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
values |
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654 | The parallelogram bounded by the lines $y=ax+c$, $y=ax+d$, $y=bx+c$, and $y=bx+d$ has area $18$. The parallelogram bounded by the lines $y=ax+c$, $y=ax-d$, $y=bx+c$, and $y=bx-d$ has area $72$. Given that $a$, $b$, $c$, and $d$ are positive integers, what is the smallest possible value of $a+b+c+d$?
$\mathrm {(A)} 13\qquad \mathrm {(B)} 14\qquad \mathrm {(C)} 15\qquad \mathrm {(D)} 16\qquad \mathrm {(E)} 17$ | 2007 AMC 12B Problem 20 | Plotting the parallelogram on the coordinate plane, the 4 corners are at $(0,c),(0,d),\left(\frac{d-c}{a-b},\frac{ad-bc}{a-b}\right),\left(\frac{c-d}{a-b},\frac{bc-ad}{a-b}\right)$. Because $72= 4\cdot 18$, we have that $4(c-d)\left(\frac{c-d}{a-b}\right) = (c+d)\left(\frac{c+d}{a-b}\right)$ or that $2(c-d)=c+d$, which gives $c=3d$ (consider a homothety, or dilation, that carries the first parallelogram to the second parallelogram; because the area increases by $4\times$, it follows that the stretch along the diagonal, or the ratio of side lengths, is $2\times$). The area of the triangular half of the parallelogram on the right side of the y-axis is given by $9 = \frac{1}{2} (c-d)\left(\frac{d-c}{a-b}\right)$, so substituting $c = 3d$:
\[\frac{1}{2} (c-d)\left(\frac{c-d}{a-b}\right) = 9 \quad \Longrightarrow \quad 2d^2 = 9(a-b)\]
Thus $3|d$, and we verify that $d = 3$, $a-b = 2 \Longrightarrow a = 3, b = 1$ will give us a minimum value for $a+b+c+d$. Then $a+b+c+d = 3 + 1 + 9 + 3 = \boxed{\mathbf{(D)} 16}$. | pathpen = linewidth(0.7);
real a = 3, b = 1, c = 9, d = 3;
D((0,c) -- ((d-c)/(a-b),(a*d-b*c)/(a-b)) -- (0,d) -- ((c-d)/(a-b),(b*c-a*d)/(a-b)) -- cycle);
D((0,c) -- ((-d-c)/(a-b),(-a*d-b*c)/(a-b)) -- (0,-d) -- ((c+d)/(a-b),-(-a*d-b*c)/(a-b)) -- cycle); | [] |
655 | Points $A$ and $B$ lie on a circle centered at $O$, and $\angle AOB = 60^\circ$. A second circle is internally tangent to the first and tangent to both $\overline{OA}$ and $\overline{OB}$. What is the ratio of the area of the smaller circle to that of the larger circle?
$\mathrm{(A)}\ \frac{1}{16}\qquad\mathrm{(B)}\ \frac{1}{9}\qquad\mathrm{(C)}\ \frac{1}{8}\qquad\mathrm{(D)}\ \frac{1}{6}\qquad\mathrm{(E)}\ \frac{1}{4}$ | 2008 AMC 10A Problem 16 | Let $P$ be the center of the small circle with radius $r$, and let $Q$ be the point where the small circle is tangent to $OA$. Also, let $C$ be the point where the small circle is tangent to the big circle with radius $R$.
Then $PQO$ is a right triangle. Angle $POQ$ is $30$ degrees because line $OP$ bisects angle $AOB$ (this can be proved by dropping a perpendicular line from $P$ to line $OB$, letting their intersection be point $S$, and proving triangles $PQO$ and $PSO$ congruent), meaning that $PQO$ is a $30-60-90$ triangle. Therefore, $OP=2PQ$.
Since $OP=OC-PC=OC-r=R-r$, we have $R-r=2PQ$, or $R-r=2r$, or $\frac{1}{3}=\frac{r}{R}$.
Ratio of areas of circles is ratio of radii squared, so the answer is $\left(\frac{1}{3}\right)^2 = \frac{1}{9} \Rightarrow \boxed{B}$ | size(200); defaultpen(fontsize(10)); pair O=(0,0), A=(3,0), B=(3/2,3/2*3^.5), C=(3^.5,1), D=(3^.5,0), F=(1.5*3^.5,1.5), G=(2*3^.5,2); picture p = new picture; draw(p,Circle(O,0.2)); clip(p,O--C--A--cycle); add(p); draw(Circle(O,3)); dot(A); dot(B); dot(C); dot(O); draw(A--O--B); draw(O--C--D); draw(C--F); draw(D-(0.2,0)--D-(0.2,-0.2)--D-(0,-0.2)); draw(Circle(C,1)); label("\(30^{\circ}\)",(0.65,0.15),O); label("\(r\)",(C+D)/2,E); label("\(2r\)",(O+C)/2,NNE); label("\(O\)",O,SW); label("\(r\)",(C+F)/2,SE); label("\(R\)",(O+A)/2-(0,0.3),S); label("\(P\)",C,NW); label("\(A\)",(3,0), A); label("\(B\)",(3/2,3/2*3^.5), B); label("\(C\)",(1.5*3^.5,1.5), G); label("\(Q\)",D,SE); | [] |
656 | An equilateral triangle has side length $6$. What is the area of the region containing all points that are outside the triangle but not more than $3$ units from a point of the triangle?
$\mathrm{(A)}\ 36+24\sqrt{3}\qquad\mathrm{(B)}\ 54+9\pi\qquad\mathrm{(C)}\ 54+18\sqrt{3}+6\pi\qquad\mathrm{(D)}\ \left(2\sqrt{3}+3\right)^2\pi\\\mathrm{(E)}\ 9\left(\sqrt{3}+1\right)^2\pi$ | 2008 AMC 10A Problem 17 | The region described contains three rectangles of dimensions $3 \times 6$, and three $120^{\circ}$ degree arcs of circles of radius $3$. Thus the answer is \[3(3 \times 6) + 3 \left( \frac{120^{\circ}}{360^{\circ}} \times 3^2 \pi\right) = 54 + 9\pi \Longrightarrow \mathrm{(B)}.\] | // Block 1
pointpen = black; pathpen = black+linewidth(0.7); pen d = linetype("6 6")+linewidth(0.7);
pair A=(0,0),B=(6,0),C=6*expi(-pi/3);
D(arc(A,3,90,210)); D(arc(B,3,-30,90)); D(arc(C,3,210,330));
D(arc(A,-3,90,210),d); D(arc(B,-3,-30,90),d); D(arc(C,-3,210,330),d);
D(D(A)--D(B)--D(C)--cycle,linewidth(1));
D(A--(0,3)--(6,3)--B); D(A--3*expi(7/6*pi)--C+3*expi(7/6*pi)--C); D(B--B+3*expi(11/6*pi)--C+3*expi(11/6*pi)--C);
MP("3",(0,1.5),W); MP("6",(3,0),NW);
// Block 2
pointpen = black; pathpen = black+linewidth(0.7); pen d = linetype("6 6")+linewidth(0.7); pair A=(0,0),B=(6,0),C=6*expi(-pi/3); D(arc(A,3,90,210)); D(arc(B,3,-30,90)); D(arc(C,3,210,330)); D(arc(A,-3,90,210),d); D(arc(B,-3,-30,90),d); D(arc(C,-3,210,330),d); D(D(A)--D(B)--D(C)--cycle,linewidth(1)); D(A--(0,3)--(6,3)--B); D(A--3*expi(7/6*pi)--C+3*expi(7/6*pi)--C); D(B--B+3*expi(11/6*pi)--C+3*expi(11/6*pi)--C); MP("3",(0,1.5),W); MP("6",(3,0),NW); | [] |
657 | Rectangle $PQRS$ lies in a plane with $PQ=RS=2$ and $QR=SP=6$. The rectangle is rotated $90^\circ$ clockwise about $R$, then rotated $90^\circ$ clockwise about the point $S$ moved to after the first rotation. What is the length of the path traveled by point $P$?
$\mathrm{(A)}\ \left(2\sqrt{3}+\sqrt{5}\right)\pi\qquad\mathrm{(B)}\ 6\pi\qquad\mathrm{(C)}\ \left(3+\sqrt{10}\right)\pi\qquad\mathrm{(D)}\ \left(\sqrt{3}+2\sqrt{5}\right)\pi\\\mathrm{(E)}\ 2\sqrt{10}\pi$ | 2008 AMC 10A Problem 19 | We let $P'Q'R'S'$ be the rectangle after the first rotation, and $P''Q''R''S''$ be the rectangle after the second rotation. Point $P$ pivots about $R$ in an arc of a circle of radius $\sqrt{2^2+6^2} = 2\sqrt{10}$, and since $\angle PRS,\, \angle P'RQ'$ are complementary, it follows that the arc has a degree measure of $90^{\circ}$ and length $\frac14$ of the circumference. Thus, $P$ travels $\frac 14 \left(4\sqrt{10}\right)\pi = \sqrt{10}\pi$ in the first rotation.
Similarly, in the second rotation, $P$ travels in a $90^{\circ}$ arc about $S'$, with the radius being $6$. It travels $\frac 14(12)\pi = 3\pi$. Therefore, the total distance it travels is $\left(3+\sqrt{10}\right)\pi\ \mathrm{(C)}$. | // Block 1
size(220);pathpen=black+linewidth(0.65);pointpen=black;
/* draw in rectangles */
D(MP("R",(0,0))--MP("Q",(-6,0))--MP("P",(-6,2),N)--MP("S",(0,2),NW)--cycle);
D((0,0)--MP("Q'",(0,6),SW)--MP("P'",(2,6),SE)--MP("S'",(2,0))--cycle);
D(MP("R''",(2,2),NE)--MP("Q''",(8,2),N)--MP("P''",(8,0))--(2,0)--cycle);
D(arc((0,0),(2,6),(-6,2)),dashed);D(arc((2,0),(8,0),(2,6)),dashed);D((2,6)--(0,0)--(-6,2),dashed);
D(rightanglemark((2,6),(0,0),(-6,2),12));D(rightanglemark((2,6),(2,0),(8,0),12));
MP("2",(-6,1),W);MP("6",(-3,0),S);
// Block 2
size(220);pathpen=black+linewidth(0.65);pointpen=black; /* draw in rectangles */ D(MP("R",(0,0))--MP("Q",(-6,0))--MP("P",(-6,2),N)--MP("S",(0,2),NW)--cycle); D((0,0)--MP("Q'",(0,6),SW)--MP("P'",(2,6),SE)--MP("S'",(2,0))--cycle); D(MP("R''",(2,2),NE)--MP("Q''",(8,2),N)--MP("P''",(8,0))--(2,0)--cycle); D(arc((0,0),(2,6),(-6,2)),dashed);D(arc((2,0),(8,0),(2,6)),dashed);D((2,6)--(0,0)--(-6,2),dashed); D(rightanglemark((2,6),(0,0),(-6,2),12));D(rightanglemark((2,6),(2,0),(8,0),12)); MP("2",(-6,1),W);MP("6",(-3,0),S); | [] |
658 | Trapezoid $ABCD$ has bases $\overline{AB}$ and $\overline{CD}$ and diagonals intersecting at $K.$ Suppose that $AB = 9$, $DC = 12$, and the area of $\triangle AKD$ is $24.$ What is the area of trapezoid $ABCD$?
$\mathrm{(A)}\ 92\qquad\mathrm{(B)}\ 94\qquad\mathrm{(C)}\ 96\qquad\mathrm{(D)}\ 98 \qquad\mathrm{(E)}\ 100$ | 2008 AMC 10A Problem 20 | Since $\overline{AB} \parallel \overline{DC}$ it follows that $\triangle ABK \sim \triangle CDK$. Thus $\frac{KA}{KC} = \frac{KB}{KD} = \frac{AB}{DC} = \frac{3}{4}$.
We now introduce the concept of area ratios: given two triangles that share the same height, the ratio of the areas is equal to the ratio of their bases. Since $\triangle AKB, \triangle AKD$ share a common altitude to $\overline{BD}$, it follows that (we let $[\triangle \ldots]$ denote the area of the triangle) $\frac{[\triangle AKB]}{[\triangle AKD]} = \frac{KB}{KD} = \frac{3}{4}$, so $[\triangle AKB] = \frac{3}{4}(24) = 18$. Similarly, we find $[\triangle DKC] = \frac{4}{3}(24) = 32$ and $[\triangle BKC] = 24$.
Therefore, the area of $ABCD = [AKD] + [AKB] + [BKC] + [CKD] = 24 + 18 + 24 + 32 = 98\ \mathrm{(D)}$. | // Block 1
pointpen = black; pathpen = black + linewidth(0.62); /* cse5 */
pen sm = fontsize(10); /* small font pen */
pair D=(0,0),C=(6,0), K=(3.5,8/3); /* note that K.x is arbitrary, as generator for A,B */
pair A=7*K/4-3*C/4, B=7*K/4-3*D/4;
D(MP("A",A,N)--MP("B",B,N)--MP("C",C)--MP("D",D)--A--C);D(B--D);D(A--MP("K",K)--D--cycle,linewidth(0.7));
MP("9",(A+B)/2,N,sm);MP("12",(C+D)/2,sm);MP("24",(A+D)/2+(1,0),E);
// Block 2
pointpen = black; pathpen = black + linewidth(0.62); /* cse5 */ pen sm = fontsize(10); /* small font pen */ pair D=(0,0),C=(6,0), K=(3.5,8/3); /* note that K.x is arbitrary, as generator for A,B */ pair A=7*K/4-3*C/4, B=7*K/4-3*D/4; D(MP("A",A,N)--MP("B",B,N)--MP("C",C)--MP("D",D)--A--C);D(B--D);D(A--MP("K",K)--D--cycle,linewidth(0.7)); MP("9",(A+B)/2,N,sm);MP("12",(C+D)/2,sm);MP("24",(A+D)/2+(1,0),E); | [] |
659 | A cube with side length $1$ is sliced by a plane that passes through two diagonally opposite vertices $A$ and $C$ and the midpoints $B$ and $D$ of two opposite edges not containing $A$ or $C$, as shown. What is the area of quadrilateral $ABCD$?
$\mathrm{(A)}\ \frac{\sqrt{6}}{2}\qquad\mathrm{(B)}\ \frac{5}{4}\qquad\mathrm{(C)}\ \sqrt{2}\qquad\mathrm{(D)}\ \frac{5}{8}\qquad\mathrm{(E)}\ \frac{3}{4}$ | 2008 AMC 10A Problem 21 | Since $AB = AD = CB = CD = \sqrt{\left(\frac{1}{2}\right)^2+1^2}$, it follows that $ABCD$ is a rhombus. The area of the rhombus can be computed by the formula $A = \frac 12 d_1d_2$, where $d_1,\,d_2$ are the diagonals of the rhombus (or of a kite in general). $BD$ has the same length as a face diagonal, or $\sqrt{1^2 + 1^2} = \sqrt{2}$. $AC$ is a space diagonal, with length $\sqrt{1^2+1^2+1^2} = \sqrt{3}$. Thus $A = \frac 12 \times \sqrt{2} \times \sqrt{3} = \frac{\sqrt{6}}{2}\ \mathrm{(A)}$. | // Block 1
import three;
unitsize(3cm);
defaultpen(fontsize(8)+linewidth(0.7));
currentprojection=obliqueX;
triple A=(0.5,0,0),C=(0,1,1),D=(0.5,1,0.5),B=(0,0,0.5);
draw((0.5,0,0)--(0,0,0)--(0,0,1)--(0,0,0)--(0,1,0),linetype("4 4"));
draw((0.5,0,1)--(0,0,1)--(0,1,1)--(0.5,1,1)--(0.5,0,1)--(0.5,0,0)--(0.5,1,0)--(0.5,1,1));
draw((0.5,1,0)--(0,1,0)--(0,1,1));
dot((0.5,0,0));
label("$A$",A,WSW);
dot((0,1,1));
label("$C$",C,NNE);
dot((0.5,1,0.5));
label("$D$",D,ESE);
dot((0,0,0.5));
label("$B$",B,NNW);
draw(B--C--A--B--D,linetype("4 4"));
draw(A--D--C);
// Block 2
import three; unitsize(3cm); defaultpen(fontsize(8)+linewidth(0.7)); currentprojection=obliqueX; triple A=(0.5,0,0),C=(0,1,1),D=(0.5,1,0.5),B=(0,0,0.5); draw((0.5,0,0)--(0,0,0)--(0,0,1)--(0,0,0)--(0,1,0),linetype("4 4")); draw((0.5,0,1)--(0,0,1)--(0,1,1)--(0.5,1,1)--(0.5,0,1)--(0.5,0,0)--(0.5,1,0)--(0.5,1,1)); draw((0.5,1,0)--(0,1,0)--(0,1,1)); dot((0.5,0,0)); label("$A$",A,WSW); dot((0,1,1)); label("$C$",C,NNE); dot((0.5,1,0.5)); label("$D$",D,ESE); dot((0,0,0.5)); label("$B$",B,NNW); draw(B--C--A--B--D,linetype("4 4")); draw(A--D--C); | [] |
660 | A round table has radius $4$. Six rectangular place mats are placed on the table. Each place mat has width $1$ and length $x$ as shown. They are positioned so that each mat has two corners on the edge of the table, these two corners being end points of the same side of length $x$. Further, the mats are positioned so that the inner corners each touch an inner corner of an adjacent mat. What is $x$?
$\mathrm{(A)}\ 2\sqrt{5}-\sqrt{3}\qquad\mathrm{(B)}\ 3\qquad\mathrm{(C)}\ \frac{3\sqrt{7}-\sqrt{3}}{2}\qquad\mathrm{(D)}\ 2\sqrt{3}\qquad\mathrm{(E)}\ \frac{5+2\sqrt{3}}{2}$ | 2008 AMC 10A Problem 25 | Let one of the mats be $ABCD$, and the center be $O$ as shown:
Since there are $6$ mats, $\Delta BOC$ is equilateral (the hexagon with side length $x$ is regular). So, $BC=CO=x$. Also, $\angle OCD = \angle OCB + \angle BCD = 60^\circ+90^\circ=150^\circ$.
By the Law of Cosines: $4^2=1^2+x^2-2\cdot1\cdot x \cdot \cos(150^\circ) \Rightarrow x^2 + x\sqrt{3} - 15 = 0 \Rightarrow x = \frac{-\sqrt{3}\pm 3\sqrt{7}}{2}$.
Since $x$ must be positive, $x = \frac{3\sqrt{7}-\sqrt{3}}{2} \Rightarrow (C)$. | unitsize(8mm); defaultpen(linewidth(.8)+fontsize(8)); draw(Circle((0,0),4)); path mat=(-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle; draw(mat); draw(rotate(60)*mat); draw(rotate(120)*mat); draw(rotate(180)*mat); draw(rotate(240)*mat); draw(rotate(300)*mat); label("\(x\)",(-1.55,2.1),E); label("\(x\)",(0.03,1.5),E); label("\(A\)",(-3.6,2.5513),E); label("\(B\)",(-3.15,1.35),E); label("\(C\)",(0.05,3.20),E); label("\(D\)",(-0.75,4.15),E); label("\(O\)",(0.00,-0.10),E); label("\(1\)",(-0.1,3.8),S); label("\(4\)",(-0.4,2.2),S); draw((0,0)--(0,3.103)); draw((0,0)--(-2.687,1.5513)); draw((0,0)--(-0.5,3.9686)); | [] |
660 | A round table has radius $4$. Six rectangular place mats are placed on the table. Each place mat has width $1$ and length $x$ as shown. They are positioned so that each mat has two corners on the edge of the table, these two corners being end points of the same side of length $x$. Further, the mats are positioned so that the inner corners each touch an inner corner of an adjacent mat. What is $x$?
$\mathrm{(A)}\ 2\sqrt{5}-\sqrt{3}\qquad\mathrm{(B)}\ 3\qquad\mathrm{(C)}\ \frac{3\sqrt{7}-\sqrt{3}}{2}\qquad\mathrm{(D)}\ 2\sqrt{3}\qquad\mathrm{(E)}\ \frac{5+2\sqrt{3}}{2}$ | 2008 AMC 10A Problem 25 | Draw $OD$ and $OC$ as in the diagram. Draw the altitude from $O$ to $DC$ and call the intersection $E$.
As proved in the first solution, $\angle OCD = 150^\circ$.
That makes $\Delta OCE$ a $30-60-90$ triangle, so $OE = \frac{x}{2}$ and $CE= \frac{x\sqrt 3}{2}$
Since $\Delta ODE$ is a right triangle,
$\left({\frac{x}{2}}\right)^2 + \left({\frac{x\sqrt 3}{2} +1}\right)^2 = 4^2 \Rightarrow x^2+x\sqrt3-15 = 0$
Solving for $x$ gives $x =\frac{3\sqrt{7}-\sqrt{3}}{2}\Rightarrow (C)$ | unitsize(8mm); defaultpen(linewidth(.8)+fontsize(8)); draw(Circle((0,0),4)); path mat=((-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle); draw(mat); draw(rotate(60)*mat); draw(rotate(120)*mat); draw(rotate(180)*mat); draw(rotate(240)*mat); draw(rotate(300)*mat); pair D = rotate(300)*(-3.687,1.5513); pair C = rotate(300)*(-2.687,1.5513); pair EE = foot((0.00,0.00),D,C); draw(D--EE--(0,0)); label("\(x\)",(-1.55,2.1),E); label("\(x\)",(0.03,1.5),E); label("\(A\)",(-3.6,2.5513),E); label("\(B\)",(-3.15,1.35),E); label("\(C\)",(0.05,3.20),E); label("\(D\)",(-0.75,4.15),E); label("\(O\)",(0.00,-0.10),E); label("\(1\)",(-0.1,3.8),S); label("\(4\)",(-0.4,2.2),S); draw((0,0)--(0,3.103)); draw((0,0)--(-2.687,1.5513)); draw((0,0)--(-0.5,3.9686)); label("\(E\)", EE,SE); | [] |
660 | A round table has radius $4$. Six rectangular place mats are placed on the table. Each place mat has width $1$ and length $x$ as shown. They are positioned so that each mat has two corners on the edge of the table, these two corners being end points of the same side of length $x$. Further, the mats are positioned so that the inner corners each touch an inner corner of an adjacent mat. What is $x$?
$\mathrm{(A)}\ 2\sqrt{5}-\sqrt{3}\qquad\mathrm{(B)}\ 3\qquad\mathrm{(C)}\ \frac{3\sqrt{7}-\sqrt{3}}{2}\qquad\mathrm{(D)}\ 2\sqrt{3}\qquad\mathrm{(E)}\ \frac{5+2\sqrt{3}}{2}$ | 2008 AMC 10A Problem 25 | By symmetry, $E$ is the midpoint of $DF$ and $OE$ is an extension of $OC$. Thus $\angle OED = 90^\circ$. Since $OD = 4$ and $DE = \frac{1}{2}$, $OE = \sqrt{16-\frac{1}{4}} = \frac{\sqrt{63}}{2} = \frac{3\sqrt{7}}{2}$. Since $\triangle CED$ is $30-60-90$, $CE = \frac{\sqrt{3}}{2}$ (this can also be deduced from Pythagoras on $\triangle CED$).
Thus $OC = \frac{3\sqrt{7}-\sqrt{3}}{2}$. As previous solutions noted, $\triangle BOC$ is equilateral, and thus the desired length is $x = OC \implies (C)$. | unitsize(8mm); defaultpen(linewidth(.8)+fontsize(8)); draw(Circle((0,0),4)); path mat=(-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle; draw(mat); draw(rotate(60)*mat); draw(rotate(120)*mat); draw(rotate(180)*mat); draw(rotate(240)*mat); draw(rotate(300)*mat); label("\(x\)",(-1.55,2.1),E); label("\(x\)",(0.03,1.5),E); label("\(A\)",(-3.6,2.5513),E); label("\(B\)",(-3.15,1.35),E); label("\(C\)",(0.05,3.20),E); label("\(D\)",(-0.75,4.15),E); label("\(E\)",(0,4.17)); label("\(F\)",(0.75,4.15),W); label("\(O\)",(0.00,-0.10),E); label("\(1\)",(-0.1,3.8),S); label("\(4\)",(-0.4,2.2),S); draw((0,0)--(0,3.103)); draw((0,0)--(-2.687,1.5513)); draw((0,0)--(-0.5,3.9686)); draw((0,0)--(-0.5,3.9686)); | [] |
660 | A round table has radius $4$. Six rectangular place mats are placed on the table. Each place mat has width $1$ and length $x$ as shown. They are positioned so that each mat has two corners on the edge of the table, these two corners being end points of the same side of length $x$. Further, the mats are positioned so that the inner corners each touch an inner corner of an adjacent mat. What is $x$?
$\mathrm{(A)}\ 2\sqrt{5}-\sqrt{3}\qquad\mathrm{(B)}\ 3\qquad\mathrm{(C)}\ \frac{3\sqrt{7}-\sqrt{3}}{2}\qquad\mathrm{(D)}\ 2\sqrt{3}\qquad\mathrm{(E)}\ \frac{5+2\sqrt{3}}{2}$ | 2008 AMC 10A Problem 25 | Looking at the diagram above, we know that $BE$ is a diameter of circle $O$ due to symmetry. Due to Thales' theorem, triangle $ABE$ is a right triangle with $A = 90 ^\circ$. $AE$ lies on $AD$ and $GE$ because $BAD$ is also a right angle. To find the length of $DG$, notice that if we draw a line from $F$ to $M$, the midpoint of line $DG$, it creates two $30$ - $60$ - $90$ triangles. Therefore, $MD = MG = \frac{\sqrt{3}x}{2} \Rightarrow DG = \sqrt{3}x$. $AE = 2 + \sqrt{3}x$
Applying the Pythagorean theorem to triangle $ABE$, we get \[(2+\sqrt{3}x)^2 + x^2 = 8^2 \Rightarrow 4 + 3x^2 + 4\sqrt{3}x + x^2 = 64 \Rightarrow x^2 + \sqrt{3}x - 15 = 0\] Using the quadratic formula to solve, we get \[x = \frac{-\sqrt{3} \pm \sqrt{3 -4(1)(-15)}}{2} = \frac{\pm 3\sqrt{7} - \sqrt{3}}{2}\] $x$ must be positive, therefore \[x = \frac{3\sqrt{7} - \sqrt{3}}{2} \Rightarrow (C)\]
~Zeric Hang | unitsize(8mm); defaultpen(linewidth(.8)+fontsize(8)); draw(Circle((0,0),4)); path mat=(-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle; draw(mat); draw(rotate(60)*mat); draw(rotate(120)*mat); draw(rotate(180)*mat); draw(rotate(240)*mat); draw(rotate(300)*mat); label("\(x\)",(-1.95,3),E); label("\(A\)",(-3.6,2.5513),E); label("\(C\)",(0.05,3.20),E); label("\(E\)",(0.40,-3.60),E); label("\(B\)",(-0.75,4.15),E); label("\(D\)",(-2.62,1.5),E); label("\(F\)",(-2.64,-1.43),E); label("\(G\)",(-0.2,-2.8),E); label("\( \sqrt{3}x\)",(-1.5,-0.5),E); label("\(M\)",(-2,-0.9),E); label("\(O\)",(0.00,-0.10),E); label("\(1\)",(-2.7,2.3),S); label("\(1\)",(0.1,-3.4),S); label("\(8\)",(-0.3,0),S); draw((0,-3.103)--(-2.687,1.5513)); draw((0.5,-3.9686)--(-0.5,3.9686)); | [] |
660 | A round table has radius $4$. Six rectangular place mats are placed on the table. Each place mat has width $1$ and length $x$ as shown. They are positioned so that each mat has two corners on the edge of the table, these two corners being end points of the same side of length $x$. Further, the mats are positioned so that the inner corners each touch an inner corner of an adjacent mat. What is $x$?
$\mathrm{(A)}\ 2\sqrt{5}-\sqrt{3}\qquad\mathrm{(B)}\ 3\qquad\mathrm{(C)}\ \frac{3\sqrt{7}-\sqrt{3}}{2}\qquad\mathrm{(D)}\ 2\sqrt{3}\qquad\mathrm{(E)}\ \frac{5+2\sqrt{3}}{2}$ | 2008 AMC 10A Problem 25 | We will let $O(0,0)$ be the origin. This way the coordinates of $C$ will be $(0,y)$. By $30-60-90$, the coordinates of $D$ will be $\left(-\frac{1}{2}, y + \frac{\sqrt{3}}{2}\right)$. The distance any point with coordinates $(x, y)$ is from the origin is $\sqrt{x^2 + y^2}$. Therefore, the distance $D$ is from the origin is $4$ and $\frac{1}{4} + x^2 + x\sqrt{3} + \frac{3}{4} = x^2 + x\sqrt{3} + 1 = 16$. We get the quadratic equation mentioned in solution 2. Using the quadratic formula, we get that $x = \frac{3\sqrt{7}-\sqrt{3}}{2} \Rightarrow (C)$
Note: Since $C$ and $D$ are not labeled in the diagram, refer to solution 1 for the location of points $C$ and $D$. | unitsize(8mm); defaultpen(linewidth(.8)+fontsize(8)); draw(Circle((0,0),4)); path mat=(-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle; draw(mat); draw(rotate(60)*mat); draw(rotate(120)*mat); draw(rotate(180)*mat); draw(rotate(240)*mat); draw(rotate(300)*mat); label("\(x\)",(-1.55,2.1),E); label("\(1\)",(-0.5,3.8),S); | [] |
660 | A round table has radius $4$. Six rectangular place mats are placed on the table. Each place mat has width $1$ and length $x$ as shown. They are positioned so that each mat has two corners on the edge of the table, these two corners being end points of the same side of length $x$. Further, the mats are positioned so that the inner corners each touch an inner corner of an adjacent mat. What is $x$?
$\mathrm{(A)}\ 2\sqrt{5}-\sqrt{3}\qquad\mathrm{(B)}\ 3\qquad\mathrm{(C)}\ \frac{3\sqrt{7}-\sqrt{3}}{2}\qquad\mathrm{(D)}\ 2\sqrt{3}\qquad\mathrm{(E)}\ \frac{5+2\sqrt{3}}{2}$ | 2008 AMC 10A Problem 25 | Notice that $\overarc{AE}$ is $\frac16$ the circumference of the circle. Therefore, $\overline{AE}$ is the side length of an inscribed hexagon with side length $4$. $\triangle AFE$ is a right triangle with $\overline{AF}=\frac12$. The length of $\overline{EF}=x+\frac{\sqrt{3}}{2}$. Using the Pythagorean Theorem, we get
$4^2 = \left(\frac{1}{2}\right)^2 + \left(x+\frac{\sqrt{3}}{2}\right)^2$
Solving for $x$, we get $x = \frac{3\sqrt{7}-\sqrt{3}}{2}\ \boxed{\text{C}}$ | unitsize(8mm); defaultpen(linewidth(.8)+fontsize(8)); draw(Circle((0,0),4)); path mat=(-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle; draw(mat); draw(rotate(60)*mat); draw(rotate(120)*mat); draw(rotate(180)*mat); draw(rotate(240)*mat); draw(rotate(300)*mat); label("\(x\)",(-1.55,2.1),E); label("\(A\)",(-3.6,2.5513),E); label("\(B\)",(-3.15,1.35),E); label("\(C\)",(0.05,3.20),E); label("\(D\)",(-0.75,4.15),E); label("\(E\)",(0.3,4.15),E); label("\(F\)",(-3.4,1.89),E); draw((0.5,3.9686)--(-3.13,2.45)); draw((0.5,3.9686)--(-2.95,2)); | [] |
661 | Points $A$ and $B$ lie on a circle centered at $O$, and $\angle AOB = 60^\circ$. A second circle is internally tangent to the first and tangent to both $\overline{OA}$ and $\overline{OB}$. What is the ratio of the area of the smaller circle to that of the larger circle?
$\mathrm{(A)}\ \frac{1}{16}\qquad\mathrm{(B)}\ \frac{1}{9}\qquad\mathrm{(C)}\ \frac{1}{8}\qquad\mathrm{(D)}\ \frac{1}{6}\qquad\mathrm{(E)}\ \frac{1}{4}$ | 2008 AMC 12A Problem 13 | Let $P$ be the center of the small circle with radius $r$, and let $Q$ be the point where the small circle is tangent to $OA$. Also, let $C$ be the point where the small circle is tangent to the big circle with radius $R$.
Then $PQO$ is a right triangle. Angle $POQ$ is $30$ degrees because line $OP$ bisects angle $AOB$ (this can be proved by dropping a perpendicular line from $P$ to line $OB$, letting their intersection be point $S$, and proving triangles $PQO$ and $PSO$ congruent), meaning that $PQO$ is a $30-60-90$ triangle. Therefore, $OP=2PQ$.
Since $OP=OC-PC=OC-r=R-r$, we have $R-r=2PQ$, or $R-r=2r$, or $\frac{1}{3}=\frac{r}{R}$.
Ratio of areas of circles is ratio of radii squared, so the answer is $\left(\frac{1}{3}\right)^2 = \frac{1}{9} \Rightarrow \boxed{B}$ | // Block 1
size(200);
defaultpen(fontsize(10));
pair O=(0,0), A=(3,0), B=(3/2,3/2*3^.5), C=(3^.5,1), D=(3^.5,0), F=(1.5*3^.5,1.5), G=(2*3^.5,2);
picture p = new picture;
draw(p,Circle(O,0.2));
clip(p,O--C--A--cycle);
add(p);
draw(Circle(O,3));
dot(A); dot(B); dot(C); dot(O);
draw(A--O--B);
draw(O--C--D);
draw(C--F);
draw(D-(0.2,0)--D-(0.2,-0.2)--D-(0,-0.2));
draw(Circle(C,1));
label("\(30^{\circ}\)",(0.65,0.15),O);
label("\(r\)",(C+D)/2,E);
label("\(2r\)",(O+C)/2,NNE);
label("\(O\)",O,SW);
label("\(r\)",(C+F)/2,SE);
label("\(R\)",(O+A)/2-(0,0.3),S);
label("\(P\)",C,NW);
label("\(A\)",(3,0), A);
label("\(B\)",(3/2,3/2*3^.5), B);
label("\(C\)",(1.5*3^.5,1.5), G);
label("\(Q\)",D,SE);
// Block 2
size(200); defaultpen(fontsize(10)); pair O=(0,0), A=(3,0), B=(3/2,3/2*3^.5), C=(3^.5,1), D=(3^.5,0), F=(1.5*3^.5,1.5), G=(2*3^.5,2); picture p = new picture; draw(p,Circle(O,0.2)); clip(p,O--C--A--cycle); add(p); draw(Circle(O,3)); dot(A); dot(B); dot(C); dot(O); draw(A--O--B); draw(O--C--D); draw(C--F); draw(D-(0.2,0)--D-(0.2,-0.2)--D-(0,-0.2)); draw(Circle(C,1)); label("\(30^{\circ}\)",(0.65,0.15),O); label("\(r\)",(C+D)/2,E); label("\(2r\)",(O+C)/2,NNE); label("\(O\)",O,SW); label("\(r\)",(C+F)/2,SE); label("\(R\)",(O+A)/2-(0,0.3),S); label("\(P\)",C,NW); label("\(A\)",(3,0), A); label("\(B\)",(3/2,3/2*3^.5), B); label("\(C\)",(1.5*3^.5,1.5), G); label("\(Q\)",D,SE); | [] |
662 | Triangle $ABC$, with sides of length $5$, $6$, and $7$, has one vertex on the positive $x$-axis, one on the positive $y$-axis, and one on the positive $z$-axis. Let $O$ be the origin. What is the volume of tetrahedron $OABC$?
$\mathrm{(A)}\ \sqrt{85}\qquad\mathrm{(B)}\ \sqrt{90}\qquad\mathrm{(C)}\ \sqrt{95}\qquad\mathrm{(D)}\ 10\qquad\mathrm{(E)}\ \sqrt{105}$ | 2008 AMC 12A Problem 18 | Without loss of generality, let $A$ be on the $x$ axis, $B$ be on the $y$ axis, and $C$ be on the $z$ axis, and let $AB, BC, CA$ have respective lengths of 5, 6, and 7. Let $a,b,c$ denote the lengths of segments $OA,OB,OC,$ respectively. Then by the Pythagorean Theorem,
\begin{align*} a^2+b^2 &=5^2 , \\ b^2+c^2&=6^2, \\ c^2+a^2 &=7^2 , \end{align*}
so $a^2 = (5^2+7^2-6^2)/2 = 19$; similarly, $b^2 = 6$ and $c^2 = 30$. Since $OA$, $OB$, and $OC$ are mutually perpendicular, the tetrahedron's volume is \[abc/6\] because we can consider the tetrahedron to be a right triangular pyramid.
\[abc/6 = \sqrt{a^2b^2c^2}/6 = \frac{\sqrt{19 \cdot 6 \cdot 30}}{6} = \sqrt{95},\]
which is answer choice $\boxed{\text{C}}$. | // Block 1
defaultpen(fontsize(8));
draw((0,10)--(0,0)--(8,0));draw((-3,-4)--(0,0));draw((0,10)--(-3,-4)--(8,0)--cycle);
label("A",(8,0),(1,0));label("B",(0,10),(0,1));label("C",(-3,-4),(-1,-1));label("O",(0,0),(1,1));
label("$a$",(4,0),(0,1));label("$b$",(0,5),(1,0));label("$c$",(-1.5,-2),(1,0));
label("$5$",(4,5),(1,1));label("$6$",(-1.5,3),(-1,0));label("$7$",(2.5,-2),(1,-1));
// Block 2
defaultpen(fontsize(8)); draw((0,10)--(0,0)--(8,0));draw((-3,-4)--(0,0));draw((0,10)--(-3,-4)--(8,0)--cycle); label("A",(8,0),(1,0));label("B",(0,10),(0,1));label("C",(-3,-4),(-1,-1));label("O",(0,0),(1,1)); label("$a$",(4,0),(0,1));label("$b$",(0,5),(1,0));label("$c$",(-1.5,-2),(1,0)); label("$5$",(4,5),(1,1));label("$6$",(-1.5,3),(-1,0));label("$7$",(2.5,-2),(1,-1)); | [] |
663 | Triangle $ABC$ has $AC=3$, $BC=4$, and $AB=5$. Point $D$ is on $\overline{AB}$, and $\overline{CD}$ bisects the right angle. The inscribed circles of $\triangle ADC$ and $\triangle BCD$ have radii $r_a$ and $r_b$, respectively. What is $r_a/r_b$?
$\mathrm{(A)}\ \frac{1}{28}\left(10-\sqrt{2}\right)\qquad\mathrm{(B)}\ \frac{3}{56}\left(10-\sqrt{2}\right)\qquad\mathrm{(C)}\ \frac{1}{14}\left(10-\sqrt{2}\right)\qquad\mathrm{(D)}\ \frac{5}{56}\left(10-\sqrt{2}\right)\\\mathrm{(E)}\ \frac{3}{28}\left(10-\sqrt{2}\right)$ | 2008 AMC 12A Problem 20 | By the Angle Bisector Theorem,
\[\frac{BD}{4} = \frac{5-BD}{3} \Longrightarrow BD = \frac{20}7\]
By Law of Sines on $\triangle BCD$,
\[\frac{BD}{\sin 45^{\circ}} = \frac{CD}{\sin \angle B} \Longrightarrow \frac{20/7}{\sqrt{2}/2} = \frac{CD}{3/5} \Longrightarrow CD=\frac{12\sqrt{2}}{7}\]
Since the area of a triangle satisfies $[\triangle]=rs$, where $r =$ the inradius and $s =$ the semiperimeter, we have
\[\frac{r_A}{r_B} = \frac{[ACD] \cdot s_B}{[BCD] \cdot s_A}\]
Since $\triangle ACD$ and $\triangle BCD$ share the altitude (to $\overline{AB}$), their areas are the ratio of their bases, or \[\frac{[ACD]}{[BCD]} = \frac{AD}{BD} = \frac{3}{4}\]
The semiperimeters are $s_A = \left(3 + \frac{15}{7} + \frac{12\sqrt{2}}{7}\right)\left/\right.2 = \frac{18+6\sqrt{2}}{7}$ and $s_B = \frac{24+ 6\sqrt{2}}{7}$. Thus,
\begin{align*} \frac{r_A}{r_B} &= \frac{[ACD] \cdot s_B}{[BCD] \cdot s_A} = \frac{3}{4} \cdot \frac{(24+ 6\sqrt{2})/7}{(18+6\sqrt{2})/7} \\ &= \frac{3(4+\sqrt{2})}{4(3+\sqrt{2})} \cdot \left(\frac{3-\sqrt{2}}{3-\sqrt{2}}\right) = \frac{3}{28}(10-\sqrt{2}) \Rightarrow \mathrm{(E)}\qquad \blacksquare \end{align*}
-(you could also use stewart's theorem to find the length of cd) | // Block 1
import olympiad;
size(300);
defaultpen(0.8);
pair C=(0,0),A=(0,3),B=(4,0),D=(4-2.28571,1.71429);
pair O=incenter(A,C,D), P=incenter(B,C,D);
picture p = new picture;
draw(p,Circle(C,0.2)); draw(p,Circle(B,0.2));
clip(p,B--C--D--cycle);
add(p);
draw(A--B--C--D--C--cycle);
draw(incircle(A,C,D));
draw(incircle(B,C,D));
dot(O);dot(P);
label("\(A\)",A,W);
label("\(B\)",B,E);
label("\(C\)",C,W);
label("\(D\)",D,NE);
label("\(O_A\)",O,W);
label("\(O_B\)",P,W);
label("\(3\)",(A+C)/2,W);
label("\(4\)",(B+C)/2,S);
label("\(\frac{15}{7}\)",(A+D)/2,NE);
label("\(\frac{20}{7}\)",(B+D)/2,NE);
label("\(45^{\circ}\)",(.2,.1),E);
label("\(\sin \theta = \frac{3}{5}\)",B-(.2,-.1),W);
// Block 2
import olympiad; size(300); defaultpen(0.8); pair C=(0,0),A=(0,3),B=(4,0),D=(4-2.28571,1.71429); pair O=incenter(A,C,D), P=incenter(B,C,D); picture p = new picture; draw(p,Circle(C,0.2)); draw(p,Circle(B,0.2)); clip(p,B--C--D--cycle); add(p); draw(A--B--C--D--C--cycle); draw(incircle(A,C,D)); draw(incircle(B,C,D)); dot(O);dot(P); label("\(A\)",A,W); label("\(B\)",B,E); label("\(C\)",C,W); label("\(D\)",D,NE); label("\(O_A\)",O,W); label("\(O_B\)",P,W); label("\(3\)",(A+C)/2,W); label("\(4\)",(B+C)/2,S); label("\(\frac{15}{7}\)",(A+D)/2,NE); label("\(\frac{20}{7}\)",(B+D)/2,NE); label("\(45^{\circ}\)",(.2,.1),E); label("\(\sin \theta = \frac{3}{5}\)",B-(.2,-.1),W); | [] |
663 | Triangle $ABC$ has $AC=3$, $BC=4$, and $AB=5$. Point $D$ is on $\overline{AB}$, and $\overline{CD}$ bisects the right angle. The inscribed circles of $\triangle ADC$ and $\triangle BCD$ have radii $r_a$ and $r_b$, respectively. What is $r_a/r_b$?
$\mathrm{(A)}\ \frac{1}{28}\left(10-\sqrt{2}\right)\qquad\mathrm{(B)}\ \frac{3}{56}\left(10-\sqrt{2}\right)\qquad\mathrm{(C)}\ \frac{1}{14}\left(10-\sqrt{2}\right)\qquad\mathrm{(D)}\ \frac{5}{56}\left(10-\sqrt{2}\right)\\\mathrm{(E)}\ \frac{3}{28}\left(10-\sqrt{2}\right)$ | 2008 AMC 12A Problem 20 | We start by finding the length of $AD$ and $BD$ as in solution 1. Using the angle bisector theorem, we see that $AD = \frac{15}{7}$ and $BD = \frac{20}{7}$. Using Stewart's Theorem gives us the equation $5d^2 + \frac{1500}{49} = \frac{240}{7} + \frac{180}{7}$, where $d$ is the length of $CD$. Solving gives us $d = \frac{12\sqrt{2}}{7}$, so $CD = \frac{12\sqrt{2}}{7}$.
