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

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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>
 
<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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== Solution ==
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No solutions yet!
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Place square <math>ABCD</math> on the coordinate plane with <math>A</math> at the origin.
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In polar form, line <math>BD</math> is <math>r\sin(\theta) = \sqrt{1 + \sqrt{3}} - r \cos(\theta)</math> and line <math>AF</math> is <math>\theta = \frac{\pi}{3}</math>.
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This means that the length from the origin to the intersection (<math>r</math>) is <math>r\frac{\sqrt{3}}{2} = \sqrt{1 + \sqrt{3}} - \frac{r}{2}</math>
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Solving for <math>r</math>, you get: <math>r = \frac{2}{\sqrt{1 + \sqrt{3}}}</math>
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Using the formula for area of a triangle (<math>A = \frac{ab\sin{C}}{2}</math>), you get <math>A = \frac{(\frac{2}{\sqrt{1 + \sqrt{3}}})(\sqrt{1 + \sqrt{3}}) \sin(\frac{\pi}{3})}{2} = \frac{\sqrt{3}}{2}</math>
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Getting <math>\textbf{C}</math> as the answer

Latest revision as of 12:33, 17 August 2025

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 $r\sin(\theta) = \sqrt{1 + \sqrt{3}} - r \cos(\theta)$ and line $AF$ is $\theta = \frac{\pi}{3}$.

This means that the length from the origin to the intersection ($r$) is $r\frac{\sqrt{3}}{2} = \sqrt{1 + \sqrt{3}} - \frac{r}{2}$

Solving for $r$, you get: $r = \frac{2}{\sqrt{1 + \sqrt{3}}}$

Using the formula for area of a triangle ($A = \frac{ab\sin{C}}{2}$), you get $A = \frac{(\frac{2}{\sqrt{1 + \sqrt{3}}})(\sqrt{1 + \sqrt{3}}) \sin(\frac{\pi}{3})}{2} = \frac{\sqrt{3}}{2}$

Getting $\textbf{C}$ as the answer

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