# 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

Since $\angle CAD$ is an angle inscribed in a $60{^\circ}$ arc, we obtain $\angle CAD =\frac{60^{\circ}}{2} = 30^{\circ}$, so $\triangle ABO$ is a $30^{\circ}$-$60^{\circ}$-$90^{\circ}$ right triangle. This gives $AO = BO\sqrt{3} = 5\sqrt{3}$ and $AB = 2BO = 10$, and now since $AD$ is a diameter, $AD = 2AO = 10\sqrt{3}$. The fact that $AD$ is a diameter also means that $\angle ACD = 90^{\circ}$, so $\triangle ACD$ is again a $30^{\circ}$-$60^{\circ}$-$90^{\circ}$ right triangle, yielding $CD = \frac{1}{2}AD = 5\sqrt{3}$ and $AC = \frac{\sqrt{3}}{2}AD = 15$, which finally gives $BC = AC-AB = 15-10 = \boxed{\text{(D)} \ 5}$.

## See Also

 1985 AHSME (Problems • Answer Key • Resources) Preceded byProblem 21 Followed byProblem 23 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 All AHSME Problems and Solutions

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