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Source: Korean National Junior Olympiad Problem 1

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#1 h Nov 2, 2014, 9:24 AM • 2 Y

Y by Adventure10, Mango247

In a triangle $\triangle ABC$ with incenter $I$ ,
Let $D$ = $AI$ $\cap$ $BC$
$E$ = incenter of $\triangle ABD$
$F$ = incenter of $\triangle ACD$
$P$ = intersection of $\odot BCE$ and $\overline {ED}$
$Q$ = intersection of $\odot BCF$ and $\overline {FD}$
$M$ = midpoint of $\overline {BC}$

Prove that $D, M, P, Q$ concycle

This post has been edited 1 time. Last edited by rkm0959, Jun 7, 2015, 2:40 AM

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#2 h Nov 2, 2014, 10:02 AM • 2 Y

Y by Adventure10, Mango247

My solution:

Let $X=EF \cap BC$ .

Since $BC$ is a common tangent of $\{ \odot (E), \odot (F) \}$ ,
so $X$ is the exsimilicenter of $\odot (E) \sim \odot (F)$ . ... $( \star )$
Since $AD$ is a common tangent of $\{ \odot (E), \odot (F) \}$ ,
so combine with $(\star )$ we get $(B, C; X, D)=-1$ .
Since $DX \cdot DM=DB \cdot DC=DE \cdot DP=DF \cdot DQ=\omega$ ,
so invert with center $D$ and factor $\omega$ we get $P, Q, D, M$ are concyclic.

Q.E.D

This post has been edited 1 time. Last edited by TelvCohl, May 9, 2015, 12:25 PM

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#3 h Nov 2, 2014, 10:06 AM • 2 Y

Y by Adventure10, Mango247

Ummmmm....
Inversion (and harmonic divison) is not really a Junior Topic in Korea.
Can you make a solution without them?
(This was the problem that made everyone cry today)

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#4 h Nov 6, 2014, 2:46 PM • 4 Y

Y by Adventure10, Mango247, endless_abyss, and 1 other user

Let $X$ be the intersection point of $EF$ with BC and let $R$ be the midpoint of the minor arc of BC.Now,cause $DA*DR=DB*DC=DE*DP=DF*DQ$ , $EFQP$ is cyclic so it is enough to prove that $QMFX$ is a cyclic(by an easy angle chase) and by radical axis we reduce it to prove $AMRX$ is cyclic which is equivalent proving $AX$ is the external angle bisector of $<BAC$ and this is obvious from menalous on $BIC$ ,so we are finished.

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jayme

#5 h May 9, 2015, 12:14 PM • 2 Y

Y by Adventure10, Mango247

Dear Mathlinkers,
a synthetic some up...

1. according to the three chords theorem, E, F, P, Q are concyclic
2. X the point of intersection of EF and BC
3. according to the MacLaurin formula, XD.XM= XB.XC
4. P’, Q’ the second points of intersection of XP, XQ wrt (BEC), (BFC)
5. XD.XM = XB.XC = XP.XP' = XQ.XQ' and we are done

Sincerely
Jean-Louis

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#6 h Mar 16, 2025, 5:21 PM

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Let $L$ be the arc midpoint. A negative inversion at $D$ with radius $\sqrt{BD\cdot CD}$ swaps $(A,L),(B,C),(E,P),(F,Q)$ . So it suffices to show $EFX$ collinear, where $X=(AML)\cap BC$ . It turns out that this is the external angle bisector cevian. Now let $\ell$ be line $EF$ , then note that $(DE,DF;DX,DI)=-1$ and $(AE,AF;AX,AE)=-1$ . So $AX\cap\ell$ is the same as $DX\cap\ell$ , so $X$ lies on $\ell$ . $\blacksquare$

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