# 1978 AHSME Problems/Problem 23

## Problem

Vertex of equilateral is in the interior of square , and is the point of intersection of diagonal and line segment . If length is then the area of is

## 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)/2=(sqrt3)/2 Getting C as the answer