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2006 AMC 10A Problems/Problem 17: Difference between revisions

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~made by AoPS  (somewhere) -put here by qkddud~
~made by AoPS  (somewhere) -put here by qkddud~
== Solution 5 (Proof Bash) ==
<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>
By symmetry, quadrilateral <math>WXYZ</math> is a square.
First step, proving that <math>\triangle BXD \sim \triangle BWC</math>.
We can tell quadrilateral <math>CDGH</math> is a parallelogram because <math>CD \parallel GH</math> and <math>CD \cong GH</math>.
By knowing that, we can say that <math>CW \parallel DX</math>.
Finally, we can now prove <math>\triangle BXD \sim \triangle BWC</math> by AA, with a ratio of 2:1.
Since <math>BD = DE = 2</math> and <math>\angle BDE = 90</math>. Then <math>\triangle BDE</math> is a 45-45-90 triangle.
This will make <math>\angle DBE = 45</math> making <math>\triangle BXD</math> and <math>\triangle BWC</math> a 45-45-90 triangle.
This will make, <math>BD = 2, BC=1, DX=BX=\sqrt 2, BW=WX=\frac{\sqrt 2}{2}</math>. Since <math>WX</math> is the length of the square, our answer will be <math>(\frac{\sqrt 2}{2})^2 = \frac{2}{4} = \boxed{\textbf{(A) }\frac{1}{2}}.</math>
~ghfhgvghj10


==Video Solution by the Beauty of Math==
==Video Solution by the Beauty of Math==

Revision as of 23:21, 1 February 2023

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?

[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]

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

Solution 1

By symmetry, $WXYZ$ 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 $\overline{BZ}$. $BZ = \frac 12AH = 1$, so $\triangle BWZ$ is a $45-45-90 \triangle$. Hence $WZ = \frac{1}{\sqrt{2}}$, and $[WXYZ] = \left(\frac{1}{\sqrt{2}}\right)^2 =\boxed{\textbf{(A) }\frac 12}$.

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]

Drawing lines as shown above and piecing together the triangles, we see that $ABCD$ is made up of $12$ squares congruent to $WXYZ$. Hence $[WXYZ] = \frac{2\cdot 3}{12} =\boxed{\textbf{(A) }\frac 12}$.

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 $(\sqrt{\frac{1^2}{2}})^2=(\sqrt{\frac{1}{2}})^2=\boxed{\textbf{(A) }\frac{1}{2}}$ which is our answer.

Solution 4

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 \[XW^2 = \left( \frac{1}{\sqrt{2}} \right)^2 = \boxed{\textbf{(A) }\frac{1}{2}}.\]

~made by AoPS (somewhere) -put here by qkddud~

Solution 5 (Proof Bash)

[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]


By symmetry, quadrilateral $WXYZ$ is a square.

First step, proving that $\triangle BXD \sim \triangle BWC$.

We can tell quadrilateral $CDGH$ is a parallelogram because $CD \parallel GH$ and $CD \cong GH$.

By knowing that, we can say that $CW \parallel DX$.

Finally, we can now prove $\triangle BXD \sim \triangle BWC$ by AA, with a ratio of 2:1.

Since $BD = DE = 2$ and $\angle BDE = 90$. Then $\triangle BDE$ is a 45-45-90 triangle.

This will make $\angle DBE = 45$ making $\triangle BXD$ and $\triangle BWC$ a 45-45-90 triangle.

This will make, $BD = 2, BC=1, DX=BX=\sqrt 2, BW=WX=\frac{\sqrt 2}{2}$. Since $WX$ is the length of the square, our answer will be $(\frac{\sqrt 2}{2})^2 = \frac{2}{4} = \boxed{\textbf{(A) }\frac{1}{2}}.$

~ghfhgvghj10

Video Solution by the Beauty of Math

https://www.youtube.com/watch?v=GX33rxlJz7s

~IceMatrix

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

These problems are copyrighted © by the Mathematical Association of America.

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