2015 AMC 10B Problem 19

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

Four concyclic points are pinned down by their center: it must lie on the perpendicular bisector of every chord. Two of those bisectors were already familiar lines (the bisectors of the triangle's own sides), which located the center without any computation. After that, "same distance to two vertices" is one equation, and the symmetric answer $a = b$ is the only non-degenerate solution. Look for chords parallel to known segments; they share bisectors.

Sanity check with the answer choices: an isosceles right triangle with hypotenuse $12$ has legs $6\sqrt2 \approx 8.49$, perimeter $\approx 28.97$. Choice (C) is the only one of that size that is not an integer perimeter, and a $30$-$60$-$90$ triangle can be ruled out by testing $a = 6$, $b = 6\sqrt3$ in $MW = MX$.

• Assuming the triangle is a 30-60-90 (choices (A), (B)) because 12 is a nice hypotenuse, instead of deriving the shape from the concyclic condition.
• Placing square $ABXY$ on the same side as $C$, which changes $X$ and $Y$ and breaks the bisector argument.