2015 AMC 10B Problem 19

The plan: find the center of the circle first, then use "all four points are the same distance from the center".

1. Set up coordinates. Put the right angle $C$ at $(0, 0)$, $A$ at $(0, a)$ on the $y$-axis and $B$ at $(b, 0)$ on the $x$-axis. By the Pythagorean theorem, $a^2 + b^2 = 12^2 = 144$.
2. Square on $AC$. It is built on the left side of the $y$-axis (away from $B$), so its other corners are $W = (-a, 0)$ and $Z = (-a, a)$.
3. Square on $AB$. Going from $A$ to $B$ means moving $b$ right and $a$ down. Turning that step a quarter turn gives the move "$a$ right, $b$ up", which also has length $12$ and points away from $C$. So the other corners are
$$X = B + (a, b) = (a + b,\ b), \qquad Y = A + (a, b) = (a,\ a + b).$$
4. Find the center. The center of a circle is the same distance from both ends of any chord, so it lies on the line that cuts the chord in half at a right angle.
- The chord $ZW$ is a vertical segment from height $0$ to height $a$, so the center is at height $\frac{a}{2}$.
- The chord $XY$ is just $AB$ slid over, so the center is on the line that cuts $AB$ in half at a right angle.

Both lines pass through the midpoint of $AB$, which is $M = \left(\frac{b}{2}, \frac{a}{2}\right)$. So $M$ is the center.
5. Use equal distances. The center is equally far from $W$ and $X$. Compare squared distances:
$$MW^2 = \left(a + \frac{b}{2}\right)^2 + \left(\frac{a}{2}\right)^2, \qquad MX^2 = \left(a + \frac{b}{2}\right)^2 + \left(b - \frac{a}{2}\right)^2.$$
The first parts match, so the second parts must match:
$$\left(\frac{a}{2}\right)^2 = \left(b - \frac{a}{2}\right)^2.$$
So $b - \frac{a}{2}$ is either $\frac{a}{2}$ or $-\frac{a}{2}$. The second choice gives $b = 0$, which is not a triangle. So $b = a$.
6. Perimeter. The legs are equal and $a^2 + a^2 = 144$, so $a^2 = 72$ and $a = 6\sqrt2$. The perimeter is
$$6\sqrt2 + 6\sqrt2 + 12 = 12 + 12\sqrt2.$$

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

The center of a circle is equally far from both ends of every chord, so it sits on the line that cuts each chord in half at a right angle. Here two chords were copies of the triangle's own sides, which put the center at a point we already knew. After that, one "equal distances" equation fixed the shape. When some points lie on a circle, look for the center first.

Place $C = (0,0)$, $A = (0, a)$, $B = (b, 0)$ with $a^2 + b^2 = 144$.

Square $ACWZ$ sits on the side of $AC$ away from $B$: $W = (-a, 0)$, $Z = (-a, a)$. Square $ABXY$ sits on the side of $AB$ away from $C$; the outward unit direction is along $(a, b)$ scaled to length $12$, i.e. the vector $(a, b)$ itself since $\sqrt{a^2+b^2} = 12$. So $X = (a + b, b)$ and $Y = (a, a + b)$.

Find the center of the circle through $X, Y, Z, W$. The chord $ZW$ is parallel to $AC$ and shares its perpendicular bisector, the line $y = \frac{a}{2}$. The chord $XY$ is a translate of $AB$, so its perpendicular bisector is that of $AB$. The two bisectors meet at the circumcenter of right triangle $ABC$, which is the midpoint of the hypotenuse, $M = \left(\frac{b}{2}, \frac{a}{2}\right)$.

Now set $MW = MX$:
$$MW^2 = \left(\frac{b}{2} + a\right)^2 + \frac{a^2}{4}, \qquad MX^2 = \left(a + \frac{b}{2}\right)^2 + \left(b - \frac{a}{2}\right)^2.$$
Cancelling the common term, $\frac{a^2}{4} = \left(b - \frac{a}{2}\right)^2$, so $b - \frac{a}{2} = \pm\frac{a}{2}$, giving $b = a$ (the option $b = 0$ is degenerate).

Thus the triangle is an isosceles right triangle with legs $\frac{12}{\sqrt2} = 6\sqrt2$, and the perimeter is $12 + 12\sqrt2$.

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