\theta)$ is a measure of the angle if $z=10$ then $arg(\dfrac{w-z}{z})=arg(w-z)$ and we can see that the question is equivalent to having a triangle $OAB$ with sides $OA =10$ $AB=1$ and $OB=t$ and trying to maximize the angle $BOA$
using the Law of Cosines we get:
$1^2=10^2+t^2-t*10*2\cos\theta$
rearranging:
\[20t\cos\theta=t^2+99\]
solving for $\cos\theta$ we get:
\[\frac{99}{20t}+\frac{t}{20}=\cos\theta\]
if we want to maximize $\theta$ we need to minimize $\cos\theta$
, using AM-GM inequality we get that the minimum value for $\cos\theta= 2\left(\sqrt{\dfrac{99}{20t}\dfrac{t}{20}}\right)=2\sqrt{\dfrac{99}{400}}=\dfrac{\sqrt{99}}{10}$
hence using the identity $\tan^2\theta=\sec^2\theta-1$
we get $\tan^2\theta=\frac{1}{99}$and our answer is $1 + 99 = \boxed{100}$.
Note : You can also realize that the max $\theta$ is when the line from $0$ is tangent to the circle of radius $1$ centered at $10.$ | // Block 1
pair O = (0,0);
pair A = (100,0);
pair B = (80,30);
pair D = (sqrt(850),sqrt(850));
draw(A--B--O--cycle);
dotfactor = 3;
dot("$A$",A,dir(45));
dot("$B$",B,dir(45));
dot("$O$",O,dir(135));
dot("$ \theta$",O,(7,1.2));
label("$1$", ( A--B ));
label("$10$",(O--A));
label("$t$",(O--B));
// Block 2
pair O = (0,0); pair A = (100,0); pair B = (80,30); pair D = (sqrt(850),sqrt(850)); draw(A--B--O--cycle); dotfactor = 3; dot("$A$",A,dir(45)); dot("$B$",B,dir(45)); dot("$O$",O,dir(135)); dot("$ \theta$",O,(7,1.2)); label("$1$", ( A--B )); label("$10$",(O--A)); label("$t$",(O--B)); | [] |
178 | A disk with radius $1$ is externally tangent to a disk with radius $5$. Let $A$ be the point where the disks are tangent, $C$ be the center of the smaller disk, and $E$ be the center of the larger disk. While the larger disk remains fixed, the smaller disk is allowed to roll along the outside of the larger disk until the smaller disk has turned through an angle of $360^\circ$. That is, if the center of the smaller disk has moved to the point $D$, and the point on the smaller disk that began at $A$ has now moved to point $B$, then $\overline{AC}$ is parallel to $\overline{BD}$. Then $\sin^2(\angle BEA)=\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. | 2014 AIME I Problem 10 | Let $F$ be the new tangency point of the two disks. The smaller disk rolled along minor arc $\overset{\frown}{AF}$ on the larger disk.
Let $\alpha = \angle AEF$, in radians. The smaller disk must then have rolled along an arc of length $5\alpha$, since the larger disk has a radius of $5$. Since all of the points on major arc $\overset{\frown}{BF}$ on the smaller disk have come into contact with the larger disk at some point during the rolling, and none of the other points on the smaller disk did,\[\overset{\frown}{BF}=\overset{\frown}{AF}=5\alpha\].
Since $\overline{AC} || \overline{BD}$, \[\angle BDF \cong \angle FEA\] so the angles of minor arc $\overset{\frown}{BF}$ and minor arc $\overset{\frown}{AF}$ are equal, so minor arc $\overset{\frown}{BF}$ has an angle of $\alpha$. Since the smaller disk has a radius of $1$, the length of minor arc $\overset{\frown}{BF}$ is $\alpha$. This means that $5\alpha + \alpha$ equals the circumference of the smaller disk, so $6\alpha = 2\pi$, or $\alpha = \frac{\pi}{3}$.
Now, to find $\sin^2{\angle BEA}$, we construct $\triangle BDE$. Also, drop a perpendicular from $D$ to $\overline{EA}$, and call this point $X$. Since $\alpha = \frac{\pi}{3}$ and $\angle DXE$ is right, \[DE = 6\] \[EX = 3\] and \[DX = 3\sqrt{3}\]
Now drop a perpendicular from $B$ to $\overline{EA}$, and call this point $Y$. Since $\overline{BD} || \overline{EA}$, \[XY = BD = 1\] and \[BY = DX = 3\sqrt{3}\] Thus, we know that \[EY = EX - XY = 3 - 1 = 2\] and by using the Pythagorean Theorem on $\triangle BEY$, we get that \[BE = \sqrt{31}\] Thus, \[\sin{\angle BEA} = \frac{\sqrt{27}}{\sqrt{31}}\] so \[\sin^2{\angle BEA} = \frac{27}{31}\] and our answer is $27 + 31 = \boxed{058}$. | // Block 1
size(150);
pair a=(5,0),b=(2,3*sqrt(3)),c=(6,0),d=(3,3*sqrt(3)),e=(0,0);
draw(circle(e,5));
draw(circle(c,1));
draw(circle(d,1));
dot(a^^b^^c^^d^^e^^(5/2,5*sqrt(3)/2));
label("$A$",a,W,fontsize(9));
label("$B$",b,NW,fontsize(9));
label("$C$",c,E,fontsize(9));
label("$D$",d,E,fontsize(9));
label("$E$",e,SW,fontsize(9));
label("$F$",(5/2,5*sqrt(3)/2),SSW,fontsize(9));
// Block 2
size(150); pair a=(5,0),b=(2,3*sqrt(3)),c=(6,0),d=(3,3*sqrt(3)),e=(0,0); draw(circle(e,5)); draw(circle(c,1)); draw(circle(d,1)); dot(a^^b^^c^^d^^e^^(5/2,5*sqrt(3)/2)); label("$A$",a,W,fontsize(9)); label("$B$",b,NW,fontsize(9)); label("$C$",c,E,fontsize(9)); label("$D$",d,E,fontsize(9)); label("$E$",e,SW,fontsize(9)); label("$F$",(5/2,5*sqrt(3)/2),SSW,fontsize(9)); | [] |
179 | On square $ABCD$, points $E,F,G$, and $H$ lie on sides $\overline{AB},\overline{BC},\overline{CD},$ and $\overline{DA},$ respectively, so that $\overline{EG} \perp \overline{FH}$ and $EG=FH = 34$. Segments $\overline{EG}$ and $\overline{FH}$ intersect at a point $P$, and the areas of the quadrilaterals $AEPH, BFPE, CGPF,$ and $DHPG$ are in the ratio $269:275:405:411.$ Find the area of square $ABCD$. | 2014 AIME I Problem 13 | Let $s$ be the side length of $ABCD$, let $Q$, and $R$ be the midpoints of $\overline{EG}$ and $\overline{FH}$, respectively, let $S$ be the foot of the perpendicular from $Q$ to $\overline{CD}$, let $T$ be the foot of the perpendicular from $R$ to $\overline{AD}$.
The fraction of the area of the square $ABCD$ which is occupied by trapezoid $BCGE$ is \[\frac{275+405}{269+275+405+411}=\frac 12,\]so $Q$ is the center of $ABCD$. Thus $R$, $Q$, $S$ are collinear, and $RT=QS=\tfrac 12 s$. Similarly, the fraction of the area occupied by trapezoid $CDHF$ is $\tfrac 35$, so $RS=\tfrac 35s$ and $RQ=\tfrac{1}{10}s$.
Because $\triangle QSG \cong \triangle RTH$, the area of $DHPG$ is the sum \[[DHPG]=[DTRS]+[RPQ].\] Rectangle $DTRS$ has area $RS\cdot RT = \tfrac 35s\cdot \tfrac 12 s = \tfrac{3}{10}s^2$. If $\angle QRP = \theta$ , then $\triangle RPQ$ has area \[[RPQ]= \tfrac 12 \cdot \tfrac 1{10}s\sin\theta \cdot \tfrac 1{10}s\cos\theta = \tfrac 1{400}s^2\sin 2\theta.\]Therefore the area of $[DHPG]$ is $s^2(\tfrac 3{10}+\tfrac 1{400}\sin 2\theta)$. Because the area of trapezoid $CDHF$ is $\tfrac 35 s^2$, the area of $CGPF$ is $s^2(\tfrac 3{10}-\tfrac 1{400}\sin 2\theta)$.
Because these areas are in the ratio $411:405=(408+3):(408-3)$, it follows that \[\frac{\frac 1{400}\sin 2\theta}{\frac 3{10}}=\frac 3{408},\]from which we get $\sin 2\theta = \tfrac {15}{17}$. Note that $\theta =\angle RHT > \angle QAT = 45^\circ$, so $\cos 2\theta = -\sqrt{1-\sin^2 2\theta}= -\tfrac 8{17}$ and $\sin^2\theta = \tfrac{1}{2}(1-\cos 2\theta) = \tfrac{25}{34}$. Then \[[ABCD]=s^2 = EG^2\sin^2\theta = 34^2 \cdot \tfrac {25}{34} = 850.\] | size(150); defaultpen(fontsize(10pt)); pair A,B,C,D,E,F,Fp,G,Gp,H,O,I,J,R,S,T; A=dir(45*3); B=dir(-45*3); C=dir(-45); D=dir(45); O = origin; real theta=15; E=extension(A,B,O,dir(180+theta)); G=extension(C,D,O,dir(theta)); I=extension(A,D,O,dir(90+theta)); J=extension(B,C,O,dir(-90+theta)); H=(A+I)/2; F=H+(J-I); R=midpoint(H--F); S=midpoint(C--D); T=(R.x, A.y); draw(A--B--C--D--cycle^^E--G^^F--H, black+0.8); draw(S--R--T, gray+0.4); dotfactor = 3; dot("$A$",A,dir(135)); dot("$B$",B,dir(215)); dot("$C$",C,dir(305)); dot("$D$",D,dir(45)); dot("$H$",H,dir(90)); dot("$F$",F,dir(270)); dot("$E$",E,dir(180)); dot("$G$",G,dir(0)); dot("$Q$",O,dir(-90)); dot("$R$",R,dir(-180)); dot("$S$",S,dir(0)); dot("$T$",T,dir(90)); pair P = extension(F,H,E,G); dot("$P$",P,dir(180+60)); | [] |
179 | On square $ABCD$, points $E,F,G$, and $H$ lie on sides $\overline{AB},\overline{BC},\overline{CD},$ and $\overline{DA},$ respectively, so that $\overline{EG} \perp \overline{FH}$ and $EG=FH = 34$. Segments $\overline{EG}$ and $\overline{FH}$ intersect at a point $P$, and the areas of the quadrilaterals $AEPH, BFPE, CGPF,$ and $DHPG$ are in the ratio $269:275:405:411.$ Find the area of square $ABCD$. | 2014 AIME I Problem 13 | Let $s$ be the side length of $ABCD$, let $[ABCD]=1360a$. Let $Q$ and $R$ be the midpoints of $\overline{EG}$ and $\overline{FH}$, respectively; because $269+411=275+405$, $Q$ is also the center of the square. Draw $\overline{IJ} \parallel \overline{HF}$ through $Q$, with $I$ on $\overline{AD}$, $J$ on $\overline{BC}$.
Segments $\overline{EG}$ and $\overline{IJ}$ divide the square into four congruent quadrilaterals, each of area $\tfrac 14 [ABCD]=340a$. Then \[[HFJI]=[ABJI]-[ABFH]=136a.\] The fraction of the total area occupied by parallelogram $HFJI$ is $\tfrac 1{10}$, so $RQ=\tfrac{s}{10}$.
Because $[HFJI]= HF\cdot PQ$, with $HF=34$, we get $PQ=4a$.
Now \[[PQR]=[HPQI]-[HRQI]= ([AEQI]-[AEPH])-\tfrac 12[IJFH] = 3a,\] and because $[PQR]=\tfrac 12 \cdot PQ\cdot PR$, with $PQ=4a$, we get $PR=\tfrac 32$.
By Pythagoras' Theorem on $\triangle PQR$, we get \[16a^2+\frac 94 =\tfrac{68}{5}a,\quad \text{i.e.}\quad 320a^2-272a+45=0,\] with roots $a=\tfrac 9{40}$ or $a=\tfrac58$. The former leads to a square with diagonal less than $34$, which can't be, since $EG=FH=34$; therefore $a=\tfrac 58$ and $[ABCD]=850$. | // Block 1
size(150);
defaultpen(fontsize(9pt));
pair A,B,C,D,E,F,G,H,I,J,L,P,Q,R,S;
Q=MP("Q",origin,down); A=MP("A",(-1,1),dir(135)); B=MP("B",(-1,-1),dir(225)); C=MP("C",(1,-1),dir(-45)); D=MP("D",(1,1),dir(45)); real theta = 20; real shift=0.4; E=MP("E",extension(A,B,Q,dir(theta)),left); J=MP("J",extension(B,C,Q,dir(90+theta)),down); F=MP("F",J+(shift*left),down); G=MP("G",extension(C,D,Q,dir(theta)),right); I=MP("I",extension(A,D,Q,dir(90+theta)),up); H=MP("H",I+(shift*left),up); P=MP("P",extension(E,G,F,H),2*dir(-110)); R=MP("R",extension(F,H,Q,left),left);
draw(A--B--C--D--cycle^^E--G^^F--H, black+1); draw(R--Q^^I--J, gray);
// Block 2
size(150); defaultpen(fontsize(9pt)); pair A,B,C,D,E,F,G,H,I,J,L,P,Q,R,S; Q=MP("Q",origin,down); A=MP("A",(-1,1),dir(135)); B=MP("B",(-1,-1),dir(225)); C=MP("C",(1,-1),dir(-45)); D=MP("D",(1,1),dir(45)); real theta = 20; real shift=0.4; E=MP("E",extension(A,B,Q,dir(theta)),left); J=MP("J",extension(B,C,Q,dir(90+theta)),down); F=MP("F",J+(shift*left),down); G=MP("G",extension(C,D,Q,dir(theta)),right); I=MP("I",extension(A,D,Q,dir(90+theta)),up); H=MP("H",I+(shift*left),up); P=MP("P",extension(E,G,F,H),2*dir(-110)); R=MP("R",extension(F,H,Q,left),left); draw(A--B--C--D--cycle^^E--G^^F--H, black+1); draw(R--Q^^I--J, gray); | [] |
179 | On square $ABCD$, points $E,F,G$, and $H$ lie on sides $\overline{AB},\overline{BC},\overline{CD},$ and $\overline{DA},$ respectively, so that $\overline{EG} \perp \overline{FH}$ and $EG=FH = 34$. Segments $\overline{EG}$ and $\overline{FH}$ intersect at a point $P$, and the areas of the quadrilaterals $AEPH, BFPE, CGPF,$ and $DHPG$ are in the ratio $269:275:405:411.$ Find the area of square $ABCD$. | 2014 AIME I Problem 13 | saw this at bottom of https://artofproblemsolving.com/community/c5h580667p3428922
Let $O$ be the center of the square. Then let $Q$ be on $HF$ susch that $OQ \perp CD$. Draw $H'F' \parallel HF$ through $O$. Let $s$ be the side length of the square, so that $s^2$ is the desired.
First, we note that as $w + z = x + y$, the area of $[ADGC]$ is half the area of $[ABCD]$, which means $O \in EG$. Similarly, $[CDHF] = y + z = \frac{3}{5}[ABCD]$ because of the ratios. But because $O \in H'F'$, we have $[CDH'F'] = \frac{1}{2}[ABCD]$. Subtracting yields $[H'HFF'] = \frac{1}{10}[ABCD]$. But as it is a parallelogram with base $34$ and height $OP$, we get:
\[340(OP) = s^2\]
Now consider quadrilateral $OEBF'$, which has area $1/4$ of the square. We can write:
\[\frac{1}{4}s^2 = [OEBF'] = [PEBF] + [FPOF'] = x + [FQOF'] - [QOP]\]However, we know the areas of $x$ and parallelogram $[FQOF'] = 17OP = \frac{s^2}{20}$ by our earlier result. Substituting these yields:
\[[QOP] = \frac{3}{1360}s^2\]
But as we know $OP = \frac{s^2}{340}$, we get $QP = 3/2$.
Because $CDHF$ is a trapezoid of height $s$ and area $\frac{3}{5}s^2$, we have $\frac{1}{2}s(CF + DH) = \frac{3}{5}s^2$. This means:
\[CF + DH = \frac{6}{5}s\]But using the same method on trapezoid $CDH'F'$ of area $\frac{1}{2}s^2$, we get
\[CF' + DH' = s\]And subtracting yields $HH' + FF' = 2OQ = \frac{1}{5}s$, which gives $OQ = \frac{s}{10}$. At this point, applying Pythagoras on $\triangle QOP$ yields:
\[\left( \frac{s^2}{340} \right)^2 + \left( \frac{3}{2} \right)^2 = \left( \frac{s}{10} \right)^2\]and the resulting quadratic is easy to solve for $s^2 = 306, 850$. But as $[ABCD] > [EFGH] = 34^2/2 > 306$, we have $\boxed{850}$ as the solution. | // Block 1
size(200);
defaultpen(linewidth(0.8)+fontsize(10.6));
pair A = (0,sqrt(850));
pair B = (0,0);
pair C = (sqrt(850),0);
pair D = (sqrt(850),sqrt(850));
draw(A--B--C--D--cycle);
dotfactor = 3;
dot("$A$",A,dir(135));
dot("$B$",B,dir(215));
dot("$C$",C,dir(305));
dot("$D$",D,dir(45));
pair H = ((2sqrt(850)-sqrt(120))/6,sqrt(850));
pair F = ((2sqrt(850)+sqrt(306)+7)/6,0);
dot("$H$",H,dir(90));
dot("$F$",F,dir(270));
draw(H--F);
pair E = (0,(sqrt(850)-6)/2 - 1);
pair G = (sqrt(850),(sqrt(850)+sqrt(100))/2 - 1);
pair O = (B + D)/2;
pair M = (D + C)/2;
pair Q = extension(M, O, H, F);
pair P = extension(H,F,E,G);
pair Hp = H + O- Q;
pair Fp = F + O - Q;
dot("$H'$ ", Hp, dir(90));
dot("$F'$ ", Fp, dir(270));
draw(Hp--Fp, dotted + defaultpen);
dot("$O$", O, dir(300));
dot("$E$",E,dir(180));
dot("$G$",G,dir(0));
dot("Q", Q, dir(60));
draw(E--G);
dot("$P$",P,dir(240));
label("$w$", (H+E)/2,fontsize(15));
label("$x$", (E+F)/2,fontsize(15));
label("$y$", (G+F)/2,fontsize(15));
label("$z$", (H+G)/2,fontsize(15));
label("$w:x:y:z=269:275:405:411$",(sqrt(850)/2,-4.5),fontsize(11));
draw(O--Q);
// Block 2
size(200); defaultpen(linewidth(0.8)+fontsize(10.6)); pair A = (0,sqrt(850)); pair B = (0,0); pair C = (sqrt(850),0); pair D = (sqrt(850),sqrt(850)); draw(A--B--C--D--cycle); dotfactor = 3; dot("$A$",A,dir(135)); dot("$B$",B,dir(215)); dot("$C$",C,dir(305)); dot("$D$",D,dir(45)); pair H = ((2sqrt(850)-sqrt(120))/6,sqrt(850)); pair F = ((2sqrt(850)+sqrt(306)+7)/6,0); dot("$H$",H,dir(90)); dot("$F$",F,dir(270)); draw(H--F); pair E = (0,(sqrt(850)-6)/2 - 1); pair G = (sqrt(850),(sqrt(850)+sqrt(100))/2 - 1); pair O = (B + D)/2; pair M = (D + C)/2; pair Q = extension(M, O, H, F); pair P = extension(H,F,E,G); pair Hp = H + O- Q; pair Fp = F + O - Q; dot("$H'$ ", Hp, dir(90)); dot("$F'$ ", Fp, dir(270)); draw(Hp--Fp, dotted + defaultpen); dot("$O$", O, dir(300)); dot("$E$",E,dir(180)); dot("$G$",G,dir(0)); dot("Q", Q, dir(60)); draw(E--G); dot("$P$",P,dir(240)); label("$w$", (H+E)/2,fontsize(15)); label("$x$", (E+F)/2,fontsize(15)); label("$y$", (G+F)/2,fontsize(15)); label("$z$", (H+G)/2,fontsize(15)); label("$w:x:y:z=269:275:405:411$",(sqrt(850)/2,-4.5),fontsize(11)); draw(O--Q); | [] |
180 | In $\triangle ABC$, $AB = 3$, $BC = 4$, and $CA = 5$. Circle $\omega$ intersects $\overline{AB}$ at $E$ and $B$, $\overline{BC}$ at $B$ and $D$, and $\overline{AC}$ at $F$ and $G$. Given that $EF=DF$ and $\frac{DG}{EG} = \frac{3}{4}$, length $DE=\frac{a\sqrt{b}}{c}$, where $a$ and $c$ are relatively prime positive integers, and $b$ is a positive integer not divisible by the square of any prime. Find $a+b+c$. | 2014 AIME I Problem 15 | First we note that $\triangle DEF$ is an isosceles right triangle with hypotenuse $\overline{DE}$ the same as the diameter of $\omega$. We also note that $\triangle DGE \sim \triangle ABC$ since $\angle EGD$ is a right angle and the ratios of the sides are $3:4:5$.
From congruent arc intersections, we know that $\angle GED \cong \angle GBC$, and that from similar triangles $\angle GED$ is also congruent to $\angle GCB$. Thus, $\triangle BGC$ is an isosceles triangle with $BG = GC$, so $G$ is the midpoint of $\overline{AC}$ and $AG = GC = 5/2$. Similarly, we can find from angle chasing that $\angle ABF = \angle EDF = \frac{\pi}4$. Therefore, $\overline{BF}$ is the angle bisector of $\angle B$. From the angle bisector theorem, we have $\frac{AF}{AB} = \frac{CF}{CB}$, so $AF = 15/7$ and $CF = 20/7$.
Lastly, we apply power of a point from points $A$ and $C$ with respect to $\omega$ and have $AE \times AB=AF \times AG$ and $CD \times CB=CG \times CF$, so we can compute that $EB = \frac{17}{14}$ and $DB = \frac{31}{14}$. From the Pythagorean Theorem, we result in $DE = \frac{25 \sqrt{2}}{14}$, so $a+b+c=25+2+14= \boxed{041}$
Also: $FG=\frac{20}{7}-\frac{5}{2}=\frac{5}{2}-\frac{15}{7}=\frac{5}{14}$. We can also use Ptolemy's Theorem on quadrilateral $DEFG$ to figure what $FG$ is in terms of $d$:
\[DE\cdot FG+DG\cdot EF=DF\cdot EG\]
\[d\cdot FG+\frac{3d}{5}\cdot \frac{d}{\sqrt{2}}=\frac{4d}{5}\cdot \frac{d}{\sqrt{2}}\]
\[d\cdot FG+\frac{3d^2}{5\sqrt{2}}=\frac{4d^2}{5\sqrt{2}}\implies FG=\frac{d}{5\sqrt{2}}\]
Thus $\frac{d}{5\sqrt{2}}=\frac{5}{14}\rightarrow d=5\sqrt{2}\cdot\frac{5}{14}=\frac{25\sqrt{2}}{14}$. $a+b+c=25+2+14= \boxed{041}$
Solution 3
Call $DE=x$ and as a result $DF=EF=\frac{x\sqrt{2}}{2}, EG=\frac{4x}{5}, GD=\frac{3x}{5}$. Since $EFGD$ is cyclic we just need to get $DG$ and using LoS(for more detail see the $2$nd paragraph of Solution $2$) we get $AG=\frac{5}{2}$ and using a similar argument(use LoS again) and subtracting you get $FG=\frac{5}{14}$ so you can use Ptolemy to get $x=\frac{25\sqrt{2}}{14} \implies \boxed{041}$.
~First | // Block 1
pair A = (0,3);
pair B = (0,0);
pair C = (4,0);
draw(A--B--C--cycle);
dotfactor = 3;
dot("$A$",A,dir(135));
dot("$B$",B,dir(215));
dot("$C$",C,dir(305));
pair D = (2.21, 0);
pair E = (0, 1.21);
pair F = (1.71, 1.71);
pair G = (2, 1.5);
dot("$D$",D,dir(270));
dot("$E$",E,dir(180));
dot("$F$",F,dir(90));
dot("$G$",G,dir(0));
draw(Circle((1.109, 0.609), 1.28));
draw(D--E);
draw(E--F);
draw(D--F);
draw(E--G);
draw(D--G);
draw(B--F);
draw(B--G);
// Block 2
pair A = (0,3); pair B = (0,0); pair C = (4,0); draw(A--B--C--cycle); dotfactor = 3; dot("$A$",A,dir(135)); dot("$B$",B,dir(215)); dot("$C$",C,dir(305)); pair D = (2.21, 0); pair E = (0, 1.21); pair F = (1.71, 1.71); pair G = (2, 1.5); dot("$D$",D,dir(270)); dot("$E$",E,dir(180)); dot("$F$",F,dir(90)); dot("$G$",G,dir(0)); draw(Circle((1.109, 0.609), 1.28)); draw(D--E); draw(E--F); draw(D--F); draw(E--G); draw(D--G); draw(B--F); draw(B--G); | [] |
181 | Arnold is studying the prevalence of three health risk factors, denoted by A, B, and C, within a population of men. For each of the three factors, the probability that a randomly selected man in the population has only this risk factor (and none of the others) is 0.1. For any two of the three factors, the probability that a randomly selected man has exactly these two risk factors (but not the third) is 0.14. The probability that a randomly selected man has all three risk factors, given that he has A and B is $\frac{1}{3}$. The probability that a man has none of the three risk factors given that he does not have risk factor A is $\frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p+q$. | 2014 AIME II Problem 2 | We first assume a population of $100$ to facilitate solving. Then we simply organize the statistics given into a Venn diagram.
Let $x$ be the number of men with all three risk factors. Since "the probability that a randomly selected man has all three risk factors, given that he has A and B is $\frac{1}{3}$," we can tell that $x = \frac{1}{3}(x+14)$, since there are $x$ people with all three factors and 14 with only A and B. Thus $x=7$.
Let $y$ be the number of men with no risk factors. It now follows that
\[y= 100 - 3 \cdot 10 - 3 \cdot 14 - 7 = 21.\]
The number of men with risk factor A is $10+2 \cdot 14+7 = 45$ (10 with only A, 28 with A and one of the others, and 7 with all three). Thus the number of men without risk factor $A$ is 55, so the desired conditional probability is $21/55$. So the answer is $21+55=\boxed{076}$. | // Block 1
pair A,B,C,D,E,F,G;
A=(0,55);
B=(60,55);
C=(60,0);
D=(0,0);
draw(A--B--C--D--A);
E=(30,35);
F=(20,20);
G=(40,20);
draw(circle(E,15));
draw(circle(F,15));
draw(circle(G,15));
draw("$A$",(30,52));
draw("$B$",(7,7));
draw("$C$",(53,7));
draw("100",(5,60));
draw("10",(30,40));
draw("10",(15,15));
draw("10",(45,15));
draw("14",(30,16));
draw("14",(38,29));
draw("14",(22,29));
draw("$x$",(30,25));
draw("$y$",(10,45));
// Block 2
pair A,B,C,D,E,F,G; A=(0,55); B=(60,55); C=(60,0); D=(0,0); draw(A--B--C--D--A); E=(30,35); F=(20,20); G=(40,20); draw(circle(E,15)); draw(circle(F,15)); draw(circle(G,15)); draw("$A$",(30,52)); draw("$B$",(7,7)); draw("$C$",(53,7)); draw("100",(5,60)); draw("10",(30,40)); draw("10",(15,15)); draw("10",(45,15)); draw("14",(30,16)); draw("14",(38,29)); draw("14",(22,29)); draw("$x$",(30,25)); draw("$y$",(10,45)); | [] |
182 | A rectangle has sides of length $a$ and 36. A hinge is installed at each vertex of the rectangle, and at the midpoint of each side of length 36. The sides of length $a$ can be pressed toward each other keeping those two sides parallel so the rectangle becomes a convex hexagon as shown. When the figure is a hexagon with the sides of length $a$ parallel and separated by a distance of 24, the hexagon has the same area as the original rectangle. Find $a^2$. | 2014 AIME II Problem 3 | When we squish the rectangle, the hexagon is composed of a rectangle and two isosceles triangles with side lengths 18, 18, and 24 as shown below.
By Heron's Formula, the area of each isosceles triangle is $\sqrt{(30)(12)(12)(6)}=\sqrt{180\times 12^2}=72\sqrt{5}$. So the area of both is $144\sqrt{5}$. From the rectangle, our original area is $36a$. The area of the rectangle in the hexagon is $24a$. So we have \[24a+144\sqrt{5}=36a\implies 12a=144\sqrt{5}\implies a=12\sqrt{5}\implies a^2=\boxed{720}.\] | // Block 1
pair R,S,T,X,Y,Z;
dotfactor = 2;
unitsize(.1cm);
R = (12*2.236 +22,0);
S = (12*2.236 + 22 - 13.4164,12);
T = (12*2.236 + 22,24);
X = (12*4.472+ 22,24);
Y = (12*4.472+ 22 + 13.4164,12);
Z = (12*4.472+ 22,0);
draw(R--S--T--X--Y--Z--cycle);
draw(T--R,red);
draw(X--Z,red);
dot(" ",R,NW);
dot(" ",S,NW);
dot(" ",T,NW);
dot(" ",X,NW);
dot(" ",Y,NW);
dot(" ",Z,NW);
// sqrt180 = 13.4164
// sqrt5 = 2.236
// Block 2
pair R,S,T,X,Y,Z; dotfactor = 2; unitsize(.1cm); R = (12*2.236 +22,0); S = (12*2.236 + 22 - 13.4164,12); T = (12*2.236 + 22,24); X = (12*4.472+ 22,24); Y = (12*4.472+ 22 + 13.4164,12); Z = (12*4.472+ 22,0); draw(R--S--T--X--Y--Z--cycle); draw(T--R,red); draw(X--Z,red); dot(" ",R,NW); dot(" ",S,NW); dot(" ",T,NW); dot(" ",X,NW); dot(" ",Y,NW); dot(" ",Z,NW); // sqrt180 = 13.4164 // sqrt5 = 2.236 | [] |
183 | Circle $C$ with radius 2 has diameter $\overline{AB}$. Circle D is internally tangent to circle $C$ at $A$. Circle $E$ is internally tangent to circle $C$, externally tangent to circle $D$, and tangent to $\overline{AB}$. The radius of circle $D$ is three times the radius of circle $E$, and can be written in the form $\sqrt{m}-n$, where $m$ and $n$ are positive integers. Find $m+n$. | 2014 AIME II Problem 8 | Using the diagram above, let the radius of $D$ be $3r$, and the radius of $E$ be $r$. Then, $EF=r$, and $CE=2-r$, so the Pythagorean theorem in $\triangle CEF$ gives $CF=\sqrt{4-4r}$. Also, $CD=CA-AD=2-3r$, so \[DF=DC+CF=2-3r+\sqrt{4-4r}.\] Noting that $DE=4r$, we can now use the Pythagorean theorem in $\triangle DEF$ to get
\[(2-3r+\sqrt{4-4r})^2+r^2=16r^2.\]
Solving this quadratic is somewhat tedious, but the constant terms cancel, so the computation isn't terrible. Solving gives $3r=\sqrt{240}-14$ for a final answer of $\boxed{254}$.
Notice that C, E and the point of tangency to circle C for circle E will be collinear because C and E intersect the tangent line at a right angle, implying they must be on the same line. | // Block 1
import graph; size(7.99cm);
real labelscalefactor = 0.5;
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps);
pen dotstyle = black;
real xmin = 4.087153740193288, xmax = 11.08175859031552, ymin = -4.938019122704778, ymax = 1.194137062512079;
draw(circle((7.780000000000009,-1.320000000000002), 2.000000000000000));
draw(circle((7.271934046987930,-1.319179731427737), 1.491933384829670));
draw(circle((9.198812158392690,-0.8249788848962227), 0.4973111282761416));
draw((5.780002606580324,-1.316771019595571)--(9.779997393419690,-1.323228980404432));
draw((9.198812158392690,-0.8249788848962227)--(9.198009254448635,-1.322289365031666));
draw((7.271934046987930,-1.319179731427737)--(9.198812158392690,-0.8249788848962227));
draw((9.198812158392690,-0.8249788848962227)--(7.780000000000009,-1.320000000000002));
dot((7.780000000000009,-1.320000000000002),dotstyle);
label("$C$", (7.707377218471464,-1.576266740352400), NE * labelscalefactor);
dot((7.271934046987930,-1.319179731427737),dotstyle);
label("$D$", (7.303064016111554,-1.276266740352400), NE * labelscalefactor);
dot((9.198812158392690,-0.8249788848962227),dotstyle);
label("$E$", (9.225301294671791,-0.7792624249832147), NE * labelscalefactor);
dot((9.198009254448635,-1.322289365031666),dotstyle);
label("$F$", (9.225301294671791,-1.276266740352400), NE * labelscalefactor);
dot((9.779997393419690,-1.323228980404432),dotstyle);
label("$B$", (9.810012253929656,-1.276266740352400), NE * labelscalefactor);
dot((5.780002606580324,-1.316771019595571),dotstyle);
label("$A$", (5.812051070003994,-1.276266740352400), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
// Block 2
import graph; size(7.99cm); real labelscalefactor = 0.5; pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); pen dotstyle = black; real xmin = 4.087153740193288, xmax = 11.08175859031552, ymin = -4.938019122704778, ymax = 1.194137062512079; draw(circle((7.780000000000009,-1.320000000000002), 2.000000000000000)); draw(circle((7.271934046987930,-1.319179731427737), 1.491933384829670)); draw(circle((9.198812158392690,-0.8249788848962227), 0.4973111282761416)); draw((5.780002606580324,-1.316771019595571)--(9.779997393419690,-1.323228980404432)); draw((9.198812158392690,-0.8249788848962227)--(9.198009254448635,-1.322289365031666)); draw((7.271934046987930,-1.319179731427737)--(9.198812158392690,-0.8249788848962227)); draw((9.198812158392690,-0.8249788848962227)--(7.780000000000009,-1.320000000000002)); dot((7.780000000000009,-1.320000000000002),dotstyle); label("$C$", (7.707377218471464,-1.576266740352400), NE * labelscalefactor); dot((7.271934046987930,-1.319179731427737),dotstyle); label("$D$", (7.303064016111554,-1.276266740352400), NE * labelscalefactor); dot((9.198812158392690,-0.8249788848962227),dotstyle); label("$E$", (9.225301294671791,-0.7792624249832147), NE * labelscalefactor); dot((9.198009254448635,-1.322289365031666),dotstyle); label("$F$", (9.225301294671791,-1.276266740352400), NE * labelscalefactor); dot((9.779997393419690,-1.323228980404432),dotstyle); label("$B$", (9.810012253929656,-1.276266740352400), NE * labelscalefactor); dot((5.780002606580324,-1.316771019595571),dotstyle); label("$A$", (5.812051070003994,-1.276266740352400), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); | [] |
184 | In $\triangle RED$, $\measuredangle DRE=75^{\circ}$ and $\measuredangle RED=45^{\circ}$. $RD=1$. Let $M$ be the midpoint of segment $\overline{RD}$. Point $C$ lies on side $\overline{ED}$ such that $\overline{RC}\perp\overline{EM}$. Extend segment $\overline{DE}$ through $E$ to point $A$ such that $CA=AR$. Then $AE=\frac{a-\sqrt{b}}{c}$, where $a$ and $c$ are relatively prime positive integers, and $b$ is a positive integer. Find $a+b+c$. | 2014 AIME II Problem 11 | Let $P$ be the foot of the perpendicular from $A$ to $\overline{CR}$, so $\overline{AP}\parallel\overline{EM}$. Since triangle $ARC$ is isosceles, $P$ is the midpoint of $\overline{CR}$, and $\overline{PM}\parallel\overline{CD}$. Thus, $APME$ is a parallelogram and $AE = PM = \frac{CD}{2}$. We can then use coordinates. Let $O$ be the foot of altitude $RO$ and set $O$ as the origin. Now we notice special right triangles! In particular, $DO = \frac{1}{2}$ and $EO = RO = \frac{\sqrt{3}}{2}$, so $D\left(\frac{1}{2}, 0\right)$, $E\left(-\frac{\sqrt{3}}{2}, 0\right)$, and $R\left(0, \frac{\sqrt{3}}{2}\right).$ $M =$ midpoint$(D, R) = \left(\frac{1}{4}, \frac{\sqrt{3}}{4}\right)$ and the slope of $ME = \frac{\frac{\sqrt{3}}{4}}{\frac{1}{4} + \frac{\sqrt{3}}{2}} = \frac{\sqrt{3}}{1 + 2\sqrt{3}}$, so the slope of $RC = -\frac{1 + 2\sqrt{3}}{\sqrt{3}}.$ Instead of finding the equation of the line, we use the definition of slope: for every $CO = x$ to the left, we go $\frac{x(1 + 2\sqrt{3})}{\sqrt{3}} = \frac{\sqrt{3}}{2}$ up. Thus, $x = \frac{\frac{3}{2}}{1 + 2\sqrt{3}} = \frac{3}{4\sqrt{3} + 2} = \frac{3(4\sqrt{3} - 2)}{44} = \frac{6\sqrt{3} - 3}{22}.$ $DC = \frac{1}{2} - x = \frac{1}{2} - \frac{6\sqrt{3} - 3}{22} = \frac{14 - 6\sqrt{3}}{22}$, and $AE = \frac{7 - \sqrt{27}}{22}$, so the answer is $\boxed{056}$. | // Block 1
unitsize(8cm);
pair a, o, d, r, e, m, cm, c,p;
o =(0,0);
d = (0.5, 0);
r = (0,sqrt(3)/2);
e = (-sqrt(3)/2,0);
m = midpoint(d--r);
draw(e--m);
cm = foot(r, e, m);
draw(L(r, cm,1, 1));
c = IP(L(r, cm, 1, 1), e--d);
clip(r--d--e--cycle);
draw(r--d--e--cycle);
draw(rightanglemark(e, cm, c, 1.5));
a = -(4sqrt(3)+9)/11+0.5;
dot(a);
draw(a--r, dashed);
draw(a--c, dashed);
pair[] PPAP = {a, o, d, r, e, m, c};
for(int i = 0; i<7; ++i) {
dot(PPAP[i]);
}
label("$A$", a, W);
label("$E$", e, SW);
label("$C$", c, S);
label("$O$", o, S);
label("$D$", d, SE);
label("$M$", m, NE);
label("$R$", r, N);
p = foot(a, r, c);
label("$P$", p, NE);
draw(p--m, dashed);
draw(a--p, dashed);
dot(p);
// Block 2
unitsize(8cm); pair a, o, d, r, e, m, cm, c,p; o =(0,0); d = (0.5, 0); r = (0,sqrt(3)/2); e = (-sqrt(3)/2,0); m = midpoint(d--r); draw(e--m); cm = foot(r, e, m); draw(L(r, cm,1, 1)); c = IP(L(r, cm, 1, 1), e--d); clip(r--d--e--cycle); draw(r--d--e--cycle); draw(rightanglemark(e, cm, c, 1.5)); a = -(4sqrt(3)+9)/11+0.5; dot(a); draw(a--r, dashed); draw(a--c, dashed); pair[] PPAP = {a, o, d, r, e, m, c}; for(int i = 0; i<7; ++i) { dot(PPAP[i]); } label("$A$", a, W); label("$E$", e, SW); label("$C$", c, S); label("$O$", o, S); label("$D$", d, SE); label("$M$", m, NE); label("$R$", r, N); p = foot(a, r, c); label("$P$", p, NE); draw(p--m, dashed); draw(a--p, dashed); dot(p); | [] |
184 | In $\triangle RED$, $\measuredangle DRE=75^{\circ}$ and $\measuredangle RED=45^{\circ}$. $RD=1$. Let $M$ be the midpoint of segment $\overline{RD}$. Point $C$ lies on side $\overline{ED}$ such that $\overline{RC}\perp\overline{EM}$. Extend segment $\overline{DE}$ through $E$ to point $A$ such that $CA=AR$. Then $AE=\frac{a-\sqrt{b}}{c}$, where $a$ and $c$ are relatively prime positive integers, and $b$ is a positive integer. Find $a+b+c$. | 2014 AIME II Problem 11 | Let $P$ be the foot of the perpendicular from $A$ to $\overline{CR}$, so $\overline{AP}\parallel\overline{EM}$. Since $\triangle ARC$ is isosceles, $P$ is the midpoint of $\overline{CR}$, and by midpoint theorem $\overline{PM}\parallel\overline{CD}$. Thus, $APME$ is a parallelogram and therefore $AE = PM = \tfrac 12 CD$.