Call the incenters of triangles $ACD$ and $BCD$ $O_a$ and $O_b$ respectively. Since $O_a$ is an incenter, it lies on the angle bisector of $\angle ACD$. Similarly, $O_b$ lies on the angle bisector of $\angle BCD$. Call the point on $CD$ tangent to $O_a$ $M$, and the point tangent to $O_b$ $N$. Since $\triangle CO_aM$ and $\triangle CO_bN$ are right, and $\angle O_aCM = \angle O_bCN$, $\triangle CO_aM \sim \triangle CO_bN$. Then, $\frac{r_a}{r_b} = \frac{CM}{CN}$.
We now use common tangents to find the length of $CM$ and $CN$. Let $CM = m$, and the length of the other tangents be $n$ and $p$. Since common tangents are equal, we can write that $m + n = \frac{12\sqrt{2}}{7}$, $n + p = \frac{15}{7}$ and $m + p = 3$. Solving gives us that $CM = m = \frac{6\sqrt{2} + 3}{7}$. Similarly, $CN = \frac{6\sqrt{2} + 4}{7}$.
We see now that $\frac{r_a}{r_b} = \frac{\frac{6\sqrt{2} + 3}{7}}{\frac{6\sqrt{2} + 4}{7}} = \frac{6\sqrt{2} + 3}{6\sqrt{2} + 4} = \frac{60-6\sqrt{2}}{56} = \frac{3}{28}(10 - \sqrt{2}) \Rightarrow \boxed{E}$ | // Block 1
import olympiad;
import geometry;
size(300);
defaultpen(0.8);
pair C=(0,0),A=(0,3),B=(4,0),D=(4-2.28571,1.71429);
pair O=incenter(A,C,D), P=incenter(B,C,D);
line cd = line(C, D);
picture p = new picture;
picture q = new picture;
picture r = new picture;
picture s = new picture;
draw(p,Circle(C,0.2));
clip(p,P--C--D--cycle);
draw(q, Circle(C, 0.3));
clip(q, O--C--D--cycle);
line l1 = perpendicular(O, cd);
draw(r, l1);
clip(r, C--D--O--cycle);
line l2 = perpendicular(P, cd);
draw(s, l2);
clip(s, C--P--D--cycle);
add(p);
add(q);
add(r);
add(s);
draw(A--B--C--D--C--cycle);
draw(incircle(A,C,D));
draw(incircle(B,C,D));
draw(C--O);
draw(C--P);
dot(O);
dot(P);
point inter1 = intersectionpoint(l1, cd);
point inter2 = intersectionpoint(l2, cd);
dot(inter1);
dot(inter2);
label("\(A\)",A,W);
label("\(B\)",B,E);
label("\(C\)",C,W);
label("\(D\)",D,NE);
label("\(O_a\)",O,W);
label("\(O_b\)",P,E);
label("\(3\)",(A+C)/2,W);
label("\(4\)",(B+C)/2,S);
label("\(\frac{15}{7}\)",(A+D)/2,NE);
label("\(\frac{20}{7}\)",(B+D)/2,NE);
label("\(M\)", inter1, 2W);
label("\(N\)", inter2, 2E);
// Block 2
import olympiad; import geometry; size(300); defaultpen(0.8); pair C=(0,0),A=(0,3),B=(4,0),D=(4-2.28571,1.71429); pair O=incenter(A,C,D), P=incenter(B,C,D); line cd = line(C, D); picture p = new picture; picture q = new picture; picture r = new picture; picture s = new picture; draw(p,Circle(C,0.2)); clip(p,P--C--D--cycle); draw(q, Circle(C, 0.3)); clip(q, O--C--D--cycle); line l1 = perpendicular(O, cd); draw(r, l1); clip(r, C--D--O--cycle); line l2 = perpendicular(P, cd); draw(s, l2); clip(s, C--P--D--cycle); add(p); add(q); add(r); add(s); draw(A--B--C--D--C--cycle); draw(incircle(A,C,D)); draw(incircle(B,C,D)); draw(C--O); draw(C--P); dot(O); dot(P); point inter1 = intersectionpoint(l1, cd); point inter2 = intersectionpoint(l2, cd); dot(inter1); dot(inter2); label("\(A\)",A,W); label("\(B\)",B,E); label("\(C\)",C,W); label("\(D\)",D,NE); label("\(O_a\)",O,W); label("\(O_b\)",P,E); label("\(3\)",(A+C)/2,W); label("\(4\)",(B+C)/2,S); label("\(\frac{15}{7}\)",(A+D)/2,NE); label("\(\frac{20}{7}\)",(B+D)/2,NE); label("\(M\)", inter1, 2W); label("\(N\)", inter2, 2E); | [] |
664 | A round table has radius $4$. Six rectangular place mats are placed on the table. Each place mat has width $1$ and length $x$ as shown. They are positioned so that each mat has two corners on the edge of the table, these two corners being end points of the same side of length $x$. Further, the mats are positioned so that the inner corners each touch an inner corner of an adjacent mat. What is $x$?
$\mathrm{(A)}\ 2\sqrt{5}-\sqrt{3}\qquad\mathrm{(B)}\ 3\qquad\mathrm{(C)}\ \frac{3\sqrt{7}-\sqrt{3}}{2}\qquad\mathrm{(D)}\ 2\sqrt{3}\qquad\mathrm{(E)}\ \frac{5+2\sqrt{3}}{2}$ | 2008 AMC 12A Problem 22 | Let one of the mats be $ABCD$, and the center be $O$ as shown:
Since there are $6$ mats, $\Delta BOC$ is equilateral (the hexagon with side length $x$ is regular). So, $BC=CO=x$. Also, $\angle OCD = \angle OCB + \angle BCD = 60^\circ+90^\circ=150^\circ$.
By the Law of Cosines: $4^2=1^2+x^2-2\cdot1\cdot x \cdot \cos(150^\circ) \Rightarrow x^2 + x\sqrt{3} - 15 = 0 \Rightarrow x = \frac{-\sqrt{3}\pm 3\sqrt{7}}{2}$.
Since $x$ must be positive, $x = \frac{3\sqrt{7}-\sqrt{3}}{2} \Rightarrow (C)$. | // Block 1
unitsize(8mm);
defaultpen(linewidth(.8)+fontsize(8));
draw(Circle((0,0),4));
path mat=(-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle;
draw(mat);
draw(rotate(60)*mat);
draw(rotate(120)*mat);
draw(rotate(180)*mat);
draw(rotate(240)*mat);
draw(rotate(300)*mat);
label("\(x\)",(-1.55,2.1),E);
label("\(x\)",(0.03,1.5),E);
label("\(A\)",(-3.6,2.5513),E);
label("\(B\)",(-3.15,1.35),E);
label("\(C\)",(0.05,3.20),E);
label("\(D\)",(-0.75,4.15),E);
label("\(O\)",(0.00,-0.10),E);
label("\(1\)",(-0.1,3.8),S);
label("\(4\)",(-0.4,2.2),S);
draw((0,0)--(0,3.103));
draw((0,0)--(-2.687,1.5513));
draw((0,0)--(-0.5,3.9686));
// Block 2
unitsize(8mm); defaultpen(linewidth(.8)+fontsize(8)); draw(Circle((0,0),4)); path mat=(-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle; draw(mat); draw(rotate(60)*mat); draw(rotate(120)*mat); draw(rotate(180)*mat); draw(rotate(240)*mat); draw(rotate(300)*mat); label("\(x\)",(-1.55,2.1),E); label("\(x\)",(0.03,1.5),E); label("\(A\)",(-3.6,2.5513),E); label("\(B\)",(-3.15,1.35),E); label("\(C\)",(0.05,3.20),E); label("\(D\)",(-0.75,4.15),E); label("\(O\)",(0.00,-0.10),E); label("\(1\)",(-0.1,3.8),S); label("\(4\)",(-0.4,2.2),S); draw((0,0)--(0,3.103)); draw((0,0)--(-2.687,1.5513)); draw((0,0)--(-0.5,3.9686)); | [] |
664 | A round table has radius $4$. Six rectangular place mats are placed on the table. Each place mat has width $1$ and length $x$ as shown. They are positioned so that each mat has two corners on the edge of the table, these two corners being end points of the same side of length $x$. Further, the mats are positioned so that the inner corners each touch an inner corner of an adjacent mat. What is $x$?
$\mathrm{(A)}\ 2\sqrt{5}-\sqrt{3}\qquad\mathrm{(B)}\ 3\qquad\mathrm{(C)}\ \frac{3\sqrt{7}-\sqrt{3}}{2}\qquad\mathrm{(D)}\ 2\sqrt{3}\qquad\mathrm{(E)}\ \frac{5+2\sqrt{3}}{2}$ | 2008 AMC 12A Problem 22 | Draw $OD$ and $OC$ as in the diagram. Draw the altitude from $O$ to $DC$ and call the intersection $E$.
As proved in the first solution, $\angle OCD = 150^\circ$.
That makes $\Delta OCE$ a $30-60-90$ triangle, so $OE = \frac{x}{2}$ and $CE= \frac{x\sqrt 3}{2}$
Since $\Delta ODE$ is a right triangle,
$\left({\frac{x}{2}}\right)^2 + \left({\frac{x\sqrt 3}{2} +1}\right)^2 = 4^2 \Rightarrow x^2+x\sqrt3-15 = 0$
Solving for $x$ gives $x =\frac{3\sqrt{7}-\sqrt{3}}{2}\Rightarrow (C)$ | // Block 1
unitsize(8mm);
defaultpen(linewidth(.8)+fontsize(8));
draw(Circle((0,0),4));
path mat=((-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle);
draw(mat);
draw(rotate(60)*mat);
draw(rotate(120)*mat);
draw(rotate(180)*mat);
draw(rotate(240)*mat);
draw(rotate(300)*mat);
pair D = rotate(300)*(-3.687,1.5513);
pair C = rotate(300)*(-2.687,1.5513);
pair EE = foot((0.00,0.00),D,C);
draw(D--EE--(0,0));
label("\(x\)",(-1.55,2.1),E);
label("\(x\)",(0.03,1.5),E);
label("\(A\)",(-3.6,2.5513),E);
label("\(B\)",(-3.15,1.35),E);
label("\(C\)",(0.05,3.20),E);
label("\(D\)",(-0.75,4.15),E);
label("\(O\)",(0.00,-0.10),E);
label("\(1\)",(-0.1,3.8),S);
label("\(4\)",(-0.4,2.2),S);
draw((0,0)--(0,3.103));
draw((0,0)--(-2.687,1.5513));
draw((0,0)--(-0.5,3.9686));
label("\(E\)", EE,SE);
// Block 2
unitsize(8mm); defaultpen(linewidth(.8)+fontsize(8)); draw(Circle((0,0),4)); path mat=((-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle); draw(mat); draw(rotate(60)*mat); draw(rotate(120)*mat); draw(rotate(180)*mat); draw(rotate(240)*mat); draw(rotate(300)*mat); pair D = rotate(300)*(-3.687,1.5513); pair C = rotate(300)*(-2.687,1.5513); pair EE = foot((0.00,0.00),D,C); draw(D--EE--(0,0)); label("\(x\)",(-1.55,2.1),E); label("\(x\)",(0.03,1.5),E); label("\(A\)",(-3.6,2.5513),E); label("\(B\)",(-3.15,1.35),E); label("\(C\)",(0.05,3.20),E); label("\(D\)",(-0.75,4.15),E); label("\(O\)",(0.00,-0.10),E); label("\(1\)",(-0.1,3.8),S); label("\(4\)",(-0.4,2.2),S); draw((0,0)--(0,3.103)); draw((0,0)--(-2.687,1.5513)); draw((0,0)--(-0.5,3.9686)); label("\(E\)", EE,SE); | [] |
664 | A round table has radius $4$. Six rectangular place mats are placed on the table. Each place mat has width $1$ and length $x$ as shown. They are positioned so that each mat has two corners on the edge of the table, these two corners being end points of the same side of length $x$. Further, the mats are positioned so that the inner corners each touch an inner corner of an adjacent mat. What is $x$?
$\mathrm{(A)}\ 2\sqrt{5}-\sqrt{3}\qquad\mathrm{(B)}\ 3\qquad\mathrm{(C)}\ \frac{3\sqrt{7}-\sqrt{3}}{2}\qquad\mathrm{(D)}\ 2\sqrt{3}\qquad\mathrm{(E)}\ \frac{5+2\sqrt{3}}{2}$ | 2008 AMC 12A Problem 22 | By symmetry, $E$ is the midpoint of $DF$ and $OE$ is an extension of $OC$. Thus $\angle OED = 90^\circ$. Since $OD = 4$ and $DE = \frac{1}{2}$, $OE = \sqrt{16-\frac{1}{4}} = \frac{\sqrt{63}}{2} = \frac{3\sqrt{7}}{2}$. Since $\triangle CED$ is $30-60-90$, $CE = \frac{\sqrt{3}}{2}$ (this can also be deduced from Pythagoras on $\triangle CED$).
Thus $OC = \frac{3\sqrt{7}-\sqrt{3}}{2}$. As previous solutions noted, $\triangle BOC$ is equilateral, and thus the desired length is $x = OC \implies (C)$. | // Block 1
unitsize(8mm);
defaultpen(linewidth(.8)+fontsize(8));
draw(Circle((0,0),4));
path mat=(-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle;
draw(mat);
draw(rotate(60)*mat);
draw(rotate(120)*mat);
draw(rotate(180)*mat);
draw(rotate(240)*mat);
draw(rotate(300)*mat);
label("\(x\)",(-1.55,2.1),E);
label("\(x\)",(0.03,1.5),E);
label("\(A\)",(-3.6,2.5513),E);
label("\(B\)",(-3.15,1.35),E);
label("\(C\)",(0.05,3.20),E);
label("\(D\)",(-0.75,4.15),E);
label("\(E\)",(0,4.17));
label("\(F\)",(0.75,4.15),W);
label("\(O\)",(0.00,-0.10),E);
label("\(1\)",(-0.1,3.8),S);
label("\(4\)",(-0.4,2.2),S);
draw((0,0)--(0,3.103));
draw((0,0)--(-2.687,1.5513));
draw((0,0)--(-0.5,3.9686));
draw((0,0)--(-0.5,3.9686));
// Block 2
unitsize(8mm); defaultpen(linewidth(.8)+fontsize(8)); draw(Circle((0,0),4)); path mat=(-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle; draw(mat); draw(rotate(60)*mat); draw(rotate(120)*mat); draw(rotate(180)*mat); draw(rotate(240)*mat); draw(rotate(300)*mat); label("\(x\)",(-1.55,2.1),E); label("\(x\)",(0.03,1.5),E); label("\(A\)",(-3.6,2.5513),E); label("\(B\)",(-3.15,1.35),E); label("\(C\)",(0.05,3.20),E); label("\(D\)",(-0.75,4.15),E); label("\(E\)",(0,4.17)); label("\(F\)",(0.75,4.15),W); label("\(O\)",(0.00,-0.10),E); label("\(1\)",(-0.1,3.8),S); label("\(4\)",(-0.4,2.2),S); draw((0,0)--(0,3.103)); draw((0,0)--(-2.687,1.5513)); draw((0,0)--(-0.5,3.9686)); draw((0,0)--(-0.5,3.9686)); | [] |
664 | A round table has radius $4$. Six rectangular place mats are placed on the table. Each place mat has width $1$ and length $x$ as shown. They are positioned so that each mat has two corners on the edge of the table, these two corners being end points of the same side of length $x$. Further, the mats are positioned so that the inner corners each touch an inner corner of an adjacent mat. What is $x$?
$\mathrm{(A)}\ 2\sqrt{5}-\sqrt{3}\qquad\mathrm{(B)}\ 3\qquad\mathrm{(C)}\ \frac{3\sqrt{7}-\sqrt{3}}{2}\qquad\mathrm{(D)}\ 2\sqrt{3}\qquad\mathrm{(E)}\ \frac{5+2\sqrt{3}}{2}$ | 2008 AMC 12A Problem 22 | Looking at the diagram above, we know that $BE$ is a diameter of circle $O$ due to symmetry. Due to Thales' theorem, triangle $ABE$ is a right triangle with $A = 90 ^\circ$. $AE$ lies on $AD$ and $GE$ because $BAD$ is also a right angle. To find the length of $DG$, notice that if we draw a line from $F$ to $M$, the midpoint of line $DG$, it creates two $30$ - $60$ - $90$ triangles. Therefore, $MD = MG = \frac{\sqrt{3}x}{2} \Rightarrow DG = \sqrt{3}x$. $AE = 2 + \sqrt{3}x$
Applying the Pythagorean theorem to triangle $ABE$, we get \[(2+\sqrt{3}x)^2 + x^2 = 8^2 \Rightarrow 4 + 3x^2 + 4\sqrt{3}x + x^2 = 64 \Rightarrow x^2 + \sqrt{3}x - 15 = 0\] Using the quadratic formula to solve, we get \[x = \frac{-\sqrt{3} \pm \sqrt{3 -4(1)(-15)}}{2} = \frac{\pm 3\sqrt{7} - \sqrt{3}}{2}\] $x$ must be positive, therefore \[x = \frac{3\sqrt{7} - \sqrt{3}}{2} \Rightarrow (C)\]
~Zeric Hang | // Block 1
unitsize(8mm);
defaultpen(linewidth(.8)+fontsize(8));
draw(Circle((0,0),4));
path mat=(-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle;
draw(mat);
draw(rotate(60)*mat);
draw(rotate(120)*mat);
draw(rotate(180)*mat);
draw(rotate(240)*mat);
draw(rotate(300)*mat);
label("\(x\)",(-1.95,3),E);
label("\(A\)",(-3.6,2.5513),E);
label("\(C\)",(0.05,3.20),E);
label("\(E\)",(0.40,-3.60),E);
label("\(B\)",(-0.75,4.15),E);
label("\(D\)",(-2.62,1.5),E);
label("\(F\)",(-2.64,-1.43),E);
label("\(G\)",(-0.2,-2.8),E);
label("\( \sqrt{3}x\)",(-1.5,-0.5),E);
label("\(M\)",(-2,-0.9),E);
label("\(O\)",(0.00,-0.10),E);
label("\(1\)",(-2.7,2.3),S);
label("\(1\)",(0.1,-3.4),S);
label("\(8\)",(-0.3,0),S);
draw((0,-3.103)--(-2.687,1.5513));
draw((0.5,-3.9686)--(-0.5,3.9686));
// Block 2
unitsize(8mm); defaultpen(linewidth(.8)+fontsize(8)); draw(Circle((0,0),4)); path mat=(-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle; draw(mat); draw(rotate(60)*mat); draw(rotate(120)*mat); draw(rotate(180)*mat); draw(rotate(240)*mat); draw(rotate(300)*mat); label("\(x\)",(-1.95,3),E); label("\(A\)",(-3.6,2.5513),E); label("\(C\)",(0.05,3.20),E); label("\(E\)",(0.40,-3.60),E); label("\(B\)",(-0.75,4.15),E); label("\(D\)",(-2.62,1.5),E); label("\(F\)",(-2.64,-1.43),E); label("\(G\)",(-0.2,-2.8),E); label("\( \sqrt{3}x\)",(-1.5,-0.5),E); label("\(M\)",(-2,-0.9),E); label("\(O\)",(0.00,-0.10),E); label("\(1\)",(-2.7,2.3),S); label("\(1\)",(0.1,-3.4),S); label("\(8\)",(-0.3,0),S); draw((0,-3.103)--(-2.687,1.5513)); draw((0.5,-3.9686)--(-0.5,3.9686)); | [] |
664 | A round table has radius $4$. Six rectangular place mats are placed on the table. Each place mat has width $1$ and length $x$ as shown. They are positioned so that each mat has two corners on the edge of the table, these two corners being end points of the same side of length $x$. Further, the mats are positioned so that the inner corners each touch an inner corner of an adjacent mat. What is $x$?
$\mathrm{(A)}\ 2\sqrt{5}-\sqrt{3}\qquad\mathrm{(B)}\ 3\qquad\mathrm{(C)}\ \frac{3\sqrt{7}-\sqrt{3}}{2}\qquad\mathrm{(D)}\ 2\sqrt{3}\qquad\mathrm{(E)}\ \frac{5+2\sqrt{3}}{2}$ | 2008 AMC 12A Problem 22 | We will let $O(0,0)$ be the origin. This way the coordinates of $C$ will be $(0,y)$. By $30-60-90$, the coordinates of $D$ will be $\left(-\frac{1}{2}, y + \frac{\sqrt{3}}{2}\right)$. The distance any point with coordinates $(x, y)$ is from the origin is $\sqrt{x^2 + y^2}$. Therefore, the distance $D$ is from the origin is $4$ and $\frac{1}{4} + x^2 + x\sqrt{3} + \frac{3}{4} = x^2 + x\sqrt{3} + 1 = 16$. We get the quadratic equation mentioned in solution 2. Using the quadratic formula, we get that $x = \frac{3\sqrt{7}-\sqrt{3}}{2} \Rightarrow (C)$
Note: Since $C$ and $D$ are not labeled in the diagram, refer to solution 1 for the location of points $C$ and $D$. | // Block 1
unitsize(8mm);
defaultpen(linewidth(.8)+fontsize(8));
draw(Circle((0,0),4));
path mat=(-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle;
draw(mat);
draw(rotate(60)*mat);
draw(rotate(120)*mat);
draw(rotate(180)*mat);
draw(rotate(240)*mat);
draw(rotate(300)*mat);
label("\(x\)",(-1.55,2.1),E);
label("\(1\)",(-0.5,3.8),S);
// Block 2
unitsize(8mm); defaultpen(linewidth(.8)+fontsize(8)); draw(Circle((0,0),4)); path mat=(-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle; draw(mat); draw(rotate(60)*mat); draw(rotate(120)*mat); draw(rotate(180)*mat); draw(rotate(240)*mat); draw(rotate(300)*mat); label("\(x\)",(-1.55,2.1),E); label("\(1\)",(-0.5,3.8),S); | [] |
664 | A round table has radius $4$. Six rectangular place mats are placed on the table. Each place mat has width $1$ and length $x$ as shown. They are positioned so that each mat has two corners on the edge of the table, these two corners being end points of the same side of length $x$. Further, the mats are positioned so that the inner corners each touch an inner corner of an adjacent mat. What is $x$?
$\mathrm{(A)}\ 2\sqrt{5}-\sqrt{3}\qquad\mathrm{(B)}\ 3\qquad\mathrm{(C)}\ \frac{3\sqrt{7}-\sqrt{3}}{2}\qquad\mathrm{(D)}\ 2\sqrt{3}\qquad\mathrm{(E)}\ \frac{5+2\sqrt{3}}{2}$ | 2008 AMC 12A Problem 22 | Notice that $\overarc{AE}$ is $\frac16$ the circumference of the circle. Therefore, $\overline{AE}$ is the side length of an inscribed hexagon with side length $4$. $\triangle AFE$ is a right triangle with $\overline{AF}=\frac12$. The length of $\overline{EF}=x+\frac{\sqrt{3}}{2}$. Using the Pythagorean Theorem, we get
$4^2 = \left(\frac{1}{2}\right)^2 + \left(x+\frac{\sqrt{3}}{2}\right)^2$
Solving for $x$, we get $x = \frac{3\sqrt{7}-\sqrt{3}}{2}\ \boxed{\text{C}}$ | // Block 1
unitsize(8mm);
defaultpen(linewidth(.8)+fontsize(8));
draw(Circle((0,0),4));
path mat=(-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle;
draw(mat);
draw(rotate(60)*mat);
draw(rotate(120)*mat);
draw(rotate(180)*mat);
draw(rotate(240)*mat);
draw(rotate(300)*mat);
label("\(x\)",(-1.55,2.1),E);
label("\(A\)",(-3.6,2.5513),E);
label("\(B\)",(-3.15,1.35),E);
label("\(C\)",(0.05,3.20),E);
label("\(D\)",(-0.75,4.15),E);
label("\(E\)",(0.3,4.15),E);
label("\(F\)",(-3.4,1.89),E);
draw((0.5,3.9686)--(-3.13,2.45));
draw((0.5,3.9686)--(-2.95,2));
// Block 2
unitsize(8mm); defaultpen(linewidth(.8)+fontsize(8)); draw(Circle((0,0),4)); path mat=(-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle; draw(mat); draw(rotate(60)*mat); draw(rotate(120)*mat); draw(rotate(180)*mat); draw(rotate(240)*mat); draw(rotate(300)*mat); label("\(x\)",(-1.55,2.1),E); label("\(A\)",(-3.6,2.5513),E); label("\(B\)",(-3.15,1.35),E); label("\(C\)",(0.05,3.20),E); label("\(D\)",(-0.75,4.15),E); label("\(E\)",(0.3,4.15),E); label("\(F\)",(-3.4,1.89),E); draw((0.5,3.9686)--(-3.13,2.45)); draw((0.5,3.9686)--(-2.95,2)); | [] |
665 | Triangle $ABC$ has $\angle C = 60^{\circ}$ and $BC = 4$. Point $D$ is the midpoint of $BC$. What is the largest possible value of $\tan{\angle BAD}$?
$\mathrm{(A)}\ \frac{\sqrt{3}}{6}\qquad\mathrm{(B)}\ \frac{\sqrt{3}}{3}\qquad\mathrm{(C)}\ \frac{\sqrt{3}}{2\sqrt{2}}\qquad\mathrm{(D)}\ \frac{\sqrt{3}}{4\sqrt{2}-3}\qquad\mathrm{(E)}\ 1$ | 2008 AMC 12A Problem 24 | Let $x = CA$. Then $\tan\theta = \tan(\angle BAF - \angle DAE)$, and since $\tan\angle BAF = \frac{2\sqrt{3}}{x-2}$ and $\tan\angle DAE = \frac{\sqrt{3}}{x-1}$, we have
\[\tan\theta = \frac{\frac{2\sqrt{3}}{x-2} - \frac{\sqrt{3}}{x-1}}{1 + \frac{2\sqrt{3}}{x-2}\cdot\frac{\sqrt{3}}{x-1}}= \frac{x\sqrt{3}}{x^2-3x+8}\]
With calculus, taking the derivative and setting equal to zero will give the maximum value of $\tan \theta$. Otherwise, we can apply AM-GM:
\begin{align*} \frac{x^2 - 3x + 8}{x} = \left(x + \frac{8}{x}\right) -3 &\geq 2\sqrt{x \cdot \frac 8x} - 3 = 4\sqrt{2} - 3\\ \frac{x}{x^2 - 3x + 8} &\leq \frac{1}{4\sqrt{2}-3}\\ \frac{x\sqrt{3}}{x^2 - 3x + 8} = \tan \theta &\leq \frac{\sqrt{3}}{4\sqrt{2}-3}\end{align*}
Thus, the maximum is at
$\frac{\sqrt{3}}{4\sqrt{2}-3} \Rightarrow \mathbf{(D)}$. | // Block 1
unitsize(12mm);
pair C=(0,0), B=(4 * dir(60)), A = (8,0), D=(2 * dir(60));
pair E=(1,0), F=(2,0);
draw(C--B--A--C);
draw(A--D);draw(D--E);draw(B--F);
dot(A);dot(B);dot(C);dot(D);dot(E);dot(F);
label("\(C\)",C,SW);
label("\(B\)",B,N);
label("\(A\)",A,SE);
label("\(D\)",D,NW);
label("\(E\)",E,S);
label("\(F\)",F,S);
label("\(60^\circ\)",C+(.1,.1),ENE);
label("\(2\)",1*dir(60),NW);
label("\(2\)",3*dir(60),NW);
label("\(\theta\)",(7,.4));
label("\(1\)",(.5,0),S);
label("\(1\)",(1.5,0),S);
label("\(x-2\)",(5,0),S);
// Block 2
unitsize(12mm); pair C=(0,0), B=(4 * dir(60)), A = (8,0), D=(2 * dir(60)); pair E=(1,0), F=(2,0); draw(C--B--A--C); draw(A--D);draw(D--E);draw(B--F); dot(A);dot(B);dot(C);dot(D);dot(E);dot(F); label("\(C\)",C,SW); label("\(B\)",B,N); label("\(A\)",A,SE); label("\(D\)",D,NW); label("\(E\)",E,S); label("\(F\)",F,S); label("\(60^\circ\)",C+(.1,.1),ENE); label("\(2\)",1*dir(60),NW); label("\(2\)",3*dir(60),NW); label("\(\theta\)",(7,.4)); label("\(1\)",(.5,0),S); label("\(1\)",(1.5,0),S); label("\(x-2\)",(5,0),S); | [] |
666 | Points $B$ and $C$ lie on $\overline{AD}$. The length of $\overline{AB}$ is $4$ times the length of $\overline{BD}$, and the length of $\overline{AC}$ is $9$ times the length of $\overline{CD}$. The length of $\overline{BC}$ is what fraction of the length of $\overline{AD}$?
$\textbf{(A)}\ \frac{1}{36}\qquad\textbf{(B)}\ \frac{1}{13}\qquad\textbf{(C)}\ \frac{1}{10}\qquad\textbf{(D)}\ \frac{5}{36}\qquad\textbf{(E)}\ \frac{1}{5}$ | 2008 AMC 10B Problem 6 | Let $x = BD$ and $y = CD$. Therefore, $AB = 4x$ and $AC = 9y$, as shown in the diagram(the labels on the bottom are for that line segment while the labels on the top are from one point to the left to one point to the right).
From this, we can see that $AD = 10y = 5x$, and since $BC = BD - CD = x-y$. Now, our ratio is $\frac{x-y}{AD}$. We can split this into 2 fractions: $\frac{x}{AD} - \frac{y}{AD} = \frac{x}{5x} - \frac{y}{10y} = \frac{1}{5} - \frac{1}{10} = \boxed{\textbf{(C)}\ \frac{1}{10}}$
~idk12345678 | // Block 1
dot((0,0));
label("A", (0,0), S);
dot((5,0));
label("B", (5,0), S);
dot((10,0));
label("C", (10,0), S);
dot((15,0));
label("D", (15,0), S);
draw((0,0)--(5,0));
draw((5,0)--(10,0));
draw((10,0)--(15,0));
draw((0,0)--(10,0));
draw((10,0)--(15,0));
label("$4x$", (0,0)--(5,0), S);
label("$9y$", (0,0)--(10,0), N);
label("$y$", (10,0)--(15,0), S);
label("$x$", (5,0)--(15,0), N);
// Block 2
dot((0,0)); label("A", (0,0), S); dot((5,0)); label("B", (5,0), S); dot((10,0)); label("C", (10,0), S); dot((15,0)); label("D", (15,0), S); draw((0,0)--(5,0)); draw((5,0)--(10,0)); draw((10,0)--(15,0)); draw((0,0)--(10,0)); draw((10,0)--(15,0)); label("$4x$", (0,0)--(5,0), S); label("$9y$", (0,0)--(10,0), N); label("$y$", (10,0)--(15,0), S); label("$x$", (5,0)--(15,0), N); | [] |
667 | An equilateral triangle of side length $10$ is completely filled in by non-overlapping equilateral triangles of side length $1$. How many small triangles are required?
$\mathrm{(A)}\ 10\qquad\mathrm{(B)}\ 25\qquad\mathrm{(C)}\ 100\qquad\mathrm{(D)}\ 250\qquad\mathrm{(E)}\ 1000$ | 2008 AMC 10B Problem 7 | The number of triangles is $1+3+\dots+19 = \boxed{100}$.
Also, another way to do it is to notice that as you go row by row (from the bottom), the number of triangles decrease by 2 from 19, so we have:
$19+17+15...+3+1 = \frac{19+1}{2}\cdot 10 = \boxed{100}$
A fourth solution is to notice that the small triangles are similar to the large triangle as they are both equilateral. Therefore, the ratio of their areas is the square of the ratio of their side lengths. Hence the ratio of their areas is $(1/10)^2=1/100$, so the answer is $\boxed{100}$. | // Block 1
unitsize(0.5cm);
defaultpen(0.8);
for (int i=0; i<10; ++i) { draw( (i*dir(60)) -- ( (10,0) + (i*dir(120)) ) ); }
for (int i=0; i<10; ++i) { draw( (i*dir(0)) -- ( 10*dir(60) + (i*dir(-60)) ) ); }
for (int i=0; i<10; ++i) { draw( ((10-i)*dir(60)) -- ((10-i)*dir(0)) ); }
// Block 2
unitsize(0.5cm); defaultpen(0.8); for (int i=0; i<10; ++i) { draw( (i*dir(60)) -- ( (10,0) + (i*dir(120)) ) ); } for (int i=0; i<10; ++i) { draw( (i*dir(0)) -- ( 10*dir(60) + (i*dir(-60)) ) ); } for (int i=0; i<10; ++i) { draw( ((10-i)*dir(60)) -- ((10-i)*dir(0)) ); } | [] |
668 | Points $A$ and $B$ are on a circle of radius $5$ and $AB=6$. Point $C$ is the midpoint of the minor arc $AB$. What is the length of the line segment $AC$?
$\mathrm{(A)}\ \sqrt{10}\qquad\mathrm{(B)}\ \frac{7}{2}\qquad\mathrm{(C)}\ \sqrt{14}\qquad\mathrm{(D)}\ \sqrt{15}\qquad\mathrm{(E)}\ 4$ | 2008 AMC 10B Problem 10 | Let the center of the circle be $O$, and let $D$ be the intersection of $\overline{AB}$ and $\overline{OC}$ (then $D$ is the midpoint of $\overline{AB}$). $OA=OB=5$, since they are both radii of the circle.
By the Pythagorean Theorem, $OD = \sqrt{OA^2 - DA^2} = 4$, and by subtraction, $CD=OC - OD = 1$.
Using the Pythagorean Theorem again, $AC= \sqrt{AD^2 + CD^2} = \sqrt{3^2+1^2}=\sqrt{10} \Longrightarrow \textbf{(A)}$. | // Block 1
pen d = linewidth(0.8); pathpen = d; pointpen = black; pen f = fontsize(9);
path p = CR((0,0),5);
pair O = (0,0), A=(5,0), B = IP(p,CR(A,6)), C = IP(p,CR(A,3)), D=IP(A--B,O--C);
D(p); D(MP("A",A,E)--D(MP("O",O))--MP("B",B,NE)--cycle); D(A--MP("C",C,ENE),dashed+d); D(O--C,dashed+d); D(rightanglemark(O,D(MP("D",D,W)),A)); MP("5",(A+O)/2); MP("3",(A+D)/2,SW);
// Block 2
pen d = linewidth(0.8); pathpen = d; pointpen = black; pen f = fontsize(9); path p = CR((0,0),5); pair O = (0,0), A=(5,0), B = IP(p,CR(A,6)), C = IP(p,CR(A,3)), D=IP(A--B,O--C); D(p); D(MP("A",A,E)--D(MP("O",O))--MP("B",B,NE)--cycle); D(A--MP("C",C,ENE),dashed+d); D(O--C,dashed+d); D(rightanglemark(O,D(MP("D",D,W)),A)); MP("5",(A+O)/2); MP("3",(A+D)/2,SW); | [] |
669 | A cylindrical tank with radius $4$ feet and height $9$ feet is lying on its side. The tank is filled with water to a depth of $2$ feet. What is the volume of water, in cubic feet?
$\mathrm{(A)}\ 24\pi - 36 \sqrt {2} \qquad \mathrm{(B)}\ 24\pi - 24 \sqrt {3} \qquad \mathrm{(C)}\ 36\pi - 36 \sqrt {3} \qquad \mathrm{(D)}\ 36\pi - 24 \sqrt {2} \qquad \mathrm{(E)}\ 48\pi - 36 \sqrt {3}$ | 2008 AMC 10B Problem 19 | Any vertical cross-section of the tank parallel with its base looks as follows:
The volume of water can be computed as the height of the tank times the area of the shaded part.
Let $\theta$ be the size of the smaller angle $DAC$. We then have $\cos\theta = \frac{AD}{AC}=\frac 12$, hence $\theta=60^\circ$.
The figure is symmetrical, so the angle $CAB$ has size $2\cdot 60^\circ = 120^\circ$. Hence the shaded part consists of $\frac{120^\circ}{360^\circ} = \frac 13$ of the circle, minus the area of the triangle $ABC$.
Using the Pythagorean theorem we can compute that $CD=\sqrt{AC^2-AD^2}=\sqrt{16-4}=2\sqrt 3$. Thus $BC=4\sqrt 3$, and the area of the triangle $ABC$ is $\frac {2 \cdot 4\sqrt 3} 2 = 4\sqrt 3$.
The area of the shaded part is then $\frac{4^2\pi}3 - 4\sqrt 3$, and the volume of water is $9\cdot\left(\frac{4^2\pi}3 - 4\sqrt 3\right) = \boxed{48\pi - 36\sqrt 3}$. The answer is $\text{E}$.
Alternatively, we can solve for DC and see it is congruent to a 30 60 90 triangle by SSS ~Williamgolly | // Block 1
unitsize(0.8cm);
defaultpen(0.8);
pair s=(0,0), bottom=(0,-4), mid=(0,-2);
pair x[]=intersectionpoints( (-10,-2)--(10,-2), circle(s,4) );
fill( arc(s,x[0],x[1]) -- cycle, lightgray );
draw( circle(s,4) );
dot(s);
draw( s -- bottom );
label( "$2$", (mid+bottom)/2, E );
draw ( s -- x[0] -- x[1] -- s );
label( "$4$", (s+x[0])/2, NW );
label( "$4$", (s+x[1])/2, NE );
label( "$A$", s, N );
label( "$B$", x[0], W );
label( "$C$", x[1], E );
label( "$D$", mid, NW );
label( "$E$", bottom, S );
// Block 2
unitsize(0.8cm); defaultpen(0.8); pair s=(0,0), bottom=(0,-4), mid=(0,-2); pair x[]=intersectionpoints( (-10,-2)--(10,-2), circle(s,4) ); fill( arc(s,x[0],x[1]) -- cycle, lightgray ); draw( circle(s,4) ); dot(s); draw( s -- bottom ); label( "$2$", (mid+bottom)/2, E ); draw ( s -- x[0] -- x[1] -- s ); label( "$4$", (s+x[0])/2, NW ); label( "$4$", (s+x[1])/2, NE ); label( "$A$", s, N ); label( "$B$", x[0], W ); label( "$C$", x[1], E ); label( "$D$", mid, NW ); label( "$E$", bottom, S ); | [] |
670 | Quadrilateral $ABCD$ has $AB = BC = CD$, $m\angle ABC = 70^\circ$ and $m\angle BCD = 170^\circ$. What is the degree measure of $\angle BAD$?
$\mathrm{(A)}\ 75\qquad\mathrm{(B)}\ 80\qquad\mathrm{(C)}\ 85\qquad\mathrm{(D)}\ 90\qquad\mathrm{(E)}\ 95$ | 2008 AMC 10B Problem 24 | First, connect the diagonal $DB$, then, draw line $DE$ such that it is congruent to $DC$ and is parallel to $AB$. Because triangle $DCB$ is isosceles and angle $DCB$ is $170^\circ$, the angles $CDB$ and $CBD$ are both $\frac{180-170}{2} = 5^\circ$. Because angle $ABC$ is $70^\circ$, we get angle $ABD$ is $65^\circ$. Next, noticing parallel lines $AB$ and $DE$ and transversal $DB$, we see that angle $BDE$ is also $65^\circ$, and subtracting off angle $CDB$ gives that angle $EDC$ is $60^\circ$.
Now, because we drew $ED = DC$, triangle $DEC$ is equilateral. We can also conclude that $EC=DC=CB$ meaning that triangle $ECB$ is isosceles, and angles $CBE$ and $CEB$ are equal.
Finally, we can set up our equation. Denote angle $BAD$ as $x^\circ$. Then, because $ABED$ is a parallelogram, the angle $DEB$ is also $x^\circ$. Then, $CEB$ is $(x-60)^\circ$. Again because $ABED$ is a parallelogram, angle $ABE$ is $(180-x)^\circ$. Subtracting angle $ABC$ gives that angle $CBE$ equals $(110-x)^\circ$. Because angle $CBE$ equals angle $CEB$, we get \[x-60=110-x\], solving into $x=\boxed{85^\circ}$.
Side note: this solution was inspired by some basic angle chasing and finding some 60 degree angles, which made me want to create equilateral triangles.
~Someonenumber011 | // Block 1
unitsize(1cm);
defaultpen(.8);
real a=4;
pair A=(0,0), B=a*dir(0), C=B+a*dir(110), D=C+a*dir(120), E=D+a*dir(0);
draw(A--B--C--D--cycle);
draw(E--C);
draw(B--D);
draw(B--E);
draw(D--E);
label("$A$",A,SW);
label("$B$",B,SE);
label("$C$",C,SE);
label("$D$",D,N);
label("$E$",E,NE);
label("$60^\circ$",C + .75*dir(360-65-115-55-30));
label("$65^\circ$",B + .75*dir(180-32.5));
label("$x^\circ$",A + .5*dir(42.5));
label("$5^\circ$",D + 2.5*dir(360-60-2.5));
label("$60^\circ$",D + .75*dir(360-30));
label("$60^\circ$",E + .5*dir(360-150));
label("$5^\circ$",B + 2.5*dir(180-65-2.5));
// Block 2
unitsize(1cm); defaultpen(.8); real a=4; pair A=(0,0), B=a*dir(0), C=B+a*dir(110), D=C+a*dir(120), E=D+a*dir(0); draw(A--B--C--D--cycle); draw(E--C); draw(B--D); draw(B--E); draw(D--E); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,SE); label("$D$",D,N); label("$E$",E,NE); label("$60^\circ$",C + .75*dir(360-65-115-55-30)); label("$65^\circ$",B + .75*dir(180-32.5)); label("$x^\circ$",A + .5*dir(42.5)); label("$5^\circ$",D + 2.5*dir(360-60-2.5)); label("$60^\circ$",D + .75*dir(360-30)); label("$60^\circ$",E + .5*dir(360-150)); label("$5^\circ$",B + 2.5*dir(180-65-2.5)); | [] |
670 | Quadrilateral $ABCD$ has $AB = BC = CD$, $m\angle ABC = 70^\circ$ and $m\angle BCD = 170^\circ$. What is the degree measure of $\angle BAD$?
$\mathrm{(A)}\ 75\qquad\mathrm{(B)}\ 80\qquad\mathrm{(C)}\ 85\qquad\mathrm{(D)}\ 90\qquad\mathrm{(E)}\ 95$ | 2008 AMC 10B Problem 24 | Let the unknown $\angle BAD$ be $x$.
First, we draw diagonal $BD$ and $AC$.
$I$ is the intersection of the two diagonals. The diagonals each form two isosceles triangles, $\triangle BCD$ and $\triangle ABC$.
Using this, we find: $\angle DBC = \angle CDB = 5^\circ$ and $\angle BAC = \angle BCA = 55^\circ$. Expanding on this, we can fill in a couple more angles.
$\angle ABD = 70^\circ - 5^\circ = 65^\circ$, $\angle ACD = 170^\circ - 55^\circ = 115^\circ$, $\angle BIA = \angle CID = 180^\circ - (65^\circ + 55^\circ) = 60^\circ$, $\angle BIC = \angle AID = 180^\circ - 60^\circ = 120^\circ$.
We can rewrite $\angle CAD$ and $\angle BDA$ in terms of $x$. $\angle CAD = x - 55^\circ$ and $\angle BDA = 180^\circ - (120^\circ + x - 55^\circ) = 115^\circ - x$.
Let us relabel $AB = BC = CD = a$ and $AD = b$.
