Difference between revisions of "2013 AMC 8 Problems/Problem 24"
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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); | ||
+ | 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> | ||
+ | |||
+ | ==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 <math>\frac{1}{3}\implies\boxed{C}</math> | ||
+ | |||
+ | ==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 <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("$A$", A, NW); | ||
label("$B$", B, NE); | label("$B$", B, NE); | ||
Line 36: | Line 119: | ||
label("$H$", H, N); | label("$H$", H, N); | ||
label("$I$", I, NE); | label("$I$", I, NE); | ||
− | label("$J$", J, SE); | + | 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>. | ||
+ | |||
+ | ==Solution 3== | ||
− | |||
<asy> | <asy> | ||
− | pair A,B,C,D,E,F,G,H,I,J, | + | pair A,B,C,D,E,F,G,H,I,J,K; |
A = (0.5,2); | A = (0.5,2); | ||
B = (1.5,2); | B = (1.5,2); | ||
Line 52: | Line 156: | ||
I = (2,1); | I = (2,1); | ||
J = (2,0); | J = (2,0); | ||
− | + | K= (1.25,1); | |
− | |||
− | |||
draw(A--B); | draw(A--B); | ||
draw(C--B); | draw(C--B); | ||
Line 65: | Line 167: | ||
filldraw(A--B--C--I--J--cycle,grey); | filldraw(A--B--C--I--J--cycle,grey); | ||
draw(E--I); | draw(E--I); | ||
+ | dot(K,red); | ||
label("$A$", A, NW); | label("$A$", A, NW); | ||
label("$B$", B, NE); | label("$B$", B, NE); | ||
Line 74: | Line 177: | ||
label("$H$", H, N); | label("$H$", H, N); | ||
label("$I$", I, NE); | label("$I$", I, NE); | ||
+ | label("$K$", K,SW,red); | ||
label("$J$", J, SE);</asy> | label("$J$", J, SE);</asy> | ||
+ | Let the intersection of <math>AJ</math> and <math>EI</math> be <math>K</math>. | ||
+ | |||
+ | Now we have <math>\triangle ADK</math> and <math>\triangle KIJ</math>. | ||
+ | |||
+ | Because both triangles has a side on congruent squares therefore <math>AD \cong IJ</math>. | ||
+ | |||
+ | Because <math>\angle AKD</math> and <math>\angle JKI</math> are vertical angles <math>\angle AKD \cong \angle JKI</math>. | ||
+ | |||
+ | Also both <math>\angle ADK</math> and <math>\angle JIK</math> are right angles so <math>\angle ADK \cong \angle JIK</math>. | ||
+ | |||
+ | Therefore by AAS(Angle, Angle, Side) <math>\triangle ADK \cong \triangle KIJ</math>. | ||
+ | |||
+ | Then translating/rotating the shaded <math>\triangle JIK</math> into the position of <math>\triangle ADK</math> | ||
− | + | So the shaded area now completely covers the square <math>ABCD</math> | |
+ | |||
+ | Set the area of a square as <math>x</math> | ||
+ | |||
+ | Therefore, <math>\frac{x}{3x}= \boxed{\textbf{(C)}\hspace{.05in}\frac{1}{3}}</math>. | ||
+ | |||
+ | ==Solution 4== | ||
+ | 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. | ||
==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 23:41, 15 October 2020
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?
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
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, .
Solution 3
Let the intersection of and be .
Now we have and .
Because both triangles has a side on congruent squares therefore .
Because and are vertical angles .
Also both and are right angles so .
Therefore by AAS(Angle, Angle, Side) .
Then translating/rotating the shaded into the position of
So the shaded area now completely covers the square
Set the area of a square as
Therefore, .
Solution 4
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.
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.