1951 AMC 12 Problem 46

Question

\overline{AB}

is a fixed diameter of a circle whose center is

O.

From

C,

any point on the circle, a chord

\overline{CD}

is drawn perpendicular to

\overline{AB}.

Then, as

C

moves over a semicircle, the bisector of angle

OCD

cuts the circle in a point that always:

Bisects the arc

AB

Trisects the arc

AB

Varies

Is as far from

\overline{AB}

as from

D

Is equidistant from

B

and

C

Extend

\overline{CO}

through the center to the opposite point

E

of the circle

Because

CE

is a diameter and

CD\perp AB,

the angle bisector at

C

bisects arc

DE

Extend

CO

to meet the circle again at

E.

Since

CE

is a diameter,

\angle CDE=90^\circ.

Also

CD\perp AB,

so

DE\parallel AB.

If the bisector of

\angle OCD=\angle ECD

meets the circle at

P,

equal inscribed angles give equal arcs

EP=PD.

Because the parallel chord

DE

has endpoints symmetrically placed relative to the fixed diameter

AB,

their arc midpoint

P

is the midpoint of arc

AB.

Thus, the correct answer is A.

Problem 46 in other years