# Difference between revisions of "1978 AHSME Problems/Problem 23"

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

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+ | <math>\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}</math> | ||

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+ | ==Solution== | ||

+ | No solutions yet! |

## Latest revision as of 16:41, 18 June 2021

## 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

No solutions yet!