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Difference between revisions of "2013 AMC 8 Problems/Problem 24"

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==Problem==
 
==Problem==
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A ball with diameter 4 inches starts at point A to roll along the track shown. The track is comprised of 3 semicircular arcs whose radii are <math>R_1 = 100</math> inches, <math>R_2 = 60</math> inches, and <math>R_3 = 80</math> inches, respectively. The ball always remains in contact with the track and does not slip. What is the distance the center of the ball travels over the course from A to B?
+
Squares <math>ABCD</math>, <math>EFGH</math>, and <math>GHIJ</math> are equal in area. Points <math>C</math> and <math>D</math> are the midpoints of sides <math>IH</math> and <math>HE</math>, respectively. What is the ratio of the area of the shaded pentagon <math>AJICB</math> to the sum of the areas of the three squares?
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<asy>
 
−
size(8cm);
 
−
draw((0,0)--(480,0),linetype("3 4"));
 
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filldraw(circle((8,0),8),black);
 
−
draw((0,0)..(100,-100)..(200,0));
 
−
draw((200,0)..(260,60)..(320,0));
 
−
draw((320,0)..(400,-80)..(480,0));
 
−
draw((100,0)--(150,-50sqrt(3)),Arrow(size=4));
 
−
draw((260,0)--(290,30sqrt(3)),Arrow(size=4));
 
−
draw((400,0)--(440,-40sqrt(3)),Arrow(size=4));
 
−
</asy>
 
  
−
<math>\textbf{(A)}\ 238\pi \qquad \textbf{(B)}\ 240\pi \qquad \textbf{(C)}\ 260\pi \qquad \textbf{(D)}\ 280\pi \qquad \textbf{(E)}\ 500\pi</math>
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<math> \textbf{(A)}\hspace{.05in}\frac{1}{4}\qquad\textbf{(B)}\hspace{.05in}\frac{7}{24}\qquad\textbf{(C)}\hspace{.05in}\frac{1}{3}\qquad\textbf{(D)}\hspace{.05in}\frac{3}{8}\qquad\textbf{(E)}\hspace{.05in}\frac{5}{12}</math>
  
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==Solution==
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<asy>
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The radius of the ball is 2 inches. If you think about the ball rolling or draw a path for the ball, you see that in A and C it loses <math>2\pi*2/2=2\pi</math> inches, and it gains <math>2\pi</math> inches on B. So, the departure from the length of the track means that the answer is <math>\frac{200+120+160}{2}*\pi+(-2-2+2)*\pi=240\pi-2\pi=\boxed{\textbf{(A)} 238\pi}</math>.
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pair A,B,C,D,E,F,G,H,I,J;
  
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==See Also==
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A = (0.5,2);
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{{AMC8 box|year=2013|num-b=23|num-a=25}}
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B = (1.5,2);
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{{MAA Notice}}
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C = (1.5,1);
 +
D = (0.5,1);
 +
E = (0,1);
 +
F = (0,0);
 +
G = (1,0);
 +
H = (1,1);
 +
I = (2,1);
 +
J = (2,0);
 +
draw(A--B);
 +
draw(C--B);
 +
draw(D--A); 
 +
draw(F--E);
 +
draw(I--J);
 +
draw(J--F);
 +
draw(G--H);
 +
draw(A--J);
 +
filldraw(A--B--C--I--J--cycle,grey);
 +
draw(E--I);
 +
dot("$A$", A, NW);
 +
dot("$B$", B, NE);
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dot("$C$", C, NE);
 +
dot("$D$", D, NW);
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dot("$E$", E, NW);
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dot("$F$", F, SW);
 +
dot("$G$", G, S);
 +
dot("$H$", H, N);
 +
dot("$I$", I, NE);
 +
dot("$J$", J, SE);
 +
</asy>

Revision as of 09:59, 16 July 2024

Problem

Squares $ABCD$, $EFGH$, and $GHIJ$ are equal in area. Points $C$ and $D$ are the midpoints of sides $IH$ and $HE$, respectively. What is the ratio of the area of the shaded pentagon $AJICB$ to the sum of the areas of the three squares?

$\textbf{(A)}\hspace{.05in}\frac{1}{4}\qquad\textbf{(B)}\hspace{.05in}\frac{7}{24}\qquad\textbf{(C)}\hspace{.05in}\frac{1}{3}\qquad\textbf{(D)}\hspace{.05in}\frac{3}{8}\qquad\textbf{(E)}\hspace{.05in}\frac{5}{12}$

[asy] pair A,B,C,D,E,F,G,H,I,J; A = (0.5,2); B = (1.5,2); C = (1.5,1); D = (0.5,1); E = (0,1); F = (0,0); G = (1,0); H = (1,1); I = (2,1); J = (2,0); draw(A--B); draw(C--B); draw(D--A); draw(F--E); draw(I--J); draw(J--F); draw(G--H); draw(A--J); filldraw(A--B--C--I--J--cycle,grey); draw(E--I); dot("$A$", A, NW); dot("$B$", B, NE); dot("$C$", C, NE); dot("$D$", D, NW); dot("$E$", E, NW); dot("$F$", F, SW); dot("$G$", G, S); dot("$H$", H, N); dot("$I$", I, NE); dot("$J$", J, SE); [/asy]

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