By Rule of Sines on $\triangle ACD$ and $\triangle ABD$ respectively, $\frac{\sin(\angle CAD)}{a} = \frac{\sin(\angle ACD)}{b}$, and $\frac{\sin(\angle ABD)}{b} = \frac{\sin(\angle BDA)}{a}$
In a more convenient form, $\frac{\sin(x-55^\circ)}{a} = \frac{\sin(115^\circ)}{b} \implies \frac{a}{b} = \frac{\sin(x-55^\circ)}{\sin(115^\circ)}$
and $\frac{\sin(65^\circ)}{b} = \frac{\sin(115^\circ-x)}{a} \implies \frac{a}{b} = \frac{\sin(115^\circ-x)}{\sin(65^\circ)}$
$\implies \frac{\sin(115^\circ-x)}{\sin(65^\circ)} = \frac{\sin(x-55^\circ)}{\sin(115^\circ)}$
Now, by identity $\sin(\theta) = \sin(180^\circ-\theta)$, $\sin(65^\circ) = \sin(115^\circ)$
Therefore, $\sin(115^\circ-x) = \sin(x-55^\circ).$ This equation is only satisfied by option $\boxed{\text {(C) } 85^\circ}$
Note: I'm pretty bad at Asymptote, if anyone could edit this and fill in the angles into the diagram, that would be pretty cool.
~Raghu9372 | // Block 1
unitsize(3 cm);
pair A, B, C, D;
A = (0,0);
B = dir(85);
C = B + dir(-25);
D = C + dir(-35);
draw(A--B--C--D--cycle);
draw(A--C);
draw(B--D);
draw(((A + B)/2 + scale(0.02)*rotate(90)*(B - A))--((A + B)/2 + scale(0.02)*rotate(90)*(A - B)));
draw(((B + C)/2 + scale(0.02)*rotate(90)*(C - B))--((B + C)/2 + scale(0.02)*rotate(90)*(B - C)));
draw(((C + D)/2 + scale(0.02)*rotate(90)*(D - C))--((C + D)/2 + scale(0.02)*rotate(90)*(C - D)));
dot("$A$", A, SW);
dot("$B$", B, N);
dot("$C$", C, NE);
dot("$D$", D, SE);
label("$I$", 6/7*C);
// Block 2
unitsize(3 cm); pair A, B, C, D; A = (0,0); B = dir(85); C = B + dir(-25); D = C + dir(-35); draw(A--B--C--D--cycle); draw(A--C); draw(B--D); draw(((A + B)/2 + scale(0.02)*rotate(90)*(B - A))--((A + B)/2 + scale(0.02)*rotate(90)*(A - B))); draw(((B + C)/2 + scale(0.02)*rotate(90)*(C - B))--((B + C)/2 + scale(0.02)*rotate(90)*(B - C))); draw(((C + D)/2 + scale(0.02)*rotate(90)*(D - C))--((C + D)/2 + scale(0.02)*rotate(90)*(C - D))); dot("$A$", A, SW); dot("$B$", B, N); dot("$C$", C, NE); dot("$D$", D, SE); label("$I$", 6/7*C); | [] |
671 | Michael walks at the rate of $5$ feet per second on a long straight path. Trash pails are located every $200$ feet along the path. A garbage truck traveling at $10$ feet per second in the same direction as Michael stops for $30$ seconds at each pail. As Michael passes a pail, he notices the truck ahead of him just leaving the next pail. How many times will Michael and the truck meet?
$\mathrm{(A)}\ 4\qquad\mathrm{(B)}\ 5\qquad\mathrm{(C)}\ 6\qquad\mathrm{(D)}\ 7\qquad\mathrm{(E)}\ 8$ | 2008 AMC 10B Problem 25 | Pick a coordinate system where Michael's starting pail is $0$ and the one where the truck starts is $200$.
Let $M(t)$ and $T(t)$ be the coordinates of Michael and the truck after $t$ seconds.
Let $D(t)=T(t)-M(t)$ be their (signed) distance after $t$ seconds.
Meetings occur whenever $D(t)=0$.
We have $D(0)=200$.
The truck always moves for $20$ seconds, then stands still for $30$. During the first $20$ seconds of the cycle the truck moves by $200$ feet and Michael by $100$, hence during the first $20$ seconds of the cycle $D(t)$ increases by $100$.
During the remaining $30$ seconds $D(t)$ decreases by $150$.
From this observation it is obvious that after four full cycles, i.e. at $t=200$, we will have $D(t)=0$ for the first time.
During the fifth cycle, $D(t)$ will first grow from $0$ to $100$, then fall from $100$ to $-50$. Hence Michael overtakes the truck while it is standing at the pail.
During the sixth cycle, $D(t)$ will first grow from $-50$ to $50$, then fall from $50$ to $-100$. Hence the truck starts moving, overtakes Michael on their way to the next pail, and then Michael overtakes the truck while it is standing at the pail.
During the seventh cycle, $D(t)$ will first grow from $-100$ to $0$, then fall from $0$ to $-150$. Hence the truck meets Michael at the moment when it arrives to the next pail.
Obviously, from this point on $D(t)$ will always be negative, meaning that Michael is already too far ahead. Hence we found all $\boxed{5 \Longrightarrow B}$ meetings.
The movement of Michael and the truck is plotted below: Michael in blue, the truck in red. | import graph; size(400,300,IgnoreAspect); real[] xt = new real[21]; real[] yt = new real[21]; for (int i=0; i<11; ++i) xt[2*i]=50*i; for (int i=0; i<10; ++i) xt[2*i+1]=50*i+20; for (int i=0; i<11; ++i) yt[2*i]=200*(i+1); for (int i=0; i<10; ++i) yt[2*i+1]=200*(i+2); real[] xm={0,500}; real[] ym={0,2500}; draw(graph(xt,yt),red); draw(graph(xm,ym),blue); xaxis("$\mathrm{time}$",Bottom,LeftTicks); yaxis("$\mathrm{position}$",Left,LeftTicks); | [] |
672 | Points $A$ and $B$ are on a circle of radius $5$ and $AB = 6$. Point $C$ is the midpoint of the minor arc $AB$. What is the length of the line segment $AC$?
$\textbf{(A)}\ \sqrt {10} \qquad \textbf{(B)}\ \frac {7}{2} \qquad \textbf{(C)}\ \sqrt {14} \qquad \textbf{(D)}\ \sqrt {15} \qquad \textbf{(E)}\ 4$ | 2008 AMC 12B Problem 9 | Solution 1
Let $\alpha$ be the angle that subtends the arc $AB$. By the law of cosines,
$6^2=5^2+5^2-2\cdot 5\cdot 5\cos(\alpha)$ implies $\cos(\alpha) = 7/25$.
The half-angle formula says that
$\cos(\alpha/2) = \frac{\sqrt{1+\cos(\alpha)}}{2} = \sqrt{\frac{32/25}{2}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$. The law of cosines tells us $AC = \sqrt{5^2+5^2-2\cdot 5\cdot 5\cdot \frac{4}{5}} = \sqrt{50-50\frac{4}{5}} = \sqrt{10}$, which is answer choice $\boxed{\text{A}}$.
Solution 2
Define $D$ as the midpoint of line segment $\overline{AB}$, and $O$ the center of the circle. Then $O$, $C$, and $D$ are collinear, and since $D$ is the midpoint of $AB$, $m\angle ODA=90\deg$ and so $OD=\sqrt{5^2-3^2}=4$. Since $OD=4$, $CD=5-4=1$, and so $AC=\sqrt{3^2+1^2}=\sqrt{10} \rightarrow \boxed{\text{A}}$. | pen d = linewidth(0.7); pathpen = d; pointpen = black; pen f = fontsize(9); path p = CR((0,0),5); pair O = (0,0), A=(5,0), B = IP(p,CR(A,6)), C = IP(p,CR(A,3)), D=IP(A--B,O--C); D(p); D(MP("A",A,E)--D(MP("O",O))--MP("B",B,NE)--cycle); D(A--MP("C",C,ENE),dashed+d); D(O--C,dashed+d); D(rightanglemark(O,D(MP("D",D,W)),A)); MP("5",(A+O)/2); MP("3",(A+D)/2,SW); | [] |
673 | Vertex $E$ of equilateral $\triangle{ABE}$ is in the interior of unit square $ABCD$. Let $R$ be the region consisting of all points inside $ABCD$ and outside $\triangle{ABE}$ whose distance from $AD$ is between $\frac{1}{3}$ and $\frac{2}{3}$. What is the area of $R$?
$\textbf{(A)}\ \frac{12-5\sqrt3}{72} \qquad \textbf{(B)}\ \frac{12-5\sqrt3}{36} \qquad \textbf{(C)}\ \frac{\sqrt3}{18} \qquad \textbf{(D)}\ \frac{3-\sqrt3}{9} \qquad \textbf{(E)}\ \frac{\sqrt3}{12}$ | 2008 AMC 12B Problem 13 | The region is the shaded area:
We can find the area of the shaded region by subtracting the pentagon from the middle third of the square. The area of the middle third of the square is $\left(\frac13\right)(1)=\frac13$. The pentagon can be split into a rectangle and an equilateral triangle.
The base of the equilateral triangle is $\frac13$ and the height is $\left(\frac13\right)\left(\frac12\right)(\sqrt{3})=\frac{\sqrt{3}}{6}$. Thus, the area is $\left(\frac{\sqrt3}{6}\right)\left(\frac13\right)\left(\frac12\right)=\frac{\sqrt3}{36}$.
The base of the rectangle is $\frac13$ and the height is the height of the equilateral triangle minus the height of the smaller equilateral triangle. This is:
$\frac{\sqrt3}{2}-\frac{\sqrt3}{6}=\frac{\sqrt3}{3}$
Therefore, the area of the shaded region is
$\frac13-\frac{\sqrt3}{9}-\frac{\sqrt3}{36}=\boxed{\text{(B) }\frac{12-5\sqrt3}{36}}.$ | // Block 1
pair A,B,C,D,E;
A=(0,1);
B=(1,1);
C=(1,0);
D=(0,0);
E=(1/2,1-sqrt(3)/2);
draw(A--B--C--D--cycle);
label("A",A,NW);
dot(A);
label("B",B,NE);
dot(B);
label("C",C,SE);
dot(C);
label("D",D,SW);
dot(D);
draw(A--E--B--cycle);
label("E",E,S);
dot(E);
draw((1/3,0)--(1/3,1));
draw((2/3,0)--(2/3,1));
fill((1/3,0)--(1/3,1-sqrt(3)/3)--E--(2/3,1-sqrt(3)/3)--(2/3,0)--cycle,Black);
// Block 2
pair A,B,C,D,E; A=(0,1); B=(1,1); C=(1,0); D=(0,0); E=(1/2,1-sqrt(3)/2); draw(A--B--C--D--cycle); label("A",A,NW); dot(A); label("B",B,NE); dot(B); label("C",C,SE); dot(C); label("D",D,SW); dot(D); draw(A--E--B--cycle); label("E",E,S); dot(E); draw((1/3,0)--(1/3,1)); draw((2/3,0)--(2/3,1)); fill((1/3,0)--(1/3,1-sqrt(3)/3)--E--(2/3,1-sqrt(3)/3)--(2/3,0)--cycle,Black); | [] |
674 | On each side of a unit square, an equilateral triangle of side length 1 is constructed. On each new side of each equilateral triangle, another equilateral triangle of side length 1 is constructed. The interiors of the square and the 12 triangles have no points in common. Let $R$ be the region formed by the union of the square and all the triangles, and $S$ be the smallest convex polygon that contains $R$. What is the area of the region that is inside $S$ but outside $R$?
$\textbf{(A)} \; \frac{1}{4} \qquad \textbf{(B)} \; \frac{\sqrt{2}}{4} \qquad \textbf{(C)} \; 1 \qquad \textbf{(D)} \; \sqrt{3} \qquad \textbf{(E)} \; 2 \sqrt{3}$ | 2008 AMC 12B Problem 15 | The equilateral triangles form trapezoids with side lengths $1, 1, 1, 2$ (half a unit hexagon) on each face of the unit square. The four triangles "in between" these trapezoids are isosceles triangles with two side lengths $1$ and an angle of $30^{\circ}$ in between them, so the total area of these triangles (which is the area of $S - R$) is, $4 \left( \frac {1}{2} \sin 30^{\circ} \right) = 1$ which makes the answer $\boxed{C}$. | // Block 1
real a = 1/2, b = sqrt(3)/2;
draw((0,0)--(1,0)--(1,1)--(0,1)--cycle);
draw((0,0)--(a,-b)--(1,0)--(1+b,a)--(1,1)--(a,1+b)--(0,1)--(-b,a)--(0,0));
draw((0,0)--(-1+a,-b)--(1+a,-b)--(1,0)--(1+b,-1+a)--(1+b,1+a)--(1,1)--(1+a,1+b)--(-1+a,1+b)--(0,1)--(-b,1+a)--(-b,-1+a)--(0,0));
filldraw((1+a,-b)--(1,0)--(1+b,-1+a)--cycle,gray(0.9));
filldraw((1+b,1+a)--(1,1)--(1+a,1+b)--cycle,gray(0.9));
filldraw((-1+a,1+b)--(0,1)--(-b,1+a)--cycle,gray(0.9));
filldraw((-b,-1+a)--(0,0)--(-1+a,-b)--cycle,gray(0.9));
// Block 2
real a = 1/2, b = sqrt(3)/2; draw((0,0)--(1,0)--(1,1)--(0,1)--cycle); draw((0,0)--(a,-b)--(1,0)--(1+b,a)--(1,1)--(a,1+b)--(0,1)--(-b,a)--(0,0)); draw((0,0)--(-1+a,-b)--(1+a,-b)--(1,0)--(1+b,-1+a)--(1+b,1+a)--(1,1)--(1+a,1+b)--(-1+a,1+b)--(0,1)--(-b,1+a)--(-b,-1+a)--(0,0)); filldraw((1+a,-b)--(1,0)--(1+b,-1+a)--cycle,gray(0.9)); filldraw((1+b,1+a)--(1,1)--(1+a,1+b)--cycle,gray(0.9)); filldraw((-1+a,1+b)--(0,1)--(-b,1+a)--cycle,gray(0.9)); filldraw((-b,-1+a)--(0,0)--(-1+a,-b)--cycle,gray(0.9)); | [] |
674 | On each side of a unit square, an equilateral triangle of side length 1 is constructed. On each new side of each equilateral triangle, another equilateral triangle of side length 1 is constructed. The interiors of the square and the 12 triangles have no points in common. Let $R$ be the region formed by the union of the square and all the triangles, and $S$ be the smallest convex polygon that contains $R$. What is the area of the region that is inside $S$ but outside $R$?
$\textbf{(A)} \; \frac{1}{4} \qquad \textbf{(B)} \; \frac{\sqrt{2}}{4} \qquad \textbf{(C)} \; 1 \qquad \textbf{(D)} \; \sqrt{3} \qquad \textbf{(E)} \; 2 \sqrt{3}$ | 2008 AMC 12B Problem 15 | The area of the largest square is $(1+\sqrt{3})^2=4+2\sqrt{3}$. The area of region $R$ is $1+12\frac{1}{2}\frac{\sqrt{3}}{2}=1+3\sqrt{3}$. The total area of four small 45-45-90 triangles at corner is $4*\frac{1}{2}(\frac{\sqrt{3}+1-2}{2})^2=2-\sqrt{3}$. $S=4+2\sqrt{3}-(2-\sqrt{3})=2+3\sqrt{3}$, $S-R=1$.
~Bran Qin | // Block 1
real a = 1/2, b = sqrt(3)/2;
draw((0,0)--(1,0)--(1,1)--(0,1)--cycle);
draw((a+b+a,a+b+a)--(a+b+a,-b)--(-b,-b)--(-b,a+a+b)--cycle,dashed);
draw((0,0)--(a,-b)--(1,0)--(1+b,a)--(1,1)--(a,1+b)--(0,1)--(-b,a)--(0,0));
draw((0,0)--(-1+a,-b)--(1+a,-b)--(1,0)--(1+b,-1+a)--(1+b,1+a)--(1,1)--(1+a,1+b)--(-1+a,1+b)--(0,1)--(-b,1+a)--(-b,-1+a)--(0,0));
filldraw((1+a,-b)--(1,0)--(1+b,-1+a)--cycle,gray(0.9));
filldraw((1+b,1+a)--(1,1)--(1+a,1+b)--cycle,gray(0.9));
filldraw((-1+a,1+b)--(0,1)--(-b,1+a)--cycle,gray(0.9));
filldraw((-b,-1+a)--(0,0)--(-1+a,-b)--cycle,gray(0.9));
// Block 2
real a = 1/2, b = sqrt(3)/2; draw((0,0)--(1,0)--(1,1)--(0,1)--cycle); draw((a+b+a,a+b+a)--(a+b+a,-b)--(-b,-b)--(-b,a+a+b)--cycle,dashed); draw((0,0)--(a,-b)--(1,0)--(1+b,a)--(1,1)--(a,1+b)--(0,1)--(-b,a)--(0,0)); draw((0,0)--(-1+a,-b)--(1+a,-b)--(1,0)--(1+b,-1+a)--(1+b,1+a)--(1,1)--(1+a,1+b)--(-1+a,1+b)--(0,1)--(-b,1+a)--(-b,-1+a)--(0,0)); filldraw((1+a,-b)--(1,0)--(1+b,-1+a)--cycle,gray(0.9)); filldraw((1+b,1+a)--(1,1)--(1+a,1+b)--cycle,gray(0.9)); filldraw((-1+a,1+b)--(0,1)--(-b,1+a)--cycle,gray(0.9)); filldraw((-b,-1+a)--(0,0)--(-1+a,-b)--cycle,gray(0.9)); | [] |
675 | Michael walks at the rate of $5$ feet per second on a long straight path. Trash pails are located every $200$ feet along the path. A garbage truck traveling at $10$ feet per second in the same direction as Michael stops for $30$ seconds at each pail. As Michael passes a pail, he notices the truck ahead of him just leaving the next pail. How many times will Michael and the truck meet?
$\mathrm{(A)}\ 4\qquad\mathrm{(B)}\ 5\qquad\mathrm{(C)}\ 6\qquad\mathrm{(D)}\ 7\qquad\mathrm{(E)}\ 8$ | 2008 AMC 12B Problem 20 | Pick a coordinate system where Michael's starting pail is $0$ and the one where the truck starts is $200$.
Let $M(t)$ and $T(t)$ be the coordinates of Michael and the truck after $t$ seconds.
Let $D(t)=T(t)-M(t)$ be their (signed) distance after $t$ seconds.
Meetings occur whenever $D(t)=0$.
We have $D(0)=200$.
The truck always moves for $20$ seconds, then stands still for $30$. During the first $20$ seconds of the cycle the truck moves by $200$ feet and Michael by $100$, hence during the first $20$ seconds of the cycle $D(t)$ increases by $100$.
During the remaining $30$ seconds $D(t)$ decreases by $150$.
From this observation it is obvious that after four full cycles, i.e. at $t=200$, we will have $D(t)=0$ for the first time.
During the fifth cycle, $D(t)$ will first grow from $0$ to $100$, then fall from $100$ to $-50$. Hence Michael overtakes the truck while it is standing at the pail.
During the sixth cycle, $D(t)$ will first grow from $-50$ to $50$, then fall from $50$ to $-100$. Hence the truck starts moving, overtakes Michael on their way to the next pail, and then Michael overtakes the truck while it is standing at the pail.
During the seventh cycle, $D(t)$ will first grow from $-100$ to $0$, then fall from $0$ to $-150$. Hence the truck meets Michael at the moment when it arrives to the next pail.
Obviously, from this point on $D(t)$ will always be negative, meaning that Michael is already too far ahead. Hence we found all $\boxed{5 \Longrightarrow B}$ meetings.
The movement of Michael and the truck is plotted below: Michael in blue, the truck in red. | // Block 1
import graph;
size(400,300,IgnoreAspect);
real[] xt = new real[21];
real[] yt = new real[21];
for (int i=0; i<11; ++i) xt[2*i]=50*i;
for (int i=0; i<10; ++i) xt[2*i+1]=50*i+20;
for (int i=0; i<11; ++i) yt[2*i]=200*(i+1);
for (int i=0; i<10; ++i) yt[2*i+1]=200*(i+2);
real[] xm={0,500};
real[] ym={0,2500};
draw(graph(xt,yt),red);
draw(graph(xm,ym),blue);
xaxis("$\mathrm{time}$",Bottom,LeftTicks);
yaxis("$\mathrm{position}$",Left,LeftTicks);
// Block 2
import graph; size(400,300,IgnoreAspect); real[] xt = new real[21]; real[] yt = new real[21]; for (int i=0; i<11; ++i) xt[2*i]=50*i; for (int i=0; i<10; ++i) xt[2*i+1]=50*i+20; for (int i=0; i<11; ++i) yt[2*i]=200*(i+1); for (int i=0; i<10; ++i) yt[2*i+1]=200*(i+2); real[] xm={0,500}; real[] ym={0,2500}; draw(graph(xt,yt),red); draw(graph(xm,ym),blue); xaxis("$\mathrm{time}$",Bottom,LeftTicks); yaxis("$\mathrm{position}$",Left,LeftTicks); | [] |
676 | Let $ABCD$ be a trapezoid with $AB||CD, AB=11, BC=5, CD=19,$ and $DA=7$. Bisectors of $\angle A$ and $\angle D$ meet at $P$, and bisectors of $\angle B$ and $\angle C$ meet at $Q$. What is the area of hexagon $ABQCDP$?
$\textbf{(A)}\ 28\sqrt{3}\qquad \textbf{(B)}\ 30\sqrt{3}\qquad \textbf{(C)}\ 32\sqrt{3}\qquad \textbf{(D)}\ 35\sqrt{3}\qquad \textbf{(E)}\ 36\sqrt{3}$ | 2008 AMC 12B Problem 25 | $P$ is the intersection of the angle bisectors of $\angle A$ and $\angle D$. By definition, angle bisectors are always equidistant from the sides of the angle, so $P$ is equidistant from $\overline{AB}$, $\overline{AD}$, and $\overline{CD}$. Likewise, point $Q$ is equidistant from $\overline{AB}$, $\overline{BC}$, and $\overline{CD}$. Because both points $P$ and $Q$ are equidistant from $\overline{AB}$ and $\overline{CD}$ and the distance between $\overline{AB}$ and $\overline{CD}$ is constant, the common distances from each of the points to the mentioned segments is equal for $P$ and $Q$. Call this distance $x$.
The distance between a point and a line is the length of the segment perpendicular to the line with one endpoint on the line and the other on the point. This means the altitude from $P$ to $\overline{AD}$ is $x$, so the area of $\triangle ADP$ is equal to $\frac12\cdot AD\cdot x=\frac72x$. Similarly, the area of $\triangle BCQ$ is $\frac12\cdot BC\cdot x=\frac52x$. The altitude of the trapezoid is $2x$, because it is the sum of the distances from either $P$ or $Q$ to $\overline{AB}$ and $\overline{CD}$. This means the area of trapezoid $ABCD$ is $\frac12(AB+CD)\cdot2x=\frac12(11+19)\cdot2x=30x$. Now, the area of hexagon $ABQCDP$ is the area of trapezoid $ABCD$, minus the areas of triangles $ADP$ and $BCQ$. This is $30x-\frac72x-\frac52x=24x$. Now it remains to find $x$.
We let $R$ and $S$ be the feet of the altitudes of $A$ and $B$, respectively, to $\overline{CD}$. We define $r=RD$ and $s=SC$. We know that $AB=RS$, so $RS=11$ and $r+s=19-11=8$. By the Pythagorean Theorem on $\triangle ADR$ and $\triangle BCS$, we get $r^2+(2x)^2=7^2$ and $s^2+(2x)^2=5^2$, respectively. Subtracting the second equation from the first gives us $r^2-s^2=49-25=24$. The left hand side of this equation is a difference of squares and factors to $(r+s)(r-s)$. We know that $r+s=8$, so $r-s=\frac{24}8=3$. Now we can solve for $r$ by adding the two equations we just got to see that $2r=11$, or $r=\frac{11}2$.
We now solve for $x$. We know that $r^2+(2x)^2=49$, so $(2x)^2=49-\left(\frac{11}2\right)^2=\frac{75}4$ and $2x=\frac{5\sqrt3}2$. We multiply both sides of this equation by $12$ to get $24x=30\sqrt3$. However, the area of hexagon $ABQCDP$ is $24x$, so the answer is $30\sqrt 3$, or answer choice $\boxed{B}$. | // Block 1
unitsize(0.6cm);
import olympiad;
pair A,B,C,D,P,Q,M,N,W,X,Y,Z;
A=(11/2,5sqrt(3)/2);
B=(33/2,5sqrt(3)/2);
C=(19,0);
D=(0,0);
P=incenter(A,D,(99999,5sqrt(3)/4));
Q=incenter(B,C,(-99999,5sqrt(3)/4));
W=P+(0,5sqrt(3)/4);
X=P-(0,5sqrt(3)/4);
Y=Q+(0,5sqrt(3)/4);
Z=Q-(0,5sqrt(3)/4);
M=reflect(A,P)*W;
N=reflect(B,Q)*Y;
draw(A--B--C--D--cycle);
draw(A--P--D);
draw(B--Q--C);
label("$A$",A,dir(135));
label("$B$",B,dir(45));
label("$C$",C,dir(315));
label("$D$",D,dir(225));
dot("$P$",P,dir(0));
dot("$Q$",Q,dir(180));
draw(W--X);
draw(Y--Z);
draw(M--P);
draw(N--Q);
label("$11$",midpoint(A--B),dir(90));
label("$5$",midpoint(B--C),dir(45));
label("$19$",midpoint(C--D),dir(270));
label("$7$",midpoint(D--A),dir(135));
label("$x$",midpoint(P--W),dir(0));
label("$x$",midpoint(P--X),dir(0));
label("$x$",midpoint(P--M),dir(225));
label("$x$",midpoint(Q--Y),dir(180));
label("$x$",midpoint(Q--Z),dir(180));
label("$x$",midpoint(Q--N),dir(315));
draw(rightanglemark(P,W,B,12.5));
draw(rightanglemark(P,X,C,12.5));
draw(rightanglemark(P,M,D,12.5));
draw(rightanglemark(Q,Y,A,12.5));
draw(rightanglemark(Q,Z,D,12.5));
draw(rightanglemark(Q,N,C,12.5));
// Block 2
unitsize(0.6cm);
import olympiad;
pair A,B,C,D,R,S;
A=(11/2,5sqrt(3)/2);
B=(33/2,5sqrt(3)/2);
C=(19,0);
D=(0,0);
R=(11/2,0);
S=(33/2,0);
draw(A--B--C--D--cycle);
draw(A--R);
draw(B--S);
label("$A$",A,dir(135));
label("$B$",B,dir(45));
label("$C$",C,dir(315));
label("$D$",D,dir(225));
label("$R$",R,dir(270));
label("$S$",S,dir(270));
label("$11$",midpoint(A--B),dir(90));
label("$5$",midpoint(B--C),dir(45));
label("$11$",midpoint(R--S),dir(270));
label("$7$",midpoint(D--A),dir(135));
label("$r$",midpoint(R--D),dir(270));
label("$s$",midpoint(C--S),dir(270));
label("$19$",midpoint(C--D),5*dir(270));
label("$2x$",midpoint(A--R),dir(0));
label("$2x$",midpoint(B--S),dir(180));
draw(rightanglemark(A,R,D,15));
draw(rightanglemark(B,S,C,15));
// Block 3
unitsize(0.6cm); import olympiad; pair A,B,C,D,P,Q,M,N,W,X,Y,Z; A=(11/2,5sqrt(3)/2); B=(33/2,5sqrt(3)/2); C=(19,0); D=(0,0); P=incenter(A,D,(99999,5sqrt(3)/4)); Q=incenter(B,C,(-99999,5sqrt(3)/4)); W=P+(0,5sqrt(3)/4); X=P-(0,5sqrt(3)/4); Y=Q+(0,5sqrt(3)/4); Z=Q-(0,5sqrt(3)/4); M=reflect(A,P)*W; N=reflect(B,Q)*Y; draw(A--B--C--D--cycle); draw(A--P--D); draw(B--Q--C); label("$A$",A,dir(135)); label("$B$",B,dir(45)); label("$C$",C,dir(315)); label("$D$",D,dir(225)); dot("$P$",P,dir(0)); dot("$Q$",Q,dir(180)); draw(W--X); draw(Y--Z); draw(M--P); draw(N--Q); label("$11$",midpoint(A--B),dir(90)); label("$5$",midpoint(B--C),dir(45)); label("$19$",midpoint(C--D),dir(270)); label("$7$",midpoint(D--A),dir(135)); label("$x$",midpoint(P--W),dir(0)); label("$x$",midpoint(P--X),dir(0)); label("$x$",midpoint(P--M),dir(225)); label("$x$",midpoint(Q--Y),dir(180)); label("$x$",midpoint(Q--Z),dir(180)); label("$x$",midpoint(Q--N),dir(315)); draw(rightanglemark(P,W,B,12.5)); draw(rightanglemark(P,X,C,12.5)); draw(rightanglemark(P,M,D,12.5)); draw(rightanglemark(Q,Y,A,12.5)); draw(rightanglemark(Q,Z,D,12.5)); draw(rightanglemark(Q,N,C,12.5));
// Block 4
unitsize(0.6cm); import olympiad; pair A,B,C,D,R,S; A=(11/2,5sqrt(3)/2); B=(33/2,5sqrt(3)/2); C=(19,0); D=(0,0); R=(11/2,0); S=(33/2,0); draw(A--B--C--D--cycle); draw(A--R); draw(B--S); label("$A$",A,dir(135)); label("$B$",B,dir(45)); label("$C$",C,dir(315)); label("$D$",D,dir(225)); label("$R$",R,dir(270)); label("$S$",S,dir(270)); label("$11$",midpoint(A--B),dir(90)); label("$5$",midpoint(B--C),dir(45)); label("$11$",midpoint(R--S),dir(270)); label("$7$",midpoint(D--A),dir(135)); label("$r$",midpoint(R--D),dir(270)); label("$s$",midpoint(C--S),dir(270)); label("$19$",midpoint(C--D),5*dir(270)); label("$2x$",midpoint(A--R),dir(0)); label("$2x$",midpoint(B--S),dir(180)); draw(rightanglemark(A,R,D,15)); draw(rightanglemark(B,S,C,15)); | [] |
676 | Let $ABCD$ be a trapezoid with $AB||CD, AB=11, BC=5, CD=19,$ and $DA=7$. Bisectors of $\angle A$ and $\angle D$ meet at $P$, and bisectors of $\angle B$ and $\angle C$ meet at $Q$. What is the area of hexagon $ABQCDP$?
$\textbf{(A)}\ 28\sqrt{3}\qquad \textbf{(B)}\ 30\sqrt{3}\qquad \textbf{(C)}\ 32\sqrt{3}\qquad \textbf{(D)}\ 35\sqrt{3}\qquad \textbf{(E)}\ 36\sqrt{3}$ | 2008 AMC 12B Problem 25 | Let $W, X, Y, Z = \overline{AP} \cap \overline{CD}, \overline{BQ} \cap \overline{CD}, \overline{CQ} \cap \overline{AB}, \overline{DP} \cap \overline{AB}$ respectively. Since $\angle{DCQ} = \angle{BCQ}, \angle{CBQ} = \angle{ABQ}$ we have $\angle{QCB} + \angle{CBQ} = 90 \iff \overline{BX} \perp \overline{CY};$ similarly we get $\overline{AW} \perp \overline{DZ}.$ Thus, $\overline{BQ}$ is both an angle bisector and altitude of $\triangle{CBY}$ so $BC = BY.$ Using the same logic on $\triangle{BCX}$ gives $BC = BX \iff BYXC$ is a rhombus; similarly, $ADWZ$ is a rhombus. Then, $[ABQCDP] = [ABCD] - \frac{1}{4}\left([BYXC] + [ADWZ]\right) = 15h - \frac{1}{4}(7h + 12h) = 12h$ where $h$ is the height of trapezoid $ABCD.$ Finding $h$ is the same as finding the altitude to the side of length $8$ in a $5-7-8$ triangle, and using Heron's, the area of such a triangle is $\sqrt{10(5)(3)(2)} = 10 \sqrt{3} = 4h \iff h = \frac{5\sqrt{3}}{2}.$ Multiply to get our answer is $\boxed{\mathrm{B}}.$ | // Block 1
import olympiad;
unitsize(0.5cm);
pair A, B, C, D;
A = 5*(Cos(120), Sin(120));
B = A + (-11, 0);
C = origin + (-19, 0);
D = origin;
label("$A$", A, dir(30));
label("$B$", B, dir(150));
label("$C$", C, dir(150));
label("$D$", D, dir(30));
pair E, F, G, H;
E = bisectorpoint(B, A, D);
F = bisectorpoint(A, B, C);
G = bisectorpoint(B, C, D);
H = bisectorpoint(C, D, A);
pair P, Q;
P = extension(A, E, D, H);
Q = extension(B, F, C, G);
dot("$P$", P, dir(20));
dot("$Q$", Q, dir(150));
pair W, X, Y, Z;
W = extension(A, P, D, C);
X = extension(B, Q, C, D);
Y = extension(C, Q, A, B);
Z = extension(D, P, A, B);
label("$W$", W, dir(100));
label("$X$", X, dir(60));
label("$Y$", Y, dir(50));
label("$Z$", Z, dir(140));
draw(A--W--X--B--Y--C--D--Z--B--C--Y--A--D);
// Block 2
import olympiad; unitsize(0.5cm); pair A, B, C, D; A = 5*(Cos(120), Sin(120)); B = A + (-11, 0); C = origin + (-19, 0); D = origin; label("$A$", A, dir(30)); label("$B$", B, dir(150)); label("$C$", C, dir(150)); label("$D$", D, dir(30)); pair E, F, G, H; E = bisectorpoint(B, A, D); F = bisectorpoint(A, B, C); G = bisectorpoint(B, C, D); H = bisectorpoint(C, D, A); pair P, Q; P = extension(A, E, D, H); Q = extension(B, F, C, G); dot("$P$", P, dir(20)); dot("$Q$", Q, dir(150)); pair W, X, Y, Z; W = extension(A, P, D, C); X = extension(B, Q, C, D); Y = extension(C, Q, A, B); Z = extension(D, P, A, B); label("$W$", W, dir(100)); label("$X$", X, dir(60)); label("$Y$", Y, dir(50)); label("$Z$", Z, dir(140)); draw(A--W--X--B--Y--C--D--Z--B--C--Y--A--D); | [] |
677 | Triangle $ABC$ has a right angle at $B$. Point $D$ is the foot of the altitude from $B$, $AD=3$, and $DC=4$. What is the area of $\triangle ABC$?
$\mathrm{(A)}\ 4\sqrt3 \qquad \mathrm{(B)}\ 7\sqrt3 \qquad \mathrm{(C)}\ 21 \qquad \mathrm{(D)}\ 14\sqrt3 \qquad \mathrm{(E)}\ 42$ | 2009 AMC 10A Problem 10 | Draw the circumcircle $\omega$ of the $\Delta ABC$. Because $\Delta ABC$ is a right angle triangle, AC is the diameter of the circumcircle. And we know that AC is the diameter because of Perpendicular Chord Bisector Theorem. By applying Power of a Point Theorem, we can have $BD=DE$ and $AD\cdot CD=BD^2$ $\Rightarrow BD=\sqrt{3\times 4}=2\sqrt{3}$. Then we have $S_{[ABC]}=\frac{1}{2}(7)(2\sqrt{3})=\boxed{7\sqrt{3}}$
~Bran_Qin
~twosetisdabest (Just proving that AC is diameter of the circle)
Solution 4 (Fakesolve)/Answer Choices
Suppose that $BD=x$. Then, the area of the triangle is $\frac{7x}{2}$. We want to find what $x$ is. Now, we try each answer choice.
$A): 4 \sqrt{3}$. I am too lazy to go over this, but we immediately see that this is very improbably due to the area being $\frac{7x}{2}$. This does not work.
$B): 7 \sqrt{3}$. This is promising. This means that $x = 2\sqrt{3}$. Now, applying Pythagorean Theorem, we have the vertical sides is $\sqrt {21}$ and the horizontal side is $\sqrt {28}$. Multiplying these and dividing by $2$ indeed gives us $7 \sqrt {3}$ as desired. Therefore, the answer is $\boxed{7 \sqrt{3}}$
~Arcticturn
Solution 5
Let $\overline{BD} = x$, and $\angle ABD = \theta$. Since $m\angle ABC = 90, DBC = 90 - \theta$. $\tan{\theta} = \frac{3}{x}$, and $\tan{90-\theta} = \frac{4}{x}$. Using trig identities, we can simplify the second one to $\cot{\theta} = \frac{4}{x}$. Let $y = \tan{\theta}$. We get $xy = 3$, and using the fact that $\cot{a} = \frac{1}{\tan{a}}$, $\frac{x}{y} = 4$. Solving for x and y(using substitution), we get $y^2 = \frac{3}{4}$. We could substitute back in and solve, but there is an easier way. Squaring $xy = 3$, we get $x^2y^2 = 9$. Since we know $y^2 = \frac{3}{4}$, $x = \sqrt{12}$. Using the Pythagorean Theorem in the original triangle, we get the legs of the triangle to be $\sqrt{9+x^2}$ and $\sqrt{16+x^2}$. Putting in $x$, we get the equation for the area, which is $\frac{\sqrt{28} \cdot \sqrt{21}}{2} = \frac{14\sqrt{3}}{2} = \boxed{\textbf{(B) } 7\sqrt{3}}$
~idk12345678 | // Block 1
unitsize(6mm);
defaultpen(linewidth(.8pt)+fontsize(8pt));
dotfactor=4;
pair B=(0,0), C=(sqrt(28),0), A=(0,sqrt(21)), E=(6*sqrt(28)/7,8*sqrt(21)/7);
pair D=foot(B,A,C);
pair[] ps={B,C,A,D};
filldraw(Circle((sqrt(28)/2,sqrt(21)/2),sqrt(49)/2),white,black);
draw(A--B--C--cycle);
draw(B--D);
draw(E--D);
draw(rightanglemark(B,D,C));
dot(ps);
label("$A$",A,NW);
label("$B$",B,SW);
label("$E$",E,NE);
label("$C$",C,SE);
label("$D$",D,NE);
label("$3$",midpoint(A--D),NE);
label("$4$",midpoint(D--C),NE);
// Block 2
unitsize(6mm); defaultpen(linewidth(.8pt)+fontsize(8pt)); dotfactor=4; pair B=(0,0), C=(sqrt(28),0), A=(0,sqrt(21)), E=(6*sqrt(28)/7,8*sqrt(21)/7); pair D=foot(B,A,C); pair[] ps={B,C,A,D}; filldraw(Circle((sqrt(28)/2,sqrt(21)/2),sqrt(49)/2),white,black); draw(A--B--C--cycle); draw(B--D); draw(E--D); draw(rightanglemark(B,D,C)); dot(ps); label("$A$",A,NW); label("$B$",B,SW); label("$E$",E,NE); label("$C$",C,SE); label("$D$",D,NE); label("$3$",midpoint(A--D),NE); label("$4$",midpoint(D--C),NE); | [] |
678 | The figures $F_1$, $F_2$, $F_3$, and $F_4$ shown are the first in a sequence of figures. For $n\ge3$, $F_n$ is constructed from $F_{n - 1}$ by surrounding it with a square and placing one more diamond on each side of the new square than $F_{n - 1}$ had on each side of its outside square. For example, figure $F_3$ has $13$ diamonds. How many diamonds are there in figure $F_{20}$?
$\textbf{(A)}\ 401 \qquad \textbf{(B)}\ 485 \qquad \textbf{(C)}\ 585 \qquad \textbf{(D)}\ 626 \qquad \textbf{(E)}\ 761$ | 2009 AMC 10A Problem 15 | Color the diamond layers alternately blue and red, starting from the outside. You'll get the following pattern:
In the figure $F_n$, the blue diamonds form a $n\times n$ square, and the red diamonds form a $(n-1)\times(n-1)$ square.
Hence the total number of diamonds in $F_{20}$ is $20^2 + 19^2 = \boxed{761}$. | unitsize(3mm); defaultpen(linewidth(.8pt)+fontsize(8pt)); path d=(1/2,0)--(0,sqrt(3)/2)--(-1/2,0)--(0,-sqrt(3)/2)--cycle; marker mred=marker(scale(5)*d,red,Fill); marker mblue=marker(scale(5)*d,blue,Fill); path f1=(0,0); path f2=(-1,1)--(1,1)--(1,-1)--(-1,-1)--cycle; path f3=(-2,-2)--(-2,0)--(-2,2)--(0,2)--(2,2)--(2,0)--(2,-2)--(0,-2)--cycle; path f4=(-3,-3)--(-3,-1)--(-3,1)--(-3,3)--(-1,3)--(1,3)--(3,3)--(3,1)--(3,-1)--(3,-3)--(1,-3)--(-1,-3)--cycle; draw((-3,-3)--(3,3)); draw((-3,3)--(3,-3)); draw(f1,mred); draw(f2,mblue); draw(f3,mred); draw(f4,mblue); | [] |
679 | Rectangle $ABCD$ has $AB=4$ and $BC=3$. Segment $EF$ is constructed through $B$ so that $EF$ is perpendicular to $DB$, and $A$ and $C$ lie on $DE$ and $DF$, respectively. What is $EF$?
$\mathrm{(A)}\ 9 \qquad \mathrm{(B)}\ 10 \qquad \mathrm{(C)}\ \frac {125}{12} \qquad \mathrm{(D)}\ \frac {103}{9} \qquad \mathrm{(E)}\ 12$ | 2009 AMC 10A Problem 17 | Solution 1
The situation is shown in the picture below.
From the Pythagorean theorem we have $BD=5$.
Triangle $EAB$ is similar to $BAD$, as they have the same angles. Segment $BA$ is perpendicular to $DA$, meaning that angle $DAB$ and $BAE$ are right angles and congruent. Also, angle $DBE$ is a right angle. Because it is a rectangle, angle $BDC$ is congruent to $DBA$ and angle $ADC$ is also a right angle. By the transitive property:
$mADB + mBDC = mDBA + mABE$
$mBDC = mDBA$
$mADB + mBDC = mBDC + mABE$
$mADB = mABE$
Next, because every triangle has a degree measure of $180$, angle $BEA$ and angle $DBA$ are similar.
Hence $BE/AB = DB/AD$, and therefore $BE = AB\cdot DB/AD = 20/3$.
Also triangle $CBF$ is similar to $ABD$. Hence $BF/BC = DB/AB$, and therefore $BF=BC\cdot DB / AB = 15/4$.
We then have $EF = EB+BF = \frac{20}3 + \frac{15}4 = \frac{80 + 45}{12} = \boxed{\frac{125}{12}}$.
Solution 2
Since $BD$ is the altitude from $D$ to $EF$, we can use the equation $BD^2 = EB\cdot BF$.
Looking at the angles, we see that triangle $BDE$ is similar to $DCB$. Because of this, $\frac{AB}{CB} = \frac{EB}{DB}$. From the given information and the Pythagorean theorem, $AB=4$, $CB=3$, and $DB=5$. Solving gives $EB=20/3$.
We can use the above formula to solve for $BF$. $BD^2 = 20/3\cdot BF$. Solve to obtain $BF=15/4$.
We now know $EB$ and $BF$. $EF = EB+BF = \frac{20}3 + \frac{15}4 = \frac{80 + 45}{12} = \boxed{\frac{125}{12}}$.
Solution 3(Coordinate Bash)
To keep things simple, we will use coordinates in only the first quadrant. The picture will look like the diagram above reflected over the $x$-axis.It is also worth noting the $F$ will lie on the $x$ axis and $E$ on the $y$. Let $D$ be the origin, $A(3,0)$, $C(4,0)$, and $B(4,3)$. We can express segment $DB$ as the line $y=\frac{3x}{4}$.
