, 2022

Solution:
By Archimedes lemma, we have $M, Q, V$ are collinear and $M, P, U$ are collinear as well. Note that inversion at $M$ with radius $M X$ shows that $P Q U V$ is cyclic. Thus, we have $M P \cdot M U = M Q \cdot M V$, so $M$ lies on the radical axis of $(P U Z)$ and $(Q V Z)$, thus $T$ must lie on the line $M Z$. Thus, we have $M Z \cdot M T = M Q \cdot M V = M N^{2}$, which implies triangles $M Z N$ and $M N T$ are similar. Thus, we have $N T \perp M N$. However, since the line through $O_{1}$ and $O_{2}$ passes through $N$ and is perpendicular to $M N$, we have $T$ lies on line $O_{1} O_{2}$. Additionally, since $M Z \cdot M T = M N^{2} = M X^{2}$, inversion at $M$ with radius $M X$ swaps $Z$ and $T$, and since $(M X Y)$ maps to line $X Y$, this means $T$ also lies on $X Y$.

Therefore, $T$ is the intersection of $P Q$ and $O_{1} O_{2}$, and thus by Monge's Theorem, we must have $T$ lies on $U V$.

Now, to finish, we will consider triangle $O U V$. Since $O_{1} O_{2} T$ is a line that cuts this triangle, by Menelaus, we have
$$\frac{O O_{1}}{O_{1} U} \cdot \frac{U T}{V T} \cdot \frac{V O_{2}}{O_{2} O} = 1$$
Using the values of the radii, this simplifies to
$$\frac{R - r_{1}}{r_{1}} \cdot \frac{U T}{V T} \cdot \frac{r_{2}}{R - r_{2}} = 1 \Longrightarrow \frac{U T}{V T} = \frac{r_{1}(R - r_{2})}{r_{2}(R - r_{1})}$$
Now, note that
$$T U \cdot T V = T P \cdot T Q = \frac{4 r_{1}^{2} r_{2}^{2}}{(r_{1} - r_{2})^{2}}$$
Now, let $T U = r_{1}(R - r_{2}) k$ and $T V = r_{2}(R - r_{1}) k$. Plugging this into the above equation gives
$$r_{1} r_{2} (R - r_{1})(R - r_{2}) k^{2} = \frac{4 (r_{1} r_{2})^{2}}{(r_{1} - r_{2})^{2}}$$
Solving gives
$$k = \frac{2 \sqrt{r_{1} r_{2}}}{|r_{1} - r_{2}| \sqrt{(R - r_{1})(R - r_{2})}}$$
To finish, note that
$$T U + T V = k (R r_{1} + R r_{2} - 2 r_{1} r_{2}) = \frac{2 (R r_{1} + R r_{2} - 2 r_{1} r_{2}) \sqrt{r_{1} r_{2}}}{|r_{1} - r_{2}| \sqrt{(R - r_{1})(R - r_{2})}}$$