USA IMO TST

First solution, elementary (original) We have $\triangle YAC \sim \triangle YBD$, from which it follows
$$\frac{d(Y, AC)}{d(Y, BD)} = \frac{AC}{BD}.$$
Moreover, if we denote by $r_1$ and $r_2$ the radii of $\Gamma_1$ and $\Gamma_2$, then
$$\frac{d(T, AC)}{d(T, BD)} = \frac{TA \sin \angle (AC, l_1)}{TB \sin \angle (BD, l_2)} = \frac{2 r_1 \sin \angle (AC, l_1)}{2 r_2 \sin \angle (BD, l_2)} = \frac{AC}{BD}$$
the last step by the law of sines.

This solves the problem up to configuration issues: we claim that $Y$ and $T$ both lie inside $\angle AXB \equiv \angle CXD$. WLOG $TA < TB$.
* The former is since $Y$ lies outside segments $BC$ and $AD$, since we assumed $ABCD$ was convex.
* For the latter, we note that $X$ lies inside both $\Gamma_1$ and $\Gamma_2$ in fact on the radical axis of the two circles (since $X$ was an interior point of both chords $AC$ and $BD$). In particular, $X$ is contained inside $\angle ATB$, and moreover $\angle ATB < 90^\circ$, and this is enough to imply the result.

Second solution, invasive This is based on the solution posted by kapilpavase on AoPS. Consider the inversion at $T$ swapping $\Gamma_1$ and $\Gamma_2$; we let it send $A$ to $E$, $B$ to $F$, $C$ to $V$, $D$ to $W$, as shown. Draw circles $ADWE$ and $BCVF$.

Claim — Points $T$ and $Y$ lie on the radical axis of $(ADE)$ and $(BCF)$.
Proof. Because $TF \cdot TB = TA \cdot TE$ and $YA \cdot YD = YC \cdot YB$.
Claim — Point $X$ has equal power to $(ADE)$ and $(BCF)$.
Proof. Since $TV \cdot TC = TA \cdot TE$, quadrilateral $VCEA$ is cyclic too, so by radical axis with $\Gamma_1$ and $\Gamma_2$ we find $X$ lies on $VE$. Similarly, $X$ lies on $FW$. Thus, $X$ is the center of negative inversion between $(ADE)$ and $(BCF)$.
But since $AE = BF$ and moreover
$$\begin{aligned} \angle BCF + \angle ADE &= (\angle BCA + \angle ACF) + (\angle ADB + \angle BDE) \\ &= (\angle BCA + \angle ADB) + (\angle ACF + \angle BDE) = 0 + 0 = 0 \end{aligned}$$
we conclude that $(ADE)$ and $(BCF)$ are congruent. As $X$ was the center of negative inversion between them, we're done.

Third solution, projective (Nikolai Beluhov) We start with some definitions. Let $\ell_1$ touch $\Gamma_2$ at $E$, $\ell_2$ touch $\Gamma_1$ at $F$, $K = \ell_1 \cap \overline{BD}$, $L = \ell_2 \cap \overline{AC}$, line $FX$ meet $\Gamma_1$ again at $M$, line $EX$ meet $\Gamma_2$ again at $N$, and lines $AB$, $AD$, and $BC$ meet line $TX$ at $Z$, $Y_1$, and $Y_2$. Thus the desired statement is equivalent to $Y_1 = Y_2$.
Claim — $(EB; ND)_{\Gamma_2} = (FA; MC)_{\Gamma_1}$.
Proof. Note that $AX \cdot XC = BX \cdot XD = EX \cdot XN$, so $AECN$ is cyclic. Likewise $BFDM$ is cyclic.
Consider the inversion with center $T$ which swaps $\Gamma_1$ and $\Gamma_2$; it also swaps the pairs $\{A, E\}$ and $\{B, F\}$. Since $AECN$ is cyclic, $C$ is on $\Gamma_1$, and $N$ is on $\Gamma_2$, it also swaps $\{C, N\}$; similarly it swaps $\{D, M\}$.
Thus $(EB; ND)_{\Gamma_2} = (AF; CM)_{\Gamma_1} = (FA; MC)_{\Gamma_1}$ as desired. $\square$
With this claim, the remainder of the proof is chasing cross-ratios:
$$(TZ; XY_1) \stackrel{A}{=} (KB; XD) \stackrel{E}{=} (EB; ND)_{\Gamma_2} = (FA; MC)_{\Gamma_1} \stackrel{F}{=} (LA; XC) \stackrel{B}{=} (TZ; XY_2)$$
implies $Y_1 = Y_2$ as desired.

Fourth solution by untethered moving points Fix $\ell_1, \ell_2, T$, $\Gamma_1$ and $\Gamma_2$, and let $\Gamma_1$ and $\Gamma_2$ meet at $U$ and $V$. By the radical axis theorem, $X$ lies on $UV$.
Thus we instead $X$ as a variable point on line $UV$ and let $C = AX \cap \Gamma_1$, $D = BX \cap \Gamma_2$. By definition, $X$ has degree 1 and $T$ has degree 0.
We apply Zack's lemma to untethered point $Y$. Note that $C$ and $D$ move projectively on conics, and therefore have degree 2. Then, lines $AD$ and $BC$ each have degree at most $\deg(A) + \deg(D) = 0 + 2 = 2$, and so their intersection $Y$ has degree at most $2 + 2 = 4$. But when $X \in AB$, the lines $AD$ and $BC$ are the same, so Zack's lemma implies that
$$\deg Y \leq 4 - 1 = 3.$$
Thus the assertion that $T, X, Y$ are collinear (which for example can be seen as a certain vanishing determinant) is a statement of degree at most $0 + 1 + 3 = 4$. Thus it suffices to find 5 values of $X$ (other than $X \in AB$, which we used already). This is remarkably easy:
1.
When $X = U$ or $X = V$, then $X = C = D = Y$ and the statement is obvious
2.
When $X \in \ell_1$, say, then $A = C$ and so $Y$ lies on $AC = \ell_1$ as well. The case $X \in \ell_2$ is symmetric.
3.
Finally, take $X$ at infinity along $UV$. Then $C$ and $D$ are the other tangency points of the circles with $\ell_1, \ell_2$, and so $AC = \ell_1, BD = \ell_2$, so $Y = T$.
This finishes the problem.