Let $ABC$ be a triangle. Consider the circle $\omega_B$ internally tangent to the sides $BC$ and $BA$, and to the circumcircle of the triangle $ABC$, let $P$ be the point of contact of the two circles. Similarly, consider the circle $\omega_C$ internally tangent to the sides $CB$ and $CA$, and to the circumcircle of the triangle $ABC$, let $Q$ be the point of contact of the two circles. Show that the incentre of the triangle $ABC$ lies on the segment $PQ$ if and only if $AB + AC = 3BC$. proposed by Luis Eduardo Garcia Hernandez, Mexico
Problem
Source: RMM Shortlist 2017 G2
Tags: geometry, incenter, mixtilinear
04.07.2019 22:23
Let $I$ be the incenter. Recall that $PI$ and $QI$ meet $(ABC)$ again at the arc midpoints $M$ and $N$ of major arcs $ABC$ and $BCA$, so $P, I, Q$ are collinear if and only if $M, I, N$ collinear. This was Sharygin 2014 (solution is given here).
08.08.2021 13:15
From properties of mixtilinear incircles,we get that if it is true ,then $P$ and $Q$ are just midpoints of arcs($ABC$) and arc(BCA),and we we have to prove $P,I,Q$ collinear with the condition and we can finish using complex numbers.
19.03.2022 07:32
Solved with Jeffrey Chen We prove both parts. Define $M_B,M_C$ as the arc midpoints of $\overarc{AC}$ and $\overarc{AB}$, $O$ as the circumcenter, and $T_B, T_C$ as the B-extouch point with $AC$ and the C-extouch point with $AB$. Part 1: If $AB + AC = 3BC$, then $P,I,Q$ are collinear. Proof: Observe that $s = 2BC$, so $BT_C = s - BC = BC$, and similarly $CT_B = BC$. Therefore, we have \[\angle QCA = \angle T_CCB = 90 - \frac12 B\]This implies $Q$ is the arc midpoint of $\overarc{ABC}$. Similarly, $P$ is the arc midpoint of $\overarc{APB}$, so $P, O, M_C$ and $Q,O,M_B$ are collinear. Observe that $M_BM_C$ is the perpendicular bisector of $AI$. Then, \[QM_C = 2R\sin \frac12 A = \frac{BC}{\sin A}\sin \frac12 A = \frac{BC}{2\cos \frac{1}{2}A}\]If $QM_C = \frac12 AI$, we're done. Observe that the length of the tangent from $A$ to the incircle is $s - BC = BC$, so \[\tan \frac12 A = \frac{r}{BC} \Rightarrow QM_C = \frac{BC}{2\cos \frac12 A} = \frac{1}{2} \frac{r\sin \frac{1}{2}A} = \frac 12 AI\]and we are done. Part 2: If $P,I,Q$ are collinear, then $AB + AC = 3BC$. If we let $S_B,S_C$ be the touch-points of $\omega_C, \omega_B$ with $AC, AB$ respectively, then by homothety, $P,S_C,M_C$ and $Q,S_B,M_B$ are collinear. Furthermore, it is well known that $(QIS_BA)$ and $(PIS_CA)$ are cyclic. If $O' = QM_B\cap PM_C$, then \[\angle AS_CO' + \angle AS_BO' = \angle AIQ + \angle AIP = 180\]which means $(AS_CO'S_B)$ is cyclic. Therefore, $\angle M_CO' M_B = 180 - \angle A$. However, since ${M_C M_B} = 90 - \frac{1}{2} A$, this means $\overarc{PQ} = 90 - \frac12 A$, so $PQ = M_CM_B$. Define $S = CQ\cap PB$. By pascals on $PBM_BQCM_C$, we have $I, S, O'$ are collinear. Since $S$ is the exsimilicenter of the incircle and circumcircle, we have $I,O,S$ are collinear. Now, I claim the only way for $O', O, I$ to be collinear is when $O = O'$. First of all, this is possible by setting $P, Q$ as the reflection of $M_C,M_B$ over $O$. Next, for any other placement of $P$, $Q$ is fixed (since $PQ = M_CM_B$. Inverting with the circumcenter, $O'$ goes to the intersection of $M_CM_B$ and $PQ$, so the locus of $O'$ is $(M_COM_B)$. However, since $M_CM_B\cap PQ$ lies outside of $M_CM_B$, this means the locus of $O'$ must also be within $(ABC)$, so the only intersection of $(M_COM_B)$ with $OI$ is $O$. Therefore, $O' = O$. Finally, this means \[\angle BCT_C = \angle QCA = 90 - \angle ACM_B = 90 - \frac12 B\]so $BT_C = BC$, and $s - BC = BC$ so $AB + AC = 3BC$.