We can now use coordinates with $D(0,0)$ as origin and $DE$ along the $x$-axis.
Let $RD=4$ instead of $1$ (in the end we will scale down by $4$). Since $\angle D = 60^\circ$, we get $R(2,2\sqrt{3})$, and therefore $M(1, \sqrt{3})$.
We use sine-law in $\triangle RED$ to find the coordinates $E(2+2\sqrt{3}, 0)$:\[DE =4\cdot \frac{\sin 75^\circ}{\sin 45^\circ} = 4(\sin 30^\circ + \cos 30^\circ) = 2+2\sqrt{3}.\]
Since slope$(ME)= -\sqrt{3}/(1+2\sqrt{3})$, and $RC\perp ME$, it follows that slope$(RC)=(1+2\sqrt{3})/\sqrt{3}$. If $C(c,0)$ then we have\[\frac{2\sqrt{3}}{2-c}=\frac{1+2\sqrt{3}}{\sqrt{3}}\qquad \Longrightarrow\qquad c=\frac{4\sqrt{3}-4}{1+2\sqrt{3}}\]
Now $\tfrac 12 CD = \tfrac 12c =(2\sqrt{3}-2)/(1+2\sqrt{3})= \tfrac 1{11}(14-6\sqrt{3})$.
Scaling down by $4$, we get $AE=\tfrac 1{22}(7-3\sqrt{3})$, so our answer is $7+27+22=056$. | // Block 1
unitsize(8cm); pair a, d, r, e, m, cm, c,p;
d=origin; r=dir(60); e=extension(d,left,r,r+dir(75)*(d-r)); m = midpoint(d--r); cm = foot(r, e, m); c=extension(r,cm,d,e); p=midpoint(r--c); a=p+(e-m);
draw(e--m); draw(L(r, cm,1, 1)); clip(r--d--e--cycle); draw(r--d--e--cycle); draw(rightanglemark(e, cm, c, 1.5)); draw(a--r, dashed); draw(a--c, dashed); draw(p--m, dashed); draw(a--p, dashed);
pair[] PPAP = {a, d, r, e, m, c, p};
for(int i = 0; i<7; ++i) { dot(PPAP[i]); }
label("$A$", a, E); label("$E$", e, S); label("$C$", c, S); label("$D$", d, SW); label("$M$", m, NW); label("$R$", r, N); label("$P$", p, NW);
MA("60^\circ",black,c,d,m,0.07, black);
// Block 2
unitsize(8cm); pair a, d, r, e, m, cm, c,p; d=origin; r=dir(60); e=extension(d,left,r,r+dir(75)*(d-r)); m = midpoint(d--r); cm = foot(r, e, m); c=extension(r,cm,d,e); p=midpoint(r--c); a=p+(e-m); draw(e--m); draw(L(r, cm,1, 1)); clip(r--d--e--cycle); draw(r--d--e--cycle); draw(rightanglemark(e, cm, c, 1.5)); draw(a--r, dashed); draw(a--c, dashed); draw(p--m, dashed); draw(a--p, dashed); pair[] PPAP = {a, d, r, e, m, c, p}; for(int i = 0; i<7; ++i) { dot(PPAP[i]); } label("$A$", a, E); label("$E$", e, S); label("$C$", c, S); label("$D$", d, SW); label("$M$", m, NW); label("$R$", r, N); label("$P$", p, NW); MA("60^\circ",black,c,d,m,0.07, black); | [] |
185 | In $\triangle{ABC}, AB=10, \angle{A}=30^\circ$ , and $\angle{C=45^\circ}$. Let $H, D,$ and $M$ be points on the line $BC$ such that $AH\perp{BC}$, $\angle{BAD}=\angle{CAD}$, and $BM=CM$. Point $N$ is the midpoint of the segment $HM$, and point $P$ is on ray $AD$ such that $PN\perp{BC}$. Then $AP^2=\dfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. | 2014 AIME II Problem 14 | Draw the $45-45-90 \triangle AHC$. Now, take the perpendicular bisector of $BC$ to intersect the circumcircle of $\triangle ABC$ and $AC$ at $F, L, G$ as shown, and denote $O$ to be the circumcenter of $\triangle ABC$. It is not difficult to see by angle chasing that $AHBGO$ is cyclic, namely with diameter $AB$. Then, by symmetry, $EH = HB$ and as $HB, OG$ are both subtended by equal arcs they are equal. Hence, $EH = GO$. Now, draw line $HL$ and intersect it at $AC$ at point $K$ in the diagram. It is not hard to use angle chase to arrive at $AEOL$ a parallelogram, and from our length condition derived earlier, $AL \parallel HG$. From here, it is clear that $AK = KG$; that is, $P$ is just the intersection of the perpendicular from $K$ down to $BC$ and $AD$! After this point, note that $AP = PF$. It is easily derived that the circumradius of $\triangle ABC$ is $\frac{10}{\sqrt{2}}$. Now, $APO$ is a $30-60-90$ triangle, and from here it is easy to arrive at the final answer of $\boxed{077}$. ~awang11's sol | // Block 1
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(15cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -8.455641974276588, xmax = 26.731282460265, ymin = -10.92318356252699, ymax = 9.023689834456471; /* image dimensions */
pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882);
draw((-1.4934334172297545,2.6953043701763835)--(-1.459546107520503,-6.96389444957376)--(1.1286284157632023,-6.954814372303504)--(4.651736947776926,-3.406898607850789)--(4.642656870506668,-0.8187240845670819)--cycle, linewidth(2) + rvwvcq);
/* draw figures */
draw((-1.4934334172297545,2.6953043701763835)--(1.1286284157632023,-6.954814372303504), linewidth(2) + wrwrwr);
draw((xmin, -0.9930079421029264*xmin + 1.2123131258653241)--(xmax, -0.9930079421029264*xmax + 1.2123131258653241), linewidth(2) + wrwrwr); /* line */
draw((xmin, 0.0035082940460819836*xmin-6.958773932654766)--(xmax, 0.0035082940460819836*xmax-6.958773932654766), linewidth(2) + wrwrwr); /* line */
draw((xmin, -285.03882139434313*xmin-422.9911967079192)--(xmax, -285.03882139434313*xmax-422.9911967079192), linewidth(2) + wrwrwr); /* line */
draw((xmin, -1.7181023895538718*xmin + 0.12943284739433739)--(xmax, -1.7181023895538718*xmax + 0.12943284739433739), linewidth(2) + wrwrwr); /* line */
draw(circle((4.642656870506668,-0.8187240845670819), 7.071067811865476), linewidth(2) + wrwrwr);
draw((xmin, -285.0388213943529*xmin + 1322.5187184230485)--(xmax, -285.0388213943529*xmax + 1322.5187184230485), linewidth(2) + wrwrwr); /* line */
draw((-1.4934334172297545,2.6953043701763835)--(4.617849638067675,6.252300211899359), linewidth(2) + wrwrwr);
draw((4.617849638067675,6.252300211899359)--(-1.459546107520503,-6.96389444957376), linewidth(2) + wrwrwr);
draw(circle((-0.18240250073327363,-2.12975500106356), 5), linewidth(2) + wrwrwr);
draw((xmin, -285.0388213943432*xmin + 449.7637608575419)--(xmax, -285.0388213943432*xmax + 449.7637608575419), linewidth(2) + wrwrwr); /* line */
draw((1.1286284157632023,-6.954814372303504)--(4.651736947776926,-3.406898607850789), linewidth(2) + wrwrwr);
draw((-1.4934334172297545,2.6953043701763835)--(4.642656870506668,-0.8187240845670819), linewidth(2) + wrwrwr);
draw((-1.4934334172297545,2.6953043701763835)--(-1.459546107520503,-6.96389444957376), linewidth(2) + rvwvcq);
draw((-1.459546107520503,-6.96389444957376)--(1.1286284157632023,-6.954814372303504), linewidth(2) + rvwvcq);
draw((1.1286284157632023,-6.954814372303504)--(4.651736947776926,-3.406898607850789), linewidth(2) + rvwvcq);
draw((4.651736947776926,-3.406898607850789)--(4.642656870506668,-0.8187240845670819), linewidth(2) + rvwvcq);
draw((4.642656870506668,-0.8187240845670819)--(-1.4934334172297545,2.6953043701763835), linewidth(2) + rvwvcq);
/* dots and labels */
dot((-1.4934334172297545,2.6953043701763835),dotstyle);
label("$A$", (-1.3954084351380491,2.9230996889873015), NE * labelscalefactor);
dot((1.1286284157632023,-6.954814372303504),dotstyle);
label("$B$", (1.2093379191072373,-6.719031552166216), NE * labelscalefactor);
dot((8.199652712229643,-6.930007139864511),linewidth(4pt) + dotstyle);
label("$C$", (8.292420110475998,-6.741880204396438), NE * labelscalefactor);
dot((-1.459546107520503,-6.96389444957376),linewidth(4pt) + dotstyle);
label("$H$", (-1.3725597829078273,-6.787577508856881), NE * labelscalefactor);
dot((-1.4686261847907602,-4.375719926290057),linewidth(4pt) + dotstyle);
label("$E$", (-1.3725597829078273,-4.182831154611618), NE * labelscalefactor);
dot((4.617849638067675,6.252300211899359),linewidth(4pt) + dotstyle);
label("$L$", (4.705181710331174,6.441792132441429), NE * labelscalefactor);
dot((4.642656870506668,-0.8187240845670819),linewidth(4pt) + dotstyle);
label("$O$", (4.728030362561396,-0.6412900589272691), NE * labelscalefactor);
dot((4.117194931218359,-6.944329602191013),linewidth(4pt) + dotstyle);
label("$D$", (4.2025113612662945,-6.764728856626659), NE * labelscalefactor);
dot((4.6674641029456625,-7.889748381033524),linewidth(4pt) + dotstyle);
label("$F$", (4.750879014791618,-7.701523598065745), NE * labelscalefactor);
dot((4.651736947776926,-3.406898607850789),linewidth(4pt) + dotstyle);
label("$G$", (4.750879014791618,-3.2231877609423107), NE * labelscalefactor);
dot((1.5791517652735851,-0.3557971188372022),linewidth(4pt) + dotstyle);
label("$K$", (1.6663109637116735,-0.18431701432283698), NE * labelscalefactor);
dot((1.5870153428579534,-2.5972220054285695),linewidth(4pt) + dotstyle);
label("$P$", (1.6891596159418953,-2.4234849328845542), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
// Block 2
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(15cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -8.455641974276588, xmax = 26.731282460265, ymin = -10.92318356252699, ymax = 9.023689834456471; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); draw((-1.4934334172297545,2.6953043701763835)--(-1.459546107520503,-6.96389444957376)--(1.1286284157632023,-6.954814372303504)--(4.651736947776926,-3.406898607850789)--(4.642656870506668,-0.8187240845670819)--cycle, linewidth(2) + rvwvcq); /* draw figures */ draw((-1.4934334172297545,2.6953043701763835)--(1.1286284157632023,-6.954814372303504), linewidth(2) + wrwrwr); draw((xmin, -0.9930079421029264*xmin + 1.2123131258653241)--(xmax, -0.9930079421029264*xmax + 1.2123131258653241), linewidth(2) + wrwrwr); /* line */ draw((xmin, 0.0035082940460819836*xmin-6.958773932654766)--(xmax, 0.0035082940460819836*xmax-6.958773932654766), linewidth(2) + wrwrwr); /* line */ draw((xmin, -285.03882139434313*xmin-422.9911967079192)--(xmax, -285.03882139434313*xmax-422.9911967079192), linewidth(2) + wrwrwr); /* line */ draw((xmin, -1.7181023895538718*xmin + 0.12943284739433739)--(xmax, -1.7181023895538718*xmax + 0.12943284739433739), linewidth(2) + wrwrwr); /* line */ draw(circle((4.642656870506668,-0.8187240845670819), 7.071067811865476), linewidth(2) + wrwrwr); draw((xmin, -285.0388213943529*xmin + 1322.5187184230485)--(xmax, -285.0388213943529*xmax + 1322.5187184230485), linewidth(2) + wrwrwr); /* line */ draw((-1.4934334172297545,2.6953043701763835)--(4.617849638067675,6.252300211899359), linewidth(2) + wrwrwr); draw((4.617849638067675,6.252300211899359)--(-1.459546107520503,-6.96389444957376), linewidth(2) + wrwrwr); draw(circle((-0.18240250073327363,-2.12975500106356), 5), linewidth(2) + wrwrwr); draw((xmin, -285.0388213943432*xmin + 449.7637608575419)--(xmax, -285.0388213943432*xmax + 449.7637608575419), linewidth(2) + wrwrwr); /* line */ draw((1.1286284157632023,-6.954814372303504)--(4.651736947776926,-3.406898607850789), linewidth(2) + wrwrwr); draw((-1.4934334172297545,2.6953043701763835)--(4.642656870506668,-0.8187240845670819), linewidth(2) + wrwrwr); draw((-1.4934334172297545,2.6953043701763835)--(-1.459546107520503,-6.96389444957376), linewidth(2) + rvwvcq); draw((-1.459546107520503,-6.96389444957376)--(1.1286284157632023,-6.954814372303504), linewidth(2) + rvwvcq); draw((1.1286284157632023,-6.954814372303504)--(4.651736947776926,-3.406898607850789), linewidth(2) + rvwvcq); draw((4.651736947776926,-3.406898607850789)--(4.642656870506668,-0.8187240845670819), linewidth(2) + rvwvcq); draw((4.642656870506668,-0.8187240845670819)--(-1.4934334172297545,2.6953043701763835), linewidth(2) + rvwvcq); /* dots and labels */ dot((-1.4934334172297545,2.6953043701763835),dotstyle); label("$A$", (-1.3954084351380491,2.9230996889873015), NE * labelscalefactor); dot((1.1286284157632023,-6.954814372303504),dotstyle); label("$B$", (1.2093379191072373,-6.719031552166216), NE * labelscalefactor); dot((8.199652712229643,-6.930007139864511),linewidth(4pt) + dotstyle); label("$C$", (8.292420110475998,-6.741880204396438), NE * labelscalefactor); dot((-1.459546107520503,-6.96389444957376),linewidth(4pt) + dotstyle); label("$H$", (-1.3725597829078273,-6.787577508856881), NE * labelscalefactor); dot((-1.4686261847907602,-4.375719926290057),linewidth(4pt) + dotstyle); label("$E$", (-1.3725597829078273,-4.182831154611618), NE * labelscalefactor); dot((4.617849638067675,6.252300211899359),linewidth(4pt) + dotstyle); label("$L$", (4.705181710331174,6.441792132441429), NE * labelscalefactor); dot((4.642656870506668,-0.8187240845670819),linewidth(4pt) + dotstyle); label("$O$", (4.728030362561396,-0.6412900589272691), NE * labelscalefactor); dot((4.117194931218359,-6.944329602191013),linewidth(4pt) + dotstyle); label("$D$", (4.2025113612662945,-6.764728856626659), NE * labelscalefactor); dot((4.6674641029456625,-7.889748381033524),linewidth(4pt) + dotstyle); label("$F$", (4.750879014791618,-7.701523598065745), NE * labelscalefactor); dot((4.651736947776926,-3.406898607850789),linewidth(4pt) + dotstyle); label("$G$", (4.750879014791618,-3.2231877609423107), NE * labelscalefactor); dot((1.5791517652735851,-0.3557971188372022),linewidth(4pt) + dotstyle); label("$K$", (1.6663109637116735,-0.18431701432283698), NE * labelscalefactor); dot((1.5870153428579534,-2.5972220054285695),linewidth(4pt) + dotstyle); label("$P$", (1.6891596159418953,-2.4234849328845542), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ | [] |
186 | In the diagram below, $ABCD$ is a square. Point $E$ is the midpoint of $\overline{AD}$. Points $F$ and $G$ lie on $\overline{CE}$, and $H$ and $J$ lie on $\overline{AB}$ and $\overline{BC}$, respectively, so that $FGHJ$ is a square. Points $K$ and $L$ lie on $\overline{GH}$, and $M$ and $N$ lie on $\overline{AD}$ and $\overline{AB}$, respectively, so that $KLMN$ is a square. The area of $KLMN$ is 99. Find the area of $FGHJ$. | 2015 AIME I Problem 7 | Let us find the proportion of the side length of $KLMN$ and $FJGH$. Let the side length of $KLMN=y$ and the side length of $FJGH=x$.
Now, examine $BC$. We know $BC=BJ+JC$, and triangles $\Delta BHJ$ and $\Delta JFC$ are similar to $\Delta EDC$ since they are $1-2-\sqrt{5}$ triangles. Thus, we can rewrite $BC$ in terms of the side length of $FJGH$.
\[BJ=\frac{1}{\sqrt{5}}HJ=\frac{x}{\sqrt{5}}=\frac{x\sqrt{5}}{5}, JC=\frac{\sqrt{5}}{2}JF=\frac{x\sqrt{5}}{2}\Rightarrow BC=\frac{7x\sqrt{5}}{10}\]
Now examine $AB$. We can express this length in terms of $x,y$ since $AB=AN+NH+HB$. By using similar triangles as in the first part, we have
\[AB=\frac{1}{\sqrt{5}}y+\frac{\sqrt{5}}{2}y+\frac{2}{\sqrt{5}}x\]
\[AB=BC\Rightarrow \frac{7y\sqrt{5}}{10}+\frac{2x\sqrt{5}}{5}=\frac{7x\sqrt{5}}{10}\Rightarrow \frac{7y\sqrt{5}}{10}=\frac{3x\sqrt{5}}{10}\Rightarrow 7y=3x\]
Now, it is trivial to see that $[FJGH]=\left(\frac{x}{y}\right)^2[KLMN]=\left(\frac{7}{3}\right)^2\cdot 99=\boxed{539}.$ | // Block 1
pair A,B,C,D,E,F,G,H,J,K,L,M,N,P;
B=(0,0);
real m=7*sqrt(55)/5;
J=(m,0);
C=(7*m/2,0);
A=(0,7*m/2);
D=(7*m/2,7*m/2);
E=(A+D)/2;
H=(0,2m);
N=(0,2m+3*sqrt(55)/2);
G=foot(H,E,C);
F=foot(J,E,C);
draw(A--B--C--D--cycle);
draw(C--E);
draw(G--H--J--F);
pair X=foot(N,E,C);
M=extension(N,X,A,D);
K=foot(N,H,G);
L=foot(M,H,G);
draw(K--N--M--L);
label("$A$",A,NW);
label("$B$",B,SW);
label("$C$",C,SE);
label("$D$",D,NE);
label("$E$",E,dir(90));
label("$F$",F,NE);
label("$G$",G,NE);
label("$H$",H,W);
label("$J$",J,S);
label("$K$",K,SE);
label("$L$",L,SE);
label("$M$",M,dir(90));
label("$N$",N,dir(180));
// Block 2
pair A,B,C,D,E,F,G,H,J,K,L,M,N,P; B=(0,0); real m=7*sqrt(55)/5; J=(m,0); C=(7*m/2,0); A=(0,7*m/2); D=(7*m/2,7*m/2); E=(A+D)/2; H=(0,2m); N=(0,2m+3*sqrt(55)/2); G=foot(H,E,C); F=foot(J,E,C); draw(A--B--C--D--cycle); draw(C--E); draw(G--H--J--F); pair X=foot(N,E,C); M=extension(N,X,A,D); K=foot(N,H,G); L=foot(M,H,G); draw(K--N--M--L); label("$A$",A,NW); label("$B$",B,SW); label("$C$",C,SE); label("$D$",D,NE); label("$E$",E,dir(90)); label("$F$",F,NE); label("$G$",G,NE); label("$H$",H,W); label("$J$",J,S); label("$K$",K,SE); label("$L$",L,SE); label("$M$",M,dir(90)); label("$N$",N,dir(180)); | [] |
186 | In the diagram below, $ABCD$ is a square. Point $E$ is the midpoint of $\overline{AD}$. Points $F$ and $G$ lie on $\overline{CE}$, and $H$ and $J$ lie on $\overline{AB}$ and $\overline{BC}$, respectively, so that $FGHJ$ is a square. Points $K$ and $L$ lie on $\overline{GH}$, and $M$ and $N$ lie on $\overline{AD}$ and $\overline{AB}$, respectively, so that $KLMN$ is a square. The area of $KLMN$ is 99. Find the area of $FGHJ$. | 2015 AIME I Problem 7 | We begin by denoting the length $ED$ $a$, giving us $DC = 2a$ and $EC = a\sqrt5$. Since angles $\angle DCE$ and $\angle FCJ$ are complementary, we have that $\triangle CDE \sim \triangle JFC$ (and similarly the rest of the triangles are $1-2-\sqrt5$ triangles). We let the sidelength of $FGHJ$ be $b$, giving us:
\[JC = \sqrt5 \cdot FC = \sqrt5 \cdot FJ/2 = \frac{b\sqrt 5}{2}\]
and
\[BJ = \frac{1}{\sqrt5} \cdot HJ = \frac{b}{\sqrt5}\]
Since $BC = CJ + BJ$,
\[2a = \frac{b\sqrt 5}{2} + \frac{b}{\sqrt5}\]
Solving for $b$ in terms of $a$ yields \[b = \frac{4a\sqrt5}{7}\]
We now use the given that $[KLMN] = 99$, implying that $KL = LM = MN = NK = 3\sqrt{11}$. We also draw the perpendicular from $E$ to $ML$ and label the point of intersection $P$ as in the diagram at the top
This gives that \[AM = 2 \cdot AN = 2 \cdot \frac{3\sqrt{11}}{\sqrt5}\]
and \[ME = \sqrt5 \cdot MP = \sqrt5 \cdot \frac{EP}{2} = \sqrt5 \cdot \frac{LG}{2} = \sqrt5 \cdot \frac{HG - HK - KL}{2} = \sqrt{5} \cdot \frac{\frac{4a\sqrt5}{7} - \frac{9\sqrt{11}}{2}}{2}\]
Since $AE$ = $AM + ME$, we get
\[2 \cdot \frac{3\sqrt{11}}{\sqrt5} + \sqrt{5} \cdot \frac{\frac{4a\sqrt5}{7} - \frac{9\sqrt{11}}{2}}{2} = a\]
\[\Rightarrow 12\sqrt{11} + 5(\frac{4a\sqrt5}{7} - \frac{9\sqrt{11}}{2}) = 2\sqrt5a\]
\[\Rightarrow \frac{-21}{2}\sqrt{11} + \frac{20a\sqrt5}{7} = 2\sqrt5a\]
\[\Rightarrow -21\sqrt{11} = 2\sqrt5a\frac{14 - 20}{7}\]
\[\Rightarrow \frac{49\sqrt{11}}{4} = \sqrt5a\]
\[\Rightarrow 7\sqrt{11} = \frac{4a\sqrt{5}}{7}\]
So our final answer is $(7\sqrt{11})^2 = \boxed{539}$. | // Block 1
pair A,B,C,D,E,F,G,H,J,K,L,M,N,P;
B=(0,0);
real m=7*sqrt(55)/5;
J=(m,0);
C=(7*m/2,0);
A=(0,7*m/2);
D=(7*m/2,7*m/2);
E=(A+D)/2;
H=(0,2m);
N=(0,2m+3*sqrt(55)/2);
G=foot(H,E,C);
F=foot(J,E,C);
draw(A--B--C--D--cycle);
draw(C--E);
draw(G--H--J--F);
pair X=foot(N,E,C);
M=extension(N,X,A,D);
K=foot(N,H,G);
L=foot(M,H,G);
draw(K--N--M--L);
P=foot(E,M,L);
draw(P--E);
label("$A$",A,NW);
label("$B$",B,SW);
label("$C$",C,SE);
label("$D$",D,NE);
label("$E$",E,dir(90));
label("$F$",F,NE);
label("$G$",G,NE);
label("$H$",H,W);
label("$J$",J,S);
label("$K$",K,SE);
label("$L$",L,SE);
label("$M$",M,dir(90));
label("$N$",N,dir(180));
label("$P$",P,dir(235));
// Block 2
pair A,B,C,D,E,F,G,H,J,K,L,M,N,P; B=(0,0); real m=7*sqrt(55)/5; J=(m,0); C=(7*m/2,0); A=(0,7*m/2); D=(7*m/2,7*m/2); E=(A+D)/2; H=(0,2m); N=(0,2m+3*sqrt(55)/2); G=foot(H,E,C); F=foot(J,E,C); draw(A--B--C--D--cycle); draw(C--E); draw(G--H--J--F); pair X=foot(N,E,C); M=extension(N,X,A,D); K=foot(N,H,G); L=foot(M,H,G); draw(K--N--M--L); P=foot(E,M,L); draw(P--E); label("$A$",A,NW); label("$B$",B,SW); label("$C$",C,SE); label("$D$",D,NE); label("$E$",E,dir(90)); label("$F$",F,NE); label("$G$",G,NE); label("$H$",H,W); label("$J$",J,S); label("$K$",K,SE); label("$L$",L,SE); label("$M$",M,dir(90)); label("$N$",N,dir(180)); label("$P$",P,dir(235)); | [] |
187 | In an isosceles trapezoid, the parallel bases have lengths $\log 3$ and $\log 192$, and the altitude to these bases has length $\log 16$. The perimeter of the trapezoid can be written in the form $\log 2^p 3^q$, where $p$ and $q$ are positive integers. Find $p + q$. | 2015 AIME II Problem 4 | Let $a=\log2$ and $b=\log3$ so that the base lengths are $\log3=b$ and $\log192=\log(3\cdot64)=\log3+\log\left(2^6\right)=6a+b$ and the altitudes are $\log16=4a$. Then we have the following picture:
Note that we have the two right triangles to the side; one of each of their bases is an altitude, which we know the length to be $4a$. The length of the other base can be calculated as $\dfrac{(6a+b)-b}2=3a$ via simple isosceles trapezoid geometry; it is clear that each right triangle is actually a $3-4-5$ triangle, so we know their hypotenuses (and the remaining unknown sides of our original trapezoid) have length $5a$ (because of the $3a$ and $4a$ bases). Our answer is therefore $5a+5a+b+6a+b=16a+2b=16\log2+2\log3=\log\left(2^{16}3^2\right)$; $p+q=16+2=\boxed{018}$. QED.
~Technodoggo | // Block 1
import graph;
unitsize(1cm);
draw((0,0)--(3,4)--(9,4)--(12,0)--cycle);
draw((3,4)--(3,0));draw((9,4)--(9,0));
label("$b$",(3,4)--(9,4),N);
label("$6a+b$",(0,0)--(12,0),S);
label("$b$",(0,0)--(12,0),N);
label("$4a$", (3,0)--(3,4),W);
label("$4a$", (9,0)--(9,4),E);
// Block 2
import graph; unitsize(1cm); draw((0,0)--(3,4)--(9,4)--(12,0)--cycle); draw((3,4)--(3,0));draw((9,4)--(9,0)); label("$b$",(3,4)--(9,4),N); label("$6a+b$",(0,0)--(12,0),S); label("$b$",(0,0)--(12,0),N); label("$4a$", (3,0)--(3,4),W); label("$4a$", (9,0)--(9,4),E); | [] |
188 | Triangle $ABC$ has side lengths $AB = 12$, $BC = 25$, and $CA = 17$. Rectangle $PQRS$ has vertex $P$ on $\overline{AB}$, vertex $Q$ on $\overline{AC}$, and vertices $R$ and $S$ on $\overline{BC}$. In terms of the side length $PQ = \omega$, the area of $PQRS$ can be expressed as the quadratic polynomial \[Area(PQRS) = \alpha \omega - \beta \omega^2.\]
Then the coefficient $\beta = \frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. | 2015 AIME II Problem 7 | Similar triangles can also solve the problem.
First, solve for the area of the triangle. $[ABC] = 90$. This can be done by Heron's Formula or placing an $8-15-17$ right triangle on $AC$ and solving. (The $8$ side would be collinear with line $AB$)
After finding the area, solve for the altitude to $BC$. Let $E$ be the intersection of the altitude from $A$ and side $BC$. Then $AE = \frac{36}{5}$.
Solving for $BE$ using the Pythagorean Formula, we get $BE = \frac{48}{5}$. We then know that $CE = \frac{77}{5}$.
Now consider the rectangle $PQRS$. Since $SR$ is collinear with $BC$ and parallel to $PQ$, $PQ$ is parallel to $BC$ meaning $\Delta APQ$ is similar to $\Delta ABC$.
Let $F$ be the intersection between $AE$ and $PQ$. By the similar triangles, we know that $\frac{PF}{FQ}=\frac{BE}{EC} = \frac{48}{77}$. Since $PF+FQ=PQ=\omega$. We can solve for $PF$ and $FQ$ in terms of $\omega$. We get that $PF=\frac{48}{125} \omega$ and $FQ=\frac{77}{125} \omega$.
Let's work with $PF$. We know that $PQ$ is parallel to $BC$ so $\Delta APF$ is similar to $\Delta ABE$. We can set up the proportion:
$\frac{AF}{PF}=\frac{AE}{BE}=\frac{3}{4}$. Solving for $AF$, $AF = \frac{3}{4} PF = \frac{3}{4} \cdot \frac{48}{125} \omega = \frac{36}{125} \omega$.
We can solve for $PS$ then since we know that $PS=FE$ and $FE= AE - AF = \frac{36}{5} - \frac{36}{125} \omega$.
Therefore, $[PQRS] = PQ \cdot PS = \omega (\frac{36}{5} - \frac{36}{125} \omega) = \frac{36}{5}\omega - \frac{36}{125} \omega^2$.
This means that $\beta = \frac{36}{125} \Rightarrow (m,n) = (36,125) \Rightarrow m+n = \boxed{161}$. | // Block 1
unitsize(20);
pair A,B,C,E,F,P,Q,R,S;
A=(48/5,36/5);
B=(0,0);
C=(25,0);
E=(48/5,0);
F=(48/5,18/5);
P=(24/5,18/5);
Q=(173/10,18/5);
S=(24/5,0);
R=(173/10,0);
draw(A--B--C--cycle);
draw(P--Q);
draw(Q--R);
draw(R--S);
draw(S--P);
draw(A--E,dashed);
label("$A$",A,N);
label("$B$",B,SW);
label("$C$",C,SE);
label("$E$",E,SE);
label("$F$",F,NE);
label("$P$",P,NW);
label("$Q$",Q,NE);
label("$R$",R,SE);
label("$S$",S,SW);
draw(rightanglemark(B,E,A,12));
dot(E);
dot(F);
// Block 2
unitsize(20); pair A,B,C,E,F,P,Q,R,S; A=(48/5,36/5); B=(0,0); C=(25,0); E=(48/5,0); F=(48/5,18/5); P=(24/5,18/5); Q=(173/10,18/5); S=(24/5,0); R=(173/10,0); draw(A--B--C--cycle); draw(P--Q); draw(Q--R); draw(R--S); draw(S--P); draw(A--E,dashed); label("$A$",A,N); label("$B$",B,SW); label("$C$",C,SE); label("$E$",E,SE); label("$F$",F,NE); label("$P$",P,NW); label("$Q$",Q,NE); label("$R$",R,SE); label("$S$",S,SW); draw(rightanglemark(B,E,A,12)); dot(E); dot(F); | [] |
189 | A cylindrical barrel with radius $4$ feet and height $10$ feet is full of water. A solid cube with side length $8$ feet is set into the barrel so that the diagonal of the cube is vertical. The volume of water thus displaced is $v$ cubic feet. Find $v^2$. | 2015 AIME II Problem 9 | We can use the same method as in Solution 2 to find the side length of the equilateral triangle, which is $4\sqrt3$. From here, its area is
\[\dfrac{\bigl(4\sqrt3\bigr)^2\sqrt3}4=12\sqrt3.\]
The leg of the isosceles right triangle is $\dfrac{4\sqrt3}{\sqrt2}=2\sqrt6$, and the horizontal distance from the vertex to the base of the tetrahedron is $4$ (the radius of the cylinder), so we can find the height, as shown in the diagram.
The height from the tip to the base is $2\sqrt2$, so the volume is $\dfrac{12\sqrt3\cdot2\sqrt2}3=8\sqrt6$, and thus the answer is $\boxed{384}$.
-integralarefun | // Block 1
import olympiad;
pair V, T, B;
V = (-4, 0);
B = origin;
T = (0, 2*sqrt(2));
draw(V--B--T--cycle);
draw(rightanglemark(V, B, T));
label("Vertex", V, W);
label("Tip", T, N);
label("Base", B, SE);
label("$4$", V--B, S);
label("$2\sqrt6$", V--T, NW);
// Block 2
import olympiad; pair V, T, B; V = (-4, 0); B = origin; T = (0, 2*sqrt(2)); draw(V--B--T--cycle); draw(rightanglemark(V, B, T)); label("Vertex", V, W); label("Tip", T, N); label("Base", B, SE); label("$4$", V--B, S); label("$2\sqrt6$", V--T, NW); | [] |
190 | The circumcircle of acute $\triangle ABC$ has center $O$. The line passing through point $O$ perpendicular to $\overline{OB}$ intersects lines $AB$ and $BC$ at $P$ and $Q$, respectively. Also $AB=5$, $BC=4$, $BQ=4.5$, and $BP=\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. | 2015 AIME II Problem 11 | Solution 1
Call $M$ and $N$ the feet of the altitudes from $O$ to $BC$ and $AB$, respectively. Let $OB = r$ . Notice that $\triangle{OMB} \sim \triangle{QOB}$ because both are right triangles, and $\angle{OBQ} \cong \angle{OBM}$. By $\frac{MB}{BO}=\frac{BO}{BQ}$, $MB = r\left(\frac{r}{4.5}\right) = \frac{r^2}{4.5}$. However, since $O$ is the circumcenter of triangle $ABC$, $OM$ is a perpendicular bisector by the definition of a circumcenter. Hence, $\frac{r^2}{4.5} = 2 \implies r = 3$. Since we know $BN=\frac{5}{2}$ and $\triangle BOP \sim \triangle BNO$, we have $\frac{BP}{3} = \frac{3}{\frac{5}{2}}$. Thus, $BP = \frac{18}{5}$. $m + n=\boxed{023}$.
Solution 2 (fastest)
Minor arc $BC = 2A$ so $\angle{BOC}=2A$. Since $\triangle{BOC}$ is isosceles ($BO$ and $OC$ are radii), $\angle{CBO}=(180-2A)/2=90-A$. $\angle{CBO}=90-A$, so $\angle{BQO}=A$. From this we get that $\triangle{BPQ}\sim \triangle{BCA}$. So $\dfrac{BP}{BC}=\dfrac{BQ}{BA}$, plugging in the given values we get $\dfrac{BP}{4}=\dfrac{4.5}{5}$, so $BP=\dfrac{18}{5}$, and $m+n=\boxed{023}$.
Solution 3
Let $r=BO$. Drawing perpendiculars, $BM=MC=2$ and $BN=NA=2.5$. From there, \[OM=\sqrt{r^2-4}\] Thus, \[OQ=\frac{\sqrt{4r^2+9}}{2}\] Using $\triangle{BOQ}$, we get $r=3$. Now let's find $NP$. After some calculations with $\triangle{BON}$ ~ $\triangle{OPN}$, ${NP=11/10}$. Therefore, \[BP=\frac{5}{2}+\frac{11}{10}=18/5\] $18+5=\boxed{023}$.
Solution 4
Let $\angle{BQO}=\alpha$. Extend $OB$ to touch the circumcircle at a point $K$. Then, note that $\angle{KAC}=\angle{CBK}=\angle{QBO}=90^\circ-\alpha$. But since $BK$ is a diameter, $\angle{KAB}=90^\circ$, implying $\angle{CAB}=\alpha$. It follows that $APCQ$ is a cyclic quadrilateral.
Let $BP=x$. By Power of a Point, \[5x=4\cdot\frac 9 2\implies x=\frac{18}{5}.\]The answer is $18+5=\boxed{023}$.
Solution 5
$\textit{Note: This is not a very good solution, but it is relatively natural and requires next to no thinking.}$
Denote the circumradius of $ABC$ to be $R$, the circumcircle of $ABC$ to be $O$, and the shortest distance from $Q$ to circle $O$ to be $x$.
