Let $ABC$ be an acute nonisosceles triangle with incenter $I$ and $(d)$ is an arbitrary line tangent to $(I)$ at $K$. The lines passes through $I$, perpendicular to $IA, IB, IC$ cut $(d)$ at $A_1, B_1,C_1$ respectively. Suppose that $(d)$ cuts $BC, CA, AB$ at $M,N, P$ respectively. The lines through $M,N,P$ and respectively parallel to the internal bisectors of $A, B, C$ in triangle $ABC$ meet each other to define a triange $XYZ$. Prove that three lines $AA_1, BB_1, CC_1$ are concurrent and $IK$ is tangent to the circle $(XY Z)$
Problem
Source: 2019 Saudi Arabia IMO TST II p3
Tags: geometry, incenter, incirle, concurrency, concurrent, tangent
tuandat03
12.08.2021 06:13
Any solutions?
GuvercinciHoca
13.12.2021 09:42
For the first part of the problem (concurrency part): Just take poles and polars of the lines WRT incircle. Problem becomes to prove simson line of $K$ WRT incircle $\triangle ABC$ $\smiley$.
nbdaaa
21.01.2022 19:36
The first part of the problem can be solved as above did: Take the poles of $AA_1, BB_1, CC_1$ wrt $(I)$ then they are all lying on the Simson line of $K$ wrt $(I)$ Proof for the second part: Take $S$ be the reflection of $I$ wrt $(d)$ Then $\angle SPN = \angle NIP = 180^{\circ} - \angle BIC = 180^{\circ}- \angle NZP \implies $ $S,P,Z,N$ are concyclic Similarly we have $S$ be the Miquel point of the complete quadrilateral $MPZYNX$ We're gonna prove that $IK$ touches $(XYZ)$ at $S$ Let $(I)$ touch $BC,CA,AB$ at $D,E,F$. Then by some easy angle chasing we have $$ \angle ISZ = \angle IKD \implies \angle ISZ = \angle KMI = \angle SMP = \angle SXP $$Which leads to $IS$ is tangent to $(XYZ)$ (Q.E.D)
teomihai
22.01.2022 09:19
nice one!
ThisNameIsNotAvailable
06.02.2022 08:18
Let $D,E,F$ be the tangent point of $(I)$ and $BC,CA,AB$.
Because the poles of $AA_1,BB_1,CC_1$ both lies on the Simson line of $\Delta DEF$, $AA_1,BB_1,CC_1$ are concurrent.
By angle chasing, easily we have $\angle AIN=\angle BIM \implies \angle AIB=\angle MIN$.
Let $T$ be the symmetrical point of $I$ through $(d)$. Then $\angle MXY=180^\circ -\angle AIB = 180^\circ -\angle MIN=180^\circ -\angle MTN$.
Similarly, we have $T$ is the Miquel point of quadrilateral $XMPY.ZN$, so $\angle ITX=\angle ITM - \angle MTX=\angle 90^\circ -\angle TMP-\angle MNX=...=\angle IPN=\angle TZX$, and we are done!