1973 AMC 12 Problem 15

Question

A sector with acute central angle

\theta

is cut from a circle of radius

6.

The radius of the circle circumscribed about the sector is

3\cos\theta

3\sec\theta

3\cos\frac12\theta

3\sec\frac12\theta

3

Treat the sector’s two radii and chord as an isosceles triangle

Express the chord using

\frac{\theta}{2},

then apply the extended sine rule

The two radii and the sector’s chord form an isosceles triangle with equal sides

6

and vertex angle

\theta.

Its base has length

12\sin(\frac{\theta}{2}).

If

R

is the triangle’s circumradius, the extended sine rule gives

2R=\frac{12\sin(\frac{\theta}{2})}{\sin\theta}.

Since

\sin\theta=2\sin(\frac{\theta}{2})\cos(\frac{\theta}{2}),

R=3\sec\frac{\theta}{2}.

Therefore, the correct answer is D.

Problem 15 in other years