Using Power of a Point on $Q$ relative to circle $O$, we get that $x(x+2r) = 0.5 \cdot 4.5 = \frac{9}{4}$. Using Pythagorean Theorem on triangle $QOB$ to get $(x + r)^2 + r^2 = \frac{81}{4}$. Subtracting the first equation from the second, we get that $2r^2 = 18$ and therefore $r = 3$. Now, set $\cos{ABC} = y$. Using law of cosines on $ABC$ to find $AC$ in terms of $y$ and plugging that into the extended law of sines, we get $\frac{\sqrt{4^2 + 5^2 - 2 \cdot 4 \cdot 5 x}}{\sqrt{1 - x^2}} = 2R = 6$. Squaring both sides and cross multiplying, we get $36x^2 - 40x + 5 = 0$. Now, we get $x = \frac{10 \pm \sqrt{55}}{18}$ using quadratic formula. If you drew a decent diagram, $B$ is acute and therefore $x = \frac{10 + \sqrt{55}}{18}$(You can also try plugging in both in the end and seeing which gives a rational solution). Note that $BP = 3\frac{1}{\sin{OPB}} = \frac{3}{\cos{\angle ABC - \angle QBO}}.$ Using the cosine addition formula and then plugging in what we know about $QBO$, we get that $BP = \frac{162}{2\cos{B} + \sqrt{5}\sin{B}}$. Now, the hard part is to find what $\sin{B}$ is. We therefore want $\frac{\sqrt{324 - (10 + \sqrt{55})^2}}{18} = \frac{\sqrt{169 - 20\sqrt{55}}}{18}$. For the numerator, by inspection $(a + b\sqrt{55})^2$ will not work for integers $a$ and $b$. The other case is if there is $(a\sqrt{5} + b\sqrt{11})^2$. By inspection, $5\sqrt{5} - 2\sqrt{11}$ works. Therefore, plugging all this in yields the answer, $\frac{18}{5} \rightarrow \boxed{23}$. Solution by hyxue
Solution 6
Reflect $A$, $P$ across $OB$ to points $A'$ and $P'$, respectively with $A'$ on the circle and $P, O, P'$ collinear. Now, $\angle A'CQ = 180^{\circ} - \angle A'CB = \angle A'AB = \angle P'PB$ by parallel lines. From here, $\angle P'PB = \angle PP'B = \angle A'P'Q$ as $P, P', Q$ collinear. From here, $A'P'QC$ is cyclic, and by power of a point we obtain $\frac{18}{5} \implies \boxed{023}$.
~awang11's sol | /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(15cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.4) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -4.7673964645097335, xmax = 9.475267639476614, ymin = -1.6884766592324019, ymax = 6.385449160754665; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); /* draw figures */ draw(circle((0.7129306199257198,2.4781596958650733), 3.000319171815248), linewidth(2) + wrwrwr); draw((0.7129306199257198,2.4781596958650733)--(3.178984115621537,0.7692140299269852), linewidth(2) + wrwrwr); draw((xmin, 1.4430262733614363*xmin + 1.4493820802284032)--(xmax, 1.4430262733614363*xmax + 1.4493820802284032), linewidth(2) + wrwrwr); /* line */ draw((xmin, -0.020161290322580634*xmin + 0.8333064516129032)--(xmax, -0.020161290322580634*xmax + 0.8333064516129032), linewidth(2) + wrwrwr); /* line */ draw((xmin, -8.047527437688247*xmin + 26.352175924366414)--(xmax, -8.047527437688247*xmax + 26.352175924366414), linewidth(2) + wrwrwr); /* line */ draw((xmin, -2.5113572383524088*xmin + 8.752778799300463)--(xmax, -2.5113572383524088*xmax + 8.752778799300463), linewidth(2) + wrwrwr); /* line */ draw((xmin, 0.12426176956126818*xmin + 2.389569675458691)--(xmax, 0.12426176956126818*xmax + 2.389569675458691), linewidth(2) + wrwrwr); /* line */ draw(circle((1.9173376033752174,4.895608471162773), 0.7842529827808445), linewidth(2) + wrwrwr); /* dots and labels */ dot((-1.82,0.87),dotstyle); label("$A$", (-1.7801363959463627,0.965838014692327), NE * labelscalefactor); dot((3.178984115621537,0.7692140299269852),dotstyle); label("$B$", (3.2140445236332655,0.8641046996638531), NE * labelscalefactor); dot((2.6857306099246263,4.738685150758791),dotstyle); label("$C$", (2.7238749148597092,4.831703985774336), NE * labelscalefactor); dot((0.7129306199257198,2.4781596958650733),linewidth(4pt) + dotstyle); label("$O$", (0.7539479965810783,2.556577122410283), NE * labelscalefactor); dot((-0.42105034508654754,0.8417953698606159),linewidth(4pt) + dotstyle); label("$P$", (-0.38361543510094825,0.9195955987702934), NE * labelscalefactor); dot((2.6239558409689123,5.235819298886746),linewidth(4pt) + dotstyle); label("$Q$", (2.6591355325688624,5.312625111363486), NE * labelscalefactor); dot((1.3292769824200672,5.414489427724579),linewidth(4pt) + dotstyle); label("$A'$", (1.3643478867519216,5.488346291867214), NE * labelscalefactor); dot((1.8469115849379867,4.11452402186953),linewidth(4pt) + dotstyle); label("$P'$", (1.8822629450786978,4.184310162865866), NE * labelscalefactor); dot((2.5624172335003985,5.731052930966743),linewidth(4pt) + dotstyle); label("$D$", (2.603644633462422,5.802794720137042), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); | [] |
191 | Circles $\mathcal{P}$ and $\mathcal{Q}$ have radii $1$ and $4$, respectively, and are externally tangent at point $A$. Point $B$ is on $\mathcal{P}$ and point $C$ is on $\mathcal{Q}$ so that line $BC$ is a common external tangent of the two circles. A line $\ell$ through $A$ intersects $\mathcal{P}$ again at $D$ and intersects $\mathcal{Q}$ again at $E$. Points $B$ and $C$ lie on the same side of $\ell$, and the areas of $\triangle DBA$ and $\triangle ACE$ are equal. This common area is $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. | 2015 AIME II Problem 15 | Call $O_1$ and $O_2$ the centers of circles $\mathcal{P}$ and $\mathcal{Q}$, respectively, and extend $CB$ and $O_2O_1$ to meet at point $N$. Call $K$ and $L$ the feet of the altitudes from $B$ to $O_1N$ and $C$ to $O_2N$, respectively. Using the fact that $\triangle{O_1BN} \sim \triangle{O_2CN}$ and setting $NO_1 = k$, we have that $\frac{k+5}{k} = \frac{4}{1} \implies k=\frac{5}{3}$. We can do some more length chasing using triangles similar to $O_1BN$ to get that $AK = AL = \frac{24}{15}$, $BK = \frac{12}{15}$, and $CL = \frac{48}{15}$. Now, consider the circles $\mathcal{P}$ and $\mathcal{Q}$ on the coordinate plane, where $A$ is the origin. If the line $\ell$ through $A$ intersects $\mathcal{P}$ at $D$ and $\mathcal{Q}$ at $E$ then $4 \cdot DA = AE$. To verify this, notice that $\triangle{AO_1D} \sim \triangle{EO_2A}$ from the fact that both triangles are isosceles with $\angle{O_1AD} \cong \angle{O_2AE}$, which are corresponding angles. Since $O_2A = 4\cdot O_1A$, we can conclude that $4 \cdot DA = AE$.
Hence, we need to find the slope $m$ of line $\ell$ such that the perpendicular distance $n$ from $B$ to $AD$ is four times the perpendicular distance $p$ from $C$ to $AE$. This will mean that the product of the bases and heights of triangles $ACE$ and $DBA$ will be equal, which in turn means that their areas will be equal. Let the line $\ell$ have the equation $y = -mx \implies mx + y = 0$, and let $m$ be a positive real number so that the negative slope of $\ell$ is preserved. Setting $A = (0,0)$, the coordinates of $B$ are $(x_B, y_B) = \left(\frac{-24}{15}, \frac{-12}{15}\right)$, and the coordinates of $C$ are $(x_C, y_C) = \left(\frac{24}{15}, \frac{-48}{15}\right)$. Using the point-to-line distance formula and the condition $n = 4p$, we have \[\frac{|mx_B + 1(y_B) + 0|}{\sqrt{m^2 + 1}} = \frac{4|mx_C + 1(y_C) + 0|}{\sqrt{m^2 + 1}}\] \[\implies |mx_B + y_B| = 4|mx_C + y_C| \implies \left|\frac{-24m}{15} + \frac{-12}{15}\right| = 4\left|\frac{24m}{15} + \frac{-48}{15}\right|.\] If $m > 2$, then clearly $B$ and $C$ would not lie on the same side of $\ell$. Thus since $m > 0$, we must switch the signs of all terms in this equation when we get rid of the absolute value signs. We then have \[\frac{6m}{15} + \frac{3}{15} = \frac{48}{15} - \frac{24m}{15} \implies 2m = 3 \implies m = \frac{3}{2}.\] Thus, the equation of $\ell$ is $y = -\frac{3}{2}x$.
Then we can find the coordinates of $D$ by finding the point $(x,y)$ other than $A = (0,0)$ where the circle $\mathcal{P}$ intersects $\ell$. $\mathcal{P}$ can be represented with the equation $(x + 1)^2 + y^2 = 1$, and substituting $y = -\frac{3}{2}x$ into this equation yields $x = 0, -\frac{8}{13}$ as solutions. Discarding $x = 0$, the $y$-coordinate of $D$ is $-\frac{3}{2} \cdot -\frac{8}{13} = \frac{12}{13}$. The distance from $D$ to $A$ is then $\frac{4}{\sqrt{13}}.$ The perpendicular distance from $B$ to $AD$ or the height of $\triangle{DBA}$ is $\frac{|\frac{3}{2}\cdot\frac{-24}{15} + \frac{-12}{15} + 0|}{\sqrt{\frac{3}{2}^2 + 1}} = \frac{\frac{48}{15}}{\frac{\sqrt{13}}{2}} = \frac{32}{5\sqrt{13}}.$ Finally, the common area is $\frac{1}{2}\left(\frac{32}{5\sqrt{13}} \cdot \frac{4}{\sqrt{13}}\right) = \frac{64}{65}$, and $m + n = 64 + 65 = \boxed{129}$. | // Block 1
unitsize(35);
draw(Circle((-1,0),1));
draw(Circle((4,0),4));
pair A,O_1, O_2, B,C,D,E,N,K,L,X,Y;
A=(0,0);O_1=(-1,0);O_2=(4,0);B=(-24/15,-12/15);D=(-8/13,12/13);E=(32/13,-48/13);C=(24/15,-48/15);N=extension(E,B,O_2,O_1);K=foot(B,O_1,N);L=foot(C,O_2,N);X=foot(B,A,D);Y=foot(C,E,A);
label("$A$",A,NE);label("$O_1$",O_1,NE);label("$O_2$",O_2,NE);label("$B$",B,SW);label("$C$",C,SW);label("$D$",D,NE);label("$E$",E,NE);label("$N$",N,W);label("$K$",(-24/15,0.2));label("$L$",(24/15,0.2));label("$n$",(-0.8,-0.12));label("$p$",((29/15,-48/15)));label("$\mathcal{P}$",(-1.6,1.1));label("$\mathcal{Q}$",(6,4));
draw(A--B--D--cycle);draw(A--E--C--cycle);draw(C--N);draw(O_2--N);draw(O_1--B,dashed);draw(O_2--C,dashed);
dot(O_1);dot(O_2);
draw(rightanglemark(O_1,B,N,5));draw(rightanglemark(O_2,C,N,5));draw(C--L,dashed);draw(B--K,dashed);draw(C--Y);draw(B--X);
// Block 2
unitsize(35); draw(Circle((-1,0),1)); draw(Circle((4,0),4)); pair A,O_1, O_2, B,C,D,E,N,K,L,X,Y; A=(0,0);O_1=(-1,0);O_2=(4,0);B=(-24/15,-12/15);D=(-8/13,12/13);E=(32/13,-48/13);C=(24/15,-48/15);N=extension(E,B,O_2,O_1);K=foot(B,O_1,N);L=foot(C,O_2,N);X=foot(B,A,D);Y=foot(C,E,A); label("$A$",A,NE);label("$O_1$",O_1,NE);label("$O_2$",O_2,NE);label("$B$",B,SW);label("$C$",C,SW);label("$D$",D,NE);label("$E$",E,NE);label("$N$",N,W);label("$K$",(-24/15,0.2));label("$L$",(24/15,0.2));label("$n$",(-0.8,-0.12));label("$p$",((29/15,-48/15)));label("$\mathcal{P}$",(-1.6,1.1));label("$\mathcal{Q}$",(6,4)); draw(A--B--D--cycle);draw(A--E--C--cycle);draw(C--N);draw(O_2--N);draw(O_1--B,dashed);draw(O_2--C,dashed); dot(O_1);dot(O_2); draw(rightanglemark(O_1,B,N,5));draw(rightanglemark(O_2,C,N,5));draw(C--L,dashed);draw(B--K,dashed);draw(C--Y);draw(B--X); | [] |
191 | Circles $\mathcal{P}$ and $\mathcal{Q}$ have radii $1$ and $4$, respectively, and are externally tangent at point $A$. Point $B$ is on $\mathcal{P}$ and point $C$ is on $\mathcal{Q}$ so that line $BC$ is a common external tangent of the two circles. A line $\ell$ through $A$ intersects $\mathcal{P}$ again at $D$ and intersects $\mathcal{Q}$ again at $E$. Points $B$ and $C$ lie on the same side of $\ell$, and the areas of $\triangle DBA$ and $\triangle ACE$ are equal. This common area is $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. | 2015 AIME II Problem 15 | Consider the homothety that takes triangle BDA onto CXY on the big circle, as plotted. Some hidden congruence angles are revealed which help reduce computation complexity. Just some angle chasing and straight forward trigs. Because $AE=XY$ and $AE \parallel XY$, $XYE$ is right angle.
First, $\frac{NO_1}{NO_1+5} = \frac{1}{4}$, so $NO_1=\frac{5}{3}$. And,
\[\cos{\angle{AO_2C}}=\cos{2\angle{AYC}} = \frac{O_2C}{NO_1+5} = \frac{3}{5}\]
\[\sin{\angle{AYC}} = \sqrt{\dfrac{1-\cos{2\angle{AYC}}}{2}}=\frac{1}{\sqrt{5}}\]
\[\cos{\angle{AYC}} = \frac{2}{\sqrt{5}}\]
Then,
\[[AEC] = \frac{1}{2}AE*CE*\sin{\angle{ACE}}=\frac{1}{2}AE*8\sin{\angle{CYE}}*\frac{1}{\sqrt{5}}\]
\[[CXY] = \frac{1}{2}CX*XY*\sin{\angle{CXY}}=\frac{1}{2}*8\sin{\angle{XYC}}*XY*\sin{\angle{CAY}}\]
Since $\angle{CAY} = 90 - \angle{AYC}$, $\angle{XYC} = 90 - \angle{CYE}$, $XY = AE$, we have
\[[CXY] = \frac{1}{2}*8AE\cos{\angle{CYE}}*\cos{\angle{AYC}}=\frac{1}{2}*8AE\cos{\angle{CYE}}*\frac{2}{\sqrt{5}}\]
Since $\triangle{CXY}$ is four times in scale to $\triangle{AEC}$, their area ratio is 16. Divide the two equations for the two areas, we have
\[\tan{\angle{CYE}} = \frac{1}{8}\]
With this angle found, everything else just follows.
\[\sin{\angle{CYE}} = \dfrac{\tan{\angle{CYE}}}{\sqrt{1+\tan^2{\angle{CYE}}}}=\dfrac{1}{\sqrt{65}}\]
\[\cos{\angle{CYE}} = \dfrac{8}{\sqrt{65}}\]
\[\sin{\angle{AYE}} = \sin({\angle{AYC}+\angle{CYE}} )= \frac{1}{\sqrt{5}}*\frac{8}{\sqrt{65}} + \frac{2}{\sqrt{5}}*\frac{1}{\sqrt{65}} = \frac{2}{\sqrt{13}}\]
\[AE = 8\sin{\angle{AYE}} = \frac{16}{\sqrt{13}}\]
\[[AEC] = \frac{1}{2}*8*\frac{16}{\sqrt{13}}*\dfrac{1}{\sqrt{65}}*\frac{1}{\sqrt{5}}=\frac{64}{65}\]
Thus, our answer is $64+65=\boxed{129}$. | // Block 1
unitsize(35);
draw(Circle((-1,0),1));
draw(Circle((4,0),4));
pair A,O_1, O_2, B,C,D,E,N,K,L,X,Y;
A=(0,0);O_1=(-1,0);O_2=(4,0);B=(-24/15,-12/15);D=(-8/13,12/13);E=(32/13,-48/13);C=(24/15,-48/15);N=extension(E,B,O_2,O_1);K=foot(B,O_1,N);L=foot(C,O_2,N);X=foot(B,A,D);Y=foot(C,E,A);
label("$A$",A,NE);label("$O_1$",O_1,NE);label("$O_2$",O_2,NE);label("$B$",B,SW);label("$C$",C,SW);label("$D$",D,NE);label("$E$",E,NE);label("$N$",N,W);label("$K$",(-24/15,0.2));label("$L$",(24/15,0.2));label("$n$",(-0.8,-0.12));label("$p$",((29/15,-48/15)));label("$\mathcal{P}$",(-1.6,1.1));label("$\mathcal{Q}$",(6,4));
draw(A--B--D--cycle);draw(A--E--C--cycle);draw(C--N);draw(O_2--N);draw(O_1--B,dashed);draw(O_2--C,dashed);
dot(O_1);dot(O_2);
draw(rightanglemark(O_1,B,N,5));draw(rightanglemark(O_2,C,N,5));
//draw(C--L,dashed);draw(B--K,dashed);draw(C--Y);draw(B--X);
path circle2 = Circle((4,0),4);
N = (-8/3,0);
pair X =rotate(180,O_2)*E;
pair Y = (8,0);
draw(X--Y,dashed); draw(E--Y,dashed);draw(E--X,dashed); draw(Y--C,dashed); draw(C--X,dashed); draw(O_2--Y);
dot("$X$", X, NE);dot("$Y$", Y, NE);
// Block 2
unitsize(35); draw(Circle((-1,0),1)); draw(Circle((4,0),4)); pair A,O_1, O_2, B,C,D,E,N,K,L,X,Y; A=(0,0);O_1=(-1,0);O_2=(4,0);B=(-24/15,-12/15);D=(-8/13,12/13);E=(32/13,-48/13);C=(24/15,-48/15);N=extension(E,B,O_2,O_1);K=foot(B,O_1,N);L=foot(C,O_2,N);X=foot(B,A,D);Y=foot(C,E,A); label("$A$",A,NE);label("$O_1$",O_1,NE);label("$O_2$",O_2,NE);label("$B$",B,SW);label("$C$",C,SW);label("$D$",D,NE);label("$E$",E,NE);label("$N$",N,W);label("$K$",(-24/15,0.2));label("$L$",(24/15,0.2));label("$n$",(-0.8,-0.12));label("$p$",((29/15,-48/15)));label("$\mathcal{P}$",(-1.6,1.1));label("$\mathcal{Q}$",(6,4)); draw(A--B--D--cycle);draw(A--E--C--cycle);draw(C--N);draw(O_2--N);draw(O_1--B,dashed);draw(O_2--C,dashed); dot(O_1);dot(O_2); draw(rightanglemark(O_1,B,N,5));draw(rightanglemark(O_2,C,N,5)); //draw(C--L,dashed);draw(B--K,dashed);draw(C--Y);draw(B--X); path circle2 = Circle((4,0),4); N = (-8/3,0); pair X =rotate(180,O_2)*E; pair Y = (8,0); draw(X--Y,dashed); draw(E--Y,dashed);draw(E--X,dashed); draw(Y--C,dashed); draw(C--X,dashed); draw(O_2--Y); dot("$X$", X, NE);dot("$Y$", Y, NE); | [] |
191 | Circles $\mathcal{P}$ and $\mathcal{Q}$ have radii $1$ and $4$, respectively, and are externally tangent at point $A$. Point $B$ is on $\mathcal{P}$ and point $C$ is on $\mathcal{Q}$ so that line $BC$ is a common external tangent of the two circles. A line $\ell$ through $A$ intersects $\mathcal{P}$ again at $D$ and intersects $\mathcal{Q}$ again at $E$. Points $B$ and $C$ lie on the same side of $\ell$, and the areas of $\triangle DBA$ and $\triangle ACE$ are equal. This common area is $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. | 2015 AIME II Problem 15 | Let the centers of circles $\mathcal{P}$ and $\mathcal{Q}$ be $P$ and $Q$, respectively. Let $F$ be the point on $\overline{QC}$ such that $\overline{PF}$ is parallel to $\overline{BC}$. We find that $BC=PF=\sqrt{(4+1)^{2}-(4-1)^2}=4$.
Now, we define $A$ to be $(0,0)$. It follows that:
\begin{align}
P &= \left(-\frac{4}{5},-\frac{3}{5}\right) \\
Q &= \left(\frac{16}{5},\frac{12}{5}\right) \\
B &= \left(-\frac{4}{5},-\frac{8}{5}\right) \\
C &= \left(\frac{16}{5},-\frac{8}{5}\right) \\
\end{align}
The equation of circle $P$ is $\left(x+\frac{4}{5}\right)^2+\left(y+\frac{3}{5}\right)^2 = 1^2$, which simplifies to $x^2+\frac{8}{5}x+y^2+\frac{6}{5}y=0$.
The equation of circle $Q$ is $\left(x-\frac{16}{5}\right)^2+\left(y-\frac{12}{5}\right)^2 = 4^2$, which simplifies to $x^2-\frac{32}{5}x+y^2-\frac{24}{5}y=0$.
Suppose points $D$ and $E$ lie on the line $y=mx$ for some slope $m$. We can now plug in $y=mx$ into the equations for the circles to find the coordinates of $D$ and $E$ in terms of $m$.
For point $D$:
\[x^2+\frac{8}{5}x+m^2x^2+\frac{6}{5}mx=0\]
\[(1+m^2)x^2+\left(\frac{8}{5}+\frac{6}{5}m\right)=0\]
Since $D \neq A$, we have $x \neq 0$, thus $x=\dfrac{-\left(\frac{8}{5}+\frac{6}{5}m\right)}{1+m^2}$.
For point $E$:
\[x^2-\frac{32}{5}x+m^2x^2-\frac{24}{5}mx=0\]
\[(1+m^2)x^2-\left(\frac{32}{5}+\frac{24}{5}m\right)=0\]
Since $E \neq A$, we have $x \neq 0$, thus $x=\dfrac{\left(\frac{32}{5}+\frac{24}{5}m\right)}{1+m^2}$.
We let $a=\dfrac{\left(\frac{8}{5}+\frac{6}{5}m\right)}{1+m^2}$ to write things quicker, and so the coordinates of $D$ are $(-a,-am)$ and the coordinates of $E$ are $(4a,4am)$.
Now, we can apply shoelace on $\triangle DBA$ and $\triangle ACE$:
$[DBA] = \frac {1}{2}\left|(0 \cdot -\frac{8}{5} + -\frac{4}{5} \cdot -am + -a \cdot 0) - (0 \cdot -\frac{4}{5} + -\frac{8}{5} \cdot -a + -am \cdot 0)\right| = \dfrac{\left|-\frac{4}{5}am + \frac{8}{5}a\right|}{2} = \dfrac{\frac{8}{5}a-\frac{4}{5}am}{2}$.
$[ACE] = \frac {1}{2}\left|(0 \cdot -\frac{8}{5} + \frac{16}{5} \cdot 4am + 4a \cdot 0) - (0 \cdot \frac{16}{5} + -\frac{8}{5} \cdot 4a + 4am \cdot 0)\right| = \dfrac{\left|-\frac{32}{5}a-\frac{64}{5}am\right|}{2} = \dfrac{\frac{32}{5}a+\frac{64}{5}am}{2}$.
Note on determining the absolute value: We can assume $-\frac{1}{2}<m<0$ because $\overline{CA}$ has a slope of $-\frac{1}{2}$ and the area of $\triangle ACE$ gets too big if $m>0$. So, we can also assume $a$ is positive.
Equating the two areas yields $\dfrac{\frac{8}{5}a-\frac{4}{5}am}{2} = \dfrac{\frac{32}{5}a+\frac{64}{5}am}{2} \implies -\frac{24}{5}a=\frac{68}{5}am \implies m=-\frac{6}{17}$.
We also know that $a=\dfrac{\left(\frac{8}{5}+\frac{6}{5}m\right)}{1+m^2}=\dfrac{\frac{8}{5}-\frac{36}{85}}{1+\frac{36}{289}}=\dfrac{\frac{100}{85}}{\frac{325}{289}}=\frac{68}{65}$.
Now, we can compute the area of $\triangle DBA$: $[DBA] = \dfrac{\frac{8}{5} \cdot \frac{68}{65} -\frac{4}{5} \cdot \frac{68}{65} \cdot \left(-\frac{6}{17}\right)}{2}=\dfrac{\frac{8\cdot68+4\cdot4\cdot6}{5\cdot65}}{2}=\frac{64}{65} \implies m+m=\boxed{129}$. | // Block 1
import graph;
size(300);
// This diagram was brought to you by ChatGPT!
// Define points
pair P = (0,0);
pair B = (0,-1);
pair C = (4,-1);
pair F = (4,0);
pair Q = (4,3);
pair A = (4/5, 3/5);
// Draw rectangle PBCF
draw(P--B--C--F--cycle);
// Draw right triangle PQF
draw(P--Q--F--P);
// Circles
draw(circle(P,1), dashed);
draw(circle(Q,4), dashed);
// Draw points
dot(P); dot(B); dot(C); dot(F); dot(Q); dot(A);
// Label points
label("$P$", P, SW);
label("$B$", B, S);
label("$C$", C, S);
label("$F$", F, E);
label("$Q$", Q, NE);
label("$A$", A, NW);
// Side length labels — rectangle
label("$1$", midpoint(P--B), W);
label("$4$", midpoint(B--C), S);
label("$1$", midpoint(C--F), E);
label("$4$", midpoint(P--F), N);
// Side length labels — triangle
label("$1$", midpoint(P--A), W);
label("$4$", midpoint(A--Q), W);
label("$3$", midpoint(Q--F), E);
label("$4$", midpoint(P--F), N);
// Block 2
import graph; size(300); // This diagram was brought to you by ChatGPT! // Define points pair P = (0,0); pair B = (0,-1); pair C = (4,-1); pair F = (4,0); pair Q = (4,3); pair A = (4/5, 3/5); // Draw rectangle PBCF draw(P--B--C--F--cycle); // Draw right triangle PQF draw(P--Q--F--P); // Circles draw(circle(P,1), dashed); draw(circle(Q,4), dashed); // Draw points dot(P); dot(B); dot(C); dot(F); dot(Q); dot(A); // Label points label("$P$", P, SW); label("$B$", B, S); label("$C$", C, S); label("$F$", F, E); label("$Q$", Q, NE); label("$A$", A, NW); // Side length labels — rectangle label("$1$", midpoint(P--B), W); label("$4$", midpoint(B--C), S); label("$1$", midpoint(C--F), E); label("$4$", midpoint(P--F), N); // Side length labels — triangle label("$1$", midpoint(P--A), W); label("$4$", midpoint(A--Q), W); label("$3$", midpoint(Q--F), E); label("$4$", midpoint(P--F), N); | [] |
192 | In $\triangle ABC$ let $I$ be the center of the inscribed circle, and let the bisector of $\angle ACB$ intersect $AB$ at $L$. The line through $C$ and $L$ intersects the circumscribed circle of $\triangle ABC$ at the two points $C$ and $D$. If $LI=2$ and $LD=3$, then $IC=\tfrac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p+q$. | 2016 AIME I Problem 6 | Suppose we label the angles as shown below.
As $\angle BCD$ and $\angle BAD$ intercept the same arc, we know that $\angle BAD=\gamma$. Similarly, $\angle ABD=\gamma$. Also, using $\triangle ICA$, we find $\angle CIA=180-\alpha-\gamma$. Therefore, $\angle AID=\alpha+\gamma=\angle DAI$, so $\triangle AID$ must be isosceles with $AD=ID=5$. Similarly, $BD=ID=5$. Then $\triangle DLB \sim \triangle ALC$, hence $\frac{AL}{AC} = \frac{3}{5}$. Also, $AI$ bisects $\angle LAC$, so by the Angle Bisector Theorem $\frac{CI}{IL} =\frac{AC}{AL}= \frac{5}{3}$. Thus $CI = \frac{10}{3}$, and the answer is $\boxed{013}$. | // Block 1
size(150);
import olympiad;
real c=8.1,a=5*(c+sqrt(c^2-64))/6,b=5*(c-sqrt(c^2-64))/6;
pair A=(0,0),B=(c,0),D=(c/2,-sqrt(25-(c/2)^2));
pair C=intersectionpoints(circle(A,b),circle(B,a))[0];
pair I=incenter(A,B,C);
pair L=extension(C,D,A,B);
dot(I^^A^^B^^C^^D);
draw(C--D);
path midangle(pair d,pair e,pair f) {return e--e+((f-e)/length(f-e)+(d-e)/length(d-e))/2;}
draw(A--B--D--cycle);
draw(circumcircle(A,B,D));
draw(A--C--B);
draw(A--I--B^^C--I);
draw(incircle(A,B,C));
label("$A$",A,SW,fontsize(8));
label("$B$",B,SE,fontsize(8));
label("$C$",C,N,fontsize(8));
label("$D$",D,S,fontsize(8));
label("$I$",I,NE,fontsize(8));
label("$L$",L,SW,fontsize(8));
label("$\alpha$",A,5*dir(midangle(C,A,I)),fontsize(8));
label("$\alpha$",A,5*dir(midangle(I,A,B)),fontsize(8));
label("$\beta$",B,12*dir(midangle(A,B,I)),fontsize(8));
label("$\beta$",B,12*dir(midangle(I,B,C)),fontsize(8));
label("$\gamma$",C,5*dir(midangle(A,C,I)),fontsize(8));
label("$\gamma$",C,5*dir(midangle(I,C,B)),fontsize(8));
// Block 2
size(150); import olympiad; real c=8.1,a=5*(c+sqrt(c^2-64))/6,b=5*(c-sqrt(c^2-64))/6; pair A=(0,0),B=(c,0),D=(c/2,-sqrt(25-(c/2)^2)); pair C=intersectionpoints(circle(A,b),circle(B,a))[0]; pair I=incenter(A,B,C); pair L=extension(C,D,A,B); dot(I^^A^^B^^C^^D); draw(C--D); path midangle(pair d,pair e,pair f) {return e--e+((f-e)/length(f-e)+(d-e)/length(d-e))/2;} draw(A--B--D--cycle); draw(circumcircle(A,B,D)); draw(A--C--B); draw(A--I--B^^C--I); draw(incircle(A,B,C)); label("$A$",A,SW,fontsize(8)); label("$B$",B,SE,fontsize(8)); label("$C$",C,N,fontsize(8)); label("$D$",D,S,fontsize(8)); label("$I$",I,NE,fontsize(8)); label("$L$",L,SW,fontsize(8)); label("$\alpha$",A,5*dir(midangle(C,A,I)),fontsize(8)); label("$\alpha$",A,5*dir(midangle(I,A,B)),fontsize(8)); label("$\beta$",B,12*dir(midangle(A,B,I)),fontsize(8)); label("$\beta$",B,12*dir(midangle(I,B,C)),fontsize(8)); label("$\gamma$",C,5*dir(midangle(A,C,I)),fontsize(8)); label("$\gamma$",C,5*dir(midangle(I,C,B)),fontsize(8)); | [] |
192 | In $\triangle ABC$ let $I$ be the center of the inscribed circle, and let the bisector of $\angle ACB$ intersect $AB$ at $L$. The line through $C$ and $L$ intersects the circumscribed circle of $\triangle ABC$ at the two points $C$ and $D$. If $LI=2$ and $LD=3$, then $IC=\tfrac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p+q$. | 2016 AIME I Problem 6 | Connect $D$ to $A$ and $D$ to $B$ to form quadrilateral $ACBD$. Since quadrilateral $ACBD$ is cyclic, we can apply Ptolemy's Theorem on the quadrilateral.
Denote the length of $BD$ and $AD$ as $z$ (they must be congruent, as $\angle ABD$ and $\angle DAB$ are both inscribed in arcs that have the same degree measure due to the angle bisector intersecting the circumcircle at $D$), and the lengths of $BC$, $AC$, $AB$, and $CI$ as $a,b,c, x$, respectively.
After applying Ptolemy's, one will get that:
\[z(a+b)=c(x+5)\]
Next, since $ACBD$ is cyclic, triangles $ALD$ and $CLB$ are similar, yielding the following equation once simplifications are made to the equation $\frac{AD}{CB}=\frac{AL}{BL}$, with the length of $BL$ written in terms of $a,b,c$ using the angle bisector theorem on triangle $ABC$:
\[zc=3(a+b)\]
Next, drawing in the bisector of $\angle BAC$ to the incenter $I$, and applying the angle bisector theorem, we have that:
\[cx=2(a+b)\]
Now, solving for $z$ in the second equation, and $x$ in the third equation and plugging them both back into the first equation, and making the substitution $w=\frac{a+b}{c}$, we get the quadratic equation:
\[3w^2-2w-5=0\]
Solving, we get $w=5/3$, which gives $z=5$ and $x=10/3$, when we rewrite the above equations in terms of $w$. Thus, our answer is $\boxed{013}$ and we're done.
-mathislife52 | // Block 1
size(150);
import olympiad;
real c=8.1,a=5*(c+sqrt(c^2-64))/6,b=5*(c-sqrt(c^2-64))/6;
pair A=(0,0),B=(c,0),D=(c/2,-sqrt(25-(c/2)^2));
pair C=intersectionpoints(circle(A,b),circle(B,a))[0];
pair I=incenter(A,B,C);
pair L=extension(C,D,A,B);
dot(I^^A^^B^^C^^D);
draw(C--D);
path midangle(pair d,pair e,pair f) {return e--e+((f-e)/length(f-e)+(d-e)/length(d-e))/2;}
draw(A--B--D--cycle);
draw(circumcircle(A,B,D));
draw(A--C--B);
draw(A--I--B^^C--I);
draw(incircle(A,B,C));
label("$A$",A,SW,fontsize(8));
label("$B$",B,SE,fontsize(8));
label("$C$",C,N,fontsize(8));
label("$D$",D,S,fontsize(8));
label("$I$",I,NE,fontsize(8));
label("$L$",L,SW,fontsize(8));
label("$\alpha$",A,5*dir(midangle(C,A,I)),fontsize(8));
label("$\alpha$",A,5*dir(midangle(I,A,B)),fontsize(8));
label("$\beta$",B,12*dir(midangle(A,B,I)),fontsize(8));
label("$\beta$",B,12*dir(midangle(I,B,C)),fontsize(8));
label("$\gamma$",C,5*dir(midangle(A,C,I)),fontsize(8));
label("$\gamma$",C,5*dir(midangle(I,C,B)),fontsize(8));
// Block 2
size(150); import olympiad; real c=8.1,a=5*(c+sqrt(c^2-64))/6,b=5*(c-sqrt(c^2-64))/6; pair A=(0,0),B=(c,0),D=(c/2,-sqrt(25-(c/2)^2)); pair C=intersectionpoints(circle(A,b),circle(B,a))[0]; pair I=incenter(A,B,C); pair L=extension(C,D,A,B); dot(I^^A^^B^^C^^D); draw(C--D); path midangle(pair d,pair e,pair f) {return e--e+((f-e)/length(f-e)+(d-e)/length(d-e))/2;} draw(A--B--D--cycle); draw(circumcircle(A,B,D)); draw(A--C--B); draw(A--I--B^^C--I); draw(incircle(A,B,C)); label("$A$",A,SW,fontsize(8)); label("$B$",B,SE,fontsize(8)); label("$C$",C,N,fontsize(8)); label("$D$",D,S,fontsize(8)); label("$I$",I,NE,fontsize(8)); label("$L$",L,SW,fontsize(8)); label("$\alpha$",A,5*dir(midangle(C,A,I)),fontsize(8)); label("$\alpha$",A,5*dir(midangle(I,A,B)),fontsize(8)); label("$\beta$",B,12*dir(midangle(A,B,I)),fontsize(8)); label("$\beta$",B,12*dir(midangle(I,B,C)),fontsize(8)); label("$\gamma$",C,5*dir(midangle(A,C,I)),fontsize(8)); label("$\gamma$",C,5*dir(midangle(I,C,B)),fontsize(8)); | [] |
193 | Triangle $ABC$ has $AB=40,AC=31,$ and $\sin{A}=\frac{1}{5}$. This triangle is inscribed in rectangle $AQRS$ with $B$ on $\overline{QR}$ and $C$ on $\overline{RS}$. Find the maximum possible area of $AQRS$. | 2016 AIME I Problem 9 | We start by drawing a diagram;
We know that $\sin \alpha = \frac{1}{5}$. Since $\sin \alpha = \cos (90- \alpha)$,
\[\cos (90- \alpha) = \frac{1}{5} \implies \cos (\beta + \gamma) = \frac{1}{5}\]
Using our angle sum identities, we expand this to $\cos \beta \cdot \cos \gamma - \sin \beta \cdot \sin \gamma = \frac{1}{5}$.
We can now use the right triangle definition of cosine and sine to rewrite this equation as;
\[\frac{l}{40} \cdot \frac{w}{31} - \frac{x}{40} \cdot \frac{y}{31} = \frac{1}{5} \implies lw- xy = 8 \cdot 31 \implies lw = xy + 31 \cdot 8\]
Hang on; $lw$ is the area we want to maximize! Therefore, to maximize this area we must maximize
$xy = 40 \sin \beta \cdot 31 \sin \gamma = 31 \cdot 40 \cdot \frac{1}{2}( \cos (\beta - \gamma) - \cos ( \beta + \gamma)) = 31 \cdot 20 \cdot (\cos(\beta-\gamma)-\frac{1}{5})$.
Since $\cos(\beta-\gamma)$ is the only variable component of this expression, to maximize the expression we must maximize $\cos(\beta-\gamma)$. The cosine function has a maximum value of 1, so our equation evaluates to $xy = 31 \cdot 20 \cdot (1-\frac{1}{5}) = 31 \cdot 20 \cdot \frac{4}{5} = 31 \cdot 16$ (Note that at this max value, since $\beta$ and $\gamma$ are both acute, $\beta-\gamma=0 \implies \beta=\gamma$).
