Does anybody have synthetic though?

Yay, trig bash: Let the feet of the altitude from points $A$ and $C$ to $BC$ and $AB$ respectively are $H_{a}$ and $H_{c}$. Also let $H$ be the orthocenter of $\triangle ABC$ and $M$ be the midpoint of $AB$. Now let's call $O_{1}=AH_{a}\cap OM$ (I chose this name because if $B_{1}=BT\cap AC$, then the intersection point of the three lines is the circumcenter of $\triangle ABB_{1}$). We will prove that the point $O_{1}$ is the intersection point. This is equivalent to showing that $O_{1}\in KT$. However, using the Menelaus theorem for $\triangle AH_{c}H$ and the points $T\in HH_{c}$ , $K\in AH_{c}$ , $O_{1}\in AH$ we know that: $$O_{1}\in KT\iff\left(\frac{HT}{TH_{c}}\right)\left(\frac{H_{c}K}{KA}\right)\left(\frac{AO_{1}}{O_{1}H}\right)=1$$Now $\frac{TH_{c}}{BH_{c}}=\tan{\gamma}=\frac{\sin{\gamma}}{\cos{\gamma}}$ and $\frac{HH_{c}}{BH_{c}}=\tan(90^{\circ}-\alpha)=\frac{\sin(90^{\circ}-\alpha)}{\cos(90^{\circ}-\alpha)}=\frac{\cos\alpha}{\sin\alpha}$, so: $$\frac{HT}{TH_{c}}=1-\frac{HH_{c}}{TH_{c}}=1-\frac{\left(\frac{HH_{c}}{BH_{c}}\right)}{\left(\frac{TH_{c}}{BH_{c}}\right)}=1-\frac{(\cos\alpha)(\cos\gamma)}{(\sin\alpha)(\sin\gamma)}=\frac{(\sin\alpha)(\sin\gamma)-(\cos\alpha)(\cos\gamma)}{(\sin\alpha)(\sin\gamma)}$$$$\Longrightarrow \boxed{\frac{HT}{TH_{c}}=\frac{(\sin\alpha)(\sin\gamma)-(\cos\alpha)(\cos\gamma)}{(\sin\alpha)(\sin\gamma)}}$$Now $\frac{KH_{c}}{CH_{c}}=\tan(\beta-\alpha)=\frac{\sin(\beta-\alpha)}{\cos(\beta-\alpha)}=\frac{\sin\beta\cos\alpha-\sin\alpha\cos\beta}{\cos\alpha\cos\beta+\sin\alpha\sin\beta}$ and $\frac{AH_{c}}{CH_{c}}=\tan(90^{\circ}-\alpha)=\frac{\sin(90^{\circ}-\alpha)}{\cos(90^{\circ}-\alpha)}=\frac{\cos\alpha}{\sin\alpha}$, so: $$\frac{KH_{c}}{AK}=\frac{KH_{c}}{AH_{c}-KH_{c}}=\frac{\left(\frac{\sin\beta\cos\alpha-\sin\alpha\cos\beta}{\cos\alpha\cos\beta+\sin\alpha\sin\beta}\right)}{\left(\frac{\cos\alpha}{\sin\alpha}\right)-\left(\frac{\sin\beta\cos\alpha-\sin\alpha\cos\beta}{\cos\alpha\cos\beta+\sin\alpha\sin\beta}\right)}=\frac{(\sin\alpha)(\sin\beta\cos\alpha-\sin\alpha\cos\beta)}{(\cos\alpha)(\cos\alpha\cos\beta+\sin\alpha\sin\beta)-(\sin\alpha)(\sin\beta\cos\alpha-\sin\alpha\cos\beta)}=$$$$=\frac{\sin\alpha(\sin\beta\cos\alpha-\sin\alpha\cos\beta)}{\cos\beta(\sin^{2}\alpha+\cos^{2}\alpha)}=\frac{\sin\alpha(\sin\beta\cos\alpha-\sin\alpha\cos\beta)}{\cos\beta}\Longrightarrow \boxed{\frac{KH_{c}}{AK}=\frac{\sin\alpha(\sin\beta\cos\alpha-\sin\alpha\cos\beta)}{\cos\beta}}$$Finally $\frac{AO_{1}}{O_{1}H}=\frac{AM}{MH_{c}}$, because $O_{1}M\perp AB\perp HH_{c}$. Now $BH_{c}=a\sin(90^{\circ}-\beta)=a\cos\beta$, so: $$\frac{AO_{1}}{O_{1}H}=\frac{AM}{MH_{c}}=\frac{\left(\frac{AB}{2}\right)}{\left(\frac{AB}{2}\right)-BH_{c}}=\frac{\left(\frac{c}{2}\right)}{\left(\frac{c}{2}\right)-a\cos\beta}=\frac{2R\sin\gamma}{2R\sin\gamma-4R\sin\alpha\cos\beta}=\frac{\sin{\gamma}}{\sin\gamma-2\sin\alpha\cos\beta}\Longrightarrow \boxed{\frac{AO_{1}}{O_{1}H}=\frac{\sin{\gamma}}{\sin\gamma-2\sin\alpha\cos\beta}}$$$$\Longrightarrow \left(\frac{HT}{TH_{c}}\right)\left(\frac{H_{c}K}{KA}\right)\left(\frac{AO_{1}}{O_{1}H}\right)=\left(\frac{(\sin\alpha)(\sin\gamma)-(\cos\alpha)(\cos\gamma)}{(\sin\alpha)(\sin\gamma)}\right)\left(\frac{\sin\alpha(\sin\beta\cos\alpha-\sin\alpha\cos\beta)}{\cos\beta}\right)\left(\frac{\sin{\gamma}}{\sin\gamma-2\sin\alpha\cos\beta}\right)=$$$$=\left(\frac{(\sin\alpha)(\sin\gamma)-(\cos\alpha)(\cos\gamma)}{\cos\beta}\right)\left(\frac{\sin\beta\cos\alpha-\sin\alpha\cos\beta}{\sin\gamma-2\sin\alpha\cos\beta}\right)$$Now it's easy, because $\cos\gamma=\cos(180^{\circ}-\alpha-\beta)=\cos(180^{\circ}-\alpha)\cos(\beta)-\sin(180^{\circ}-\alpha)\sin\beta=\sin\alpha\sin\beta-\cos\alpha\cos\beta$, so $\frac{(\sin\alpha)(\sin\gamma)-(\cos\alpha)(\cos\gamma)}{\cos\beta}=1$ and $\sin\gamma=\sin\beta\cos\alpha+\sin\alpha\cos\beta$, so $\sin\gamma-2\sin\alpha\cos\beta=(\sin\beta\cos\alpha+\sin\alpha\cos\beta)-2\sin\alpha\cos\beta=\sin\beta\cos\alpha-\sin\alpha\cos\beta$, so the second fraction is also $1$, so we've solved the problem.

