Let $ABC$ be a triangle with incenter $I$. The line $AI$ intersects $BC$ and the circumcircle of $ABC$ at the points $T$ and $S$, respectively. Let $K$ and $L$ be the incenters of $SBT$ and $SCT$, respectively, $M$ be the midpoint of $BC$ and $P$ be the reflection of $I$ with respect to $KL$. a) Prove that $M$, $T$, $K$ and $L$ are concyclic. b) Determine the measure of $\angle BPC$.
Problem
Source: Bulgaria EGMO TST 2017 Day 1 Problem 2
Tags: geometry, incenter, incircle, excircle, circumcircle, geometric transformation, reflection
PROF65
04.02.2023 15:32
known fact : Let $ABC$ be a triangle , $D\in BC$. $(I),(J) ,(K)$ the incircles of $ABC,ADC,ADB$ .If $(I)$ touches $BC$ at $G$ then $JKDG$ are cyclic . Proof: Let $(J), (K)$ touch resp. $BC$ at $E,F$ it obvious that $\angle JDK=90^{\circ}$ plus we easily get $ED=GF$ hence $EF$ and $GD$ has the same perpendicular bisector which bisects $JK$ at $O$ then $OD=OG$ besides $OD=OJ=OK$ hence $DGKJ$ are cyclic . Lemma : $ABC$ is a triangle with incenter $I$ ; $A'$ the midpoint of arc $BC$ not containing $A$.Let the bissectors of $\angle AA'B,\angle AA'C$ cut respectively $BA,CA$ at $U,V$ then $I,U$ and $V$ are collinear and $UV\parallel BC$ proved previously . Back to the problem The fact leads $MTKL$ are cyclic Let $SK,SL$ hit resp. $BA,CA$ at $ U,V$ ; the lemma leads $I\in UV\parallel BC$ then $IUB$ isoceles and since $SI=SB$ we deduces $SU$ is the bisector of $BI$ thus $ KI=KB$ hence the circle of $PIB$ has $K$ as center whence $2. \angle BPI=\angle BKI$ idem we get $2. \angle CPI=\angle CLI$ computation of angles leads to $\angle BPC=90^{\circ}$
euclides05
04.02.2023 18:45
Assume WLOG that AB<AC.
For part a) Let the incircles of triangles SBT and SCT be tangent to BC at D and E respectively. We claim that $TD= ME$. Indeed, we have $TD= \frac{TB+TS- BS}{2}$, $TE= \frac{TS+TC- SC}{2}$ and by incenter- excenter lemma $SB= SC= SI$. Therefore, $TE- TD= \frac{TC- TB}{2}= TM$ or $DT= EM$ and the claim is proved. Also, $\angle KTL= 90^\circ$.
Now, we can easily finish either via perpendicular bisectors or using the Pythagorean Theorem.
For part b): Let $\omega$ be the circle (M, T, K, L) and let TS intersect $\omega$ for the second time at N. Then, $\angle KLN= \angle KTN= \angle KTB= \angle KLM$, which implies that M and and N are symmetric with respect to KL. Therefore by symmetry we are left to prove that $NI= \frac{BC}{2}$. This can be shown easily by length- chasing, computing NT with Ptolemy's Theorem.
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