Difference between revisions of "2006 AMC 10A Problems/Problem 17"
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== Problem == | == Problem == | ||
− | In rectangle <math>ADEH</math>, points <math>B</math> and <math>C</math> trisect <math>\overline{AD}</math>, and points <math>G</math> and <math>F</math> trisect <math>\overline{HE}</math>. In addition, <math>AH=AC=2</math>. What is the area of quadrilateral <math>WXYZ</math> shown in the figure? | + | In [[rectangle]] <math>ADEH</math>, points <math>B</math> and <math>C</math> [[trisect]] <math>\overline{AD}</math>, and points <math>G</math> and <math>F</math> trisect <math>\overline{HE}</math>. In addition, <math>AH=AC=2</math>. What is the area of [[quadrilateral]] <math>WXYZ</math> shown in the figure? |
<math>\mathrm{(A) \ } \frac{1}{2}\qquad\mathrm{(B) \ } \frac{\sqrt{2}}{2}\qquad\mathrm{(C) \ } \frac{\sqrt{3}}{2}\qquad\mathrm{(D) \ } \frac{2\sqrt{2}}{2}\qquad\mathrm{(E) \ } \frac{2\sqrt{3}}{3}\qquad</math> | <math>\mathrm{(A) \ } \frac{1}{2}\qquad\mathrm{(B) \ } \frac{\sqrt{2}}{2}\qquad\mathrm{(C) \ } \frac{\sqrt{3}}{2}\qquad\mathrm{(D) \ } \frac{2\sqrt{2}}{2}\qquad\mathrm{(E) \ } \frac{2\sqrt{3}}{3}\qquad</math> | ||
− | [[Image:2006_AMC10A-17.png]] | + | <!-- [[Image:2006_AMC10A-17.png]] --> |
− | + | <asy> | |
+ | size(7cm); pathpen = linewidth(0.7); pointpen = black; pointfontpen = fontsize(10); | ||
+ | pair A,B,C,D,E,F,G,H,W,X,Y,Z; | ||
+ | A=(0,2); B=(1,2); C=(2,2); D=(3,2); | ||
+ | H=(0,0); G=(1,0); F=(2,0); E=(3,0); | ||
+ | D('A',A, N); D('B',B,N); D('C',C,N); D('D',D,N); D('E',E,SE); D('F',F,SE); D('G',G,SW); D('H',H,SW); | ||
+ | D(A--F); D(B--E); D(D--G); D(C--H); | ||
+ | Z=IP(A--F, C--H); Y=IP(A--F, D--G); X=IP(B--E,D--G); W=IP(B--E,C--H); | ||
+ | D('W',W,1.6*N); D('X',X,1.6*plain.E); D('Y',Y,1.6*S); D('Z',Z,1.6*plain.W); | ||
+ | D(A--D--E--H--cycle); | ||
+ | </asy> | ||
__TOC__ | __TOC__ | ||
== Solution == | == Solution == | ||
=== Solution 1 === | === Solution 1 === | ||
It is not difficult to see by [[symmetry]] that <math>WXYZ</math> is a [[square]]. | It is not difficult to see by [[symmetry]] that <math>WXYZ</math> is a [[square]]. | ||
+ | <!-- [[Image:2006_AMC10A-17a.png]] --> | ||
+ | <asy> | ||
+ | size(7cm); pathpen = linewidth(0.7); pointpen = black; pointfontpen = fontsize(10); | ||
+ | pair A,B,C,D,E,F,G,H,W,X,Y,Z; | ||
+ | A=(0,2); B=(1,2); C=(2,2); D=(3,2); | ||
+ | H=(0,0); G=(1,0); F=(2,0); E=(3,0); | ||
+ | D('A',A, N); D('B',B,N); D('C',C,N); D('D',D,N); D('E',E,SE); D('F',F,SE); D('G',G,SW); D('H',H,SW); | ||
+ | D(A--F); D(B--E); D(D--G); D(C--H); | ||
+ | Z=IP(A--F, C--H); Y=IP(A--F, D--G); X=IP(B--E,D--G); W=IP(B--E,C--H); | ||
+ | D('W',W,1.6*N); D('X',X,1.6*plain.E); D('Y',Y,1.6*S); D('Z',Z,1.6*plain.W); | ||
+ | MP("3",(A+B)/2,2*N); MP("2",(A+H)/2,plain.W); D(B--Z); MP("1",(B+Z)/2,plain.W); MP("\frac{\sqrt{2}}{2}",(W+Z)/2,plain.SE); | ||
+ | D(A--D--E--H--cycle); | ||
+ | </asy> | ||
+ | Draw <math>\overline{BZ}</math>. Clearly <math>BZ = \frac 12AH = 1</math>. Then <math>\triangle BWZ</math> is [[isosceles]], and is a <math>45-45-90 \triangle</math>. Hence <math>WZ = \frac{1}{\sqrt{2}}</math>, and <math>[WXYZ] = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac 12\ \mathrm{(A)}</math>. | ||
− | + | There are many different similar ways to come to the same conclusion using different [[right triangle|45-45-90 triangles]]. | |
− | |||
− | |||
− | |||
− | There are many different similar ways to come to the same conclusion using different 45-45-90 triangles. | ||
=== Solution 2 === | === Solution 2 === | ||
− | [[Image:2006_AMC10A-17b.png]] | + | <!-- [[Image:2006_AMC10A-17b.png]] --> |
+ | <asy> | ||
+ | size(7cm); pathpen = linewidth(0.7); pointpen = black; pointfontpen = fontsize(10); | ||
+ | pair A,B,C,D,E,F,G,H,W,X,Y,Z; | ||
+ | A=(0,2); B=(1,2); C=(2,2); D=(3,2); | ||
+ | H=(0,0); G=(1,0); F=(2,0); E=(3,0); | ||
+ | D('A',A, N); D('B',B,N); D('C',C,N); D('D',D,N); D('E',E,SE); D('F',F,SE); D('G',G,SW); D('H',H,SW); | ||
+ | D(A--F); D(B--E); D(D--G); D(C--H); | ||
+ | Z=IP(A--F, C--H); Y=IP(A--F, D--G); X=IP(B--E,D--G); W=IP(B--E,C--H); | ||
+ | D('W',W,1.6*N); D('X',X,1.6*plain.E); D('Y',Y,1.6*S); D('Z',Z,1.6*plain.W); | ||
+ | D(B--D((A+H)/2)--G);D(C--D((E+D)/2)--F); | ||
+ | D(A--D--E--H--cycle); | ||
+ | </asy> | ||
Draw the lines as shown above, and count the squares. There are 12, so we have <math>\frac{2\cdot 3}{12} = \frac 12</math>. | Draw the lines as shown above, and count the squares. There are 12, so we have <math>\frac{2\cdot 3}{12} = \frac 12</math>. | ||
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[[Category:Introductory Geometry Problems]] | [[Category:Introductory Geometry Problems]] | ||
+ | [[Category:Area Problems]] |
Revision as of 14:55, 20 August 2011
Problem
In rectangle , points and trisect , and points and trisect . In addition, . What is the area of quadrilateral shown in the figure?
Contents
[hide]Solution
Solution 1
It is not difficult to see by symmetry that is a square. Draw . Clearly . Then is isosceles, and is a . Hence , and .
There are many different similar ways to come to the same conclusion using different 45-45-90 triangles.
Solution 2
Draw the lines as shown above, and count the squares. There are 12, so we have .
See Also
2006 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 16 |
Followed by Problem 18 | |
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 AMC 10 Problems and Solutions |