Finally, $lw = xy + 31 \cdot 8 = 31 \cdot 16 + 31\cdot 8 = 31 \cdot 24 = \boxed{744}$
~KingRavi | // Block 1
size(400);
import olympiad;
import geometry;
pair A = (0, 20) ,B=(30,10) ,C=(15,0), Q=(30,20) ,R=(30,0), S=(0,0);
draw(A--B--C--cycle);
draw(A--Q);
draw(Q--R);
draw(R--S);
draw(S--A);
label("$A$", A, W);
label("$B$", B, E);
label("$C$", C, N);
label("$Q$", Q, E);
label("$R$", R, E);
label("$S$", S, W);
label("$w$", (-1,10));
label("$l$", (15,21));
label("$y$", (7.5,-1));
label("$x$", (31,15));
label("$31$",(7.5,10), E);
label("$40$",(15,15), N);
markangle(Label("$\alpha$", Relative(0.5)), n=1, C, A, B);
markangle(Label("$\beta$", Relative(0.5)), n=1, B, A, Q);
markangle(Label("$\gamma$", Relative(0.5)), n=1, S, A, C);
// Block 2
size(400); import olympiad; import geometry; pair A = (0, 20) ,B=(30,10) ,C=(15,0), Q=(30,20) ,R=(30,0), S=(0,0); draw(A--B--C--cycle); draw(A--Q); draw(Q--R); draw(R--S); draw(S--A); label("$A$", A, W); label("$B$", B, E); label("$C$", C, N); label("$Q$", Q, E); label("$R$", R, E); label("$S$", S, W); label("$w$", (-1,10)); label("$l$", (15,21)); label("$y$", (7.5,-1)); label("$x$", (31,15)); label("$31$",(7.5,10), E); label("$40$",(15,15), N); markangle(Label("$\alpha$", Relative(0.5)), n=1, C, A, B); markangle(Label("$\beta$", Relative(0.5)), n=1, B, A, Q); markangle(Label("$\gamma$", Relative(0.5)), n=1, S, A, C); | [] |
194 | Centered at each lattice point in the coordinate plane are a circle radius $\frac{1}{10}$ and a square with sides of length $\frac{1}{5}$ whose sides are parallel to the coordinate axes. The line segment from $(0,0)$ to $(1001, 429)$ intersects $m$ of the squares and $n$ of the circles. Find $m + n$. | 2016 AIME I Problem 14 | See if you can solve the problem with the following.
Solution to Solution 2
This is mostly a clarification to Solution 1, but let's take the diagram for the origin to $(7,3)$. We have the origin circle and square intersected, then two squares, then the circle and square at $(7,3)$. If we take the circle and square at the origin out of the diagram, we will be able to repeat the resulting segment (with its circles and squares) end to end from $(0,0)$ to $(1001,429)$, which forms the line we need without overlapping. Since $143$ of these segments are needed to do this, and $3$ squares and $1$ circle are intersected with each, there are $143 \cdot (3+1) = 572$ squares and circles intersected. Adding the circle and square that are intersected at the origin back into the picture, we get that there are $572+2=\boxed{574}$ squares and circles intersected in total.
Solution to Solution 2 without a diagram
This solution is a more systematic approach for finding when the line intersects the squares and circles. Because $1001 = 7*11*13$ and $429=3*11*13$, the slope of our line is $\frac{3}{7}$, and we only need to consider the line in the rectangle from the origin to $(7,3)$, and we can iterate the line $11*13=143$ times. First, we consider how to figure out if the line intersects a square. Given a lattice point $(x_1, y_1)$, we can think of representing a square centered at that lattice point as all points equal to $(x_1 \pm a, y_1 \pm b)$ s.t. $0 \leq a,b \leq \frac{1}{10}$. If the line $y = \frac{3}{7}x$ intersects the square, then we must have $\frac{y_1 + b}{x_1 + a} = \frac{3}{7}$. The line with the least slope that intersects the square intersects at the bottom right corner and the line with the greatest slope that intersects the square intersects at the top left corner; thus we must have that $\frac{3}{7}$ lies in between these slopes, or that $\frac{y_1-\frac{1}{10}}{x_1+\frac{1}{10}} \leq \frac{3}{7} \leq \frac{y_1+\frac{1}{10}}{x_1-\frac{1}{10}}$. Simplifying, $3x_1 - 1 \leq 7y_1 \leq 3x_1 + 1$. Because $y$ can only equal $0, 1, 2, 3$, we just do casework based on the values of $y$ and find that the points $(2, 1)$ and $(5, 2)$ are intersected just at the corner of the square and $(0, 0), (7, 3)$ are intersected through the center of the square. However, we disregard one of $(0, 0)$ and $(7, 3)$, WLOG $(0, 0)$, since we just use it in our count for the next of the 143 segments. Therefore, in one of our "segments", 3 squares are intersected and 1 circle is intersected giving 4 total. Thus our answer is $143*4 = 572$. HOWEVER, we cannot forget that we ignored $(0, 0)$, which contributes another square and circle to our count, making the final answer $572 + 2 = \boxed{574}$.
-Patrick4President | // Block 1
size(12cm);draw((0,0)--(7,3));draw(box((0,0),(7,3)),dotted);
for(int i=0;i<8;++i)for(int j=0; j<4; ++j){dot((i,j),linewidth(1));draw(box((i-.1,j-.1),(i+.1,j+.1)),linewidth(.5));draw(circle((i,j),.1),linewidth(.5));}
// Block 2
size(12cm);draw((0,0)--(7,3));draw(box((0,0),(7,3)),dotted); for(int i=0;i<8;++i)for(int j=0; j<4; ++j){dot((i,j),linewidth(1));draw(box((i-.1,j-.1),(i+.1,j+.1)),linewidth(.5));draw(circle((i,j),.1),linewidth(.5));} | [] |
195 | Circles $\omega_1$ and $\omega_2$ intersect at points $X$ and $Y$. Line $\ell$ is tangent to $\omega_1$ and $\omega_2$ at $A$ and $B$, respectively, with line $AB$ closer to point $X$ than to $Y$. Circle $\omega$ passes through $A$ and $B$ intersecting $\omega_1$ again at $D \neq A$ and intersecting $\omega_2$ again at $C \neq B$. The three points $C$, $Y$, $D$ are collinear, $XC = 67$, $XY = 47$, and $XD = 37$. Find $AB^2$. | 2016 AIME I Problem 15 | First, we note that as $\triangle XDY$ and $\triangle XYC$ have bases along the same line, $\frac{[\triangle XDY]}{[\triangle XYC]}=\frac{DY}{YC}$. We can also find the ratio of their areas using the circumradius area formula. If $R_1$ is the radius of $\omega_1$ and if $R_2$ is the radius of $\omega_2$, then
\[\frac{[\triangle XDY]}{[\triangle XYC]}=\frac{(37\cdot 47\cdot DY)/(4R_1)}{(47\cdot 67\cdot YC)/(4R_2)}=\frac{37\cdot DY\cdot R_2}{67\cdot YC\cdot R_1}.\]
Since we showed this to be $\frac{DY}{YC}$, we see that $\frac{R_2}{R_1}=\frac{67}{37}$.
We extend $AD$ and $BC$ to meet at point $P$, and we extend $AB$ and $CD$ to meet at point $Q$ as shown below.
As $ABCD$ is cyclic, we know that $\angle BCD=180-\angle DAB=\angle BAP$. But then as $AB$ is tangent to $\omega_2$ at $B$, we see that $\angle BCD=\angle ABY$. Therefore, $\angle ABY=\angle BAP$, and $BY\parallel PD$. A similar argument shows $AY\parallel PC$. These parallel lines show $\triangle PDC\sim\triangle ADY\sim\triangle BYC$. Also, we showed that $\frac{R_2}{R_1}=\frac{67}{37}$, so the ratio of similarity between $\triangle ADY$ and $\triangle BYC$ is $\frac{37}{67}$, or rather
\[\frac{AD}{BY}=\frac{DY}{YC}=\frac{YA}{CB}=\frac{37}{67}.\]
We can now use the parallel lines to find more similar triangles. As $\triangle AQD\sim \triangle BQY$, we know that
\[\frac{QA}{QB}=\frac{QD}{QY}=\frac{AD}{BY}=\frac{37}{67}.\]
Setting $QA=37x$, we see that $QB=67x$, hence $AB=30x$, and the problem simplifies to finding $30^2x^2$. Setting $QD=37^2y$, we also see that $QY=37\cdot 67y$, hence $DY=37\cdot 30y$. Also, as $\triangle AQY\sim \triangle BQC$, we find that
\[\frac{QY}{QC}=\frac{YA}{CB}=\frac{37}{67}.\]
As $QY=37\cdot 67y$, we see that $QC=67^2y$, hence $YC=67\cdot30y$.
Applying Power of a Point to point $Q$ with respect to $\omega_2$, we find
\[67^2x^2=37\cdot 67^3 y^2,\]
or $x^2=37\cdot 67 y^2$. We wish to find $AB^2=30^2x^2=30^2\cdot 37\cdot 67y^2$.
Applying Stewart's Theorem to $\triangle XDC$, we find
\[37^2\cdot (67\cdot 30y)+67^2\cdot(37\cdot 30y)=(67\cdot 30y)\cdot (37\cdot 30y)\cdot (104\cdot 30y)+47^2\cdot (104\cdot 30y).\]
We can cancel $30\cdot 104\cdot y$ from both sides, finding $37\cdot 67=30^2\cdot 67\cdot 37y^2+47^2$. Therefore,
\[AB^2=30^2\cdot 37\cdot 67y^2=37\cdot 67-47^2=\boxed{270}.\] | // Block 1
size(200);
import olympiad;
real R1=45,R2=67*R1/37;
real m1=sqrt(R1^2-23.5^2);
real m2=sqrt(R2^2-23.5^2);
pair o1=(0,0),o2=(m1+m2,0),x=(m1,23.5),y=(m1,-23.5);
draw(circle(o1,R1));
draw(circle(o2,R2));
pair q=(-R1/(R2-R1)*o2.x,0);
pair a=tangent(q,o1,R1,2);
pair b=tangent(q,o2,R2,2);
pair d=intersectionpoints(circle(o1,R1),q--y+15*(y-q))[0];
pair c=intersectionpoints(circle(o2,R2),q--y+15*(y-q))[1];
pair p=extension(a,d,b,c);
dot(q^^a^^b^^x^^y^^c^^d^^p);
draw(q--b^^q--c);
draw(p--d^^p--c^^x--y);
draw(a--y^^b--y);
draw(d--x--c);
label("$A$",a,NW,fontsize(8));
label("$B$",b,NE,fontsize(8));
label("$C$",c,SE,fontsize(8));
label("$D$",d,SW,fontsize(8));
label("$X$",x,2*WNW,fontsize(8));
label("$Y$",y,3*S,fontsize(8));
label("$P$",p,N,fontsize(8));
label("$Q$",q,W,fontsize(8));
// Block 2
size(200); import olympiad; real R1=45,R2=67*R1/37; real m1=sqrt(R1^2-23.5^2); real m2=sqrt(R2^2-23.5^2); pair o1=(0,0),o2=(m1+m2,0),x=(m1,23.5),y=(m1,-23.5); draw(circle(o1,R1)); draw(circle(o2,R2)); pair q=(-R1/(R2-R1)*o2.x,0); pair a=tangent(q,o1,R1,2); pair b=tangent(q,o2,R2,2); pair d=intersectionpoints(circle(o1,R1),q--y+15*(y-q))[0]; pair c=intersectionpoints(circle(o2,R2),q--y+15*(y-q))[1]; pair p=extension(a,d,b,c); dot(q^^a^^b^^x^^y^^c^^d^^p); draw(q--b^^q--c); draw(p--d^^p--c^^x--y); draw(a--y^^b--y); draw(d--x--c); label("$A$",a,NW,fontsize(8)); label("$B$",b,NE,fontsize(8)); label("$C$",c,SE,fontsize(8)); label("$D$",d,SW,fontsize(8)); label("$X$",x,2*WNW,fontsize(8)); label("$Y$",y,3*S,fontsize(8)); label("$P$",p,N,fontsize(8)); label("$Q$",q,W,fontsize(8)); | [] |
195 | Circles $\omega_1$ and $\omega_2$ intersect at points $X$ and $Y$. Line $\ell$ is tangent to $\omega_1$ and $\omega_2$ at $A$ and $B$, respectively, with line $AB$ closer to point $X$ than to $Y$. Circle $\omega$ passes through $A$ and $B$ intersecting $\omega_1$ again at $D \neq A$ and intersecting $\omega_2$ again at $C \neq B$. The three points $C$, $Y$, $D$ are collinear, $XC = 67$, $XY = 47$, and $XD = 37$. Find $AB^2$. | 2016 AIME I Problem 15 | First of all, since quadrilaterals $ADYX$ and $XYCB$ are cyclic, we can let $\angle DAX = \angle XYC = \theta$, and $\angle XYD = \angle CBX = 180 - \theta$, due to the properties of cyclic quadrilaterals. In addition, let $\angle BAX = x$ and $\angle ABX = y$. Thus, $\angle ADX = \angle AYX = x$ and $\angle XYB = \angle XCB = y$. Then, since quadrilateral $ABCD$ is cyclic as well, we have the following sums:
\[\theta + x +\angle XCY + y = 180^{\circ}\]
\[180^{\circ} - \theta + y + \angle XDY + x = 180^{\circ}\]
Cancelling out $180^{\circ}$ in the second equation and isolating $\theta$ yields $\theta = y + \angle XDY + x$. Substituting $\theta$ back into the first equation, we obtain
\[2x + 2y + \angle XCY + \angle XDY = 180^{\circ}\]
Since
\[x + y +\angle XAY + \angle XCY + \angle DAY = 180^{\circ}\]
\[x + y + \angle XDY + \angle XCY + \angle DAY = 180^{\circ}\]
we can then imply that $\angle DAY = x + y$. Similarly, $\angle YBC = x + y$. So then $\angle DXY = \angle YXC = x + y$, so since we know that $XY$ bisects $\angle DXC$, we can solve for $DY$ and $YC$ with Stewart’s Theorem. Let $DY = 37n$ and $YC = 67n$. Then
\[37n \cdot 67n \cdot 104n + 47^2 \cdot 104n = 37^2 \cdot 67n + 67^2 \cdot 37n\]
\[37n \cdot 67n + 47^2 = 37 \cdot 67\]
\[n^2 = \frac{270}{2479}\]
Now, since $\angle AYX = x$ and $\angle BYX = y$, $\angle AYB = x + y$. From there, let $\angle AYD = \alpha$ and $\angle BYC = \beta$. From angle chasing we can derive that $\angle YDX = \angle YAX = \beta - x$ and $\angle YCX = \angle YBX = \alpha - y$. From there, since $\angle ADX = x$, it is quite clear that $\angle ADY = \beta$, and $\angle YAB = \beta$ can be found similarly. From there, since $\angle ADY = \angle YAB = \angle BYC = \beta$ and $\angle DAY = \angle AYB = \angle YBC = x + y$, we have $AA$ similarity between $\triangle DAY$, $\triangle AYB$, and $\triangle YBC$. Therefore the length of $AY$ is the geometric mean of the lengths of $DA$ and $YB$ (from $\triangle DAY \sim \triangle AYB$). However, $\triangle DAY \sim \triangle AYB \sim \triangle YBC$ yields the proportion $\frac{AD}{DY} = \frac{YA}{AB} = \frac{BY}{YC}$; hence, the length of $AB$ is the geometric mean of the lengths of $DY$ and $YC$.
We can now simply use arithmetic to calculate $AB^2$.
\[AB^2 = DY \cdot YC\]
\[AB^2 = 37 \cdot 67 \cdot \frac{270}{2479}\]
\[AB^2 = \boxed{270}\]
-Solution by TheBoomBox77 | // Block 1
size(9cm);
import olympiad;
real R1=45,R2=67*R1/37;
real m1=sqrt(R1^2-23.5^2);
real m2=sqrt(R2^2-23.5^2);
pair o1=(0,0),o2=(m1+m2,0),x=(m1,23.5),y=(m1,-23.5);
draw(circle(o1,R1));
draw(circle(o2,R2));
pair q=(-R1/(R2-R1)*o2.x,0);
pair a=tangent(q,o1,R1,2);
pair b=tangent(q,o2,R2,2);
pair d=intersectionpoints(circle(o1,R1),q--y+15*(y-q))[0];
pair c=intersectionpoints(circle(o2,R2),q--y+15*(y-q))[1];
dot(a^^b^^x^^y^^c^^d);
draw(x--y);
draw(a--y^^b--y);
draw(d--x--c);
draw(a--b--c--d--cycle);
draw(x--a^^x--b);
label("$A$",a,NW,fontsize(9));
label("$B$",b,NE,fontsize(9));
label("$C$",c,SE,fontsize(9));
label("$D$",d,SW,fontsize(9));
label("$X$",x,2*N,fontsize(9));
label("$Y$",y,3*S,fontsize(9));
// Block 2
size(9cm); import olympiad; real R1=45,R2=67*R1/37; real m1=sqrt(R1^2-23.5^2); real m2=sqrt(R2^2-23.5^2); pair o1=(0,0),o2=(m1+m2,0),x=(m1,23.5),y=(m1,-23.5); draw(circle(o1,R1)); draw(circle(o2,R2)); pair q=(-R1/(R2-R1)*o2.x,0); pair a=tangent(q,o1,R1,2); pair b=tangent(q,o2,R2,2); pair d=intersectionpoints(circle(o1,R1),q--y+15*(y-q))[0]; pair c=intersectionpoints(circle(o2,R2),q--y+15*(y-q))[1]; dot(a^^b^^x^^y^^c^^d); draw(x--y); draw(a--y^^b--y); draw(d--x--c); draw(a--b--c--d--cycle); draw(x--a^^x--b); label("$A$",a,NW,fontsize(9)); label("$B$",b,NE,fontsize(9)); label("$C$",c,SE,fontsize(9)); label("$D$",d,SW,fontsize(9)); label("$X$",x,2*N,fontsize(9)); label("$Y$",y,3*S,fontsize(9)); | [] |
196 | Triangle $ABC_0$ has a right angle at $C_0$. Its side lengths are pairwise relatively prime positive integers, and its perimeter is $p$. Let $C_1$ be the foot of the altitude to $\overline{AB}$, and for $n \geq 2$, let $C_n$ be the foot of the altitude to $\overline{C_{n-2}B}$ in $\triangle C_{n-2}C_{n-1}B$. The sum $\sum_{n=2}^\infty C_{n-2}C_{n-1} = 6p$. Find $p$. | 2016 AIME II Problem 5 | Let $a = BC_0$, $b = AC_0$, and $c = AB$.
Note that the total length of the red segments in the figure above is equal to the length of the blue segment times $\frac{a+c}{b}$.
The desired sum is equal to the total length of the infinite path $C_0 C_1 C_2 C_3 \cdots$, shown in red in the figure below.
Since each of the triangles $\triangle C_0 C_1 C_2, \triangle C_2 C_3 C_4, \dots$ on the left are similar, it follows that the total length of the red segments in the figure below is equal to the length of the blue segment times $\frac{a+c}{b}$.
In other words, we have that $a\left(\frac{a+c}{b}\right) = 6p$.
Guessing and checking Pythagorean triples reveals that $a = 84$, $b=13$, $c = 85$, and $p = a + b + c = \boxed{182}$ satisfies this equation. | // Block 1
size(10cm);
// Setup
pair A, B;
pair C0, C1, C2, C3, C4, C5, C6, C7, C8;
A = (5, 0);
B = (0, 3);
C0 = (0, 0);
C1 = foot(C0, A, B);
C2 = foot(C1, C0, B);
C3 = foot(C2, C1, B);
C4 = foot(C3, C2, B);
C5 = foot(C4, C3, B);
C6 = foot(C5, C4, B);
C7 = foot(C6, C5, B);
C8 = foot(C7, C6, B);
// Labels
label("$A$", A, SE);
label("$B$", B, NW);
label("$C_0$", C0, SW);
label("$C_1$", C1, NE);
label("$C_2$", C2, W);
label("$C_3$", C3, NE);
label("$C_4$", C4, W);
label("$a$", (B+C0)/2, W);
label("$b$", (A+C0)/2, S);
label("$c$", (A+B)/2, NE);
// Drawings
draw(A--B--C0--cycle);
draw(C0--C1--C2, red);
draw(C2--C3--C4--C5--C6--C7--C8);
draw(C0--C2, blue);
// Block 2
size(10cm);
// Setup
pair A, B;
pair C0, C1, C2, C3, C4, C5, C6, C7, C8;
A = (5, 0);
B = (0, 3);
C0 = (0, 0);
C1 = foot(C0, A, B);
C2 = foot(C1, C0, B);
C3 = foot(C2, C1, B);
C4 = foot(C3, C2, B);
C5 = foot(C4, C3, B);
C6 = foot(C5, C4, B);
C7 = foot(C6, C5, B);
C8 = foot(C7, C6, B);
// Labels
label("$A$", A, SE);
label("$B$", B, NW);
label("$C_0$", C0, SW);
label("$a$", (B+C0)/2, W);
label("$b$", (A+C0)/2, S);
label("$c$", (A+B)/2, NE);
// Drawings
draw(A--B--C0--cycle);
draw(C0--C1--C2--C3--C4--C5--C6--C7--C8, red);
draw(C0--B, blue);
// Block 3
size(10cm); // Setup pair A, B; pair C0, C1, C2, C3, C4, C5, C6, C7, C8; A = (5, 0); B = (0, 3); C0 = (0, 0); C1 = foot(C0, A, B); C2 = foot(C1, C0, B); C3 = foot(C2, C1, B); C4 = foot(C3, C2, B); C5 = foot(C4, C3, B); C6 = foot(C5, C4, B); C7 = foot(C6, C5, B); C8 = foot(C7, C6, B); // Labels label("$A$", A, SE); label("$B$", B, NW); label("$C_0$", C0, SW); label("$C_1$", C1, NE); label("$C_2$", C2, W); label("$C_3$", C3, NE); label("$C_4$", C4, W); label("$a$", (B+C0)/2, W); label("$b$", (A+C0)/2, S); label("$c$", (A+B)/2, NE); // Drawings draw(A--B--C0--cycle); draw(C0--C1--C2, red); draw(C2--C3--C4--C5--C6--C7--C8); draw(C0--C2, blue);
// Block 4
size(10cm); // Setup pair A, B; pair C0, C1, C2, C3, C4, C5, C6, C7, C8; A = (5, 0); B = (0, 3); C0 = (0, 0); C1 = foot(C0, A, B); C2 = foot(C1, C0, B); C3 = foot(C2, C1, B); C4 = foot(C3, C2, B); C5 = foot(C4, C3, B); C6 = foot(C5, C4, B); C7 = foot(C6, C5, B); C8 = foot(C7, C6, B); // Labels label("$A$", A, SE); label("$B$", B, NW); label("$C_0$", C0, SW); label("$a$", (B+C0)/2, W); label("$b$", (A+C0)/2, S); label("$c$", (A+B)/2, NE); // Drawings draw(A--B--C0--cycle); draw(C0--C1--C2--C3--C4--C5--C6--C7--C8, red); draw(C0--B, blue); | [] |
197 | Squares $ABCD$ and $EFGH$ have a common center and $\overline{AB} || \overline{EF}$. The area of $ABCD$ is 2016, and the area of $EFGH$ is a smaller positive integer. Square $IJKL$ is constructed so that each of its vertices lies on a side of $ABCD$ and each vertex of $EFGH$ lies on a side of $IJKL$. Find the difference between the largest and smallest positive integer values for the area of $IJKL$. | 2016 AIME II Problem 7 | Letting $AI=a$ and $IB=b$, we have \[IJ^{2}=a^{2}+b^{2} \geq 1008\] by AM-GM inequality. Also, since $EFGH||ABCD$, the angles that each square cuts another are equal, so all the triangles are formed by a vertex of a larger square and $2$ adjacent vertices of a smaller square are similar. Therefore, the areas form a geometric progression, so since \[2016=12^{2} \cdot 14\] we have the maximum area is \[2016 \cdot \dfrac{11}{12} = 1848\] (the areas of the squares from largest to smallest are $12^{2} \cdot 14, 11 \cdot 12 \cdot 14, 11^{2} \cdot 14$ forming a geometric progression).
The minimum area is $1008$ (every square is half the area of the square whose sides its vertices touch), so the desired answer is \[1848-1008=\boxed{840}\] | // Block 1
pair A,B,C,D,E,F,G,H,I,J,K,L;
A=(0,0);
B=(2016,0);
C=(2016,2016);
D=(0,2016);
I=(1008,0);
J=(2016,1008);
K=(1008,2016);
L=(0,1008);
E=(504,504);
F=(1512,504);
G=(1512,1512);
H=(504,1512);
draw(A--B--C--D--A);
draw(I--J--K--L--I);
draw(E--F--G--H--E);
label("$A$",A,SW);
label("$B$",B,SE);
label("$C$",C,NE);
label("$D$",D,NW);
label("$E$",E,SW);
label("$F$",F,SE);
label("$G$",G,NE);
label("$H$",H,NW);
label("$I$",I,S);
label("$J$",J,NE);
label("$K$",K,N);
label("$L$",L,NW);
// Block 2
pair A,B,C,D,E,F,G,H,I,J,K,L; A=(0,0); B=(2016,0); C=(2016,2016); D=(0,2016); I=(1008,0); J=(2016,1008); K=(1008,2016); L=(0,1008); E=(504,504); F=(1512,504); G=(1512,1512); H=(504,1512); draw(A--B--C--D--A); draw(I--J--K--L--I); draw(E--F--G--H--E); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,NE); label("$D$",D,NW); label("$E$",E,SW); label("$F$",F,SE); label("$G$",G,NE); label("$H$",H,NW); label("$I$",I,S); label("$J$",J,NE); label("$K$",K,N); label("$L$",L,NW); | [] |
198 | Triangle $ABC$ is inscribed in circle $\omega$. Points $P$ and $Q$ are on side $\overline{AB}$ with $AP<AQ$. Rays $CP$ and $CQ$ meet $\omega$ again at $S$ and $T$ (other than $C$), respectively. If $AP=4,PQ=3,QB=6,BT=5,$ and $AS=7$, then $ST=\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. | 2016 AIME II Problem 10 | Let $\angle ACP=\alpha$, $\angle PCQ=\beta$, and $\angle QCB=\gamma$. Note that since $\triangle ACQ\sim\triangle TBQ$ we have $\tfrac{AC}{CQ}=\tfrac56$, so by the Ratio Lemma \[\dfrac{AP}{PQ}=\dfrac{AC}{CQ}\cdot\dfrac{\sin\alpha}{\sin\beta}\quad\implies\quad \dfrac{\sin\alpha}{\sin\beta}=\dfrac{24}{15}.\]Similarly, we can deduce $\tfrac{PC}{CB}=\tfrac47$ and hence $\tfrac{\sin\beta}{\sin\gamma}=\tfrac{21}{24}$.
Now Law of Sines on $\triangle ACS$, $\triangle SCT$, and $\triangle TCB$ yields \[\dfrac{AS}{\sin\alpha}=\dfrac{ST}{\sin\beta}=\dfrac{TB}{\sin\gamma}.\]Hence \[\dfrac{ST^2}{\sin^2\beta}=\dfrac{TB\cdot AS}{\sin\alpha\sin\gamma},\]so \[TS^2=TB\cdot AS\left(\dfrac{\sin\beta}{\sin\alpha}\dfrac{\sin\beta}{\sin\gamma}\right)=\dfrac{15\cdot 21}{24^2}\cdot 5\cdot 7=\dfrac{35^2}{8^2}.\]Hence $ST=\tfrac{35}8$ and the requested answer is $35+8=\boxed{43}$.
Edit: Note that the finish is much simpler. Once you get $\dfrac{AS}{\sin\alpha}=\dfrac{ST}{\sin\beta}$, you can solve quickly from there getting $ST=\dfrac{AS \sin(\beta)}{\sin(\alpha)}=7\cdot \dfrac{15}{24}=\dfrac{35}{8}$. | // Block 1
import cse5;
pathpen = black; pointpen = black;
pointfontsize = 9;
size(8cm);
pair A = origin, B = (13,0), P = (4,0), Q = (7,0),
T = B + 5 dir(220), C = IP(circumcircle(A,B,T),Line(T,Q,-0.1,10)),
S = IP(circumcircle(A,B,C),Line(C,P,-0.1,10));
Drawing(A--B--C--cycle);
D(circumcircle(A,B,C),rgb(0,0.6,1));
DrawPathArray(C--S^^C--T,rgb(1,0.4,0.1));
DrawPathArray(A--S^^B--T,rgb(0,0.4,0));
D(S--T,rgb(1,0.2,0.4));
D("A",A,dir(215));
D("B",B,dir(330));
D("P",P,dir(240));
D("Q",Q,dir(240));
D("T",T,dir(290));
D("C",C,dir(120));
D("S",S,dir(250));
MP("4",(A+P)/2,dir(90));
MP("3",(P+Q)/2,dir(90));
MP("6",(Q+B)/2,dir(90));
MP("5",(B+T)/2,dir(140));
MP("7",(A+S)/2,dir(40));
// Block 2
import cse5; pathpen = black; pointpen = black; pointfontsize = 9; size(8cm); pair A = origin, B = (13,0), P = (4,0), Q = (7,0), T = B + 5 dir(220), C = IP(circumcircle(A,B,T),Line(T,Q,-0.1,10)), S = IP(circumcircle(A,B,C),Line(C,P,-0.1,10)); Drawing(A--B--C--cycle); D(circumcircle(A,B,C),rgb(0,0.6,1)); DrawPathArray(C--S^^C--T,rgb(1,0.4,0.1)); DrawPathArray(A--S^^B--T,rgb(0,0.4,0)); D(S--T,rgb(1,0.2,0.4)); D("A",A,dir(215)); D("B",B,dir(330)); D("P",P,dir(240)); D("Q",Q,dir(240)); D("T",T,dir(290)); D("C",C,dir(120)); D("S",S,dir(250)); MP("4",(A+P)/2,dir(90)); MP("3",(P+Q)/2,dir(90)); MP("6",(Q+B)/2,dir(90)); MP("5",(B+T)/2,dir(140)); MP("7",(A+S)/2,dir(40)); | [] |
199 | The figure below shows a ring made of six small sections which you are to paint on a wall. You have four paint colors available and you will paint each of the six sections a solid color. Find the number of ways you can choose to paint the sections if no two adjacent sections can be painted with the same color. | 2016 AIME II Problem 12 | Let's number the regions $1,2,\dots 6$. Suppose we color regions $1,2,3$. Then, how many ways are there to color $4,5,6$?
Note: the numbers are numbered as shown:
$\textbf{Case 1:}$ The colors of $1,2,3$ are $BAB$, in that order.
Then the colors of $6,5,4$ can be $AXA$, $AXC$, $CXA$, $CXC$, or $CXD$ in that order, where $X$ is any color not equal to its surroundings. Then there are $4$ choices for $A$, $3$ choices for $B$ (it cannot be $A$), $2$ choices for $C$, and $1$ for $D$, the last color. So, summing up, we have \[4*3(1*3+2*2+2*2+2*3+2*1*2)=12*21=252\] colorings.
$\textbf{Case 2:}$ The colors of $1,2,3$ are $BAC$, in that order.
Again, we list out the possible arrangements of $6,5,4$: $AXA$, $AXB$, $CXA$, $AXD$, $DXA$, $CXB$, $CXD$, $DXB$, or $DXD$. (Easily simplified; listed here for clarity.) Then there are $4$ choices for $A$ as usual, $3$ choices for $B$, and so on. Hence we have \[4*3*2(1*3+1*2+1*2+1*2+1*2+1*2+1*2+1*2+1*3)=24*20=480\] colorings in this case.
Adding up, we have $252+480=\boxed{732}$ as our answer.
This solution was brought to you by LEONARD_MY_DUDE | // Block 1
draw(Circle((0,0), 4));
draw(Circle((0,0), 3));
draw((0,4)--(0,3));
draw((0,-4)--(0,-3));
draw((-2.598, 1.5)--(-3.4641, 2));
draw((-2.598, -1.5)--(-3.4641, -2));
draw((2.598, -1.5)--(3.4641, -2));
draw((2.598, 1.5)--(3.4641, 2));
label("1",(-3.5,0));
label("2",(-1.6,3));
label("3",(1.6,3));
label("4",(3.5,0));
label("5",(1.6,-3));
label("6",(-1.6,-3));
// Block 2
draw(Circle((0,0), 4)); draw(Circle((0,0), 3)); draw((0,4)--(0,3)); draw((0,-4)--(0,-3)); draw((-2.598, 1.5)--(-3.4641, 2)); draw((-2.598, -1.5)--(-3.4641, -2)); draw((2.598, -1.5)--(3.4641, -2)); draw((2.598, 1.5)--(3.4641, 2)); label("1",(-3.5,0)); label("2",(-1.6,3)); label("3",(1.6,3)); label("4",(3.5,0)); label("5",(1.6,-3)); label("6",(-1.6,-3)); | [] |
200 | Beatrix is going to place six rooks on a $6 \times 6$ chessboard where both the rows and columns are labeled $1$ to $6$; the rooks are placed so that no two rooks are in the same row or the same column. The $value$ of a square is the sum of its row number and column number. The $score$ of an arrangement of rooks is the least value of any occupied square. The average score over all valid configurations is $\frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p+q$. | 2016 AIME II Problem 13 | If the score is $n+1$, then one of the rooks must appear in the $n$th antidiagonal, and this is the first antidiagonal in which a rook can appear. To demonstrate this, we draw the following diagram when $n=4$.
We first count the number of arrangements that avoid the squares above the $n$th diagonal, and then we subtract from these the number of arrangements that avoid all squares above the $(n+1)$th diagonal. In the first column, there are $7-n$ rows in which to place the rook. In the second column, there is one more possible row, but one of the rows is used up by the rook in the first column, hence there are still $7-n$ places to place the rook. This pattern continues through the $n$th column, so there are $(7-n)^n$ ways to place the first $n$ rooks while avoiding the crossed out squares. We can similarly compute that there are $(6-n)^n$ ways to place the rooks in the first $n$ columns that avoid both the crossed out and shaded squares. Therefore, there are $(7-n)^n-(6-n)^n$ ways to place the first $n$ rooks such that at least one of them appears in a shaded square.
After this, there are $(6-n)$ rows and $(6-n)$ columns in which to place the remaining rooks, and we can do this in $(6-n)!$ ways. Hence the number of arrangements with a score of $n$ is $((7-n)^n-(6-n)^n)\cdot(6-n)!$. We also know that $n$ can range from from $1$ to $6$, so the average score is given by
\[\frac{2\cdot(6^1-5^1)\cdot5!+3\cdot(5^2-4^2)\cdot4!+4\cdot(4^3-3^3)\cdot3!+5\cdot(3^4-2^4)\cdot 2!+6\cdot(2^5-1^5)\cdot 1!+7\cdot(1^6-0^6)\cdot 0!}{6!}=\frac{291}{80}.\]
Thus the answer is $\boxed{371}$. | // Block 1
for (int i=0;i<7;++i) {draw((0,10*i)--(60,10*i));draw((10*i,0)--(10*i,60));}
path x=(1,1)--(9,9),y=(1,9)--(9,1);
for (int i=0;i<3;++i) {for (int j=3+i;j<6;++j) {draw(shift(10*i,10*j)*x,linewidth(1));draw(shift(10*i,10*j)*y,linewidth(1));}}
for (int i=0;i<4;++i) {filldraw((10*i,20+10*i)--(10*i+10,20+10*i)--(10*i+10,20+10*i+10)--(10*i,20+10*i+10)--cycle,lightgray);}
// Block 2
for (int i=0;i<7;++i) {draw((0,10*i)--(60,10*i));draw((10*i,0)--(10*i,60));} path x=(1,1)--(9,9),y=(1,9)--(9,1); for (int i=0;i<3;++i) {for (int j=3+i;j<6;++j) {draw(shift(10*i,10*j)*x,linewidth(1));draw(shift(10*i,10*j)*y,linewidth(1));}} for (int i=0;i<4;++i) {filldraw((10*i,20+10*i)--(10*i+10,20+10*i)--(10*i+10,20+10*i+10)--(10*i,20+10*i+10)--cycle,lightgray);} | [] |
201 | Equilateral $\triangle ABC$ has side length $600$. Points $P$ and $Q$ lie outside the plane of $\triangle ABC$ and are on opposite sides of the plane. Furthermore, $PA=PB=PC$, and $QA=QB=QC$, and the planes of $\triangle PAB$ and $\triangle QAB$ form a $120^{\circ}$ dihedral angle (the angle between the two planes). There is a point $O$ whose distance from each of $A,B,C,P,$ and $Q$ is $d$. Find $d$. | 2016 AIME II Problem 14 | To make numbers more feasible, we'll scale everything down by a factor of $100$ so that $\overline{AB}=\overline{BC}=\overline{AC}=6$. We should also note that $P$ and $Q$ must lie on the line that is perpendicular to the plane of $ABC$ and also passes through the circumcenter of $ABC$ (due to $P$ and $Q$ being equidistant from $A$, $B$, $C$), let $D$ be the altitude from $C$ to $AB$. We can draw a vertical cross-section of the figure then: We let $\angle PDI=\alpha$ so $\angle QDI=120^{\circ}-\alpha$, also note that $\overline{PO}=\overline{QO}=\overline{CO}=d$. Because $I$ is the centroid of $ABC$, we know that ratio of $\overline{CI}$ to $\overline{DI}$ is $2:1$. Since we've scaled the figure down, the length of $CD$ is $3\sqrt{3}$, from this it's easy to know that $\overline{CI}=2\sqrt{3}$ and $\overline{DI}=\sqrt{3}$. The following two equations arise: \begin{align} \sqrt{3}\tan{\left(\alpha\right)}+\sqrt{3}\tan{\left(120^{\circ}-\alpha\right)}&=2d \\ \sqrt{3}\tan{\left(\alpha\right)} - d &= \sqrt{d^{2}-12} \end{align} Using trig identities for the tangent, we find that \begin{align*} \sqrt{3}\tan{\left(120^{\circ}-\alpha\right)}&=\sqrt{3}\left(\frac{\tan{\left(120^{\circ}\right)}+\tan{\left(\text{-}\alpha\right)}}{1-\tan{\left(120^{\circ}\right)}\tan{\left(\text{-}\alpha\right)}}\right) \\ &= \sqrt{3}\left(\frac{\text{-}\sqrt{3}+\tan{\left(\text{-}\alpha\right)}}{1+\sqrt{3}\tan{\left(\text{-}\alpha\right)}}\right) \\ &= \sqrt{3}\left(\frac{\text{-}\sqrt{3}-\tan{\left(\alpha\right)}}{1-\sqrt{3}\tan{\left(\alpha\right)}}\right) \\ &= \frac{\sqrt{3}\tan{\left(\alpha\right)}+3}{\sqrt{3}\tan{\left(\alpha\right)}-1}.\end{align*} Okay, now we can plug this into $\left(1\right)$ to get: \begin{align}\sqrt{3}\tan{\left(\alpha\right)}+\frac{\sqrt{3}\tan{\left(\alpha\right)}+3}{\sqrt{3}\tan{\left(\alpha\right)}-1}&=2d \\ \sqrt{3}\tan{\left(\alpha\right)} - d &= \sqrt{d^{2}-12} \end{align} Notice that $\alpha$ only appears in the above system of equations in the form of $\sqrt{3}\tan{\left(\alpha\right)}$, we can set $\sqrt{3}\tan{\left(\alpha\right)}=a$ for convenience since we really only care about $d$. Now we have \begin{align}a+\frac{a+3}{a-1}&=2d \\ a - d &= \sqrt{d^{2}-12} \end{align} Looking at $\left(2\right)$, it's tempting to square it to get rid of the square-root so now we have: \begin{align*}a^{2}-2ad+d^{2}&=d^{2}-12 \\ a - 2ad &= \text{-}12 \end{align*} See the sneaky $2d$ in the above equation? That we means we can substitute it for $a+\frac{a+3}{a-1}$: \begin{align*}a^{2}-2ad+d^{2}&=d^{2}-12 \\ a^{2} - a\left(a+\frac{a+3}{a-1}\right) &= \text{-}12 \\ a^{2}-a^{2}-\frac{a^{2}+3a}{a-1} &=\text{-}12 \\ -\frac{a^{2}+3a}{a-1}&=\text{-}12 \\ \text{-}a^{2}-3a&=\text{-}12a+12 \\ 0 &= a^{2}-9a+12 \end{align*} Use the quadratic formula, we find that $a=\frac{9\pm\sqrt{9^{2}-4\left(1\right)\left(12\right)}}{2\left(1\right)}=\frac{9\pm\sqrt{33}}{2}$ - the two solutions were expected because $a$ can be $\angle PDI$ or $\angle QDI$. We can plug this into $\left(1\right)$: \begin{align*}a+\frac{a+3}{a-1}&=2d \\ \frac{9\pm\sqrt{33}}{2}+\frac{\frac{9\pm\sqrt{33}}{2}+3}{\frac{9\pm\sqrt{33}}{2}-1}=2d \\ \frac{9\pm\sqrt{33}}{2}+\frac{15\pm\sqrt{33}}{7\pm\sqrt{33}} &= 2d\end{align*} I'll use $a=\frac{9+\sqrt{33}}{2}$ because both values should give the same answer for $d$. \begin{align*} \frac{9+\sqrt{33}}{2}+\frac{15+\sqrt{33}}{7+\sqrt{33}} &= 2d \\ \frac{\left(9+\sqrt{33}\right)\left(7+\sqrt{33}\right)+\left(2\right)\left(15+\sqrt{33}\right)}{\left(2\right)\left(7+\sqrt{33}\right)} &= 2d \\ \frac{63+33+16\sqrt{33}+30+2\sqrt{33}}{14+2\sqrt{33}} &= 2d \\ \frac{126+18\sqrt{33}}{14+2\sqrt{33}} &= 2d \\ 9 &= 2d \\ \frac{9}{2} &= d\end{align*} Wait! Before you get excited, remember that we scaled the entire figure by $100$?? That means that the answer is $d=100\times\frac{9}{2}=\boxed{450}$.
