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2013 AMC 8 Problems/Problem 24: Difference between revisions

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Therefore, <math>\dfrac{[AJICB]}{[ABCIJFED]}= \boxed{\textbf{(C)}\hspace{.05in}\frac{1}{3}}</math>.
Therefore, <math>\dfrac{[AJICB]}{[ABCIJFED]}= \boxed{\textbf{(C)}\hspace{.05in}\frac{1}{3}}</math>.
==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 <math>AJ</math> and <math>EI</math> be <math>K</math>.
Now we have <math>\triangle ADK</math> and <math>\triangle KIJ</math>.
Because both triangles has a side on congruent squares therefore <math>AD \cong IJ</math>.
Because <math>\angle AKD</math> and <math>\angle JKI</math> are vertical angles <math>\angle AKD \cong \angle JKI</math>.
Also both <math>\angle ADK</math> and <math>\angle JIK</math> are right angles so <math>\angle ADK \cong \angle JIK</math>.
Therefore by AAS(Angle, Angle, Side) <math>\triangle ADK \cong \triangle KIJ</math>.
Then translating/rotating the shaded <math>\triangle JIK</math> into the position of <math>\triangle ADK</math>
So the shaded area now completely covers the square ABCD
Set the area of a square as <math>x</math>


==See Also==
==See Also==
{{AMC8 box|year=2013|num-b=23|num-a=25}}
{{AMC8 box|year=2013|num-b=23|num-a=25}}
{{MAA Notice}}
{{MAA Notice}}

Revision as of 08:27, 28 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); 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

[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 $x$

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

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