AIME 1994 · 第 8 题

Solution

Solution 1

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:

$$(a+11i)\left(\mathrm{cis}\,60^{\circ}\right) = (a+11i)\left(\frac 12+\frac{\sqrt{3}i}2\right)=b+37i.$$ Equating the real and imaginary parts, we have:

$$\begin{aligned}b&=\frac{a}{2}-\frac{11\sqrt{3}}{2}\\37&=\frac{11}{2}+\frac{a\sqrt{3}}{2} \end{aligned}$$ 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. The product $ab$ is unchanged.

Solution 2

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, $M=\left(\frac{a+b}{2}, 24\right)$. The vector from $O$ to $M$ is $\left[\frac{a+b}{2}, 24\right]$. 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 $\left[8\sqrt{3}, \frac{(a+b)\sqrt{3}}{6}\right]$. (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 $\left[\frac{a-b}{2}, 13\right]$ 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 3

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 $\left(\frac{a+b}{3}, 16i\right)$. The new coordinates of the equilateral triangle are $\left(-\frac{a+b}{3}-16i\right), \left(a-\frac{a+b}{3}-5i\right), \left(b-\frac{a+b}{3}+21i\right)$. 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 $r_1, r_2,$ and $r_3,$$\, r_1r_2 + r_2r_3 + r_3r_1 = 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 $\frac{21}{5}\sqrt{3}$. Multiplying $a$ and $b$ yields $\boxed{315}$.

Solution 4 (gigantic numbers)

Just using the Pythagorean Theorem, we get that $a^2 + 11^2 = (b-a)^2 + 26^2 = b^2 + 37^2$.

$a^2 + 121 = b^2 + 1369 \longrightarrow a^2 = b^2 + 1248$. Expanding the second and subtracting the first equation from it we get $b^2 = 2ab - 555$.

$b^2 = 2ab - 555 \longrightarrow a^2 = 2ab + 693$.

We have $b^2 + 1248 = 2b\sqrt{b^2+1248} + 693$.

Moving the square root to one side and non square roots to the other we eventually get $b^4 + 1110b^2 + 308025 = 4b^4 + 4992b^2$.

$3b^4 + 3882b^2 - 308025 = 0$.

This factors to $(3b^2-225)(b^2+1369)$, so $3b^2 = 225, b = 5\sqrt 3$.

Plugging it back in, we find that a = $\sqrt{1323}$ which is $21\sqrt3$, so the product $ab$ is $\boxed {315}$.

~Arcticturn