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| − | ==Easiest Solution== | + | == Solution 1 == |
| | + | We notice that ABCX is a trapezoid with the bases AB and CX. Assuming AB is 1, we find that CX is 0.25 since it is half of CH which is 0.5. Using the area of the trapezoid formula, we calculate the area of ABCX to be 0.625 and XCIJ to be 0.375. The combined areas equal 1. Therefore, the ratio of the area of the hexagon to the three squares is 1:3 because the area of the three squares is 3. The answer is <math>\boxed{\textbf{(C)}\ \frac {1}{3}}</math>-~TheNerdwhoIsNerdy. (I don't know if this is correct, pls check). |
| | | | |
| − | 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 <math>\frac{1}{3}\implies\boxed{C}</math>
| + | ==Solution 2 (extremely simple)== |
| | + | Extend line AD such that it intersects line FG. Call the intersection K. If we assign an arbitrary value to the side lengths of the squares, such as 2, the base of the now formed triangle AJK would have a length of 3, while the height would be 4. The area of triangle AJK would then be 6, and the rectangle EDKF would have an area of 2. 6 + 2 = 8, and since all of the squares' areas add up to a total of 12 (if each has side length 2), we obtain (12 - 8)/12 for the shaded part of the diagram which yields 4/12 = <math>\boxed{\textbf{(C)}\ \frac {1}{3}}</math> |
| | | | |
| − | ==Solution 1==
| + | ~ martianrunner |
| − | <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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| | | | |
| − | | + | ==See Also== |
| − | 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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| − | | |
| − | <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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| − | | |
| − | 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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| − | | |
| − | 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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| − | | |
| − | 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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| − | | |
| − | ==Solution 3==
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| − | | |
| − | <asy>
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| − | pair A,B,C,D,E,F,G,H,I,J,K;
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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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| − | K= (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(K,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("$K$", K,SW,red);
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| − | label("$J$", J, SE);</asy>
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| − | | |
| − | Let the intersection of <math>AJ</math> and <math>EI</math> be <math>K</math>.
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| − | | |
| − | Now we have <math>\triangle ADK</math> and <math>\triangle KIJ</math>.
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| − | | |
| − | Because both triangles has a side on congruent squares therefore <math>AD \cong IJ</math>.
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| − | | |
| − | Because <math>\angle AKD</math> and <math>\angle JKI</math> are vertical angles <math>\angle AKD \cong \angle JKI</math>.
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| − | | |
| − | Also both <math>\angle ADK</math> and <math>\angle JIK</math> are right angles so <math>\angle ADK \cong \angle JIK</math>.
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| − | | |
| − | Therefore by AAS(Angle, Angle, Side) <math>\triangle ADK \cong \triangle KIJ</math>.
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| − | | |
| − | Then translating/rotating the shaded <math>\triangle JIK</math> into the position of <math>\triangle ADK</math>
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| − | | |
| − | So the shaded area now completely covers the square <math>ABCD</math>
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| − | | |
| − | Set the area of a square as <math>x</math>
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| − | | |
| − | Therefore, <math>\frac{x}{3x}= \boxed{\textbf{(C)}\hspace{.05in}\frac{1}{3}}</math>.
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| − | | |
| − | ==Solution 4==
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| − | Given the information in the problem, we see that the black area is congruent to ADHGJ. Since it is half of 2/3, it takes up 1/3 of the area.
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| − | | |
| − | ==See Also These Resources...Trust Us, They Will Help!== | |
| | {{AMC8 box|year=2013|num-b=23|num-a=25}} | | {{AMC8 box|year=2013|num-b=23|num-a=25}} |
| | {{MAA Notice}} | | {{MAA Notice}} |
Problem
Squares 
, 
, 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]](//latex.artofproblemsolving.com/1/6/9/169139bda69e81f8342f246f95cc3f636c1f9abe.png)
Solution 1
We notice that ABCX is a trapezoid with the bases AB and CX. Assuming AB is 1, we find that CX is 0.25 since it is half of CH which is 0.5. Using the area of the trapezoid formula, we calculate the area of ABCX to be 0.625 and XCIJ to be 0.375. The combined areas equal 1. Therefore, the ratio of the area of the hexagon to the three squares is 1:3 because the area of the three squares is 3. The answer is 
-~TheNerdwhoIsNerdy. (I don't know if this is correct, pls check).
Solution 2 (extremely simple)
Extend line AD such that it intersects line FG. Call the intersection K. If we assign an arbitrary value to the side lengths of the squares, such as 2, the base of the now formed triangle AJK would have a length of 3, while the height would be 4. The area of triangle AJK would then be 6, and the rectangle EDKF would have an area of 2. 6 + 2 = 8, and since all of the squares' areas add up to a total of 12 (if each has side length 2), we obtain (12 - 8)/12 for the shaded part of the diagram which yields 4/12 =
~ martianrunner
See Also
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
, 
, 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)
-~TheNerdwhoIsNerdy. (I don't know if this is correct, pls check).
