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| dot("$J$", J, SE); | | dot("$J$", J, SE); |
| </asy> | | </asy> |
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− | ==Solution 1==
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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,red);
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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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− | dot("$A$", A, NW);
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− | dot("$B$", B, NE);
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− | dot("$C$", C, NE);
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− | dot("$D$", D, NW);
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− | dot("$E$", E, NW);
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− | dot("$F$", F, SW);
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− | dot("$G$", G, S);
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− | dot("$H$", H, N);
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− | dot("$I$", I, NE);
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− | label("$X$", X,SE);
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− | dot("$J$", 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>.
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− | ==Solution 2==
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− |
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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= (1.25,1);
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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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− | dot(X,red);
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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("$X$", X,SW,red);
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− | label("$J$", J, SE);</asy>
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− | Let the side length of each square be <math>1</math>.
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− |
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− | Let the intersection of <math>AJ</math> and <math>EI</math> be <math>X</math>.
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− | Since <math>[ABCD]=[GHIJ]</math>, <math>AD=IJ</math>. Since <math>\angle IXJ</math> and <math>\angle AXD</math> are vertical angles, they are congruent. We also have <math>\angle JIH\cong\angle ADC</math> by definition.
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− | So we have <math>\triangle ADX\cong\triangle JIX</math> by <math>\textit{AAS}</math> congruence. Therefore, <math>DX=JX</math>.
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− |
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− | Since <math>C</math> and <math>D</math> are midpoints of sides, <math>DH=CJ=\dfrac{1}{2}</math>. This combined with <math>DX=JX</math> yields <math>HX=CX=\dfrac{1}{2}\times \dfrac{1}{2}=\dfrac{1}{4}</math>.
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− |
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− | The area of trapezoid <math>ABCX</math> is <math>\dfrac{1}{2}(AB+CX)(BC)=\dfrac{1}{2}\times \dfrac{5}{4}\times 1=\dfrac{5}{8}</math>.
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− | The area of triangle <math>JIX</math> is <math>\dfrac{1}{2}\times XJ\times IJ=\dfrac{1}{2}\times \dfrac{3}{4}\times 1=\dfrac{3}{8}</math>.
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− | So the area of the pentagon <math>AJICB</math> is <math>\dfrac{3}{8}+\dfrac{5}{8}=1</math>.
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− | The area of the <math>3</math> squares is <math>1\times 3=3</math>.
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− | Therefore, <math>\dfrac{[AJICB]}{[ABCIJFED]}= \boxed{\textbf{(C)}\hspace{.05in}\frac{1}{3}}</math>.
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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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