IMO Team Selection Test

Let $T'$ be the point on $AB$ such that $\angle ACT' = \angle BCT'$. Let $\omega$ be the circle centered at $T'$ tangent to segments $AC$ and $BC$, and denote the points of tangency to $AC$ and $BC$ by $M$ and $N$ respectively. Note that $\omega$ exists because $CT'$ is an angle bisector.

Lemma 1. If $RDT'S$ is cyclic and $R$ and $S$ are on the boundary of $\omega$, then $BR$, $AS$, and $CD$ concur.

Proof. Let $P$ be the intersection of the lines tangent to $\omega$ that intersect $R$ and $S$. Since $\angle T'SP = \angle T'PP = 90^\circ$, $PRST'$ is cyclic. Since we assumed $RDT'S$ is cyclic, then $PRDT'S$ is cyclic. Thus $\angle T'DP = 90^\circ$. Thus $P$ is on $CD$. Now, let $M$ and $N$ meet at $C$. If $MN$ and $RS$ meet at $Q$, $PC$ is the polar of $Q$, so $MS$ and $RN$ lie on $PQ$. Let $MN$ and $RN$ meet at $G$.

By the law of sines, we have
$$\begin{aligned} \left( \frac{\sin \angle CAS}{\sin \angle SAB} \right) \cdot \left( \frac{\sin \angle RBA}{\sin \angle RBC} \right) &= \frac{\frac{MS \sin \angle AMG}{AS}}{\frac{SD \sin \angle ADS}{AS}} \cdot \frac{\frac{RD \sin \angle RDB}{RB}}{\frac{RN \sin \angle BNG}{RB}} \\ &= \frac{MS}{SD} \cdot \frac{RD}{RN} \cdot \frac{\sin \angle AMG}{\sin \angle BNG} \cdot \frac{\sin \angle RDB}{\sin \angle SDA} \\ &= \frac{\sin \angle MDS}{\sin \angle SMD} \cdot \frac{\sin \angle RND}{\sin \angle RDN} \cdot \frac{\sin \angle CMG}{\sin \angle CNG} \cdot \frac{\sin \angle RDA}{\sin \angle SDB} \\ &= \frac{\sin \angle RND}{\sin \angle SMD} \cdot \frac{\sin \angle CMG}{\sin \angle CNG} \cdot \frac{\sin \angle MDS}{\sin \angle RDN} \cdot \frac{\sin \angle RDA}{\sin \angle SDB}. \end{aligned}$$

Denote this expression by $S$. For convenience, let $x = \frac{\sin \angle MDS}{\sin \angle RDN} \cdot \frac{\sin \angle RDA}{\sin \angle SDB}$. Then, we have
$$\begin{aligned} S &= \frac{\sin \angle RND}{\sin \angle SMD} \cdot \frac{\sin \angle CMG}{\sin \angle CNG} x = \frac{\frac{GD \cdot \sin \angle CDN}{GN}}{\frac{GD \sin \angle CDM}{GM}} \cdot \frac{\frac{CG \cdot \sin \angle MCD}{MG}}{\frac{CG \sin \angle NCD}{NG}} x = \frac{\sin \angle CDN}{\sin \angle NCD} \cdot \frac{\sin \angle MCD}{\sin \angle CDM} x \\ &= \frac{CN}{DN} \cdot \frac{DM}{CM} x = \frac{DM}{DN} x = \frac{\frac{CD \sin \angle ACD}{\sin \angle DMC}}{\frac{CD \sin \angle BCD}{\sin \angle DNC}} x = \frac{\sin \angle ACD}{\sin \angle BCD} \cdot \frac{\sin \angle DNC}{\sin \angle DMC} x. \end{aligned}$$
Since $\angle T'NC = \angle T'MC = \angle T'DC = 90^\circ$, $CMDN$ is cyclic, so $\angle DNC = 180^\circ - \angle DMC$, and
$$\angle MDC = \angle MNC = \angle NMC = \angle NDC.$$
Further, notice that
$$\angle RDA = \angle RST' = \angle SRT' = \angle SDB$$
because $T'R = T'S$. Hence, we conclude that
$$\angle MDS = \angle MDC + 90^\circ - \angle SDB = \angle NDC + 90^\circ - \angle RDA = \angle RDN.$$
Therefore $x = \frac{\sin \angle MDS}{\sin \angle RDN} \cdot \frac{\sin \angle RDA}{\sin \angle SDB} = 1$, so we have
$$\left( \frac{\sin \angle CAS}{\sin \angle SAB} \right) \cdot \left( \frac{\sin \angle RBA}{\sin \angle RBC} \right) \cdot \left( \frac{\sin \angle BCD}{\sin \angle ACD} \right) = 1,$$
which implies that lines $AS$, $RB$, and $CD$ concur by trig Ceva. This concludes the proof of the lemma. $\square$

Denote the intersection point of $BX$ and $\omega$ farther from $B$ by $R$, and let the circumcircle of $RDT'$ meet $\omega$ again at $S$. By Lemma 1, $AS$ and $BR$ meet on $CD$, so $A$, $S$, and $K$ are collinear. Let $\omega_A$ and $\omega_B$ be the circles centered at $A$, $B$ with radii $AC$ and $BC$ respectively. Let $AK$ meet $\omega_B$ again at $S'$, and let $BL$ meet $\omega_A$ again at $R'$.

A homothety centered at $B$ takes $\omega$ to $\omega_A$, so
$$\frac{BR}{BR'} = \frac{BN}{BC} = \frac{BT'}{BA} = \frac{a}{a+b}.$$
Thus $BL \cdot BR = \frac{a}{a+b} \cdot BL \cdot BR' = \frac{a}{a+b} \cdot BC^2 = \frac{a^3}{a+b}$ and $BT' \cdot BD = c\left(\frac{a}{a+b}\right) \cdot \left(c \cdot \frac{a^2}{a^2+b^2}\right) = \frac{a^3}{a+b}$. Thus $BLRDT'$ is cyclic, and similarly $SKDT'$ is cyclic. Since $LRDT'$, $KDT'S$, and $RDT'S$ are cyclic, then $LRDT'SK$ is cyclic, so $RDT'S$ is cyclic. Therefore $T = T'$, as desired.