Since $EF$ is perpendicular to $DB$, and we know that $(4,3)$ lies on it, we can use this information to find that segment $EF$
is on the line $y=\frac{-4x}{3}+\frac{25}{3}$. Since $E$ and $F$ are on the $y$ and $x$ axis, respectively, we plug in $0$ for $x$
and $y$, we find that point $E$ is at $(0,\frac{25}{3})$, and point $F$ is at $(\frac{25}{4},0)$. Applying the distance formula,
we obtain that $EF$= $\boxed{\frac{125}{12}}$. | unitsize(0.6cm); defaultpen(0.8); pair A=(0,0), B=(4,0), C=(4,3), D=(0,3); pair EF=rotate(90)*(D-B); pair E=intersectionpoint( (0,-100)--(0,100), (B-100*EF)--(B+100*EF) ); pair F=intersectionpoint( (-100,3)--(100,3), (B-100*EF)--(B+100*EF) ); draw(A--B--C--D--cycle); draw(B--D, dashed); draw(E--F); draw(A--E, dashed); draw(C--F, dashed); label("$A$",A,W); label("$B$",B,SE); label("$C$",C,N); label("$D$",D,NW); label("$E$",E,SW); label("$F$",F,NE); label("$3$",A--D,W); label("$4$",C--D,N); | [] |
680 | Many Gothic cathedrals have windows with portions containing a ring of congruent circles that are circumscribed by a larger circle, In the figure shown, the number of smaller circles is four. What is the ratio of the sum of the areas of the four smaller circles to the area of the larger circle?
$\mathrm{(A)}\ 3-2\sqrt2 \qquad \mathrm{(B)}\ 2-\sqrt2 \qquad \mathrm{(C)}\ 4(3-2\sqrt2) \qquad \mathrm{(D)}\ \frac12(3-\sqrt2) \qquad \mathrm{(E)}\ 2\sqrt2-2$ | 2009 AMC 10A Problem 21 | Draw some of the radii of the small circles as in the picture below.
Out of symmetry, the quadrilateral in the center must be a square. Its side is $2r$, and therefore its diagonal is $2r\sqrt{2}$. We can now compute the length of the vertical diameter of the large circle as $2r + 2r\sqrt{2}$. Hence $2R=2r + 2r\sqrt{2}$, and thus $R=r+r\sqrt{2}=r(1+\sqrt{2})$.
Then the area of the large circle is $L = \pi R^2 = \pi r^2 (1+\sqrt 2)^2 = \pi r^2 (3+2\sqrt 2)$.
The area of four small circles is $S = 4\pi r^2$. Hence their ratio is:
\begin{align*} \frac SL & = \frac{4\pi r^2}{\pi r^2 (3+2\sqrt 2)} \\ & = \frac 4{3+2\sqrt 2} \\ & = \frac 4{3+2\sqrt 2} \cdot \frac{3-2\sqrt 2}{3 - 2\sqrt 2} \\ & = \frac{4(3 - 2\sqrt 2)}{3^2 - (2\sqrt 2)^2} \\ & = \frac{4(3 - 2\sqrt 2)}1 \\ & = \boxed{4(3 - 2\sqrt 2)} \end{align*} | // Block 1
unitsize(12mm);
defaultpen(linewidth(.8pt));
draw(Circle((0,0),1+sqrt(2)));
draw(Circle((sqrt(2),0),1));
draw(Circle((0,sqrt(2)),1));
draw(Circle((-sqrt(2),0),1));
draw(Circle((0,-sqrt(2)),1));
draw( (sqrt(2),0) -- (0,sqrt(2)) -- (-sqrt(2),0) -- (0,-sqrt(2)) -- cycle );
draw( (0,sqrt(2)) -- (0,1+sqrt(2)) );
draw( (0,-sqrt(2)) -- (0,-1-sqrt(2)) );
draw( (0,sqrt(2)) -- (0,-sqrt(2)), dashed );
// Block 2
unitsize(12mm); defaultpen(linewidth(.8pt)); draw(Circle((0,0),1+sqrt(2))); draw(Circle((sqrt(2),0),1)); draw(Circle((0,sqrt(2)),1)); draw(Circle((-sqrt(2),0),1)); draw(Circle((0,-sqrt(2)),1)); draw( (sqrt(2),0) -- (0,sqrt(2)) -- (-sqrt(2),0) -- (0,-sqrt(2)) -- cycle ); draw( (0,sqrt(2)) -- (0,1+sqrt(2)) ); draw( (0,-sqrt(2)) -- (0,-1-sqrt(2)) ); draw( (0,sqrt(2)) -- (0,-sqrt(2)), dashed ); | [] |
681 | Convex quadrilateral $ABCD$ has $AB = 9$ and $CD = 12$. Diagonals $AC$ and $BD$ intersect at $E$, $AC = 14$, and $\triangle AED$ and $\triangle BEC$ have equal areas. What is $AE$?
$\textbf{(A)}\ \frac {9}{2}\qquad \textbf{(B)}\ \frac {50}{11}\qquad \textbf{(C)}\ \frac {21}{4}\qquad \textbf{(D)}\ \frac {17}{3}\qquad \textbf{(E)}\ 6$ | 2009 AMC 10A Problem 23 | Let $[ABC]$ denote the area of triangle $ABC$. $[AED] = [BEC]$, so $[ABD] = [AED] + [AEB] = [BEC] + [AEB] = [ABC]$. Since triangles $ABD$ and $ABC$ share a base, they also have the same height and thus $\overline{AB}||\overline{CD}$ and $\triangle{AEB}\sim\triangle{CED}$ with a ratio of $3: 4$. $AE = \frac {3}{7}\times AC$, so $AE = \frac {3}{7}\times 14 = 6\ \boxed{\textbf{(E)}}$. | pathpen = linewidth(0.7);pointpen = black; pair D=MP("D",(0,0)),C=MP("C",(12,0)),A=MP("A",C+14*expi(145*pi/180),N),B=MP("B",A+(9,0),N),E=IP(A--C,B--D);MP("9",(A+B)/2,N);MP("12",(C+D)/2); fill(A--D--E--cycle,rgb(0.8,0.8,0.8));fill(B--C--E--cycle,rgb(0.8,0.8,0.8));D(A--B--C--D--cycle);D(A--C);D(B--D);D(E); | [] |
682 | The figures $F_1$, $F_2$, $F_3$, and $F_4$ shown are the first in a sequence of figures. For $n\ge3$, $F_n$ is constructed from $F_{n - 1}$ by surrounding it with a square and placing one more diamond on each side of the new square than $F_{n - 1}$ had on each side of its outside square. For example, figure $F_3$ has $13$ diamonds. How many diamonds are there in figure $F_{20}$?
$\textbf{(A)}\ 401 \qquad \textbf{(B)}\ 485 \qquad \textbf{(C)}\ 585 \qquad \textbf{(D)}\ 626 \qquad \textbf{(E)}\ 761$ | 2009 AMC 12A Problem 11 | Color the diamond layers alternately blue and red, starting from the outside. You'll get the following pattern:
In the figure $F_n$, the blue diamonds form a $n\times n$ square, and the red diamonds form a $(n-1)\times(n-1)$ square.
Hence the total number of diamonds in $F_{20}$ is $20^2 + 19^2 = \boxed{761}$. | // Block 1
unitsize(3mm);
defaultpen(linewidth(.8pt)+fontsize(8pt));
path d=(1/2,0)--(0,sqrt(3)/2)--(-1/2,0)--(0,-sqrt(3)/2)--cycle;
marker mred=marker(scale(5)*d,red,Fill);
marker mblue=marker(scale(5)*d,blue,Fill);
path f1=(0,0);
path f2=(-1,1)--(1,1)--(1,-1)--(-1,-1)--cycle;
path f3=(-2,-2)--(-2,0)--(-2,2)--(0,2)--(2,2)--(2,0)--(2,-2)--(0,-2)--cycle;
path f4=(-3,-3)--(-3,-1)--(-3,1)--(-3,3)--(-1,3)--(1,3)--(3,3)--(3,1)--(3,-1)--(3,-3)--(1,-3)--(-1,-3)--cycle;
draw((-3,-3)--(3,3));
draw((-3,3)--(3,-3));
draw(f1,mred);
draw(f2,mblue);
draw(f3,mred);
draw(f4,mblue);
// Block 2
unitsize(3mm); defaultpen(linewidth(.8pt)+fontsize(8pt)); path d=(1/2,0)--(0,sqrt(3)/2)--(-1/2,0)--(0,-sqrt(3)/2)--cycle; marker mred=marker(scale(5)*d,red,Fill); marker mblue=marker(scale(5)*d,blue,Fill); path f1=(0,0); path f2=(-1,1)--(1,1)--(1,-1)--(-1,-1)--cycle; path f3=(-2,-2)--(-2,0)--(-2,2)--(0,2)--(2,2)--(2,0)--(2,-2)--(0,-2)--cycle; path f4=(-3,-3)--(-3,-1)--(-3,1)--(-3,3)--(-1,3)--(1,3)--(3,3)--(3,1)--(3,-1)--(3,-3)--(1,-3)--(-1,-3)--cycle; draw((-3,-3)--(3,3)); draw((-3,3)--(3,-3)); draw(f1,mred); draw(f2,mblue); draw(f3,mred); draw(f4,mblue); | [] |
683 | A ship sails $10$ miles in a straight line from $A$ to $B$, turns through an angle between $45^{\circ}$ and $60^{\circ}$, and then sails another $20$ miles to $C$. Let $AC$ be measured in miles. Which of the following intervals contains $AC^2$?
$\textbf{(A)}\ [400,500] \qquad \textbf{(B)}\ [500,600] \qquad \textbf{(C)}\ [600,700] \qquad \textbf{(D)}\ [700,800]$
$\textbf{(E)}\ [800,900]$ | 2009 AMC 12A Problem 13 | Answering the question
To answer the question we are asked, it is enough to compute $AC^2$ for two different angles, preferably for both extremes ($45$ and $60$ degrees). You can use the law of cosines to do so.
Alternately, it is enough to compute $AC^2$ for one of the extreme angles. In case it falls inside one of the given intervals, we are done. In case it falls on the boundary between two options, we also have to argue whether our $AC^2$ is the minimal or the maximal possible value of $AC^2$.
Below we show a complete solution in which we also show that all possible values of $AC^2$ do indeed lie in the given interval.
Complete solution
Let $C_1$ be the point the ship would reach if it turned $45^\circ$, and $C_2$ the point it would reach if it turned $60^\circ$. Obviously, $C_1$ is the furthest possible point from $A$, and $C_2$ is the closest possible point to $A$.
Hence the interval of possible values for $AC^2$ is $[AC_2^2,AC_1^2]$.
We can find $AC_1^2$ and $AC_2^2$ as follows:
Let $D_1$ and $D_2$ be the feet of the heights from $C_1$ and $C_2$ onto $AB$. The angles in the triangle $BD_1C_1$ are $45^\circ$, $45^\circ$, and $90^\circ$, hence $BD_1 = D_1C_1 = BC_1 / \sqrt 2$. Similarly, the angles in the triangle $BD_2C_2$ are $30^\circ$, $60^\circ$, and $90^\circ$, hence $BD_2 = BC_2 / 2$ and $D_2C_2 = BC_2 \sqrt 3 / 2$.
Hence we get:
\[AC_2^2 = AD_2^2 + D_2C_2^2 = 20^2 + (10\sqrt 3)^2 = 400 + 300 = 700\]
\[AC_1^2 = AD_1^2 + D_1C_1^2 = (10 + 20/\sqrt 2)^2 + (20\sqrt 2)^2 = 100 + 400/\sqrt 2 + 200 + 200 = 500 + 200\sqrt 2 < 500 + 200\cdot 1.5 = 800\]
Therefore for any valid $C$ the value $AC^2$ is surely in the interval $\boxed{ \textbf{(D)}[700,800] }$.
Alternate Solution
From the law of cosines, $500-400\cos120^\circ<AC^2<500-400\cos135^\circ\implies700<AC^2<500+200\sqrt{2}$. This is essentially the same solution as above. The answer is $\boxed{\textbf{(D)}}$. | // Block 1
unitsize(3mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; pair B=(0,0), A=(-10,0), C1=20*dir(45), C2=20*dir(60); draw(A--B--C1); draw(A--C1,linetype("4 4")); draw(A--B--C2); draw(A--C2,linetype("4 4")); draw( arc(A, length(C1-A), 0, 55 ), dotted ); draw( arc(A, length(C2-A), 0, 55 ), dotted ); draw( arc(B, C1, C2) ); dot(A); dot(B); dot(C1); dot(C2); label("$10$",midpoint(A--B),S); label("$20$",midpoint(B--C1),SE); label("$A$",A,SW); label("$B$",B,SE); label("$C_1$",C1,NE); label("$C_2$",C2,N);
// Block 2
unitsize(3mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; pair B=(0,0), A=(-10,0), C1=20*dir(45), C2=20*dir(60); draw(A--B--C1); draw(A--C1,linetype("4 4")); dot(A); dot(B); dot(C1); pair D1 = (C1.x,0); dot(D1); draw(B--D1--C1); label("$10$",midpoint(A--B),S); label("$20$",midpoint(B--C1),SE); label("$20/\sqrt 2$",midpoint(B--D1),S); label("$20/\sqrt 2$",midpoint(D1--C1),E); label("$A$",A,SW); label("$B$",B,SE); label("$C_1$",C1,NE); label("$D_1$",D1,S);
// Block 3
unitsize(3mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; pair B=(0,0), A=(-10,0), C1=20*dir(45), C2=20*dir(60); draw(A--B--C2); draw(A--C2,linetype("4 4")); dot(A); dot(B); dot(C2); pair D2 = (C2.x,0); dot(D2); draw(B--D2--C2); label("$10$",midpoint(A--B),S); label("$20$",midpoint(B--C2),SE); label("$10$",midpoint(B--D2),S); label("$10\sqrt 3$",midpoint(D2--C2),E); label("$A$",A,SW); label("$B$",B,SE); label("$C_2$",C2,NE); label("$D_2$",D2,S); | [] |
684 | A triangle has vertices $(0,0)$, $(1,1)$, and $(6m,0)$, and the line $y = mx$ divides the triangle into two triangles of equal area. What is the sum of all possible values of $m$?
$\textbf{A} - \!\frac {1}{3} \qquad \textbf{(B)} - \!\frac {1}{6} \qquad \textbf{(C)}\ \frac {1}{6} \qquad \textbf{(D)}\ \frac {1}{3} \qquad \textbf{(E)}\ \frac {1}{2}$ | 2009 AMC 12A Problem 14 | Let's label the three points as $A=(0,0)$, $B=(1,1)$, and $C=(6m,0)$.
Clearly, whenever the line $y=mx$ intersects the inside of the triangle, it will intersect the side $BC$. Let $D$ be the point of intersection.
The triangles $ABD$ and $ACD$ have the same height, which is the distance between the point $A$ and the line $BC$.
Hence they have equal areas if and only if $D$ is the midpoint of $BC$.
The midpoint of the segment $BC$ has coordinates $\left( \frac{6m+1}2, \frac 12 \right)$. This point lies on the line $y=mx$ if and only if $\frac 12 = m \cdot \frac{6m+1}2$. This simplifies to $6m^2 + m - 1 = 0$. Using Vieta's formulas, we find that the sum of the roots is $\boxed{\textbf{(B)} - \!\frac {1}{6}}$.
For illustration, below are pictures of the situation for $m=1.5$, $m=0.5$, $m=1/3$, and $m=-1/2$.
Note : In contests when you get problems like these, your AOPS knowledge will not be enough. AOPs books are just the basic, always prep for more. | // Block 1
unitsize(1cm);
defaultpen(0.8);
real m=1.5;
draw( (0,0)--(1,1)--(6*m,0)--cycle );
draw( ((-0.5)*(1,m)) -- ((1.5)*(1,m)), dashed );
label("$A$",(0,0),NW);
label("$B$",(1,1),NE);
label("$C$",(6*m,0),E);
// Block 2
unitsize(2cm);
defaultpen(0.8);
real m=0.5;
draw( (0,0)--(1,1)--(6*m,0)--cycle );
draw( ((-1)*(1,m)) -- (3*(1,m)), dashed );
label("$A$",(0,0),NW);
label("$B$",(1,1),N);
label("$C$",(6*m,0),E);
pair D = intersectionpoint( (1,1)--(6*m,0), ((-1)*(1,m)) -- (3*(1,m)) );
dot(D); label("$D$",D,S);
// Block 3
unitsize(2cm);
defaultpen(0.8);
real m=1/3;
draw( (0,0)--(1,1)--(6*m,0)--cycle );
draw( ((-1)*(1,m)) -- (3*(1,m)), dashed );
label("$A$",(0,0),NW);
label("$B$",(1,1),N);
label("$C$",(6*m,0),E);
pair D = intersectionpoint( (1,1)--(6*m,0), ((-1)*(1,m)) -- (3*(1,m)) );
dot(D); label("$D$",D,S);
// Block 4
unitsize(2cm);
defaultpen(0.8);
real m=-1/2;
draw( (0,0)--(1,1)--(6*m,0)--cycle );
draw( ((-2)*(1,m)) -- (1*(1,m)), dashed );
label("$A$",(0,0),S);
label("$B$",(1,1),NE);
label("$C$",(6*m,0),W);
pair D = intersectionpoint( (1,1)--(6*m,0), ((-1)*(1,m)) -- (3*(1,m)) );
dot(D); label("$D$",D,S);
// Block 5
unitsize(1cm); defaultpen(0.8); real m=1.5; draw( (0,0)--(1,1)--(6*m,0)--cycle ); draw( ((-0.5)*(1,m)) -- ((1.5)*(1,m)), dashed ); label("$A$",(0,0),NW); label("$B$",(1,1),NE); label("$C$",(6*m,0),E);
// Block 6
unitsize(2cm); defaultpen(0.8); real m=0.5; draw( (0,0)--(1,1)--(6*m,0)--cycle ); draw( ((-1)*(1,m)) -- (3*(1,m)), dashed ); label("$A$",(0,0),NW); label("$B$",(1,1),N); label("$C$",(6*m,0),E); pair D = intersectionpoint( (1,1)--(6*m,0), ((-1)*(1,m)) -- (3*(1,m)) ); dot(D); label("$D$",D,S);
// Block 7
unitsize(2cm); defaultpen(0.8); real m=1/3; draw( (0,0)--(1,1)--(6*m,0)--cycle ); draw( ((-1)*(1,m)) -- (3*(1,m)), dashed ); label("$A$",(0,0),NW); label("$B$",(1,1),N); label("$C$",(6*m,0),E); pair D = intersectionpoint( (1,1)--(6*m,0), ((-1)*(1,m)) -- (3*(1,m)) ); dot(D); label("$D$",D,S);
// Block 8
unitsize(2cm); defaultpen(0.8); real m=-1/2; draw( (0,0)--(1,1)--(6*m,0)--cycle ); draw( ((-2)*(1,m)) -- (1*(1,m)), dashed ); label("$A$",(0,0),S); label("$B$",(1,1),NE); label("$C$",(6*m,0),W); pair D = intersectionpoint( (1,1)--(6*m,0), ((-1)*(1,m)) -- (3*(1,m)) ); dot(D); label("$D$",D,S); | [] |
685 | A circle with center $C$ is tangent to the positive $x$ and $y$-axes and externally tangent to the circle centered at $(3,0)$ with radius $1$. What is the sum of all possible radii of the circle with center $C$?
$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 9$ | 2009 AMC 12A Problem 16 | Let $r$ be the radius of our circle. For it to be tangent to the positive $x$ and $y$ axes, we must have $C=(r,r)$. For the circle to be externally tangent to the circle centered at $(3,0)$ with radius $1$, the distance between $C$ and $(3,0)$ must be exactly $r+1$.
By the Pythagorean theorem the distance between $(r,r)$ and $(3,0)$ is $\sqrt{ (r-3)^2 + r^2 }$, hence we get the equation $(r-3)^2 + r^2 = (r+1)^2$.
Simplifying, we obtain $r^2 - 8r + 8 = 0$. By Vieta's formulas the sum of the two roots of this equation is $\boxed{8}$.
(We should actually solve for $r$ to verify that there are two distinct positive roots. In this case we get $r=4\pm 2\sqrt 2$. This is generally a good rule of thumb, but is not necessary as all of the available answers are integers, and the equation obviously doesn't factor as integers. You can also tell that there are two positive roots based on the visual interpretation.) | // Block 1
unitsize(0.5cm);
defaultpen(0.8);
filldraw( Circle( (3,0), 1 ), lightgray, black );
draw( (0,0) -- (15,0), Arrow );
draw( (0,0) -- (0,15), Arrow );
draw( (0,0) -- (15,15), dashed );
real r1 = 4 - 2*sqrt(2), r2 = 4 + 2*sqrt(2);
pair S1=(r1,r1), S2=(r2,r2);
dot(S1); dot(S2); dot((3,0));
draw( Circle(S1,r1) );
draw( Circle(S2,r2) );
// Block 2
unitsize(0.5cm); defaultpen(0.8); filldraw( Circle( (3,0), 1 ), lightgray, black ); draw( (0,0) -- (15,0), Arrow ); draw( (0,0) -- (0,15), Arrow ); draw( (0,0) -- (15,15), dashed ); real r1 = 4 - 2*sqrt(2), r2 = 4 + 2*sqrt(2); pair S1=(r1,r1), S2=(r2,r2); dot(S1); dot(S2); dot((3,0)); draw( Circle(S1,r1) ); draw( Circle(S2,r2) ); | [] |
686 | Convex quadrilateral $ABCD$ has $AB = 9$ and $CD = 12$. Diagonals $AC$ and $BD$ intersect at $E$, $AC = 14$, and $\triangle AED$ and $\triangle BEC$ have equal areas. What is $AE$?
$\textbf{(A)}\ \frac {9}{2}\qquad \textbf{(B)}\ \frac {50}{11}\qquad \textbf{(C)}\ \frac {21}{4}\qquad \textbf{(D)}\ \frac {17}{3}\qquad \textbf{(E)}\ 6$ | 2009 AMC 12A Problem 20 | Let $[ABC]$ denote the area of triangle $ABC$. $[AED] = [BEC]$, so $[ABD] = [AED] + [AEB] = [BEC] + [AEB] = [ABC]$. Since triangles $ABD$ and $ABC$ share a base, they also have the same height and thus $\overline{AB}||\overline{CD}$ and $\triangle{AEB}\sim\triangle{CED}$ with a ratio of $3: 4$. $AE = \frac {3}{7}\times AC$, so $AE = \frac {3}{7}\times 14 = 6\ \boxed{\textbf{(E)}}$. | // Block 1
pathpen = linewidth(0.7);pointpen = black;
pair D=MP("D",(0,0)),C=MP("C",(12,0)),A=MP("A",C+14*expi(145*pi/180),N),B=MP("B",A+(9,0),N),E=IP(A--C,B--D);MP("9",(A+B)/2,N);MP("12",(C+D)/2);
fill(A--D--E--cycle,rgb(0.8,0.8,0.8));fill(B--C--E--cycle,rgb(0.8,0.8,0.8));D(A--B--C--D--cycle);D(A--C);D(B--D);D(E);
// Block 2
pathpen = linewidth(0.7);pointpen = black; pair D=MP("D",(0,0)),C=MP("C",(12,0)),A=MP("A",C+14*expi(145*pi/180),N),B=MP("B",A+(9,0),N),E=IP(A--C,B--D);MP("9",(A+B)/2,N);MP("12",(C+D)/2); fill(A--D--E--cycle,rgb(0.8,0.8,0.8));fill(B--C--E--cycle,rgb(0.8,0.8,0.8));D(A--B--C--D--cycle);D(A--C);D(B--D);D(E); | [] |
687 | Functions $f$ and $g$ are quadratic, $g(x) = - f(100 - x)$, and the graph of $g$ contains the vertex of the graph of $f$. The four $x$-intercepts on the two graphs have $x$-coordinates $x_1$, $x_2$, $x_3$, and $x_4$, in increasing order, and $x_3 - x_2 = 150$. Then $x_4 - x_1 = m + n\sqrt p$, where $m$, $n$, and $p$ are positive integers, and $p$ is not divisible by the square of any prime. What is $m + n + p$?
$\textbf{(A)}\ 602\qquad \textbf{(B)}\ 652\qquad \textbf{(C)}\ 702\qquad \textbf{(D)}\ 752 \qquad \textbf{(E)}\ 802$ | 2009 AMC 12A Problem 23 | The two quadratics are $180^{\circ}$ rotations of each other about $(50,0)$. Since we are only dealing with differences of roots, we can translate them to be symmetric about $(0,0)$. Now $x_3 = - x_2 = 75$ and $x_4 = - x_1$. Say our translated versions of $f$ and $g$ are $p$ and $q$, respectively, so that $p(x) = - q( - x)$. Let $x_3 = 75$ be a root of $p$ and $x_2 = - 75$ a root of $q$ by symmetry. Note that since they each contain each other's vertex, $x_1$, $x_2$, $x_3$, and $x_4$ must be roots of alternating polynomials, so $x_1$ is a root of $p$ and $x_4$ a root of $q$
\[p(x) = a(x - 75)(x - x_1)\]
\[q(x) = - a(x + 75)(x + x_1)\]
The vertex of $p(x)$ is half the sum of its roots, or $\frac {75 + x_1}{2}$. We are told that the vertex of one quadratic lies on the other, so
\begin{eqnarray*} p\left(\frac {75 + x_1}{2}\right) & = & a\left(\frac {75 - x_1}{2}\right)\left(\frac { - 75 + x_1}{2}\right) \\ & = & - \frac {a}{4}(x_1 - 75)^2 \\ - \frac {a}{4}(x_1 - 75)^2 & = & q\left(\frac {75 + x_1}{2}\right) \\ & = & - a\left(\frac {x_1 + 225}{2}\right)\left(\frac {3x_1 + 75}{2}\right) \\ & = & - \frac {a}{4}(x_1 + 225)(3x_1 + 75) \end{eqnarray*}
Let $x_1 = 75u$ and divide through by $75^2$, since it will drastically simplify computations. We know $u < - 1$ and that $(u - 1)^2 = (3u + 1)(u + 3)$, or
\begin{eqnarray*} 0 & = & (3u + 1)(u + 3) - (u - 1)^2 \\ & = & 3u^2 + 10u + 3 - (u^2 - 2u + 1) \\ & = & 2u^2 + 12u + 2 \\ & = & u^2 + 6u + 1 \end{eqnarray*}
So $u = \frac { - 6\pm\sqrt {32}}{2} = - 3\pm2\sqrt2$. Since $u < - 1$, $u = - 3 - 2\sqrt2$.
The answer is $x_4 - x_1 = (-x_1) - x_1 = - 150u = 450 + 300\sqrt {2}$, and $450 + 300 + 2 = 752\ \mathbf{(D)}$. | // Block 1
import graph; size(250);Label k; k.p=fontsize(6); int ymax = 400, ymin = -400; real rt = 175+150*2^.5;
real f(real x){return 1/400*(x-125)*(x+rt);} real g(real x){return -f(100-x);}
xaxis(-600,600,Ticks(k, 5),Arrows(6));yaxis(ymin,ymax,Ticks(k, 5),Arrows(6));
draw(graph(f,-450,300),red+linewidth(0.8),Arrows(6));draw(graph(g,-200,550),blue+linewidth(0.8),Arrows(6));
draw((50,ymax)--(50,ymin),linetype("4 4"),Arrows(4));dot((-rt,0));dot((100+rt,0));dot((-25,0));dot((125,0));
// Block 2
import graph; size(250);Label k; k.p=fontsize(6); int ymax = 400, ymin = -400; real rt = 175+150*2^.5; real f(real x){return 1/400*(x-125)*(x+rt);} real g(real x){return -f(100-x);} xaxis(-600,600,Ticks(k, 5),Arrows(6));yaxis(ymin,ymax,Ticks(k, 5),Arrows(6)); draw(graph(f,-450,300),red+linewidth(0.8),Arrows(6));draw(graph(g,-200,550),blue+linewidth(0.8),Arrows(6)); draw((50,ymax)--(50,ymin),linetype("4 4"),Arrows(4));dot((-rt,0));dot((100+rt,0));dot((-25,0));dot((125,0)); | [] |
688 | Points $A$ and $C$ lie on a circle centered at $O$, each of $\overline{BA}$ and $\overline{BC}$ are tangent to the circle, and $\triangle ABC$ is equilateral. The circle intersects $\overline{BO}$ at $D$. What is $\frac{BD}{BO}$?
$\text{(A) } \frac {\sqrt2}{3} \qquad \text{(B) } \frac {1}{2} \qquad \text{(C) } \frac {\sqrt3}{3} \qquad \text{(D) } \frac {\sqrt2}{2} \qquad \text{(E) } \frac {\sqrt3}{2}$ | 2009 AMC 10B Problem 16 | Solution 1
As $\triangle ABC$ is equilateral, we have $\angle BAC = \angle BCA = 60^\circ$, hence $\angle OAC = \angle OCA = 30^\circ$. Then $\angle AOC = 120^\circ$, and from symmetry we have $\angle AOB = \angle COB = 60^\circ$. Thus, this gives us $\angle ABO = \angle CBO = 30^\circ$.
We know that $DO = AO$, as $D$ lies on the circle. From $\triangle ABO$ we also have $AO = BO \sin 30^\circ = \frac{BO}2$, Hence $DO = \frac{BO}2$, therefore $BD = BO - DO = \frac{BO}2$, and $\frac{BD}{BO} = \boxed{\frac 12 \Longrightarrow B}$.
Solution 2
As in the previous solution, we find out that $\angle AOB = \angle COB = 60^\circ$. Hence $\triangle AOD$ and $\triangle COD$ are both equilateral.
We then have $\angle SCD = \angle SAD = 30^\circ$, hence $D$ is the incenter of $\triangle ABC$, and as $\triangle ABC$ is equilateral, $D$ is also its centroid. Hence $2 \cdot SD = BD$, and as $SD = SO$, we have $2\cdot SD = SD + SO = OD$, therefore $BD=OD$, and as before we conclude that $\frac{BD}{BO} = \boxed{\frac 12}$. | // Block 1
unitsize(1.5cm); defaultpen(0.8); pair B=(0,0), A=(3,0), C=3*dir(60), O=intersectionpoint( C -- (C+3*dir(-30)), A -- (A+3*dir(90)) ); pair D=intersectionpoint(B--O, circle(O,length(A-O))); draw(circle(O,length(A-O))); draw(A--B--C--O--A); draw(B--O); draw(rightanglemark(B,A,O)); draw(rightanglemark(B,C,O)); draw(A--C, dashed ); label("$B$",B,SW); label("$A$",A,S); label("$C$",C,NW); label("$O$",O,NE); label("$D$",D,(S+SSW));
// Block 2
unitsize(1.5cm); defaultpen(0.8); pair B=(0,0), A=(3,0), C=3*dir(60), O=intersectionpoint( C -- (C+3*dir(-30)), A -- (A+3*dir(90)) ); pair D=intersectionpoint(B--O, circle(O,length(A-O))); draw(circle(O,length(A-O))); draw(A--B--C--O--A); draw(B--O); draw(rightanglemark(B,A,O)); draw(rightanglemark(B,C,O)); draw(A--C--D--A, dashed ); pair Sp = intersectionpoint(D--O,A--C); label("$B$",B,SW); label("$A$",A,S); label("$C$",C,NW); label("$O$",O,NE); label("$D$",D,(S+SSW)); label("$S$",Sp,(S+SSW)); | [] |
689 | Rectangle $ABCD$ has $AB=8$ and $BC=6$. Point $M$ is the midpoint of diagonal $\overline{AC}$, and $E$ is on $AB$ with $\overline{ME}\perp\overline{AC}$. What is the area of $\triangle AME$?
$\text{(A) } \frac{65}{8} \qquad \text{(B) } \frac{25}{3} \qquad \text{(C) } 9 \qquad \text{(D) } \frac{75}{8} \qquad \text{(E) } \frac{85}{8}$ | 2009 AMC 10B Problem 18 | By the Pythagorean theorem we have $AC=10$, hence $AM=5$.
The triangles $AME$ and $ABC$ have the same angle at $A$ and a right angle, thus all their angles are equal, and therefore these two triangles are similar.
The ratio of their sides is $\frac{AM}{AB} = \frac 58$, hence the ratio of their areas is $\left( \frac 58 \right)^2 = \frac{25}{64}$.
And as the area of triangle $ABC$ is $\frac{6\cdot 8}2 = 24$, the area of triangle $AME$ is $24\cdot \frac{25}{64} = \boxed{ \frac{75}8 }$. | // Block 1
unitsize(0.75cm);
defaultpen(0.8);
pair A=(0,0), B=(8,0), C=(8,6), D=(0,6), M=(A+C)/2;
path ortho = shift(M)*rotate(-90)*(A--C);
pair Ep = intersectionpoint(ortho, A--B);
draw( A--B--C--D--cycle );
draw( A--C );
draw( M--Ep );
filldraw( A--M--Ep--cycle, lightgray, black );
draw( rightanglemark(A,M,Ep) );
draw( C--Ep );
label("$A$",A,SW);
label("$B$",B,SE);
label("$C$",C,NE);
label("$D$",D,NW);
label("$E$",Ep,S);
label("$M$",M,NW);
// Block 2
unitsize(0.75cm); defaultpen(0.8); pair A=(0,0), B=(8,0), C=(8,6), D=(0,6), M=(A+C)/2; path ortho = shift(M)*rotate(-90)*(A--C); pair Ep = intersectionpoint(ortho, A--B); draw( A--B--C--D--cycle ); draw( A--C ); draw( M--Ep ); filldraw( A--M--Ep--cycle, lightgray, black ); draw( rightanglemark(A,M,Ep) ); draw( C--Ep ); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,NE); label("$D$",D,NW); label("$E$",Ep,S); label("$M$",M,NW); | [] |
689 | Rectangle $ABCD$ has $AB=8$ and $BC=6$. Point $M$ is the midpoint of diagonal $\overline{AC}$, and $E$ is on $AB$ with $\overline{ME}\perp\overline{AC}$. What is the area of $\triangle AME$?
$\text{(A) } \frac{65}{8} \qquad \text{(B) } \frac{25}{3} \qquad \text{(C) } 9 \qquad \text{(D) } \frac{75}{8} \qquad \text{(E) } \frac{85}{8}$ | 2009 AMC 10B Problem 18 | Draw $EC$ as shown from the diagram. Since $AC$ is of length $10$, we have that $AM$ is of length $5$, because of the midpoint $M$. Through the Pythagorean theorem, we know that $AE^2 = AM^2 + ME^2 \implies 25 + ME^2$, which means $AE = \sqrt{25 + ME^2}$. Define $ME$ to be $x$ for the sake of clarity. We know that $EB = 8 - \sqrt{25 + x^2}$. From here, we know that $CE^2 = CB^2 + BE^2 = ME^2 + MC^2$. From here, we can write the expression $6^2 + (8 - \sqrt{25 + x^2})^2 = 5^2 + x^2 \implies 36 + (64 - 16\sqrt{25 + x^2} + 25 + x^2) = 25 + x^2 \implies 100 - 16\sqrt{25 + x^2} \implies \sqrt{25 + x^2} = \frac{25}{4} \implies 25 + x^2 = \frac{625}{16} \implies 400 + 16x^2 = 625 \implies 16x^2 = 225 \implies x = \frac{15}{4}$. Now, remember $CE \neq \frac{15}{4}$. $x = \frac{15}{4} = ME$, since we set $x = ME$ in the start of the solution. Now to find the area $\frac{15}{4} \cdot 5 \cdot \frac{1}{2} = \frac{75}{8} = CE$ | // Block 1
unitsize(0.75cm);
defaultpen(0.8);
pair A=(0,0), B=(8,0), C=(8,6), D=(0,6), M=(A+C)/2;
path ortho = shift(M)*rotate(-90)*(A--C);
pair Ep = intersectionpoint(ortho, A--B);
draw( A--B--C--D--cycle );
draw( A--C );
draw( M--Ep );
filldraw( A--M--Ep--cycle, lightgray, black );
draw( rightanglemark(A,M,Ep) );
draw( C--Ep );
label("$A$",A,SW);
label("$B$",B,SE);
label("$C$",C,NE);
label("$D$",D,NW);
label("$E$",Ep,S);
label("$M$",M,NW);
// Block 2
unitsize(0.75cm); defaultpen(0.8); pair A=(0,0), B=(8,0), C=(8,6), D=(0,6), M=(A+C)/2; path ortho = shift(M)*rotate(-90)*(A--C); pair Ep = intersectionpoint(ortho, A--B); draw( A--B--C--D--cycle ); draw( A--C ); draw( M--Ep ); filldraw( A--M--Ep--cycle, lightgray, black ); draw( rightanglemark(A,M,Ep) ); draw( C--Ep ); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,NE); label("$D$",D,NW); label("$E$",Ep,S); label("$M$",M,NW); | [] |
689 | Rectangle $ABCD$ has $AB=8$ and $BC=6$. Point $M$ is the midpoint of diagonal $\overline{AC}$, and $E$ is on $AB$ with $\overline{ME}\perp\overline{AC}$. What is the area of $\triangle AME$?
$\text{(A) } \frac{65}{8} \qquad \text{(B) } \frac{25}{3} \qquad \text{(C) } 9 \qquad \text{(D) } \frac{75}{8} \qquad \text{(E) } \frac{85}{8}$ | 2009 AMC 10B Problem 18 | By the Pythagorean Theorem, we claim that $AC = 10$. It then follows that $AM \cong MC = 5.$
Because we have $AM \cong MC, \angle AME \cong \angle CME,$ and reflexive side $EM$, it follows that $\triangle AME \cong \triangle CME.$ By CPCTC, we have $AE \cong EC.$ For the sake of simplicity, we'll call those side lengths $x$. Also, since $AE = x,$ we get $BE = 8 - x.$ We can now set up the Pythagorean theorem on $\triangle EBC$: \[(8 - x)^2 + 6^2 = x^2.\] Combining like terms and simplifying gives $-16x + 100 = 0$ so $x = \frac{25}{4}.$
It helps to think that in order to find $[AME],$ we must have $\overline{MC}$ and $\overline{EM}.$ Let $EM = y.$ Applying the Pythagorean Theorem to $\triangle CME$ gives \[5^2 + y^2 = \left(\frac{25}{4} \right)^2.\] Solving for $y$ (this is not that difficult) gives $y = \frac{15}{4}.$ So, the area of $\triangle AME$ is $\frac{\frac{15}{4} \cdot 5}{2} = \frac{75}{8} \implies \boxed{D}.$ | // Block 1
unitsize(0.75cm);
defaultpen(0.8);
pair A=(0,0), B=(8,0), C=(8,6), D=(0,6), M=(A+C)/2;
path ortho = shift(M)*rotate(-90)*(A--C);
pair Ep = intersectionpoint(ortho, A--B);
draw( A--B--C--D--cycle );
draw( A--C );
draw( M--Ep );
filldraw( A--M--Ep--cycle, lightgray, black );
draw( rightanglemark(A,M,Ep) );
draw( C--Ep );
label("$A$",A,SW);
label("$B$",B,SE);
label("$C$",C,NE);
label("$D$",D,NW);
label("$E$",Ep,S);
label("$M$",M,NW);
// Block 2
unitsize(0.75cm); defaultpen(0.8); pair A=(0,0), B=(8,0), C=(8,6), D=(0,6), M=(A+C)/2; path ortho = shift(M)*rotate(-90)*(A--C); pair Ep = intersectionpoint(ortho, A--B); draw( A--B--C--D--cycle ); draw( A--C ); draw( M--Ep ); filldraw( A--M--Ep--cycle, lightgray, black ); draw( rightanglemark(A,M,Ep) ); draw( C--Ep ); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,NE); label("$D$",D,NW); label("$E$",Ep,S); label("$M$",M,NW); | [] |
690 | A cubical cake with edge length $2$ inches is iced on the sides and the top. It is cut vertically into three pieces as shown in this top view, where $M$ is the midpoint of a top edge. The piece whose top is triangle $B$ contains $c$ cubic inches of cake and $s$ square inches of icing. What is $c+s$?
$\text{(A) } \frac{24}{5} \qquad \text{(B) } \frac{32}{5} \qquad \text{(C) } 8+\sqrt5 \qquad \text{(D) } 5+\frac{16\sqrt5}{5} \qquad \text{(E) } 10+5\sqrt5$ | 2009 AMC 10B Problem 22 | Let's label the points as in the picture above. Let $[RNQ]$ be the area of $\triangle RNQ$. Then the volume of the corresponding piece is $c=2[RNQ]$. This cake piece has icing on the top and on the vertical side that contains the edge $QR$. Hence the total area with icing is $[RNQ]+2^2 = [RNQ]+4$. Thus the answer to our problem is $3[RNQ]+4$, and all we have to do now is determine $[RNQ]$.
Solution 1
Introduce a coordinate system where $Q=(0,0)$, $P=(2,0)$ and $R=(0,2)$.
In this coordinate system we have $M=(2,1)$, and the line $QM$ has the equation $2y-x=0$.
As the line $RN$ is orthogonal to $QM$, it must have the equation $y+2x+q=0$ for some suitable constant $q$. As this line contains the point $R=(0,2)$, we have $q=-2$.
Substituting $x=2y$ into $y+2x-2=0$, we get $y=\frac 25$, and then $x=\frac 45$.
We can note that in $\triangle RNQ$ $x$ is the height from $N$ onto $RQ$, hence its area is $[RNQ] = \frac{x \cdot RQ} 2 = \frac{2x}2 = x = \frac 45$, and therefore the answer is $3[RNQ]+4 = 3\cdot \frac 45 + 4 = \boxed{\frac{32}5 \Longrightarrow B}$.
Solution 2
Extend $RN$ to intersect $PQ$ at $O$:
It is now obvious that $O$ is the midpoint of $PQ$. (Imagine rotating the square $PQRS$ by $90^\circ$ clockwise around its center. This rotation will map the segment $MQ$ to a segment that is orthogonal to $MQ$, contains $R$ and contains the midpoint of $PQ$.
From $\triangle PQM$ we can compute that $QM = \sqrt{1^2 + 2^2} = \sqrt 5$.
Observe that $\triangle PQM$ and $\triangle NQO$ have the same angles and therefore they are similar. The ratio of their sides is $\frac{QM}{OQ} = \frac{\sqrt 5}1 = \sqrt 5$.
Hence we have $ON = \frac{PM}{\sqrt 5} = \frac 1{\sqrt 5}$, and $NQ = \frac{PQ}{\sqrt 5} = \frac 2{\sqrt 5}$.
Knowing this, we can compute the area of $\triangle NQO$ as $[NQO] = \frac{ON \cdot NQ}2 = \frac 15$.
Finally, we compute $[RNQ] = [ROQ] - [NQO] = 1 - \frac 15 = \frac 45,$ and conclude that the answer is $3[RNQ]+4 = 3\cdot \frac 45 + 4 = \boxed{\frac{32}5}.$
You could also notice that the two triangles $\triangle PMQ$ and $\triangle NQR$ in the original figure are similar.
