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Difference between revisions of "2015 AMC 12B Problems/Problem 19"

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First, we should find the center and radius of this circle. We can find the center by drawing the perpendicular bisectors of <math>WZ</math> and <math>XY</math> and finding their intersection point. This point happens to be the midpoint of <math>AB</math>, the hypotenuse. Let this point be <math>M</math>. To find the radius, determine <math>MY</math>, where <math>MY^{2} = MA^2 + AY^2</math>, <math>MA = \frac{12}{2} = 6</math>, and <math>AY = AB = 12</math>. Thus, the radius <math>=r =MY = 6\sqrt5</math>.
 
First, we should find the center and radius of this circle. We can find the center by drawing the perpendicular bisectors of <math>WZ</math> and <math>XY</math> and finding their intersection point. This point happens to be the midpoint of <math>AB</math>, the hypotenuse. Let this point be <math>M</math>. To find the radius, determine <math>MY</math>, where <math>MY^{2} = MA^2 + AY^2</math>, <math>MA = \frac{12}{2} = 6</math>, and <math>AY = AB = 12</math>. Thus, the radius <math>=r =MY = 6\sqrt5</math>.
  
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Next we let <math>AC = b</math> and <math>BC = a</math>. Consider the right triangle <math>ACB</math> first. Using the pythagorean theorem, we find that <math>a^2 + b^2 = 12^2 = 144</math>. Next, we let <math>M'</math> to be the midpoint of <math>WZ</math>, and we consider right triangle <math>ZM'M</math>. By the pythagorean theorem, we have that <math>\left(\frac{b}{2}\right)^2 + \left(b + \frac{a}{2}\right)^2 = r^2 = 180</math>. Expanding this equation, we get that <cmath>\frac{1}{4}(a^2+b^2) + b^2 + ab = 180</cmath> <cmath>\frac{144}{4} + b^2 + ab = 180</cmath> <cmath>b^2 + ab = 144 = a^2 + b^2</cmath> <cmath>ab = a^2</cmath> <cmath>b = a</cmath> This means that <math>ABC</math> is a <math>45-45-90</math> triangle, so <math>a = b = \frac{12}{\sqrt2} = 6\sqrt2</math>. Thus the perimeter is <math>a + b + AB = 12\sqrt2 + 12</math> which is answer <math>\boxed{\textbf{(C)} 12 + 12\sqrt2}</math>
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Next we let <math>AC = b</math> and <math>BC = a</math>. Consider the right triangle <math>ACB</math> first. Using the pythagorean theorem, we find that <math>a^2 + b^2 = 12^2 = 144</math>. Next, we let <math>M'</math> to be the midpoint of <math>WZ</math>, and we consider right triangle <math>ZM'M</math>. By the pythagorean theorem, we have that <math>\left(\frac{b}{2}\right)^2 + \left(b + \frac{a}{2}\right)^2 = r^2 = 180</math>. Expanding this equation, we get that
 +
 
 +
<cmath>\frac{1}{4}(a^2+b^2) + b^2 + ab = 180</cmath>
 +
<cmath>\frac{144}{4} + b^2 + ab = 180</cmath>
 +
<cmath>b^2 + ab = 144 = a^2 + b^2</cmath>
 +
<cmath>ab = a^2</cmath>
 +
<cmath>b = a</cmath>
 +
 
 +
This means that <math>ABC</math> is a 45-45-90 triangle, so <math>a = b = \frac{12}{\sqrt2} = 6\sqrt2</math>. Thus the perimeter is <math>a + b + AB = 12\sqrt2 + 12</math> which is answer <math>\boxed{\textbf{(C)}\; 12 + 12\sqrt2}</math>.
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2015|ab=B|num-a=20|num-b=18}}
 
{{AMC12 box|year=2015|ab=B|num-a=20|num-b=18}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 09:20, 5 March 2015

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

First, we should find the center and radius of this circle. We can find the center by drawing the perpendicular bisectors of $WZ$ and $XY$ and finding their intersection point. This point happens to be the midpoint of $AB$, the hypotenuse. Let this point be $M$. To find the radius, determine $MY$, where $MY^{2} = MA^2 + AY^2$, $MA = \frac{12}{2} = 6$, and $AY = AB = 12$. Thus, the radius $=r =MY = 6\sqrt5$.

Next we let $AC = b$ and $BC = a$. Consider the right triangle $ACB$ first. Using the pythagorean theorem, we find that $a^2 + b^2 = 12^2 = 144$. Next, we let $M'$ to be the midpoint of $WZ$, and we consider right triangle $ZM'M$. By the pythagorean theorem, we have that $\left(\frac{b}{2}\right)^2 + \left(b + \frac{a}{2}\right)^2 = r^2 = 180$. Expanding this equation, we get that

\[\frac{1}{4}(a^2+b^2) + b^2 + ab = 180\] \[\frac{144}{4} + b^2 + ab = 180\] \[b^2 + ab = 144 = a^2 + b^2\] \[ab = a^2\] \[b = a\]

This means that $ABC$ is a 45-45-90 triangle, so $a = b = \frac{12}{\sqrt2} = 6\sqrt2$. Thus the perimeter is $a + b + AB = 12\sqrt2 + 12$ which is answer $\boxed{\textbf{(C)}\; 12 + 12\sqrt2}$.

See Also

2015 AMC 12B (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 12 Problems and Solutions

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