# Difference between revisions of "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 1

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 $\boxed{\textbf{(C)}\; 12 + 12\sqrt2}$. image needed

## Solution 2

The center of the circle on which $X$, $Y$, $Z$, and $W$ lie must be equidistant from each of these four points. Draw the perpendicular bisectors of $\overline{XY}$ and of $\overline{WZ}$. Note that the perpendicular bisector of $\overline{WZ}$ is parallel to $\overline{CW}$ and passes through the midpoint of $\overline{AC}$. Therefore, the triangle that is formed by $A$, the midpoint of $\overline{AC}$, and the point at which this perpendicular bisector intersects $\overline{AB}$ must be similar to $\triangle ABC$, and the ratio of a side of the smaller triangle to a side of $\triangle ABC$ is 1:2. Consequently, the perpendicular bisector of $\overline{XY}$ passes through the midpoint of $\overline{AB}$. The perpendicular bisector of $\overline{WZ}$ must include the midpoint of $\overline{AB}$ as well. Since all points on a perpendicular bisector of any two points $M$ and $N$ are equidistant from $M$ and $N$, the center of the circle must be the midpoint of $\overline{AB}$.

Now the distance between the midpoint of $\overline{AB}$ and $Z$, which is equal to the radius of this circle, is $\sqrt{12^2 + 6^2} = \sqrt{180}$. Let $a=AC$. Then the distance between the midpoint of $\overline{AB}$ and $Y$, also equal to the radius of the circle, is given by $\sqrt{\left(\frac{a}{2}\right)^2 + \left(a + \frac{\sqrt{144 - a^2}}{2}\right)^2}$ (the ratio of the similar triangles is involved here). Squaring these two expressions for the radius and equating the results, we have

$$\left(\frac{a}{2}\right)^2+\left(a+\frac{\sqrt{144-a^2}}{2}\right)^{2} = 180$$ $$144 - a^2 = a\sqrt{144-a^2}$$ $$(144-a^2)^2 = a^2(144-a^2)$$

Since $a$ cannot be equal to 12, the length of the hypotenuse of the right triangle, we can divide by $(144-a^2)$, and arrive at $a = 6\sqrt{2}$. The length of other leg of the triangle must be $\sqrt{144-72} = 6\sqrt{2}$. Thus, the perimeter of the triangle is $12+2(6\sqrt{2}) = \boxed{\textbf{(C)}\; 12+12\sqrt{2}}$.

## Solution 3

In order to solve this problem, we can search for similar triangles. Begin by drawing triangle $ABC$ and squares $ABXY$ and $ACWZ$. Draw segments $\overline{YZ}$ and $\overline{WX}$. Because we are given points $X$, $Y$, $Z$, and $W$ lie on a circle, we can conclude that $WXYZ$ forms a cyclic quadrilateral. Take $\overline{AC}$ and extend it through a point $P$ on $\overline{YZ}$. Now, we must do some angle chasing to prove that $\triangle WBX$ is similar to $\triangle YAZ$.

Let $\alpha$ denote the measure of $\angle ABC$. Following this, $\angle BAC$ measures $90 - \alpha$. By our construction, $\overline{CAP}$ is a straight line, and we know $\angle YAB$ is a right angle. Therefore, $\angle PAY$ measures $\alpha$. Also, $\angle CAZ$ is a right angle and thus, $\angle ZAP$ is a right angle. Sum $\angle ZAP$ and $\angle PAY$ to find $\angle ZAY$, which measures $90 + \alpha$. We also know that $\angle WBY$ measures $90 + \alpha$. Therefore, $\angle ZAY = \angle WBX$.

Let $\beta$ denote the measure of $\angle AZY$. It follows that $\angle WZY$ measures $90 + \beta^\circ$. Because $WXYZ$ is a cyclic quadrilateral, $\angle WZY + \angle YXW = 180^\circ$. Therefore, $\angle YXW$ must measure $90 - \beta$, and $\angle BXW$ must measure $\beta$. Therefore, $\angle AZY = \angle BXW$.

$\angle ZAY = \angle WBX$ and $\angle AZY = \angle BXW$, so $\triangle AZY \sim \triangle BXW$! Let $x = AC = WC$. By Pythagorean theorem, $BC = \sqrt{144-x^2}$. Now we have $WB = WC + BC = x + \sqrt{144-x^2}$, $BX = 12$, $YA = 12$, and $AZ = x$. We can set up an equation:

$$\frac{YA}{AZ} = \frac{WB}{BX}$$ $$\frac{12}{x} = \frac{x+\sqrt{144-x^2}}{12}$$ $$144 = x^2 + x\sqrt{144-x^2}$$ $$12^2 - x^2 = x\sqrt{144-x^2}$$ $$12^4 - 2*12^2*x^2 + x^4 = 144x^2 - x^4$$ $$2x^4 - 3(12^2)x^2 + 12^4 = 0$$ $$(2x^2 - 144)(x^2 - 144) = 0$$

Solving for $x$, we find that $x = 6\sqrt{2}$ or $x = 12$, which we omit. The perimeter of the triangle is $12 + x + \sqrt{144-x^2}$. Plugging in $x = 6\sqrt{2}$, we get $\boxed{\textbf{(C)}\; 12+12\sqrt{2}}$.

## Solution 4

We claim that $X$, $Y$, $Z$, and $W$ lie on a circle if $\triangle ACB$ is an isosceles right triangle.

Proof: If $\triangle ACB$ is a right triangle, then $\angle WAY=180º$. Therefore, $W$, $A$, and $Y$ are collinear. Since $WY$ and $YX$ form a right angle, $WX$ is the diameter of the circumcircle of $\triangle WYX$. Similarly, $Z$, $A$, and $X$ are collinear, and $ZX$ forms a right angle with $ZW$. Thus, $WX$ is also the diameter of the circumcircle of $\triangle WZX$. Therefore, since $\triangle WYX$ and $\triangle WZX$ share a circumcircle, $X$, $Y$, $Z$, and $W$ lie on a circle if $\triangle ACB$ is an isosceles triangle.

If $\triangle ACB$ is isosceles, then its legs have length $6\sqrt{2}$. The perimeter of $\triangle ACB$ is $\boxed{\textbf{(C) }12+12\sqrt{2}}$.