Solution 3 (Pythagorean Theorem only)
Since $PQ = SR = 2$ and $PM = MS = 1$, we know that $MQ = MR = \sqrt{2^{2} + 1^{2}} = \sqrt{5}$. If we let $NQ = x$, then $MN = \sqrt{5} - x$. Now, by the Pythagorean Theorem, we have:
\[x^{2} + NR^{2} = 2^{2} = 4\]
\[(\sqrt{5} - x)^{2} + NR^{2} = (\sqrt{5})^{2} = 5\]
Expanding and rearranging the second equation gives:
\[5 - 2x\sqrt{5} + x^{2} + NR^{2} = 5\]
\[x^{2} + NR^{2} - 2x\sqrt{5} = 0\]
\[x^{2} + NR^{2} = 2x\sqrt{5}\]
Since $x^{2} + NR^{2} = 4$, we have that:
\[2x\sqrt{5} = 4\]
\[x\sqrt{5} = 2\]
\[x = \frac{2}{\sqrt{5}} = \frac{2\sqrt{5}}{5}\]
Knowing $x$, we can solve for the height $NR$:
\[NR^{2} = 2^{2} - x^{2} = 4 - ({\frac{2\sqrt{5}}{5}})^{2} = 4 - \frac{4}{5} = \frac{16}{5}\]
\[NR = \sqrt{\frac{16}{5}} = \frac{4}{\sqrt{5}} = \frac{4\sqrt{5}}{5}\]
Therefore, the area of triangle $RNQ$ is $\frac{1}{2} \cdot \frac{2\sqrt{5}}{5} \cdot\frac{4\sqrt{5}}{5} = \frac{1}{2} \cdot\frac{40}{25} = \frac{4}{5}$. Since the solution to the problem is $3[RNQ] + 4$, the answer is $3(\frac{4}{5}) + 4 = \frac{12}{5} + \frac{20}{5} = \boxed{(B) \frac{32}{5}}$.
Solution 4
$MQ = \sqrt{2^{2} + 1^{2}} = \sqrt{5}$
since $\angle PQM + \angle PMQ = 90 = \angle PQM + \angle NQR$
therefore $\angle PMQ = \angle NQR$
and since $\angle MPQ = \angle QNR = 90$
therefore $\triangle MPQ \sim \triangle QNR$
therefore $\frac {[QNR]}{[MPQ]} = (\frac{QR}{QM})^{2}, [QNR] = [MPQ] \cdot (\frac{QR}{QM})^{2}$
$[MPQ] = \frac{1}{2} \cdot 2 \cdot 1 = 1$
$[QNR] = 1 \cdot (\frac{2}{\sqrt{5}})^{2} = 1 \cdot \frac{4}{5} = \frac{4}{5}$
Since the solution to the problem is $3[QNR] + 4$, the answer is $3(\frac{4}{5}) + 4 = \frac{12}{5} + \frac{20}{5} = \boxed{(B) \frac{32}{5}}$.
Solution 5 (Similarity)
All units of length in the following solution are in inches, or inches squared, or inches cubed. Units of angles are in degrees.
$PQ = 2$. Since $M$ is the midpoint of $\overline{SP}$ which measures $2$, $MP = 1$.
Since angle MNR is right, angle QNR is also right. Let $m\angle PQM = x$. Then $m\angle PMQ = 90 - x$. Notice also since $\angle PQR$ is right, $m \angle NQR = 90 - x$. Since $\angle QNR$ is right, $m\angle QRN = x$. Therefore, $\triangle PQM \sim \triangle NRQ$.
Let $QN = a$. By the Pythagorean theorem, $MQ = \sqrt{5}$. By similarity, $\frac{PM}{MQ} = \frac{NQ}{QR} \longrightarrow \frac{1}{\sqrt{5}} = \frac{a}{2}$, so $a=\frac{2}{\sqrt{5}}=\frac{2\sqrt{5}}{5}$. By the Pythagorean theorem, $NR^2+NQ^2=QR^2$. Substituting known values in and solving for $NR$, we get $NR=\frac{4\sqrt{5}}{5}$. (Alternatively, use the fact that $\triangle PQM \sim \triangle NRQ$). Since $\triangle NQR$ is a right triangle, the area is just $NQ \cdot NR \cdot \frac{1}{2}$ which, substituting values, is equal to $\frac{4}{5}$. But remember that $s$ also consists of the side of the cake, so we have to add $2^2=4$. So $s=\frac{4}{5}+4=\frac{24}{5}$.
Meanwhile, $c$ is the volume of the slice (a triangular prism) which is found by the base area times height. We already calculated the base area to be $\frac{4}{5}$, so simply multiply by $2$ to get the volume $=\frac{8}{5}$. This is the value of $c$.
Sum $c+s$: $c+s=\frac{8}{5}+\frac{24}{5}=\frac{32}{5} \Longrightarrow \boxed{\textbf{(B) } \frac{32}{5}}$.
~JH. L
Solution 6 (only Pythagorean Theorem, no algebra)
Label the vertices of the square, $P$, $Q$, $R$, and $S$ and draw line segment $MA$ (as shown below):
$PM=MS=1$ and $SR=2$, so by the pythagorean theorem, \[MR = \sqrt{1^2+2^2} = \sqrt{5}.\] By the same logic, $MQ=\sqrt{5}$. The area of $\triangle QMR$ is the area of the whole square, minus the combined areas of $\triangle PMQ$ and $\triangle MSR$, so \[[QMR] = 4-1-1=2.\] Since $QM$ is the base of $\triangle QMR$ and $RN$ is the height, \[\frac{QM(RN)}{2} = 2\] \[\frac{\sqrt5(RN)}{2} = 2,\] so $RN=\frac{3\sqrt5}{5}$. We also know that \[QN = QM-MN = \sqrt{5} - \frac{3\sqrt5}{5} = \frac{2\sqrt5}{5}.\] Now, we can find the area of $\triangle QNR$. \[[QNR] = \frac{1}{2}(QN)(RN) = \frac{1}{2}(\frac{4\sqrt5}{5})(\frac{2\sqrt5}{5}) = \frac{4}{5}.\] The area of icing is the area of $\triangle QNR$, plus the area of the 2x2 square on $QR$, so \[s=\frac{4}{5}+4 =\frac{4}{5} + \frac{20}{5} = \frac{24}{5}.\] The cubic inches of cake is the volume of the piece, which is the area of $\triangle QNR$ times $2$ (the height of the cake), so $c = \frac{8}{5}$. Hence, $c+s = \frac{32}{5}$, and the answer is $\boxed{\textbf{(B) } \frac{32}{5}}$ ~azc1027 | // Block 1
unitsize(2cm);
defaultpen(linewidth(.8pt)+fontsize(8pt));
draw((-1,-1)--(1,-1)--(1,1)--(-1,1)--cycle);
draw((1,1)--(-1,0));
pair P=foot((1,-1),(1,1),(-1,0));
draw((1,-1)--P);
draw(rightanglemark((-1,0),P,(1,-1),4));
label("$M$",(-1,0),W);
label("$C$",(-0.1,-0.3));
label("$A$",(-0.4,0.7));
label("$B$",(0.7,0.4));
label("$P$",(-1,1),NW);
label("$Q$",(1,1),NE);
label("$R$",(1,-1),SE);
label("$S$",(-1,-1),SW);
label("$N$",P,NW);
// Block 2
unitsize(2cm); defaultpen(linewidth(.8pt)+fontsize(8pt)); draw((-1,-1)--(1,-1)--(1,1)--(-1,1)--cycle); draw((1,1)--(-1,0)); pair P=foot((1,-1),(1,1),(-1,0)); draw((1,-1)--P); draw(rightanglemark((-1,0),P,(1,-1),4)); label("$M$",(-1,0),W); label("$C$",(-0.1,-0.3)); label("$A$",(-0.4,0.7)); label("$B$",(0.7,0.4)); label("$P$",(-1,1),NW); label("$Q$",(1,1),NE); label("$R$",(1,-1),SE); label("$S$",(-1,-1),SW); label("$N$",P,NW);
// Block 3
unitsize(2cm); defaultpen(linewidth(.8pt)+fontsize(8pt)); draw((-1,-1)--(1,-1)--(1,1)--(-1,1)--cycle); draw((1,1)--(-1,0)); pair P=foot((1,-1),(1,1),(-1,0)); draw((1,-1)--P); draw(rightanglemark((-1,0),P,(1,-1),4)); draw(P -- (0,1)); label("$M$",(-1,0),W); label("$C$",(-0.1,-0.3)); label("$A$",(-0.4,0.7)); label("$B$",(0.7,0.4)); label("$P$",(-1,1),NW); label("$Q$",(1,1),NE); label("$R$",(1,-1),SE); label("$S$",(-1,-1),SW); label("$N$",P,1.5*WNW); label("$O$",(0,1),N);
// Block 4
unitsize(2cm); defaultpen(linewidth(.8pt)+fontsize(8pt)); draw((-1,-1)--(1,-1)--(1,1)--(-1,1)--cycle); draw((1,1)--(-1,0)); pair P=foot((1,-1),(1,1),(-1,0)); draw((1,-1)--P); draw(rightanglemark((-1,0),P,(1,-1),4)); draw((-1,0)--(1,-1)); label("$M$",(-1,0),W); label("$C$",(-0.1,-0.3)); label("$A$",(-0.4,0.7)); label("$B$",(0.7,0.4)); label("$P$",(-1,1),NW); label("$Q$",(1,1),NE); label("$R$",(1,-1),SE); label("$S$",(-1,-1),SW); label("$N$",P,NW); label("$x$", (0.65, 0.7)); label("$\sqrt{5} - x$", (-0.3, 0.15));
// Block 5
unitsize(2cm); defaultpen(linewidth(.8pt)+fontsize(8pt)); draw((-1,-1)--(1,-1)--(1,1)--(-1,1)--cycle); draw((1,1)--(-1,0)); pair P=foot((1,-1),(1,1),(-1,0)); draw((1,-1)--P); draw(rightanglemark((-1,0),P,(1,-1),4)); label("$M$",(-1,0),W); label("$C$",(-0.1,-0.3)); label("$A$",(-0.4,0.7)); label("$B$",(0.7,0.4)); label("$P$",(-1,1),NW); label("$Q$",(1,1),NE); label("$R$",(1,-1),SE); label("$S$",(-1,-1),SW); label("$N$",P,NW);
// Block 6
unitsize(2cm); defaultpen(linewidth(.8pt)+fontsize(8pt)); draw((-1,-1)--(1,-1)--(1,1)--(-1,1)--cycle); draw((1,1)--(-1,0)); pair P=foot((1,-1),(1,1),(-1,0)); draw((1,-1)--P); draw(rightanglemark((-1,0),P,(1,-1),4)); label("$M$",(-1,0),W); label("$C$",(-0.1,-0.3)); label("$A$",(-0.4,0.7)); label("$B$",(0.7,0.4)); label("$P$",(-1,1),NW); label("$Q$",(1,1),NE); label("$R$",(1,-1),SE); label("$S$",(-1,-1),SW); label("$N$",P,NW);
// Block 7
unitsize(2cm); defaultpen(linewidth(.8pt)+fontsize(8pt)); draw((-1,-1)--(1,-1)--(1,1)--(-1,1)--cycle); draw((1,1)--(-1,0)); pair P=foot((1,-1),(1,1),(-1,0)); draw((1,-1)--P); draw((1, 1)--(1,0)); draw(rightanglemark((-1,0),P,(1,-1),4)); draw((-1,0)--(1,-1)); label("$M$",(-1,0),W); label("$C$",(-0.1,-0.3)); label("$A$",(-0.4,0.7)); label("$B$",(0.7,0.4)); label("$P$",(-1,1),NW); label("$Q$",(1,1),NE); label("$R$",(1,-1),SE); label("$S$",(-1,-1),SW); label("$N$",P,NW); | [] |
691 | The keystone arch is an ancient architectural feature. It is composed of congruent isosceles trapezoids fitted together along the non-parallel sides, as shown. The bottom sides of the two end trapezoids are horizontal. In an arch made with $9$ trapezoids, let $x$ be the angle measure in degrees of the larger interior angle of the trapezoid. What is $x$?
$\text{(A) } 100 \qquad \text{(B) } 102 \qquad \text{(C) } 104 \qquad \text{(D) } 106 \qquad \text{(E) } 108$ | 2009 AMC 10B Problem 24 | Extend all the legs of the trapezoids. They will all intersect in the middle of the bottom side of the picture, forming the situation shown below.
Each of the angles at $X$ is $\frac{180^\circ}9 = 20^\circ$. From $\triangle XYZ$, the degree measure of the smaller interior angle of the trapezoid is $\frac{180^\circ - 20^\circ}2 = 80^\circ$, hence the degree measure of the larger interior angle is $180^\circ - 80^\circ = \boxed{100^\circ \Longrightarrow A}$.
Proof that all the extended trapezoid legs intersect at the same point: It is sufficient to prove this for any pair of neighboring trapezoids. For two neighboring trapezoids, the situation is symmetric according to their common leg, therefore the extensions of both outside legs intersect the extension of the common leg at the same point, Q.E.D.
Knowing this, we can now easily see that the intersection point must be on the bottom side of our picture, as it lies on the bottom leg of the rightmost trapezoid. And by symmetry the point must be in the center of this side. | // Block 1
unitsize(6mm);
defaultpen(linewidth(.8pt));
int i;
real r=5, R=6;
path t=r*dir(0)--r*dir(20)--R*dir(20)--R*dir(0);
for(i=0; i<9; ++i)
{
draw(rotate(20*i)*t);
}
draw((-r,0)--(R+1,0));
draw((-R,0)--(-R-1,0));
for (int i=1; i<9; ++i) draw( (0,0) -- (rotate(20*i)*(r,0)), dotted );
label("$X$",(0,0),S);
label("$Y$",(R,0),SE);
label("$Z$",rotate(20)*(R,0),ENE);
draw( arc( (0,0), (1.5,0), rotate(20)*(1.5,0) ) );
label("$20^\circ$", rotate(10)*(1.75,0), E );
// Block 2
unitsize(6mm); defaultpen(linewidth(.8pt)); int i; real r=5, R=6; path t=r*dir(0)--r*dir(20)--R*dir(20)--R*dir(0); for(i=0; i<9; ++i) { draw(rotate(20*i)*t); } draw((-r,0)--(R+1,0)); draw((-R,0)--(-R-1,0)); for (int i=1; i<9; ++i) draw( (0,0) -- (rotate(20*i)*(r,0)), dotted ); label("$X$",(0,0),S); label("$Y$",(R,0),SE); label("$Z$",rotate(20)*(R,0),ENE); draw( arc( (0,0), (1.5,0), rotate(20)*(1.5,0) ) ); label("$20^\circ$", rotate(10)*(1.75,0), E ); | [] |
692 | Triangle $ABC$ has vertices $A = (3,0)$, $B = (0,3)$, and $C$, where $C$ is on the line $x + y = 7$. What is the area of $\triangle ABC$?
$\mathrm{(A)}\ 6\qquad \mathrm{(B)}\ 8\qquad \mathrm{(C)}\ 10\qquad \mathrm{(D)}\ 12\qquad \mathrm{(E)}\ 14$ | 2009 AMC 12B Problem 9 | Solution 1
Because the line $x + y = 7$ is parallel to $\overline {AB}$, the area of $\triangle ABC$ is independent of the location of $C$ on that line. Therefore it may be assumed that $C$ is $(7,0)$. In that case the triangle has base $AC = 4$ and altitude $3$, so its area is $\frac 12 \cdot 4 \cdot 3 = \boxed {6}$.
Solution 2
The base of the triangle is $AB = \sqrt{3^2 + 3^2} = 3\sqrt 2$. Its altitude is the distance between the point $A$ and the parallel line $x + y = 7$, which is $\frac 4{\sqrt 2} = 2\sqrt 2$. Therefore its area is $\frac 12 \cdot 3\sqrt 2 \cdot 2\sqrt 2 = \boxed{6}$. The answer is $\mathrm{(A)}$.
Solution 3
By Shoelace, our area is:
\[\frac {1}{2} \cdot |(3\cdot 3 + 0 \cdot y+ 0 \cdot x)-(0*0+3(x+y))|.\]
We know $x+y=7$ so we get:
\[\frac {1}{2} \cdot |9-21|=\boxed 6\]
Solution 4
WLOG, let the coordinates of $C$ be $(3,4)$ , or any coordinate, for that matter. Applying the shoelace formula, we get the area as $\boxed 6$.
~coolmath2017 | unitsize(0.75cm); defaultpen(0.8); pair A=(3,0), B=(0,3); draw ( (-1,0) -- (9,0), dashed ); draw ( (0,-1) -- (0,9), dashed ); dot(A); dot(B); draw(A--B); draw ( (-1,8) -- (8,-1) ); label( "$A$", A, S ); label( "$B$", B, W ); label( "$3$", A--(0,0), S ); label( "$3$", B--(0,0), W ); label( "$x+y=7$", (8,-1), SE ); pair C = intersectionpoint(A--(10,7),(7,0)--(0,7)); draw( A--C, dashed ); draw(rightanglemark(A,C,(7,0))); draw(rightanglemark(C,A,B)); label( "$4$", A--(7,0), S ); label( "$3\sqrt 2$", 0.67*B+0.33*A, NE ); label( "$\frac 4{\sqrt 2}$", A--C, NW ); label( "$\frac 4{\sqrt 2}$", C--(7,0), NE ); | [] |
693 | A convex polyhedron $Q$ has vertices $V_1,V_2,\ldots,V_n$, and $100$ edges. The polyhedron is cut by planes $P_1,P_2,\ldots,P_n$ in such a way that plane $P_k$ cuts only those edges that meet at vertex $V_k$. In addition, no two planes intersect inside or on $Q$. The cuts produce $n$ pyramids and a new polyhedron $R$. How many edges does $R$ have?
$\mathrm{(A)}\ 200\qquad \mathrm{(B)}\ 2n\qquad \mathrm{(C)}\ 300\qquad \mathrm{(D)}\ 400\qquad \mathrm{(E)}\ 4n$ | 2009 AMC 12B Problem 20 | Solution 1
Each edge of $Q$ is cut by two planes, so $R$ has $200$ vertices. Three edges of $R$ meet at each vertex, so $R$ has a total of $\frac 12 \cdot 3 \cdot 200 = \boxed {300}$ edges.
Solution 2
At each vertex, as many new edges are created by this process as there are original edges meeting at that vertex. Thus the total number of new edges is the total number of endpoints of the original edges, which is $200$. A middle portion of each original edge is also present in $R$, so $R$ has a total of $100 + 200 = \boxed {300}$ edges.
Solution 3
Euler's Polyhedron Formula applied to $Q$ gives $n - 100 + F = 2$, where F is the number of faces of $Q$. Each edge of $Q$ is cut by two planes, so $R$ has $200$ vertices. Each cut by a plane $P_k$ creates an additional face on $R$, so Euler's Polyhedron Formula applied to $R$ gives $200 - E + (F+n) = 2$, where $E$ is the number of edges of $R$. Subtracting the first equation from the second gives $300 - E = 0$, whence $E = \boxed {(C)300}$.
Solution 4
Each edge connects two points. The plane cuts that edge so it splits into $2$ at each end (like two legs) for a total of $4$ new edges.
But because each new edge is shared by an adjacent original edge cut similarly, the additional edges are overcounted $\times 2$.
Since there are $100$ edges to start with, $400/2=200$ new edges result. So there are $100+200=\boxed{\textbf{(C) }300}$ edges in the figure.
Solution 5
The question specifies the slices create as many pyramids as there are vertices, implying each vertex owns 4 edge ends. There are twice as many edge-ends as there are edges, and $2 \cdot 100 = 200$.
$\frac{200}{4} = 50$, so there are $50$ vertices.
The base of a pyramid has 4 edges, so each sliced vertex would add four edges to $R$.
$100 + 4 \cdot 50$ = $\boxed{\textbf{(C) } 300}$ | // Block 1
pair A,B,C,D,E,F; A=(0,50); B=(0,-50); C=(86.602540378443864676372317075294,0); D=(300,0); E=(370.7,70.7); F=(370.7,-70.7); draw(A--C--B); draw(C--D); draw(E--D--F);
// Block 2
pair A,B,C,D,E,F,G,H,I,J,K; A=(0,50); B=(0,-50); C=(86.602540378443864676372317075294,0); D=(300,0); E=(370.7,70.7); F=(370.7,-70.7); G=(-107.5,236.2); H=(-107.5,-236.2); I=(370.7,285); J=(370.7,-285); K=(441.4,0); draw(A--C--B); draw(C--D); draw(E--D--F); draw(G--A, red); draw(H--B, red); draw(I--E, red); draw(J--F, red); draw(A--B, dashed); draw(E--K, dashed); draw(F--K, dashed); | [] |
694 | The set $G$ is defined by the points $(x,y)$ with integer coordinates, $3\le|x|\le7$, $3\le|y|\le7$. How many squares of side at least $6$ have their four vertices in $G$?
$\textbf{(A)}\ 125\qquad \textbf{(B)}\ 150\qquad \textbf{(C)}\ 175\qquad \textbf{(D)}\ 200\qquad \textbf{(E)}\ 225$ | 2009 AMC 12B Problem 25 | We need to find a reasonably easy way to count the squares.
First, obviously the maximum distance between two points in the same quadrant is $4\sqrt 2 < 6$, hence each square has exactly one vertex in each quadrant.
Given any square, we can circumscribe another axes-parallel square around it. In the picture below, the original square is red and the circumscribed one is blue.
Let's now consider the opposite direction. Assume that we picked the blue square, how many different red squares do share it?
Answering this question is not as simple as it may seem. Consider the picture below. It shows all three red squares that share the same blue square. In addition, the picture shows a green square that is not valid, as two of its vertices are in bad locations.
The size of the blue square can range from $6\times 6$ to $14\times 14$, and for the intermediate sizes, there is more than one valid placement. We will now examine the cases one after another. Also, we can use symmetry to reduce the number of cases.
size upper_right solutions symmetries total
6 (3,3) 1 1 1
7 (3,3) 1 4 4
8 (3,3) 1 4 4
8 (3,4) 1 4 4
8 (4,4) 3 1 3
9 (3,3) 1 4 4
9 (3,4) 1 8 8
9 (4,4) 3 4 12
10 (3,3) 1 4 4
10 (3,4) 1 8 8
10 (3,5) 1 4 4
10 (4,4) 3 4 12
10 (4,5) 3 4 12
10 (5,5) 5 1 5
11 (4,4) 3 4 12
11 (4,5) 3 8 24
11 (5,5) 5 4 20
12 (5,5) 5 4 20
12 (5,6) 5 4 20
12 (6,6) 7 1 7
13 (6,6) 7 4 28
14 (7,7) 9 1 9
Summing the last column, we get that the answer is $\boxed{225}$. | // Block 1
defaultpen(black+0.75bp+fontsize(8pt));
size(7.5cm);
path p = scale(.15)*unitcircle;
draw((-8,0)--(8.5,0),Arrow(HookHead,1mm));
draw((0,-8)--(0,8.5),Arrow(HookHead,1mm));
int i,j;
for (i=-7;i<8;++i) {
for (j=-7;j<8;++j) {
if (((-7 <= i) && (i <= -3)) || ((3 <= i) && (i<= 7))) { if (((-7 <= j) && (j <= -3)) || ((3 <= j) && (j<= 7))) { fill(shift(i,j)*p,black); }}}} draw((-7,-.2)--(-7,.2),black+0.5bp);
draw((-3,-.2)--(-3,.2),black+0.5bp);
draw((3,-.2)--(3,.2),black+0.5bp);
draw((7,-.2)--(7,.2),black+0.5bp);
draw((-.2,-7)--(.2,-7),black+0.5bp);
draw((-.2,-3)--(.2,-3),black+0.5bp);
draw((-.2,3)--(.2,3),black+0.5bp);
draw((-.2,7)--(.2,7),black+0.5bp);
label("$-7$",(-7,0),S);
label("$-3$",(-3,0),S);
label("$3$",(3,0),S);
label("$7$",(7,0),S);
label("$-7$",(0,-7),W);
label("$-3$",(0,-3),W);
label("$3$",(0,3),W);
label("$7$",(0,7),W);
draw( (5,3) -- (-3,4) -- (-4,-4) -- (4,-5) -- cycle, red );
draw( (5,4) -- (-4,4) -- (-4,-5) -- (5,-5) -- cycle, dashed + blue );
// Block 2
defaultpen(black+0.75bp+fontsize(8pt));
size(7.5cm);
path p = scale(.15)*unitcircle;
draw((-8,0)--(8.5,0),Arrow(HookHead,1mm));
draw((0,-8)--(0,8.5),Arrow(HookHead,1mm));
int i,j;
for (i=-7;i<8;++i) {
for (j=-7;j<8;++j) {
if (((-7 <= i) && (i <= -3)) || ((3 <= i) && (i<= 7))) { if (((-7 <= j) && (j <= -3)) || ((3 <= j) && (j<= 7))) { fill(shift(i,j)*p,black); }}}} draw((-7,-.2)--(-7,.2),black+0.5bp);
draw((-3,-.2)--(-3,.2),black+0.5bp);
draw((3,-.2)--(3,.2),black+0.5bp);
draw((7,-.2)--(7,.2),black+0.5bp);
draw((-.2,-7)--(.2,-7),black+0.5bp);
draw((-.2,-3)--(.2,-3),black+0.5bp);
draw((-.2,3)--(.2,3),black+0.5bp);
draw((-.2,7)--(.2,7),black+0.5bp);
label("$-7$",(-7,0),S);
label("$-3$",(-3,0),S);
label("$3$",(3,0),S);
label("$7$",(7,0),S);
label("$-7$",(0,-7),W);
label("$-3$",(0,-3),W);
label("$3$",(0,3),W);
label("$7$",(0,7),W);
draw( (5,4) -- (-4,4) -- (-4,-5) -- (5,-5) -- cycle, red );
draw( (5,3) -- (-3,4) -- (-4,-4) -- (4,-5) -- cycle, red );
draw( (5,-3) -- (-2,-5) -- (-4,2) -- (3,4) -- cycle, dashed + green );
draw( (5,-4) -- (-3,-5) -- (-4,3) -- (4,4) -- cycle, red );
draw( scale(1.05)*((5,4) -- (-4,4) -- (-4,-5) -- (5,-5) -- cycle), dashed + blue );
// Block 3
defaultpen(black+0.75bp+fontsize(8pt)); size(7.5cm); path p = scale(.15)*unitcircle; draw((-8,0)--(8.5,0),Arrow(HookHead,1mm)); draw((0,-8)--(0,8.5),Arrow(HookHead,1mm)); int i,j; for (i=-7;i<8;++i) { for (j=-7;j<8;++j) { if (((-7 <= i) && (i <= -3)) || ((3 <= i) && (i<= 7))) { if (((-7 <= j) && (j <= -3)) || ((3 <= j) && (j<= 7))) { fill(shift(i,j)*p,black); }}}} draw((-7,-.2)--(-7,.2),black+0.5bp); draw((-3,-.2)--(-3,.2),black+0.5bp); draw((3,-.2)--(3,.2),black+0.5bp); draw((7,-.2)--(7,.2),black+0.5bp); draw((-.2,-7)--(.2,-7),black+0.5bp); draw((-.2,-3)--(.2,-3),black+0.5bp); draw((-.2,3)--(.2,3),black+0.5bp); draw((-.2,7)--(.2,7),black+0.5bp); label("$-7$",(-7,0),S); label("$-3$",(-3,0),S); label("$3$",(3,0),S); label("$7$",(7,0),S); label("$-7$",(0,-7),W); label("$-3$",(0,-3),W); label("$3$",(0,3),W); label("$7$",(0,7),W); draw( (5,3) -- (-3,4) -- (-4,-4) -- (4,-5) -- cycle, red ); draw( (5,4) -- (-4,4) -- (-4,-5) -- (5,-5) -- cycle, dashed + blue );
// Block 4
defaultpen(black+0.75bp+fontsize(8pt)); size(7.5cm); path p = scale(.15)*unitcircle; draw((-8,0)--(8.5,0),Arrow(HookHead,1mm)); draw((0,-8)--(0,8.5),Arrow(HookHead,1mm)); int i,j; for (i=-7;i<8;++i) { for (j=-7;j<8;++j) { if (((-7 <= i) && (i <= -3)) || ((3 <= i) && (i<= 7))) { if (((-7 <= j) && (j <= -3)) || ((3 <= j) && (j<= 7))) { fill(shift(i,j)*p,black); }}}} draw((-7,-.2)--(-7,.2),black+0.5bp); draw((-3,-.2)--(-3,.2),black+0.5bp); draw((3,-.2)--(3,.2),black+0.5bp); draw((7,-.2)--(7,.2),black+0.5bp); draw((-.2,-7)--(.2,-7),black+0.5bp); draw((-.2,-3)--(.2,-3),black+0.5bp); draw((-.2,3)--(.2,3),black+0.5bp); draw((-.2,7)--(.2,7),black+0.5bp); label("$-7$",(-7,0),S); label("$-3$",(-3,0),S); label("$3$",(3,0),S); label("$7$",(7,0),S); label("$-7$",(0,-7),W); label("$-3$",(0,-3),W); label("$3$",(0,3),W); label("$7$",(0,7),W); draw( (5,4) -- (-4,4) -- (-4,-5) -- (5,-5) -- cycle, red ); draw( (5,3) -- (-3,4) -- (-4,-4) -- (4,-5) -- cycle, red ); draw( (5,-3) -- (-2,-5) -- (-4,2) -- (3,4) -- cycle, dashed + green ); draw( (5,-4) -- (-3,-5) -- (-4,3) -- (4,4) -- cycle, red ); draw( scale(1.05)*((5,4) -- (-4,4) -- (-4,-5) -- (5,-5) -- cycle), dashed + blue ); | [] |
694 | The set $G$ is defined by the points $(x,y)$ with integer coordinates, $3\le|x|\le7$, $3\le|y|\le7$. How many squares of side at least $6$ have their four vertices in $G$?
$\textbf{(A)}\ 125\qquad \textbf{(B)}\ 150\qquad \textbf{(C)}\ 175\qquad \textbf{(D)}\ 200\qquad \textbf{(E)}\ 225$ | 2009 AMC 12B Problem 25 | Consider any square that meets the requirements described in the problem. Then, take the vertices of the square and translate them to the first quadrant (This is the "mapping" described in Solution 2). For example, consider a square with vertices $(7, 7), (-7, 7), (-7, -7),$ and $(7, -7)$:
After following the mapping described in Solution 2, the square looks like this:
The position of each vertex within their corresponding grid has not changed.For example, the point $(-7, 7)$ is still the top-left point in a grid, albeit a change in quadrant. Trying this out with a couple of other squares, we see that the following property holds:
\[\text{"For any and all squares with vertices in each of the four quadrants, the 'mapped' version is also a square."}\]
Therefore the logical inverse is true:
\[\text{"For any 'mapped' square with all four vertices in the first quadrant,}\]
\[\text{there exists at least one corresponding 'unmapped' square with a vertex in each of the four quadrants."}\]
But how many "unmapped" squares, to be exact?
This might seem complicated at first, but with some intuitive thinking, we realize that there are exactly $4$ "unmapped" squares that correspond with a "mapped" square. This is because given a "mapped" square, there are $4$ choices for the vertex that will remain in the first quadrant; but once that point is chosen, there is only $1$ distribution of the other $3$ vertices that will result in a square. So, we want four times the number of squares we can make in the first quadrant grid.
We divide our counting method into two cases: squares with side length $0$ after mapping (which means all four vertices are in the same position relative to their own grids) and squares with side length $1-4$ after mapping.
Case 1: There are $25$ such squares of length $0$ (this is equivalent to counting the number of points on the grid). However, in this scenario, all of the vertices have been mapped onto the same point. So instead of $4$ choices for the first quadrant vertex, there is only one. Subsequently there are only $25$ such squares that correspond to them.
Case 2: Let a square with sides parallel to the axes be known as $A$ squares. These $A$ squares can have side length $1, 2, 3,$ or $4$. However, the number of $A$ squares possible depends on the side length. For example, there is only $1$ possible $A$ square of side length $4$, but $16$ squares of side length $1$. To be exact, there are $(5-s)^2$ possible $A$ squares of side length $s$. So, the total number of $A$ squares is
$\sum_{s=1}^4 (5-s)^2$
But what about "tilted" squares? Notice that "tilted squares" can always be inscribed (drawn within) another, bigger square. Let a square inscribed within an $A$ square be called a $B$ square. How many $B$ squares are there? Well, this also depends on the side length. We only want squares whose vertices are lattice points (integer value coordinates), so the number of $B$ squares should increase along with side length. We defined $B$ squares to be inscribed within $A$ squares, so we can say that all $B$ squares have their vertices on the side on an $A$ square. Consider an $A$ square with side length $4$. There are $3$ other lattice points along the side of the $A$ square, not counting the vertices. Therefore, we can say that there are $s-1$ possible $B$ squares for every $A$ square with side length $s$. We can multiply $(s-1)$ times the number of $A$ squares to get the number of $B$ squares. This is
$\sum_{s=1}^4 (s-1)(5-s)^2$
total $B$ squares. But we need to add these two quantities to get the number of squares for Case 2:
$\sum_{s=1}^4 (s-1)(5-s)^2 + \sum_{s=1}^4 (5-s)^2$
By distributive property, the expression becomes
$\sum_{s=1}^4 s(5-s)^2$
Solving, we get $50$ "mapped" squares, both $A$ and $B$. Multiplying this by $4$ to get the corresponding number of "unmapped" squares, then adding to get the number of squares for Case 1, we get
$50*4 + 25 = 225 \Rightarrow \boxed{\text{E}}$. | // Block 1
defaultpen(black+0.75bp+fontsize(8pt));
size(5cm);
path p = scale(.15)*unitcircle;
draw((-8,0)--(8.5,0),Arrow(HookHead,1mm));
draw((0,-8)--(0,8.5),Arrow(HookHead,1mm));
int i,j;
for (i=-7;i<8;++i) {
for (j=-7;j<8;++j) {
if (((-7 <= i) && (i <= -3)) || ((3 <= i) && (i<= 7))) { if (((-7 <= j) && (j <= -3)) || ((3 <= j) && (j<= 7))) { fill(shift(i,j)*p,black); }}}} draw((-7,-.2)--(-7,.2),black+0.5bp);
draw((-3,-.2)--(-3,.2),black+0.5bp);
draw((3,-.2)--(3,.2),black+0.5bp);
draw((7,-.2)--(7,.2),black+0.5bp);
draw((-.2,-7)--(.2,-7),black+0.5bp);
draw((-.2,-3)--(.2,-3),black+0.5bp);
draw((-.2,3)--(.2,3),black+0.5bp);
draw((-.2,7)--(.2,7),black+0.5bp);
label("$-7$",(-7,0),S);
label("$-3$",(-3,0),S);
label("$3$",(3,0),S);
label("$7$",(7,0),S);
label("$-7$",(0,-7),W);
label("$-3$",(0,-3),W);
label("$3$",(0,3),W);
label("$7$",(0,7),W);
draw( (7,7) -- (-7,7) -- (-7,-7) -- (7,-7) -- cycle, black );
// Block 2
defaultpen(black+0.75bp+fontsize(8pt));
size(5cm);
path p = scale(.15)*unitcircle;
draw((-8,0)--(8.5,0),Arrow(HookHead,1mm));
draw((0,-8)--(0,8.5),Arrow(HookHead,1mm));
int i,j;
for (i=-7;i<8;++i) {
for (j=-7;j<8;++j) {
if (((-7 <= i) && (i <= -3)) || ((3 <= i) && (i<= 7))) { if (((-7 <= j) && (j <= -3)) || ((3 <= j) && (j<= 7))) { fill(shift(i,j)*p,black); }}}} draw((-7,-.2)--(-7,.2),black+0.5bp);
draw((-3,-.2)--(-3,.2),black+0.5bp);
draw((3,-.2)--(3,.2),black+0.5bp);
draw((7,-.2)--(7,.2),black+0.5bp);
draw((-.2,-7)--(.2,-7),black+0.5bp);
draw((-.2,-3)--(.2,-3),black+0.5bp);
draw((-.2,3)--(.2,3),black+0.5bp);
draw((-.2,7)--(.2,7),black+0.5bp);
label("$-7$",(-7,0),S);
label("$-3$",(-3,0),S);
label("$3$",(3,0),S);
label("$7$",(7,0),S);
label("$-7$",(0,-7),W);
label("$-3$",(0,-3),W);
label("$3$",(0,3),W);
label("$7$",(0,7),W);
draw( (3,3) -- (3,7) -- (7,7) -- (7,3) -- cycle, black );
// Block 3
defaultpen(black+0.75bp+fontsize(8pt)); size(5cm); path p = scale(.15)*unitcircle; draw((-8,0)--(8.5,0),Arrow(HookHead,1mm)); draw((0,-8)--(0,8.5),Arrow(HookHead,1mm)); int i,j; for (i=-7;i<8;++i) { for (j=-7;j<8;++j) { if (((-7 <= i) && (i <= -3)) || ((3 <= i) && (i<= 7))) { if (((-7 <= j) && (j <= -3)) || ((3 <= j) && (j<= 7))) { fill(shift(i,j)*p,black); }}}} draw((-7,-.2)--(-7,.2),black+0.5bp); draw((-3,-.2)--(-3,.2),black+0.5bp); draw((3,-.2)--(3,.2),black+0.5bp); draw((7,-.2)--(7,.2),black+0.5bp); draw((-.2,-7)--(.2,-7),black+0.5bp); draw((-.2,-3)--(.2,-3),black+0.5bp); draw((-.2,3)--(.2,3),black+0.5bp); draw((-.2,7)--(.2,7),black+0.5bp); label("$-7$",(-7,0),S); label("$-3$",(-3,0),S); label("$3$",(3,0),S); label("$7$",(7,0),S); label("$-7$",(0,-7),W); label("$-3$",(0,-3),W); label("$3$",(0,3),W); label("$7$",(0,7),W); draw( (7,7) -- (-7,7) -- (-7,-7) -- (7,-7) -- cycle, black );
// Block 4
defaultpen(black+0.75bp+fontsize(8pt)); size(5cm); path p = scale(.15)*unitcircle; draw((-8,0)--(8.5,0),Arrow(HookHead,1mm)); draw((0,-8)--(0,8.5),Arrow(HookHead,1mm)); int i,j; for (i=-7;i<8;++i) { for (j=-7;j<8;++j) { if (((-7 <= i) && (i <= -3)) || ((3 <= i) && (i<= 7))) { if (((-7 <= j) && (j <= -3)) || ((3 <= j) && (j<= 7))) { fill(shift(i,j)*p,black); }}}} draw((-7,-.2)--(-7,.2),black+0.5bp); draw((-3,-.2)--(-3,.2),black+0.5bp); draw((3,-.2)--(3,.2),black+0.5bp); draw((7,-.2)--(7,.2),black+0.5bp); draw((-.2,-7)--(.2,-7),black+0.5bp); draw((-.2,-3)--(.2,-3),black+0.5bp); draw((-.2,3)--(.2,3),black+0.5bp); draw((-.2,7)--(.2,7),black+0.5bp); label("$-7$",(-7,0),S); label("$-3$",(-3,0),S); label("$3$",(3,0),S); label("$7$",(7,0),S); label("$-7$",(0,-7),W); label("$-3$",(0,-3),W); label("$3$",(0,3),W); label("$7$",(0,7),W); draw( (3,3) -- (3,7) -- (7,7) -- (7,3) -- cycle, black ); | [] |
695 | Crystal has a running course marked out for her daily run. She starts this run by heading due north for one mile. She then runs northeast for one mile, then southeast for one mile. The last portion of her run takes her on a straight line back to where she started. How far, in miles is this last portion of her run?
$\mathrm{(A)}\ 1 \qquad \mathrm{(B)}\ \sqrt{2} \qquad \mathrm{(C)}\ \sqrt{3} \qquad \mathrm{(D)}\ 2 \qquad \mathrm{(E)}\ 2\sqrt{2}$ | 2010 AMC 10A Problem 7 | Crystal first runs north for one mile. Changing directions, she runs northeast for another mile. The angle difference between north and northeast is 45 degrees. She then switches directions to southeast, meaning a 90 degree angle change. The distance now from traveling north for one mile, and her current destination is $\sqrt{2}$ miles, because it is the hypotenuse of a 45-45-90 triangle with side length one (mile). Therefore, Crystal's distance from her starting position, x, is equal to $\sqrt{((\sqrt{2})^2+1^2)}$, which is equal to $\sqrt{3}$. The answer is $\boxed{C}$ | // Block 1
import olympiad;
draw((0,0)--(0,1));
draw((0,1)--(0,1.7071067811865476), dotted);
draw((0,1)--(0.7071067811865476, 1.7071067811865476));
draw((0.7071067811865476, 1.7071067811865476)--(1.4142135623730951,1));
draw(anglemark((0.7071067811865476, 1.7071067811865476),(0,1),(0,1.7071067811865476)));
label("$45^{\circ}$", (0,1.25), NE);
draw((0, 1)--(1.4142135623730951,1), dotted);
draw((1.4142135623730951,1)--(0,0), green);
// Block 2
import olympiad; draw((0,0)--(0,1)); draw((0,1)--(0,1.7071067811865476), dotted); draw((0,1)--(0.7071067811865476, 1.7071067811865476)); draw((0.7071067811865476, 1.7071067811865476)--(1.4142135623730951,1)); draw(anglemark((0.7071067811865476, 1.7071067811865476),(0,1),(0,1.7071067811865476))); label("$45^{\circ}$", (0,1.25), NE); draw((0, 1)--(1.4142135623730951,1), dotted); draw((1.4142135623730951,1)--(0,0), green); | [] |
696 | Equiangular hexagon $ABCDEF$ has side lengths $AB=CD=EF=1$ and $BC=DE=FA=r$. The area of $\triangle ACE$ is $70\%$ of the area of the hexagon. What is the sum of all possible values of $r$?
$\textbf{(A)}\ \frac{4\sqrt{3}}{3} \qquad \textbf{(B)} \frac{10}{3} \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ \frac{17}{4} \qquad \textbf{(E)}\ 6$ | 2010 AMC 10A Problem 19 | Drawing not to scale.*
To find r, we'll form an equation by finding the area of hexagon $ABCDEF$ and $\triangle{ACE}$ separately. First, connect a directly opposite diagonal such as $CF$. Because the hexagon is equiangular, edges $AF$ and $BC$ protrude from points $A$ and $B$ at the same angle but in opposite directions and equal length. This places points $C$ and $F$ at the same "height" relative to segment $AB$, so segment $CF$ is parallel to $AB$ and also $DE$ (opposite sides are parallel since you rotated through three $120^\circ$ angles between each other).
Next, we'll calculate the area of isosceles trapezoids $ABCF$ and $CDEF$. Drop a perpendicular down from $B$ to $CF$, and call the intersection to $CF$, $P$. Because adjacent angles between parallel sides in a trapezoid sum to $180^\circ$ and $\angle ABC = 120^\circ$, $\angle BCP = 60^\circ$ meaning $\triangle{BPC}$ is a 30-60-90 right triangle. Now we get $BP = \frac{\sqrt{3}}{2}r$ and $CP = \frac{r}{2}$. $CF$ is equal to $AB + 2CP = r+1$. Hence, the area of trapezoid $ABCF$ is $\frac{1}{2} \cdot \frac{\sqrt{3}}{2}r \cdot (r+1+1) = \frac{\sqrt{3}}{4} r(r+2)$. Similarly, on trapezoid $CDEF$ we find $\triangle EQF$ is 30-60-90, and $EQ = \frac{\sqrt{3}}{2}, FQ = \frac{1}{2}$. Then, area of $CDEF$ is $\frac{\sqrt{3}}{4} (2r + 1)$. Thus, the area of the hexagon is the sum of the two areas, $\frac{\sqrt{3}}{4} (r^2 + 4r + 1)$.
Next, notice that $\triangle ACE$ is equilateral. And by Pythagorean theorem in $\triangle CQE$, side $CE = \sqrt{r^2 + r + 1}$. Applying area of equilateral triangle formula, area $\triangle ACE = \frac{\sqrt{3}}{4} (r^2 + r + 1)$. Using this area and the area given from 70% of the area of the hexagon, we get the equation $\frac{7}{10} \cdot \frac{\sqrt{3}}{4} (r^2 + 4r + 1) = \frac{\sqrt{3}}{4} (r^2 + r + 1)$. Simplifying, we get the quadratic $r^2-6r+1=0$, which by Vieta's yields the answer $6, \boxed{\textbf{E}}$.
