1963 AMC 12 Problem 22

Question

Acute-angled triangle

ABC

is inscribed in a circle with center at

O;

\overset{\frown}{AB}=120^\circ

and

\overset{\frown}{BC}=72^\circ.

A point

E

is taken in minor arc

AC

such that

OE

is perpendicular to

AC.

Then the ratio of the magnitudes of angles

OBE

and

BAC

is:

\dfrac5{18}

\dfrac29

\dfrac14

\dfrac13

\dfrac49

Find minor arc

AC,

then use

OE\perp AC

to locate

E

at its midpoint

Compute the central angle

BOE

and the inscribed angle

BAC

Minor arc

AC

has measure

360^\circ-120^\circ-72^\circ=168^\circ.

Since

OE\perp AC,

E

bisects that arc, so

\angle BOE=72^\circ+84^\circ=156^\circ.

Isosceles triangle

OBE

gives

\angle OBE=12^\circ,

while

\angle BAC=\frac{72^\circ}{2}=36^\circ.

Their ratio is

\frac{1}{3}.

Thus, the correct answer is D.

Problem 22 in other years