Question

Isosceles $\triangle ABC$ has a right angle at $C.$ Point $P$ is inside $\triangle ABC,$ such that $PA = 11,$ $PB = 7,$ and $PC = 6.$ Legs $\overline{AC}$ and $\overline{BC}$ have length $s = \sqrt{a + b\sqrt{2}},$ where $a$ and $b$ are positive integers. What is $a + b?$

$85$

$91$

$108$

$121$

$127$

Solution:

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 $\triangle BPC,$ $BC^2 = 6^2 + 7^2 - 2 \cdot 6 \cdot 7 \cos 135^\circ = 85 + 42\sqrt2.$

So $s^2 = 85 + 42\sqrt2,$ giving $a = 85,$ $b = 42,$ and $a + b = 127.$

Thus, the correct answer is E.

2006 AMC 12B Problem 23

Solution:

Problem 23 in Other Years