# 1957 AHSME Problems/Problem 47

## Problem

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

$[asy] defaultpen(linewidth(.8pt)); unitsize(2.5cm); real m = 0; real b = 0; pair O = origin; pair X = (-1,0); pair Y = (1,0); pair Q = midpoint(O--X); pair A = (Q.x, -1*sqrt(3)/2); pair B = (Q.x, -1*A.y); pair M = (Q.x + sqrt(3)/2,0); m = (B.y - M.y)/(B.x - M.x); b = (B.y - m*B.x); pair D = intersectionpoint(Circle(O,1),M--(1.5,1.5*m + b)); m = (A.y - M.y)/(A.x - M.x); b = (A.y - m*A.x); pair C = intersectionpoint(Circle(O,1),M--(1.5,1.5*m + b)); draw(Circle(O,1)); draw(Arc(Q,sqrt(3)/2,-90,90)); draw(A--B); draw(X--Y); draw(B--D); draw(A--C); draw(A--D); dot(O);dot(M); label("B",B,NW); label("C",C,NE); label("Y",Y,E); label("D",D,SE); label("A",A,SW); label("X",X,W); label("Q",Q,SW); label("O",O,SW); label("M",M,NE+2N);[/asy]$

$\textbf{(A)}\ r\sqrt{2}\qquad\textbf{(B)}\ r\qquad\textbf{(C)}\ \text{not a side of an inscribed regular polygon}\qquad\textbf{(D)}\ \frac{r\sqrt{3}}{2}\qquad\textbf{(E)}\ r\sqrt{3}$

## Solution

Because $\overline{XY}$ is a diameter of the circle and $\overline{AB} \perp \overline{XY}$, we know that $\overline{XY}$ bisects $\overline{AB}$, so $AQ=BQ$. Thus, $M$ is on the perpendicular bisector of $\overline{AB}$, and so $AM=BM$. Furthermore, by Thales' Theorem, $\measuredangle AMB = 90^{\circ}$. Thus, because $\triangle AMB$ is a right isosceles triangle with $AM=BM$, it is a 45-45-90 triangle. Thus, $\measuredangle ABM = 45^{\circ}$. Now, draw $\overline{OA}$ and $\overline{OD}$. Because $\angle ABD \cong \angle ABM$ is an inscribed angle which intercepts minor arc $AD$, the measure of central angle $\angle AOD$ must be $2\measuredangle ABD = 2 \cdot 45^{\circ} = 90^{\circ}$. Because $OA=OD=r$, $\triangle AOD$ is also a $45$-$45$-$90$ triangle, so $AD=\boxed{\textbf{(A) }r\sqrt2}$.