HMMT February, 2024

Solution:

WLOG let the side length of the hexagon be 1. Let $O$ be the center of the hexagon. Consider drawing in the circles $(A P D)$, $(B P E)$, and $(C P F)$. Note that $O$ lies on the radical axis of all three circles, since $A O \cdot O D = B O \cdot O E = C O \cdot O F$. Since $P$ also lies on the radical axis, all three circles are coaxial.
Let $X$, $Y$, and $Z$ be the centers of $(A P D)$, $(B P E)$, and $(C P F)$, respectively. Since the circles are coaxial, $X$, $Y$, and $Z$ are collinear. WLOG $Y$ lies on segment $X Z$. Note that $X O \perp A D$, $Y O \perp B E \Longrightarrow \angle X O Y = 60^\circ$. Similarly, we have $\angle Y O Z = 60^\circ$. Now, inverting at $O$ and using van Schooten's Theorem gives that $1 / O Y = 1 / O X + 1 / O Z$.
Furthermore, we have
$$\angle A P D = 180^\circ - \frac{1}{2} \angle A X D = 180^\circ - \angle A X O \Longrightarrow \tan \angle A P D = -\tan \angle A X O = -\frac{A O}{O X} = -\frac{1}{O X}$$
Similarly, we have $\tan \angle B P E = -\frac{1}{O Y}$ and $\tan \angle C P F = -\frac{1}{O Z}$. Therefore, two of these tangent values sum to the third, as desired.

Solution:
Firstly, note that $A D$, $B E$, and $C F$ are diameters of the circle $(A B C D E F)$, so the angles $\angle A P D$, $\angle B P E$, $\angle C P F$ are all obtuse. Therefore, the desired tangents are well-defined.
WLOG let the side length of the hexagon be 1. Let $O$ be the center of the hexagon, and let $O P = x$. Finally, let $\angle A O P = \theta$.
Now, by Law of Cosines, we have $A P = \sqrt{x^{2} + 1 - 2x \cos \theta}$ and $D P = \sqrt{x^{2} + 1 + 2x \cos \theta}$. Now, by Law of Cosines again we have
$$\begin{gathered} \cos \angle A P D = \frac{2(x^{2} + 1) - 4}{2 \sqrt{(x^{2} + 1)^{2} - 4x^{2} \cos^{2} \theta}} = \frac{x^{2} - 1}{\sqrt{(1 - x^{2})^{2} + 4x^{2} \sin^{2} \theta}} \\ \Longrightarrow \tan \angle A P D = \frac{2x|\sin \theta|}{x^{2} - 1}. \end{gathered}$$
Similarly, $\angle (B E, O P)$ and $\angle (C F, O P)$ are $\theta + 60^\circ$ and $\theta + 120^\circ$, respectively (here we are using directed angles). Therefore, the desired three values are
$$\{\tan \angle A P D, \tan \angle B P E, \tan \angle C P F\} = \left\{\frac{2x|\sin \theta|}{x^{2} - 1}, \frac{2x|\sin(\theta + 60^\circ)|}{x^{2} - 1}, \frac{2x|\sin(\theta + 120^\circ)|}{x^{2} - 1}\right\}.$$
We can scale down the tangent values by $\frac{2x}{x^{2} - 1}$ to get $\{|\sin \theta|, |\sin(\theta + 60^\circ)|, |\sin(\theta + 120^\circ)|\}$. Now, consider an equilateral triangle with vertices at the third roots of unity rotated by $\theta$ degrees counterclockwise. The three values represent the distances from the three vertices to the real axis. Since the centroid of this triangle is the origin (lying on the real axis), two of these quantities must sum to the third, as desired.

Solution:
We will show either the three sum to 0 or two of them sum to the third one; since they're all negative, the former case is actually impossible.
Let $(A B C D E F)$ be the unit circle, with $a = 1$, $b = \omega$, and so on, where $\omega = e^{\pi i / 3}$. Then $\angle A P D$ is the argument of
$$\frac{1 - p}{\omega^{3} - p} = \frac{(1 - p)(\omega^{3} - \bar{p})}{(\omega^{3} - p)(\omega^{3} - \bar{p})} = \frac{\omega^{3} - \bar{p} - \omega^{3} p + |p|^{2}}{1 + |p|^{2} - \omega^{3} p - \omega^{3} \bar{p}}$$
Then $\tan \angle A P D$ is the imaginary part divided by the real part of this, which is
$$-\frac{1}{i} \cdot \frac{\bar{p} - p}{|p|^{2} - 1} = c \cdot \operatorname{dist}(P, A D)$$
for some constant $c$. (Note that $\tan A P D$ might actually be the negative of this, depending on direction; this is why we added the remark at the beginning about them possibly summing to 0.)
Similarly, $\tan \angle B P E = c \cdot \operatorname{dist}(P, B E)$ and $\tan \angle C P F = c \cdot \operatorname{dist}(P, C F)$. It suffices to show that two $\operatorname{dist}(P, A D)$, $\operatorname{dist}(P, B E)$, and $\operatorname{dist}(P, C F)$ sum to the third. However, this is easy; without loss of generality let $P$ be inside $O A B$, and let the hexagon have side length 1. Then
$$\operatorname{dist}(P, A D) + \operatorname{dist}(P, B E) = \frac{\sqrt{3}}{2} - \operatorname{dist}(P, A B) = \operatorname{dist}(P, C F)$$
as desired.