-fatant | pair C, D, I, P, Q, O; D=(0,0); C=(5.196152,0); P=(1.732051,7.37228); I=(1.732051,0); Q=(1.732051,-1.62772); O=(1.732051,2.87228); draw(C--Q--D--P--cycle); draw(C--D, dashed); draw(P--Q, dotted); draw(O--C, dotted); label("$C$", C, E); label("$D$", D, W); label("$I$", I, NW); label("$P$", P, N); label("$Q$", Q, S); label("$O$", O, SW); dot(O); dot(I); | [] |
201 | Equilateral $\triangle ABC$ has side length $600$. Points $P$ and $Q$ lie outside the plane of $\triangle ABC$ and are on opposite sides of the plane. Furthermore, $PA=PB=PC$, and $QA=QB=QC$, and the planes of $\triangle PAB$ and $\triangle QAB$ form a $120^{\circ}$ dihedral angle (the angle between the two planes). There is a point $O$ whose distance from each of $A,B,C,P,$ and $Q$ is $d$. Find $d$. | 2016 AIME II Problem 14 | Let $M$ be the midpoint of $\overline{AB}$ and $X$ the center of $\triangle ABC$. Then \[P, O, Q, M, X, C\] all lie in the same vertical plane. We can make the following observations:
The equilateral triangle has side length $600$, so $MC=300\sqrt{3}$ and $X$ divides $MC$ so that $MX=100\sqrt{3}$ and $XC=200\sqrt{3}$;
$O$ is the midpoint of $PQ$ since $O$ is equidistant from $A, B, C, P, Q$ – it is also the circumcenter of $\triangle PCQ$;
$\angle PMQ=120^{\circ}$, the dihedral angle.
To make calculations easier, we will denote $100\sqrt{3}=m$, so that $MX=m$ and $XC=2m$.
Denote $PX=p$ and $QX=q$, where the tangent addition formula on $\triangle PMQ$ yields \[\frac{\tan\measuredangle PMX+\tan\measuredangle QMX}{1-\tan\measuredangle PMX\tan\measuredangle QMX}=\tan(120^{\circ})=-\sqrt{3}.\] Using $\tan\measuredangle PMX=\frac{p}{m}$ and $\tan\measuredangle QMX=\frac{q}{m}$, we have \[\frac{\frac{p}{m}+\frac{q}{m}}{1-\frac{p}{m}\cdot\frac{q}{m}}=-\sqrt{3}.\] After multiplying both numerator and denominator by $m^{2}$ we have \[\frac{(p+q)m}{m^{2}-pq}=-\sqrt{3}.\] But note that $pq=(2m)(2m)=4m^{2}$ by power of a point at $X$, where we deduce by symmetry that $MM^{\prime}=MX=m$ on the diagram below:
Thus \begin{align*} \frac{(p+q)m}{m^{2}-4m^{2}}=-\sqrt{3} \\ \frac{p+q}{-3m}=-\sqrt{3} \\ p+q=\left(-\sqrt{3}\right)\left(-3m\right) \\ p+q=3\sqrt{3}\cdot m.\end{align*} Earlier we assigned the variable $m$ to the length $100\sqrt{3}$ which implies $PQ=\left(3\sqrt{3}\right)\left(100\sqrt{3}\right)=900$. Thus the distance $d$ is equal to $\frac{PQ}{2}=\boxed{450}$. | // Block 1
unitsize(20);
pair P = (0, 12);
pair Q = (0, -3);
pair O = (P+Q)/2;
pair M = (-3, 0);
pair X = (0, 0);
pair C = (6, 0);
draw(P--O--Q);
draw(M--X--C);
draw(P--M--Q, blue);
draw(Q--C--P);
draw(circle((0, 4.5), 7.5));
label("$P$", P, N);
label("$Q$", Q, S);
label("$O$", O, E);
dot(O);
label("$M$", M, W);
label("$X$", X, NE);
label("$C$", C, E);
label("$m$", (M+X)/2, N);
label("$2m$", (X+C)/2, N);
// Block 2
unitsize(20);
pair P = (0, 12);
pair Q = (0, -3);
pair O = (P+Q)/2;
pair M = (-3, 0);
pair Mprime = (-6, 0);
pair X = (0, 0);
pair C = (6, 0);
draw(P--O--Q);
draw(Mprime--M--X--C);
draw(P--M--Q, blue);
draw(Q--C--P);
draw(circle((0, 4.5), 7.5));
label("$P$", P, N);
label("$Q$", Q, S);
label("$O$", O, E);
dot(O);
label("$M$", M, S);
label("$M^{\prime}$", Mprime, W);
label("$X$", X, SE);
label("$C$", C, E);
label("$m$", (Mprime+M)/2, N);
label("$m$", (M+X)/2, N);
label("$2m$", (X+C)/2, N);
// Block 3
unitsize(20); pair P = (0, 12); pair Q = (0, -3); pair O = (P+Q)/2; pair M = (-3, 0); pair X = (0, 0); pair C = (6, 0); draw(P--O--Q); draw(M--X--C); draw(P--M--Q, blue); draw(Q--C--P); draw(circle((0, 4.5), 7.5)); label("$P$", P, N); label("$Q$", Q, S); label("$O$", O, E); dot(O); label("$M$", M, W); label("$X$", X, NE); label("$C$", C, E); label("$m$", (M+X)/2, N); label("$2m$", (X+C)/2, N);
// Block 4
unitsize(20); pair P = (0, 12); pair Q = (0, -3); pair O = (P+Q)/2; pair M = (-3, 0); pair Mprime = (-6, 0); pair X = (0, 0); pair C = (6, 0); draw(P--O--Q); draw(Mprime--M--X--C); draw(P--M--Q, blue); draw(Q--C--P); draw(circle((0, 4.5), 7.5)); label("$P$", P, N); label("$Q$", Q, S); label("$O$", O, E); dot(O); label("$M$", M, S); label("$M^{\prime}$", Mprime, W); label("$X$", X, SE); label("$C$", C, E); label("$m$", (Mprime+M)/2, N); label("$m$", (M+X)/2, N); label("$2m$", (X+C)/2, N); | [] |
202 | Two real numbers $a$ and $b$ are chosen independently and uniformly at random from the interval $(0, 75)$. Let $O$ and $P$ be two points on the plane with $OP = 200$. Let $Q$ and $R$ be on the same side of line $OP$ such that the degree measures of $\angle POQ$ and $\angle POR$ are $a$ and $b$ respectively, and $\angle OQP$ and $\angle ORP$ are both right angles. The probability that $QR \leq 100$ is equal to $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$. | 2017 AIME I Problem 8 | Let $QR=x.$ Since we are given many angles in the problem, we can compute the lengths of some of the lines in terms of trigonometric functions: $OQ = 200 \cos a, PQ = 200 \sin a, PR = 200 \sin b, OR = 200 \cos b.$ Now observe that quadrilateral $OQRP$ is a cyclic quadrilateral. Thus, we are able to apply Ptolemy's Theorem to it:
\[200 x + (200 \cos a) (200 \sin b) = (200 \sin a) (200 \cos b),\]
\[x + 200 (\cos a \sin b) = 200 (\sin a \cos b),\]
\[x = 200(\sin a \cos b - \sin b \cos a),\]
\[x = 200 \sin(a-b).\]
We want $|x| \le 100$ (the absolute value comes from the fact that $a$ is not necessarily greater than $b,$ so we cannot assume that $Q$ is to the right of $R$ as in the diagram), so we substitute:
\[|200 \sin(a-b)| \le 100,\]
\[|\sin(a-b)| \le \frac{1}{2},\]
\[|a-b| \le 30 ^\circ,\]
\[-30 \le a-b \le 30.\]
By simple geometric probability (see Solution 2 for complete explanation), $\frac{m}{n} = 1 - \frac{2025}{5625} = 1 - \frac{9}{25} = \frac{16}{25},$ so $m+n = \boxed{041}.$
~burunduchok | // Block 1
pair O, P, Q, R;
draw(circle(O, 10));
O = (10, 0);
P = (-10, 0);
Q = (10*cos(pi/3), 10*sin(pi/3));
R = (10*cos(5*pi/6), 10*sin(5*pi/6));
dot(Q);
dot(O);
dot(P);
dot(R);
draw(P--O--Q--P--R--O);
draw(Q--R, red);
label("$O$", O, 2*E);
label("$P$", P, 2*W);
label("$Q$", Q, NE);
label("$R$", R, NW);
label("$200$", (0,0), 2*S);
label("$x$", (Q+R)/2, N);
draw(rightanglemark(O, Q, P, 38));
draw(rightanglemark(O, R, P, 38));
// Block 2
pair O, P, Q, R; draw(circle(O, 10)); O = (10, 0); P = (-10, 0); Q = (10*cos(pi/3), 10*sin(pi/3)); R = (10*cos(5*pi/6), 10*sin(5*pi/6)); dot(Q); dot(O); dot(P); dot(R); draw(P--O--Q--P--R--O); draw(Q--R, red); label("$O$", O, 2*E); label("$P$", P, 2*W); label("$Q$", Q, NE); label("$R$", R, NW); label("$200$", (0,0), 2*S); label("$x$", (Q+R)/2, N); draw(rightanglemark(O, Q, P, 38)); draw(rightanglemark(O, R, P, 38)); | [] |
203 | Let $z_1=18+83i,~z_2=18+39i,$ and $z_3=78+99i,$ where $i=\sqrt{-1}.$ Let $z$ be the unique complex number with the properties that $\frac{z_3-z_1}{z_2-z_1}~\cdot~\frac{z-z_2}{z-z_3}$ is a real number and the imaginary part of $z$ is the greatest possible. Find the real part of $z$. | 2017 AIME I Problem 10 | This problem is pretty obvious how to bash, and indeed many of the solutions below explain how to do that. But there’s no fun in that, and let’s see if we can come up with a slicker solution that will be more enjoyable.
Instead of thinking of complex numbers as purely a real plus a constant times $i$, let’s graph them and hope that the geometric visualization adds insight to the problem.
Note that when we subtract two vectors, the geometric result is the line segment between the two endpoints of the vectors. Thus we can fill in $z_3 - z_1, z_2 - z_1, z-z_2$ and $z-z_3$ as so;
$\angle Z_3Z_1Z_2$ looks similar to $\angle Z_3ZZ_2$, so let’s try to prove that they are congruent. We can show this in two ways;
Solution 1.1
Let’s look back at the information given to us. The problem states that $\frac{z_3-z_1}{z_2-z_1}~\cdot~\frac{z-z_2}{z-z_3}$ is a real number. Let the real number be some constant, $c$. Rearranging yields $\frac{z_3-z_1}{z_2-z_1} = c~\cdot~\frac{z-z_3}{z-z_2}$.
But how do we relate this expression to our angles? Well, let’s take a look at the divisions themselves.
The subtraction of two vectors yields a vector, and we can write any vector division as $\frac{V_1}{V_2}= X$ where $X$ is a complex number, as the division of two vectors also yields a vector. We can rewrite this as $V_1 = V_2 \cdot X$. We can think of this expression as transforming $V_2$ directly on to $V_1$, and $X$ is the transformation function. However, this transformation must be some kind of rotation, which means that the degree measure of $X$ is equal to the angle between the two vectors since we need to rotate by that angle to lay $V_2$ flat on $V_1$.
Thus we can rewrite our previous equation as $X_1 = c \cdot X_2$, where the angle of $X_1$ equals the angle between $z_3-z_1$ and $z_2-z_1$ and likewise for $X_2$. More precisely, we can write $X_1, X_2$ as $r_1 \cdot e^{i \theta_1}$ and $r_2 \cdot e^{i\theta_2}$, respectively by Euler’s formula. Then $\theta_1= \theta_2$ is the claim we wish to prove.
We can now do some simple algebra to prove this;
$r_1 \cdot e^{i \theta_1} = c \cdot r_2 \cdot e^{i\theta_2} \implies \frac{c \cdot r_2}{r_1} = e^{i(\theta_1-\theta_2)} = \cos{(\theta_1-\theta_2)} + i \cdot \sin{(\theta_1-\theta_2)}$
$\frac{c \cdot r_2}{r_1}$ is obviously real, so $\cos{(\theta_1-\theta_2)} + i \cdot \sin{(\theta_1-\theta_2)}$ must be real as well. But the only way that can happen is if $\sin{(\theta_1-\theta_2)} = 0 \implies \theta_1-\theta_2 = 0 \implies \boxed{\theta_1 = \theta_2}$.
Food For Thought: Let a, b, c, and d be pairwise distinct complex numbers. Then, a, b, c, and d are concyclic if and only if d-b/d-a : c-b/c-a is a real number. How can we use the above theorem to prove this?
Solution 1.2
Let us write $\frac{z_3 - z_1}{z_2 - z_1}$ as some complex number with form $r_1 (\cos \theta_1 + i \sin \theta_1).$ Similarly, we can write $\frac{z-z_2}{z-z_3}$ as some $r_2 (\cos \theta_2 + i \sin \theta_2).$
The product must be real, so we have that $r_1 r_2 (\cos \theta_1 + i \sin \theta_1) (\cos \theta_2 + i \sin \theta_2)$ is real. $r_1 r_2$ is real by definition, so dividing the real number above by $r_1 r_2$ will still yield a real number. (Note that we can see that $r_1 r_2 \not= 0$ from the definitions of $z_1,$ $z_2,$ and $z_3$). Thus we have
\[(\cos \theta_1 + i \sin \theta_1) (\cos \theta_2 + i \sin \theta_2) = \cos \theta_1 \cos \theta_2 - \sin \theta_1 \sin \theta_2 + i(\cos \theta_1 \sin \theta_2 + \cos \theta_2 \sin \theta_1)\]
is real. The imaginary part of this is $(\cos \theta_1 \sin \theta_2 + \cos \theta_2 \sin \theta_1),$ which we recognize as $\sin(\theta_1 + \theta_2).$ This is only $0$ when $\theta_1 + \theta_2 = k\pi$ for some integer $k$. Here $\theta_2$ represents the major angle$\angle Z_3ZZ_2$, and the angle we want is the minor angle, so we can rewrite the equation as $\theta_1 + 2\pi - \theta_2 = k\pi \implies \theta_1 - \theta_2 = (k-2)\pi$. We can see from the diagram that both $\theta_1$ and $\theta_2$ are obtuse, so therefore $\theta_1 - \theta_2 = 0 \implies \boxed{\theta_1 = \theta_2}$.
Solution 1 Rejoined
Now that we’ve proven this fact, we know that all four points lie on a circle (intuitively one can also observe this because $z=z_1$ and $z=z_2$ satisfy the property in the question, and $z=z_3$ techincally gives no imaginary part), so let’s draw that in;
While $Z_1, Z_2, Z_3$ are fixed, $Z$ can be anywhere on the circle because those are the only values of $Z$ that satisfy the problem requirements. However, we want to find the real part of the $Z$ with the maximum imaginary part. This Z would lie directly above the center of the circle, and thus the real part would be the same as the x-value of the center of the circle. So all we have to do is find this value and we’re done.
Consider the perpendicular bisectors of $Z_1Z_2$ and $Z_2Z_3$. Since any chord can be perpendicularly bisected by a radius of a circle, these two lines both intersect at the center. Since $Z_1Z_2$ is vertical, the perpendicular bisector will be horizontal and pass through the midpoint of this line, which is (18, 61). Therefore the equation for this line is $y=61$.
$Z_2Z_3$ is nice because it turns out the differences in the x and y values are both equal (60) which means that the slope of the line is 1. The slope of the perpendicular bisector is therefore -1 and it passes through the midpoint, (48,69), so the equation of this line is $y=-x+117$. Finally, equating the two yields \[61=-x+117 \implies x = \boxed{056}\]
~KingRavi
~Anonymous(Solution 1.2) | // Block 1
size(1200,300);
real xMin = 0;
real xMax = 100;
real yMin = 0;
real yMax = 120;
draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5));
draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5));
label("$R$",(xMax,0),(2,0));
label("$Im$",(0,yMax),(0,2));
pair A,B,C,D;
A = (18,83); B = (18,39); C = (78,99); D = (56, 104.908997);
dot(A);
dot(B);
dot(C);
dot(D);
label("$Z_1$", A, N);
label("$Z_2$", B, S);
label("$Z_3$", C, N);
label("$Z$", D, N);
// Block 2
size(1200,300);
real xMin = 0;
real xMax = 100;
real yMin = 0;
real yMax = 120;
draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5));
draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5));
label("$R$",(xMax,0),(2,0));
label("$Im$",(0,yMax),(0,2));
pair A,B,C,D;
A = (18,83); B = (18,39); C = (78,99); D = (56, 104.908997);
dot(A);
dot(B);
dot(C);
dot(D);
draw(A--C);
draw(A--B);
draw(D--C);
draw(D--B);
label("$Z_1$", A, N);
label("$Z_2$", B, S);
label("$Z_3$", C, N);
label("$Z$", D, N);
// Block 3
size(1200,300); real xMin = 0; real xMax = 100; real yMin = 0; real yMax = 120; draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); label("$R$",(xMax,0),(2,0)); label("$Im$",(0,yMax),(0,2)); pair A,B,C,D; A = (18,83); B = (18,39); C = (78,99); D = (56, 104.908997); dot(A); dot(B); dot(C); dot(D); label("$Z_1$", A, N); label("$Z_2$", B, S); label("$Z_3$", C, N); label("$Z$", D, N);
// Block 4
size(1200,300); real xMin = 0; real xMax = 100; real yMin = 0; real yMax = 120; draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); label("$R$",(xMax,0),(2,0)); label("$Im$",(0,yMax),(0,2)); pair A,B,C,D; A = (18,83); B = (18,39); C = (78,99); D = (56, 104.908997); dot(A); dot(B); dot(C); dot(D); draw(A--C); draw(A--B); draw(D--C); draw(D--B); label("$Z_1$", A, N); label("$Z_2$", B, S); label("$Z_3$", C, N); label("$Z$", D, N);
// Block 5
size(1200,300); real xMin = 0; real xMax = 100; real yMin = 0; real yMax = 120; draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); label("$R$",(xMax,0),(2,0)); label("$Im$",(0,yMax),(0,2)); pair A,B,C,D; A = (18,83); B = (18,39); C = (78,99); D = (56,104.908997); dot(A); dot(B); dot(C); dot(D); draw(A--C); draw(A--B); draw(D--C); draw(C--B); draw(D--B); label("$Z_1$", A, W); label("$Z_2$", B, W); label("$Z_3$", C, N); label("$Z$", D, N); draw(circle((56,61), 43.9089968)); | [] |
204 | A triangle has vertices $A(0,0)$, $B(12,0)$, and $C(8,10)$. The probability that a randomly chosen point inside the triangle is closer to vertex $B$ than to either vertex $A$ or vertex $C$ can be written as $\frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p+q$. | 2017 AIME II Problem 3 | The set of all points closer to point $B$ than to point $A$ lie to the right of the perpendicular bisector of $AB$ (line $PZ$ in the diagram), and the set of all points closer to point $B$ than to point $C$ lie below the perpendicular bisector of $BC$ (line $PX$ in the diagram). Therefore, the set of points inside the triangle that are closer to $B$ than to either vertex $A$ or vertex $C$ is bounded by quadrilateral $BXPZ$. Because $X$ is the midpoint of $BC$ and $Z$ is the midpoint of $AB$, $X=(10,5)$ and $Z=(6,0)$. The coordinates of point $P$ is the solution to the system of equations defined by lines $PX$ and $PZ$. Using the point-slope form of a linear equation and the fact that the slope of the line perpendicular to a line with slope $m$ is $-\frac{1}{m}$, the equation for line $PX$ is $y=\frac{2}{5}x+1$ and the equation for line $PZ$ is $x=6$. The solution of this system is $P=\left(6,\frac{17}{5}\right)$. Using the shoelace formula on quadrilateral $BXPZ$ and triangle $ABC$, the area of quadrilateral $BXPZ$ is $\frac{109}{5}$ and the area of triangle $ABC$ is $60$. Finally, the probability that a randomly chosen point inside the triangle is closer to vertex $B$ than to vertex $A$ or vertex $C$ is the ratio of the area of quadrilateral $BXPZ$ to the area of $ABC$, which is $\frac{\frac{109}{5}}{60}=\frac{109}{300}$. The answer is $109+300=\boxed{409}$. | // Block 1
pair A,B,C,D,X,Z,P;
A=(0,0); B=(12,0); C=(8,10); X=(10,5); Z=(6,0); P=(6,3.4);
fill(B--X--P--Z--cycle,lightgray);
draw(A--B--C--cycle);
dot(A);
label("$A$",A,SW);
dot(B);
label("$B$",B,SE);
dot(C);
label("$C$",C,N);
draw(X--P,dashed);
draw(Z--P,dashed);
dot(X);
label("$X$",X,NE);
dot(Z);
label("$Z$",Z,S);
dot(P);
label("$P$",P,NW);
// Block 2
pair A,B,C,D,X,Z,P; A=(0,0); B=(12,0); C=(8,10); X=(10,5); Z=(6,0); P=(6,3.4); fill(B--X--P--Z--cycle,lightgray); draw(A--B--C--cycle); dot(A); label("$A$",A,SW); dot(B); label("$B$",B,SE); dot(C); label("$C$",C,N); draw(X--P,dashed); draw(Z--P,dashed); dot(X); label("$X$",X,NE); dot(Z); label("$Z$",Z,S); dot(P); label("$P$",P,NW); | [] |
204 | A triangle has vertices $A(0,0)$, $B(12,0)$, and $C(8,10)$. The probability that a randomly chosen point inside the triangle is closer to vertex $B$ than to either vertex $A$ or vertex $C$ can be written as $\frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p+q$. | 2017 AIME II Problem 3 | Since we know the coordinates of all three vertices of the triangle, we can find the side lengths: $AB=12$, $AC=2\sqrt{41}$, and $BC=2\sqrt{29}$. We notice that the point where the three distances are the same is the circumcenter - so we use one of the triangle area formulas to find the circumradius, since we know what the area is.
\[\frac{12 \cdot 2\sqrt{41} \cdot 2\sqrt{29}}{4 \cdot R}=\frac{12 \cdot 10}{2}.\]
We rearrange to get
\[R=\frac{\sqrt{41} \cdot \sqrt{29}}{5}.\]
We know that $AP=\frac{\sqrt{41} \cdot \sqrt{29}}{5}$, and $AS=6$, so using the Pythagorean Theorem gives $SP=\frac{17}{5}$. This means $[ASP]=[BSP]=\frac{17}{5} \cdot 6 \cdot \frac{1}{2} = \frac{51}{5}$.
Similarly, we know that $BP=\frac{\sqrt{41} \cdot \sqrt{29}}{5}$, and $BR=\sqrt{29}$, so we get that $PR=\frac{4\sqrt{29}}{5}$, and so $[BRP]=[CRP]=\frac{4\sqrt{29}}{5} \cdot \sqrt{29} \cdot \frac{1}{2} = \frac{58}{5}$.
Lastly, we know that $CP=\frac{\sqrt{41} \cdot \sqrt{29}}{5}$, and $CT=\sqrt{41}$, so we get that $PT=\frac{2\sqrt{41}}{5}$, and $[ATP]=[CTP]=\frac{2\sqrt{41}}{5} \cdot \sqrt{41} \cdot \frac{1}{2} = \frac{41}{5}$.
Therefore, our answer is $\frac{51+58}{2(51+58+41)}=\frac{109}{300} \rightarrow \boxed{409}$. | // Block 1
draw((0,0)--(12,0)--(8,10)--(0,0));
draw((6,0)--(6,3.4)--(10,5));
draw((6,3.4)--(4,5));
label("$A$", (0,0), SW);
label("$B$", (12,0), SE);
label("$C$", (8, 10), N);
label("$P$", (6, 3.4), NNE);
label("$R$", (10, 5), NE);
label("$S$", (6, 0), S);
label("$T$", (4, 5), NW);
// Block 2
draw((0,0)--(12,0)--(8,10)--(0,0)); draw((6,0)--(6,3.4)--(10,5)); draw((6,3.4)--(4,5)); label("$A$", (0,0), SW); label("$B$", (12,0), SE); label("$C$", (8, 10), N); label("$P$", (6, 3.4), NNE); label("$R$", (10, 5), NE); label("$S$", (6, 0), S); label("$T$", (4, 5), NW); | [] |
205 | Find the number of integer values of $k$ in the closed interval $[-500,500]$ for which the equation $\log(kx)=2\log(x+2)$ has exactly one real solution. | 2017 AIME II Problem 7 | Note the equation $\log(kx)=2\log(x+2)$ is valid for $kx>0$ and $x>-2$.
Now, we simplify the given expression:
$\log(kx)=2\log(x+2)=\log((x+2)^2) \implies kx=(x+2)^2$. The last equation is derived by taking away the outside logs from the previous equation.
Because $(x+2)^2$ is always non-negative, $kx$ must also be non-negative; therefore this takes care of the $kx>0$ condition as long as $k\neq0$, i.e. $k$ cannot be $0$. Now, we graph both $(x+2)^2$ (the green graph) and $kx$ (the red graph for $k=-1,k=-2,k=-3,k=8$) for $x>-2$. It is easy to see that all negative values of $k$ make the equation $\log(kx)=2\log(x+2)$ have only one solution.
However, there is also one positive value of $k$ that makes the equation only have one solution, as shown by the steepest line in the diagram. We can show that the slope of this line is a positive integer by setting the discriminant of the equation $(x+2)^2=kx$ to be $0$ and solving for $k$, which yields $k=8$.
When $1\leq k \leq 7$, $y=kx$ will intersect $y(x+2)^2$ twice, and when $k \geq 9$, $y=kx$ will never intersect $y(x+2)^2$.
Therefore, there are $500$ negative solutions and $1$ positive solution, for a total of $\boxed{501}$. | // Block 1
Label f;
f.p=fontsize(5);
xaxis(-3,3,Ticks(f,1.0));
yaxis(-3,26,Ticks(f,1.0));
real f(real x){return (x+2)^2;}
real g(real x){return x*-1;}
real h(real x){return x*-2;}
real i(real x){return x*-3;}
real j(real x){return x*8;}
draw(graph(f,-2,3),green);
draw(graph(g,-2,2),red);
draw(graph(h,-2,1),red);
draw(graph(i,-2,1/3),red);
draw(graph(j,-0.25,3),red);
// Block 2
Label f; f.p=fontsize(5); xaxis(-3,3,Ticks(f,1.0)); yaxis(-3,26,Ticks(f,1.0)); real f(real x){return (x+2)^2;} real g(real x){return x*-1;} real h(real x){return x*-2;} real i(real x){return x*-3;} real j(real x){return x*8;} draw(graph(f,-2,3),green); draw(graph(g,-2,2),red); draw(graph(h,-2,1),red); draw(graph(i,-2,1/3),red); draw(graph(j,-0.25,3),red); | [] |
206 | Rectangle $ABCD$ has side lengths $AB=84$ and $AD=42$. Point $M$ is the midpoint of $\overline{AD}$, point $N$ is the trisection point of $\overline{AB}$ closer to $A$, and point $O$ is the intersection of $\overline{CM}$ and $\overline{DN}$. Point $P$ lies on the quadrilateral $BCON$, and $\overline{BP}$ bisects the area of $BCON$. Find the area of $\triangle CDP$. | 2017 AIME II Problem 10 | Impose a coordinate system on the diagram where point $D$ is the origin. Therefore $A=(0,42)$, $B=(84,42)$, $C=(84,0)$, and $D=(0,0)$. Because $M$ is a midpoint and $N$ is a trisection point, $M=(0,21)$ and $N=(28,42)$. The equation for line $DN$ is $y=\frac{3}{2}x$ and the equation for line $CM$ is $\frac{1}{84}x+\frac{1}{21}y=1$, so their intersection, point $O$, is $(12,18)$. Using the shoelace formula on quadrilateral $BCON$, or drawing diagonal $\overline{BO}$ and using $\frac12bh$, we find that its area is $2184$. Therefore the area of triangle $BCP$ is $\frac{2184}{2} = 1092$. Using $A = \frac 12 bh$, we get $2184 = 42h$. Simplifying, we get $h = 52$. This means that the x-coordinate of $P = 84 - 52 = 32$. Since P lies on $\frac{1}{84}x+\frac{1}{21}y=1$, you can solve and get that the y-coordinate of $P$ is $13$. Therefore the area of $CDP$ is $\frac{1}{2} \cdot 13 \cdot 84=\boxed{546}$. | // Block 1
pair A,B,C,D,M,n,O,P;
A=(0,42);B=(84,42);C=(84,0);D=(0,0);M=(0,21);n=(28,42);O=(12,18);P=(32,13);
fill(C--D--P--cycle,blue);
draw(A--B--C--D--cycle);
draw(C--M);
draw(D--n);
draw(B--P);
draw(D--P);
label("$A$",A,NW);
label("$B$",B,NE);
label("$C$",C,SE);
label("$D$",D,SW);
label("$M$",M,W);
label("$N$",n,N);
label("$O$",O,(-0.5,1));
label("$P$",P,N);
dot(A);
dot(B);
dot(C);
dot(D);
dot(M);
dot(n);
dot(O);
dot(P);
label("28",(0,42)--(28,42),N);
label("56",(28,42)--(84,42),N);
// Block 2
pair A,B,C,D,M,n,O,P; A=(0,42);B=(84,42);C=(84,0);D=(0,0);M=(0,21);n=(28,42);O=(12,18);P=(32,13); fill(C--D--P--cycle,blue); draw(A--B--C--D--cycle); draw(C--M); draw(D--n); draw(B--P); draw(D--P); label("$A$",A,NW); label("$B$",B,NE); label("$C$",C,SE); label("$D$",D,SW); label("$M$",M,W); label("$N$",n,N); label("$O$",O,(-0.5,1)); label("$P$",P,N); dot(A); dot(B); dot(C); dot(D); dot(M); dot(n); dot(O); dot(P); label("28",(0,42)--(28,42),N); label("56",(28,42)--(84,42),N); | [] |
207 | Circle $C_0$ has radius $1$, and the point $A_0$ is a point on the circle. Circle $C_1$ has radius $r<1$ and is internally tangent to $C_0$ at point $A_0$. Point $A_1$ lies on circle $C_1$ so that $A_1$ is located $90^{\circ}$ counterclockwise from $A_0$ on $C_1$. Circle $C_2$ has radius $r^2$ and is internally tangent to $C_1$ at point $A_1$. In this way a sequence of circles $C_1,C_2,C_3,\ldots$ and a sequence of points on the circles $A_1,A_2,A_3,\ldots$ are constructed, where circle $C_n$ has radius $r^n$ and is internally tangent to circle $C_{n-1}$ at point $A_{n-1}$, and point $A_n$ lies on $C_n$ $90^{\circ}$ counterclockwise from point $A_{n-1}$, as shown in the figure below. There is one point $B$ inside all of these circles. When $r = \frac{11}{60}$, the distance from the center $C_0$ to $B$ is $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. | 2017 AIME II Problem 12 | Using the invariance again as in Solution 3, assume $B$ is $d$ away from the origin. The locus of possible points is a circle with radius $d$. Consider the following diagram.
Let the distance from $B$ to $(1,0)$ be $x$. As $B$ is invariant, $x = r(BB' + x) \implies x = r\frac{d\sqrt{2}}{1-r}$. Then by Power of a Point, $x(BB' + x) = (1-d)(1+d) \implies xr(BB' + x) = r(1-d)(1+d) \implies x^2 = r(1-d^2) \implies d^2 = \left(1 + \frac{2r}{(1-r)^2}\right)$. Solving, $d = \frac{49}{61}$. Our answer is $49+61=\boxed{110}$
-Isogonal | // Block 1
size(7cm);
draw(circle((0,0), 49/61));
draw((0,0)--(0.790110185, 0.144853534));
draw((0,0)--(-0.144853534, 0.790110185));
draw((-0.144853534, 0.790110185)--(1,0));
draw((0,0)--(1,0));
draw(rightanglemark((-0.144853534, 0.790110185), (0,0), (0.790110185, 0.144853534), 3));
label("$O$", (0,0), SW);
label("$(1,0)$", (1,0), E);
label("$B$", (0.790110185, 0.144853534), NE);
label("$B'$", (-0.144853534, 0.790110185), N);
label("$d$", (0.5 * 49/61, 0), S);
// Block 2
size(7cm); draw(circle((0,0), 49/61)); draw((0,0)--(0.790110185, 0.144853534)); draw((0,0)--(-0.144853534, 0.790110185)); draw((-0.144853534, 0.790110185)--(1,0)); draw((0,0)--(1,0)); draw(rightanglemark((-0.144853534, 0.790110185), (0,0), (0.790110185, 0.144853534), 3)); label("$O$", (0,0), SW); label("$(1,0)$", (1,0), E); label("$B$", (0.790110185, 0.144853534), NE); label("$B'$", (-0.144853534, 0.790110185), N); label("$d$", (0.5 * 49/61, 0), S); | [] |
208 | A $10\times10\times10$ grid of points consists of all points in space of the form $(i,j,k)$, where $i$, $j$, and $k$ are integers between $1$ and $10$, inclusive. Find the number of different lines that contain exactly $8$ of these points. | 2017 AIME II Problem 14 | A line through the point $(a,b,c)$ must contain $(a \pm 1, b \pm 1, c \pm 1)$ on it as well, as otherwise, the line would not pass through more than 5 points. This corresponds to the 4 diagonals of the cube. With this information, we proceed with casework:
$\textbf{Case 1}:$ The lines are not parallel to the faces
We look at the one from $(1,1,1)$ to $(10,10,10)$. The lower endpoint of the desired lines must contain both a 1 and a 3, so it can be $(1,1,3), (1,2,3), (1,3,3)$. If $\textrm{min} > 0$ then the point $(a-1,b-1,c-1)$ will also be on the line for example, 3 applies to the other end.
Accounting for permutations, there are $12$ ways, so there are $12 \cdot 4 = 48$ different lines for this case.
A nice visualization for this case is the following:
The black vertex is the corner $(1,1,1)$. If we draw a line straight down through this point (black) you will hit $10$ points, if we draw a line straight down through the blue points you will hit $9$ points, and if we draw a line straight down through the red points you will hit $8$ points. Thus, for each pair of opposite vertices, we have $12$ lines corresponding to the $12$ red dots of length $8$, for a total of $12\cdot4=48$ points.
$\textbf{Case 2}:$ The lines where the $x$, $y$, or $z$ is the same for all the points on the line.
WLOG, let the $x$ value stay the same throughout. Let the line be parallel to the diagonal from $(1,1,1)$ to $(1,10,10)$. For the line to have 8 points, the $y$ and $z$ must be 1 and 3 in either order, and the $x$ value can be any value from 1 to 10. In addition, this line can be parallel to 6 face diagonals. So we get $2 \cdot 10 \cdot 6 = 120$ possible lines for this case.
The answer is, therefore, $120 + 48 = \boxed{168}$
Solution by stephcurry added to the wiki by Thedoge edited by Rapurt9 and phoenixfire | // Block 1
import graph;
size(8cm);
// This diagram was brought to you by ChatGPT!!!
