Belarusian Mathematical Olympiad

Let $x=\angle CQI$, $y=\angle QCI$, $2\alpha =\angle BAC$, $2\beta =\angle ABC$, $2\gamma =\angle ACB$. Let $r$ be the inradius of the triangle $ABC$. It is easy to see that



$$\angle CIQ=360^{\circ }-90^{\circ }-2\alpha -\gamma =$$ $$=270^{\circ }-(\alpha +\beta +\gamma )-\alpha +\beta =180^{\circ }-\alpha +\beta .$$ So, $$x+y=180^{\circ }-\angle CIQ=\alpha -\beta .(1)$$ Further, $$\angle QTL=\angle NTB=180^{\circ }-2(90^{\circ }-x)=2x,$$

hence $$\begin{array}{rl}\angle ILT& =0.5\angle ALT=0.5(2\alpha -2x)=\alpha -x,\angle INT=0.5\angle CNT=\\ & =0.5(2\beta +2x)=\beta +x.\end{array}$$ Then $$\begin{array}{rl}LT& =LK-KT=r\cot \angle ILT-r\tan x=r(\cot (\alpha -x)-\tan x),\\ NT& =NK+KT=r\cot \angle INT+r\tan x=r(\cot (\beta +x)+\tan x).\end{array}$$ Therefore, from $$\cot (\alpha -x)-\tan x=\cot (\beta +x)+\tan x(2)$$ it will follow the problem condition. Indeed, since $$\begin{array}{rl}\cot (\alpha -x)-\tan x& =\frac{1+\tan \alpha \tan x}{\tan \alpha -\tan x}-\tan x=\\ & =\frac{1+\tan \alpha \tan x-\tan \alpha \tan x+{\tan }^{2}x}{\tan \alpha -\tan x}=\frac{1+{\tan }^{2}x}{\tan \alpha -\tan x},\\ \cot (\beta +x)-\tan x& =\frac{1-\tan \beta \tan x}{\tan \beta +\tan x}+\tan x=\\ & =\frac{1-\tan \beta \tan x+\tan \beta \tan x+{\tan }^{2}x}{\tan \beta +\tan x}=\frac{1+{\tan }^{2}x}{\tan \beta +\tan x},\end{array}$$ we see that (2) is equivalent to $$\tan \alpha -\tan x=\tan \beta +\tan x.(3)$$ By law of sines for $△CQT$, $\frac{r}{\sin y}=\frac{CI}{\sin x}$. Since $CI=r/\sin \gamma$, we have $\sin y=\sin \gamma \cdot \sin x$, i.e. from (1) and the equality $\alpha +\beta +\gamma =90^{\circ }$ it follows that $\sin (\alpha -\beta -x)=\sin x\cdot \cos (\alpha +\beta )$. Then $\sin x\cdot \cos (\alpha +\beta )=\sin (\alpha -\beta )\cos x-\cos (\alpha -\beta )\sin x,$ so $\sin x(\cos (\alpha +\beta )+\cos (\alpha -\beta ))=\sin (\alpha -\beta )\cos x$, which is equivalent to $2\sin x\cos \alpha \cos \beta =(\sin \alpha \cos \beta -\cos \alpha \sin \beta )\cos x$, $⟺2\tan x=\tan \alpha -\tan \beta$, i.e. is equivalent to (3), as required.