1993 AMC 12 Problem 23

Attempt Problem 23 of the 1993 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 1993 AMC 12 solutions, or check the answer key.

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23.

Points

A,

B,

C

and

D

are on a circle of diameter

1,

and

X

is on diameter

\overline{AD}.

If

BX=CX

and

3\angle BAC=\angle BXC=36^\circ,

then

AX=

\cos6^\circ\cos12^\circ\sec18^\circ

\cos6^\circ\sin12^\circ\csc18^\circ

\cos6^\circ\sin12^\circ\sec18^\circ

\sin6^\circ\sin12^\circ\csc18^\circ

\sin6^\circ\sin12^\circ\sec18^\circ

Answer: B

Concepts:circle geometry angle bisector law of sines

Difficulty rating: 2280

Small hint:

Because

BX=CX,

the diameter through

X

bisects

\angle BXC

Big hint:

Find

AB

using right triangle

ABD,

then apply the Law of Sines in

\triangle ABX

Solution:

Since

BX=CX,

the line

AD

bisects

\angle BXC,

and symmetry also gives

\angle BAD=6^\circ.

Because

AD=1

is a diameter,

\angle ABD=90^\circ,

so

AB=\cos6^\circ.

Also

\begin{gathered} \angle AXB=180^\circ-18^\circ=162^\circ, \\ \angle ABX=180^\circ-162^\circ\\ {}-6^\circ=12^\circ. \end{gathered}

The Law of Sines in

\triangle ABX

gives

\begin{aligned} AX&=\frac{AB\sin12^\circ}{\sin162^\circ}\\ &=\cos6^\circ\sin12^\circ\\ &\qquad{}\csc18^\circ. \end{aligned}

Thus the correct answer is B.

Problem 23 in other years