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

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==Problem==
 
==Problem==
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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==
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Place square ABCD on the coordinate plane with A at the origin.
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In polar form, line BD is rsinθ=sqrt(1+sqrt(3))-rcosθ and line AF is θ=(pi)/3
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This means that the length of the intersection (r) is
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r*(sqrt3)/2=sqrt(1+sqrt(3))-r/2
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Solving for r you get: r=2/sqrt(1+sqrt(3))
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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
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Getting C as the answer

Latest revision as of 09:29, 20 March 2023

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

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