Latest revision as of 22:15, 20 December 2023

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$", J, SE); [/asy]$

=Video Solution for Problems 21-25

https://youtu.be/-mi3qziCuec

Video Solution

https://youtu.be/sbBjMvq5GG4 ~savannahsolver

Easy Solution1-
let side of 1 square be= 2 (doesn't matter if we are calculating ratios)
Then, total area= 3x2^2=12
Shaded area=Total area - Unshaded area
Unshaded area = [AJFEDA]
We can now extend AD to the centre point of FG. This can be called y.
△AJY[{(2+2) x 2+(2/2)}/2]=6 + [DEFY]=2x1=2
Unshaded area=6+2=8
Shaded area = 12-8=4
Hence, ratio= 4/12 = 1/3
~aarnavSOni

Easiest Solution

We can see that the Pentagon is made of two congruent shapes. 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, $\frac{x}{3x}= \boxed{\textbf{(C)}\hspace{.05in}\frac{1}{3}}$ .

Solution 4

Given the information in the problem, we see that the ABCIJ is congruent to ADHGJ. Since it is half of 2/3, it takes up 1/3 of the area.

Video Solution by OmegaLearn

https://youtu.be/j3QSD5eDpzU?t=346

2013 AMC 8 (Problems • Answer Key • Resources)
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