1983 AMC 12 Problem 30

Attempt Problem 30 of the 1983 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 1983 AMC 12 solutions, or check the answer key.

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30.

Distinct points

A

and

B

are on a semicircle with diameter

MN

and center

C.

The point

P

is on

CN

and

\angle CAP=\angle CBP=10^\circ.

If

\overset{\frown}{MA}=40^\circ,

then

\overset{\frown}{BN}

equals

10^\circ

15^\circ

20^\circ

25^\circ

30^\circ

Answer: C

Concepts:circle arc law of sines angle chasing

Difficulty rating: 2430

Small hint:

The

40^\circ

arc gives

\angle ACP=140^\circ

Big hint:

Compare triangles

ACP

and

BCP

using

AC=BC

and the law of sines

Solution:

Since

\overset{\frown}{MA}=40^\circ,

the central angle

\angle MCA=40^\circ,

so

\angle ACP=140^\circ.

Hence

\angle APC=180^\circ-140^\circ

{}-10^\circ=30^\circ.

Applying the law of sines in triangles

ACP

and

BCP,

and using

AC=BC,

gives

\sin\angle BPC=\sin30^\circ.

Because

A

and

B

are distinct,

\angle BPC=150^\circ.

Therefore

\begin{aligned} \angle BCN &=180^\circ-150^\circ-10^\circ\\ &=20^\circ, \end{aligned}

so

\overset{\frown}{BN}=20^\circ.

Therefore, the correct answer is C.