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

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dot("$J$", J, SE);
 
dot("$J$", J, SE);
 
</asy>
 
</asy>
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==Solution 1 (shortcut)==
 +
It can be proven that <math>\Delta ADX \cong \Delta JIX</math> (where <math>X</math> is the point where <math>\overline {AJ}</math> intersects <math>\overline {HC}</math>) which also means quadrilaterals <math>ABCX \cong JGHX</math> (due to the squares being equal in area which means the squares are congruent, and since the triangles earlier mentioned are congruent). The area of the shaded region is equal to the area of one square since the quadrilaterals and triangles are congruent. The total area of the shape is the area of three squares. Putting these two pieces of information together, <math>\boxed{\textbf{(C)}\ \frac {1}{3}}</math>
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~ julia333

Revision as of 20:21, 29 August 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]

Solution 1 (shortcut)

It can be proven that $\Delta ADX \cong \Delta JIX$ (where $X$ is the point where $\overline {AJ}$ intersects $\overline {HC}$) which also means quadrilaterals $ABCX \cong JGHX$ (due to the squares being equal in area which means the squares are congruent, and since the triangles earlier mentioned are congruent). The area of the shaded region is equal to the area of one square since the quadrilaterals and triangles are congruent. The total area of the shape is the area of three squares. Putting these two pieces of information together, $\boxed{\textbf{(C)}\ \frac {1}{3}}$

~ julia333

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