1951 AHSME Problems/Problem 46

Problem

$AB$ is a fixed diameter of a circle whose center is $O$. From $C$, any point on the circle, a chord $CD$ is drawn perpendicular to $AB$. Then, as $C$ moves over a semicircle, the bisector of angle $OCD$ cuts the circle in a point that always:

$\textbf{(A)}\ \text{bisects the arc }AB\qquad\textbf{(B)}\ \text{trisects the arc }AB\qquad\textbf{(C)}\ \text{varies}$ $\textbf{(D)}\ \text{is as far from }AB\text{ as from }D\qquad\textbf{(E)}\ \text{is equidistant from }B\text{ and }C$

Solution

[asy] pair O=(0,0), A=(-1,0), B=(1,0), C=(-0.5,0.5sqrt(3)),  D=(-0.5,-0.5sqrt(3)), E=(0.5,-0.5sqrt(3)), P=(0,-1); draw(circle(O,1));  label("$A$",A,W); label("$B$",B,E); label("$O$",O,NE); label("$P$",P,S); label("$C$",C,NW); label("$D$",D,SW); label ("$E$",E,SE); dot(A); dot(B); dot(C); dot(D); dot(E); dot(O); dot(P); draw(A--B); draw(C--D--E--cycle); draw(C--P,dashed); [/asy] Draw in diameter $CE$. Note that, as it is inscribed in a semicircle, $\bigtriangleup CDE$ always has a right angle at $D$. Since it is given that $AB \perp CD$, $AB \parallel DE$ by congruent corresponding angles. $CP$ bisects $DE$, as well as the arc subtended by $DE$, $\widehat{DPE}$. Because $AB \parallel DE$, $CP$ always $\boxed{\textbf{(A)}\ \text{bisects the arc }AB}$.

See Also

1951 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 45
Followed by
Problem 47
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