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".

Solution Three

Plot this equilateral triangle on the complex plane. Translate the equilateral triangle so that its centroid is located at the origin. (The centroid can be found by taking the average of the three vertices of the triangle, which gives $(\frac{a+b}{3}, 16i)$ . The new coordinates of the equilateral triangle are $(-\frac{a+b}{3}-16i), (a-\frac{a+b}{3}-5i), (b-\frac{a+b}{3}+21i)$ . These three vertices are solutions of a cubic polynomial of form $x^3 + C$ . By Vieta's Formulas, the sum of the paired roots of the cubic polynomial are zero. (Or for the three roots r1, r2, and r3, r1r2 + r2r3 + r3r1 = 0.) The vertices of the equilateral triangle represent the roots of a polynomial, so the vertices can be plugged into the above equation. Because both the real and complex components of the equation have to sum to zero, you really have two equations. Multiply out the equation given by Vieta's Formulas and isolate the ones with imaginary components. Simplify that equation, and that gives the equation $5a = 21b.$ Now use the equation with only real parts. This should give you a quadratic $a^2 - ab + b^2 = 1083$ . Use your previously obtained equation to plug in for a and solve for b, which should yield $5\sqrt{3}$ . a is then $21/5\sqrt{3}$ . Multiplying a and b yields $\boxed{315}$ .

1994 AIME (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

Solution Four

Using distance formula, we get the following system of 3 equations: a^2+121=b^2+1369 a^2+121=b^2-2ba+676 b^2-2ba+676=b^2+1369. The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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