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 rsinθ=sqrt(1+sqrt(3))-rcosθ and line AF is θ=(pi)/3 This means that the length of the intersection (r) is r*(sqrt3)/2=sqrt(1+sqrt(3))-r/2 Solving for r you get: r=2/sqrt(1+sqrt(3)) Using the firmly for area of a triangle (A=ab(Sinθ)/2) you get A=2/sqrt(1+sqrt(3))*sqrt(1+sqrt(3))*sin(pi/3)/3=(sqrt3)/2 Getting C as the answer

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1978 AHSME Problems/Problem 23

Problem

Solution