// Settings for dots
real dotSize = 5pt;
// Ring 1: 1 Point (Black)
dot((0,0), black + dotSize);
// Ring 2: 6 Points (Blue)
// Angles: 30, 90, 150, 210, 270, 330
for(int i=0; i<6; ++i) {
dot(dir(30 + 60*i), blue + dotSize);
}
// Ring 3: 12 Points (Red)
// 6 Vertices at distance 2
for(int i=0; i<6; ++i) {
dot(2*dir(30 + 60*i), red + dotSize);
}
// 6 Mid-points at distance sqrt(3)
for(int i=0; i<6; ++i) {
dot(sqrt(3)*dir(60*i), red + dotSize);
}
// Cube Lines
// Line 1: y = (sqrt(3)/3)x for x > 0 (30 degrees)
draw((0,0) -- 2*dir(30), black + linewidth(1pt));
// Line 2: y = (-sqrt(3)/3)x for x < 0 (150 degrees)
draw((0,0) -- 2*dir(150), black + linewidth(1pt));
// Line 3: y-axis for y < 0 (270 degrees)
draw((0,0) -- 2*dir(270), black + linewidth(1pt));
// Block 2
import graph; size(8cm); // This diagram was brought to you by ChatGPT!!! // Settings for dots real dotSize = 5pt; // Ring 1: 1 Point (Black) dot((0,0), black + dotSize); // Ring 2: 6 Points (Blue) // Angles: 30, 90, 150, 210, 270, 330 for(int i=0; i<6; ++i) { dot(dir(30 + 60*i), blue + dotSize); } // Ring 3: 12 Points (Red) // 6 Vertices at distance 2 for(int i=0; i<6; ++i) { dot(2*dir(30 + 60*i), red + dotSize); } // 6 Mid-points at distance sqrt(3) for(int i=0; i<6; ++i) { dot(sqrt(3)*dir(60*i), red + dotSize); } // Cube Lines // Line 1: y = (sqrt(3)/3)x for x > 0 (30 degrees) draw((0,0) -- 2*dir(30), black + linewidth(1pt)); // Line 2: y = (-sqrt(3)/3)x for x < 0 (150 degrees) draw((0,0) -- 2*dir(150), black + linewidth(1pt)); // Line 3: y-axis for y < 0 (270 degrees) draw((0,0) -- 2*dir(270), black + linewidth(1pt)); | [] |
209 | In $\triangle ABC, AB = AC = 10$ and $BC = 12$. Point $D$ lies strictly between $A$ and $B$ on $\overline{AB}$ and point $E$ lies strictly between $A$ and $C$ on $\overline{AC}$ so that $AD = DE = EC$. Then $AD$ can be expressed in the form $\dfrac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p+q$. | 2018 AIME I Problem 4 | We draw the altitude from $B$ to $\overline{AC}$ to get point $F$. We notice that the triangle's height from $A$ to $\overline{BC}$ is 8 because it is a $3-4-5$ Right Triangle. To find the length of $\overline{BF}$, we let $h$ represent $\overline{BF}$ and set up an equation by finding two ways to express the area. The equation is $(8)(12)=(10)(h)$, which leaves us with $h=9.6$. We then solve for the length $\overline{AF}$, which is done through pythagorean theorm and get $\overline{AF}$ = $2.8$. We can now see that $\triangle AFB$ is a $7-24-25$ Right Triangle. Thus, we set $\overline{AG}$ as $5-$$\tfrac{x}{2}$, and yield that $\overline{AD}$ $=$ $($ $5-$ $\tfrac{x}{2}$ $)$ $($ $\tfrac{25}{7}$ $)$. Now, we can see $x$ = $($ $5-$ $\tfrac{x}{2}$ $)$ $($ $\tfrac{25}{7}$ $)$. Solving this equation, we yield $39x=250$, or $x=$ $\tfrac{250}{39}$. Thus, our final answer is $250+39=\boxed{289}$.
~bluebacon008
Diagram edited by Afly | // Block 1
import olympiad;
import cse5;
unitsize(10mm);
pathpen=black;
dotfactor=3;
pair B = (0,0), A = (6,8), C = (12,0), D = intersectionpoints(circle(A,250/39),A--B)[0], E = intersectionpoints(circle(D,250/39),A--C)[0], F=intersectionpoints(circle(B,9.6),A--C)[0], G=A/2+E/2;
pair[] dotted = {A,B,C,D,E,F,G};
D(A--B);
D(C--B);
D(A--C);
D(D--E);
pathpen=dashed;
D(B--F);
D(D--G);
dot(dotted);
label("$A$",A,N);
label("$B$",B,SW);
label("$C$",C,SE);
label("$D$",D,NW);
label("$E$",E,NE);
label("$F$",F,NE);
label("$G$",G,NE);
label("$x$",A--D,NW);
label("$x$",D--E,NW);
label("$x$",E--C,NE);
draw(rightanglemark(D,G,E));
draw(rightanglemark(B,F,E));
// Block 2
import olympiad; import cse5; unitsize(10mm); pathpen=black; dotfactor=3; pair B = (0,0), A = (6,8), C = (12,0), D = intersectionpoints(circle(A,250/39),A--B)[0], E = intersectionpoints(circle(D,250/39),A--C)[0], F=intersectionpoints(circle(B,9.6),A--C)[0], G=A/2+E/2; pair[] dotted = {A,B,C,D,E,F,G}; D(A--B); D(C--B); D(A--C); D(D--E); pathpen=dashed; D(B--F); D(D--G); dot(dotted); label("$A$",A,N); label("$B$",B,SW); label("$C$",C,SE); label("$D$",D,NW); label("$E$",E,NE); label("$F$",F,NE); label("$G$",G,NE); label("$x$",A--D,NW); label("$x$",D--E,NW); label("$x$",E--C,NE); draw(rightanglemark(D,G,E)); draw(rightanglemark(B,F,E)); | [] |
210 | Suppose that $x$, $y$, and $z$ are complex numbers such that $xy = -80 - 320i$, $yz = 60$, and $zx = -96 + 24i$, where $i$ $=$ $\sqrt{-1}$. Then there are real numbers $a$ and $b$ such that $x + y + z = a + bi$. Find $a^2 + b^2$. | 2018 AIME II Problem 5 | First we evaluate the magnitudes. $|xy|=80\sqrt{17}$, $|yz|=60$, and $|zx|=24\sqrt{17}$. Therefore, $|x^2y^2z^2|=17\cdot80\cdot60\cdot24$, or $|xyz|=240\sqrt{34}$. Divide to find that $|z|=3\sqrt{2}$, $|x|=4\sqrt{34}$, and $|y|=10\sqrt{2}$.
This allows us to see that the argument of $y$ is $\frac{\pi}{4}$, and the argument of $z$ is $-\frac{\pi}{4}$. We need to convert the polar form to a standard form. Simple trig identities show $y=10+10i$ and $z=3-3i$. More division is needed to find what $x$ is. \[x=-20-12i\] \[x+y+z=-7-5i\] \[(-7)^2+(-5)^2=\boxed{74}\]
\[QED\blacksquare\]
Written by a1b2 | // Block 1
draw((0,0)--(4,0));
dot((4,0),red);
draw((0,0)--(-4,0));
draw((0,0)--(0,-4));
draw((0,0)--(-4,1));
dot((-4,1),red);
draw((0,0)--(-1,-4));
dot((-1,-4),red);
draw((0,0)--(4,4),red);
draw((0,0)--(4,-4),red);
// Block 2
draw((0,0)--(4,0)); dot((4,0),red); draw((0,0)--(-4,0)); draw((0,0)--(0,-4)); draw((0,0)--(-4,1)); dot((-4,1),red); draw((0,0)--(-1,-4)); dot((-1,-4),red); draw((0,0)--(4,4),red); draw((0,0)--(4,-4),red); | [] |
211 | A frog is positioned at the origin of the coordinate plane. From the point $(x, y)$, the frog can jump to any of the points $(x + 1, y)$, $(x + 2, y)$, $(x, y + 1)$, or $(x, y + 2)$. Find the number of distinct sequences of jumps in which the frog begins at $(0, 0)$ and ends at $(4, 4)$. | 2018 AIME II Problem 8 | We solve this problem by working backwards. Notice, the only points the frog can be on to jump to $(4,4)$ in one move are $(2,4),(3,4),(4,2),$ and $(4,3)$. This applies to any other point, thus we can work our way from $(0,0)$ to $(4,4)$, recording down the number of ways to get to each point recursively.
$(0,0): 1$
$(1,0)=(0,1)=1$
$(2,0)=(0, 2)=2$
$(3,0)=(0, 3)=3$
$(4,0)=(0, 4)=5$
$(1,1)=2$, $(1,2)=(2,1)=5$, $(1,3)=(3,1)=10$, $(1,4)=(4,1)= 20$
$(2,2)=14, (2,3)=(3,2)=32, (2,4)=(4,2)=71$
$(3,3)=84, (3,4)=(4,3)=207$
$(4,4)=2\cdot \left( (4,2)+(4,3)\right) = 2\cdot \left( 207+71\right)=2\cdot 278=\boxed{556}$
A diagram of the numbers:
~First | // Block 1
import graph;
add(shift(0,0)*grid(4,4));
label((0,0), "1", SW);
label((1,0), "1", SW);
label((2,0), "2", SW);
label((3,0), "3", SW);
label((4,0), "5", SW);
label((0,1), "1", SW);
label((1,1), "2", SW);
label((2,1), "5", SW);
label((3,1), "10", SW);
label((4,1), "20", SW);
label((0,2), "2", SW);
label((1,2), "5", SW);
label((2,2), "14", SW);
label((3,2), "32", SW);
label((4,2), "71", SW);
label((0,3), "3", SW);
label((1,3), "10", SW);
label((2,3), "32", SW);
label((3,3), "84", SW);
label((4,3), "207", SW);
label((0,4), "5", SW);
label((1,4), "20", SW);
label((2,4), "71", SW);
label((3,4), "207", SW);
label((4,4), "556", SW);
// Block 2
import graph; add(shift(0,0)*grid(4,4)); label((0,0), "1", SW); label((1,0), "1", SW); label((2,0), "2", SW); label((3,0), "3", SW); label((4,0), "5", SW); label((0,1), "1", SW); label((1,1), "2", SW); label((2,1), "5", SW); label((3,1), "10", SW); label((4,1), "20", SW); label((0,2), "2", SW); label((1,2), "5", SW); label((2,2), "14", SW); label((3,2), "32", SW); label((4,2), "71", SW); label((0,3), "3", SW); label((1,3), "10", SW); label((2,3), "32", SW); label((3,3), "84", SW); label((4,3), "207", SW); label((0,4), "5", SW); label((1,4), "20", SW); label((2,4), "71", SW); label((3,4), "207", SW); label((4,4), "556", SW); | [] |
212 | In convex quadrilateral $KLMN$ side $\overline{MN}$ is perpendicular to diagonal $\overline{KM}$, side $\overline{KL}$ is perpendicular to diagonal $\overline{LN}$, $MN = 65$, and $KL = 28$. The line through $L$ perpendicular to side $\overline{KN}$ intersects diagonal $\overline{KM}$ at $O$ with $KO = 8$. Find $MO$. | 2019 AIME I Problem 6 | NOTE: this solution is wrong. The equation is correct due to similar triangles as described in solution 8, not PoP.
Because $\angle KLN = \angle KMN = 90^{\circ}$, $KLMN$ is a cyclic quadrilateral. (THE FOLLOWING SENTENCE IS WRONG) Hence, by Power of Point, \[KO\cdot KM = KL^2 \implies KM=\dfrac{28^2}{8}=98 \implies MO=98-8=\boxed{090}\] as desired.
~Mathkiddie | // Block 1
size(250);
real h = sqrt(98^2+65^2);
real l = sqrt(h^2-28^2);
pair K = (0,0);
pair N = (h, 0);
pair M = ((98^2)/h, (98*65)/h);
pair L = ((28^2)/h, (28*l)/h);
pair P = ((28^2)/h, 0);
pair O = ((28^2)/h, (8*65)/h);
draw(K--L--N);
draw(K--M--N--cycle);
draw(L--M);
label("K", K, SW);
label("L", L, NW);
label("M", M, NE);
label("N", N, SE);
draw(L--P);
label("P", P, S);
dot(O);
label("O", shift((1,1))*O, NNE);
label("28", scale(1/2)*L, W);
label("65", ((M.x+N.x)/2, (M.y+N.y)/2), NE);
// Block 2
size(250); real h = sqrt(98^2+65^2); real l = sqrt(h^2-28^2); pair K = (0,0); pair N = (h, 0); pair M = ((98^2)/h, (98*65)/h); pair L = ((28^2)/h, (28*l)/h); pair P = ((28^2)/h, 0); pair O = ((28^2)/h, (8*65)/h); draw(K--L--N); draw(K--M--N--cycle); draw(L--M); label("K", K, SW); label("L", L, NW); label("M", M, NE); label("N", N, SE); draw(L--P); label("P", P, S); dot(O); label("O", shift((1,1))*O, NNE); label("28", scale(1/2)*L, W); label("65", ((M.x+N.x)/2, (M.y+N.y)/2), NE); | [] |
212 | In convex quadrilateral $KLMN$ side $\overline{MN}$ is perpendicular to diagonal $\overline{KM}$, side $\overline{KL}$ is perpendicular to diagonal $\overline{LN}$, $MN = 65$, and $KL = 28$. The line through $L$ perpendicular to side $\overline{KN}$ intersects diagonal $\overline{KM}$ at $O$ with $KO = 8$. Find $MO$. | 2019 AIME I Problem 6 | First, let $P$ be the intersection of $LO$ and $KN$ as shown above. Note that $m\angle KPL = 90^{\circ}$ as given in the problem. Since $\angle KPL \cong \angle KLN$ and $\angle PKL \cong \angle LKN$, $\triangle PKL \sim \triangle LKN$ by AA similarity. Similarly, $\triangle KMN \sim \triangle KPO$. Using these similarities we see that
\[\frac{KP}{KL} = \frac{KL}{KN}\]
\[KP = \frac{KL^2}{KN} = \frac{28^2}{KN} = \frac{784}{KN}\]
and
\[\frac{KP}{KO} = \frac{KM}{KN}\]
\[KP = \frac{KO \cdot KM}{KN} = \frac{8\cdot KM}{KN}\]
Combining the two equations, we get
\[\frac{8\cdot KM}{KN} = \frac{784}{KN}\]
\[8 \cdot KM = 28^2\]
\[KM = 98\]
Since $KM = KO + MO$, we get $MO = 98 -8 = \boxed{090}$.
Solution by vedadehhc | // Block 1
size(250);
real h = sqrt(98^2+65^2);
real l = sqrt(h^2-28^2);
pair K = (0,0);
pair N = (h, 0);
pair M = ((98^2)/h, (98*65)/h);
pair L = ((28^2)/h, (28*l)/h);
pair P = ((28^2)/h, 0);
pair O = ((28^2)/h, (8*65)/h);
draw(K--L--N);
draw(K--M--N--cycle);
draw(L--M);
label("K", K, SW);
label("L", L, NW);
label("M", M, NE);
label("N", N, SE);
draw(L--P);
label("P", P, S);
dot(O);
label("O", shift((1,1))*O, NNE);
label("28", scale(1/2)*L, W);
label("65", ((M.x+N.x)/2, (M.y+N.y)/2), NE);
// Block 2
size(250); real h = sqrt(98^2+65^2); real l = sqrt(h^2-28^2); pair K = (0,0); pair N = (h, 0); pair M = ((98^2)/h, (98*65)/h); pair L = ((28^2)/h, (28*l)/h); pair P = ((28^2)/h, 0); pair O = ((28^2)/h, (8*65)/h); draw(K--L--N); draw(K--M--N--cycle); draw(L--M); label("K", K, SW); label("L", L, NW); label("M", M, NE); label("N", N, SE); draw(L--P); label("P", P, S); dot(O); label("O", shift((1,1))*O, NNE); label("28", scale(1/2)*L, W); label("65", ((M.x+N.x)/2, (M.y+N.y)/2), NE); | [] |
212 | In convex quadrilateral $KLMN$ side $\overline{MN}$ is perpendicular to diagonal $\overline{KM}$, side $\overline{KL}$ is perpendicular to diagonal $\overline{LN}$, $MN = 65$, and $KL = 28$. The line through $L$ perpendicular to side $\overline{KN}$ intersects diagonal $\overline{KM}$ at $O$ with $KO = 8$. Find $MO$. | 2019 AIME I Problem 6 | Notice that $KLMN$ is inscribed in the circle with diameter $\overline{KN}$ and $XOMN$ is inscribed in the circle with diameter $\overline{ON}$. Furthermore, $(XLN)$ is tangent to $\overline{KL}$. Then, \[KO\cdot KM=KX\cdot KN=KL^2\implies KM=\frac{28^2}{8}=98,\]and $MO=KM-KO=\boxed{090}$.
(Solution by TheUltimate123)
If you're wondering why $KX \cdot KN=KL^2,$ it's because PoP on $(XLN)$ or by $KX \cdot KN=KX \cdot (KX+XN)=KX^2+KX \cdot XN=KX^2+LX^2=KL^2$ (last part by geometric mean theorem / similarity).
Note: the "semicircle" circumscribing points XOMN is not a semicircle. That is just there to tell you that X, O, M, N are indeed concyclic, so ignore the subtlety of the diagram that makes O seems slightly off the marks than it should be. | // Block 1
size(8cm);
pair K, L, M, NN, X, O;
K=(-sqrt(98^2+65^2)/2, 0);
NN=(sqrt(98^2+65^2)/2, 0);
L=sqrt(98^2+65^2)/2*dir(180-2*aSin(28/sqrt(98^2+65^2)));
M=sqrt(98^2+65^2)/2*dir(2*aSin(65/sqrt(98^2+65^2)));
X=foot(L, K, NN);
O=extension(L, X, K, M);
draw(K -- L -- M -- NN -- K -- M); draw(L -- NN); draw(arc((K+NN)/2, NN, K));
draw(L -- X, dashed); draw(arc((O+NN)/2, NN, X), dashed);
draw(rightanglemark(K, L, NN, 100));
draw(rightanglemark(K, M, NN, 100));
draw(rightanglemark(L, X, NN, 100));
dot("$K$", K, SW);
dot("$L$", L, unit(L));
dot("$M$", M, unit(M));
dot("$N$", NN, SE);
dot("$X$", X, S);
// Block 2
size(8cm); pair K, L, M, NN, X, O; K=(-sqrt(98^2+65^2)/2, 0); NN=(sqrt(98^2+65^2)/2, 0); L=sqrt(98^2+65^2)/2*dir(180-2*aSin(28/sqrt(98^2+65^2))); M=sqrt(98^2+65^2)/2*dir(2*aSin(65/sqrt(98^2+65^2))); X=foot(L, K, NN); O=extension(L, X, K, M); draw(K -- L -- M -- NN -- K -- M); draw(L -- NN); draw(arc((K+NN)/2, NN, K)); draw(L -- X, dashed); draw(arc((O+NN)/2, NN, X), dashed); draw(rightanglemark(K, L, NN, 100)); draw(rightanglemark(K, M, NN, 100)); draw(rightanglemark(L, X, NN, 100)); dot("$K$", K, SW); dot("$L$", L, unit(L)); dot("$M$", M, unit(M)); dot("$N$", NN, SE); dot("$X$", X, S); | [] |
212 | In convex quadrilateral $KLMN$ side $\overline{MN}$ is perpendicular to diagonal $\overline{KM}$, side $\overline{KL}$ is perpendicular to diagonal $\overline{LN}$, $MN = 65$, and $KL = 28$. The line through $L$ perpendicular to side $\overline{KN}$ intersects diagonal $\overline{KM}$ at $O$ with $KO = 8$. Find $MO$. | 2019 AIME I Problem 6 | (Diagram by vedadehhc)
Call the base of the altitude from $L$ to $NK$ point $P$. Let $PO=x$. Now, we have that $KP=\sqrt{64-x^2}$ by the Pythagorean Theorem. Once again by Pythagorean, $LO=\sqrt{720+x^2}-x$. Using Power of a Point, we have
\[(KO)(OM)=(LO)(OQ)\] ($Q$ is the intersection of $OL$ with the circle $\neq L$)
\[8(MO)=(\sqrt{720+x^2}-x)(\sqrt{720+x^2}+x)\]
\[8(MO)=720\]
\[MO=\boxed{090}\].
(Solution by RootThreeOverTwo)
\[\]
Remark: Length of OQ
Since $P$ is on the circle’s diameter, $QP = LP = \sqrt{720+x^2}$. So, $OQ = PQ + PO = x + \sqrt{720+x^2}$.
~diyarv | // Block 1
size(250);
real h = sqrt(98^2+65^2);
real l = sqrt(h^2-28^2);
pair K = (0,0);
pair N = (h, 0);
pair M = ((98^2)/h, (98*65)/h);
pair L = ((28^2)/h, (28*l)/h);
pair P = ((28^2)/h, 0);
pair O = ((28^2)/h, (8*65)/h);
draw(K--L--N);
draw(K--M--N--cycle);
draw(L--M);
label("K", K, SW);
label("L", L, NW);
label("M", M, NE);
label("N", N, SE);
draw(L--P);
label("P", P, S);
dot(O);
label("O", shift((1,1))*O, NNE);
label("28", scale(1/2)*L, W);
label("65", ((M.x+N.x)/2, (M.y+N.y)/2), NE);
// Block 2
size(250); real h = sqrt(98^2+65^2); real l = sqrt(h^2-28^2); pair K = (0,0); pair N = (h, 0); pair M = ((98^2)/h, (98*65)/h); pair L = ((28^2)/h, (28*l)/h); pair P = ((28^2)/h, 0); pair O = ((28^2)/h, (8*65)/h); draw(K--L--N); draw(K--M--N--cycle); draw(L--M); label("K", K, SW); label("L", L, NW); label("M", M, NE); label("N", N, SE); draw(L--P); label("P", P, S); dot(O); label("O", shift((1,1))*O, NNE); label("28", scale(1/2)*L, W); label("65", ((M.x+N.x)/2, (M.y+N.y)/2), NE); | [] |
212 | In convex quadrilateral $KLMN$ side $\overline{MN}$ is perpendicular to diagonal $\overline{KM}$, side $\overline{KL}$ is perpendicular to diagonal $\overline{LN}$, $MN = 65$, and $KL = 28$. The line through $L$ perpendicular to side $\overline{KN}$ intersects diagonal $\overline{KM}$ at $O$ with $KO = 8$. Find $MO$. | 2019 AIME I Problem 6 | Note that since $\angle KLN = \angle KMN$, quadrilateral $KLMN$ is cyclic. Therefore, we have \[\angle LMK = \angle LNK = 90^{\circ} - \angle LKN = \angle KLP,\]so $\triangle KLO \sim \triangle KML$, giving \[\frac{KM}{28} = \frac{28}{8} \implies KM = 98.\] Therefore, $OM = 98-8 = \boxed{90}$. | // Block 1
size(250);
real h = sqrt(98^2+65^2);
real l = sqrt(h^2-28^2);
pair K = (0,0);
pair N = (h, 0);
pair M = ((98^2)/h, (98*65)/h);
pair L = ((28^2)/h, (28*l)/h);
pair P = ((28^2)/h, 0);
pair O = ((28^2)/h, (8*65)/h);
draw(K--L--N);
draw(K--M--N--cycle);
draw(L--M);
label("K", K, SW);
label("L", L, NW);
label("M", M, NE);
label("N", N, SE);
draw(L--P);
label("P", P, S);
dot(O);
label("O", shift((1,1))*O, NNE);
label("28", scale(1/2)*L, W);
label("65", ((M.x+N.x)/2, (M.y+N.y)/2), NE);
// Block 2
size(250); real h = sqrt(98^2+65^2); real l = sqrt(h^2-28^2); pair K = (0,0); pair N = (h, 0); pair M = ((98^2)/h, (98*65)/h); pair L = ((28^2)/h, (28*l)/h); pair P = ((28^2)/h, 0); pair O = ((28^2)/h, (8*65)/h); draw(K--L--N); draw(K--M--N--cycle); draw(L--M); label("K", K, SW); label("L", L, NW); label("M", M, NE); label("N", N, SE); draw(L--P); label("P", P, S); dot(O); label("O", shift((1,1))*O, NNE); label("28", scale(1/2)*L, W); label("65", ((M.x+N.x)/2, (M.y+N.y)/2), NE); | [] |
212 | In convex quadrilateral $KLMN$ side $\overline{MN}$ is perpendicular to diagonal $\overline{KM}$, side $\overline{KL}$ is perpendicular to diagonal $\overline{LN}$, $MN = 65$, and $KL = 28$. The line through $L$ perpendicular to side $\overline{KN}$ intersects diagonal $\overline{KM}$ at $O$ with $KO = 8$. Find $MO$. | 2019 AIME I Problem 6 | By Pythagorean Theorem, $KM^2+65^2 = KN^2 = 28^2 + LN^2$. Thus, $LN^2 = KM^2 + 65^2 - 28^2$.
By Pythagorean Theorem, $KP^2 + LP^2 = 28^2$, and $PN^2 + LP^2 = LN^2$.
\[PN^2 = (KN - KP)^2 = (\sqrt{KM^2 + 65^2} - KP)^2\]
It follows that
\[(\sqrt{KM^2 + 65^2} - KP)^2 + LP^2 = KM^2 + 65^2 - 28^2\]
\[KM^2 + 65^2 - 2\sqrt{KM^2 + 65^2}(KP) + KP^2 + LP^2 = KM^2 + 65^2 - 28^2\]
Since $KP^2 + LP^2 = 28^2$,
\[KM^2 + 65^2 - 2\sqrt{KM^2 + 65^2}(KP) + 28^2 = KM^2 + 65^2 - 28^2\]
\[-2\sqrt{KM^2 + 65^2}(KP) = -2 \times 28^2\]
\[KP = \frac{28^2}{\sqrt{KM^2 + 65^2}}\]
$\angle OKP = \angle NKM$ (it's the same angle) and $\angle OPK = \angle KMN = 90^{\circ}$. Thus, $\triangle KOP \sim \triangle KNM$.
Thus,
\[\frac{KO}{KN} = \frac{KP}{KM}\]
\[\frac{8}{\sqrt{KM^2 + 65^2}} = \frac{\frac{28^2}{\sqrt{KM^2 + 65^2}}}{KM}\]
Multiplying both sides by $\sqrt{KM^2 + 65^2}$:
\[8 = \frac{28^2}{KM}\]
\[KM = 98\]
Therefore, $OM = 98-8 = \boxed{90}$
~ Solution by adam_zheng | // Block 1
size(250);
real h = sqrt(98^2+65^2);
real l = sqrt(h^2-28^2);
pair K = (0,0);
pair N = (h, 0);
pair M = ((98^2)/h, (98*65)/h);
pair L = ((28^2)/h, (28*l)/h);
pair P = ((28^2)/h, 0);
pair O = ((28^2)/h, (8*65)/h);
draw(K--L--N);
draw(K--M--N--cycle);
draw(L--M);
label("K", K, SW);
label("L", L, NW);
label("M", M, NE);
label("N", N, SE);
draw(L--P);
label("P", P, S);
dot(O);
label("O", shift((1,1))*O, NNE);
label("28", scale(1/2)*L, W);
label("65", ((M.x+N.x)/2, (M.y+N.y)/2), NE);
// Block 2
size(250); real h = sqrt(98^2+65^2); real l = sqrt(h^2-28^2); pair K = (0,0); pair N = (h, 0); pair M = ((98^2)/h, (98*65)/h); pair L = ((28^2)/h, (28*l)/h); pair P = ((28^2)/h, 0); pair O = ((28^2)/h, (8*65)/h); draw(K--L--N); draw(K--M--N--cycle); draw(L--M); label("K", K, SW); label("L", L, NW); label("M", M, NE); label("N", N, SE); draw(L--P); label("P", P, S); dot(O); label("O", shift((1,1))*O, NNE); label("28", scale(1/2)*L, W); label("65", ((M.x+N.x)/2, (M.y+N.y)/2), NE); | [] |
213 | In $\triangle ABC$, the sides have integer lengths and $AB=AC$. Circle $\omega$ has its center at the incenter of $\triangle ABC$. An excircle of $\triangle ABC$ is a circle in the exterior of $\triangle ABC$ that is tangent to one side of the triangle and tangent to the extensions of the other two sides. Suppose that the excircle tangent to $\overline{BC}$ is internally tangent to $\omega$, and the other two excircles are both externally tangent to $\omega$. Find the minimum possible value of the perimeter of $\triangle ABC$. | 2019 AIME I Problem 11 | First, assume $BC=2$ and $AB=AC=x$. The triangle can be scaled later if necessary. Let $I$ be the incenter and let $r$ be the inradius. Let the points at which the incircle intersects $AB$, $BC$, and $CA$ be denoted $M$, $N$, and $O$, respectively.
Next, we calculate $r$ in terms of $x$. Note the right triangle formed by $A$, $I$, and $M$. The length $IM$ is equal to $r$. Using the Pythagorean Theorem, the length $AN$ is $\sqrt{x^2-1}$, so the length $AI$ is $\sqrt{x^2-1}-r$. Note that $BN$ is half of $BC=2$, and by symmetry caused by the incircle, $BN=BM$ and $BM=1$, so $MA=x-1$. Applying the Pythagorean Theorem to $AIM$, we get
\[r^2+(x-1)^2=\left(\sqrt{x^2-1}-r\right)^2.\]
Expanding yields
\[r^2+x^2-2x+1=x^2-1-2r\sqrt{x^2-1}+r^2,\]
which can be simplified to
\[2r\sqrt{x^2-1}=2x-2.\]
Dividing by $2$ and then squaring results in
\[r^2(x^2-1)=(x-1)^2,\]
and isolating $r^2$ gets us
\[r^2=\frac{(x-1)^2}{x^2-1}=\frac{(x-1)^2}{(x+1)(x-1)}=\frac{x-1}{x+1},\]
so $r=\sqrt{\frac{x-1}{x+1}}$.
We then calculate the radius of the excircle tangent to $BC$. We denote the center of the excircle $E_N$ and the radius $r_N$.
Consider the quadrilateral formed by $M$, $I$, $E_N$, and the point at which the excircle intersects the extension of $AB$, which we denote $H$. By symmetry caused by the excircle, $BN=BH$, so $BH=1$.
Note that triangles $MBI$ and $NBI$ are congruent, and $HBE$ and $NBE$ are also congruent. Denoting the measure of angles $MBI$ and $NBI$ measure $\alpha$ and the measure of angles $HBE$ and $NBE$ measure $\beta$, straight angle $MBH=2\alpha+2\beta$, so $\alpha + \beta=90^\circ$. This means that angle $IBE$ is a right angle, so it forms a right triangle.
Setting the base of the right triangle to $IE$, the height is $BN=1$ and the base consists of $IN=r$ and $EN=r_N$. Triangles $INB$ and $BNE$ are similar to $IBE$, so $\frac{IN}{BN}=\frac{BN}{EN}$, or $\frac{r}{1}=\frac{1}{r_N}$. This makes $r_N$ the reciprocal of $r$, so $r_N=\sqrt{\frac{x+1}{x-1}}$.
Circle $\omega$'s radius can be expressed by the distance from the incenter $I$ to the bottom of the excircle with center $E_N$. This length is equal to $r+2r_N$, or $\sqrt{\frac{x-1}{x+1}}+2\sqrt{\frac{x+1}{x-1}}$. Denote this value $r_\omega$.
Finally, we calculate the distance from the incenter $I$ to the closest point on the excircle tangent to $AB$, which forms another radius of circle $\omega$ and is equal to $r_\omega$. We denote the center of the excircle $E_M$ and the radius $r_M$. We also denote the points where the excircle intersects $AB$ and the extension of $BC$ using $J$ and $K$, respectively. In order to calculate the distance, we must find the distance between $I$ and $E_M$ and subtract off the radius $r_M$.
We first must calculate the radius of the excircle. Because the excircle is tangent to both $AB$ and the extension of $AC$, its center must lie on the angle bisector formed by the two lines, which is parallel to $BC$. This means that the distance from $E_M$ to $K$ is equal to the length of $AN$, so the radius is also $\sqrt{x^2-1}$.
Next, we find the length of $IE_M$. We can do this by forming the right triangle $IAE_M$. The length of leg $AI$ is equal to $AN$ minus $r$, or $\sqrt{x^2-1}-\sqrt{\frac{x-1}{x+1}}$. In order to calculate the length of leg $AE_M$, note that right triangles $AJE_M$ and $BNA$ are congruent, as $JE_M$ and $NA$ share a length of $\sqrt{x^2-1}$, and angles $E_MAJ$ and $NAB$ add up to the right angle $NAE_M$. This means that $AE_M=BA=x$.
Using Pythagorean Theorem, we get
\[IE_M=\sqrt{\left(\sqrt{x^2-1}-\sqrt{\frac{x-1}{x+1}}\right)^2+x^2}.\]
Bringing back
\[r_\omega=IE_M-r_M\]
and substituting in some values, the equation becomes
\[r_\omega=\sqrt{\left(\sqrt{x^2-1}-\sqrt{\frac{x-1}{x+1}}\right)^2+x^2}-\sqrt{x^2-1}.\]
Rearranging and squaring both sides gets
\[\left(r_\omega+\sqrt{x^2-1}\right)^2=\left(\sqrt{x^2-1}-\sqrt{\frac{x-1}{x+1}}\right)^2+x^2.\]
Distributing both sides yields
\[r_\omega^2+2r_\omega\sqrt{x^2-1}+x^2-1=x^2-1-2\sqrt{x^2-1}\sqrt{\frac{x-1}{x+1}}+\frac{x-1}{x+1}+x^2.\]
Canceling terms results in
\[r_\omega^2+2r_\omega\sqrt{x^2-1}=-2\sqrt{x^2-1}\sqrt{\frac{x-1}{x+1}}+\frac{x-1}{x+1}+x^2.\]
Since
\[-2\sqrt{x^2-1}\sqrt{\frac{x-1}{x+1}}=-2\sqrt{(x+1)(x-1)\frac{x-1}{x+1}}=-2(x-1),\]
We can further simplify to
\[r_\omega^2+2r_\omega\sqrt{x^2-1}=-2(x-1)+\frac{x-1}{x+1}+x^2.\]
Substituting out $r_\omega$ gets
\[\left(\sqrt{\frac{x-1}{x+1}}+2\sqrt{\frac{x+1}{x-1}}\right)^2+2\left(\sqrt{\frac{x-1}{x+1}}+2\sqrt{\frac{x+1}{x-1}}\right)\sqrt{x^2-1}=-2(x-1)+\frac{x-1}{x+1}+x^2\]
which when distributed yields
\[\frac{x-1}{x+1}+4+4\left(\frac{x+1}{x-1}\right)+2(x-1+2(x+1))=-2(x-1)+\frac{x-1}{x+1}+x^2.\]
After some canceling, distributing, and rearranging, we obtain
\[4\left(\frac{x+1}{x-1}\right)=x^2-8x-4.\]
Multiplying both sides by $x-1$ results in
\[4x+4=x^3-x^2-8x^2+8x-4x+4,\]
which can be rearranged into
\[x^3-9x^2=0\]
and factored into
\[x^2(x-9)=0.\]
This means that $x$ equals $0$ or $9$, and since a side length of $0$ cannot exist, $x=9$.
As a result, the triangle must have sides in the ratio of $9:2:9$. Since the triangle must have integer side lengths, and these values share no common factors greater than $1$, the triangle with the smallest possible perimeter under these restrictions has a perimeter of $9+2+9=\boxed{020}$. ~emerald_block | // Block 1
unitsize(1cm);
var x = 9;
pair A = (0,sqrt(x^2-1));
pair B = (-1,0);
pair C = (1,0);
dot(Label("$A$",A,NE),A);
dot(Label("$B$",B,SW),B);
dot(Label("$C$",C,SE),C);
draw(A--B--C--cycle);
var r = sqrt((x-1)/(x+1));
pair I = (0,r);
dot(Label("$I$",I,SE),I);
draw(circle(I,r));
draw(Label("$r$"),I--I+r*SSW,dashed);
pair M = intersectionpoint(A--B,circle(I,r));
pair N = (0,0);
pair O = intersectionpoint(A--C,circle(I,r));
dot(Label("$M$",M,W),M);
dot(Label("$N$",N,S),N);
dot(Label("$O$",O,E),O);
var rN = sqrt((x+1)/(x-1));
pair EN = (0,-rN);
dot(Label("$E_N$",EN,SE),EN);
draw(circle(EN,rN));
draw(Label("$r_N$"),EN--EN+rN*SSW,dashed);
pair AB = (-1-2/(x-1),-2rN);
pair AC = (1+2/(x-1),-2rN);
draw(B--AB,EndArrow);
draw(C--AC,EndArrow);
pair H = intersectionpoint(B--AB,circle(EN,rN));
dot(Label("$H$",H,W),H);
var rM = sqrt(x^2-1);
pair EM = (-x,rM);
dot(Label("$E_M$",EM,SW),EM);
draw(Label("$r_M$"),EM--EM+rM*SSE,dashed);
pair CB = (-x-1,0);
pair CA = (-2/x,sqrt(x^2-1)+2(sqrt(x^2-1)/x));
draw(B--CB,EndArrow);
draw(A--CA,EndArrow);
pair J = intersectionpoint(A--B,circle(EM,rM));
pair K = intersectionpoint(B--CB,circle(EM,rM));
dot(Label("$J$",J,W),J);
dot(Label("$K$",K,S),K);
draw(arc(EM,rM,-100,15),Arrows);
// Block 2
unitsize(1cm); var x = 9; pair A = (0,sqrt(x^2-1)); pair B = (-1,0); pair C = (1,0); dot(Label("$A$",A,NE),A); dot(Label("$B$",B,SW),B); dot(Label("$C$",C,SE),C); draw(A--B--C--cycle); var r = sqrt((x-1)/(x+1)); pair I = (0,r); dot(Label("$I$",I,SE),I); draw(circle(I,r)); draw(Label("$r$"),I--I+r*SSW,dashed); pair M = intersectionpoint(A--B,circle(I,r)); pair N = (0,0); pair O = intersectionpoint(A--C,circle(I,r)); dot(Label("$M$",M,W),M); dot(Label("$N$",N,S),N); dot(Label("$O$",O,E),O); var rN = sqrt((x+1)/(x-1)); pair EN = (0,-rN); dot(Label("$E_N$",EN,SE),EN); draw(circle(EN,rN)); draw(Label("$r_N$"),EN--EN+rN*SSW,dashed); pair AB = (-1-2/(x-1),-2rN); pair AC = (1+2/(x-1),-2rN); draw(B--AB,EndArrow); draw(C--AC,EndArrow); pair H = intersectionpoint(B--AB,circle(EN,rN)); dot(Label("$H$",H,W),H); var rM = sqrt(x^2-1); pair EM = (-x,rM); dot(Label("$E_M$",EM,SW),EM); draw(Label("$r_M$"),EM--EM+rM*SSE,dashed); pair CB = (-x-1,0); pair CA = (-2/x,sqrt(x^2-1)+2(sqrt(x^2-1)/x)); draw(B--CB,EndArrow); draw(A--CA,EndArrow); pair J = intersectionpoint(A--B,circle(EM,rM)); pair K = intersectionpoint(B--CB,circle(EM,rM)); dot(Label("$J$",J,W),J); dot(Label("$K$",K,S),K); draw(arc(EM,rM,-100,15),Arrows); | [] |
213 | In $\triangle ABC$, the sides have integer lengths and $AB=AC$. Circle $\omega$ has its center at the incenter of $\triangle ABC$. An excircle of $\triangle ABC$ is a circle in the exterior of $\triangle ABC$ that is tangent to one side of the triangle and tangent to the extensions of the other two sides. Suppose that the excircle tangent to $\overline{BC}$ is internally tangent to $\omega$, and the other two excircles are both externally tangent to $\omega$. Find the minimum possible value of the perimeter of $\triangle ABC$. | 2019 AIME I Problem 11 | Before we start thinking about the problem, let’s draw it out;
For the sake of space, I've drawn only 2 of the 3 excircles because the third one looks the same as the second large one because the triangle is isosceles. By the incenter-excenter lemma, $AII_A$ and $BII_B$ are collinear, $E$ is the tangent of circle $I_B$ to $AC$, $F$ is the tangent of that circle to the extension of $BC$, and $J$ is the tangent of the circle to the extension of $BA$. The interesting part of the diagram is circle $\omega$, which is internally tangent to circle $I_A$ yet externally tangent to circle $I_B$. Therefore, perhaps we can relate the radius of this circle to the semiperimeter of triangle $ABC$.
