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Difference between revisions of "2015 AMC 10B Problems/Problem 19"

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==Solution 2==
 
==Solution 2==
  
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Let <math>AC = b</math> and <math>AB = c</math>. Draw line segments <math>YZ</math> and <math>WX</math>. Now we have cyclic quadrilateral <math>WXYZ</math>. This means that opposite angles sum to <math>180^{\circ}</math>. Therefore, <math>90 + m\angle YZA + 90 - m\angle WXB = 180</math>. Simplifying carefully, we get <math>m\angle YZA = m\angle WXB</math>. Similarly, you can find that <math>m\angle{ZYA}</math> = <math>m\angle{XWB}</math>. So, we know that <math>\triangle ZYA</math> is similar to <math>\triangle WBX</math>. So <math>BC = \sqrt{c^2-b^2}</math>. Now, we can use the proportion <math>\dfrac{c}{b} = \dfrac{b + \sqrt{c^2-b^2}}{c}</math>. Simplifying, we get <math>c^2 - b^2 = b \sqrt{c^2-b^2}</math>, which gives us <math>c = b\sqrt{2}</math>. Therefore, <math>\triangle ABC</math> is an isosceles right triangle, so <math>AC=BC=\dfrac{12}{\sqrt{2}}=6\sqrt{2}</math>, and the perimeter is <math>\boxed{\textbf{(C)}\ 12+12\sqrt{2}}</math>.
+
Let <math>AC = b</math> and <math>BC = a</math> (and we're given that <math>AB=12</math>). Draw line segments <math>YZ</math> and <math>WX</math>. Now we have cyclic quadrilateral <math>WXYZ.</math>  
 +
 
 +
This means that opposite angles sum to <math>180^{\circ}</math>. Therefore, <math>90 + m\angle YZA + 90 - m\angle WXB = 180</math>. Simplifying carefully, we get <math>m\angle YZA = m\angle WXB</math>. Similarly, <math>m\angle{ZYA}</math> = <math>m\angle{XWB}</math>.  
 +
 
 +
That means <math>\triangle ZYA \sim \triangle XWB</math>.  
 +
 
 +
Setting up proportions,
 +
<math>\dfrac{b}{12}=\dfrac{12}{a+b}.</math>
 +
Cross-multiplying we get:
 +
<math>b^2+ab=12^2</math>
 +
 
 +
But also, by Pythagoras,  
 +
<math>b^2+a^2=12^2</math>, so <math>ab=a^2 \Rightarrow a=b</math>
 +
 
 +
Therefore, <math>\triangle ABC</math> is an isosceles right triangle. <math>AC=BC=\dfrac{12}{\sqrt{2}}=6\sqrt{2}</math>, so the perimeter is <cmath>\boxed{\textbf{(C)}\ 12+12\sqrt{2}}</cmath>
 +
 
 +
~BakedPotato66
  
 
~LegionOfAvatars
 
~LegionOfAvatars

Revision as of 17:01, 6 September 2020

Problem

In $\triangle{ABC}$, $\angle{C} = 90^{\circ}$ and $AB = 12$. Squares $ABXY$ and $ACWZ$ are constructed outside of the triangle. The points $X, Y, Z$, and $W$ lie on a circle. What is the perimeter of the triangle?

$\textbf{(A) }12+9\sqrt{3}\qquad\textbf{(B) }18+6\sqrt{3}\qquad\textbf{(C) }12+12\sqrt{2}\qquad\textbf{(D) }30\qquad\textbf{(E) }32$

