1983 AHSME Problems/Problem 10

Problem

Segment $AB$ is both a diameter of a circle of radius $1$ and a side of an equilateral triangle $ABC$ . The circle also intersects $AC$ and $BD$ at points $D$ and $E$ , respectively. The length of $AE$ is

$\textbf{(A)} \ \frac{3}{2} \qquad \textbf{(B)} \ \frac{5}{3} \qquad \textbf{(C)} \ \frac{\sqrt 3}{2} \qquad \textbf{(D)}\ \sqrt{3}\qquad \textbf{(E)}\ \frac{2+\sqrt 3}{2}$

Solution

Note that since $AB$ is a diameter, $\angle AEB = 90^{\circ}$ , which means $AB$ is an altitude of equilateral triangle $ABC$ . It follows that $\triangle ABE$ is a $30^{\circ},60^{\circ},90^{\circ}$ triangle, and so $AE = AB \cdot \frac{\sqrt{3}}{2} = (2 \cdot 1) (\frac{\sqrt{3}}{2}) = \boxed{\textbf{(D)}\ \sqrt{3}}$ .

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1983 AHSME Problems/Problem 10

Problem

Solution