The sides of [rectangle](https://artofproblemsolving.com/wiki/index.php/Rectangle) $ABCD$ have lengths $10$ and $11$. An [equilateral triangle](https://artofproblemsolving.com/wiki/index.php/Equilateral_triangle) is drawn so that no point of the triangle lies outside $ABCD$. The maximum possible [area](https://artofproblemsolving.com/wiki/index.php/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 $p+q+r$.
Consider points on the [complex plane](https://artofproblemsolving.com/wiki/index.php/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](https://artofproblemsolving.com/wiki/index.php/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 $\left(\frac {11}{2} - \frac {a\sqrt {3}}{2}\right) + \left(\frac {11\sqrt {3}}{2} + \frac {a}{2}\right)i = b + 10i$.

[1997 AIME-15a.PNG](https://artofproblemsolving.com/wiki/index.php/File:1997_AIME-15a.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 $p + q + r = 221 + 3 + 330 = \boxed{554}$.