# 1985 AHSME Problems/Problem 22

## Problem

In a circle with center $O$, $AD$ is a diameter, $ABC$ is a chord, $BO=5$ and $\angle ABO= \ \stackrel{\frown}{CD} \ =60^\circ$. Then the length of $BC$ is

$[asy] defaultpen(linewidth(0.7)+fontsize(10)); pair O=origin, A=dir(35), C=dir(155), D=dir(215), B=intersectionpoint(dir(125)--O, A--C); draw(C--A--D^^B--O^^Circle(O,1)); pair point=O; label("A", A, dir(point--A)); label("B", B, dir(point--B)); label("C", C, dir(point--C)); label("D", D, dir(point--D)); label("O", O, dir(305)); label("5", B--O, dir(O--B)*dir(90)); label("60^\circ", dir(185), dir(185)); label("60^\circ", B+0.05*dir(-25), dir(-25));[/asy]$

$\mathrm{(A)\ } 3 \qquad \mathrm{(B) \ }3+\sqrt{3} \qquad \mathrm{(C) \ } 5-\frac{\sqrt{3}}{2} \qquad \mathrm{(D) \ } 5 \qquad \mathrm{(E) \ }\text{none of the above}$

## Solution

$[asy] defaultpen(linewidth(0.7)+fontsize(10)); pair O=origin, A=dir(35), C=dir(155), D=dir(215), B=intersectionpoint(dir(125)--O, A--C); draw(C--A--D^^B--O^^Circle(O,1)); pair point=O; draw(C--D); label("A", A, dir(point--A)); label("B", B, dir(point--B)); label("C", C, dir(point--C)); label("D", D, dir(point--D)); label("O", O, dir(305)); label("5", B--O, dir(O--B)*dir(90)); label("60^\circ", dir(185), dir(185)); label("60^\circ", B+0.05*dir(-25), dir(-25));[/asy]$

Since $\angle CAD$ is inscribed and intersects an arc of length $60^\circ$, $\angle CAD=30^\circ$. Thus, $\triangle ABO$ is a $30-60-90$ right triangle. Thus, $AO=BO\sqrt{3}=5\sqrt{3}$ and $AB=2BO=10$. Since $AO$ and $DO$ are both radii, $DO=AO=5\sqrt{3}$ and $AD=10\sqrt{3}$. Since $\angle ACD$ is inscribed in a semicircle, it's a right angle, and $\triangle ACD$ is also a $30-60-90$ right triangle. Thus, $CD=\frac{AD}{2}=5\sqrt{3}$ and $AC=CD\sqrt{3}=15$. Finally, $BC=AC-AB=15-10=5, \boxed{\text{D}}$.