Equilateral Triangle in a Square

We will use complex numbers. The equilateral triangle with the largest area is the one shown above.

Let $A$ be the origin and one corner of the equilateral triangle, and let the other two points of the equilateral triangle be $11+ai$ and $b+10i$ .

Let $\omega = e^{\frac{\pi}{3}i}$

This implies that $b+ 10i = \omega(11+ai) = (\frac{1}{2} + \frac{\sqrt{3}}{2}i)(11+ai)$

$b+10i = (\frac{11}{2} - \frac{a\sqrt{3}}{2}) + (\frac{a}{2} + \frac{11\sqrt{3}}{2})i \implies a = 20 - 11\sqrt{3}$

$s^2 = |11+ai|^2 = a^2 + 11^2 = 4(-221 + 110\sqrt{3})$

It is well known that

$[ABC] = \frac{s^2\sqrt{3}}{4} = 330 - 221\sqrt{3} \implies p+q+r = \boxed{554}$

This question is from AIME 1997.

solve

121+x^2=(11-√(121+x^2-100))^2+(10-x)^2,

y^2=121+x^2,

w=y^2*√3/4, x>0,y>0

Area of Triangle

w=221(3)^0.5-330, p=221,q=3,r=330

Answer =221+3+330=554