~henry
Proof Triangle ACE is Equilateral.
We know $\triangle{ABC}$, $\triangle{CDE}$, and $\triangle{EFA}$ are congruent by SAS, so the side opposite the 120 degree angle is also the same (since the triangles are congruent). Thus $\triangle{ACE}$ is equilateral.
Q.E.D.
~mathboy282 | import graph; size(8cm); pen dps = fontsize(10); defaultpen(dps); real r = 0.7; // Define hexagon vertices clockwise with AB on top pair A = (0, 0); pair B = (1, 0); pair C = B + r * dir(-60); pair D = C + dir(-120); pair E = D + r * dir(180); pair F = E + dir(120); // Draw the hexagon draw(A--B--C--D--E--F--cycle); // Draw triangle ACE draw(A--C--E--cycle); // Draw line CF draw(C--F); // Drop perpendiculars from B and E to CF pair foot_B = foot(B, C, F); pair foot_E = foot(E, C, F); // Draw perpendiculars (dashed) draw(B--foot_B, dashed); draw(E--foot_E, dashed); // Right angle markers (smaller) draw(rightanglemark(B, foot_B, C, 4)); draw(rightanglemark(E, foot_E, C, 4)); // Vertex labels label("$A$", A, NW); label("$B$", B, NE); label("$C$", C, SE); label("$D$", D, S); label("$E$", E, SW); label("$F$", F, NW); // Feet labels label("$P$", foot_B, S); label("$Q$", foot_E, N); // Side length labels label("$1$", midpoint(A--B), N); label("$r$", midpoint(B--C), dir(45)); label("$1$", midpoint(C--D), E); label("$r$", midpoint(D--E), S); label("$1$", midpoint(E--F), W); label("$r$", midpoint(F--A), W); | [] |
697 | A fly trapped inside a cubical box with side length $1$ meter decides to relieve its boredom by visiting each corner of the box. It will begin and end in the same corner and visit each of the other corners exactly once. To get from a corner to any other corner, it will either fly or crawl in a straight line. What is the maximum possible length, in meters, of its path?
$\textbf{(A)}\ 4+4\sqrt{2} \qquad \textbf{(B)}\ 2+4\sqrt{2}+2\sqrt{3} \qquad \textbf{(C)}\ 2+3\sqrt{2}+3\sqrt{3} \qquad \textbf{(D)}\ 4\sqrt{2}+4\sqrt{3} \qquad \textbf{(E)}\ 3\sqrt{2}+5\sqrt{3}$ | 2010 AMC 10A Problem 20 | The path of the fly consists of eight line segments, where each line segment goes from one corner to another corner. The distance of each such line segment is $1$, $\sqrt{2}$, or $\sqrt{3}$.
The only way to obtain a line segment of length $\sqrt{3}$ is to go from one corner of the cube to the opposite corner. Since the fly visits each corner exactly once, it cannot traverse such a line segment twice. Also, the cube has exactly four such diagonals, so the path of the fly can contain at most four segments of length $\sqrt{3}$. Hence, the length of the fly's path can be at most $\boxed{4 \sqrt{3} + 4 \sqrt{2}\text{, or D.}}$. This length can be achieved by taking the path
\[A \rightarrow G \rightarrow B \rightarrow H \rightarrow C \rightarrow E \rightarrow D \rightarrow F \rightarrow A.\] | // Block 1
import graph;
unitsize(2 cm);
draw((0,0)--(1,0)--(1,1)--(0,1)--cycle);
draw((1,0)--(1.3,0.3));
draw((1,1)--(1.3,1.3));
draw((0,1)--(0.3,1.3));
draw((1.3,0.3)--(1.3,1.3)--(0.3,1.3));
draw((0,0)--(0.3,0.3),dashed);
draw((0.3,0.3)--(1.3,0.3),dashed);
draw((0.3,0.3)--(0.3,1.3),dashed);
label("$A$", (0,0), SW);
label("$B$", (1,0), SE);
label("$C$", (1.3,0.3), E);
label("$D$", (0.3,0.3), SE);
label("$E$", (0,1), W);
label("$F$", (1,1), E);
label("$G$", (1.3,1.3), NE);
label("$H$", (0.3,1.3), N);
// Block 2
import graph; unitsize(2 cm); draw((0,0)--(1,0)--(1,1)--(0,1)--cycle); draw((1,0)--(1.3,0.3)); draw((1,1)--(1.3,1.3)); draw((0,1)--(0.3,1.3)); draw((1.3,0.3)--(1.3,1.3)--(0.3,1.3)); draw((0,0)--(0.3,0.3),dashed); draw((0.3,0.3)--(1.3,0.3),dashed); draw((0.3,0.3)--(0.3,1.3),dashed); label("$A$", (0,0), SW); label("$B$", (1,0), SE); label("$C$", (1.3,0.3), E); label("$D$", (0.3,0.3), SE); label("$E$", (0,1), W); label("$F$", (1,1), E); label("$G$", (1.3,1.3), NE); label("$H$", (0.3,1.3), N); | [] |
698 | Rectangle $ABCD$, pictured below, shares $50\%$ of its area with square $EFGH$. Square $EFGH$ shares $20\%$ of its area with rectangle $ABCD$. What is $\frac{AB}{AD}$?
$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 10$ | 2010 AMC 12A Problem 3 | If we shift $A$ to coincide with $E$, and add new horizontal lines to divide $EFGH$ into five equal parts:
This helps us to see that $AD=a/5$ and $AB=2a$, where $a=EF$.
Hence $\dfrac{AB}{AD}=\dfrac{2a}{a/5}=10$. | // Block 1
unitsize(1mm);
defaultpen(linewidth(.8pt)+fontsize(8pt));
draw((0,0)--(0,25)--(25,25)--(25,0)--cycle);
fill((0,25)--(0,20)--(25,20)--(25,25)--cycle,gray);
draw((25,20)--(25,25)--(50,25)--(50,20)--cycle);
draw((0,5)--(25,5));
draw((0,10)--(25,10));
draw((0,15)--(25,15));
label("$A=E$",(0,25),W);
label("$B$",(50,25),E);
label("$C$",(50,20),E);
label("$D$",(0,20),W);
label("$F$",(25,25),NE);
label("$G$",(25,0),SE);
label("$H$",(0,0),SW);
// Block 2
unitsize(1mm); defaultpen(linewidth(.8pt)+fontsize(8pt)); draw((0,0)--(0,25)--(25,25)--(25,0)--cycle); fill((0,25)--(0,20)--(25,20)--(25,25)--cycle,gray); draw((25,20)--(25,25)--(50,25)--(50,20)--cycle); draw((0,5)--(25,5)); draw((0,10)--(25,10)); draw((0,15)--(25,15)); label("$A=E$",(0,25),W); label("$B$",(50,25),E); label("$C$",(50,20),E); label("$D$",(0,20),W); label("$F$",(25,25),NE); label("$G$",(25,0),SE); label("$H$",(0,0),SW); | [] |
699 | For how many integer values of $k$ do the graphs of $x^2+y^2=k^2$ and $xy = k$ not intersect?
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 8$ | 2010 AMC 12A Problem 13 | The image below shows the two curves for $k=4$. The blue curve is $x^2+y^2=k^2$, which is clearly a circle with radius $k$, and the red curve is a part of the curve $xy=k$.
In the special case $k=0$ the blue curve is just the point $(0,0)$, and as $0\cdot 0=0$, this point is on the red curve as well, hence they intersect.
The case $k<0$ is symmetric to $k>0$: the blue curve remains the same and the red curve is flipped according to the $x$ axis. Hence we just need to focus on $k>0$.
Clearly, on the red curve there will always be points arbitrarily far from the origin: for example, as $x$ approaches 0, $y$ approaches $\infty$. Hence the red curve intersects the blue one if and only if it contains a point whose distance from the origin is at most $k$.
At this point we can guess that on the red curve the point where $x=y$ is always closest to the origin, and skip the rest of this solution.
For an exact solution, fix $k$ and consider any point $(x,y)$ on the red curve. Its distance from the origin is $\sqrt{ x^2 + (k/x)^2 }$. To minimize this distance, it is enough to minimize $x^2 + (k/x)^2$. By the Arithmetic Mean-Geometric Mean Inequality we get that this value is at least $2k$, and that equality holds whenever $x^2 = (k/x)^2$, i.e., $x=\pm\sqrt k$.
Now recall that the red curve intersects the blue one if and only if its closest point is at most $k$ from the origin. We just computed that the distance between the origin and the closest point on the red curve is $\sqrt{2k}$. Therefore, we want to find all positive integers $k$ such that $\sqrt{2k} > k$.
Clearly the only such integer is $k=1$, hence the two curves are only disjoint for $k=1$ and $k=-1$.
This is a total of $\boxed{2\ \textbf{(C)}}$ values. | // Block 1
import graph;
size(200);
real f(real x) {return 4/x;};
real g1(real x) {return sqrt(4*4-x*x);};
real g2(real x) {return -sqrt(4*4-x*x);};
draw(graph(f,-20./3,-0.6),red);
draw(graph(f,0.6,20./3),red);
draw(graph(g1,-4,4),blue);
draw(graph(g2,-4,4),blue);
axes("$x$","$y$");
// Block 2
import graph; size(200); real f(real x) {return 4/x;}; real g1(real x) {return sqrt(4*4-x*x);}; real g2(real x) {return -sqrt(4*4-x*x);}; draw(graph(f,-20./3,-0.6),red); draw(graph(f,0.6,20./3),red); draw(graph(g1,-4,4),blue); draw(graph(g2,-4,4),blue); axes("$x$","$y$"); | [] |
700 | Equiangular hexagon $ABCDEF$ has side lengths $AB=CD=EF=1$ and $BC=DE=FA=r$. The area of $\triangle ACE$ is $70\%$ of the area of the hexagon. What is the sum of all possible values of $r$?
$\textbf{(A)}\ \frac{4\sqrt{3}}{3} \qquad \textbf{(B)} \frac{10}{3} \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ \frac{17}{4} \qquad \textbf{(E)}\ 6$ | 2010 AMC 12A Problem 17 | Drawing not to scale.*
To find r, we'll form an equation by finding the area of hexagon $ABCDEF$ and $\triangle{ACE}$ separately. First, connect a directly opposite diagonal such as $CF$. Because the hexagon is equiangular, edges $AF$ and $BC$ protrude from points $A$ and $B$ at the same angle but in opposite directions and equal length. This places points $C$ and $F$ at the same "height" relative to segment $AB$, so segment $CF$ is parallel to $AB$ and also $DE$ (opposite sides are parallel since you rotated through three $120^\circ$ angles between each other).
Next, we'll calculate the area of isosceles trapezoids $ABCF$ and $CDEF$. Drop a perpendicular down from $B$ to $CF$, and call the intersection to $CF$, $P$. Because adjacent angles between parallel sides in a trapezoid sum to $180^\circ$ and $\angle ABC = 120^\circ$, $\angle BCP = 60^\circ$ meaning $\triangle{BPC}$ is a 30-60-90 right triangle. Now we get $BP = \frac{\sqrt{3}}{2}r$ and $CP = \frac{r}{2}$. $CF$ is equal to $AB + 2CP = r+1$. Hence, the area of trapezoid $ABCF$ is $\frac{1}{2} \cdot \frac{\sqrt{3}}{2}r \cdot (r+1+1) = \frac{\sqrt{3}}{4} r(r+2)$. Similarly, on trapezoid $CDEF$ we find $\triangle EQF$ is 30-60-90, and $EQ = \frac{\sqrt{3}}{2}, FQ = \frac{1}{2}$. Then, area of $CDEF$ is $\frac{\sqrt{3}}{4} (2r + 1)$. Thus, the area of the hexagon is the sum of the two areas, $\frac{\sqrt{3}}{4} (r^2 + 4r + 1)$.
Next, notice that $\triangle ACE$ is equilateral. And by Pythagorean theorem in $\triangle CQE$, side $CE = \sqrt{r^2 + r + 1}$. Applying area of equilateral triangle formula, area $\triangle ACE = \frac{\sqrt{3}}{4} (r^2 + r + 1)$. Using this area and the area given from 70% of the area of the hexagon, we get the equation $\frac{7}{10} \cdot \frac{\sqrt{3}}{4} (r^2 + 4r + 1) = \frac{\sqrt{3}}{4} (r^2 + r + 1)$. Simplifying, we get the quadratic $r^2-6r+1=0$, which by Vieta's yields the answer $6, \boxed{\textbf{E}}$.
~henry
Proof Triangle ACE is Equilateral.
We know $\triangle{ABC}$, $\triangle{CDE}$, and $\triangle{EFA}$ are congruent by SAS, so the side opposite the 120 degree angle is also the same (since the triangles are congruent). Thus $\triangle{ACE}$ is equilateral.
Q.E.D.
~mathboy282 | // Block 1
import graph;
size(8cm);
pen dps = fontsize(10); defaultpen(dps);
real r = 0.7;
// Define hexagon vertices clockwise with AB on top
pair A = (0, 0);
pair B = (1, 0);
pair C = B + r * dir(-60);
pair D = C + dir(-120);
pair E = D + r * dir(180);
pair F = E + dir(120);
// Draw the hexagon
draw(A--B--C--D--E--F--cycle);
// Draw triangle ACE
draw(A--C--E--cycle);
// Draw line CF
draw(C--F);
// Drop perpendiculars from B and E to CF
pair foot_B = foot(B, C, F);
pair foot_E = foot(E, C, F);
// Draw perpendiculars (dashed)
draw(B--foot_B, dashed);
draw(E--foot_E, dashed);
// Right angle markers (smaller)
draw(rightanglemark(B, foot_B, C, 4));
draw(rightanglemark(E, foot_E, C, 4));
// Vertex labels
label("$A$", A, NW);
label("$B$", B, NE);
label("$C$", C, SE);
label("$D$", D, S);
label("$E$", E, SW);
label("$F$", F, NW);
// Feet labels
label("$P$", foot_B, S);
label("$Q$", foot_E, N);
// Side length labels
label("$1$", midpoint(A--B), N);
label("$r$", midpoint(B--C), dir(45));
label("$1$", midpoint(C--D), E);
label("$r$", midpoint(D--E), S);
label("$1$", midpoint(E--F), W);
label("$r$", midpoint(F--A), W);
// Block 2
import graph; size(8cm); pen dps = fontsize(10); defaultpen(dps); real r = 0.7; // Define hexagon vertices clockwise with AB on top pair A = (0, 0); pair B = (1, 0); pair C = B + r * dir(-60); pair D = C + dir(-120); pair E = D + r * dir(180); pair F = E + dir(120); // Draw the hexagon draw(A--B--C--D--E--F--cycle); // Draw triangle ACE draw(A--C--E--cycle); // Draw line CF draw(C--F); // Drop perpendiculars from B and E to CF pair foot_B = foot(B, C, F); pair foot_E = foot(E, C, F); // Draw perpendiculars (dashed) draw(B--foot_B, dashed); draw(E--foot_E, dashed); // Right angle markers (smaller) draw(rightanglemark(B, foot_B, C, 4)); draw(rightanglemark(E, foot_E, C, 4)); // Vertex labels label("$A$", A, NW); label("$B$", B, NE); label("$C$", C, SE); label("$D$", D, S); label("$E$", E, SW); label("$F$", F, NW); // Feet labels label("$P$", foot_B, S); label("$Q$", foot_E, N); // Side length labels label("$1$", midpoint(A--B), N); label("$r$", midpoint(B--C), dir(45)); label("$1$", midpoint(C--D), E); label("$r$", midpoint(D--E), S); label("$1$", midpoint(E--F), W); label("$r$", midpoint(F--A), W); | [] |
701 | A 16-step path is to go from $(-4,-4)$ to $(4,4)$ with each step increasing either the $x$-coordinate or the $y$-coordinate by 1. How many such paths stay outside or on the boundary of the square $-2 \le x \le 2$, $-2 \le y \le 2$ at each step?
$\textbf{(A)}\ 92 \qquad \textbf{(B)}\ 144 \qquad \textbf{(C)}\ 1568 \qquad \textbf{(D)}\ 1698 \qquad \textbf{(E)}\ 12,800$ | 2010 AMC 12A Problem 18 | We can draw an $8$ by $8$ square with a hollow $4$ by $4$ center. The ways to get a to a point or corner of a coordinate point is equal to the ways of getting to the point to the left of the desired point and the bottom of the desired point, since we can only move right and up. Using basic addition and symmetry (along the $y=x$) to speed things up, in the end, you sum up $849$ and $849,$ giving us our answer of $\boxed{1698}.$ | // Block 1
// Set up the Asymptote units
unitsize(1cm);
// Draw the 8x8 outer grid
for (int i = 0; i <= 8; ++i) {
// Vertical lines of the grid
draw((i, 0)--(i, 8));
// Horizontal lines of the grid
draw((0, i)--(8, i));
}
// Draw the hollow 4x4 center (white rectangle over the center part)
filldraw(box((2, 2), (6, 6)), white, white); // Hollow center (4x4 region)
// Block 2
// Set up the Asymptote units unitsize(1cm); // Draw the 8x8 outer grid for (int i = 0; i <= 8; ++i) { // Vertical lines of the grid draw((i, 0)--(i, 8)); // Horizontal lines of the grid draw((0, i)--(8, i)); } // Draw the hollow 4x4 center (white rectangle over the center part) filldraw(box((2, 2), (6, 6)), white, white); // Hollow center (4x4 region) | [] |
701 | A 16-step path is to go from $(-4,-4)$ to $(4,4)$ with each step increasing either the $x$-coordinate or the $y$-coordinate by 1. How many such paths stay outside or on the boundary of the square $-2 \le x \le 2$, $-2 \le y \le 2$ at each step?
$\textbf{(A)}\ 92 \qquad \textbf{(B)}\ 144 \qquad \textbf{(C)}\ 1568 \qquad \textbf{(D)}\ 1698 \qquad \textbf{(E)}\ 12,800$ | 2010 AMC 12A Problem 18 | Let $N(A,C,B)$ denote the number of paths from $A$ to $C$ then to $B$, $N(A,D,B)$ denote the number of paths from $A$ to $D$ then to $B$, $N(A,E,B)$ denote the number of paths from $A$ to $E$ then to $B$, $N(A,C,D, B)$ denote the number of paths from $A$ to $C$ to $D$ then to $B$, $N(A,C,D,E,B)$ denote the number of paths from $A$ to $C$ to $D$ to $E$ then to $B$, also notice we have $N(A,C,E,B)=N(A,C,D,E,B)$.
then the answer = $N(A,C,B)+N(A,D,B)+N(A,E,B)-N(A,C,D,B)-N(A,C,E,B)-N(A,D,E,B)+N(A,C,D,E,B)=N(A,C,B)+N(A,D,B)+N(A,E,B)-N(A,C,D,B)-N(A,D,E,B)$, $N(A,C,B)=N(A,C)\cdot N(C,B)={8\choose 2}\cdot {8\choose 2}$, $N(A, D, B)=N(A,D)\cdot N(D, B)={9\choose 2}\cdot {7\choose 1}$, $N(A,E, B)={10\choose 2}\cdot 1$, $N(A, C, D, B)={8\choose 2}\cdot {7\choose 1}, N(A, D, E, B)={9\choose 2}\cdot1,$ since if the path passes quadrant II, it will pass either $C$ or $D$ or $E$. Consider
the paths pass the quadrant IV together, we have ans=$2\times({8\choose 2}\cdot {8\choose 2}+{9\choose 2}\cdot {7\choose 1}+{10\choose 2}\cdot 1-{8\choose 2}\cdot {7\choose 1}-{9\choose 2}\cdot1)=2\times 849=1698$,$\boxed{(D)}$
~szhangmath | // Block 1
unitsize(1cm);
for (int i = 0; i <= 8; ++i) {
draw((i, 0)--(i, 8));
draw((0, i)--(8, i));
}
filldraw(box((2, 2), (6, 6)), white, white);
dot((0,0));
dot((2,6));
dot((2,7));
dot((2,8));
dot((8,8));
label("A",(0,0), align=W);
label("B",(8,8), align=E);
label("C",(2,6), align=NE);
label("D",(2,7), align=NE);
label("E",(2,8), align=NE);
// Block 2
unitsize(1cm); for (int i = 0; i <= 8; ++i) { draw((i, 0)--(i, 8)); draw((0, i)--(8, i)); } filldraw(box((2, 2), (6, 6)), white, white); dot((0,0)); dot((2,6)); dot((2,7)); dot((2,8)); dot((8,8)); label("A",(0,0), align=W); label("B",(8,8), align=E); label("C",(2,6), align=NE); label("D",(2,7), align=NE); label("E",(2,8), align=NE); | [] |
702 | A circle is centered at $O$, $\overline{AB}$ is a diameter and $C$ is a point on the circle with $\angle COB = 50^\circ$. What is the degree measure of $\angle CAB$?
$\textbf{(A)}\ 20 \qquad \textbf{(B)}\ 25 \qquad \textbf{(C)}\ 45 \qquad \textbf{(D)}\ 50 \qquad \textbf{(E)}\ 65$ | 2010 AMC 10B Problem 6 | Note that $\angle AOC = 180^\circ - 50^\circ = 130^\circ$. Because triangle $AOC$ is isosceles, $\angle CAB = (180^\circ - 130^\circ)/2 = \boxed{25^\circ}$. | // Block 1
import graph;
unitsize(2 cm);
pair O, A, B, C;
O = (0,0);
A = (-1,0);
B = (1,0);
C = dir(50);
draw(Circle(O,1));
draw(B--A--C--O);
label("$A$", A, W);
label("$B$", B, E);
label("$C$", C, NE);
label("$O$", O, S);
// Block 2
import graph; unitsize(2 cm); pair O, A, B, C; O = (0,0); A = (-1,0); B = (1,0); C = dir(50); draw(Circle(O,1)); draw(B--A--C--O); label("$A$", A, W); label("$B$", B, E); label("$C$", C, NE); label("$O$", O, S); | [] |
703 | A square of side length $1$ and a circle of radius $\dfrac{\sqrt{3}}{3}$ share the same center. What is the area inside the circle, but outside the square?
$\textbf{(A)}\ \dfrac{\pi}{3}-1 \qquad \textbf{(B)}\ \dfrac{2\pi}{9}-\dfrac{\sqrt{3}}{3} \qquad \textbf{(C)}\ \dfrac{\pi}{18} \qquad \textbf{(D)}\ \dfrac{1}{4} \qquad \textbf{(E)}\ \dfrac{2\pi}{9}$ | 2010 AMC 10B Problem 16 | The radius of the circle is $\frac{\sqrt{3}}{3} = \sqrt{\frac{1}{3}}$. Half the diagonal of the square is $\frac{\sqrt{1^2+1^2}}{2} = \frac{\sqrt{2}}{2} = \sqrt{\frac12}$. We can see that the circle passes outside the square, but the square is NOT completely contained in the circle. Therefore the picture will look something like this:
Then we proceed to find: 4 $\cdot$ (area of sector marked off by the two radii - area of the triangle with sides on the square and the two radii).
First we realize that the radius perpendicular to the side of the square between the two radii marking off the sector splits $AB$ in half. Let this half-length be $a$. Also note that $OX=\frac12$ because it is half the side length of the square. Because this is a right triangle, we can use the Pythagorean Theorem to solve for $a.$
\[a^2+\left( \frac12 \right) ^2 = \left( \frac{\sqrt{3}}{3} \right) ^2\]
Solving, $a= \frac{\sqrt{3}}{6}$ and $2a=\frac{\sqrt{3}}{3}$. Since $AB=AO=BO$, $\triangle AOB$ is an equilateral triangle and the central angle is $60^{\circ}$. Therefore the sector has an area $\pi \left( \frac{\sqrt{3}}{3} \right) ^2 \left( \frac{60}{360} \right) = \frac{\pi}{18}$.
Now we turn to the triangle. Since it is equilateral, we can use the formula for the area of an equilateral triangle which is
\[\frac{s^2\sqrt{3}}{4} = \frac{\frac13 \sqrt{3}}{4} = \frac{\sqrt{3}}{12}\]
Putting it together, we get the answer to be $4\cdot\left( \frac{\pi}{18}-\frac{\sqrt{3}}{12} \right)= \boxed{\textbf{(B)}\ \frac{2\pi}{9}-\frac{\sqrt{3}}{3}}$ | // Block 1
unitsize(5cm);
defaultpen(linewidth(.8pt)+fontsize(10pt));
dotfactor=3;
real r=sqrt(1/3);
pair O=(0,0);
pair W=(0.5,0.5), X=(0.5,-0.5), Y=(-0.5,-0.5), Z=(-0.5,0.5);
pair A=(-sqrt(1/12),0.5), B=(sqrt(1/12),0.5);
pair V=(0,0.5);
path outer=Circle(O,r);
draw(outer);
draw(W--X--Y--Z--cycle);
draw(O--A);
draw(O--B);
draw(V--O);
pair[] ps={A,B,V,O};
dot(ps);
label("$O$",O,SW);
label("$\frac{\sqrt{3}}{3}$",O--B,SE);
label("$A$",A,NW);
label("$B$",B,NE);
label("$X$",V,NW);
label("$a$",B--V,S);
label("$\frac12$",O--V,W);
// Block 2
unitsize(5cm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=3; real r=sqrt(1/3); pair O=(0,0); pair W=(0.5,0.5), X=(0.5,-0.5), Y=(-0.5,-0.5), Z=(-0.5,0.5); pair A=(-sqrt(1/12),0.5), B=(sqrt(1/12),0.5); pair V=(0,0.5); path outer=Circle(O,r); draw(outer); draw(W--X--Y--Z--cycle); draw(O--A); draw(O--B); draw(V--O); pair[] ps={A,B,V,O}; dot(ps); label("$O$",O,SW); label("$\frac{\sqrt{3}}{3}$",O--B,SE); label("$A$",A,NW); label("$B$",B,NE); label("$X$",V,NW); label("$a$",B--V,S); label("$\frac12$",O--V,W); | [] |
704 | A circle with center $O$ has area $156\pi$. Triangle $ABC$ is equilateral, $\overline{BC}$ is a chord on the circle, $OA = 4\sqrt{3}$, and point $O$ is outside $\triangle ABC$. What is the side length of $\triangle ABC$?
$\textbf{(A)}\ 2\sqrt{3} \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ 4\sqrt{3} \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 18$ | 2010 AMC 10B Problem 19 | The formula for the area of a circle is $\pi r^2$ so the radius of this circle is $\sqrt{156}.$
Because $OA=4\sqrt{3}=\sqrt{48} < \sqrt{156}, A$ must be in the interior of circle $O.$
Let $s$ be the side length of the triangle, the unknown value, and let $X$ be the point on $BC$ where $OX \perp BC.$ Since $\triangle ABC$ is equilateral, $BX=\frac{s}{2}$ and $AX=\frac{s\sqrt{3}}{2}.$ We are given $AO=4\sqrt{3}.$ Use the Pythagorean Theorem and solve for $s.$
\begin{align*} (\sqrt{156})^2 &= \left(\frac{s}{2}\right)^2 + \left( \frac{s\sqrt{3}}{2} + 4\sqrt{3} \right)^2\\ 156 &= \frac14s^2 + \frac34s^2 + 12s + 48\\ 0 &= s^2 + 12s - 108\\ 0 &= (s-6)(s+18)\\ s &= \boxed{\textbf{(B)}\ 6} \end{align*} | // Block 1
unitsize(3mm);
defaultpen(linewidth(.8pt)+fontsize(11pt));
dotfactor=3;
real r=sqrt(156);
pair A=(0,sqrt(48)), B=(-3,sqrt(147)), C=(3,sqrt(147));
pair O=(0,0);
pair X=(0,7sqrt(3));
path outer=Circle(O,r);
draw(outer);
draw(A--B--C--cycle);
draw(O--X); draw(O--B);
pair[] ps={A,B,C,O,X};
dot(ps);
label("$A$",A,SE);
label("$B$",B,NW);
label("$C$",C,NE);
label("$O$",O,S);
label("$X$",X,N);
label("$s$",A--C,SE);
label("$\frac{s}{2}$",B--X,N);
label("$\frac{s\sqrt{3}}{2}$",A--X,NE);
label("$\sqrt{156}$",O--B,SW);
label("$4\sqrt{3}$",A--O,E);
// Block 2
unitsize(3mm); defaultpen(linewidth(.8pt)+fontsize(11pt)); dotfactor=3; real r=sqrt(156); pair A=(0,sqrt(48)), B=(-3,sqrt(147)), C=(3,sqrt(147)); pair O=(0,0); pair X=(0,7sqrt(3)); path outer=Circle(O,r); draw(outer); draw(A--B--C--cycle); draw(O--X); draw(O--B); pair[] ps={A,B,C,O,X}; dot(ps); label("$A$",A,SE); label("$B$",B,NW); label("$C$",C,NE); label("$O$",O,S); label("$X$",X,N); label("$s$",A--C,SE); label("$\frac{s}{2}$",B--X,N); label("$\frac{s\sqrt{3}}{2}$",A--X,NE); label("$\sqrt{156}$",O--B,SW); label("$4\sqrt{3}$",A--O,E); | [] |
705 | Circles $A, B,$ and $C$ each has radius $1$. Circles $A$ and $B$ share one point of tangency. Circle $C$ has a point of tangency with the midpoint of $\overline{AB}.$ What is the area inside circle $C$ but outside circle $A$ and circle $B?$
$\textbf{(A)}\ 3 - \frac{\pi}{2} \qquad \textbf{(B)}\ \frac{\pi}{2} \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ \frac{3\pi}{4} \qquad \textbf{(E)}\ 1+\frac{\pi}{2}$ | 2011 AMC 10A Problem 18 | The requested area is the area of $C$ minus the area shared between circles $A$, $B$ and $C$.
Let $M$ be the midpoint of $\overline{AB}$ and $D$ be the other intersection of circles $C$ and $B$.
The area shared between $C$, $A$ and $B$ is $4$ of the regions between arc $\widehat {MD}$ and line $\overline{MD}$, which is (considering the arc on circle $B$) a quarter of the circle $B$ minus $\triangle MDB$:
$\frac{\pi r^2}{4}-\frac{bh}{2}$
$b = h = r = 1$
(We can assume this because $\angle DBM$ is 90 degrees, since $CDBM$ is a square, due to the application of the tangent chord theorem at point $M$)
So the area of the small region is
$\frac{\pi}{4}-\frac{1}{2}$
The requested area is area of circle $C$ minus 4 of this area:
$\pi 1^2 - 4\left(\frac{\pi}{4}-\frac{1}{2}\right) = \pi - \pi + 2 = 2$
$\boxed{\textbf{C}}$. | unitsize(1.1cm); defaultpen(linewidth(.8pt)); dotfactor=4; pair A=(0,0), B=(2,0), C=(1,-1); pair M=(1,0); pair D=(2,-1); dot (A); dot (B); dot (C); dot (D); dot (M); draw(Circle(A,1)); draw(Circle(B,1)); draw(Circle(C,1)); draw(A--B); draw(M--D); draw(D--B); label("$A$",A,W); label("$B$",B,E); label("$C$",C,W); label("$M$",M,NE); label("$D$",D,SE); | [] |
705 | Circles $A, B,$ and $C$ each has radius $1$. Circles $A$ and $B$ share one point of tangency. Circle $C$ has a point of tangency with the midpoint of $\overline{AB}.$ What is the area inside circle $C$ but outside circle $A$ and circle $B?$
$\textbf{(A)}\ 3 - \frac{\pi}{2} \qquad \textbf{(B)}\ \frac{\pi}{2} \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ \frac{3\pi}{4} \qquad \textbf{(E)}\ 1+\frac{\pi}{2}$ | 2011 AMC 10A Problem 18 | We can move the area above the part of the circle above the segment $EF$ down, and similarly for the other side. Then, we have a square, whose diagonal is $2$, so the area is then just $\left(\frac{2}{\sqrt{2}}\right)^2 = \boxed{\textbf{2 = C}}$.
~ Minor Edits, Challengees24 | unitsize(1.1cm); defaultpen(linewidth(.8pt)); dotfactor=4; pair A=(0,0), B=(2,0), C=(1,-1); pair M=(1,0); pair D=(2,-1); dot (A); dot (B); dot (C); dot (D); dot (M); draw(Circle(A,1)); draw(Circle(B,1)); draw(Circle(C,1)); draw(A--B); draw(M--D); draw(D--B); label("$A$",A,W); label("$B$",B,E); label("$C$",C,W); label("$M$",M,NE); label("$D$",D,SE); | [] |
706 | Circles $A, B,$ and $C$ each has radius $1$. Circles $A$ and $B$ share one point of tangency. Circle $C$ has a point of tangency with the midpoint of $\overline{AB}.$ What is the area inside circle $C$ but outside circle $A$ and circle $B?$
$\textbf{(A)}\ 3 - \frac{\pi}{2} \qquad \textbf{(B)}\ \frac{\pi}{2} \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ \frac{3\pi}{4} \qquad \textbf{(E)}\ 1+\frac{\pi}{2}$ | 2011 AMC 12A Problem 11 | The requested area is the area of $C$ minus the area shared between circles $A$, $B$ and $C$.
Let $M$ be the midpoint of $\overline{AB}$ and $D$ be the other intersection of circles $C$ and $B$.
The area shared between $C$, $A$ and $B$ is $4$ of the regions between arc $\widehat {MD}$ and line $\overline{MD}$, which is (considering the arc on circle $B$) a quarter of the circle $B$ minus $\triangle MDB$:
$\frac{\pi r^2}{4}-\frac{bh}{2}$
$b = h = r = 1$
(We can assume this because $\angle DBM$ is 90 degrees, since $CDBM$ is a square, due to the application of the tangent chord theorem at point $M$)
So the area of the small region is
$\frac{\pi}{4}-\frac{1}{2}$
The requested area is area of circle $C$ minus 4 of this area:
$\pi 1^2 - 4\left(\frac{\pi}{4}-\frac{1}{2}\right) = \pi - \pi + 2 = 2$
$\boxed{\textbf{C}}$. | // Block 1
unitsize(1.1cm);
defaultpen(linewidth(.8pt));
dotfactor=4;
pair A=(0,0), B=(2,0), C=(1,-1);
pair M=(1,0);
pair D=(2,-1);
dot (A);
dot (B);
dot (C);
dot (D);
dot (M);
draw(Circle(A,1));
draw(Circle(B,1));
draw(Circle(C,1));
draw(A--B);
draw(M--D);
draw(D--B);
label("$A$",A,W);
label("$B$",B,E);
label("$C$",C,W);
label("$M$",M,NE);
label("$D$",D,SE);
// Block 2
unitsize(1.1cm); defaultpen(linewidth(.8pt)); dotfactor=4; pair A=(0,0), B=(2,0), C=(1,-1); pair M=(1,0); pair D=(2,-1); dot (A); dot (B); dot (C); dot (D); dot (M); draw(Circle(A,1)); draw(Circle(B,1)); draw(Circle(C,1)); draw(A--B); draw(M--D); draw(D--B); label("$A$",A,W); label("$B$",B,E); label("$C$",C,W); label("$M$",M,NE); label("$D$",D,SE); | [] |
706 | Circles $A, B,$ and $C$ each has radius $1$. Circles $A$ and $B$ share one point of tangency. Circle $C$ has a point of tangency with the midpoint of $\overline{AB}.$ What is the area inside circle $C$ but outside circle $A$ and circle $B?$
$\textbf{(A)}\ 3 - \frac{\pi}{2} \qquad \textbf{(B)}\ \frac{\pi}{2} \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ \frac{3\pi}{4} \qquad \textbf{(E)}\ 1+\frac{\pi}{2}$ | 2011 AMC 12A Problem 11 | We can move the area above the part of the circle above the segment $EF$ down, and similarly for the other side. Then, we have a square, whose diagonal is $2$, so the area is then just $\left(\frac{2}{\sqrt{2}}\right)^2 = \boxed{\textbf{2 = C}}$.
~ Minor Edits, Challengees24 | // Block 1
unitsize(1.1cm);
defaultpen(linewidth(.8pt));
dotfactor=4;
pair A=(0,0), B=(2,0), C=(1,-1);
pair M=(1,0);
pair D=(2,-1);
dot (A);
dot (B);
dot (C);
dot (D);
dot (M);
draw(Circle(A,1));
draw(Circle(B,1));
draw(Circle(C,1));
draw(A--B);
draw(M--D);
draw(D--B);
label("$A$",A,W);
label("$B$",B,E);
label("$C$",C,W);
label("$M$",M,NE);
label("$D$",D,SE);
// Block 2
unitsize(1.1cm); defaultpen(linewidth(.8pt)); dotfactor=4; pair A=(0,0), B=(2,0), C=(1,-1); pair M=(1,0); pair D=(2,-1); dot (A); dot (B); dot (C); dot (D); dot (M); draw(Circle(A,1)); draw(Circle(B,1)); draw(Circle(C,1)); draw(A--B); draw(M--D); draw(D--B); label("$A$",A,W); label("$B$",B,E); label("$C$",C,W); label("$M$",M,NE); label("$D$",D,SE); | [] |
707 | A power boat and a raft both left dock $A$ on a river and headed downstream. The raft drifted at the speed of the river current. The power boat maintained a constant speed with respect to the river. The power boat reached dock $B$ downriver, then immediately turned and traveled back upriver. It eventually met the raft on the river 9 hours after leaving dock $A.$ How many hours did it take the power boat to go from $A$ to $B$?
$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 3.5 \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ 4.5 \qquad \textbf{(E)}\ 5$ | 2011 AMC 12A Problem 12 | Solution 1
WLOG, let the speed of the river be 0. This is allowed because the problem never states that the speed of the current has to have a magnitude greater than 0. In this case, when the powerboat travels from $A$ to $B$, the raft remains at $A$. Thus the trip from $A$ to $B$ takes the same time as the trip from $B$ to the raft. Since these times are equal and sum to $9$ hours, the trip from $A$ to $B$ must take half this time, or $4.5$ hours. The answer is thus $\boxed{\textbf{D}}$.
Remark: This is equivalent to viewing the problem from the raft's perspective.
Solution 2
What's important in this problem is to consider everything in terms of the power boat and the raft, since that is how the problem is given to us. Think of the blue arrow as the power boat and the red arrow as the raft in the following three diagrams, which represent different time intervals of the problem.
Thinking about the distance covered as their distances with respect to each other, they are $0$ distance apart in the first diagram when they haven't started to move yet, some distance $d$ apart in the second diagram when the power boat reaches $B$, and again $0$ distance apart in the third diagram when they meet. Therefore, with respect to each other, the boat and the raft cover a distance of $d$ on the way there, and again cover a distance of $d$ on when drawing closer. This makes sense, because from the 1st diagram to the second, the raft moves in the same direction as the boat, while from the 2nd to the 3rd, the boat and raft move in opposite directions.
Let $b$ denote the speed of the power boat (only the power boat, not factoring in current) and $r$ denote the speed of the raft, which, as given by the problem, is also equal to the speed of the current. Thus, from $A$ to $B$, the boat travels at a velocity of $b+r$, and on the way back, travels at a velocity of $-(b-r)=r-b$, since the current aids the boat on the way there, and goes against the boat on the way back. With respect to the raft then, the boat's velocity from $A$ to $B$ becomes $(r+b)-r=b$, and on the way back it becomes $(r-b)-r=-b$. Since the boat's velocities with respect to the raft are exact opposites, $b$ and $-b$, we therefore know that the boat and raft travel apart from each other at the same rate that they travel toward each other.
From this, we have that the boat travels a distance $d$ at rate $b$ with respect to the raft both on the way to $B$ and on the way back. Thus, using $\dfrac{distance}{speed}=time$, we have $\dfrac{2d}{b}=9\text{ hours}$, and to see how long it took to travel half the distance, we have $\dfrac{d}{b}=4.5\text{ hours}\implies\boxed{\textbf{D}}$
Solution 3
Let $t$ be the time it takes the power boat to go from $A$ to $B$ in hours, $r$ be the speed of the river current (and thus also the raft), and $p$ to be the speed of the power boat with respect to the river.
Using $d = rt$, the raft covers a distance of $9r$, the distance from $A$ to $B$ is $(p + r)t$, and the distance from $B$ to where the raft and power boat met up is $(9 - t)(p - r)$.
Then, $9r + (9 - t)(p - r) = (p + r)t$. Solving for $t$, we get $t = 4.5$, which is $\boxed{\textbf{D}}$. | size(8cm,8cm); pair A, B; A=(-3,4); B=(3,4); draw(A--B); label("$A$",A,S); label("$B$",B,S); arrow((-2.5,6.4),dir(180),blue); arrow((-2.5,6.2),dir(180),red); pair A, B; A=(-3,5); B=(3,5); draw(A--B); label("$A$",A,S); label("$B$",B,S); arrow((2.6,5.4),dir(360),blue); arrow((-0.5,5.2),dir(180),red); pair A, B; A=(-3,6); B=(3,6); draw(A--B); label("$A$",A,S); label("$B$",B,S); arrow((.3,4.4),dir(360),blue); arrow((1.1,4.2),dir(180),red); | [] |
708 | Circles with radii $1$, $2$, and $3$ are mutually externally tangent. What is the area of the triangle determined by the points of tangency?
$\textbf{(A)}\ \frac{3}{5} \qquad \textbf{(B)}\ \frac{4}{5} \qquad \textbf{(C)}\ 1 \qquad \textbf{(D)}\ \frac{6}{5} \qquad \textbf{(E)}\ \frac{4}{3}$ | 2011 AMC 12A Problem 17 | The centers of these circles form a 3-4-5 triangle, which has an area equal to 6.
The areas of the three triangles determined by the center and the two points of tangency of each circle are, using Triangle Area by Sine,
$\frac{1}{2} \cdot 1 \cdot 1 \cdot 1 = \frac{1}{2}$
$\frac{1}{2} \cdot 2 \cdot 2 \cdot \frac{4}{5} = \frac{8}{5}$
$\frac{1}{2} \cdot 3 \cdot 3 \cdot \frac{3}{5} = \frac{27}{10}$
which add up to $4.8$. The area we're looking for is the large 3-4-5 triangle minus the three smaller triangles, or $6 - 4.8 = 1.2 = \frac{6}{5} \rightarrow \boxed{(D)}$. | // Block 1
unitsize(.5cm);
defaultpen(linewidth(.8pt));
dotfactor=4;
pair A=(0,0), B=(3,0), C=(0,4);
dot (A);
dot (B);
dot (C);
draw(A--B);
draw(A--C);
draw(B--C);
draw(Circle(A,1));
draw(Circle(B,2));
draw(Circle(C,3));
// Block 2
unitsize(.5cm); defaultpen(linewidth(.8pt)); dotfactor=4; pair A=(0,0), B=(3,0), C=(0,4); dot (A); dot (B); dot (C); draw(A--B); draw(A--C); draw(B--C); draw(Circle(A,1)); draw(Circle(B,2)); draw(Circle(C,3)); | [] |
709 | Triangle $ABC$ has $\angle BAC = 60^{\circ}$, $\angle CBA \leq 90^{\circ}$, $BC=1$, and $AC \geq AB$. Let $H$, $I$, and $O$ be the orthocenter, incenter, and circumcenter of $\triangle ABC$, respectively. Assume that the area of pentagon $BCOIH$ is the maximum possible. What is $\angle CBA$?
$\textbf{(A)}\ 60^{\circ} \qquad \textbf{(B)}\ 72^{\circ} \qquad \textbf{(C)}\ 75^{\circ} \qquad \textbf{(D)}\ 80^{\circ} \qquad \textbf{(E)}\ 90^{\circ}$ | 2011 AMC 12A Problem 25 | By the Inscribed Angle Theorem, \[\angle BOC = 2\angle BAC = 120^\circ .\]Let $D$ and $E$ be the feet of the altitudes of $\triangle ABC$ from $B$ and $C$, respectively. In $\triangle ACE$ we get $\angle ACE = 30^\circ$, and as exterior angle \[\angle BHC = 90^\circ + \angle ACE = 120^\circ .\]Because the lines $BI$ and $CI$ are bisectors of $\angle CBA$ and $\angle ACB$, respectively, it follows that\[\angle BIC = 90^\circ + \tfrac 12\angle A = 120^\circ .\]Thus the points $B, C, O, I$, and $H$ are all on a circle.
