1979 AMC 12 Problem 12

Question

In the adjoining figure,

CD

is the diameter of a semicircle with center

O.

Point

A

lies on the extension of

DC

past

C;

point

E

lies on the semicircle, and

B

is the point of intersection (distinct from

E

) of line segment

AE

with the semicircle. If length

AB

equals length

OD,

and the measure of

\angle EOD

is

45^\circ,

then the measure of

\angle BAO

is

10^\circ

15^\circ

20^\circ

25^\circ

30^\circ

Let

y=\angle BAO

and use

AB=OB=OE

Express

\angle BOE

in two ways using the isosceles triangles and the

45^\circ

central angle

Let

y=\angle BAO.

Since

AB=OB,

\angle AOB=y,

so

\angle EBO=2y.

Also

OB=OE,

hence

\angle BEO=2y

and

\angle BOE=180^\circ-4y.

On the other hand,

\angle BOD=180^\circ-y,

so

\angle BOE=135^\circ-y.

Therefore

180^\circ-4y=135^\circ-y,

giving

y=15^\circ.

Therefore, the correct answer is B.

Problem 12 in other years