Question

In circle $O,$ the midpoint of radius $\overline{OX}$ is $Q;$ at $Q,$ $\overline{AB}\perp\overline{XY}.$ The semicircle with $\overline{AB}$ as diameter intersects $\overline{XY}$ in $M.$ Line $AM$ intersects circle $O$ in $C,$ and line $BM$ intersects circle $O$ in $D.$ Line $AD$ is drawn. Then, if the radius of circle $O$ is $r,$ $AD$ is:

$r\sqrt2$

$r$

not a side of an inscribed regular polygon

$\dfrac{r\sqrt3}{2}$

$r\sqrt3$

Answer: A

Concepts:circle inscribed angle special right triangle

Difficulty rating: 2070

Small Hint:

Since $\overline{XY}$ perpendicularly bisects $\overline{AB},$ compare $MA$ and $MB$

Big Hint:

Use the semicircle to find $\angle AMB,$ then use that $B,$ $M,$ and $D$ are collinear

Solution:

Line $XY$ is the perpendicular bisector of $AB,$ so $MA=MB.$ Since $M$ lies on the semicircle with diameter $AB,$ $\angle AMB=90^\circ.$ Thus triangle $AMB$ is an isosceles right triangle and $\angle ABM=45^\circ.$ Because $B,M,D$ are collinear, the inscribed angle $\angle ABD=45^\circ,$ so its intercepted arc $AD$ measures $90^\circ.$ Therefore chord $AD$ is a side of an inscribed square and has length $AD=r\sqrt2.$

Thus, the correct answer is A.

1957 AMC 12 Problem 47

Answer: A

Difficulty rating: 2070

Small Hint:

Big Hint:

Solution:

Problem 47 in Other Years