concyclic wanted, line passing through incenter related

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Source: 2006 Rusanovsky Lyceum Olympiad p133

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#1 h Jun 29, 2022, 12:59 PM

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Let $I$ be the center of the inscribed circle of triangle $ABC$ . The line $\ell$ passing through point $I$ perpendicular to line $AI$ intersects the sides $AB$ and $AC$ at points $F$ and $E$ , respectively. Let $M$ be the midpoint of the segment $FI$ , $N$ be the midpoint of the segment $EI$ , $P$ be the intersection point of the lines $CF$ and $BM$ , $Q$ be the intersection of of the lines $BE$ and $CN$ . Prove that the points $B, P, Q, C$ lie on the same circle.

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#2 h Jun 30, 2022, 9:22 AM

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We have $\angle{BFI} = 180^{\circ} - \angle{AFI} = 180^{\circ} - \angle{AEI} = \angle{CEI}$ and $\angle{BIF} = \angle{AIB} - \angle{AIF} = \angle{ACI}$ . Then $\triangle BFI$ $\sim$ $\triangle IEC$ . So we have $\dfrac{BF}{FM} = \dfrac{2 BF}{FI} = \dfrac{2 IE}{EC} = \dfrac{FE}{EC}$ . Hence $\triangle BFM$ $\sim$ $\triangle FEC$ . Then $\angle{CFE} = \angle{MBF}$ . So $\triangle PFM$ $\sim$ $\triangle FBM$ . Hence $\angle{BPC} = \angle{FPM} = \angle{BFM} = 90^{\circ} + \angle{IAB} = \angle{BIC}$ or $P$ $\in$ $(BIC)$ . Similarly, we have $Q$ $\in$ $(BIC)$

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#3 h Jun 30, 2022, 9:50 AM

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Soution using symmetry:
I want to prove $\angle MBE = \angle NCF$ , thus as $FN = ME$ it's sufficient to prove that $BME$ and $FNC$ have the same circumradius.
Claim:
$(BME)$ and $(FCN)$ are symmetric wrt. $AI$ .
Proof:
Define $C'$ to be the reflection of $C$ across $AI$ . Just need to prove $BC'ME$ is cyclic. By PoP, I need
$FC'\cdot FB = FM\cdot FE$ but $FM\cdot FE = (FI)^2$ thus it's enough to prove $(FC'I)$ is tangent to $FI$ , which is trivial by angle chasing.

This post has been edited 1 time. Last edited by MathSaiyan, Jun 30, 2022, 9:51 AM
Reason: typo

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