A geometry problem by Sanchit Mittal

We can argue, by symmetry, that the largest possible equilateral triangle that will fit must have a side collinear with one of the sides of the pool. Then, we reduce to three cases to check.

• The first, that one of the sides of the triangle is 30m, is clearly not the maximum as we could form a larger triangle with a side collinear to the 40m edge.
• The second, that we can form a triangle with a side length of 40m is also incorrect as $\sqrt{40^2-20^2}=20\sqrt{3}\approx 1.7\cdot 20 = 34 > 30$ .
• Thus the last solution is that we are bounded by the side that is 30m long, hence this becomes the height, giving us: $\frac{1}{2}s\sqrt{3}=h$ : $s=\frac{h\cdot 2\sqrt{3}}{3}=20\sqrt{3}$ This gives us for the area: $\frac{1}{2}sh=\boxed{300\sqrt{3}}$ Which is our solution.