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

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==Problem== | ==Problem== | ||

+ | [asy] size(100); draw((0,0)--(1,0)--(1,1)--(0,1)--cycle); draw((0,1)--(1,0)); draw((0,0)--(.5,sqrt(3)/2)--(1,0)); label("<math>A</math>",(0,0),SW); label("<math>B</math>",(1,0),SE); label("<math>C</math>",(1,1),NE); label("<math>D</math>",(0,1),NW); label("<math>E</math>",(.5,sqrt(3)/2),E); label("<math>F</math>",intersectionpoint((0,0)--(.5,sqrt(3)/2),(0,1)--(1,0)),2W); //Credit to chezbgone2 for the diagram [/asy] | ||

+ | 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 | ||

+ | |||

+ | <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> | ||

+ | |||

+ | ==Solution== | ||

+ | No solutions yet! |

## Revision as of 16:41, 18 June 2021

## Problem

[asy] size(100); draw((0,0)--(1,0)--(1,1)--(0,1)--cycle); draw((0,1)--(1,0)); draw((0,0)--(.5,sqrt(3)/2)--(1,0)); label("",(0,0),SW); label("",(1,0),SE); label("",(1,1),NE); label("",(0,1),NW); label("",(.5,sqrt(3)/2),E); label("",intersectionpoint((0,0)--(.5,sqrt(3)/2),(0,1)--(1,0)),2W); //Credit to chezbgone2 for the diagram [/asy] 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!