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Geometry Difficulty 7.2 National olympiad, round 2 Prove it Belarus

Let II be the incenter of the acute-angled non-isosceles triangle ABCABC. Let the incircle touch the side ABAB at point QQ. Point TT is marked on the side ABAB so that IT∥CQIT \parallel CQ. The line through TT touches the incircle at point KK (different from QQ) and meets the lines CACA and CBCB at points LL and NN, respectively.

Prove that TT is the midpoint of the segment LNLN.

(Ya. Konstantinovski)

Solution

Let x=∠CQIx = \angle CQI, y=∠QCIy = \angle QCI, 2α=∠BAC2\alpha = \angle BAC, 2β=∠ABC2\beta = \angle ABC, 2γ=∠ACB2\gamma = \angle ACB. Let rr be the inradius of the triangle ABCABC. It is easy to see that

Figure 1

∠CIQ=360∘−90∘−2α−γ= \angle CIQ = 360^\circ - 90^\circ - 2\alpha - \gamma =
=270∘−(α+β+γ)−α+β=180∘−α+β. = 270^\circ - (\alpha + \beta + \gamma) - \alpha + \beta = 180^\circ - \alpha + \beta.
So,
x+y=180∘−∠CIQ=α−β.(1) x + y = 180^\circ - \angle CIQ = \alpha - \beta. \quad (1)
Further,
∠QTL=∠NTB=180∘−2(90∘−x)=2x, \angle QTL = \angle NTB = 180^\circ - 2(90^\circ - x) = 2x,

hence
∠ILT=0.5∠ALT=0.5(2α−2x)=α−x,∠INT=0.5∠CNT==0.5(2β+2x)=β+x. \begin{aligned} \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{aligned}
Then
LT=LK−KT=rcot⁡∠ILT−rtan⁡x=r(cot⁡(α−x)−tan⁡x),NT=NK+KT=rcot⁡∠INT+rtan⁡x=r(cot⁡(β+x)+tan⁡x). \begin{aligned} 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{aligned}
Therefore, from
cot⁡(α−x)−tan⁡x=cot⁡(β+x)+tan⁡x(2) \cot(\alpha - x) - \tan x = \cot(\beta + x) + \tan x \quad (2)
it will follow the problem condition.
Indeed, since
cot⁡(α−x)−tan⁡x=1+tan⁡αtan⁡xtan⁡α−tan⁡x−tan⁡x==1+tan⁡αtan⁡x−tan⁡αtan⁡x+tan⁡2xtan⁡α−tan⁡x=1+tan⁡2xtan⁡α−tan⁡x,cot⁡(β+x)−tan⁡x=1−tan⁡βtan⁡xtan⁡β+tan⁡x+tan⁡x==1−tan⁡βtan⁡x+tan⁡βtan⁡x+tan⁡2xtan⁡β+tan⁡x=1+tan⁡2xtan⁡β+tan⁡x, \begin{aligned} \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{aligned}
we see that (2) is equivalent to
tan⁡α−tan⁡x=tan⁡β+tan⁡x.(3) \tan \alpha - \tan x = \tan \beta + \tan x. \quad (3)
By law of sines for △CQT\triangle CQT, rsin⁡y=CIsin⁡x\frac{r}{\sin y} = \frac{CI}{\sin x}. Since CI=r/sin⁡γCI = r/\sin \gamma, we have sin⁡y=sin⁡γ⋅sin⁡x\sin y = \sin \gamma \cdot \sin x, i.e. from (1) and the equality α+β+γ=90∘\alpha + \beta + \gamma = 90^\circ it follows that sin⁡(α−β−x)=sin⁡x⋅cos⁡(α+β)\sin(\alpha - \beta - x) = \sin x \cdot \cos(\alpha + \beta). Then
sin⁡x⋅cos⁡(α+β)=sin⁡(α−β)cos⁡x−cos⁡(α−β)sin⁡x,\sin x \cdot \cos(\alpha + \beta) = \sin(\alpha - \beta) \cos x - \cos(\alpha - \beta) \sin x,
so sin⁡x(cos⁡(α+β)+cos⁡(α−β))=sin⁡(α−β)cos⁡x\sin x(\cos(\alpha + \beta) + \cos(\alpha - \beta)) = \sin(\alpha - \beta) \cos x, which is equivalent to
2sin⁡xcos⁡αcos⁡β=(sin⁡αcos⁡β−cos⁡αsin⁡β)cos⁡x2 \sin x \cos \alpha \cos \beta = (\sin \alpha \cos \beta - \cos \alpha \sin \beta) \cos x,   ⟺  2tan⁡x=tan⁡α−tan⁡β\iff 2 \tan x = \tan \alpha - \tan \beta,
i.e. is equivalent to (3), as required.

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