Let $D,E,F$ denote the tangent points of the incircle of $ABC$ with sides $BC,AC,AB$ respectively. Let $M$ be the midpoint of the segment $EF$. Let $L$ be the intersection point of the circle passing through $D,M,F$ and the segment $AB$, $K$ be the intersection point of the circle passing through $D,M,E$ and the segment $AC$. Prove that the circle passing through $A,K,L$ touches the line $BC$
Problem
Source: 2016 Belarus Team Selection Test 3.3
Tags: geometry
MP8148
13.06.2020 04:35
It suffices to show $(ALD)$ is tangent to $\overline{BC}$. From $\angle DFL = \angle DEF = \angle DEM$ and $\angle FLD = 180^\circ - \angle FMD = \angle EMD$, we have $\triangle DFL \sim \triangle DEM$. We then deduce $\angle FDL = \angle EDM = \angle FDA$ since $\overline{DA}$ is a symmedian. Lastly we have $$\angle LDB = \angle FDB - \angle FDL = \angle DFB - \angle FDA = \angle DAL$$which implies the conclusion.
primesarespecial
21.09.2021 16:28
This is angle chase and to incorporate the midpoint ,we see that by construction $DA$ is a symmedian....
MathLuis
16.02.2022 18:30
This is a nice problem Lemma: On any triangle $\triangle ABC$ with incenter $I$ and $M$ a midpoint of the arc $BAC$ we have that $T_A$, the intouch point of its $A$-mixtilinear incircle with $(ABC)$, lies on $IM$. Proof of the lemma: Use $\sqrt{bc}$ inversion. Now we start solving the problem with some claims. Claim 1: $D$ lies on $(AKL)$ and $DM$ bisects $\angle LDK$ Proof: By angle chasing $$90-\frac{\angle BAC}{2}=\angle AFM=\angle LDM \; \text{and} \; 90-\frac{\angle BAC}{2}=\angle AEM=\angle KDM$$Adding both of these results we get our desired claim 1. Claim 2: $M$ is the incenter of $\triangle AKL$ Proof: Since $M$ lies on $AI$ its enough to show that $\angle KML=\angle BIC$ but by angle chasing $$\angle KML=\angle KMD+\angle DML=\angle BFD+\angle CED=\angle BID+\angle CID=\angle BIC$$Hence our claim 2 is also proven. Finishing: Using Claim 1 and Claim 2 and the Lemma we have that $D$ is the $A$-mixtilinear intouch point on $\triangle AKL$ and since by Claim 2 and that $EM=MF$ and $AM \perp EF$ we have that the incircle of $\triangle ABC$ is the $A$-mixtilinear incircle of $\triangle AKL$ hence $(AKL)$ and the incircle of $\triangle ABC$ are tangent at $D$ but since the incircle is tangent to $BC$ so is $(AKL)$ hence we are done