Let $I$ be the incenter of triangle $ABC$, tangent to sides $AB$ and $AC$ at points $E$ and $F$, respectively. The lines through $E$ and $F$ parallel to $AI$ intersect lines $BI$ and $CI$ at points $P$ and $Q$, respectively. Prove that the center of the circumcircle of triangle $IPQ$ lies on line $BC$.
Problem
Source: Oral Moscow geometry olympiad 2023 8-9.4
Tags: geometry
khina
21.04.2023 00:08
Cool problem.
Rephrasing in terms of $IBC$ and after a few initial observations, the problem becomes as follows:
rephrased wrote:
Let $ABC$ be a triangle with circumcenter $O$ and foot $D$ from $A$ to $BC$. Say $X \in AB, Y \in AC$ such that $DX // OB, DY // OC$. Prove that the circumcenter of $AXY$ lies on $BC$.
Proof: Note that $\angle{AXD} = \angle{ABO} = \angle{DAC}$, so thus $D$ is the $A$-humpty point of $AXY$. Thus if the circumcenter of $AXY$ is $O'$, then it is well-known $\angle{ADO'} = 90^\circ$; therefore $O' \in BC$ as desired.
NO_SQUARES
21.04.2023 00:17
By counting angles, it is easy to prove that the triangles $EIP$ and $FIQ$ are similar. Let $D$ be the tangent point of the inscribed circle with $BC.$ By counting, it can be shown that the $DPEI$ and $DIFQ$ quadrilaterals differ in a rotary homotety at point $D,$ therefore, with this rotary homotety, the vector $PI$ turns into the vector $IQ,$ that is, $D$ is the $I-$Dumpty point of triangle $PIQ,$ therefore the center $(PIQ)$ lies on a straight line passing through $D$ perpendicular to $ID,$ that is, on a straight $BC.$
KST2003
24.04.2023 22:06
Let the incircle touch $\overline{BC}$ at $D$, and let $(PQD)$ intersect $\overline{BC}$ again at $K$. Then angle chasing shows that $\overline{BC}$ is the external angle bisector of $\angle QDP$, so $QK = KP$. Then \[ \angle QIP + \frac{\angle QKP}{2} = 90^\circ + \frac{\angle A}{2} + 90^\circ - \frac{\angle A}{2} = 180^\circ\]shows that $K$ is the circumcenter of $\triangle QIP$.