We can see that the radius of circle $\omega$ is $2r_{I_A}+r$ using the incenter and A-excenter of our main triangle. This radius is also equal to $BI_B - BI - r_{I_B}$ from the incenter and B-excenter of our triangle. Thus, we can solve for each of these separately in terms of the lengths of the triangle and set them equal to each other to form an equation.
To find the left hand side of the equation, we have to first find $r$ and $r_{I_A}$. Let $a = AB = AC, b = BD = DC,$ and $h = AD$. Then since the perimeter of the triangle is $2a+2b$, the semiperimeter is $a+b$.
Now let's take a look at triangle $BDI$. Because $BI$ is the angle bisector of $\angle B$, by the angle bisector theorem, $\frac{AI}{ID} = \frac{BA}{BD} \implies \frac{h-r}{r} = \frac{a}{b}$. Rearranging, we get $r = \frac{hb}{a+b}$.
Take a look at triangle $AGI$. $AG = a - GB = a - BD = a-b$, $AI = h-r = \frac{ha}{a+b}$ (angle bisector theorem), and $GI = r = \frac{hb}{a+b}$. Now let's analyze triangle $AHI_A$. $AH = AB + BH = AB+ BD = a+b$, $AI_A = h+r_{I_A}$, and $HI_A = r_{I_A}$. Since $\angle GAI = \angle HAI_A$ and $\angle IGA = \angle I_AHA = 90^{\circ}$, triangle $AGI$ and $AHI_A$ are similar by AA. Then $\frac{r_{I_A}}{r} = \frac{h+r_{I_A}}{h-r} \implies r_{I_A} = r \cdot \frac{h+r_{I_A}}{h-r} = \frac{hb}{a+b} \cdot \frac{h+r_{I_A}}{\frac{ha}{a+b}} = \frac{b(h+r_{I_A})}{a}$. Now, solving yields $r_{I_A} = \frac{hb}{a-b}$.
Finally, the left hand side of our equation is \[\frac{2hb}{a-b} + \frac{hb}{a+b}\]
Now let's look at triangle $BFI_B$. How will we find $BI_B$? Let's first try to find $BF$ and $I_BF$ in terms of the lengths of the triangle. We recognize:
$BF = BC + CF = BC + CK$. We really want to have $CA$ instead of $CK$, and $AK$ looks very similar in length to $DC$, so let's try to prove that they are equal.
$BJ = BF$, so we can try to add these two and see if we get anything interesting. We have:
$BJ + BF = BA + AJ + BC + CF = BA + AE + BC + CE = BA + BC + CA$, which is our perimeter. Thus, $BF = a+b$.
Triangle $BDI$ is similar to triangle $BFI_B$ by AA, and we know that $BD = b$, and $ID = r = \frac{hb}{a+b}$, so thus $I_BF = BF \cdot \frac{ID}{BD} = (a+b) \cdot \frac{\frac{hb}{a+b}}{b} = \frac{hb}{b} = h$. Thus, the height of this triangle is $h$ by similarity ratios, the same height as vertex $A$. By the Pythagorean Theorem, $BI_B = \sqrt{(a+b)^2 + h^2}$ and by similarity ratios, $BI = \frac{b}{a+b} \cdot \sqrt{(a+b)^2 + h^2}$. Finally, $r_{I_B} = I_BF = h$, and thus the right hand side of our equation is \[\sqrt{(a+b)^2 + h^2} - \frac{b}{a+b} \cdot \sqrt{(a+b)^2 + h^2} - h = \sqrt{(a+b)^2 + h^2}(1 - \frac{b}{a+b}) - h = \sqrt{(a+b)^2 + h^2} \cdot \frac{a}{a+b} - h\].
Setting the two equal, we have \[\frac{2hb}{a-b} + \frac{hb}{a+b} = \sqrt{(a+b)^2 + h^2} \cdot \frac{a}{a+b} - h\]
Multiplying both sides by $(a+b)(a-b)$ we have $2hb(a+b) + hb(a-b) = \sqrt{(a+b)^2 + h^2} \cdot a(a-b) - h(a^2 - b^2)$
From here, let $b = 1$ arbitrarily; note that we can always scale this value to fit the requirements later. Thus our equation is $2h(a+1) + h(a-1) = \sqrt{(a+1)^2 + h^2} \cdot a(a-1) - h(a^2 - 1)$. Now since $h = \sqrt{a^2 - b^2}$, we can plug into our equation:
$2h(a+1) + h(a-1) = \sqrt{a² + 2a + 1 + a^2 - b^2 } \cdot a(a-1) - h(a^2 - 1)$. Remembering $b = 1$;
$\implies 2h(a+1) + h(a-1) = \sqrt{2a^2 + 2a} \cdot a(a-1) - h(a^2 - 1)$
$\implies 3ha + h + ha^2 - h = \sqrt{2a^2 + 2a} \cdot a(a-1)$
$\implies ah(3+a) = a(a-1) \cdot \sqrt{2a^2 + 2a}$
$\implies h^2(3+a)^2 = (a-1)^2 \cdot 2a(a+1)$
$\implies (a^2 - 1) (3 + a)^2 = 2a(a+1)(a - 1)^2$
$\implies (3+ a)^2 = 2a(a-1)$
$\implies 9 + 6a + a^2 = 2a^2 - 2a$
$\implies a^2 - 8a - 9 = 0$
$\implies (a-9)(a+1) = 0$
$\implies a = 9$ because the side lengths have to be positive numbers. Furthermore, because our values for $a$ and $b$ are relatively prime, we don't have to scale down our triangle further, and we are done. Therefore, our answer is $2a + 2b = 18 + 2 = \boxed{020}$
~KingRavi | // Block 1
unitsize(1cm);
var x = 9;
pair A = (0,sqrt(x^2-1));
pair B = (-1,0);
pair C = (1,0);
dot(Label("$A$",A,NE),A);
dot(Label("$B$",B,SW),B);
dot(Label("$C$",C,SE),C);
draw(A--B--C--cycle);
var r = sqrt((x-1)/(x+1));
pair I = (0,r);
dot(Label("$I$",I,SE),I);
draw(circle(I,r));
pair G = intersectionpoint(A--B,circle(I,r));
pair D = (0,0);
dot(Label("$G$",G,W),G);
dot(Label("$D$",D,SSE),D);
draw(Label("$r$"),I--G,dashed);
var rA = sqrt((x+1)/(x-1));
pair IA = (0,-rA);
dot(Label("$I_A$",IA,SE),IA);
draw(circle(IA,rA));
pair AB = (-1-2/(x-1),-2rA);
pair AC = (1+2/(x-1),-2rA);
draw(B--AB,EndArrow);
draw(C--AC,EndArrow);
pair H = intersectionpoint(B--AB,circle(IA,rA));
dot(Label("$H$",H,W),H);
draw(Label("$r_{I_A}$"),IA--H,dashed);
var rB = sqrt(x^2-1);
pair IB = (x,rB);
dot(Label("$I_B$",IB,SE),IB);
pair BC = (x+1,0);
pair BA = (2/x,sqrt(x^2-1)+2(sqrt(x^2-1)/x));
draw(C--BC,EndArrow);
draw(A--BA,EndArrow);
pair E = intersectionpoint(A--C,circle(IB,rB));
pair F = intersectionpoint(C--BC,circle(IB,rB));
dot(Label("$E$",E,SE),E);
dot(Label("$F$",F,S),F);
draw(Label("$r_{I_B}$"),IB--F,dashed);
draw(circle(IB,rB));
draw(A--IA);
draw(B--IB);
pair J = intersectionpoint(B--BA,circle(IB,rB));
dot(Label("$J$",J,W),J);
draw(circle(I,r+2rA));
pair W = intersectionpoint(B--F,circle(I,r+2rA));
dot(Label("$\omega$",W,SSE),W);
// Block 2
unitsize(1cm); var x = 9; pair A = (0,sqrt(x^2-1)); pair B = (-1,0); pair C = (1,0); dot(Label("$A$",A,NE),A); dot(Label("$B$",B,SW),B); dot(Label("$C$",C,SE),C); draw(A--B--C--cycle); var r = sqrt((x-1)/(x+1)); pair I = (0,r); dot(Label("$I$",I,SE),I); draw(circle(I,r)); pair G = intersectionpoint(A--B,circle(I,r)); pair D = (0,0); dot(Label("$G$",G,W),G); dot(Label("$D$",D,SSE),D); draw(Label("$r$"),I--G,dashed); var rA = sqrt((x+1)/(x-1)); pair IA = (0,-rA); dot(Label("$I_A$",IA,SE),IA); draw(circle(IA,rA)); pair AB = (-1-2/(x-1),-2rA); pair AC = (1+2/(x-1),-2rA); draw(B--AB,EndArrow); draw(C--AC,EndArrow); pair H = intersectionpoint(B--AB,circle(IA,rA)); dot(Label("$H$",H,W),H); draw(Label("$r_{I_A}$"),IA--H,dashed); var rB = sqrt(x^2-1); pair IB = (x,rB); dot(Label("$I_B$",IB,SE),IB); pair BC = (x+1,0); pair BA = (2/x,sqrt(x^2-1)+2(sqrt(x^2-1)/x)); draw(C--BC,EndArrow); draw(A--BA,EndArrow); pair E = intersectionpoint(A--C,circle(IB,rB)); pair F = intersectionpoint(C--BC,circle(IB,rB)); dot(Label("$E$",E,SE),E); dot(Label("$F$",F,S),F); draw(Label("$r_{I_B}$"),IB--F,dashed); draw(circle(IB,rB)); draw(A--IA); draw(B--IB); pair J = intersectionpoint(B--BA,circle(IB,rB)); dot(Label("$J$",J,W),J); draw(circle(I,r+2rA)); pair W = intersectionpoint(B--F,circle(I,r+2rA)); dot(Label("$\omega$",W,SSE),W); | [] |
214 | Triangle $ABC$ has side lengths $AB=4$, $BC=5$, and $CA=6$. Points $D$ and $E$ are on ray $AB$ with $AB<AD<AE$. The point $F \neq C$ is a point of intersection of the circumcircles of $\triangle ACD$ and $\triangle EBC$ satisfying $DF=2$ and $EF=7$. Then $BE$ can be expressed as $\tfrac{a+b\sqrt{c}}{d}$, where $a$, $b$, $c$, and $d$ are positive integers such that $a$ and $d$ are relatively prime, and $c$ is not divisible by the square of any prime. Find $a+b+c+d$. | 2019 AIME I Problem 13 | Notice that \[\angle DFE=\angle CFE-\angle CFD=\angle CBE-\angle CAD=180-B-A=C.\]By the Law of Cosines, \[\cos C=\frac{AC^2+BC^2-AB^2}{2\cdot AC\cdot BC}=\frac34.\]Then, \[DE^2=DF^2+EF^2-2\cdot DF\cdot EF\cos C=32\implies DE=4\sqrt2.\]Let $X=\overline{AB}\cap\overline{CF}$, $a=XB$, and $b=XD$. Then, \[XA\cdot XD=XC\cdot XF=XB\cdot XE\implies b(a+4)=a(b+4\sqrt2)\implies b=a\sqrt2.\]However, since $\triangle XFD\sim\triangle XAC$, $XF=\tfrac{4+a}3$, but since $\triangle XFE\sim\triangle XBC$, \[\frac75=\frac{4+a}{3a}\implies a=\frac54\implies BE=a+a\sqrt2+4\sqrt2=\frac{5+21\sqrt2}4,\]and the requested sum is $5+21+2+4=\boxed{032}$.
(Solution by TheUltimate123) | // Block 1
unitsize(20);
pair A, B, C, D, E, F, X, O1, O2;
A = (0, 0); B = (4, 0);
C = intersectionpoints(circle(A, 6), circle(B, 5))[0];
D = B + (5/4 * (1 + sqrt(2)), 0); E = D + (4 * sqrt(2), 0);
F = intersectionpoints(circle(D, 2), circle(E, 7))[1];
X = extension(A, E, C, F);
O1 = circumcenter(C, A, D);
O2 = circumcenter(C, B, E);
filldraw(A--B--C--cycle, lightcyan, deepcyan);
filldraw(D--E--F--cycle, lightmagenta, deepmagenta);
draw(B--D, gray(0.6));
draw(C--F, gray(0.6));
draw(circumcircle(C, A, D), dashed);
draw(circumcircle(C, B, E), dashed);
dot("$A$", A, dir(A-O1));
dot("$B$", B, dir(240));
dot("$C$", C, dir(120));
dot("$D$", D, dir(40));
dot("$E$", E, dir(E-O2));
dot("$F$", F, dir(270));
dot("$X$", X, dir(140));
label("$6$", (C+A)/2, dir(C-A)*I, deepcyan);
label("$5$", (C+B)/2, dir(B-C)*I, deepcyan);
label("$4$", (A+B)/2, dir(A-B)*I, deepcyan);
label("$7$", (F+E)/2, dir(F-E)*I, deepmagenta);
label("$2$", (F+D)/2, dir(D-F)*I, deepmagenta);
label("$4\sqrt{2}$", (D+E)/2, dir(E-D)*I, deepmagenta);
label("$a$", (B+X)/2, dir(B-X)*I, gray(0.3));
label("$a\sqrt{2}$", (D+X)/2, dir(D-X)*I, gray(0.3));
// Block 2
unitsize(20); pair A, B, C, D, E, F, X, O1, O2; A = (0, 0); B = (4, 0); C = intersectionpoints(circle(A, 6), circle(B, 5))[0]; D = B + (5/4 * (1 + sqrt(2)), 0); E = D + (4 * sqrt(2), 0); F = intersectionpoints(circle(D, 2), circle(E, 7))[1]; X = extension(A, E, C, F); O1 = circumcenter(C, A, D); O2 = circumcenter(C, B, E); filldraw(A--B--C--cycle, lightcyan, deepcyan); filldraw(D--E--F--cycle, lightmagenta, deepmagenta); draw(B--D, gray(0.6)); draw(C--F, gray(0.6)); draw(circumcircle(C, A, D), dashed); draw(circumcircle(C, B, E), dashed); dot("$A$", A, dir(A-O1)); dot("$B$", B, dir(240)); dot("$C$", C, dir(120)); dot("$D$", D, dir(40)); dot("$E$", E, dir(E-O2)); dot("$F$", F, dir(270)); dot("$X$", X, dir(140)); label("$6$", (C+A)/2, dir(C-A)*I, deepcyan); label("$5$", (C+B)/2, dir(B-C)*I, deepcyan); label("$4$", (A+B)/2, dir(A-B)*I, deepcyan); label("$7$", (F+E)/2, dir(F-E)*I, deepmagenta); label("$2$", (F+D)/2, dir(D-F)*I, deepmagenta); label("$4\sqrt{2}$", (D+E)/2, dir(E-D)*I, deepmagenta); label("$a$", (B+X)/2, dir(B-X)*I, gray(0.3)); label("$a\sqrt{2}$", (D+X)/2, dir(D-X)*I, gray(0.3)); | [] |
215 | Let $\overline{AB}$ be a chord of a circle $\omega$, and let $P$ be a point on the chord $\overline{AB}$. Circle $\omega_1$ passes through $A$ and $P$ and is internally tangent to $\omega$. Circle $\omega_2$ passes through $B$ and $P$ and is internally tangent to $\omega$. Circles $\omega_1$ and $\omega_2$ intersect at points $P$ and $Q$. Line $PQ$ intersects $\omega$ at $X$ and $Y$. Assume that $AP=5$, $PB=3$, $XY=11$, and $PQ^2 = \frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. | 2019 AIME I Problem 15 | Let $O_1$ and $O_2$ be the centers of $\omega_1$ and $\omega_2$, respectively. There is a homothety at $A$ sending $\omega$ to $\omega_1$ that sends $B$ to $P$ and $O$ to $O_1$, so $\overline{OO_2}\parallel\overline{O_1P}$. Similarly, $\overline{OO_1}\parallel\overline{O_2P}$, so $OO_1PO_2$ is a parallelogram. Moreover, \[\angle O_1QO_2=\angle O_1PO_2=\angle O_1OO_2,\]hence $OO_1O_2Q$ is cyclic. However, \[OO_1=O_2P=O_2Q,\]so $OO_1O_2Q$ is an isosceles trapezoid. Since $\overline{O_1O_2}\perp\overline{XY}$, $\overline{OQ}\perp\overline{XY}$, so $Q$ is the midpoint of $\overline{XY}$.
By Power of a Point, $PX\cdot PY=PA\cdot PB=15$. Since $PX+PY=XY=11$ and $XQ=11/2$, \[XP=\frac{11-\sqrt{61}}2\implies PQ=XQ-XP=\frac{\sqrt{61}}2\implies PQ^2=\frac{61}4,\]
and the requested sum is $61+4=\boxed{065}$.
(Solution by TheUltimate123)
Note
One may solve for $PX$ first using PoAP, $PX = \frac{11}{2} - \frac{\sqrt{61}}{2}$. Then, notice that $PQ^2$ is rational but $PX^2$ is not, also $PX = \frac{XY}{2} - \frac{\sqrt{61}}{2}$. The most likely explanation for this is that $Q$ is the midpoint of $XY$, so that $XQ = \frac{11}{2}$ and $PQ=\frac{\sqrt{61}}{2}$. Then our answer is $m+n=61+4=\boxed{065}$. One can rigorously prove this using the methods above | // Block 1
size(8cm);
pair O, A, B, P, O1, O2, Q, X, Y;
O=(0, 0);
A=dir(140); B=dir(40);
P=(3A+5B)/8;
O1=extension((A+P)/2, (A+P)/2+(0, 1), A, O);
O2=extension((B+P)/2, (B+P)/2+(0, 1), B, O);
Q=intersectionpoints(circle(O1, length(A-O1)), circle(O2, length(B-O2)))[1];
X=intersectionpoint(P -- (P+(P-Q)*100), circle(O, 1));
Y=intersectionpoint(Q -- (Q+(Q-P)*100), circle(O, 1));
draw(circle(O, 1));
draw(circle(O1, length(A-O1)));
draw(circle(O2, length(B-O2)));
draw(A -- B); draw(X -- Y); draw(A -- O -- B); draw(O1 -- P -- O2);
dot("$O$", O, S);
dot("$A$", A, A);
dot("$B$", B, B);
dot("$P$", P, dir(70));
dot("$Q$", Q, dir(200));
dot("$O_1$", O1, SW);
dot("$O_2$", O2, SE);
dot("$X$", X, X);
dot("$Y$", Y, Y);
// Block 2
size(8cm); pair O, A, B, P, O1, O2, Q, X, Y; O=(0, 0); A=dir(140); B=dir(40); P=(3A+5B)/8; O1=extension((A+P)/2, (A+P)/2+(0, 1), A, O); O2=extension((B+P)/2, (B+P)/2+(0, 1), B, O); Q=intersectionpoints(circle(O1, length(A-O1)), circle(O2, length(B-O2)))[1]; X=intersectionpoint(P -- (P+(P-Q)*100), circle(O, 1)); Y=intersectionpoint(Q -- (Q+(Q-P)*100), circle(O, 1)); draw(circle(O, 1)); draw(circle(O1, length(A-O1))); draw(circle(O2, length(B-O2))); draw(A -- B); draw(X -- Y); draw(A -- O -- B); draw(O1 -- P -- O2); dot("$O$", O, S); dot("$A$", A, A); dot("$B$", B, B); dot("$P$", P, dir(70)); dot("$Q$", Q, dir(200)); dot("$O_1$", O1, SW); dot("$O_2$", O2, SE); dot("$X$", X, X); dot("$Y$", Y, Y); | [] |
215 | Let $\overline{AB}$ be a chord of a circle $\omega$, and let $P$ be a point on the chord $\overline{AB}$. Circle $\omega_1$ passes through $A$ and $P$ and is internally tangent to $\omega$. Circle $\omega_2$ passes through $B$ and $P$ and is internally tangent to $\omega$. Circles $\omega_1$ and $\omega_2$ intersect at points $P$ and $Q$. Line $PQ$ intersects $\omega$ at $X$ and $Y$. Assume that $AP=5$, $PB=3$, $XY=11$, and $PQ^2 = \frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. | 2019 AIME I Problem 15 | $PX \cdot PY=AP \cdot PB=5 \cdot 3=15$ by power of a point. Also, $PX+PY=XY=11$, so $PX$ and $PY$ are solutions to the quadratic $x^2-11x+15=0$ so $PX$ and $PY$ is $\frac{11\pm\sqrt{61}}{2}$ in some order. Now, because we want $PQ^2$ and it is known to be rational, we can guess that $PQ$ is irrational or the problem would simply ask for $PQ$. We can also figure out that since $PQ^2$ is rational, $PQ$ is $\sqrt{\text{[something]}}$. $PQ=QX-PX$, and chances are low that $QX$ is some number with a square root plus or minus $\frac{\sqrt{61}}{2}$ to cancel out the $\frac{\sqrt{61}}{2}$ in $PX$, so one can see that $PQ^2$ is most likely to be $\left(\frac{\sqrt{61}}{2}\right)^2=\frac{61}{4}$, and our answer is $61+4=\boxed{065}$
Note : If our answer is correct, then $QX=\frac{11}{2}$, which made $Q$ the midpoint of $XY$, a feature that occurs often in AIME problems, so that again made our answer probable. Midpoints have many properties and there is a lot of ways to show if a point is the midpoint of a segment. Even if the answer is wrong, it's still the same as leaving it blank and 065 is a good guess. ~Ddk001 | // Block 1
size(8cm);
pair O, A, B, P, O1, O2, Q, X, Y;
O=(0, 0);
A=dir(140); B=dir(40);
P=(3A+5B)/8;
O1=extension((A+P)/2, (A+P)/2+(0, 1), A, O);
O2=extension((B+P)/2, (B+P)/2+(0, 1), B, O);
Q=intersectionpoints(circle(O1, length(A-O1)), circle(O2, length(B-O2)))[1];
X=intersectionpoint(P -- (P+(P-Q)*100), circle(O, 1));
Y=intersectionpoint(Q -- (Q+(Q-P)*100), circle(O, 1));
draw(circle(O, 1));
draw(circle(O1, length(A-O1)));
draw(circle(O2, length(B-O2)));
draw(A -- B,red); draw(X -- Y,green);
dot("$A$", A, A);
dot("$B$", B, B);
dot("$P$", P, dir(70),blue);
dot("$Q$", Q, dir(200));
dot("$X$", X, X);
dot("$Y$", Y, Y);
label("$3$", (A+P)/2, N, red);
label("$5$", (B+P)/2, N, red);
draw(brace(X,Y));
label("$11$",brace(X,Y),dir(20));
// Block 2
size(8cm); pair O, A, B, P, O1, O2, Q, X, Y; O=(0, 0); A=dir(140); B=dir(40); P=(3A+5B)/8; O1=extension((A+P)/2, (A+P)/2+(0, 1), A, O); O2=extension((B+P)/2, (B+P)/2+(0, 1), B, O); Q=intersectionpoints(circle(O1, length(A-O1)), circle(O2, length(B-O2)))[1]; X=intersectionpoint(P -- (P+(P-Q)*100), circle(O, 1)); Y=intersectionpoint(Q -- (Q+(Q-P)*100), circle(O, 1)); draw(circle(O, 1)); draw(circle(O1, length(A-O1))); draw(circle(O2, length(B-O2))); draw(A -- B,red); draw(X -- Y,green); dot("$A$", A, A); dot("$B$", B, B); dot("$P$", P, dir(70),blue); dot("$Q$", Q, dir(200)); dot("$X$", X, X); dot("$Y$", Y, Y); label("$3$", (A+P)/2, N, red); label("$5$", (B+P)/2, N, red); draw(brace(X,Y)); label("$11$",brace(X,Y),dir(20)); | [] |
216 | Two different points, $C$ and $D$, lie on the same side of line $AB$ so that $\triangle ABC$ and $\triangle BAD$ are congruent with $AB = 9$, $BC=AD=10$, and $CA=DB=17$. The intersection of these two triangular regions has area $\tfrac mn$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. | 2019 AIME II Problem 1 | - Diagram by Brendanb4321
Extend $AB$ to form a right triangle with legs $6$ and $8$ such that $AD$ is the hypotenuse and connect the points $CD$ so
that you have a rectangle. (We know that $\triangle ADE$ is a $6-8-10$, since $\triangle DEB$ is an $8-15-17$.) The base $CD$ of the rectangle will be $9+6+6=21$. Now, let $O$ be the intersection of $BD$ and $AC$. This means that $\triangle ABO$ and $\triangle DCO$ are similar with ratio $\frac{21}{9}=\frac73$. Set up a proportion, knowing that the two heights add up to 8. We will let $y$ be the height from $O$ to $DC$, and $x$ be the height of $\triangle ABO$.
\[\frac{7}{3}=\frac{y}{x}\]
\[\frac{7}{3}=\frac{8-x}{x}\]
\[7x=24-3x\]
\[10x=24\]
\[x=\frac{12}{5}\]
This means that the area is $A=\tfrac{1}{2}(9)(\tfrac{12}{5})=\tfrac{54}{5}$. This gets us $54+5=\boxed{059}.$
-Solution by the Math Wizard, Number Magician of the Second Order, Head of the Council of the Geometers | // Block 1
unitsize(10);
pair A = (0,0);
pair B = (9,0);
pair C = (15,8);
pair D = (-6,8);
pair E = (-6,0);
draw(A--B--C--cycle);
draw(B--D--A);
label("$A$",A,dir(-120));
label("$B$",B,dir(-60));
label("$C$",C,dir(60));
label("$D$",D,dir(120));
label("$E$",E,dir(-135));
label("$9$",(A+B)/2,dir(-90));
label("$10$",(D+A)/2,dir(-150));
label("$10$",(C+B)/2,dir(-30));
label("$17$",(D+B)/2,dir(60));
label("$17$",(A+C)/2,dir(120));
draw(D--E--A,dotted);
label("$8$",(D+E)/2,dir(180));
label("$6$",(A+E)/2,dir(-90));
// Block 2
unitsize(10); pair A = (0,0); pair B = (9,0); pair C = (15,8); pair D = (-6,8); pair E = (-6,0); draw(A--B--C--cycle); draw(B--D--A); label("$A$",A,dir(-120)); label("$B$",B,dir(-60)); label("$C$",C,dir(60)); label("$D$",D,dir(120)); label("$E$",E,dir(-135)); label("$9$",(A+B)/2,dir(-90)); label("$10$",(D+A)/2,dir(-150)); label("$10$",(C+B)/2,dir(-30)); label("$17$",(D+B)/2,dir(60)); label("$17$",(A+C)/2,dir(120)); draw(D--E--A,dotted); label("$8$",(D+E)/2,dir(180)); label("$6$",(A+E)/2,dir(-90)); | [] |
216 | Two different points, $C$ and $D$, lie on the same side of line $AB$ so that $\triangle ABC$ and $\triangle BAD$ are congruent with $AB = 9$, $BC=AD=10$, and $CA=DB=17$. The intersection of these two triangular regions has area $\tfrac mn$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. | 2019 AIME II Problem 1 | - Diagram by Brendanb4321 extended by Duoquinquagintillion
Begin with the first step of solution 1, seeing $AD$ is the hypotenuse of a $6-8-10$ triangle and calling the intersection of $DB$ and $AC$ point $E$. Next, notice $DB$ is the hypotenuse of an $8-15-17$ triangle. Drop an altitude from $E$ with length $h$, so the other leg of the new triangle formed has length $4.5$. Notice we have formed similar triangles, and we can solve for $h$.
\[\frac{h}{4.5} = \frac{8}{15}\]
\[h = \frac{36}{15} = \frac{12}{5}\]
So $\triangle ABE$ has area \[\frac{ \frac{12}{5} \cdot 9}{2} = \frac{54}{5}\]
And $54+5=\boxed{059}.$
- Solution by Duoquinquagintillion | // Block 1
unitsize(10);
pair A = (0,0);
pair B = (9,0);
pair C = (15,8);
pair D = (-6,8);
draw(A--B--C--cycle);
draw(B--D--A);
label("$A$",A,dir(-120));
label("$B$",B,dir(-60));
label("$C$",C,dir(60));
label("$D$",D,dir(120));
label("$9$",(A+B)/2,dir(-90));
label("$10$",(D+A)/2,dir(-150));
label("$10$",(C+B)/2,dir(-30));
label("$17$",(D+B)/2,dir(60));
label("$17$",(A+C)/2,dir(120));
draw(D--(-6,0)--A,dotted);
label("$8$",(D+(-6,0))/2,dir(180));
label("$6$",(A+(-6,0))/2,dir(-90));
draw((4.5,0)--(4.5,2.4),dotted);
label("$h$", (4.5,1.2), dir(180));
label("$4.5$", (6,0), dir(90));
// Block 2
unitsize(10); pair A = (0,0); pair B = (9,0); pair C = (15,8); pair D = (-6,8); draw(A--B--C--cycle); draw(B--D--A); label("$A$",A,dir(-120)); label("$B$",B,dir(-60)); label("$C$",C,dir(60)); label("$D$",D,dir(120)); label("$9$",(A+B)/2,dir(-90)); label("$10$",(D+A)/2,dir(-150)); label("$10$",(C+B)/2,dir(-30)); label("$17$",(D+B)/2,dir(60)); label("$17$",(A+C)/2,dir(120)); draw(D--(-6,0)--A,dotted); label("$8$",(D+(-6,0))/2,dir(180)); label("$6$",(A+(-6,0))/2,dir(-90)); draw((4.5,0)--(4.5,2.4),dotted); label("$h$", (4.5,1.2), dir(180)); label("$4.5$", (6,0), dir(90)); | [] |
216 | Two different points, $C$ and $D$, lie on the same side of line $AB$ so that $\triangle ABC$ and $\triangle BAD$ are congruent with $AB = 9$, $BC=AD=10$, and $CA=DB=17$. The intersection of these two triangular regions has area $\tfrac mn$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. | 2019 AIME II Problem 1 | First, let's define H as the intersection of CB and DA, and define G as the midpoint of AB. Next, let E and F be the feet of the perpendicular lines from C and D to line AB respectively. We get a pleasing line of symmetry HG where A maps to B, C maps to D, and E maps to F. We notice that 8-15-17 and 8-6-10 are both pythagorean triples and we test that theory. Then, we find AB = 15 - 6 = 9 and GB = 4.5. Since △CEB~△HGB, we find that HG = 2.4. Then, the area of △HGB is 5.4 and the total overlap is 10.8 = 54/5. Noting that GCD(54,5)=1, we add them to get 59.
Note: I omitted some computation
~ Afly (talk) | //Made by Afly. I used some resources. //Took me 10 min to get everything right. import olympiad; unitsize(18); pair A = (0,0); pair B = (0,8); pair C = (6,0); pair D = (15,0); pair E = (21,0); pair F = (21,8); pair G = (21/2,0); pair H = intersectionpoints(B--D,C--F)[0]; pen dash1 = linetype(new real [] {9,9})+linewidth(1); pen solid1 = linetype(new real [] {9,0})+linewidth(1); pen dash2 = linetype(new real [] {3,3})+linewidth(1); fill(C--G--H--cycle,rgb(3/4,1/4,1/4)); fill(D--G--H--cycle,rgb(3/4,3/4,1/4)); draw(C--A--B,dash1); draw(C--B--D--C,solid1); draw(F--E--D,dash1); draw(F--D--C--F,solid1); draw(G--H,dash2); draw(brace(D+dir(270),A+dir(270)),solid1); draw(brace(D,C),solid1); draw(A--A+2*dir(180),dash1,EndArrow); draw(E--E+2*dir(0),dash1,EndArrow); pair L1 = (15/2,-7/2); pair L2 = (21/2,-13/8); label("15",L1); label("8",A--B,W); label("6",A--C,S); label("10",B--C,SW); label("17",B--D,NE); label("9",L2); label("4.5",G--D,S); label("2.4",G--H,W); markscalefactor = 1/16; draw(rightanglemark(H,G,D)); draw(rightanglemark(B,A,C)); draw(rightanglemark(D,E,F)); label("A",C,SW); label("B",D,SE); label("C",B,NW); label("D",F,NE); label("E",A,SW); label("F",E,SE); label("G",G,NW); label("H",H,N); | [] |
217 | Triangle $ABC$ has side lengths $AB=120,BC=220$, and $AC=180$. Lines $\ell_A,\ell_B$, and $\ell_C$ are drawn parallel to $\overline{BC},\overline{AC}$, and $\overline{AB}$, respectively, such that the intersections of $\ell_A,\ell_B$, and $\ell_C$ with the interior of $\triangle ABC$ are segments of lengths $55,45$, and $15$, respectively. Find the perimeter of the triangle whose sides lie on lines $\ell_A,\ell_B$, and $\ell_C$. | 2019 AIME II Problem 7 | Let's squish a triangle with side lengths 15, 22.5, and 27.5 into a equilateral triangle with side length 1. Then, the original triangle gets turned into a equilateral triangle with side length 8. Since 15 is one eighth of 120, it has a length of one. Since 45 and 55 are one fourth of 180 and 220 respectively, they are both two long. We extend the three segments to form a big equilateral triangle shown in black. Notice it has a side length of 11. Now that our task is done, let's undo the distortion. We get 11(15+22.5+27.5)=11(65)=715.
~ Afly (talk) | // Block 1
pair A,B,C; B = (0,0); C = (1,0); A = intersectionpoints(circle(B,3/2),circle(C,11/6))[0]; draw(A--B--C--cycle); draw((3/2,3/4)--(5/2,3/4)); draw((3/2,1/4)--(5/2,1/4)); draw((9/4,1)--(11/4,1/2)--(9/4,0)); draw(shift(dir(0)*13/4)*shift(dir(30))*polygon(3));
// Block 2
for (int i=0; i<8; ++i) { for (int j=0; j<i+1; ++j) { draw(shift(dir(30))*shift(dir(0)*i*sqrt(3))*shift(dir(120)*j*sqrt(3))*polygon(3)); } } pair A = origin+2*sqrt(3)*dir(0)+7*sqrt(3)*dir(120); pair B = origin+13*sqrt(3)*dir(0)+7*sqrt(3)*dir(120); pair C = origin+2*sqrt(3)*dir(0)-4*sqrt(3)*dir(120); pair D = origin + 2*sqrt(3)*dir(0); pair E = origin + 2*sqrt(3)*dir(60); pair F = origin + 7*sqrt(3)*dir(60); pair G = origin + 7*sqrt(3)*dir(60) + 1*sqrt(3)*dir(0); pair H = origin + 6*sqrt(3)*dir(0) + 2*sqrt(3)*dir(60); pair I = origin + 6*sqrt(3)*dir(0); draw(A--B--C--cycle,linewidth(3)); draw(D--E,linewidth(3)+rgb(3/4,1/4,1/4)); draw(F--G,linewidth(3)+rgb(1/4,3/4,1/4)); draw(H--I,linewidth(3)+rgb(1/4,1/4,3/4)); | [] |
218 | Triangle $ABC$ has side lengths $AB=7, BC=8,$ and $CA=9.$ Circle $\omega_1$ passes through $B$ and is tangent to line $AC$ at $A.$ Circle $\omega_2$ passes through $C$ and is tangent to line $AB$ at $A.$ Let $K$ be the intersection of circles $\omega_1$ and $\omega_2$ not equal to $A.$ Then $AK=\tfrac mn,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | 2019 AIME II Problem 11 | -Diagram by Brendanb4321
Note that from the tangency condition that the supplement of $\angle CAB$ with respects to lines $AB$ and $AC$ are equal to $\angle AKB$ and $\angle AKC$, respectively, so from tangent-chord, \[\angle AKC=\angle AKB=180^{\circ}-\angle BAC\] Also note that $\angle ABK=\angle KAC$$^{(*)}$, so $\triangle AKB\sim \triangle CKA$. Using similarity ratios, we can easily find \[AK^2=BK*KC\] However, since $AB=7$ and $CA=9$, we can use similarity ratios to get \[BK=\frac{7}{9}AK, CK=\frac{9}{7}AK\]
Now we use Law of Cosines on $\triangle AKB$: From reverse Law of Cosines, $\cos{\angle BAC}=\frac{11}{21}\implies \cos{(180^{\circ}-\angle BAC)}=\angle AKB=-\frac{11}{21}$
Giving us \[AK^2+\frac{49}{81}AK^2+\frac{22}{27}AK^2=49\] \[\implies \frac{196}{81}AK^2=49\] \[AK=\frac{9}{2}\] so our answer is $9+2=\boxed{011}$.
