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>, and <math>AD=3</math>. What is the area of [[quadrilateral]] <math>WXYZ</math> shown in the figure? |
+ | |||
+ | <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> | ||
+ | |||
+ | <math>\mathrm{(A) \ } \frac{1}{2}\qquad\mathrm{(B) \ } \frac{\sqrt{2}}{2}\qquad\mathrm{(C) \ } \frac{\sqrt{3}}{2}\qquad\mathrm{(D)} \sqrt{2} \qquad\mathrm{(E) \ } \frac{2\sqrt{3}}{3}\qquad</math> | ||
− | |||
== Solution == | == Solution == | ||
− | == | + | === Solution 1 === |
− | *[[ | + | By [[symmetry]], <math>WXYZ</math> is a [[square]]. |
+ | |||
+ | <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("1",(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>. <math>BZ = \frac 12AH = 1</math>, so <math>\triangle BWZ</math> 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]]. | ||
+ | |||
+ | === Solution 2 === | ||
+ | <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> | ||
+ | |||
+ | Drawing lines as shown above and piecing together the triangles, we see that <math>ABCD</math> is made up of 12 squares congruent to <math>WXYZ</math>. Hence <math>WXYZ = \frac{2\cdot 3}{12} = \frac 12</math>. | ||
+ | |||
+ | === Solution 3 === | ||
+ | We see that if we draw a line to <math>BZ</math> it is half the width of the rectangle so that length would be <math>1</math>, and the resulting triangle is a <math>45-45-90</math> so using the Pythagorean Theorem we can get that each side is <math>\sqrt{\frac{1^2}{2}}</math> so the area of the middle square would be <math>(\sqrt{\frac{1^2}{2}})^2=(\sqrt{\frac{1}{2}})^2=\frac{1}{2}</math> which is our answer. | ||
+ | |||
+ | === Solution 4 === | ||
+ | |||
+ | Since <math>B</math> and <math>C</math> are trisection points and <math>AC = 2</math>, we see that <math>AD = 3</math>. Also, <math>AC = AH</math>, so triangle <math>ACH</math> is a right isosceles triangle, i.e. <math>\angle ACH = \angle AHC = 45^\circ</math>. By symmetry, triangles <math>AFH</math>, <math>DEG</math>, and <math>BED</math> are also right isosceles triangles. Therefore, <math>\angle WAD = \angle WDA = 45^\circ</math>, which means triangle <math>AWD</math> is also a right isosceles triangle. Also, triangle <math>AXC</math> is a right isosceles triangle. | ||
+ | |||
+ | Then <math>AW = AD/\sqrt{2} = 3/\sqrt{2}</math>, and <math>AX = AC/\sqrt{2} = 2/\sqrt{2}</math>. Hence, <math>XW = AW - AX = 3/\sqrt{2} - 2/\sqrt{2} = 1/\sqrt{2}</math>. | ||
+ | |||
+ | By symmetry, quadrilateral <math>WXYZ</math> is a square, so its area is | ||
+ | <cmath>XW^2 = \left( \frac{1}{\sqrt{2}} \right)^2 = \boxed{\frac{1}{2}}.</cmath> | ||
− | + | ~made by AoPS (somewhere) -put here by qkddud~ | |
− | + | == See Also == | |
+ | {{AMC10 box|year=2006|num-b=16|num-a=18|ab=A}} | ||
[[Category:Introductory Geometry Problems]] | [[Category:Introductory Geometry Problems]] | ||
+ | [[Category:Area Problems]] | ||
+ | {{MAA Notice}} |
Revision as of 23:18, 22 October 2020
Contents
Problem
In rectangle , points and trisect , and points and trisect . In addition, , and . What is the area of quadrilateral shown in the figure?
Solution
Solution 1
Draw . , so is a . Hence , and .
There are many different similar ways to come to the same conclusion using different 45-45-90 triangles.
Solution 2
Drawing lines as shown above and piecing together the triangles, we see that is made up of 12 squares congruent to . Hence .
Solution 3
We see that if we draw a line to it is half the width of the rectangle so that length would be , and the resulting triangle is a so using the Pythagorean Theorem we can get that each side is so the area of the middle square would be which is our answer.
Solution 4
Since and are trisection points and , we see that . Also, , so triangle is a right isosceles triangle, i.e. . By symmetry, triangles , , and are also right isosceles triangles. Therefore, , which means triangle is also a right isosceles triangle. Also, triangle is a right isosceles triangle.
Then , and . Hence, .
By symmetry, quadrilateral is a square, so its area is
~made by AoPS (somewhere) -put here by qkddud~
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 |
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