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

(Solution)
m (Solution)
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==Solution==
 
==Solution==
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<asy>
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pair A,B,C,D,E,F,G,H,I,J,X;
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A = (0.5,2);
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B = (1.5,2);
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C = (1.5,1);
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D = (0.5,1);
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E = (0,1);
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F = (0,0);
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G = (1,0);
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H = (1,1);
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I = (2,1);
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J = (2,0);
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X= extension(I,J,A,B);
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dot(X);
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draw(I--X--B,red);
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draw(A--B);
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draw(C--B);
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draw(D--A);
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draw(F--E);
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draw(I--J);
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draw(J--F);
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draw(G--H);
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draw(A--J);
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filldraw(A--B--C--I--J--cycle,grey);
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draw(E--I);
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label("$A$", A, NW);
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label("$B$", B, NE);
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label("$C$", C, NE);
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label("$D$", D, NW);
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label("$E$", E, NW);
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label("$F$", F, SW);
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label("$G$", G, S);
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label("$H$", H, N);
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label("$I$", I, NE);
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label("$J$", J, SE);</asy>
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First let <math>s=2</math> (where <math>s</math> is the side length of the squares) for simplicity. We can extend <math>\overline{IJ}</math> until it hits the extension of <math>\overline{AB}</math>. Call this point <math>X</math>. The area of triangle <math>AXJ</math> then is <math>\dfrac{3 \cdot 4}{2}</math> The area of rectangle <math>BXIC</math> is <math>2 \cdot 1 = 2</math>. Thus, our desired area is <math>6-2 = 4</math>. Now, the ratio of the shaded area to the combined area of the three squares is <math>\frac{4}{3\cdot 2^2} = \boxed{\textbf{(C)}\hspace{.05in}\frac{1}{3}}</math>.
 
First let <math>s=2</math> (where <math>s</math> is the side length of the squares) for simplicity. We can extend <math>\overline{IJ}</math> until it hits the extension of <math>\overline{AB}</math>. Call this point <math>X</math>. The area of triangle <math>AXJ</math> then is <math>\dfrac{3 \cdot 4}{2}</math> The area of rectangle <math>BXIC</math> is <math>2 \cdot 1 = 2</math>. Thus, our desired area is <math>6-2 = 4</math>. Now, the ratio of the shaded area to the combined area of the three squares is <math>\frac{4}{3\cdot 2^2} = \boxed{\textbf{(C)}\hspace{.05in}\frac{1}{3}}</math>.

Revision as of 18:32, 27 November 2013

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); 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("$J$", 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 (ProblemsAnswer KeyResources)
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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