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Difference between revisions of "2006 AMC 10A Problems/Problem 17"
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==Solution 3==
 
==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.
 
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.
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==Solution 4==
 +
Solution:
 +
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>
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 +
~made by AoPS HW-put here by qkddud~
  
 
== See Also ==
 
== See Also ==

Revision as of 21:14, 6 August 2019

Problem

In rectangle $ADEH$, points $B$ and $C$ trisect $\overline{AD}$, and points $G$ and $F$ trisect $\overline{HE}$. In addition, $AH=AC=2$, and $AD=3$. What is the area of quadrilateral $WXYZ$ shown in the figure?

$\mathrm{(A) \ } \frac{1}{2}\qquad\mathrm{(B) \ } \frac{\sqrt{2}}{2}\qquad\mathrm{(C) \ } \frac{\sqrt{3}}{2}\qquad\mathrm{(D) \ } \frac{2\sqrt{2}}{2}\qquad\mathrm{(E) \ } \frac{2\sqrt{3}}{3}\qquad$

[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]

Solution

Solution 1

It is not difficult to see by symmetry that $WXYZ$ 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 $\overline{BZ}$. Clearly $BZ = \frac 12AH = 1$. Then $\triangle BWZ$ is isosceles, and is a $45-45-90 \triangle$. Hence $WZ = \frac{1}{\sqrt{2}}$, and $[WXYZ] = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac 12\ \mathrm{(A)}$.

There are many different similar ways to come to the same conclusion using different 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]

Draw the lines as shown above, and count the squares. There are 12, so we have $\frac{2\cdot 3}{12} = \frac 12$.

Solution 3

We see that if we draw a line to $BZ$ it is half the width of the rectangle so that length would be $1$, and the resulting triangle is a $45-45-90$ so using the Pythagorean Theorem we can get that each side is $\sqrt{\frac{1^2}{2}}$ so the area of the middle square would be $(\sqrt{\frac{1^2}{2}})^2=(\sqrt{\frac{1}{2}})^2=\frac{1}{2}$ which is our answer.

Solution 4

Solution: Since $B$ and $C$ are trisection points and $AC = 2$, we see that $AD = 3$. Also, $AC = AH$, so triangle $ACH$ is a right isosceles triangle, i.e. $\angle ACH = \angle AHC = 45^\circ$. By symmetry, triangles $AFH$, $DEG$, and $BED$ are also right isosceles triangles. Therefore, $\angle WAD = \angle WDA = 45^\circ$, which means triangle $AWD$ is also a right isosceles triangle. Also, triangle $AXC$ is a right isosceles triangle.

Then $AW = AD/\sqrt{2} = 3/\sqrt{2}$, and $AX = AC/\sqrt{2} = 2/\sqrt{2}$. Hence, $XW = AW - AX = 3/\sqrt{2} - 2/\sqrt{2} = 1/\sqrt{2}$.

By symmetry, quadrilateral $WXYZ$ is a square, so its area is \[XW^2 = \left( \frac{1}{\sqrt{2}} \right)^2 = \boxed{\frac{1}{2}}.\]

~made by AoPS HW-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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