2006 AMC 12B Problem 23

Attempt Problem 23 of the 2006 AMC 12B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2006 AMC 12B solutions, or check the answer key.

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23.

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

Answer: E

Concepts:transformation law of cosines Pythagorean Theorem

Difficulty rating: 2390

Small hint:

Rotate

\triangle ABC

by

90^\circ

about

C

so that

A

maps onto

B

Big hint:

The image of

P

forms an isosceles right triangle with

P,

which pins down

\angle BPC

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,

\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.

Problem 23 in other years