# 1969 AHSME Problems/Problem 15

## Problem

In a circle with center $O$ and radius $r$, chord $AB$ is drawn with length equal to $r$ (units). From $O$, a perpendicular to $AB$ meets $AB$ at $M$. From $M$ a perpendicular to $OA$ meets $OA$ at $D$. In terms of $r$ the area of triangle $MDA$, in appropriate square units, is: $\text{(A) } \frac{3r^2}{16}\quad \text{(B) } \frac{\pi r^2}{16}\quad \text{(C) } \frac{\pi r^2\sqrt{2}}{8}\quad \text{(D) } \frac{r^2\sqrt{3}}{32}\quad \text{(E) } \frac{r^2\sqrt{6}}{48}$

## Solution $[asy] pair O = (0,0); pair A = (-8.660,5); pair B = (-8.660,-5); pair M = (-8.660,0); pair D = (-8.660*0.75,5*0.75); draw(circle(O,10)); dot(O); label("O",O,E); dot(A); label("A",A,NW); dot(B); label("B",B,SW); dot(M); label("M",M,SE); dot(D); label("D",D,NE); draw(A--B--O--A); draw(O--M); draw(M--D); [/asy]$

Because $AO = OB = AB$, $\triangle AOB$ is an equilateral triangle, and $\angle OAB = 60^\circ$. Using 30-60-90 triangles, $AM = \tfrac{r}{2}$, $AD = \tfrac{r}{4}$, and $DM = \tfrac{r\sqrt{3}}{4}$. Thus, the area of $\triangle ADM$ is $\tfrac{1}{2} \cdot \tfrac{r}{4} \cdot \tfrac{r\sqrt{3}}{4} = \boxed{\textbf{(D) } \tfrac{r^2 \sqrt{3}}{32}}$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 