Solution 1

The center of the circle lies on the perpendicular bisectors of both chords $ZW$ and $YX$. Therefore we know the center of the circle must also be the midpoint of the hypotenuse. Let this point be $O$. Draw perpendiculars to $ZW$ and $YX$ from $O$, and connect $OZ$ and $OY$. $OY^2=6^2+12^2=180$. Let $AC=a$ and $BC=b$. Then $\left(\dfrac{a}{2}\right)^2+\left(a+\dfrac{b}{2}\right)^2=OZ^2=OY^2=180$. Simplifying this gives $\dfrac{a^2}{4}+\dfrac{b^2}{4}+a^2+ab=180$. But by Pythagorean Theorem on $\triangle ABC$, we know $a^2+b^2=144$, because $AB=12$. Thus $\dfrac{a^2}{4}+\dfrac{b^2}{4}=\dfrac{144}{4}=36$. So our equation simplifies further to $a^2+ab=144$. However $a^2+b^2=144$, so $a^2+ab=a^2+b^2$, which means $ab=b^2$, or $a=b$. Aha! This means $\triangle ABC$ is just an isosceles right triangle, so $AC=BC=\dfrac{12}{\sqrt{2}}=6\sqrt{2}$, and thus the perimeter is $\boxed{\textbf{(C)}\ 12+12\sqrt{2}}$. [asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(11.5cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -4.3, xmax = 18.7, ymin = -5.26, ymax = 6.3; /* image dimensions */ draw((3.46,0.96)--(3.44,-3.36)--(8.02,-3.44)--cycle); draw((3.46,0.96)--(8.02,-3.44)--(12.42,1.12)--(7.86,5.52)--cycle); /* draw figures */ draw((3.46,0.96)--(3.44,-3.36)); draw((3.44,-3.36)--(8.02,-3.44)); draw((8.02,-3.44)--(3.46,0.96)); draw((3.46,0.96)--(-0.86,0.98)); draw((-0.86,0.98)--(-0.88,-3.34)); draw((-0.88,-3.34)--(3.44,-3.36)); draw((3.46,0.96)--(8.02,-3.44)); draw((8.02,-3.44)--(12.42,1.12)); draw((12.42,1.12)--(7.86,5.52)); draw((7.86,5.52)--(3.46,0.96)); draw((5.74,-1.24)--(-0.86,0.98)); draw((5.74,-1.24)--(-0.87,-1.18), linetype("4 4")); draw((5.74,-1.24)--(7.86,5.52)); draw((5.74,-1.24)--(10.14,3.32), linetype("4 4")); draw(shift((5.82,-1.21))*xscale(6.99920709795045)*yscale(6.99920709795045)*arc((0,0),1,19.44457562540183,197.63600413408128), linetype("2 2")); /* dots and labels */ dot((3.46,0.96),dotstyle); label("$A$", (3.2,1.06), NE * labelscalefactor); dot((3.44,-3.36),dotstyle); label("$C$", (3.14,-3.86), NE * labelscalefactor); dot((8.02,-3.44),dotstyle); label("$B$", (8.06,-3.8), NE * labelscalefactor); dot((-0.86,0.98),dotstyle); label("$Z$", (-1.34,1.12), NE * labelscalefactor); dot((-0.88,-3.34),dotstyle); label("$W$", (-1.48,-3.54), NE * labelscalefactor); dot((12.42,1.12),dotstyle); label("$X$", (12.5,1.24), NE * labelscalefactor); dot((7.86,5.52),dotstyle); label("$Y$", (7.94,5.64), NE * labelscalefactor); dot((5.74,-1.24),dotstyle); label("$O$", (5.52,-1.82), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]

Solution 2

Let $AC = b$ and $BC = a$ (and we're given that $AB=12$). Draw line segments $YZ$ and $WX$. Now we have cyclic quadrilateral $WXYZ.$

This means that opposite angles sum to $180^{\circ}$. Therefore, $90 + m\angle YZA + 90 - m\angle WXB = 180$. Simplifying carefully, we get $m\angle YZA = m\angle WXB$. Similarly, $m\angle{ZYA}$ = $m\angle{XWB}$.

That means $\triangle ZYA \sim \triangle XWB$.

Setting up proportions, $\dfrac{b}{12}=\dfrac{12}{a+b}.$ Cross-multiplying we get: $b^2+ab=12^2$

But also, by Pythagoras, $b^2+a^2=12^2$, so $ab=a^2 \Rightarrow a=b$

Therefore, $\triangle ABC$ is an isosceles right triangle. $AC=BC=\dfrac{12}{\sqrt{2}}=6\sqrt{2}$, so the perimeter is \[\boxed{\textbf{(C)}\ 12+12\sqrt{2}}\]

~BakedPotato66

~LegionOfAvatars

See Also

2015 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
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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