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 $C$

Answer: A

Concepts:circle angle bisector arc

Difficulty rating: 1880

Small Hint:

Extend $\overline{CO}$ through the center to the opposite point $E$ of the circle

Big Hint:

Because $CE$ is a diameter and $CD\perp AB,$ the angle bisector at $C$ bisects arc $DE$

Solution:

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.

1951 AMC 12 Problem 46

Answer: A

Difficulty rating: 1880

Small Hint:

Big Hint:

Solution:

Problem 46 in Other Years