1990 AMC 12 Problem 14

Attempt Problem 14 of the 1990 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 1990 AMC 12 solutions, or check the answer key.

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14.

An acute isosceles triangle,

ABC,

is inscribed in a circle. Through

B

and

C,

tangents to the circle are drawn, meeting at point

D.

\angle ABC=\angle ACB=2\angle D

and

x

is the radian measure of

\angle A,

then

x=

\frac37\pi

\frac49\pi

\frac5{11}\pi

\frac6{13}\pi

\frac7{15}\pi

Express each base angle in terms of

x

The angle between the tangents is

\pi-2x

The two base angles are each

\frac{\pi-x}{2}.

The minor arc

BC

has central angle

2x,

so the angle between the tangents is

\pi-2x.

The given relation yields

\frac{\pi-x}{2}=2(\pi-2x).

Therefore

7x=3\pi,

x=\frac{3\pi}{7}.

Thus the correct answer is A.

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