1978 AHSME Problems/Problem 23

Problem

Vertex $E$ of equilateral $\triangle ABE$ is in the interior of square $ABCD$ , and $F$ is the point of intersection of diagonal $BD$ and line segment $AE$ . If length $AB$ is $\sqrt{1+\sqrt{3}}$ then the area of $\triangle ABF$ is

$\textbf{(A) }1\qquad \textbf{(B) }\frac{\sqrt{2}}{2}\qquad \textbf{(C) }\frac{\sqrt{3}}{2}\qquad \textbf{(D) }4-2\sqrt{3}\qquad \textbf{(E) }\frac{1}{2}+\frac{\sqrt{3}}{4}$

Solution

Place square $ABCD$ on the coordinate plane with $A$ at the origin.

In polar form, line $BD$ is $r\sin(\theta) = \sqrt{1 + \sqrt{3}} - r \cos(\theta)$ and line $AF$ is $\theta = \frac{\pi}{3}$ .

This means that the length from the origin to the intersection ( $r$ ) is $r\frac{\sqrt{3}}{2} = \sqrt{1 + \sqrt{3}} - \frac{r}{2}$

Solving for $r$ , you get: $r = \frac{2}{\sqrt{1 + \sqrt{3}}}$

Using the formula for area of a triangle ( $A = \frac{ab\sin{C}}{2}$ ), you get $A = \frac{(\frac{2}{\sqrt{1 + \sqrt{3}}})(\sqrt{1 + \sqrt{3}}) \sin(\frac{\pi}{3})}{2} = \frac{\sqrt{3}}{2}$

Getting $\textbf{C}$ as the answer

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1978 AHSME Problems/Problem 23

Problem

Solution