Question

Point $P$ lies on the diagonal $AC$ of square $ABCD$ with $AP \gt CP.$ Let $O_1$ and $O_2$ be the circumcenters of triangles $ABP$ and $CDP,$ respectively. Given that $AB = 12$ and $\angle O_1PO_2 = 120^\circ,$ then $AP = \sqrt{a} + \sqrt{b},$ where $a$ and $b$ are positive integers. Find $a + b.$

答案

Solution:

Place $A = (0, 0),$ $B = (12, 0),$ $C = (12, 12),$ $D = (0, 12),$ and $P = (p, p)$ with $p \gt 6.$ Since $O_1$ is equidistant from $A$ and $B,$ it lies on $x = 6;$ setting $O_1 = (6, k)$ and equating $O_1A^2 = O_1P^2$ gives $k = p - 6,$ so $O_1 = (6,\, p - 6).$ Similarly $O_2,$ equidistant from $C$ and $D,$ is $O_2 = (6,\, p + 6).$

The vectors from $P$ to the centers are $(6 - p,\, -6)$ and $(6 - p,\, 6),$ so $\cos 120^\circ = \frac{(6 - p)^2 - 36}{(6 - p)^2 + 36} = -\frac{1}{2},$ which gives $3(p - 6)^2 = 36,$ so $(p - 6)^2 = 12$ and $p = 6 + 2\sqrt{3}.$

Then $AP = p\sqrt{2} = 6\sqrt{2} + 2\sqrt{6} = \sqrt{72} + \sqrt{24},$ so $a + b = 72 + 24 = 96.$

2011 AIME II Problem 13

Solution:

Problem 13 in Other Years