1997 AIME Problems/Problem 15

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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 $(11 + ai)(\frac {1}{2} + \frac {\sqrt {3}}{2}i) = b + 10i$, and expanding we get $(\frac {11}{2} - \frac {a\sqrt {3}}{2}) + (\frac {11\sqrt {3}}{2} + \frac {a}{2})i = b + 10i$.

File:1997 AIME-15.png

We can then set the real and imaginary parts equal, and solve for $(a,b) = (20 - 11\sqrt {3},22 - 10\sqrt {3})$. We can then use Shoelace Theorem to find the area to be $221\sqrt {3} - 330$, So $p + q + r = 221 + 3 + 330 = 554$.

See also

1997 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 14
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