1994 AMC 12 Problem 29

Question

Points

A,

B

and

C

on a circle of radius

r

are situated so that

AB=AC,

AB\gt r,

and the length of minor arc

\overset{\frown}{BC}

is

r.

If angles are measured in radians, then

\frac{AB}{BC}=

\frac12\csc\frac14

2\cos\frac12

4\sin\frac12

\csc\frac12

2\sec\frac12

The minor arc

\overset{\frown}{BC}

subtends a central angle of

1

radian

Use the chord formula

2r\sin(\frac{\theta}{2})

for both

BC

and

AB

The central angle subtending minor arc

\overset{\frown}{BC}

is

\frac{r}{r}=1,

so

BC=2r\sin\frac12.

Since

AB=AC

and

AB\gt r,

point

A

is the midpoint of the major arc

BC.

The minor central angle from

A

to

B

is

\pi-\frac12,

so

AB=2r

\sin\left(\frac{\pi-\frac{1}{2}}{2}\right)=2r\cos\frac14.

Therefore

\frac{AB}{BC}=\frac{\cos(\frac{1}{4})}{\sin(\frac{1}{2})}

=\frac1{2\sin(\frac{1}{4})}=\frac12\csc\frac14.

Thus the correct answer is A.

Problem 29 in other years