Difference between revisions of "2013 AMC 8 Problems/Problem 24"
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− | ==Solution== | + | ==Solution 1== |
<asy> | <asy> | ||
pair A,B,C,D,E,F,G,H,I,J,X; | pair A,B,C,D,E,F,G,H,I,J,X; | ||
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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>. | 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>. | ||
+ | |||
+ | ==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 <math>1</math>. | ||
+ | |||
+ | Let the intersection of <math>AJ</math> and <math>EI</math> be <math>X</math>. | ||
+ | |||
+ | 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. | ||
+ | |||
+ | So we have <math>\triangle ADX\cong\triangle JIX</math> by <math>\textit{AAS}</math> congruence. Therefore, <math>DX=JX</math>. | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | So the area of the pentagon <math>AJICB</math> is <math>\dfrac{3}{8}+\dfrac{5}{8}=1</math>. | ||
+ | |||
+ | The area of the <math>3</math> squares is <math>1\times 3=3</math>. | ||
+ | |||
+ | Therefore, <math>\dfrac{[AJICB]}{[ABCIJFED]}= \boxed{\textbf{(C)}\hspace{.05in}\frac{1}{3}}</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 19:08, 27 November 2013
Contents
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?
Solution 1
First let (where is the side length of the squares) for simplicity. We can extend until it hits the extension of . Call this point . The area of triangle then is The area of rectangle is . Thus, our desired area is . Now, the ratio of the shaded area to the combined area of the three squares is .
Solution 2
Let the side length of each square be .
Let the intersection of and be .
Since , . Since and are vertical angles, they are congruent. We also have by definition.
So we have by congruence. Therefore, .
Since and are midpoints of sides, . This combined with yields .
The area of trapezoid is .
The area of triangle is .
So the area of the pentagon is .
The area of the squares is .
Therefore, .
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.