2006 AMC 12B Problem 23

Rotate $\triangle ABC$ by $90^\circ$ about $C,$ sending $A$ to $B$ and $P$ to $P'.$ Then $CP' = CP = 6$ and $\angle PCP' = 90^\circ,$ so $\triangle PCP'$ is an isosceles right triangle with $PP' = 6\sqrt2.$ Also $BP' = AP = 11.$ Since $(6\sqrt2)^2 + 7^2 = 72 + 49$ $= 121 = 11^2,$ triangle $BPP'$ has a right angle at $P.$ Hence $\angle BPC = \angle BPP'$ $+ \angle P'PC = 90^\circ$ $+ 45^\circ = 135^\circ.$ By the Law of Cosines in $\triangle BPC,$ $$\begin{aligned} &BC^2 = 6^2 + 7^2 \\ &\quad {}- 2 \cdot 6 \cdot 7 \cos 135^\circ \\ &\quad = 85 + 42\sqrt2. \end{aligned}$$ So $s^2 = 85 + 42\sqrt2,$ giving $a = 85,$ $b = 42,$ and $a + b = 127.$ Thus, the correct answer is E.