2015 AMC 12B Problems/Problem 19

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 $\boxed{\textbf{(C)} 12 + 12\sqrt2}$

2015 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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2015 AMC 12B Problems/Problem 19

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