In a circle with center O, AD is a diameter, ABC is a chord, BO=5 and ∠ ABO=CD=60°. Then the length of BC is

1985 AMC 12 Problem 22

Attempt Problem 22 of the 1985 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 1985 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

22.

In a circle with center

O,

AD

is a diameter,

ABC

is a chord,

BO=5

and

\angle ABO=\overset{\frown}{CD}=60^\circ.

Then the length of

BC

3

3+\sqrt3

5-\frac{\sqrt3}{2}

5

none of the above

The

60^\circ

arc

CD

makes

\angle CAD=30^\circ

Use the resulting

30^\circ

60^\circ

90^\circ

triangles to compare

AB

and

AC

Arc

CD

gives

\angle CAD=30^\circ.

Since

A,B,C

are collinear and

A,O,D

are collinear,

\angle BAO=30^\circ.

With

\angle ABO=60^\circ,

triangle

ABO

is a

30^\circ

60^\circ

90^\circ

triangle. Because

BO=5,

we get

AB=10

and

AO=5\sqrt3.

Also

\angle ACD=90^\circ

because

AD

is a diameter. In the

30^\circ

60^\circ

90^\circ

triangle

ACD,

the hypotenuse is

AD=10\sqrt3,

AC=15.

Therefore

BC=AC-AB=5.

Thus the correct answer is D.