Further, since
\[\angle OCI = \angle OCB - \angle ICB = 30^\circ - \tfrac 12\angle C\]
\[\angle ICH = \angle ACE - \angle ACI = 30^\circ - \tfrac 12\angle C\]
we have $OI=IH$.
Because $[BCOIH]=[BCO]+[BOIH]$, it is sufficient to maximize the area of quadrilateral $BOIH$. If $P_1$, $P_2$ are two points in an arc of circle $BO$ with $BP_1<BP_2$, then the maximum area of $BOP_1P_2$ occurs when $BP_1=P_1P_2=P_2O$. Indeed, if $BP_1\neq P_1P_2$, then replacing $P_1$ by the point $P_1’$ located halfway in the arc of the circle $BP_2$ yields a triangle $BP_1’P_2$ with larger area than $\triangle BP_1P_2$, and the area of $\triangle BOP_2$ remains the same. Similarly, if $P_1P_2\neq P_2O$.
Therefore the maximum is achieved when $OI=IH=HB$, that is, when \[\angle OCI = \angle ICH = \angle HCB = \tfrac 13 \angle OCB = 10^\circ.\]Thus $\angle ACB = 40^\circ$ and $\angle CBA = 80^\circ$. | // Block 1
import geometry;
size(200);
defaultpen(fontsize(12)+0.8);
pair O,A,B,C,D,E,I,H;
real h=2*sqrt(3);
O=(0,1/h); B=(-0.5,0); C=(0.5,0);
path c1=CR(O,length(O-B));
// A=IP(c1,O--O+5*dir(120));
A=IP(c1,B--B+5*dir(80));
I=incenter(A,B,C);
H=orthocenter(A,B,C);
D=extension(A,C,B,H); E=extension(A,B,C,H);
path c2=circumcircle(O,I,H);
pair o2=circumcenter(O,I,H);
draw(c1, royalblue); draw(A--B--C--A); draw(B--D^^C--E, dotted);
draw(arc(o2,length(O-o2),10,170), dotted+red); draw(B--I--C--O--B, black+0.3);
pen p =black+3.25;
dot("$O$", O, N,p); dot("$A$", A, dir(110),p); dot("$B$", B, dir(210),p); dot("$C$", C, dir(-30),p); dot("$I$", I, N,p); dot("$H$", H, N,p); dot("$D$", D, D-H,p); dot("$E$", E, (E-C),p);
// Block 2
import geometry; size(200); defaultpen(fontsize(12)+0.8); pair O,A,B,C,D,E,I,H; real h=2*sqrt(3); O=(0,1/h); B=(-0.5,0); C=(0.5,0); path c1=CR(O,length(O-B)); // A=IP(c1,O--O+5*dir(120)); A=IP(c1,B--B+5*dir(80)); I=incenter(A,B,C); H=orthocenter(A,B,C); D=extension(A,C,B,H); E=extension(A,B,C,H); path c2=circumcircle(O,I,H); pair o2=circumcenter(O,I,H); draw(c1, royalblue); draw(A--B--C--A); draw(B--D^^C--E, dotted); draw(arc(o2,length(O-o2),10,170), dotted+red); draw(B--I--C--O--B, black+0.3); pen p =black+3.25; dot("$O$", O, N,p); dot("$A$", A, dir(110),p); dot("$B$", B, dir(210),p); dot("$C$", C, dir(-30),p); dot("$I$", I, N,p); dot("$H$", H, N,p); dot("$D$", D, D-H,p); dot("$E$", E, (E-C),p); | [] |
710 | A dart board is a regular octagon divided into regions as shown. Suppose that a dart thrown at the board is equally likely to land anywhere on the board. What is the probability that the dart lands within the center square?
$\textbf{(A)}\ \frac{\sqrt{2} - 1}{2} \qquad\textbf{(B)}\ \frac{1}{4} \qquad\textbf{(C)}\ \frac{2 - \sqrt{2}}{2} \qquad\textbf{(D)}\ \frac{\sqrt{2}}{4} \qquad\textbf{(E)}\ 2 - \sqrt{2}$ | 2011 AMC 10B Problem 16 | If the side lengths of the dart board and the side lengths of the center square are all $\sqrt{2},$ then the side length of the legs of the triangles are $1$.
\begin{align*} \text{area of center square} &: \sqrt{2} \times \sqrt{2} = 2\\ \text{total area} &: (\sqrt{2})^2 + 4(1 \times \sqrt{2}) + 4(1 \times 1 \times \frac{1}{2}) = 2 + 4\sqrt{2} + 2 = 4 + 4\sqrt{2} \end{align*}
Use Geometric probability by putting the area of the desired region over the area of the entire region.
\[\frac{2}{4+4\sqrt{2}} = \frac{1}{2+2\sqrt{2}} \times \frac{2-2\sqrt{2}}{2-2\sqrt{2}} = \frac{2-2\sqrt{2}}{-4} = \boxed{\textbf{(A)} \frac{\sqrt{2}-1}{2}}\]
Explanation:
The area of the octagon consists of the area of the triangles, the rectangles, and the square in the middle. Assume the octagon has side length $1$. The triangles are right isosceles triangles with the hypotenuse 1, so their side length is $\frac{\sqrt{2}}{2}$ and the area of one triangle is $\frac{\sqrt{2}}{2}\cdot\frac{\sqrt{2}}{2}\cdot\frac{1}{2}=\frac{1}{4}$. The area of all 4 triangles is then just $1$. The rectangles share one side with the octagon, and another with the triangle. The octagon has side length 1, and the triangle has side length $\frac{\sqrt{2}}{2}$ as found earlier. So the area of one rectangle is $\frac{\sqrt{2}}{2}$. The area of all 4 is $4\cdot\frac{\sqrt{2}}{2}=2\sqrt{2}$. Finally, the area of the square in the middle is $1\cdot1=1$. The total area is $1+1+2\sqrt{2}=2+2\sqrt{2}$. We want the area of the square over the area of the octagon, which is $\frac{1}{2+2\sqrt{2}}$. Rationalize by multiplying both numerator and denominator by $2-2\sqrt{2}$: $\frac{1}{2+2\sqrt{2}}\cdot\frac{2-2\sqrt{2}}{2-2\sqrt{2}}=\frac{2-2\sqrt{2}}{\left(2+2\sqrt{2}\right)\left(2-2\sqrt{2}\right)}$. By the difference of squares, the denominator reduces to $-4$ and the fraction is $\frac{2-2\sqrt{2}}{-4}=\frac{\sqrt{2}-1}{2}$ which is $\boxed{\textbf{(A) } \frac{\sqrt{2}-1}{2}}$.
~Explanation by JH. L | // Block 1
unitsize(10mm);
defaultpen(linewidth(.8pt)+fontsize(10pt));
dotfactor=1;
pair A=(0,1), B=(1,0), C=(1+sqrt(2),0), D=(2+sqrt(2),1), E=(2+sqrt(2),1+sqrt(2)), F=(1+sqrt(2),2+sqrt(2)), G=(1,2+sqrt(2)), H=(0,1+sqrt(2));
pair I=(1,1), J=(1+sqrt(2),1), K=(1+sqrt(2),1+sqrt(2)), L=(1,1+sqrt(2));
draw(A--B--C--D--E--F--G--H--cycle);
draw(A--D);
draw(B--G);
draw(C--F);
draw(E--H);
pair[] ps={A,B,C,D,E,F,G,H,I,J,K,L};
dot(ps);
label("$A$",A,W);
label("$B$",B,S);
label("$C$",C,S);
label("$D$",D,E);
label("$E$",E,E);
label("$F$",F,N);
label("$G$",G,N);
label("$H$",H,W);
label("$I$",I,NE);
label("$J$",J,NW);
label("$K$",K,SW);
label("$L$",L,SE);
label("$\sqrt{2}$",midpoint(B--C),S);
label("$1$",midpoint(A--I),N);
// Block 2
unitsize(10mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=1; pair A=(0,1), B=(1,0), C=(1+sqrt(2),0), D=(2+sqrt(2),1), E=(2+sqrt(2),1+sqrt(2)), F=(1+sqrt(2),2+sqrt(2)), G=(1,2+sqrt(2)), H=(0,1+sqrt(2)); pair I=(1,1), J=(1+sqrt(2),1), K=(1+sqrt(2),1+sqrt(2)), L=(1,1+sqrt(2)); draw(A--B--C--D--E--F--G--H--cycle); draw(A--D); draw(B--G); draw(C--F); draw(E--H); pair[] ps={A,B,C,D,E,F,G,H,I,J,K,L}; dot(ps); label("$A$",A,W); label("$B$",B,S); label("$C$",C,S); label("$D$",D,E); label("$E$",E,E); label("$F$",F,N); label("$G$",G,N); label("$H$",H,W); label("$I$",I,NE); label("$J$",J,NW); label("$K$",K,SW); label("$L$",L,SE); label("$\sqrt{2}$",midpoint(B--C),S); label("$1$",midpoint(A--I),N); | [] |
711 | Rectangle $ABCD$ has $AB = 6$ and $BC = 3$. Point $M$ is chosen on side $AB$ so that $\angle AMD = \angle CMD$. What is the degree measure of $\angle AMD$?
$\textbf{(A)}\ 15 \qquad\textbf{(B)}\ 30 \qquad\textbf{(C)}\ 45 \qquad\textbf{(D)}\ 60 \qquad\textbf{(E)}\ 75$ | 2011 AMC 10B Problem 18 | It is given that $\angle AMD \sim \angle CMD$. Since $\angle AMD$ and $\angle CDM$ are alternate interior angles and $\overline{AB} \parallel \overline{DC}$, $\angle AMD \cong \angle CDM \longrightarrow \angle CMD \cong \angle CDM$. Use the Base Angle Theorem to show $\overline{DC} \cong \overline{MC}$. We know that $ABCD$ is a rectangle, so it follows that $\overline{MC} = 6$. We notice that $\triangle BMC$ is a $30-60-90$ triangle, and $\angle BMC = 30^{\circ}$. If we let $x$ be the measure of $\angle AMD,$ then
\begin{align*} 2x + 30 &= 180\\ 2x &= 150\\ x &= \boxed{\textbf{(E)} 75} \end{align*}
Easier Way to Continue
After finding $MC = 6,$ we can continue using trigonometry as follows.
We know that $\angle{BMC} = 180-2x$ and so $\sin (180-2x) = \frac{3}{6} = \frac{1}{2}$
It is obvious that $\sin (30) = \frac{1}{2}$ and so $180-2x=30.$
Solving, we have $x = \boxed{75}$
~mathboy282 | // Block 1
unitsize(10mm);
defaultpen(linewidth(.5pt)+fontsize(10pt));
dotfactor=3;
pair A=(0,3), B=(6,3), C=(6,0), D=(0,0);
pair M=(0.80385,3);
draw(A--B--C--D--cycle);
draw(M--C);
draw(M--D);
draw(anglemark(A,M,D));
draw(anglemark(D,M,C));
pair[] ps={A,B,C,D,M};
dot(ps);
label("$A$",A,NW);
label("$B$",B,NE);
label("$C$",C,SE);
label("$D$",D,SW);
label("$M$",M,N);
label("$6$",midpoint(C--M),SW);
label("$3$",midpoint(B--C),E);
label("$6$",midpoint(C--D),S);
// Block 2
unitsize(10mm); defaultpen(linewidth(.5pt)+fontsize(10pt)); dotfactor=3; pair A=(0,3), B=(6,3), C=(6,0), D=(0,0); pair M=(0.80385,3); draw(A--B--C--D--cycle); draw(M--C); draw(M--D); draw(anglemark(A,M,D)); draw(anglemark(D,M,C)); pair[] ps={A,B,C,D,M}; dot(ps); label("$A$",A,NW); label("$B$",B,NE); label("$C$",C,SE); label("$D$",D,SW); label("$M$",M,N); label("$6$",midpoint(C--M),SW); label("$3$",midpoint(B--C),E); label("$6$",midpoint(C--D),S); | [] |
712 | Rhombus $ABCD$ has side length $2$ and $\angle B = 120$°. Region $R$ consists of all points inside the rhombus that are closer to vertex $B$ than any of the other three vertices. What is the area of $R$?
$\textbf{(A)}\ \frac{\sqrt{3}}{3} \qquad\textbf{(B)}\ \frac{\sqrt{3}}{2} \qquad\textbf{(C)}\ \frac{2\sqrt{3}}{3} \qquad\textbf{(D)}\ 1 + \frac{\sqrt{3}}{3} \qquad\textbf{(E)}\ 2$ | 2011 AMC 10B Problem 20 | Suppose that $P$ is a point in the rhombus $ABCD$ and let $\ell_{BC}$ be the perpendicular bisector of $\overline{BC}$. Then $PB < PC$ if and only if $P$ is on the same side of $\ell_{BC}$ as $B$. The line $\ell_{BC}$ divides the plane into two half-planes; let $S_{BC}$ be the half-plane containing $B$. Let us define similarly $\ell_{BD},S_{BD}$ and $\ell_{BA},S_{BA}$. Then $R$ is equal to $ABCD \cap S_{BC} \cap S_{BD} \cap S_{BA}$. The region turns out to be an irregular pentagon. We can make it easier to find the area of this region by dividing it into four triangles:
Since $\triangle BCD$ and $\triangle BAD$ are equilateral, $\ell_{BC}$ contains $D$, $\ell_{BD}$ contains $A$ and $C$, and $\ell_{BA}$ contains $D$. Then $\triangle BEF \cong \triangle BGF \cong \triangle BGH \cong \triangle BIH$ with $BE = 1$ and $EF = \frac{1}{\sqrt{3}}$ so $[BEF] = \frac{1}{2}\cdot 1 \cdot \frac{\sqrt{3}}{3}$. Multiply this by 4 and it turns out that the pentagon has area $\boxed{(C)\frac{2\sqrt{3}}{3}}$. | // Block 1
unitsize(8mm);
defaultpen(linewidth(0.8pt)+fontsize(10pt));
dotfactor=4;
pair A=(4,0), B=(2,2sqrt(3)), C=(-2,2sqrt(3)), D=(0,0), E=(B+C)/2, F=(B+C+D)/3, G=(A+C)/2, H=(A+B+D)/3, I=(A+B)/2;
fill((0,2sqrt(3))--B--(3,sqrt(3))--(2,(2sqrt(3))/3)--(0,(4sqrt(3))/3)--cycle,gray);
draw(A--B--C--D--cycle);
draw(D--(0,2sqrt(3))); draw(D--(3,sqrt(3))); draw(A--C); draw(F--B--H); draw(B--G);
label("$A$",A,SE);label("$B$",B,NE);label("$C$",C,NW);label("$D$",D,SW);
label("$E$",E,N);label("$F$",F,SW);label("$G$",G,SW);label("$H$",H,S);label("$I$",I,NE);
label("$2$",(D--C),SW);
// Block 2
unitsize(8mm); defaultpen(linewidth(0.8pt)+fontsize(10pt)); dotfactor=4; pair A=(4,0), B=(2,2sqrt(3)), C=(-2,2sqrt(3)), D=(0,0), E=(B+C)/2, F=(B+C+D)/3, G=(A+C)/2, H=(A+B+D)/3, I=(A+B)/2; fill((0,2sqrt(3))--B--(3,sqrt(3))--(2,(2sqrt(3))/3)--(0,(4sqrt(3))/3)--cycle,gray); draw(A--B--C--D--cycle); draw(D--(0,2sqrt(3))); draw(D--(3,sqrt(3))); draw(A--C); draw(F--B--H); draw(B--G); label("$A$",A,SE);label("$B$",B,NE);label("$C$",C,NW);label("$D$",D,SW); label("$E$",E,N);label("$F$",F,SW);label("$G$",G,SW);label("$H$",H,S);label("$I$",I,NE); label("$2$",(D--C),SW); | [] |
712 | Rhombus $ABCD$ has side length $2$ and $\angle B = 120$°. Region $R$ consists of all points inside the rhombus that are closer to vertex $B$ than any of the other three vertices. What is the area of $R$?
$\textbf{(A)}\ \frac{\sqrt{3}}{3} \qquad\textbf{(B)}\ \frac{\sqrt{3}}{2} \qquad\textbf{(C)}\ \frac{2\sqrt{3}}{3} \qquad\textbf{(D)}\ 1 + \frac{\sqrt{3}}{3} \qquad\textbf{(E)}\ 2$ | 2011 AMC 10B Problem 20 | Since $H$ and $E$ are halfway between $AB$ and $CB$, respectively, we know that $\overline{BH}=\overline{BE}=1$. By symmetry, $\Delta BFG$ is equilateral, so $\angle FBG=60^\circ\implies\angle EBF=\angle HBG=30^\circ$ and therefore $\Delta EBF$ and $\Delta HBG$ are 30-60-90 right triangles.
Thus, $[\Delta EBF]=[\Delta BFG]=\dfrac1{2\sqrt3}$.
We know that $\overline{FB}=\overline{GB}=\dfrac2{\sqrt3}$, so therefore $[\Delta BFG]=\dfrac{\sqrt3}4\left(\dfrac2{\sqrt3}\right)^2=\dfrac1{\sqrt3}$.
Summing these three regions, we get $\dfrac1{2\sqrt3}+\dfrac1{2\sqrt3}+\dfrac1{\sqrt3}=\boxed{\textbf{(C)}~\dfrac{2\sqrt3}3}$.
~ Technodoggo, Asymptote diagram modified from Solution 1 | // Block 1
unitsize(8mm);
defaultpen(linewidth(0.8pt)+fontsize(10pt));
dotfactor=4;
pair A=(4,0), B=(2,2sqrt(3)), C=(-2,2sqrt(3)), D=(0,0), E=(B+C)/2, F=(B+C+D)/3, G=(A+B+D)/3, H=(A+B)/2;
fill((0,2sqrt(3))--B--(3,sqrt(3))--(2,(2sqrt(3))/3)--(0,(4sqrt(3))/3)--cycle,gray);
draw(A--B--C--D--cycle);
draw(D--(0,2sqrt(3))); draw(D--(3,sqrt(3))); draw(A--C); draw(F--B--H); draw(B--G);
label("$A$",A,SE);label("$B$",B,NE);label("$C$",C,NW);label("$D$",D,SW);
label("$E$",E,N);label("$F$",F,SW);label("$G$",G,S);label("$H$",H,E);
label("$2$",(D--C),SW);
// Block 2
unitsize(8mm); defaultpen(linewidth(0.8pt)+fontsize(10pt)); dotfactor=4; pair A=(4,0), B=(2,2sqrt(3)), C=(-2,2sqrt(3)), D=(0,0), E=(B+C)/2, F=(B+C+D)/3, G=(A+B+D)/3, H=(A+B)/2; fill((0,2sqrt(3))--B--(3,sqrt(3))--(2,(2sqrt(3))/3)--(0,(4sqrt(3))/3)--cycle,gray); draw(A--B--C--D--cycle); draw(D--(0,2sqrt(3))); draw(D--(3,sqrt(3))); draw(A--C); draw(F--B--H); draw(B--G); label("$A$",A,SE);label("$B$",B,NE);label("$C$",C,NW);label("$D$",D,SW); label("$E$",E,N);label("$F$",F,SW);label("$G$",G,S);label("$H$",H,E); label("$2$",(D--C),SW); | [] |
712 | Rhombus $ABCD$ has side length $2$ and $\angle B = 120$°. Region $R$ consists of all points inside the rhombus that are closer to vertex $B$ than any of the other three vertices. What is the area of $R$?
$\textbf{(A)}\ \frac{\sqrt{3}}{3} \qquad\textbf{(B)}\ \frac{\sqrt{3}}{2} \qquad\textbf{(C)}\ \frac{2\sqrt{3}}{3} \qquad\textbf{(D)}\ 1 + \frac{\sqrt{3}}{3} \qquad\textbf{(E)}\ 2$ | 2011 AMC 10B Problem 20 | To keep it simple, break rhombus $ABCD$ into two triangles, $ABD$ and $BCD$. To see the area closest to the point $B$, notice that a third of each triangle, which contains all the points nearest to $B$ in each triangle, is easily visualizable. Thus, a third of rhombus $ABCD$ must be found.
We find the total area of rhombus $ABCD$, which we can again split into two congruent equilateral triangles with side length $2$. Using the formula of equilateral triangles and then multiplying by $\dfrac{1}{3}$:
\[\dfrac{\sqrt{3}}{4}\cdot 2^2 \cdot 2 \cdot \dfrac{1}{3} = 2\sqrt{3} \cdot \dfrac{1}{3} = \boxed{\textbf{(C)}~\dfrac{2\sqrt{3}}{3}}\]
-NSAoPS, diagram modified from Solution 1. | // Block 1
unitsize(8mm);
defaultpen(linewidth(0.8pt)+fontsize(10pt));
dotfactor=4;
pair A=(4,0), B=(2,2sqrt(3)), C=(-2,2sqrt(3)), D=(0,0), E=(B+C)/2, F=(B+C+D)/3, G=(A+B+D)/3, H=(A+B)/2;
fill((0,2sqrt(3))--B--(3,sqrt(3))--(2,(2sqrt(3))/3)--(0,(4sqrt(3))/3)--cycle,gray);
draw(A--B--C--D--cycle);
draw(D--F); draw(D--G); draw(F--C); draw(A--G); draw(F--B); draw(B--G);
draw(B--D);
label("$A$",A,SE);label("$B$",B,NE);label("$C$",C,NW);label("$D$",D,SW);
label("$E$",E,N);label("$F$",F,SW);label("$G$",G,S);label("$H$",H,E);
label("$2$",(D--C),SW);
// Block 2
unitsize(8mm); defaultpen(linewidth(0.8pt)+fontsize(10pt)); dotfactor=4; pair A=(4,0), B=(2,2sqrt(3)), C=(-2,2sqrt(3)), D=(0,0), E=(B+C)/2, F=(B+C+D)/3, G=(A+B+D)/3, H=(A+B)/2; fill((0,2sqrt(3))--B--(3,sqrt(3))--(2,(2sqrt(3))/3)--(0,(4sqrt(3))/3)--cycle,gray); draw(A--B--C--D--cycle); draw(D--F); draw(D--G); draw(F--C); draw(A--G); draw(F--B); draw(B--G); draw(B--D); label("$A$",A,SE);label("$B$",B,NE);label("$C$",C,NW);label("$D$",D,SW); label("$E$",E,N);label("$F$",F,SW);label("$G$",G,S);label("$H$",H,E); label("$2$",(D--C),SW); | [] |
713 | Brian writes down four integers $w > x > y > z$ whose sum is $44$. The pairwise positive differences of these numbers are $1, 3, 4, 5, 6,$ and $9$. What is the sum of the possible values for $w$?
$\textbf{(A)}\ 16 \qquad\textbf{(B)}\ 31 \qquad\textbf{(C)}\ 48 \qquad\textbf{(D)}\ 62 \qquad\textbf{(E)}\ 93$ | 2011 AMC 10B Problem 21 | The largest difference, $9,$ must be between $w$ and $z.$
The smallest difference, $1,$ must be directly between two integers. This also means the differences directly between the other two should add up to $8.$ The only remaining differences that would make this possible are $3$ and $5.$ However, those two differences can't be right next to each other because they would make a difference of $8,$ which isn't given as a possibility in the problem. This means $1$ must be the difference between $y$ and $x.$ We can express the possible configurations as the lines.
If we look at the first number line, you can express $x$ as $w-5,$ $y$ as $w-6,$ and $z$ as $w-9.$ Since the sum of all these integers equal $44$,
\begin{align*} w+w-5+w-6+w-9&=44\\ 4w&=64\\ w&=16 \end{align*}
You can do something similar to this with the second number line to find the other possible value of $w.$
\begin{align*} w+w-3+w-4+w-9&=44\\ 4w&=60\\ w&=15 \end{align*}
The sum of the possible values of $w$ is $16+15 = \boxed{\textbf{(B) }31}$ | // Block 1
unitsize(14mm);
defaultpen(linewidth(.8pt)+fontsize(10pt));
dotfactor=4;
pair Z1=(0,1), Y1=(1,1), X1=(2,1), W1=(3,1);
pair Z4=(4,1), Y4=(5,1), X4=(6,1), W4=(7,1);
draw(Z1--W1); draw(Z4--W4);
pair[] ps={W1,W4,X1,X4,Y1,Y4,Z1,Z4};
dot(ps);
label("$z$",Z1,N); label("$y$",Y1,N); label("$x$",X1,N); label("$w$",W1,N);
label("$z$",Z4,N); label("$y$",Y4,N); label("$x$",X4,N); label("$w$",W4,N);
label("$1$",(X1--Y1),N); label("$1$",(X4--Y4),N);
label("$3$",(Y1--Z1),N); label("$3$",(W4--X4),N); label("$5$",(X1--W1),N); label("$5$",(Y4--Z4),N);
// Block 2
unitsize(14mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; pair Z1=(0,1), Y1=(1,1), X1=(2,1), W1=(3,1); pair Z4=(4,1), Y4=(5,1), X4=(6,1), W4=(7,1); draw(Z1--W1); draw(Z4--W4); pair[] ps={W1,W4,X1,X4,Y1,Y4,Z1,Z4}; dot(ps); label("$z$",Z1,N); label("$y$",Y1,N); label("$x$",X1,N); label("$w$",W1,N); label("$z$",Z4,N); label("$y$",Y4,N); label("$x$",X4,N); label("$w$",W4,N); label("$1$",(X1--Y1),N); label("$1$",(X4--Y4),N); label("$3$",(Y1--Z1),N); label("$3$",(W4--X4),N); label("$5$",(X1--W1),N); label("$5$",(Y4--Z4),N); | [] |
714 | A pyramid has a square base with sides of length $1$ and has lateral faces that are equilateral triangles. A cube is placed within the pyramid so that one face is on the base of the pyramid and its opposite face has all its edges on the lateral faces of the pyramid. What is the volume of this cube?
$\textbf{(A)}\ 5\sqrt{2} - 7 \qquad\textbf{(B)}\ 7 - 4\sqrt{3} \qquad\textbf{(C)}\ \frac{2\sqrt{2}}{27} \qquad\textbf{(D)}\ \frac{\sqrt{2}}{9} \qquad\textbf{(E)}\ \frac{\sqrt{3}}{9}$ | 2011 AMC 10B Problem 22 | It is often easier to first draw a diagram for such a problem.
[Image: images/amc/2011_AMC_10B_Problem_22_0.png]
Sometimes, it may also be easier to think of the problem in 2D. Take a cross section of the pyramid through the apex and two points from the base that are opposite to each other. Place it in two dimensions.
The dimensions of this triangle are $1, 1,$ and $\sqrt{2}$ because the sidelengths of the pyramid are $1$ and the base of the triangle is the diagonal of the pyramid's base. This is a $45-45-90$ triangle. Also, we can let the dimensions of the rectangle be $s$ and $s\sqrt{2}$ because the longer length was the diagonal of the cube's base and the shorter length was a side of the cube.
The two triangles on the right and left of the rectangle are also $45-45-90$ triangles because the rectangle is perpendicular to the base, and they share a $45^\circ$ angle with the larger triangle. Therefore, the legs of the right triangles can be expressed as $s.$
Now we can just use segment addition to find the value of $s.$
\[\sqrt{2}=s+s\sqrt{2}+s=2s+s\sqrt{2}=(2+\sqrt{2})s\]
\[s=\frac{\sqrt{2}}{2+\sqrt{2}}=\frac{1}{\sqrt{2}+1}=\frac{\sqrt{2}-1}{2-1}=\sqrt{2}-1\]
The volume of the cube is $s^3 = (\sqrt{2}-1)^3 = (\sqrt{2}-1)(3-2\sqrt{2}) = 3\sqrt{2}-3-4+2\sqrt{2} = \boxed{\textbf{(A)} 5\sqrt{2}-7}$ | // Block 1
unitsize(35mm);
defaultpen(linewidth(2pt)+fontsize(10pt));
pair A=(0,0), B=(sqrt(2),0), C=(0.5sqrt(2),0.5sqrt(2));
pair W=(sqrt(2)-1,0), X=(1,0), Y=(1,sqrt(2)-1), Z=(sqrt(2)-1,sqrt(2)-1);
draw(A--B--C--cycle);
draw(W--X--Y--Z--cycle,red);
// Block 2
unitsize(35mm);
defaultpen(linewidth(2pt)+fontsize(12pt));
pair A=(0,0), B=(sqrt(2),0), C=(0.5sqrt(2),0.5sqrt(2));
pair W=(sqrt(2)-1,0), X=(1,0), Y=(1,sqrt(2)-1), Z=(sqrt(2)-1,sqrt(2)-1);
draw(A--B--C--cycle);
draw(W--X--Y--Z--cycle,red);
label("$1$",(A--C),NW); label("$1$",(B--C),NE); label("$\sqrt{2}$",(A--B),S);
label("$s$",(W--Z),E,red); label("$s$",(X--Y),W,red); label("$s\sqrt{2}$",(W--X),N,red);
// Block 3
unitsize(35mm);
defaultpen(linewidth(2pt)+fontsize(12pt));
pair A=(0,0), B=(sqrt(2),0), C=(0.5sqrt(2),0.5sqrt(2));
pair W=(sqrt(2)-1,0), X=(1,0), Y=(1,sqrt(2)-1), Z=(sqrt(2)-1,sqrt(2)-1);
draw(A--B--C--cycle);
draw(W--X--Y--Z--cycle,red);
label("$1$",(A--C),NW); label("$1$",(B--C),NE); label("$\sqrt{2}$",(A--B),S);
label("$s$",(W--Z),E,red); label("$s$",(X--Y),W,red); label("$s\sqrt{2}$",(W--X),N,red);
label("$s$",(A--W),N); label("$s$",(X--B),N);
// Block 4
unitsize(35mm); defaultpen(linewidth(2pt)+fontsize(10pt)); pair A=(0,0), B=(sqrt(2),0), C=(0.5sqrt(2),0.5sqrt(2)); pair W=(sqrt(2)-1,0), X=(1,0), Y=(1,sqrt(2)-1), Z=(sqrt(2)-1,sqrt(2)-1); draw(A--B--C--cycle); draw(W--X--Y--Z--cycle,red);
// Block 5
unitsize(35mm); defaultpen(linewidth(2pt)+fontsize(12pt)); pair A=(0,0), B=(sqrt(2),0), C=(0.5sqrt(2),0.5sqrt(2)); pair W=(sqrt(2)-1,0), X=(1,0), Y=(1,sqrt(2)-1), Z=(sqrt(2)-1,sqrt(2)-1); draw(A--B--C--cycle); draw(W--X--Y--Z--cycle,red); label("$1$",(A--C),NW); label("$1$",(B--C),NE); label("$\sqrt{2}$",(A--B),S); label("$s$",(W--Z),E,red); label("$s$",(X--Y),W,red); label("$s\sqrt{2}$",(W--X),N,red);
// Block 6
unitsize(35mm); defaultpen(linewidth(2pt)+fontsize(12pt)); pair A=(0,0), B=(sqrt(2),0), C=(0.5sqrt(2),0.5sqrt(2)); pair W=(sqrt(2)-1,0), X=(1,0), Y=(1,sqrt(2)-1), Z=(sqrt(2)-1,sqrt(2)-1); draw(A--B--C--cycle); draw(W--X--Y--Z--cycle,red); label("$1$",(A--C),NW); label("$1$",(B--C),NE); label("$\sqrt{2}$",(A--B),S); label("$s$",(W--Z),E,red); label("$s$",(X--Y),W,red); label("$s\sqrt{2}$",(W--X),N,red); label("$s$",(A--W),N); label("$s$",(X--B),N); | ["https://artofproblemsolving.com/wiki/images/c/ce/2011AMC10B22.png"] |
715 | Rhombus $ABCD$ has side length $2$ and $\angle B = 120$°. Region $R$ consists of all points inside the rhombus that are closer to vertex $B$ than any of the other three vertices. What is the area of $R$?
$\textbf{(A)}\ \frac{\sqrt{3}}{3} \qquad\textbf{(B)}\ \frac{\sqrt{3}}{2} \qquad\textbf{(C)}\ \frac{2\sqrt{3}}{3} \qquad\textbf{(D)}\ 1 + \frac{\sqrt{3}}{3} \qquad\textbf{(E)}\ 2$ | 2011 AMC 12B Problem 16 | Suppose that $P$ is a point in the rhombus $ABCD$ and let $\ell_{BC}$ be the perpendicular bisector of $\overline{BC}$. Then $PB < PC$ if and only if $P$ is on the same side of $\ell_{BC}$ as $B$. The line $\ell_{BC}$ divides the plane into two half-planes; let $S_{BC}$ be the half-plane containing $B$. Let us define similarly $\ell_{BD},S_{BD}$ and $\ell_{BA},S_{BA}$. Then $R$ is equal to $ABCD \cap S_{BC} \cap S_{BD} \cap S_{BA}$. The region turns out to be an irregular pentagon. We can make it easier to find the area of this region by dividing it into four triangles:
Since $\triangle BCD$ and $\triangle BAD$ are equilateral, $\ell_{BC}$ contains $D$, $\ell_{BD}$ contains $A$ and $C$, and $\ell_{BA}$ contains $D$. Then $\triangle BEF \cong \triangle BGF \cong \triangle BGH \cong \triangle BIH$ with $BE = 1$ and $EF = \frac{1}{\sqrt{3}}$ so $[BEF] = \frac{1}{2}\cdot 1 \cdot \frac{\sqrt{3}}{3}$. Multiply this by 4 and it turns out that the pentagon has area $\boxed{(C)\frac{2\sqrt{3}}{3}}$. | unitsize(8mm); defaultpen(linewidth(0.8pt)+fontsize(10pt)); dotfactor=4; pair A=(4,0), B=(2,2sqrt(3)), C=(-2,2sqrt(3)), D=(0,0), E=(B+C)/2, F=(B+C+D)/3, G=(A+C)/2, H=(A+B+D)/3, I=(A+B)/2; fill((0,2sqrt(3))--B--(3,sqrt(3))--(2,(2sqrt(3))/3)--(0,(4sqrt(3))/3)--cycle,gray); draw(A--B--C--D--cycle); draw(D--(0,2sqrt(3))); draw(D--(3,sqrt(3))); draw(A--C); draw(F--B--H); draw(B--G); label("$A$",A,SE);label("$B$",B,NE);label("$C$",C,NW);label("$D$",D,SW); label("$E$",E,N);label("$F$",F,SW);label("$G$",G,SW);label("$H$",H,S);label("$I$",I,NE); label("$2$",(D--C),SW); | [] |
715 | Rhombus $ABCD$ has side length $2$ and $\angle B = 120$°. Region $R$ consists of all points inside the rhombus that are closer to vertex $B$ than any of the other three vertices. What is the area of $R$?
$\textbf{(A)}\ \frac{\sqrt{3}}{3} \qquad\textbf{(B)}\ \frac{\sqrt{3}}{2} \qquad\textbf{(C)}\ \frac{2\sqrt{3}}{3} \qquad\textbf{(D)}\ 1 + \frac{\sqrt{3}}{3} \qquad\textbf{(E)}\ 2$ | 2011 AMC 12B Problem 16 | Since $H$ and $E$ are halfway between $AB$ and $CB$, respectively, we know that $\overline{BH}=\overline{BE}=1$. By symmetry, $\Delta BFG$ is equilateral, so $\angle FBG=60^\circ\implies\angle EBF=\angle HBG=30^\circ$ and therefore $\Delta EBF$ and $\Delta HBG$ are 30-60-90 right triangles.
Thus, $[\Delta EBF]=[\Delta BFG]=\dfrac1{2\sqrt3}$.
We know that $\overline{FB}=\overline{GB}=\dfrac2{\sqrt3}$, so therefore $[\Delta BFG]=\dfrac{\sqrt3}4\left(\dfrac2{\sqrt3}\right)^2=\dfrac1{\sqrt3}$.
Summing these three regions, we get $\dfrac1{2\sqrt3}+\dfrac1{2\sqrt3}+\dfrac1{\sqrt3}=\boxed{\textbf{(C)}~\dfrac{2\sqrt3}3}$.
~ Technodoggo, Asymptote diagram modified from Solution 1 | unitsize(8mm); defaultpen(linewidth(0.8pt)+fontsize(10pt)); dotfactor=4; pair A=(4,0), B=(2,2sqrt(3)), C=(-2,2sqrt(3)), D=(0,0), E=(B+C)/2, F=(B+C+D)/3, G=(A+B+D)/3, H=(A+B)/2; fill((0,2sqrt(3))--B--(3,sqrt(3))--(2,(2sqrt(3))/3)--(0,(4sqrt(3))/3)--cycle,gray); draw(A--B--C--D--cycle); draw(D--(0,2sqrt(3))); draw(D--(3,sqrt(3))); draw(A--C); draw(F--B--H); draw(B--G); label("$A$",A,SE);label("$B$",B,NE);label("$C$",C,NW);label("$D$",D,SW); label("$E$",E,N);label("$F$",F,SW);label("$G$",G,S);label("$H$",H,E); label("$2$",(D--C),SW); | [] |
715 | Rhombus $ABCD$ has side length $2$ and $\angle B = 120$°. Region $R$ consists of all points inside the rhombus that are closer to vertex $B$ than any of the other three vertices. What is the area of $R$?
$\textbf{(A)}\ \frac{\sqrt{3}}{3} \qquad\textbf{(B)}\ \frac{\sqrt{3}}{2} \qquad\textbf{(C)}\ \frac{2\sqrt{3}}{3} \qquad\textbf{(D)}\ 1 + \frac{\sqrt{3}}{3} \qquad\textbf{(E)}\ 2$ | 2011 AMC 12B Problem 16 | To keep it simple, break rhombus $ABCD$ into two triangles, $ABD$ and $BCD$. To see the area closest to the point $B$, notice that a third of each triangle, which contains all the points nearest to $B$ in each triangle, is easily visualizable. Thus, a third of rhombus $ABCD$ must be found.
We find the total area of rhombus $ABCD$, which we can again split into two congruent equilateral triangles with side length $2$. Using the formula of equilateral triangles and then multiplying by $\dfrac{1}{3}$:
\[\dfrac{\sqrt{3}}{4}\cdot 2^2 \cdot 2 \cdot \dfrac{1}{3} = 2\sqrt{3} \cdot \dfrac{1}{3} = \boxed{\textbf{(C)}~\dfrac{2\sqrt{3}}{3}}\]
-NSAoPS, diagram modified from Solution 1. | unitsize(8mm); defaultpen(linewidth(0.8pt)+fontsize(10pt)); dotfactor=4; pair A=(4,0), B=(2,2sqrt(3)), C=(-2,2sqrt(3)), D=(0,0), E=(B+C)/2, F=(B+C+D)/3, G=(A+B+D)/3, H=(A+B)/2; fill((0,2sqrt(3))--B--(3,sqrt(3))--(2,(2sqrt(3))/3)--(0,(4sqrt(3))/3)--cycle,gray); draw(A--B--C--D--cycle); draw(D--F); draw(D--G); draw(F--C); draw(A--G); draw(F--B); draw(B--G); draw(B--D); label("$A$",A,SE);label("$B$",B,NE);label("$C$",C,NW);label("$D$",D,SW); label("$E$",E,N);label("$F$",F,SW);label("$G$",G,S);label("$H$",H,E); label("$2$",(D--C),SW); | [] |
716 | A bug travels in the coordinate plane, moving only along the lines that are parallel to the $x$-axis or $y$-axis. Let $A = (-3, 2)$ and $B = (3, -2)$. Consider all possible paths of the bug from $A$ to $B$ of length at most $20$. How many points with integer coordinates lie on at least one of these paths?
$\textbf{(A)}\ 161 \qquad \textbf{(B)}\ 185 \qquad \textbf{(C)}\ 195 \qquad \textbf{(D)}\ 227 \qquad \textbf{(E)}\ 255$ | 2011 AMC 12B Problem 23 | (Diagram for Solution 2 by Technodoggo also don't ask about the patterns, I just figured out how to use the patterns module)
Notice that the bug is basically moving from A to B (length 10) but optionally going on a detour anywhere along its route.
Specifically, the detour would be of total length 10, 5 to some point and then going back by retracing its path (also length 5). Just to simplify things we can observe that the coordinates don't matter here and all we need to remember is that A and B are the diagonally opposite vertices of a 4 by 6 rectangle (5 by 7 lattice points). The bug can start the "detour" from any point on or inside the rectangle. Notice that the bug can go 5 steps in the y direction, 5 steps in the x direction, or anything in between, so the points covered by possible detours from any point would look like a rhombus or square rotated 45 degrees (with centre at a point on or inside the rectangle). Drawing this out we would get an octagon.
Finding the final answer is then easy, for this solution I will slice the octagon into 4 rectangles (2 of which are squares) and 4 isosceles triangles. There are $(4+1)\cdot (6+1) = 35$ points on or inside the original triangle, $5\cdot 7$ points covered by the rectangles above and below the original one and $5\cdot 5$ points for the squares to the right and left of the original triangle. Lastly each of the four isosceles triangles cover $4+3+2+1 = 10$ points. (Notice that although the length of the detour is 5, the points on the edge of the triangles were already counted).
Adding these up, we get $3\cdot 35 + 2\cdot 25 + 4\cdot 10 = 195 => \boxed{(C)}$
Another way to get to the octagon would be to note that aside from the 4x6/5x7 box of the bare absolutely necessary route (the checkered region), we can go on a 5-step detour anywhere (as above); we need only examine how far we can get through the edges (the border), because that takes us the farthest. Starting from any point on the border, the farthest that it can go is 5 units in some cardinal direction, which gives us the tiled region in the diagram above. Finally, from our corners, we can also go 4-1, 3-2, 2,3, or 1-4, which gives us the rest of the region enclosed in blue. This gives us our octagon.
tl;dr the interior of the octagon in blue is a result of taking all the points that can be reached from the boundary of the box as described in no more than 5 moves. We can then proceed to find the area in any simple way. (Extra way & diagram by Technodoggo) | // Block 1
import graph;
import patterns;
unitsize(1cm);
add("tile",tile());
add("checker",checker());
filldraw((-3,2)--(3,2)--(3,-2)--(-3,-2)--cycle,pattern("checker"));
filldraw((-3,2)--(-3,7)--(3,7)--(3,2)--cycle,pattern("tile"));
filldraw((-3,-2)--(-3,-7)--(3,-7)--(3,-2)--cycle,pattern("tile"));
filldraw((-8,2)--(-3,2)--(-3,-2)--(-8,-2)--cycle,pattern("tile"));
filldraw((8,2)--(3,2)--(3,-2)--(8,-2)--cycle,pattern("tile"));
draw((8,2)--(3,7)--(-3,7)--(-8,2)--(-8,-2)--(-3,-7)--(3,-7)--(8,-2)--cycle,blue);
xaxis(-9,9,Ticks(NoZero,Step=1));
yaxis(-8,8,Ticks(NoZero,Step=1));
dot((-3,2));label("$A$",(-3,2),N);
dot((3,-2));label("$B$",(3,-2),N);
// Asymptote by Technodoggo; August 16, 2024
// Block 2
import graph; import patterns; unitsize(1cm); add("tile",tile()); add("checker",checker()); filldraw((-3,2)--(3,2)--(3,-2)--(-3,-2)--cycle,pattern("checker")); filldraw((-3,2)--(-3,7)--(3,7)--(3,2)--cycle,pattern("tile")); filldraw((-3,-2)--(-3,-7)--(3,-7)--(3,-2)--cycle,pattern("tile")); filldraw((-8,2)--(-3,2)--(-3,-2)--(-8,-2)--cycle,pattern("tile")); filldraw((8,2)--(3,2)--(3,-2)--(8,-2)--cycle,pattern("tile")); draw((8,2)--(3,7)--(-3,7)--(-8,2)--(-8,-2)--(-3,-7)--(3,-7)--(8,-2)--cycle,blue); xaxis(-9,9,Ticks(NoZero,Step=1)); yaxis(-8,8,Ticks(NoZero,Step=1)); dot((-3,2));label("$A$",(-3,2),N); dot((3,-2));label("$B$",(3,-2),N); // Asymptote by Technodoggo; August 16, 2024 | [] |
717 | For every $m$ and $k$ integers with $k$ odd, denote by $\left[\frac{m}{k}\right]$ the integer closest to $\frac{m}{k}$. For every odd integer $k$, let $P(k)$ be the probability that
\[\left[\frac{n}{k}\right] + \left[\frac{100 - n}{k}\right] = \left[\frac{100}{k}\right]\]
for an integer $n$ randomly chosen from the interval $1 \leq n \leq 99!$. What is the minimum possible value of $P(k)$ over the odd integers $k$ in the interval $1 \leq k \leq 99$?
