2023 AIME I Problem 12

Decompose each vector from $E,$ $F,$ $D$ to $P$ into a part along the side and the distance to that side; the along-part equals the distance times $\cot\theta.$

Add the three relations: the unit directions $C \to A,$ $A \to B,$ $B \to C$ sum to zero, and the three distances sum to the height (Viviani).

Place $B = (0, 0),$ $C = (55, 0),$ $A = \left(\frac{55}{2}, \frac{55\sqrt{3}}{2}\right),$ so that $D = (7, 0),$ $E = (40, 15\sqrt{3}),$ and $F = \left(\frac{15}{2}, \frac{15\sqrt{3}}{2}\right).$ Let $\theta$ be the common angle and let $\mathbf{u}_1, \mathbf{u}_2, \mathbf{u}_3$ be the unit vectors in the directions $C \to A,$ $A \to B,$ $B \to C$ — the directions from $E$ toward $A,$ from $F$ toward $B,$ and from $D$ toward $C.$ Splitting $P - E$ into its component along $\mathbf{u}_1$ and its component perpendicular to side $CA,$ whose length is the distance $d_1$ from $P$ to line $CA,$ the angle condition gives $(P - E)\cdot\mathbf{u}_1 = d_1\cot\theta;$ similarly $(P - F)\cdot\mathbf{u}_2 = d_2\cot\theta$ and $(P - D)\cdot\mathbf{u}_3 = d_3\cot\theta.$

Now add all three relations. Since $\mathbf{u}_1 + \mathbf{u}_2 + \mathbf{u}_3 = \mathbf{0}$ (the directed sides of a triangle close up), $P$ drops out, and by Viviani’s theorem $d_1 + d_2 + d_3$ equals the height $\frac{55\sqrt{3}}{2}.$ With $\mathbf{u}_1 = \left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right),$ $\mathbf{u}_2 = \left(-\frac{1}{2}, -\frac{\sqrt{3}}{2}\right),$ and $\mathbf{u}_3 = (1, 0),$ we get $E \cdot \mathbf{u}_1 = \frac{5}{2},$ $F \cdot \mathbf{u}_2 = -15,$ and $D \cdot \mathbf{u}_3 = 7,$ so $$\begin{aligned} \frac{55\sqrt{3}}{2}\,\cot\theta &= -\left(\frac{5}{2} - 15 + 7\right) \\ &= \frac{11}{2}. \end{aligned}$$

Hence $\cot\theta = \frac{11}{55\sqrt{3}} = \frac{1}{5\sqrt{3}},$ so $\tan^2(\angle AEP) = \left(5\sqrt{3}\right)^2 = 75.$