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

Revision as of 15:41, 18 June 2021

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

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 ==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

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

Revision as of 15:41, 18 June 2021

Problem

Solution