1994 AMC 12 Problem 13

Attempt Problem 13 of the 1994 AMC 12 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 1994 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

13.

In triangle

ABC,

AB=AC.

If there is a point

P

strictly between

A

and

B

such that

AP=PC=CB,

then

\angle A=

30^\circ

36^\circ

48^\circ

60^\circ

72^\circ

Let

\angle A=\theta

and use

AP=PC

\triangle APC

Use

PC=CB

to relate

\angle PCB

to a base angle of

\triangle ABC

Let

\angle A=\theta.

Since

AP=PC,

triangle

APC

has

\angle PAC=\angle PCA=\theta.

Since

AB=AC,

each base angle of

\triangle ABC

\frac{180^\circ-\theta}{2}.

Also

PC=CB,

so the base angles of

\triangle PCB

P

and

B

are equal; the one at

B

\frac{180^\circ-\theta}{2}.

Hence

\angle PCB=\theta.

C,

2\theta=\frac{180^\circ-\theta}{2},

giving

5\theta=180^\circ

and

\theta=36^\circ.

Thus the correct answer is B.

Problem 13 in other years