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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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==Easiest Solution==
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 +
We can obviously see that the pentagon is made of two congruent triangles. We can fit one triangle into the gap in the upper square. Therefore, the answer is just <math>\frac{1}{3}\implies\boxed{C}</math>
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==Solution 1==
 
==Solution 1==

Revision as of 00:55, 2 February 2015

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]

Easiest Solution

We can obviously see that the pentagon is made of two congruent triangles. We can fit one triangle into the gap in the upper square. Therefore, the answer is just $\frac{1}{3}\implies\boxed{C}$


Solution 1

[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); 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); label("$X$", X,SE); dot("$J$", 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 $\frac{4}{3\cdot 2^2} = \boxed{\textbf{(C)}\hspace{.05in}\frac{1}{3}}$.

Solution 2

[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= (1.25,1); 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(X,red); 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,SW,red); label("$J$", J, SE);[/asy]

Let the side length of each square be $1$.

Let the intersection of $AJ$ and $EI$ be $X$.

Since $[ABCD]=[GHIJ]$, $AD=IJ$. Since $\angle IXJ$ and $\angle AXD$ are vertical angles, they are congruent. We also have $\angle JIH\cong\angle ADC$ by definition.

So we have $\triangle ADX\cong\triangle JIX$ by $\textit{AAS}$ congruence. Therefore, $DX=JX$.

Since $C$ and $D$ are midpoints of sides, $DH=CJ=\dfrac{1}{2}$. This combined with $DX=JX$ yields $HX=CX=\dfrac{1}{2}\times \dfrac{1}{2}=\dfrac{1}{4}$.

The area of trapezoid $ABCX$ is $\dfrac{1}{2}(AB+CX)(BC)=\dfrac{1}{2}\times \dfrac{5}{4}\times 1=\dfrac{5}{8}$.

The area of triangle $JIX$ is $\dfrac{1}{2}\times XJ\times IJ=\dfrac{1}{2}\times \dfrac{3}{4}\times 1=\dfrac{3}{8}$.

So the area of the pentagon $AJICB$ is $\dfrac{3}{8}+\dfrac{5}{8}=1$.

The area of the $3$ squares is $1\times 3=3$.

Therefore, $\dfrac{[AJICB]}{[ABCIJFED]}= \boxed{\textbf{(C)}\hspace{.05in}\frac{1}{3}}$.

Solution 3

[asy] pair A,B,C,D,E,F,G,H,I,J,K; 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); K= (1.25,1); 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(K,red); 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("$K$", K,SW,red); label("$J$", J, SE);[/asy]

Let the intersection of $AJ$ and $EI$ be $K$.

Now we have $\triangle ADK$ and $\triangle KIJ$.

Because both triangles has a side on congruent squares therefore $AD \cong IJ$.

Because $\angle AKD$ and $\angle JKI$ are vertical angles $\angle AKD \cong \angle JKI$.

Also both $\angle ADK$ and $\angle JIK$ are right angles so $\angle ADK \cong \angle JIK$.

Therefore by AAS(Angle, Angle, Side) $\triangle ADK \cong \triangle KIJ$.

Then translating/rotating the shaded $\triangle JIK$ into the position of $\triangle ADK$

So the shaded area now completely covers the square $ABCD$

Set the area of a square as $x$

Therefore, $\frac{x}{3x}= \boxed{\textbf{(C)}\hspace{.05in}\frac{1}{3}}$.


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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