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| | filldraw(A--B--C--I--J--cycle,grey); | | filldraw(A--B--C--I--J--cycle,grey); |
| | draw(E--I); | | draw(E--I); |
| − | label("$A$", A, NW);
| + | dot("$A$", A, NW); |
| − | label("$B$", B, NE);
| + | dot("$B$", B, NE); |
| − | label("$C$", C, NE);
| + | dot("$C$", C, NE); |
| − | label("$D$", D, NW);
| + | dot("$D$", D, NW); |
| − | label("$E$", E, NW);
| + | dot("$E$", E, NW); |
| − | label("$F$", F, SW);
| + | dot("$F$", F, SW); |
| − | label("$G$", G, S);
| + | dot("$G$", G, S); |
| − | label("$H$", H, N);
| + | dot("$H$", H, N); |
| − | label("$I$", I, NE);
| + | dot("$I$", I, NE); |
| − | label("$J$", J, SE);
| + | dot("$J$", J, SE); |
| | </asy> | | </asy> |
| − |
| |
| − | ==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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| − |
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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>.
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| − |
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| − | ==See Also==
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| − | {{AMC8 box|year=2013|num-b=23|num-a=25}}
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| − | {{MAA Notice}}
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Latest revision as of 09:59, 16 July 2024
, 
, and 
are equal in area. Points 
and 
are the midpoints of sides 
and 
, respectively. What is the ratio of the area of the shaded pentagon 
to the sum of the areas of the three squares?
![[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]](http://latex.artofproblemsolving.com/1/6/9/169139bda69e81f8342f246f95cc3f636c1f9abe.png)
