1997 AIME Problems/Problem 15

Problem

The sides of rectangle $ABCD$ have lengths $10$ and $11$ . An equilateral triangle is drawn so that no point of the triangle lies outside $ABCD$ . The maximum possible area of such a triangle can be written in the form $p\sqrt{q}-r$ , where $p$ , $q$ , and $r$ are positive integers, and $q$ is not divisible by the square of any prime number. Find $p+q+r$ .

Solution

Consider points on the complex plane $A (0,0),\ B (11,0),\ C (11,10),\ D (0,10)$ . Since the rectangle is quite close to a square, we figure that the area of the equilateral triangle is maximized when a vertex of the triangle coincides with that of the rectangle. Set one vertex of the triangle at $A$ , and the other two points $E$ and $F$ on $BC$ and $CD$ , respectively. Let $E (11,a)$ and $F (b, 10)$ . Since it's equilateral, then $E\cdot\text{cis}60^{\circ} = F$ , so $(11 + ai)\left(\frac {1}{2} + \frac {\sqrt {3}}{2}i\right) = b + 10i$ , and expanding we get $\left(\frac {11}{2} - \frac {a\sqrt {3}}{2}\right) + \left(\frac {11\sqrt {3}}{2} + \frac {a}{2}\right)i = b + 10i$ .

File:1997 AIME-15.png

We can then set the real and imaginary parts equal, and solve for $(a,b) = (20 - 11\sqrt {3},22 - 10\sqrt {3})$ . Hence a side $s$ of the equilateral triangle can be found by $s^2 = AE^2 = a^2 + AB^2 = 884 - 440\sqrt{3}$ . Using the area formula $\frac{s^2\sqrt{3}}{4}$ , the area of the equilateral triangle is $\frac{(884-440\sqrt{3})\sqrt{3}}{4} = 221\sqrt{3} - 330$ . Thus $p + q + r = 221 + 3 + 330 = \boxed{554}$ .

1997 AIME (Problems • Answer Key • Resources)
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1997 AIME Problems/Problem 15

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