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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>, and <math>AD=3</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?  
  
−
<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]] -->
 
 
<asy>
 
<asy>
 
size(7cm); pathpen = linewidth(0.7); pointpen = black; pointfontpen = fontsize(10);
 
size(7cm); pathpen = linewidth(0.7); pointpen = black; pointfontpen = fontsize(10);
Line 16: Line 13:
 
D(A--D--E--H--cycle);
 
D(A--D--E--H--cycle);
 
</asy>
 
</asy>
−
__TOC__
 
  
−
== Solution ==
+
<math>\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</math>
−
=== Solution 1 ===
+
 
−
It is not difficult to see by [[symmetry]] that <math>WXYZ</math> is a [[square]].
+
== Solution 1 ==
−
<!-- [[Image:2006_AMC10A-17a.png]] -->
+
By [[symmetry]], <math>WXYZ</math> is a square.
 +
 
 
<asy>
 
<asy>
 
size(7cm); pathpen = linewidth(0.7); pointpen = black; pointfontpen = fontsize(10);
 
size(7cm); pathpen = linewidth(0.7); pointpen = black; pointfontpen = fontsize(10);
Line 31: Line 28:
 
Z=IP(A--F, C--H); Y=IP(A--F, D--G); X=IP(B--E,D--G); W=IP(B--E,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('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);
+
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);
 
D(A--D--E--H--cycle);
 
</asy>
 
</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>.
+
 
 +
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 =\boxed{\textbf{(A) }\frac 12}</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 [[right triangle|45-45-90 triangles]].
  
 
=== Solution 2 ===
 
=== Solution 2 ===
−
<!-- [[Image:2006_AMC10A-17b.png]] -->
 
 
<asy>
 
<asy>
 
size(7cm); pathpen = linewidth(0.7); pointpen = black; pointfontpen = fontsize(10);
 
size(7cm); pathpen = linewidth(0.7); pointpen = black; pointfontpen = fontsize(10);
Line 53: Line 50:
 
</asy>
 
</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>.
+
Drawing lines as shown above and piecing together the triangles, we see that <math>ABCD</math> is made up of <math>12</math> squares congruent to <math>WXYZ</math>. Hence <math>[WXYZ] = \frac{2\cdot 3}{12} =\boxed{\textbf{(A) }\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=\boxed{\textbf{(A) }\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{\textbf{(A) }\frac{1}{2}}.</cmath>
 +
 
 +
~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
 +
==Solution 6 (Educated guess)==
 +
Since we know that quadrilateral <math>WXYZ</math> is a square, we conclude that its area is a rational number because the side lengths may be irrational. Therefore, the answer is <math>\boxed{\textbf(A)}.</math>
 +
~elpianista227
 +
 
 +
==Video Solution by the Beauty of Math==
 +
https://www.youtube.com/watch?v=GX33rxlJz7s
 +
 
 +
~IceMatrix
  
 
== See Also ==
 
== See Also ==

Latest revision as of 09:20, 4 November 2024

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

Solution 6 (Educated guess)

Since we know that quadrilateral $WXYZ$ is a square, we conclude that its area is a rational number because the side lengths may be irrational. Therefore, the answer is $\boxed{\textbf(A)}.$ ~elpianista227

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

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