that looks like 2 hours and a lot of pain

I doubt there is a nice synthetic solution for this problem. (I might be very wrong.) Let $E$ and $F$ be the foots of perpendiculars from $C$ and $A$. Let $M$ be the midpoint of $AB$, and let $\overline{OM}$ intersect $\overline{AF}$ at $X$. It suffices to show that $CT:TE = OX:XM$. Scale the diagram so that $(ABC)$ becomes the unit circle. Then by the extended law of sines, it is evident that $AM=\sin C$. Now some easy angle chasing shows that \[\angle CBT = \angle B - \angle C = (90^\circ-\angle C)-(90^\circ-\angle B) = \angle OCF.\]By the ratio lemma, we have \[\frac{CT}{TE}=\frac{\sin\angle BEC}{\sin\angle BCE}\cdot\frac{\sin\angle CBT}{\sin\angle TBE}=\frac{1}{\sin\angle XAM}\cdot\frac{\sin\angle OAM}{\sin C} = \frac{OA}{AM}\cdot\frac{\sin\angle OAM}{\sin\angle XAM}=\frac{OX}{XM},\]and we're done!

Olympedias wrote: that looks like 2 hours and a lot of pain It did indeed take me about 2 hours in the competition, but the idea is straightforward, so it wasn't that bad, it just looks painful. Also, it was really clean: first, the denominator of the second ratio clears nicely into just $\cos\beta$ and after that, if you get the idea to turn the sides into trig functions using the sine law you are pretty much set as two cosines cancel and you finally get two fractions which are 1s, so I really enjoyed bashing this one actually.

Bruh, I also agree that it looks rather ugly , I tried to take a point $B'$ on $BT$ such that $AB=AB'$ as then $B'\in (ABC)$ even though I'm not sure if $(ABC)$ is involved at all !

Let me bump this nice config geo with the completely synthetic solution I found today.

DDIT on XOCT from B.

Here is a trig free solution, pretty much as VicKmath7's first solution. Let $M$ be the midpoint of $AB$ and let the altitude through $A$ intersect the perpendicular bisector $OM$ of $AB$ at $P$ - it suffices to show that $K$, $P$, $T$ are collinear. An efficient way to get rid of $K$ is as follows - we have $\triangle OKM \sim \triangle CKD$ and (as $OM \parallel CD$) the desired collinearity holds if and only if $P$ and $T$ are corresponding points in this similarity -- that is, we have reduced to showing $\displaystyle \frac{PM}{OM} = \frac{TD}{CD}$. Since $\angle APM = 90^{\circ} - \angle PAM = \angle ABC = \beta$, we have $\triangle APM \sim \triangle CBD$ and so $\frac{PM}{AM} = \frac{BD}{CD}$. On the other hand, $\angle AOM = \frac{1}{2}\angle AOB = \angle ACB = \gamma = \angle TBD$ and so $\triangle AOM \sim \triangle TBD$, which gives $\frac{OM}{AM} = \frac{DT}{CD}$. Dividing the latter two equalities of ratios gives the result.

Thanks,guys!