Let $\Delta ABC$ be a triangle with an inscribed circle centered at $I$. The line perpendicular to $AI$ at $I$ intersects $\odot (ABC)$ at $P,Q$ such that, $P$ lies closer to $B$ than $C$. Let $\odot (BIP) \cap \odot (CIQ) =S$. Prove that, $SI$ is the angle bisector of $\angle PSQ$
Problem
Source: Netherlands IMO TST 2019 P2
Tags: geometry, angle bisector, circumcircle
Synthetic_Potato
30.03.2019 16:04
It suffices to show that $\angle PBI = \angle QCI$. Let $M,N$ be the intersection of $BI,CI$ with $\odot (ABC)$. Then since $MN\perp AI$ we see that $PQ\parallel MN$. Also, since $MNPQ$ is cyclic $MNQP$ is an isosceles trapezium. So $\angle PBI = \angle PBM = \angle MQP = \angle NPQ = \angle NCQ = \angle ICQ$. $\blacksquare$.
Pluto1708
30.03.2019 16:09
Trivial for a TST Let $E,F$ be the intersection of $BI,CI$ with $(ABC)$ respectively.Clearly $EF\parallel PQ \implies EFPQ$ cyclic trapeziod.$\therefore \widehat{PE}=\widehat{FQ} \implies \angle FCQ=\angle EBP \implies \angle PBI=\angle QCI \implies \angle PSI=\angle QSI$ as desired.$\blacksquare$
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XbenX
30.03.2019 16:16
This is only angle chasing. Let $X$ be the intersection of $AB$ with $PQ$ . We need to show $\angle PBI =\angle QCI$ . We have $\angle AXI=\angle PXB=90^{\circ}-\frac{\angle BAC}{2}$ and we need to show $\angle PBA+\angle ABI=\angle QCA+\angle ACI$ In $\triangle PBX$ we have $\angle PBA=\angle PBX=180^{\circ}-(90^{\circ}-\frac{\angle BAC}{2})-\angle BPI=\angle QCB-(90^{\circ}-\frac{\angle BAC}{2})$ $\angle PBA=\angle QCA+2\angle ACI-(90^{\circ}-\frac{\angle BAC}{2})$ and from the well known fact that $\angle ABI+\angle ACI=90^{\circ}-\frac{\angle BAC}{2}$ we have $\angle PBA+\angle ABI=\angle QCA+\angle ACI$ . $\blacksquare$
ELMOliveslong
30.03.2019 16:24
AlastorMoody wrote: Let $\Delta ABC$ be a triangle with an inscribed circle centered at $I$. The line perpendicular to $AI$ at $I$ intersects $\odot (ABC)$ at $P,Q$ such that, $P$ lies closer to $B$ than $C$. Let $\odot (BIP) \cap \odot (CIQ) =S$. Prove that, $SI$ is the angle bisector of $\angle PSQ$ From where did you get this?
XbenX
30.03.2019 16:27
Here are Netherlands IMO TST papers (In Dutch).
zuss77
06.04.2019 19:15
Line perpendicular to $AI$ does not have to pass through $I$.
AlastorMoody
06.04.2019 21:15
zuss77 wrote: Line perpendicular to $AI$ does not have to pass through $I$. AlastorMoody wrote: Let $\Delta ABC$ be a triangle with an inscribed circle centered at $I$. The line perpendicular to $AI$ at $I$ intersects $\odot (ABC)$ at $P,Q$ such that, $P$ lies closer to $B$ than $C$. Let $\odot (BIP) \cap \odot (CIQ) =S$. Prove that, $SI$ is the angle bisector of $\angle PSQ$ @below Yeah! You should have written that way!
zuss77
07.04.2019 11:22
@ Above. I meant that this condition is not necessary. Sollution not depends on it.
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