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2013 AMC 8 Problems/Problem 24

Revision as of 18:33, 27 November 2013 by Fadebekun (talk | contribs) (→‎Solution)

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); label("$A$", A, NW); label("$B$", B, NE); label("$C$", C, NE); label("$D$", D, NW); label("$E$", E, NW); label("$F$", F, SW); label("$G$", G, S); label("$H$", H, N); label("$I$", I, NE); label("$J$", J, SE); [/asy]

Solution

[asy] pair A,B,C,D,E,F,G,H,I,J,X; 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); X= extension(I,J,A,B); dot(X,red); draw(I--X--B,red); 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); label("$A$", A, NW); label("$B$", B, NE); label("$C$", C, NE); label("$D$", D, NW); label("$E$", E, NW); label("$F$", F, SW); label("$G$", G, S); label("$H$", H, N); label("$I$", I, NE); label("$X$", X,SE); label("$J$", J, SE);[/asy]


First let $s=2$ (where $s$ is the side length of the squares) for simplicity. We can extend $\overline{IJ}$ until it hits the extension of $\overline{AB}$. Call this point $X$. The area of triangle $AXJ$ then is $\dfrac{3 \cdot 4}{2}$ The area of rectangle $BXIC$ is $2 \cdot 1 = 2$. Thus, our desired area is $6-2 = 4$. Now, the ratio of the shaded area to the combined area of the three squares is $\frac{4}{3\cdot 2^2} = \boxed{\textbf{(C)}\hspace{.05in}\frac{1}{3}}$.

See Also

2013 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

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