# Difference between revisions of "1994 AIME Problems/Problem 8"

## Problem

The points $(0,0)\,$, $(a,11)\,$, and $(b,37)\,$ are the vertices of an equilateral triangle. Find the value of $ab\,$.

## Solution

Consider the points on the complex plane. The point $b+37i$ is then a rotation of $60$ degrees of $a+11i$ about the origin, so:

$$(a+11i)\left(\text{cis}\,60^{\circ}\right) = (a+11i)\left(\frac 12+\frac{\sqrt{3}i}2\right)=b+37i.$$

Equating the real and imaginary parts, we have:

\begin{align*}b&=\frac{a}{2}-\frac{11\sqrt{3}}{2}\\37&=\frac{11}{2}+\frac{a\sqrt{3}}{2} \end{align*}

Solving this system, we find that $a=21\sqrt{3}, b=5\sqrt{3}$. Thus, the answer is $\boxed{315}$.

Note: There is another solution where the point $b+37i$ is a rotation of $-60$ degrees of $a+11i$; however, this triangle is just a reflection of the first triangle by the $y$-axis, and the signs of $a$ and $b$ are flipped. However, the product $ab$ is unchanged.

## Solution Two

Using the Pythagorean theorem with these beastly numbers doesn't seem promising. How about properties of equilateral triangles? $\sqrt{3}$ and perpendiculars inspires this solution:

First, drop a perpendicular from $O$ to $AB$. Call this midpoint of $AB M$. Thus, $M=(\frac{a+b}{2}, 24)$. The vector from $O$ to $M$ is $[\frac{a+b}{2}, 24]$. Meanwhile from point $M$ we can use a vector with $\frac{\sqrt{3}}{3}$ the distance; we have to switch the $x$ and $y$ and our displacement is $[8\sqrt{3}, \frac{(a+b)\sqrt{3}}{6}]$. (Do you see why we switched $x$ and $y$ due to the rotation of 90 degrees?)

We see this displacement from $M$ to $A$ is $[\frac{a-b}{2}, 13]$ as well. Equating the two vectors, we get $a+b=26\sqrt{3}$ and $a-b=16\sqrt{3}$. Therefore, $a=21\sqrt{3}$ and $b=5\sqrt{3}$. And the answer is $\boxed{315}$.

Note: This solution was also present in Titu Andreescu and Zuming Feng's "103 Trigonometry Problems".