# 2015 AMC 10B 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

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 $\boxed{\textbf{(C)}\ 12+12\sqrt{2}}$.

/* 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((3.34,-3.86)--(3.14,-3.66)--(3.34,-3.66))
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);
(Error compiling LaTeX. 24cd435845f36aa6be473672bc758b9562db7c3c.asy: 15.1: syntax error
error: could not load module '24cd435845f36aa6be473672bc758b9562db7c3c.asy')