7 circles, Vietnam TST 2015 (VNTST) P5

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7 circles, Vietnam TST 2015 (VNTST) P5

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Source: Vietnam TST 2015 (VNTST) P5

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#1 h Aug 27, 2018, 11:08 PM • 3 Y

Y by AlastorMoody, Adventure10, Mango247

Let $ABC$ be a triangle with an interior point $P$ such that $\angle APB = \angle APC = \alpha$ and $\alpha > 180^o-\angle BAC$ . The circumcircle of triangle $APB$ cuts $AC$ at $E$ , the circumcircle of triangle $APC$ cuts $AB$ at $F$ . Let $Q$ be the point in the triangle $AEF$ such that $\angle AQE = \angle AQF =\alpha$ . Let $D$ be the symmetric point of $Q$ wrt $EF$ . Angle bisector of $\angle EDF$ cuts $AP$ at $T$ .
a) Prove that $\angle DET = \angle ABC, \angle DFT = \angle ACB$ .
b) Straight line $PA$ cuts straight lines $DE, DF$ at $M, N$ respectively. Denote $I, J$ the incenters of the triangles $PEM, PFN$ , and $K$ the circumcenter of the triangle $DIJ$ . Straight line $DT$ cut $(K)$ at $H$ . Prove that $HK$ passes through the incenter of the triangle $DMN$ .

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#2 h May 25, 2019, 8:39 AM • 4 Y

Y by Adventure10, Mango247, Mango247, Mango247

Here is my solution for this problem
Solution
a) We have: $(EB; EA) \equiv (PB; PA) \equiv (PA; PC) \equiv (FA; FC)$ (mod $\pi$ )
So: $B$ , $C$ , $F$ , $E$ lie on a circle
Then: $(PA; PE) \equiv (BA; BE) \equiv (CF; CA) \equiv (PF; PA)$ (mod $\pi$ ) or $PA$ is internal bisector of $\angle{EAF}$
Since the hypothesis, it's easy to prove that: $\triangle ABC \cup P$ $\stackrel{-}{\sim}$ $\triangle AEF \cup Q$
Hence: $(EA; EQ) \equiv (BP; BA) \equiv (EP; EA)$ (mod $\pi$ ) or $EA$ is internal bisector of $\angle{PEQ}$
Similarly: $FA$ is internal bisector of $\angle{PFQ}$
So: $PEQF$ is tangential quadrilateral or $PF + EQ = PE + FQ$
But: $D$ is reflection of $Q$ through $EF$ then: $PF + ED = PE + FD$
Hence: $DEPF$ is tangential quadrilateral
So: $T$ is incenter of tangential quadrilateral $DEPF$
Then: $ET$ is internal bisector of $\angle{DEP}$ or $(ET; ED) \equiv \dfrac{(EP; EA)}{2} \equiv (EA; EF) \equiv (BC; BA)$ (mod $\pi$ )
Similarly, we have: $(FD; FT) \equiv (CA; CB)$ (mod $\pi$ )
b) Let $L$ be incenter of $\triangle DMN$ ; $W$ be the intersection of a tangent from $D$ to incircle of $\triangle EMP$ which is different from $ED$ with $AP$
We have: $DW + FP = DW + EP + FP - EP = ED + WP + FP - EP = FD + EP + WP - EP = FD + WP$
So: $FDWP$ is tangential quadrilateral
Then: $(DI; DJ) \equiv (DI; DW) + (DW; DJ) \equiv \dfrac{(DE; DW)}{2} + \dfrac{(DW; DF)}{2} \equiv \dfrac{(DE; DF)}{2}$ $\equiv \dfrac{\pi}{2} - \dfrac{(DM; DF)}{2} \equiv \dfrac{\pi}{2} + \dfrac{(DM; DF)}{2} \equiv (LM; LJ) \equiv (LI; LJ)$ (mod $\pi$ )
Hence: $L$ $\in$ $(DIJ)$
But: $DT$ $\perp$ $DL$ so: $K$ , $H$ , $L$ are collinear or $KH$ passes through incenter $L$ of $\triangle DMN$

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#3 h May 1, 2021, 9:13 PM • 3 Y

Y by parmenides51, Illuzion, m4thbl3nd3r

As a past contestant of the Vietnam TST 2015, I wanna commentate a little bit about this problem.

Generally, the problem was not too hard but incredibly technical due to its construction. For part a, the key insight is to show that ET, FT is the bisector of DEP, DFP respectively, which is equivalent to DEPF is tangential. So since D is the reflection of Q w.r.t EF we have $PF + ED = PE + FD$ , and this is also my solution during the TST (similar to khanhnx). But interestingly, I was told by a grader of the TST that I am the only one who used this approach while most used Ceva sine to finish the problem. That being said, this problem was incredibly technical and was a deciding problem of the TST.

For part b, I was quick to notice that this is IMO Shortlist 2009. The key insight is to show two smaller tangential quads in DEPF by edge chasing. IMO Shortlist 2009 also motivated another Vietnamese Olympiad problem that is VMO 2011 P6. Needless to say, the configuration is very familiar, but part a being very technical probably eliminated a lot of students.

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