$\textbf{(A)}\ \frac{1}{2} \qquad \textbf{(B)}\ \frac{50}{99} \qquad \textbf{(C)}\ \frac{44}{87} \qquad \textbf{(D)}\ \frac{34}{67} \qquad \textbf{(E)}\ \frac{7}{13}$ | 2011 AMC 12B Problem 25 | Solution 1
Answer: $(D) \frac{34}{67}$
First of all, you have to realize that
if $\left[\frac{n}{k}\right] + \left[\frac{100 - n}{k}\right] = \left[\frac{100}{k}\right]$
then $\left[\frac{n - k}{k}\right] + \left[\frac{100 - (n - k)}{k}\right] = \left[\frac{100}{k}\right]$
So, we can consider what happen in $1\le n \le k$ and it will repeat. Also since range of $n$ is $1$ to $99!$, it is always a multiple of $k$. So we can just consider $P(k)$ for $1\le n \le k$.
Let $\text{fpart}(x)$ be the fractional part function
This is an AMC exam, so use the given choices wisely. With the given choices, and the previous explanation, we only need to consider $k = 99$, $87$, $67$, $13$. $1\le n \le k$
For $k > \frac{200}{3}$, $\left[\frac{100}{k}\right] = 1$. 3 of the $k$ that we should consider land in here.
For $n < \frac{k}{2}$, $\left[\frac{n}{k}\right] = 0$, then we need $\left[\frac{100 - n}{k}\right] = 1$
else for $\frac{k}{2}< n < k$, $\left[\frac{n}{k}\right] = 1$, then we need $\left[\frac{100 - n}{k}\right] = 0$
For $n < \frac{k}{2}$, $\left[\frac{100 - n}{k}\right] = \left[\frac{100}{k} - \frac{n}{k}\right]= 1$
So, for the condition to be true, $100 - n > \frac{k}{2}$ . ( $k > \frac{200}{3}$, no worry for the rounding to be $> 1$)
$100 > k > \frac{k}{2} + n$, so this is always true.
For $\frac{k}{2}< n < k$, $\left[\frac{100 - n}{k}\right] = 0$, so we want $100 - n < \frac{k}{2}$, or $100 < \frac{k}{2} + n$
$100 <\frac{k}{2} + n < \frac{3k}{2}$
For k = 67, $67 > n > 100 - \frac{67}{2} = 66.5$
For k = 69, $69 > n > 100 - \frac{69}{2} = 67.5$
etc.
We can clearly see that for this case, $k = 67$ has the minimum $P(k)$, which is $\frac{34}{67}$. Also, $\frac{7}{13} > \frac{34}{67}$ .
So for AMC purpose, answer is $\boxed{\textbf{(D) }\frac{34}{67}}$.
Proof:
Notice that for these integers $99,87,67$:
$0\rightarrow 49,50,51\rightarrow 98$
$100\rightarrow 51,50,49\rightarrow 2$
$P=\frac{98}{99}$
$0\rightarrow 43,44\rightarrow 56,57\rightarrow 86$
$87\rightarrow 57,56\rightarrow 44,43\rightarrow 14$
$P=\frac{74}{87}$
$0\rightarrow 33,34\rightarrow 66$
$100\rightarrow 67,66\rightarrow 34$
$P=\frac{34}{67}$
That the probability is $\frac{2k-100}{k}=2-\frac{100}{k}$. Even for $k=13$, $P(13)=\frac{9}{13}=\frac{100}{13}-7$. And $P(11)=\frac{10}{11}=10-\frac{100}{11}$.
Perhaps the probability for a given $k$ is $\left\lceil{\frac{100}{k}}\right\rceil-\frac{100}{k}$ if $\left[\frac{100}{k}\right]=\left\lfloor{\frac{100}{k}}\right\rfloor$ and $\frac{100}{k}-\left\lfloor{\frac{100}{k}}\right\rfloor$ if $\left[\frac{100}{k}\right]=\left\lceil{\frac{100}{k}}\right\rceil$.
So $P>\frac{1}{2}$ and $P_\text{min}=\frac{k_\text{min}+1}{2k_\text{min}}=\frac{101}{201}$. Because $201=3\cdot 67\mid 99!$ !
Now, let's say we are not given any answer, we need to consider $k < \frac{200}{3}$.
I claim that $P(k) \ge \frac{1}{2} + \frac{1}{2k}$
If $\left[\frac{100}{k}\right]$ got round down, then $1 \le n \le \frac{k}{2}$ all satisfy the condition along with $n = k$
because if $\text{fpart}\left(\frac{100}{k}\right) < \frac{1}{2}$ and $\text{fpart} \left(\frac{n}{k}\right) < \frac{1}{2}$, so must $\text{fpart} \left(\frac{100 - n}{k}\right) < \frac{1}{2}$
and for $n = k$, it is the same as $n = 0$.
, which makes
$P(k) \ge \frac{1}{2} + \frac{1}{2k}$.
If $\left[\frac{100}{k}\right]$ got round up, then $\frac{k}{2} \le n \le k$ all satisfy the condition along with $n = 1$
because if $\text{fpart}\left(\frac{100}{k}\right) > \frac{1}{2}$ and $\text{fpart} \left(\frac{n}{k}\right) > \frac{1}{2}$
Case 1)
$\text{fpart} \left(\frac{100 - n}{k}\right) < \frac{1}{2}$
-> $\text{fpart}\left(\frac{100}{k}\right) = \text{fpart} \left(\frac{n}{k}\right) +\text{fpart} \left(\frac{100 - n}{k}\right)$
Case 2)
$\text{fpart} \left(\frac{100 - n}{k}\right) > \frac{1}{2}$
-> $\text{fpart}\left(\frac{100}{k}\right) + 1 = \text{fpart} \left(\frac{n}{k}\right) +\text{fpart} \left(\frac{100 - n}{k}\right)$
and for $n = 1$, since $k$ is odd, $\left[\frac{99}{k}\right] \neq \left[\frac{100}{k}\right]$
-> $99.5 = k (p + .5)$ -> $199 = k (2p + 1)$, and $199$ is prime so $k = 1$ or $k =199$, which is not in this set
, which makes
$P(k) \ge \frac{1}{2} + \frac{1}{2k}$.
Now the only case without rounding, $k = 1$. It must be true.
Solution 2
It suffices to consider $0\le n \le k-1.$ Now for each of these $n,$ let $f(n)=\left[\frac{n}{k}\right], g(n)=\left[\frac{100-n}{k}\right]-\left[\frac{100}{k}\right].$ If we let $k=67,$ then the following graphs result for $f$ and $g.$
$f:$
$g:$
Our probability is the number of $0\le i \le k-1$ such that $f(i)+g(i)=0$ over $k.$ Of course, this always holds for $i=0.$ If we let $k$ vary, then the graph of $f$ is always very similar to what it looks like above (groups of $\frac{k+1}{2},\frac{k-1}{2}$ dots). However, the graph of $g$ can vary greatly. In the above diagram, $g(i)=0$ for all $i,$ while it is possible for $g(i)=-1$ for all $i\neq 0.$ In order to minimize the number of $i$ which satisfy $f(i)+g(i)=0,$ we either want $g(i)=0$ for $0\le i<k,$ or $g(i)=-1$ for $1\le i<k.$ This way, we see that at least half of the numbers from $1$ to $k-1$ satisfy the given equation. So, our desired probability is at least $\frac{k+1}{2k}.$ As shown by the diagram above, the probability is $\frac{34}{67}$ for $k=67.$ Clearly no better solutions can exist when $k<67.$ On the other hand, for $k>67$ $87$ and $99$ do not yield better probabilities. Therefore, our answer is $\boxed{\frac{34}{67}}.$ | // Block 1
size(10cm); for (int i = 0; i < 67; ++i) { if (i<=33) dot((i,0)); else dot((i,1)); } label("(0,0)",(0,0),SW); label("(66,1)",(66,1),NE);
// Block 2
size(10cm); for (int i = 0; i < 67; ++i) { dot((i,0)); } label("(0,0)",(0,0),NW); label("(66,0)",(66,0),SE); | [] |
718 | A bug crawls along a number line, starting at $-2$. It crawls to $-6$, then turns around and crawls to $5$. How many units does the bug crawl altogether?
$\textbf{(A)}\ 9\qquad\textbf{(B)}\ 11\qquad\textbf{(C)}\ 13\qquad\textbf{(D)}\ 14\qquad\textbf{(E)}\ 15$ | 2012 AMC 10A Problem 3 | Crawling from $-2$ to $-6$ takes it a distance of $4$ units. Crawling from $-6$ to $5$ takes it a distance of $11$ units. Add $4$ and $11$ to get $\boxed{\textbf{(E)}\ 15}$ | // Block 1
draw((-2,1)--(-6,1),red+dashed,EndArrow);
draw((-6,2)--(5,2),blue+dashed,EndArrow);
dot((-2,0));
dot((-6,0));
dot((5,0));
label("$-2$",(-2,0),dir(270));
label("$-6$",(-6,0),dir(270));
label("$5$",(5,0),dir(270));
label("$4$",(-4,0.9),dir(270));
label("$11$",(-1.5,2.5),dir(90));
// Block 2
draw((-2,1)--(-6,1),red+dashed,EndArrow); draw((-6,2)--(5,2),blue+dashed,EndArrow); dot((-2,0)); dot((-6,0)); dot((5,0)); label("$-2$",(-2,0),dir(270)); label("$-6$",(-6,0),dir(270)); label("$5$",(5,0),dir(270)); label("$4$",(-4,0.9),dir(270)); label("$11$",(-1.5,2.5),dir(90)); | [] |
719 | Externally tangent circles with centers at points $A$ and $B$ have radii of lengths $5$ and $3$, respectively. A line externally tangent to both circles intersects ray $AB$ at point $C$. What is $BC$?
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 4.8\qquad\textbf{(C)}\ 10.2\qquad\textbf{(D)}\ 12\qquad\textbf{(E)}\ 14.4$ | 2012 AMC 10A Problem 11 | Let $D$ and $E$ be the points of tangency on circles $A$ and $B$ with line $CD$. $AB=8$. Also, let $BC=x$. As $\angle ADC$ and $\angle BEC$ are right angles (a radius is perpendicular to a tangent line at the point of tangency) and both triangles share $\angle ACD$, $\triangle ADC \sim \triangle BEC$. From this we can get a proportion.
$\frac{BC}{AC}=\frac{BE}{AD} \rightarrow \frac{x}{x+8}=\frac{3}{5} \rightarrow 5x=3x+24 \rightarrow x=\boxed{\textbf{(D)}\ 12}$ | // Block 1
unitsize(3.5mm);
defaultpen(linewidth(.8pt)+fontsize(10pt));
dotfactor=4;
pair A=(0,0), B=(8,0); pair C=(20,0); pair D=(1.25,-0.25sqrt(375)); pair E=(8.75,-0.15sqrt(375));
path a=Circle(A,5);
path b=Circle(B,3);
draw(a); draw(b);
draw(C--D);
draw(A--C);
draw(A--D);
draw(B--E);
pair[] ps={A,B,C,D,E};
dot(ps);
label("$A$",A,N); label("$B$",B,N); label("$C$",C,N);
label("$D$",D,SE); label("$E$",E,SE);
label("$5$",(A--D),SW);
label("$3$",(B--E),SW);
label("$8$",(A--B),N);
label("$x$",(C--B),N);
// Block 2
unitsize(3.5mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; pair A=(0,0), B=(8,0); pair C=(20,0); pair D=(1.25,-0.25sqrt(375)); pair E=(8.75,-0.15sqrt(375)); path a=Circle(A,5); path b=Circle(B,3); draw(a); draw(b); draw(C--D); draw(A--C); draw(A--D); draw(B--E); pair[] ps={A,B,C,D,E}; dot(ps); label("$A$",A,N); label("$B$",B,N); label("$C$",C,N); label("$D$",D,SE); label("$E$",E,SE); label("$5$",(A--D),SW); label("$3$",(B--E),SW); label("$8$",(A--B),N); label("$x$",(C--B),N); | [] |
720 | Three unit squares and two line segments connecting two pairs of vertices are shown. What is the area of $\triangle ABC$?
$\textbf{(A)}\ \frac16 \qquad\textbf{(B)}\ \frac15 \qquad\textbf{(C)}\ \frac29 \qquad\textbf{(D)}\ \frac13 \qquad\textbf{(E)}\ \frac{\sqrt{2}}{4}$ | 2012 AMC 10A Problem 15 | $AC$ intersects $BC$ at a right angle, (this can be proved by noticing that the slopes of the two lines are negative reciprocals of each other) so $\triangle ABC \sim \triangle BED$. The hypotenuse of right triangle $BED$ is $\sqrt{1^2+2^2}=\sqrt{5}$.
\[\frac{AC}{BC}=\frac{BD}{ED} \Rightarrow \frac{AC}{BC} = \frac21 \Rightarrow AC=2BC\]
\[\frac{AC}{AB}=\frac{BD}{BE} \Rightarrow \frac{AC}{1}=\frac{2}{\sqrt{5}} \Rightarrow AC=\frac{2}{\sqrt{5}}\]
Since $AC=2BC$, $BC=\frac{1}{\sqrt{5}}$. $\triangle ABC$ is a right triangle so the area is just $\frac12 \cdot AC \cdot BC = \frac12 \cdot \frac{2}{\sqrt{5}} \cdot \frac{1}{\sqrt{5}} = \boxed{\textbf{(B)}\ \frac15}$ | // Block 1
unitsize(2cm);
defaultpen(linewidth(.8pt)+fontsize(10pt));
dotfactor=4;
pair A=(0,0), B=(1,0); pair C=(0.8,-0.4);
pair D=(1,-2), E=(0,-2);
draw(A--(2,0)); draw((0,-1)--(2,-1)); draw(E--D);
draw(A--E); draw(B--D); draw((2,0)--(2,-1));
draw(A--(2,-1)); draw(B--E);
pair[] ps={A,B,C,D,E};
dot(ps);
label("$A$",A,N);
label("$B$",B,N);
label("$C$",C,W);
label("$D$",D,S);
label("$E$",E,S);
label("$1$",(D--E),S);
label("$1$",(A--B),N);
label("$2$",(A--E),W);
label("$\sqrt{5}$",(B--E),NW);
// Block 2
unitsize(2cm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; pair A=(0,0), B=(1,0); pair C=(0.8,-0.4); pair D=(1,-2), E=(0,-2); draw(A--(2,0)); draw((0,-1)--(2,-1)); draw(E--D); draw(A--E); draw(B--D); draw((2,0)--(2,-1)); draw(A--(2,-1)); draw(B--E); pair[] ps={A,B,C,D,E}; dot(ps); label("$A$",A,N); label("$B$",B,N); label("$C$",C,W); label("$D$",D,S); label("$E$",E,S); label("$1$",(D--E),S); label("$1$",(A--B),N); label("$2$",(A--E),W); label("$\sqrt{5}$",(B--E),NW); | [] |
720 | Three unit squares and two line segments connecting two pairs of vertices are shown. What is the area of $\triangle ABC$?
$\textbf{(A)}\ \frac16 \qquad\textbf{(B)}\ \frac15 \qquad\textbf{(C)}\ \frac29 \qquad\textbf{(D)}\ \frac13 \qquad\textbf{(E)}\ \frac{\sqrt{2}}{4}$ | 2012 AMC 10A Problem 15 | Let $\text{E}$ be the origin. Then,
$\text{D}=(1, 0)$
$\text{A}=(0, 2)$
$\text{B}=(1, 2)$
$\text{F}=(2, 1)$
${EB}$ can be represented by the line $y=2x$
Also, ${AF}$ can be represented by the line $y=-\frac{1}{2}x+2$
Subtracting the second equation from the first gives us $\frac{5}{2}x-2=0$.
Thus, $x=\frac{4}{5}$.
Plugging this into the first equation gives us $y=\frac{8}{5}$.
Since $\text{C} (0.8, 1.6)$, $G$ is $(0.8, 2)$,
${AB}=1$ and ${CG}=0.4$.
Thus, $[ABC]=\frac{1}{2} \cdot {AB} \cdot {CG}=\frac{1}{2} \cdot 1 \cdot 0.4=0.2=\frac{1}{5}$. The answer is $\boxed{\textbf{(B)}\ \frac15}$.
Note: Once you have determined that $\text{C} (0.8, 1.6)$, you can use Shoelace Theorem to determine the area of triangle ABC. | // Block 1
unitsize(2cm);
defaultpen(linewidth(.8pt)+fontsize(10pt));
dotfactor=4;
pair A=(0,0), B=(1,0); pair C=(0.8,-0.4);
pair D=(1,-2), E=(0,-2); pair F=(2,-1); pair G=(0.8,0);
draw(A--(2,0)); draw((0,-1)--F); draw(E--D);
draw(A--E); draw(B--D); draw((2,0)--F);
draw(A--F); draw(B--E); draw(C--G);
pair[] ps={A,B,C,D,E,F,G};
dot(ps);
label("$A$",A,N);
label("$B$",B,N);
label("$C$",C,W);
label("$D$",D,S);
label("$E$",E,S);
label("$F$",F,E);
label("$G$",G,N);
// Block 2
unitsize(2cm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; pair A=(0,0), B=(1,0); pair C=(0.8,-0.4); pair D=(1,-2), E=(0,-2); pair F=(2,-1); pair G=(0.8,0); draw(A--(2,0)); draw((0,-1)--F); draw(E--D); draw(A--E); draw(B--D); draw((2,0)--F); draw(A--F); draw(B--E); draw(C--G); pair[] ps={A,B,C,D,E,F,G}; dot(ps); label("$A$",A,N); label("$B$",B,N); label("$C$",C,W); label("$D$",D,S); label("$E$",E,S); label("$F$",F,E); label("$G$",G,N); | [] |
720 | Three unit squares and two line segments connecting two pairs of vertices are shown. What is the area of $\triangle ABC$?
$\textbf{(A)}\ \frac16 \qquad\textbf{(B)}\ \frac15 \qquad\textbf{(C)}\ \frac29 \qquad\textbf{(D)}\ \frac13 \qquad\textbf{(E)}\ \frac{\sqrt{2}}{4}$ | 2012 AMC 10A Problem 15 | Triangle $EAB$ is similar to triangle $EHI$; line $HI = 1/2$
Triangle $ACB$ is similar to triangle $FCI$ and the ratio of line $AB$ to line $IF = 1 : \frac{3}{2} = 2: 3$.
Based on similarity the length of the height of $GC$ is thus $\frac{2}{5}\cdot1 = \frac{2}{5}$.
Thus, $[ABC]=\frac{1}{2} \cdot {AB} \cdot {CG}=\frac{1}{2} \cdot 1 \cdot \frac{2}{5}=\frac{1}{5}$.
The answer is $\boxed{\textbf{(B)}\ \frac15}$ | // Block 1
unitsize(2cm);
defaultpen(linewidth(.8pt)+fontsize(10pt));
dotfactor=4;
pair A=(0,0), B=(1,0); pair C=(0.8,-0.4);
pair D=(1,-2), E=(0,-2); pair F=(2,-1); pair G=(0.8,0);
pair H=(0,-1), I=(0.5,-1);
draw(A--(2,0)); draw((0,-1)--F); draw(E--D);
draw(A--E); draw(B--D); draw((2,0)--F);
draw(A--F); draw(B--E); draw(C--G);
draw(H--I);
pair[] ps={A,B,C,D,E,F,G, H, I};
dot(ps);
label("$A$",A,N);
label("$B$",B,N);
label("$C$",C,W);
label("$D$",D,S);
label("$E$",E,S);
label("$F$",F,E);
label("$G$",G,N);
label("$H$",H,W);
label("$I$",I,E);
// Block 2
unitsize(2cm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; pair A=(0,0), B=(1,0); pair C=(0.8,-0.4); pair D=(1,-2), E=(0,-2); pair F=(2,-1); pair G=(0.8,0); pair H=(0,-1), I=(0.5,-1); draw(A--(2,0)); draw((0,-1)--F); draw(E--D); draw(A--E); draw(B--D); draw((2,0)--F); draw(A--F); draw(B--E); draw(C--G); draw(H--I); pair[] ps={A,B,C,D,E,F,G, H, I}; dot(ps); label("$A$",A,N); label("$B$",B,N); label("$C$",C,W); label("$D$",D,S); label("$E$",E,S); label("$F$",F,E); label("$G$",G,N); label("$H$",H,W); label("$I$",I,E); | [] |
721 | The closed curve in the figure is made up of 9 congruent circular arcs each of length $\frac{2\pi}{3}$, where each of the centers of the corresponding circles is among the vertices of a regular hexagon of side 2. What is the area enclosed by the curve?
$\textbf{(A)}\ 2\pi+6\qquad\textbf{(B)}\ 2\pi+4\sqrt{3}\qquad\textbf{(C)}\ 3\pi+4\qquad\textbf{(D)}\ 2\pi+3\sqrt{3}+2\qquad\textbf{(E)}\ \pi+6\sqrt{3}$ | 2012 AMC 10A Problem 18 | We can draw the hexagon between the centers of the circles, and compute its area. The hexagon is made of $6$ equilateral triangles each with length $2$, so the area is:
\[\frac{\sqrt{3}}{4} \cdot 2^2 \cdot 6=6 \sqrt{3}.\]
Then, we add the areas of the three sectors outside the hexagon:
\[\frac 23 \pi \cdot 3=2\pi.\]
We now subtract the areas of the three sectors inside the hexagon but outside the figure (which is $\pi$) to get the area enclosed in the curved figure:
\[6 \sqrt{3}+2\pi-\pi=\pi+6\sqrt{3}.\]
Hence, our answer is $\boxed{\textbf{(E)}\ \pi+6\sqrt{3}},$ and we are done.
\[\]
(Minor edits, made by dbnl.) | // Block 1
unitsize(2cm);
defaultpen(fontsize(6pt));
dotfactor=4;
label("$\circ$",(0,1));
label("$\circ$",(0.865,0.5));
label("$\circ$",(-0.865,0.5));
label("$\circ$",(0.865,-0.5));
label("$\circ$",(-0.865,-0.5));
label("$\circ$",(0,-1));
dot((0,1.5));
dot((-0.4325,0.75));
dot((0.4325,0.75));
dot((-0.4325,-0.75));
dot((0.4325,-0.75));
dot((-0.865,0));
dot((0.865,0));
dot((-1.2975,-0.75));
dot((1.2975,-0.75));
draw(Arc((0,1),0.5,210,-30));
draw(Arc((0.865,0.5),0.5,150,270));
draw(Arc((0.865,-0.5),0.5,90,-150));
draw(Arc((0.865,-0.5),0.5,90,-150));
draw(Arc((0,-1),0.5,30,150));
draw(Arc((-0.865,-0.5),0.5,330,90));
draw(Arc((-0.865,0.5),0.5,-90,30));
draw((0,1)--(0.865, 0.5)--(0.865,-0.5)--(0,-1)--(-0.865,-0.5)--(-0.865,0.5)--(0,1));
// Block 2
unitsize(2cm); defaultpen(fontsize(6pt)); dotfactor=4; label("$\circ$",(0,1)); label("$\circ$",(0.865,0.5)); label("$\circ$",(-0.865,0.5)); label("$\circ$",(0.865,-0.5)); label("$\circ$",(-0.865,-0.5)); label("$\circ$",(0,-1)); dot((0,1.5)); dot((-0.4325,0.75)); dot((0.4325,0.75)); dot((-0.4325,-0.75)); dot((0.4325,-0.75)); dot((-0.865,0)); dot((0.865,0)); dot((-1.2975,-0.75)); dot((1.2975,-0.75)); draw(Arc((0,1),0.5,210,-30)); draw(Arc((0.865,0.5),0.5,150,270)); draw(Arc((0.865,-0.5),0.5,90,-150)); draw(Arc((0.865,-0.5),0.5,90,-150)); draw(Arc((0,-1),0.5,30,150)); draw(Arc((-0.865,-0.5),0.5,330,90)); draw(Arc((-0.865,0.5),0.5,-90,30)); draw((0,1)--(0.865, 0.5)--(0.865,-0.5)--(0,-1)--(-0.865,-0.5)--(-0.865,0.5)--(0,1)); | [] |
721 | The closed curve in the figure is made up of 9 congruent circular arcs each of length $\frac{2\pi}{3}$, where each of the centers of the corresponding circles is among the vertices of a regular hexagon of side 2. What is the area enclosed by the curve?
$\textbf{(A)}\ 2\pi+6\qquad\textbf{(B)}\ 2\pi+4\sqrt{3}\qquad\textbf{(C)}\ 3\pi+4\qquad\textbf{(D)}\ 2\pi+3\sqrt{3}+2\qquad\textbf{(E)}\ \pi+6\sqrt{3}$ | 2012 AMC 10A Problem 18 | As you can see, this diagram looks like a fidget spinner ;). Fidget spinners aside, we need to add stuff to our diagram to make it look easier. In the directions, they were talking about the centers of each arc create a hexagon, so let's add that to our diagram.
The side length of the hexagon is 2 and if we plug it in to the area of a regular hexagon formula $\frac{3\sqrt 3}{2}s^2$ we get $6\sqrt 3$.
Note that the interior angles of a regular hexagon is 120 because of the formula $\frac{180(n-2)}{n}$ where n is the number of sides. Knowing that, each sector is $\frac{1}{3}$ of a circle. This would mean the three sectors inside the hexagon altogether equal a full circle. Knowing that the radius is 1, we can use the area of a circle $\pi r^2$ and subtract it to $6\sqrt 3$. Thus we get the total area of $6\sqrt 3 - \pi$.
Notice that we have three sectors exterior to the hexagon. Realize that the central angles of a circle always sum up to 360. Since we know one of the central angles is equal to 120, then we subtract 360 to 120 which gives us 240. Knowing that, each sector is $\frac{2}{3}$ of a circle and since there is 3 of them, $\frac{2}{3}*3=2$ circles. To find the area of those circles, we have to use $\pi r^2$ again, but since there are 2 circles, then it is $2\pi r^2$, which gives us $2\pi$.
Now we have enough information to find the total area,
$(6\sqrt 3 -\pi+2\pi)=\textbf{(E)}\ \pi+6\sqrt{3}$
~ghfhgvghj10 | // Block 1
defaultpen(fontsize(6pt));
dotfactor=4;
label("$\circ$",(0,1));
label("$\circ$",(0.865,0.5));
label("$\circ$",(-0.865,0.5));
label("$\circ$",(0.865,-0.5));
label("$\circ$",(-0.865,-0.5));
label("$\circ$",(0,-1));
dot((0,1.5));
dot((-0.432,0.75));
dot((0.4325,0.75));
dot((-0.4325,-0.75));
dot((0.4325,-0.75));
dot((-0.865,0));
dot((0.865,0));
dot((-1.2975,-0.75));
dot((1.2975,-0.75));
draw(Arc((0,1),0.5,210,-30));
draw(Arc((0.865,0.5),0.5,150,270));
draw(Arc((0.865,-0.5),0.5,90,-150));
draw(Arc((0.865,-0.5),0.5,90,-150));
draw(Arc((0,-1),0.5,30,150));
draw(Arc((-0.865,-0.5),0.5,330,90));
draw(Arc((-0.865,0.5),0.5,-90,30));
draw((0,1)--(0.865, 0.5)--(0.865,-0.5)--(0,-1)--(-0.865,-0.5)--(-0.865,0.5)--(0,1));
// Block 2
defaultpen(fontsize(6pt)); dotfactor=4; label("$\circ$",(0,1)); label("$\circ$",(0.865,0.5)); label("$\circ$",(-0.865,0.5)); label("$\circ$",(0.865,-0.5)); label("$\circ$",(-0.865,-0.5)); label("$\circ$",(0,-1)); dot((0,1.5)); dot((-0.432,0.75)); dot((0.4325,0.75)); dot((-0.4325,-0.75)); dot((0.4325,-0.75)); dot((-0.865,0)); dot((0.865,0)); dot((-1.2975,-0.75)); dot((1.2975,-0.75)); draw(Arc((0,1),0.5,210,-30)); draw(Arc((0.865,0.5),0.5,150,270)); draw(Arc((0.865,-0.5),0.5,90,-150)); draw(Arc((0.865,-0.5),0.5,90,-150)); draw(Arc((0,-1),0.5,30,150)); draw(Arc((-0.865,-0.5),0.5,330,90)); draw(Arc((-0.865,0.5),0.5,-90,30)); draw((0,1)--(0.865, 0.5)--(0.865,-0.5)--(0,-1)--(-0.865,-0.5)--(-0.865,0.5)--(0,1)); | [] |
722 | Let points $A$ = $(0 ,0 ,0)$, $B$ = $(1, 0, 0)$, $C$ = $(0, 2, 0)$, and $D$ = $(0, 0, 3)$. Points $E$, $F$, $G$, and $H$ are midpoints of line segments $\overline{BD},\text{ } \overline{AB}, \text{ } \overline {AC},$ and $\overline{DC}$ respectively. What is the area of $EFGH$?
$\textbf{(A)}\ \sqrt{2}\qquad\textbf{(B)}\ \frac{2\sqrt{5}}{3}\qquad\textbf{(C)}\ \frac{3\sqrt{5}}{4}\qquad\textbf{(D)}\ \sqrt{3}\qquad\textbf{(E)}\ \frac{2\sqrt{7}}{3}$ | 2012 AMC 10A Problem 21 | Consider a tetrahedron with vertices at $A,B,C,D$ in the $xyz$-space. The length of $EF$ is just one-half of $AD$ because it is the midsegment of $\triangle ABD.$ The same concept applies to the other side lengths. $AD=3$ and $BC=\sqrt{1^2+2^2}=\sqrt{5}$. Then $EF=HG=\frac32$ and $EH=FG=\frac{\sqrt{5}}{2}$. The line segments lie on perpendicular planes so quadrilateral $EFGH$ is a rectangle. The area is
\[EF \cdot FG = \frac 32 \cdot \frac{\sqrt{5}}{2} = \frac{3\sqrt 5} 4\implies \boxed{\textbf C}.\] | // Block 1
import three;
draw((0,0,0)--(1,0,0)--(0,0,3)--cycle);
draw((0,0,0)--(0,2,0));
draw((0,2,0)--(0,0,3));
//EFGH
draw((0.5,0,1.5)--(0.5,0,0)--(0,1,0)--(0,1,1.5)--(0.5,0,1.5),red);
//Points
label("$E$",(0.5,0,1.5),NW);
label("$F$",(0.5,0,0),S);
label("$G$",(0,1,0),S);
label("$H$",(0,1,1.5),NE);
label("$A$",(0,0,0),NE);
label("$B$",(1,0,0),S);
label("$C$",(0,2,0),S);
label("$D$",(0,0,3),N);
// Block 2
import three; draw((0,0,0)--(1,0,0)--(0,0,3)--cycle); draw((0,0,0)--(0,2,0)); draw((0,2,0)--(0,0,3)); //EFGH draw((0.5,0,1.5)--(0.5,0,0)--(0,1,0)--(0,1,1.5)--(0.5,0,1.5),red); //Points label("$E$",(0.5,0,1.5),NW); label("$F$",(0.5,0,0),S); label("$G$",(0,1,0),S); label("$H$",(0,1,1.5),NE); label("$A$",(0,0,0),NE); label("$B$",(1,0,0),S); label("$C$",(0,2,0),S); label("$D$",(0,0,3),N); | [] |
723 | Real numbers $x$, $y$, and $z$ are chosen independently and at random from the interval $[0,n]$ for some positive integer $n$. The probability that no two of $x$, $y$, and $z$ are within 1 unit of each other is greater than $\frac {1}{2}$. What is the smallest possible value of $n$?
$\textbf{(A)}\ 7\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 11$ | 2012 AMC 10A Problem 25 | Since $x,y,z$ are all reals located in $[0, n]$, the number of choices for each one is continuous so we use geometric probability.
WLOG(Without loss of generality), assume that $n\geq x \geq y \geq z \geq 0$. Then the set of points $(x,y,z)$ is a tetrahedron, or a triangular pyramid. The point $(x,y,z)$ distributes uniformly in this region. If this is not easy to understand, read Solution II.
The altitude of the tetrahedron is $n$ and the base is an isosceles right triangle with a leg length $n$. The volume is $V_1=\dfrac{n^3}{6}$, as shown in the first figure in red.
Now we will find the region with points satisfying $|x-y|\geq1$, $|y-z|\geq1$, $|z-x|\geq1$.
Since $n\geq x \geq y \geq z \geq 0$, we have $x-y\geq1$, $y-z\geq1$.
The region of points $(x,y,z)$ satisfying the condition is shown in the second figure in black. It is a tetrahedron, too.
The volume of this region is $V_2=\dfrac{(n-2)^3}{6}$.
So the probability is $p=\dfrac{V_2}{V_1}=\dfrac{(n-2)^3}{n^3}$.
Substituting $n$ with the values in the choices, we find that when $n=10$, $p=\frac{512}{1000}>\frac{1}{2}$, when $n=9$, $p=\frac{343}{729}<\frac{1}{2}$. So $n\geq 10$.
So the answer is $\boxed{\textbf{(D)}\ 10}$. | // Block 1
import three;
unitsize(10cm);
size(150);
currentprojection=orthographic(1/2,-1,2/3);
// three - currentprojection, orthographic
draw((1,1,0)--(0,1,0)--(0,0,0),dashed);
draw((0,0,0)--(0,0,1));
draw((0,1,0)--(0,1,1),dashed);
draw((1,1,0)--(1,1,1));
draw((1,0,0)--(1,0,1));
draw((0,0,1)--(1,0,1)--(1,1,1)--(0,1,1)--cycle);
draw((0,0,0)--(1,0,0)--(1,1,0)--(1,1,1), red);
draw((1,1,0)--(0,0,0)--(1,1,1), dashed+red);
draw((1,1,1)--(1,0,0), red);
// Block 2
import three;
unitsize(10cm);
size(150);
currentprojection=orthographic(1/2, -1, 2/3);
// three - currentprojection, orthographic
draw((1, 1, 0)--(0, 1, 0)--(0, 0, 0), dashed+green);
draw((0, 0, 0)--(0, 0, 1), green);
draw((0, 1, 0)--(0, 1, 1), dashed+green);
draw((1, 0, 0)--(1, 0, 1), green);
draw((0, 0, 1)--(1, 0, 1)--(1, 1, 1)--(0, 1, 1)--cycle, green);
draw((1,0,0)--(1,1,0)--(0,0,0)--(1,1,1), dashed+red);
draw((0,0,0)--(1,0,0)--(1,1,1), red);
draw((1,1,1)--(1,1,0)--(1,0.9,0), red);
draw((1, 0.1, 0)--(1, 0.9, 0)--(1, 0.9, 0.8)--cycle);
draw((0.2, 0.1, 0)--(1, 0.9, 0.8),dashed);
draw((1, 0.1, 0)--(0.2, 0.1, 0)--(1, 0.9, 0),dashed);
// Block 3
import three; unitsize(10cm); size(150); currentprojection=orthographic(1/2,-1,2/3); // three - currentprojection, orthographic draw((1,1,0)--(0,1,0)--(0,0,0),dashed); draw((0,0,0)--(0,0,1)); draw((0,1,0)--(0,1,1),dashed); draw((1,1,0)--(1,1,1)); draw((1,0,0)--(1,0,1)); draw((0,0,1)--(1,0,1)--(1,1,1)--(0,1,1)--cycle); draw((0,0,0)--(1,0,0)--(1,1,0)--(1,1,1), red); draw((1,1,0)--(0,0,0)--(1,1,1), dashed+red); draw((1,1,1)--(1,0,0), red);
// Block 4
import three; unitsize(10cm); size(150); currentprojection=orthographic(1/2, -1, 2/3); // three - currentprojection, orthographic draw((1, 1, 0)--(0, 1, 0)--(0, 0, 0), dashed+green); draw((0, 0, 0)--(0, 0, 1), green); draw((0, 1, 0)--(0, 1, 1), dashed+green); draw((1, 0, 0)--(1, 0, 1), green); draw((0, 0, 1)--(1, 0, 1)--(1, 1, 1)--(0, 1, 1)--cycle, green); draw((1,0,0)--(1,1,0)--(0,0,0)--(1,1,1), dashed+red); draw((0,0,0)--(1,0,0)--(1,1,1), red); draw((1,1,1)--(1,1,0)--(1,0.9,0), red); draw((1, 0.1, 0)--(1, 0.9, 0)--(1, 0.9, 0.8)--cycle); draw((0.2, 0.1, 0)--(1, 0.9, 0.8),dashed); draw((1, 0.1, 0)--(0.2, 0.1, 0)--(1, 0.9, 0),dashed); | [] |
724 | A bug crawls along a number line, starting at $-2$. It crawls to $-6$, then turns around and crawls to $5$. How many units does the bug crawl altogether?
$\textbf{(A)}\ 9\qquad\textbf{(B)}\ 11\qquad\textbf{(C)}\ 13\qquad\textbf{(D)}\ 14\qquad\textbf{(E)}\ 15$ | 2012 AMC 12A Problem 1 | Crawling from $-2$ to $-6$ takes it a distance of $4$ units. Crawling from $-6$ to $5$ takes it a distance of $11$ units. Add $4$ and $11$ to get $\boxed{\textbf{(E)}\ 15}$ | draw((-2,1)--(-6,1),red+dashed,EndArrow); draw((-6,2)--(5,2),blue+dashed,EndArrow); dot((-2,0)); dot((-6,0)); dot((5,0)); label("$-2$",(-2,0),dir(270)); label("$-6$",(-6,0),dir(270)); label("$5$",(5,0),dir(270)); label("$4$",(-4,0.9),dir(270)); label("$11$",(-1.5,2.5),dir(90)); | [] |
725 | A triangle has area $30$, one side of length $10$, and the median to that side of length $9$. Let $\theta$ be the acute angle formed by that side and the median. What is $\sin{\theta}$?
$\textbf{(A)}\ \frac{3}{10}\qquad\textbf{(B)}\ \frac{1}{3}\qquad\textbf{(C)}\ \frac{9}{20}\qquad\textbf{(D)}\ \frac{2}{3}\qquad\textbf{(E)}\ \frac{9}{10}$ | 2012 AMC 12A Problem 10 | $AB$ is the side of length $10$, and $CD$ is the median of length $9$. The altitude of $C$ to $AB$ is $6$ because the 0.5(altitude)(base)=Area of the triangle. $\theta$ is $\angle CDE$. To find $\sin{\theta}$, just use opposite over hypotenuse with the right triangle $\triangle DCE$. This is equal to $\frac69=\boxed{\textbf{(D)}\ \frac23}$. | // Block 1
unitsize(5mm);
defaultpen(linewidth(.8pt)+fontsize(10pt));
dotfactor=4;
pair A=(-5,0), B=(5,0), C=(sqrt(45),6), D=(0,0), E=(sqrt(45),0);
draw(A--B--C--cycle);
draw(D--C); draw(E--C);
pair[] ps={A,B,C,D,E};
dot(ps);
label("$A$",A,SW);
label("$B$",B,SE);
label("$C$",C,N);
label("$D$",D,SW);
label("$E$",E,S);
label("$9$",(D--C),NW);
label("$6$",(E--C));
label("$10$",(A--B),SE);
// Block 2
unitsize(5mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; pair A=(-5,0), B=(5,0), C=(sqrt(45),6), D=(0,0), E=(sqrt(45),0); draw(A--B--C--cycle); draw(D--C); draw(E--C); pair[] ps={A,B,C,D,E}; dot(ps); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,N); label("$D$",D,SW); label("$E$",E,S); label("$9$",(D--C),NW); label("$6$",(E--C)); label("$10$",(A--B),SE); | [] |
726 | A square region $ABCD$ is externally tangent to the circle with equation $x^2+y^2=1$ at the point $(0,1)$ on the side $CD$. Vertices $A$ and $B$ are on the circle with equation $x^2+y^2=4$. What is the side length of this square?
$\textbf{(A)}\ \frac{\sqrt{10}+5}{10}\qquad\textbf{(B)}\ \frac{2\sqrt{5}}{5}\qquad\textbf{(C)}\ \frac{2\sqrt{2}}{3}\qquad\textbf{(D)}\ \frac{2\sqrt{19}-4}{5}\qquad\textbf{(E)}\ \frac{9-\sqrt{17}}{5}$ | 2012 AMC 12A Problem 12 | The circles have radii of $1$ and $2$. Draw the triangle shown in the figure above and write expressions in terms of $s$ (length of the side of the square) for the sides of the triangle. Because $AO$ is the radius of the larger circle, which is equal to $2$, we can write the Pythagorean Theorem.
\begin{align*} \left( \frac{s}{2} \right) ^2 + (s+1)^2 &= 2^2\\ \frac14 s^2 + s^2 + 2s + 1 &= 4\\ \frac54 s^2 +2s - 3 &= 0\\ 5s^2 + 8s - 12 &=0 \end{align*}
Use the quadratic formula.
\[s = \frac{-8+\sqrt{8^2-4(5)(-12)}}{10} = \frac{-8+\sqrt{304}}{10} = \frac{-8+4\sqrt{19}}{10} = \boxed{\textbf{(D)}\ \frac{2\sqrt{19}-4}{5}}\] | // Block 1
unitsize(15mm);
defaultpen(linewidth(.8pt)+fontsize(10pt));
dotfactor=4;
real a=1; real b=2;
pair O=(0,0);
pair A=(-(sqrt(19)-2)/5,1);
pair B=((sqrt(19)-2)/5,1);
pair C=((sqrt(19)-2)/5,1+2(sqrt(19)-2)/5);
pair D=(-(sqrt(19)-2)/5,1+2(sqrt(19)-2)/5);
pair E=(-(sqrt(19)-2)/5,0);
path inner=Circle(O,a);
path outer=Circle(O,b);
draw(outer); draw(inner);
draw(A--B--C--D--cycle);
draw(O--D--E--cycle);
label("$A$",D,NW);
label("$E$",E,SW);
label("$O$",O,SE);
label("$s+1$",(D--E),W);
label("$\frac{s}{2}$",(E--O),S);
pair[] ps={A,B,C,D,E,O};
dot(ps);
// Block 2
unitsize(15mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; real a=1; real b=2; pair O=(0,0); pair A=(-(sqrt(19)-2)/5,1); pair B=((sqrt(19)-2)/5,1); pair C=((sqrt(19)-2)/5,1+2(sqrt(19)-2)/5); pair D=(-(sqrt(19)-2)/5,1+2(sqrt(19)-2)/5); pair E=(-(sqrt(19)-2)/5,0); path inner=Circle(O,a); path outer=Circle(O,b); draw(outer); draw(inner); draw(A--B--C--D--cycle); draw(O--D--E--cycle); label("$A$",D,NW); label("$E$",E,SW); label("$O$",O,SE); label("$s+1$",(D--E),W); label("$\frac{s}{2}$",(E--O),S); pair[] ps={A,B,C,D,E,O}; dot(ps); | [] |
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