$^{(*)}$ Let $O$ be the center of $\omega_1$. Then $\angle KAC = 90 - \angle OAK = 90 - \frac{1}{2}(180 - \angle AOK) = \frac{\angle AOK}{2} = \angle ABK$. Thus, $\angle ABK = \angle KAC$
-franchester; $^{(*)}$ by firebolt360
Supplement
In order to get to the Law of Cosines first, we first apply the LOC to $\triangle{ABC},$ obtaining $\angle{BAC}.$
We angle chase before applying the law of cosines to $\angle{AKB}.$
Note that $\angle{ABK}=\angle{KAC}$ and $\angle{KCA}=\angle{KAB}$ from tangent-chord.
Thus, $\angle{AKC}=\angle{AKB}=180^{\circ}-(\angle{ABK}+\angle{KAB}).$
However from our tangent chord, note that:
\[\angle{ABK}+\angle{KAB}=\angle{KAC}+\angle{KAB}=\angle{BAC}.\]
Thus, $\angle{AKB}=180^\circ-\angle{BAC}.$
As an alternative approach, note that the sum of the angles in quadrilateral $ABKC$ is $360^{\circ}$ and we can find $\angle{AKB}=\frac12$ of convex $\angle{BKC},$ which is just:
\[\frac12 \left(360^{\circ}-2(\angle{KAB}+\angle{KBA}\right) = 180^\circ - \angle{BAC}.\]
~mathboy282 | // Block 1
unitsize(20);
pair B = (0,0);
pair A = (2,sqrt(45));
pair C = (8,0);
draw(circumcircle(A,B,(-17/8,0)),rgb(.7,.7,.7));
draw(circumcircle(A,C,(49/8,0)),rgb(.7,.7,.7));
draw(B--A--C--cycle);
label("$A$",A,dir(105));
label("$B$",B,dir(-135));
label("$C$",C,dir(-75));
dot((2.68,2.25));
label("$K$",(2.68,2.25),2*down);
label("$\omega_1$",(-4.5,1));
label("$\omega_2$",(12.75,6));
label("$7$",(A+B)/2,dir(140));
label("$8$",(B+C)/2,dir(-90));
label("$9$",(A+C)/2,dir(60));
// Block 2
unitsize(20); pair B = (0,0); pair A = (2,sqrt(45)); pair C = (8,0); draw(circumcircle(A,B,(-17/8,0)),rgb(.7,.7,.7)); draw(circumcircle(A,C,(49/8,0)),rgb(.7,.7,.7)); draw(B--A--C--cycle); label("$A$",A,dir(105)); label("$B$",B,dir(-135)); label("$C$",C,dir(-75)); dot((2.68,2.25)); label("$K$",(2.68,2.25),2*down); label("$\omega_1$",(-4.5,1)); label("$\omega_2$",(12.75,6)); label("$7$",(A+B)/2,dir(140)); label("$8$",(B+C)/2,dir(-90)); label("$9$",(A+C)/2,dir(60)); | [] |
219 | Regular octagon $A_1A_2A_3A_4A_5A_6A_7A_8$ is inscribed in a circle of area $1.$ Point $P$ lies inside the circle so that the region bounded by $\overline{PA_1},\overline{PA_2},$ and the minor arc $\widehat{A_1A_2}$ of the circle has area $\tfrac{1}{7},$ while the region bounded by $\overline{PA_3},\overline{PA_4},$ and the minor arc $\widehat{A_3A_4}$ of the circle has area $\tfrac{1}{9}.$ There is a positive integer $n$ such that the area of the region bounded by $\overline{PA_6},\overline{PA_7},$ and the minor arc $\widehat{A_6A_7}$ of the circle is equal to $\tfrac{1}{8}-\tfrac{\sqrt2}{n}.$ Find $n.$ | 2019 AIME II Problem 13 | Instead of considering the actual values of the areas, consider only the changes in the areas that result from moving point $P$ from the center of the circle. We will proceed by coordinates. Set the origin at the center of the circle and refer to the following diagram, where the octagon is oriented so as $A_1A_2$ is horizontal (and therefore $A_3A_4$ is vertical). Note that the area bounded by $\overline{A_iA_j}$ and the arc $\widehat{A_iA_j}$ is fixed, so we only need to consider the relevant triangles.
Define one arbitrary unit as the distance that you need to move $P$ from $A_1A_2$ to change the area of $\triangle PA_1A_2$ by $1$. We can see that $P$ was moved down by $\tfrac{1}{7}-\tfrac{1}{8}=\tfrac{1}{56}$ units to make the area defined by $P$, $A_1$, and $A_2$ $\tfrac{1}{7}$. Similarly, $P$ was moved right by $\tfrac{1}{8}-\tfrac{1}{9}=\tfrac{1}{72}$ to make the area defined by $P$, $A_3$, and $A_4$ $\tfrac{1}{9}$. This means that $P$ has coordinates $(\tfrac{1}{72},-\tfrac{1}{56})$.
Now, we need to consider how this displacement in $P$ affected the area defined by $P$, $A_6$, and $A_7$. This is equivalent to finding the shortest distance between $P$ and the blue line in the diagram (as $K=\tfrac{1}{2}bh$ and the blue line represents $h$ while $b$ is fixed). Using an isosceles right triangle, one can find the that shortest distance between $P$ and this line is $\tfrac{\sqrt{2}}{2}(\tfrac{1}{56}-\tfrac{1}{72})=\tfrac{\sqrt{2}}{504}$.
Remembering the definition of our unit, this yields a final area of
\[\frac{1}{8}-\frac{\sqrt{2}}{\boxed{504}}.\]
-Archeon | // Block 1
size(7cm);
draw(Circle((0,0),1));
pair P = (0.1,-0.15);
filldraw(P--dir(112.5)--dir(112.5-45)--cycle,yellow,red);
filldraw(P--dir(112.5-90)--dir(112.5-135)--cycle,yellow,red);
filldraw(P--dir(112.5-225)--dir(112.5-270)--cycle,green,red);
dot(P);
for(int i=0; i<8; ++i)
{
draw(dir(22.5+45i)--dir(67.5+45i));
draw((0,0)--dir(22.5+45i),gray+dashed);
}
draw(dir(135)--dir(-45),blue+linewidth(1));
label("$P$", P, dir(-75));
label("$A_1$", dir(112.5), dir(112.5));
label("$A_2$", dir(112.5-45), dir(112.5-45));
label("$A_3$", dir(112.5-90), dir(112.5-90));
label("$A_4$", dir(112.5-135), dir(112.5-135));
label("$A_5$", dir(112.5-180), dir(112.5-180));
label("$A_6$", dir(112.5-225), dir(112.5-225));
label("$A_7$", dir(112.5-270), dir(112.5-270));
label("$A_8$", dir(112.5-315), dir(112.5-315));
dot(dir(112.5)^^dir(112.5-45)^^dir(112.5-90)^^dir(112.5-135)^^dir(112.5-180)^^dir(112.5-225)^^dir(112.5-270)^^dir(112.5-315));
// Block 2
size(7cm); draw(Circle((0,0),1)); pair P = (0.1,-0.15); filldraw(P--dir(112.5)--dir(112.5-45)--cycle,yellow,red); filldraw(P--dir(112.5-90)--dir(112.5-135)--cycle,yellow,red); filldraw(P--dir(112.5-225)--dir(112.5-270)--cycle,green,red); dot(P); for(int i=0; i<8; ++i) { draw(dir(22.5+45i)--dir(67.5+45i)); draw((0,0)--dir(22.5+45i),gray+dashed); } draw(dir(135)--dir(-45),blue+linewidth(1)); label("$P$", P, dir(-75)); label("$A_1$", dir(112.5), dir(112.5)); label("$A_2$", dir(112.5-45), dir(112.5-45)); label("$A_3$", dir(112.5-90), dir(112.5-90)); label("$A_4$", dir(112.5-135), dir(112.5-135)); label("$A_5$", dir(112.5-180), dir(112.5-180)); label("$A_6$", dir(112.5-225), dir(112.5-225)); label("$A_7$", dir(112.5-270), dir(112.5-270)); label("$A_8$", dir(112.5-315), dir(112.5-315)); dot(dir(112.5)^^dir(112.5-45)^^dir(112.5-90)^^dir(112.5-135)^^dir(112.5-180)^^dir(112.5-225)^^dir(112.5-270)^^dir(112.5-315)); | [] |
220 | In $\triangle ABC$ with $AB=AC,$ point $D$ lies strictly between $A$ and $C$ on side $\overline{AC},$ and point $E$ lies strictly between $A$ and $B$ on side $\overline{AB}$ such that $AE=ED=DB=BC.$ The degree measure of $\angle ABC$ is $\tfrac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | 2020 AIME I Problem 1 | If we set $\angle{BAC}$ to $x$, we can find all other angles through these two properties:
1. Angles in a triangle sum to $180^{\circ}$.
2. The base angles of an isosceles triangle are congruent.
Now we angle chase. $\angle{ADE}=\angle{EAD}=x$, $\angle{AED} = 180-2x$, $\angle{BED}=\angle{EBD}=2x$, $\angle{EDB} = 180-4x$, $\angle{BDC} = \angle{BCD} = 3x$, $\angle{CBD} = 180-6x$. Since $AB = AC$ as given by the problem, $\angle{ABC} = \angle{ACB}$, so $180-4x=3x$. Therefore, $x = 180/7^{\circ}$, and our desired angle is \[180-4\left(\frac{180}{7}\right) = \frac{540}{7}\] for an answer of $\boxed{547}$.
See here for a video solution: https://youtu.be/4e8Hk04Ax_E | // Block 1
size(10cm);
pair A, B, C, D, F;
A = (0, tan(3 * pi / 7));
B = (1, 0);
C = (-1, 0);
F = rotate(90/7, A) * (A - (0, 2));
D = rotate(900/7, F) * A;
draw(A -- B -- C -- cycle);
draw(F -- D);
draw(D -- B);
label("$A$", A, N);
label("$B$", B, E);
label("$C$", C, W);
label("$D$", D, W);
label("$E$", F, E);
// Block 2
size(10cm); pair A, B, C, D, F; A = (0, tan(3 * pi / 7)); B = (1, 0); C = (-1, 0); F = rotate(90/7, A) * (A - (0, 2)); D = rotate(900/7, F) * A; draw(A -- B -- C -- cycle); draw(F -- D); draw(D -- B); label("$A$", A, N); label("$B$", B, E); label("$C$", C, W); label("$D$", D, W); label("$E$", F, E); | [] |
220 | In $\triangle ABC$ with $AB=AC,$ point $D$ lies strictly between $A$ and $C$ on side $\overline{AC},$ and point $E$ lies strictly between $A$ and $B$ on side $\overline{AB}$ such that $AE=ED=DB=BC.$ The degree measure of $\angle ABC$ is $\tfrac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | 2020 AIME I Problem 1 | Let $x = \angle ABC = \angle ACB$. Because $\triangle BCD$ is isosceles, $\angle CBD = 180^\circ - 2x$. Then
\[\angle DBE = x - \angle CBD = x - (180^\circ - 2x) = 3x - 180^\circ\!.\]Because $\triangle EDA$ and $\triangle DBE$ are also isosceles,
\[\angle BAC =\frac12(\angle EAD + \angle ADE) = \frac12(\angle BED)= \frac12(\angle DBE)\]
\[= \frac12 (3x - 180^\circ) = \frac32x-90^\circ\!.\]
Because $\triangle ABC$ is isosceles, $\angle BAC$ is also $180^\circ-2x$, so $\frac32x - 90^\circ = 180^\circ - 2x$, and it follows that
$\angle ABC = x = \left(\frac{540}7\right)^\circ$. The requested sum is $540+7 = 547$.
https://artofproblemsolving.com/wiki/index.php/1961_AHSME_Problems/Problem_25
(Almost Mirrored)
See here for a video solution:
https://youtu.be/4XkA0DwuqYk | // Block 1
unitsize(4 cm);
pair A, B, C, D, E;
real a = 180/7;
A = (0,0);
B = dir(180 - a/2);
C = dir(180 + a/2);
D = extension(B, B + dir(270 + a), A, C);
E = extension(D, D + dir(90 - 2*a), A, B);
draw(A--B--C--cycle);
draw(B--D--E);
label("$A$", A, dir(0));
label("$B$", B, NW);
label("$C$", C, SW);
label("$D$", D, S);
label("$E$", E, N);
// Block 2
unitsize(4 cm); pair A, B, C, D, E; real a = 180/7; A = (0,0); B = dir(180 - a/2); C = dir(180 + a/2); D = extension(B, B + dir(270 + a), A, C); E = extension(D, D + dir(90 - 2*a), A, B); draw(A--B--C--cycle); draw(B--D--E); label("$A$", A, dir(0)); label("$B$", B, NW); label("$C$", C, SW); label("$D$", D, S); label("$E$", E, N); | [] |
221 | Six cards numbered $1$ through $6$ are to be lined up in a row. Find the number of arrangements of these six cards where one of the cards can be removed leaving the remaining five cards in either ascending or descending order. | 2020 AIME I Problem 5 | We start with five cards in ascending order, then insert the sixth card to obtain a valid arrangement.
Based on the card to be inserted, we have six cases. As shown below, the red squares indicate the possible positions to insert the sixth card.
Note that all arrangements of the six cards are distinct.
There are $26$ arrangements in which removing one card will leave the remaining five cards in ascending order. By symmetry, there are $26$ arrangements in which removing one card will leave the remaining five cards in descending order. So, the answer is $26\cdot2=\boxed{052}.$
~MRENTHUSIASM | // Block 1
/* Made by MRENTHUSIASM */
unitsize(7mm);
void drawSquare(real x, real y)
{
draw((x+0.5,y+0.5)--(x-0.5,y+0.5)--(x-0.5,y-0.5)--(x+0.5,y-0.5)--cycle,red);
}
label("$2$",(0.5,8));
label("$3$",(2.5,8));
label("$4$",(4.5,8));
label("$5$",(6.5,8));
label("$6$",(8.5,8));
label("Insert $1.$",(13.5,8),red);
drawSquare(-0.5,8);
drawSquare(1.5,8);
drawSquare(3.5,8);
drawSquare(5.5,8);
drawSquare(7.5,8);
drawSquare(9.5,8);
label("$1$",(0.5,6.5));
label("$3$",(2.5,6.5));
label("$4$",(4.5,6.5));
label("$5$",(6.5,6.5));
label("$6$",(8.5,6.5));
label("Insert $2.$",(13.5,6.5),red);
drawSquare(3.5,6.5);
drawSquare(5.5,6.5);
drawSquare(7.5,6.5);
drawSquare(9.5,6.5);
label("$1$",(0.5,5));
label("$2$",(2.5,5));
label("$4$",(4.5,5));
label("$5$",(6.5,5));
label("$6$",(8.5,5));
label("Insert $3.$",(13.5,5),red);
drawSquare(-0.5,5);
drawSquare(5.5,5);
drawSquare(7.5,5);
drawSquare(9.5,5);
label("$1$",(0.5,3.5));
label("$2$",(2.5,3.5));
label("$3$",(4.5,3.5));
label("$5$",(6.5,3.5));
label("$6$",(8.5,3.5));
label("Insert $4.$",(13.5,3.5),red);
drawSquare(-0.5,3.5);
drawSquare(1.5,3.5);
drawSquare(7.5,3.5);
drawSquare(9.5,3.5);
label("$1$",(0.5,2));
label("$2$",(2.5,2));
label("$3$",(4.5,2));
label("$4$",(6.5,2));
label("$6$",(8.5,2));
label("Insert $5.$",(13.5,2),red);
drawSquare(-0.5,2);
drawSquare(1.5,2);
drawSquare(3.5,2);
drawSquare(9.5,2);
label("$1$",(0.5,0.5));
label("$2$",(2.5,0.5));
label("$3$",(4.5,0.5));
label("$4$",(6.5,0.5));
label("$5$",(8.5,0.5));
label("Insert $6.$",(13.5,0.5),red);
drawSquare(-0.5,0.5);
drawSquare(1.5,0.5);
drawSquare(3.5,0.5);
drawSquare(5.5,0.5);
// Block 2
/* Made by MRENTHUSIASM */ unitsize(7mm); void drawSquare(real x, real y) { draw((x+0.5,y+0.5)--(x-0.5,y+0.5)--(x-0.5,y-0.5)--(x+0.5,y-0.5)--cycle,red); } label("$2$",(0.5,8)); label("$3$",(2.5,8)); label("$4$",(4.5,8)); label("$5$",(6.5,8)); label("$6$",(8.5,8)); label("Insert $1.$",(13.5,8),red); drawSquare(-0.5,8); drawSquare(1.5,8); drawSquare(3.5,8); drawSquare(5.5,8); drawSquare(7.5,8); drawSquare(9.5,8); label("$1$",(0.5,6.5)); label("$3$",(2.5,6.5)); label("$4$",(4.5,6.5)); label("$5$",(6.5,6.5)); label("$6$",(8.5,6.5)); label("Insert $2.$",(13.5,6.5),red); drawSquare(3.5,6.5); drawSquare(5.5,6.5); drawSquare(7.5,6.5); drawSquare(9.5,6.5); label("$1$",(0.5,5)); label("$2$",(2.5,5)); label("$4$",(4.5,5)); label("$5$",(6.5,5)); label("$6$",(8.5,5)); label("Insert $3.$",(13.5,5),red); drawSquare(-0.5,5); drawSquare(5.5,5); drawSquare(7.5,5); drawSquare(9.5,5); label("$1$",(0.5,3.5)); label("$2$",(2.5,3.5)); label("$3$",(4.5,3.5)); label("$5$",(6.5,3.5)); label("$6$",(8.5,3.5)); label("Insert $4.$",(13.5,3.5),red); drawSquare(-0.5,3.5); drawSquare(1.5,3.5); drawSquare(7.5,3.5); drawSquare(9.5,3.5); label("$1$",(0.5,2)); label("$2$",(2.5,2)); label("$3$",(4.5,2)); label("$4$",(6.5,2)); label("$6$",(8.5,2)); label("Insert $5.$",(13.5,2),red); drawSquare(-0.5,2); drawSquare(1.5,2); drawSquare(3.5,2); drawSquare(9.5,2); label("$1$",(0.5,0.5)); label("$2$",(2.5,0.5)); label("$3$",(4.5,0.5)); label("$4$",(6.5,0.5)); label("$5$",(8.5,0.5)); label("Insert $6.$",(13.5,0.5),red); drawSquare(-0.5,0.5); drawSquare(1.5,0.5); drawSquare(3.5,0.5); drawSquare(5.5,0.5); | [] |
222 | A flat board has a circular hole with radius $1$ and a circular hole with radius $2$ such that the distance between the centers of the two holes is $7.$ Two spheres with equal radii sit in the two holes such that the spheres are tangent to each other. The square of the radius of the spheres is $\tfrac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | 2020 AIME I Problem 6 | Set the common radius to $r$. First, take the cross section of the sphere sitting in the hole of radius $1$. If we draw the perpendicular bisector of the chord (the hole) through the circle, this line goes through the center. Connect the center also to where the chord hits the circle, for a right triangle with hypotenuse $r$ and base $1$. Therefore, the height of this circle outside of the hole is $\sqrt{r^2-1}$.
The other circle follows similarly for a height (outside the hole) of $\sqrt{r^2-4}$. Now, if we take the cross section of the entire board, essentially making it two-dimensional, we can connect the centers of the two spheres, then form another right triangle with base $7$, as given by the problem. The height of this triangle is the difference between the heights of the parts of the two spheres outside the holes, which is $\sqrt{r^2-1} - \sqrt{r^2-4}$. Now we can set up an equation in terms of $r$ with the Pythagorean theorem: \[\left(\sqrt{r^2-1} - \sqrt{r^2-4}\right)^2 + 7^2 = (2r)^2.\] Simplifying a few times,
\begin{align*} r^2 - 1 - 2\left(\sqrt{(r^2-1)(r^2-4)}\right) + r^2 - 4 + 49 &= 4r^2 \\ 2r^2-44 &= -2\left(\sqrt{(r^2-1)(r^2-4)}\right) \\ 22-r^2 &= \left(\sqrt{r^4 - 5r^2 + 4}\right) \\ r^4 -44r^2 + 484 &= r^4 - 5r^2 + 4 \\ 39r^2&=480 \\ r^2&=\frac{480}{39} = \frac{160}{13}. \end{align*}
Therefore, our answer is $\boxed{173}$.
~molocyxu | // Block 1
size(10cm);
pair A, B, C, D, O, P, H, L, X, Y;
A = (-1, 0);
B = (1, 0);
H = (0, 0);
C = (5, 0);
D = (9, 0);
L = (7, 0);
O = (0, sqrt(160/13 - 1));
P = (7, sqrt(160/13 - 4));
X = (0, sqrt(160/13 - 4));
Y = (O + P) / 2;
draw(A -- O -- B -- cycle);
draw(C -- P -- D -- cycle);
draw(B -- C);
draw(O -- P);
draw(P -- X, dashed);
draw(O -- H, dashed);
draw(P -- L, dashed);
draw(circle(O, sqrt(160/13)));
draw(circle(P, sqrt(160/13)));
path b = brace(L-(0,1), H-(0,1),0.5);
draw(b);
label("$r$", O -- Y, N);
label("$r$", Y -- P, N);
label("$r$", O -- A, NW);
label("$r$", P -- D, NE);
label("$1$", A -- H, N);
label("$2$", L -- D, N);
label("$7$", b, S);
dot(A^^B^^C^^D^^O^^P^^H^^L^^X^^Y,linewidth(4));
// Block 2
size(10cm); pair A, B, C, D, O, P, H, L, X, Y; A = (-1, 0); B = (1, 0); H = (0, 0); C = (5, 0); D = (9, 0); L = (7, 0); O = (0, sqrt(160/13 - 1)); P = (7, sqrt(160/13 - 4)); X = (0, sqrt(160/13 - 4)); Y = (O + P) / 2; draw(A -- O -- B -- cycle); draw(C -- P -- D -- cycle); draw(B -- C); draw(O -- P); draw(P -- X, dashed); draw(O -- H, dashed); draw(P -- L, dashed); draw(circle(O, sqrt(160/13))); draw(circle(P, sqrt(160/13))); path b = brace(L-(0,1), H-(0,1),0.5); draw(b); label("$r$", O -- Y, N); label("$r$", Y -- P, N); label("$r$", O -- A, NW); label("$r$", P -- D, NE); label("$1$", A -- H, N); label("$2$", L -- D, N); label("$7$", b, S); dot(A^^B^^C^^D^^O^^P^^H^^L^^X^^Y,linewidth(4)); | [] |
223 | A club consisting of $11$ men and $12$ women needs to choose a committee from among its members so that the number of women on the committee is one more than the number of men on the committee. The committee could have as few as $1$ member or as many as $23$ members. Let $N$ be the number of such committees that can be formed. Find the sum of the prime numbers that divide $N.$ | 2020 AIME I Problem 7 | Applying [1] by setting $m=12$, $n=11$, and $r=11$, we obtain $\binom{23}{11}\implies$ $\boxed{081}$.
~Lcz
Short Proof
Consider the following setup:
The dots to the left represent the men, and the dots to the right represent the women. Now, suppose we put a mark on $11$ people (the $*$). Those to the left of the dashed line get to be "in" on the committee if they have a mark. Those on the right side of the dashed line are already on the committee, but if they're marked they get forcibly evicted from it. If there were $x$ people marked on the left, there ends up being $12-(11-x) = x+1$ people not marked on the right. Circles represent those in the committee.
We have our bijection, so the number of ways will be $\binom{23}{11}$.
~programjames1 | // Block 1
size(1000, 100);
for(int i=0; i<23; ++i){
dot((i, 0));
}
draw((10.5, -1.5)--(10.5, 1.5), dashed);
// Block 2
size(1000, 100);
for(int i=0; i<23; ++i){
dot((i, 0));
}
for(int i=0; i<23; ++i){
if(i%2==0){
if(i >= 11){
draw(circle((i, 0), 0.25));
}
continue;
}
label("$*$", (i,0.5), N);
if(i < 11){
draw(circle((i, 0), 0.25));
}
}
draw((10.5, -1.5)--(10.5, 1.5), dashed);
// Block 3
size(1000, 100); for(int i=0; i<23; ++i){ dot((i, 0)); } draw((10.5, -1.5)--(10.5, 1.5), dashed);
// Block 4
size(1000, 100); for(int i=0; i<23; ++i){ dot((i, 0)); } for(int i=0; i<23; ++i){ if(i%2==0){ if(i >= 11){ draw(circle((i, 0), 0.25)); } continue; } label("$*$", (i,0.5), N); if(i < 11){ draw(circle((i, 0), 0.25)); } } draw((10.5, -1.5)--(10.5, 1.5), dashed); | [] |
224 | A bug walks all day and sleeps all night. On the first day, it starts at point $O$, faces east, and walks a distance of $5$ units due east. Each night the bug rotates $60^\circ$ counterclockwise. Each day it walks in this new direction half as far as it walked the previous day. The bug gets arbitrarily close to the point $P$. Then $OP^2=\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. | 2020 AIME I Problem 8 | We notice that the moves cycle every 6 moves, so we plot out the first 6 moves on the coordinate grid with point $O$ as the origin. We will keep a tally of the x-coordinate and y-coordinate separately. Then, we will combine them and account for the cycling using the formula for an infinite geometric series.
First move: The ant moves right $5$.
Second move: We use properties of a $30-60-90$ triangle to get $\frac{5}{4}$ right, $\frac{5\sqrt{3}}{4}$ up.
Third move: $\frac{5}{8}$ left, $\frac{5\sqrt{3}}{8}$ up.
Fourth move: $\frac{5}{8}$ left.
Fifth move: $\frac{5}{32}$ left, $\frac{5\sqrt{3}}{32}$ down.
Sixth move: $\frac{5}{64}$ right, $\frac{5\sqrt{3}}{64}$ down.
Total of x-coordinate: $5 + \frac{5}{4} - \frac{5}{8} - \frac{5}{8} - \frac{5}{32} + \frac{5}{64} = \frac{315}{64}$.
Total of y-coordinate: $0 + \frac{5\sqrt{3}}{4} + \frac{5\sqrt{3}}{8} + 0 - \frac{5\sqrt{3}}{32} - \frac{5\sqrt{3}}{64} = \frac{105\sqrt{3}}{64}$.
After this cycle of six moves, all moves repeat with a factor of $(\frac{1}{2})^6 = \frac{1}{64}$. Using the formula for a geometric series, multiplying each sequence by $\frac{1}{1-\frac{1}{64}} = \frac{64}{63}$ will give us the point $P$.
Now, knowing the initial $x$ and $y,$ we plug this into the geometric series formula ($\frac{a}{1-r}$), and we get $\frac{315}{64} \cdot \frac{64}{63} = 5$, $\frac{105\sqrt{3}}{64} \cdot \frac{64}{63} = \frac{5\sqrt{3}}{3}$.
Therefore, the coordinates of point $P$ are $(5,\frac{5\sqrt{3}}{3})$, so using the Pythagorean Theorem, $OP^2 = \frac{100}{3}$, for an answer of $\boxed{103}$.
-molocyxu | // Block 1
size(8cm);
pair O, A, B, C, D, F, G, H, I, P, X;
O = (0, 0);
A = (5, 0);
X = (8, 0);
P = (5, 5 / sqrt(3));
B = rotate(-120, A) * ((O + A) / 2);
C = rotate(-120, B) * ((A + B) / 2);
D = rotate(-120, C) * ((B + C) / 2);
F = rotate(-120, D) * ((C + D) / 2);
G = rotate(-120, F) * ((D + F) / 2);
H = rotate(-120, G) * ((F + G) / 2);
I = rotate(-120, H) * ((G + H) / 2);
draw(O -- A -- B -- C -- D -- F -- G -- H -- I);
draw(A -- X, dashed);
markscalefactor = 0.05;
path angle = anglemark(X, A, B);
draw(angle);
dot(P);
dot(O);
dot(A);
dot(B);
dot(C);
dot(D);
label("$O$", O, W);
label("$P$", P, E);
label("$A$", A, S);
label("$B$", B, E);
label("$C$", C, E);
label("$D$", D, W);
label("$60^\circ$", angle, ENE*3);
// Block 2
size(8cm); pair O, A, B, C, D, F, G, H, I, P, X; O = (0, 0); A = (5, 0); X = (8, 0); P = (5, 5 / sqrt(3)); B = rotate(-120, A) * ((O + A) / 2); C = rotate(-120, B) * ((A + B) / 2); D = rotate(-120, C) * ((B + C) / 2); F = rotate(-120, D) * ((C + D) / 2); G = rotate(-120, F) * ((D + F) / 2); H = rotate(-120, G) * ((F + G) / 2); I = rotate(-120, H) * ((G + H) / 2); draw(O -- A -- B -- C -- D -- F -- G -- H -- I); draw(A -- X, dashed); markscalefactor = 0.05; path angle = anglemark(X, A, B); draw(angle); dot(P); dot(O); dot(A); dot(B); dot(C); dot(D); label("$O$", O, W); label("$P$", P, E); label("$A$", A, S); label("$B$", B, E); label("$C$", C, E); label("$D$", D, W); label("$60^\circ$", angle, ENE*3); | [] |
225 | Let $S$ be the set of positive integer divisors of $20^9.$ Three numbers are chosen independently and at random with replacement from the set $S$ and labeled $a_1,a_2,$ and $a_3$ in the order they are chosen. The probability that both $a_1$ divides $a_2$ and $a_2$ divides $a_3$ is $\tfrac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m.$ | 2020 AIME I Problem 9 | First, prime factorize $20^9$ as $2^{18} \cdot 5^9$. Denote $a_1$ as $2^{b_1} \cdot 5^{c_1}$, $a_2$ as $2^{b_2} \cdot 5^{c_2}$, and $a_3$ as $2^{b_3} \cdot 5^{c_3}$.
In order for $a_1$ to divide $a_2$, and for $a_2$ to divide $a_3$, $b_1\le b_2\le b_3$, and $c_1\le c_2\le c_3$. We will consider each case separately. Note that the total amount of possibilities is $190^3$, as there are $(18+1)(9+1)=190$ choices for each factor.
We notice that if we add $1$ to $b_2$ and $2$ to $b_3$, then we can reach the stronger inequality $0\le b_1<b_2+1<b_3+2\le 20$. Therefore, if we pick $3$ integers from $0$ to $20$, they will correspond to a unique solution, forming a 1-1 correspondence between the numbers $b_1$, $b_2+1$, and $b_3+2$. This is also equivalent to applying stars and bars on distributing the powers of 2 and 5 through differences. The amount of solutions to this inequality is $\dbinom{21}{3}$.
The case for $c_1$,$c_2$, and $c_3$ proceeds similarly for a result of $\dbinom{12}{3}$. Therefore, the probability of choosing three such factors is \[\frac{\dbinom{21}{3} \cdot \dbinom{12}{3}}{190^3}.\] Simplification gives $\frac{77}{1805}$, and therefore the answer is $\boxed{077}$.
-molocyxu | // Block 1
size(12cm);
for (int x = 1; x < 18; ++x) {
draw((x, 0) -- (x, 9), dotted);
}
for (int y = 1; y < 9; ++y) {
draw((0, y) -- (18, y), dotted);
}
draw((0, 0) -- (18, 0) -- (18, 9) -- (0, 9) -- cycle);
pair b1, b2, b3;
pair c1, c2, c3;
pair a1, a2, a3;
b1 = (3, 0); b2 = (12, 0); b3 = (16, 0);
c1 = (0, 2); c2 = (0, 4); c3 = (0, 8);
a1 = b1 + c1; a2 = b2 + c2; a3 = b3 + c3;
draw(b1 -- a1 -- c1);
draw(b2 -- a2 -- c2);
draw(b3 -- a3 -- c3);
dot(a1); dot(a2); dot(a3);
label("$a_1$", a1, NE);
label("$a_2$", a2, NE);
label("$a_3$", a3, NE);
label("$b_1$", b1, S);
label("$b_2$", b2, S);
label("$b_3$", b3, S);
label("$c_1$", c1, W);
label("$c_2$", c2, W);
label("$c_3$", c3, W);
// Block 2
size(12cm); for (int x = 1; x < 18; ++x) { draw((x, 0) -- (x, 9), dotted); } for (int y = 1; y < 9; ++y) { draw((0, y) -- (18, y), dotted); } draw((0, 0) -- (18, 0) -- (18, 9) -- (0, 9) -- cycle); pair b1, b2, b3; pair c1, c2, c3; pair a1, a2, a3; b1 = (3, 0); b2 = (12, 0); b3 = (16, 0); c1 = (0, 2); c2 = (0, 4); c3 = (0, 8); a1 = b1 + c1; a2 = b2 + c2; a3 = b3 + c3; draw(b1 -- a1 -- c1); draw(b2 -- a2 -- c2); draw(b3 -- a3 -- c3); dot(a1); dot(a2); dot(a3); label("$a_1$", a1, NE); label("$a_2$", a2, NE); label("$a_3$", a3, NE); label("$b_1$", b1, S); label("$b_2$", b2, S); label("$b_3$", b3, S); label("$c_1$", c1, W); label("$c_2$", c2, W); label("$c_3$", c3, W); | [] |
226 | Point $D$ lies on side $\overline{BC}$ of $\triangle ABC$ so that $\overline{AD}$ bisects $\angle BAC.$ The perpendicular bisector of $\overline{AD}$ intersects the bisectors of $\angle ABC$ and $\angle ACB$ in points $E$ and $F,$ respectively. Given that $AB=4,BC=5,$ and $CA=6,$ the area of $\triangle AEF$ can be written as $\tfrac{m\sqrt{n}}p,$ where $m$ and $p$ are relatively prime positive integers, and $n$ is a positive integer not divisible by the square of any prime. Find $m+n+p.$ | 2020 AIME I Problem 13 | Let $M_A$, $M_B$, $M_C$ be the midpoints of arcs $BC$, $CA$, $AB$. By Fact 5, we know that $M_AB = M_AC = M_AI$, and so by Ptolmey's theorem, we deduce that \[AB\cdot M_AC + AC\cdot M_AB = BC\cdot M_AA \implies M_AA = 2M_AI.\]
In particular, we have $AI = IM_A$.
Now the key claim is that:
Claim: $\triangle DEF$ and $\triangle M_AM_BM_C$ are homothetic at $I$ with ratio $2$.
Proof. First, we show that $D$ is the midpoint of $M_AI$. Indeed, we have
\[\frac{ID}{DM_A} = \frac{BI}{BM_A}\cdot \frac{\sin\angle IBC}{\sin \angle CBM_A} = \frac{BI}{AI}\cdot\frac{\sin \angle B/2}{\sin \angle A/2} = 1\]
by Ratio lemma and Law of Sines.
Now observe that:
$\overline{M_BM_C}$ is the perpendicular bisector of $\overline{AI}$,
$\overline{EF}$ is the perpendicular bisector of $\overline{AD}$, and
$ID = AI/2$.
Combining these facts gives that $\overline{EF}$ is a midline in $\triangle IM_BM_C$, which proves the claim. $\blacksquare$
To finish, we compute $[M_AM_BM_C]$, noting that $[AEF] = [DEF] = \tfrac{1}{4}[M_AM_BM_C]$.
By Heron's, we can calculate the circumradius $R = 8/\sqrt{7}$, and by Law of Cosines, we get
\begin{align*}\cos A &= \frac{9}{16}\implies \cos A/2 = \frac{5}{\sqrt{32}} \\ \cos B &= \frac{1}{8} \implies \cos B/2 = \frac{3}{4} \\ \cos C &= \frac{3}{4} \implies \cos C/2 = \sqrt{\frac{7}{8}}.\end{align*}
Then using $[XYZ] = 2R^2\sin X\sin Y\sin Z$, we can compute
\[[M_AM_BM_C] = 2\cdot \frac{64}{7}\cdot \frac{5}{\sqrt{32}}\cdot \frac{3}{4}\cdot \frac{\sqrt{7}}{\sqrt{8}} = \frac{30\sqrt{7}}{7}.\]
Thus $[AEF] = 15\sqrt{7}/14$, which gives a final answer of $\boxed{036}$.
~pinetree1 | // Block 1
defaultpen(fontsize(10pt));
size(200);
pair A, B, C, D, E, F, I, P, MA, MB, MC;
B = (0,0);
C = (5,0);
A = IP(Circle(B, 4), Circle(C, 6), 0);
I = incenter(A, B, C);
D = extension(A, I, B, C);
P = midpoint(A--D);
E = extension(P, rotate(90, P)*A, B, I);
F = extension(P, rotate(90, P)*A, C, I);
MA = IP(Line(A, I, 20), circumcircle(A, B, C), 1);
MB = IP(Line(B, I, 20), circumcircle(A, B, C), 1);
MC = IP(Line(C, I, 20), circumcircle(A, B, C), 1);
draw(A--B--C--cycle, orange);
draw(circumcircle(A, B, C), red);
draw(A--E--F--cycle, lightblue);
draw(E--D--F, lightblue);
draw(A--MA^^B--MB^^C--MC, heavygreen);
draw(MA--MB--MC--cycle, magenta);
dot("$A$", A, dir(120));
dot("$B$", B, dir(220));
dot("$C$", C, dir(320));
dot("$D$", D, dir(230));
dot("$E$", E, dir(330));
dot("$F$", F, dir(250));
dot("$I$", I, dir(80));
dot("$M_A$", MA, dir(270));
dot("$M_B$", MB, dir(60));
dot("$M_C$", MC, dir(150));
// Block 2
defaultpen(fontsize(10pt)); size(200); pair A, B, C, D, E, F, I, P, MA, MB, MC; B = (0,0); C = (5,0); A = IP(Circle(B, 4), Circle(C, 6), 0); I = incenter(A, B, C); D = extension(A, I, B, C); P = midpoint(A--D); E = extension(P, rotate(90, P)*A, B, I); F = extension(P, rotate(90, P)*A, C, I); MA = IP(Line(A, I, 20), circumcircle(A, B, C), 1); MB = IP(Line(B, I, 20), circumcircle(A, B, C), 1); MC = IP(Line(C, I, 20), circumcircle(A, B, C), 1); draw(A--B--C--cycle, orange); draw(circumcircle(A, B, C), red); draw(A--E--F--cycle, lightblue); draw(E--D--F, lightblue); draw(A--MA^^B--MB^^C--MC, heavygreen); draw(MA--MB--MC--cycle, magenta); dot("$A$", A, dir(120)); dot("$B$", B, dir(220)); dot("$C$", C, dir(320)); dot("$D$", D, dir(230)); dot("$E$", E, dir(330)); dot("$F$", F, dir(250)); dot("$I$", I, dir(80)); dot("$M_A$", MA, dir(270)); dot("$M_B$", MB, dir(60)); dot("$M_C$", MC, dir(150)); | [] |
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