prove that $A$ is the midpoint of $XY$

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prove that $A$ is the midpoint of $XY$

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Source: 19-th Korean Mathematical Olympiad 2006, Final Round

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3595 posts

#1 h Apr 25, 2007, 3:05 PM • 6 Y

Y by canhhoang30011999, Adventure10, ImSh95, Mango247, and 2 other users

In a triangle $ABC$ with $AB\not = AC$ , the incircle touches the sides $BC, CA, AB$ at $D, E, F$ , respectively. Line $AD$ meets the incircle again at $P$ . The line $EF$ and the line through $P$ perpendicular to $AD$ meet at $Q$ . Line $AQ$ intersects $DE$ at $X$ and $DF$ at $Y$ . Prove that $A$ is the midpoint of $XY$ .

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yetti

#2 h Apr 25, 2007, 4:18 PM • 2 Y

Y by Adventure10, ImSh95

See the problem Looks easy.

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#3 h Mar 13, 2009, 7:14 AM • 3 Y

Y by ImSh95, Adventure10, Mango247

We can clearly see that $AQ||BC$ . Now, consider the shaded triangles.
consider the tangents to the incircle, call them a',b',c' coming from A,B,C resp.
Then, $XA/AY=XA/BD*BD/DC*DC/AY=a'/b'*b'/c'*c'/a'=1$ so A is the midpoint as desired.
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#4 h Mar 13, 2009, 3:23 PM • 4 Y

Y by ImSh95, Adventure10, Mango247, and 1 other user

I have another solution to this problem:
Suppose the line through P perpendicular to AD meets the line through A perpendicular to ID at Q.
We need to prove that E, F, Q are collinear. Let H be the intersection of PQ and (I). Therefore we have D, I, H are collinear. Hence, DH is perpendicular to AQ. Invert around D, with power DP.DA=DH.DK (K is the intersection of DI and AQ), that maps A to P, K to H. Therefore AQ is mapped into (DHK) or in the other word, AQ is maped in to (I). So Q is maped into Q* lies on (I). E and F are maped into itself.
DFEQ* is concyclic. So E, F, Q are collinear.
This leads to the fact that XY are parrallel to BC.(1)
We have PFDE are harmonic. So (DP,DC,DF,DE)=-1 or (DA,DC,DY,DX)=-1(2)
Combine (1) and (2) we have A is the midpoint of XY.
Proof is completed

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forevercontinue

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#5 h Mar 17, 2009, 5:24 AM • 2 Y

Y by ImSh95, Adventure10

N.T.TUAN wrote:

In a triangle $ABC$ with $AB\not = AC$ , the incircle touches the sides $BC, CA, AB$ at $D, E, F$ , respectively. Line $AD$ meets the incircle again at $P$ . The line $EF$ and the line through $P$ perpendicular to $AD$ meet at $Q$ . Line $AQ$ intersects $DE$ at $X$ and $DF$ at $Y$ . Prove that $A$ is the midpoint of $XY$ .

I am sorry if having anyone have posted the same solution with me before.
Here is my solution for this problem :
Let $M$ is midpoint of $FE$ .
Easy to prove that quarilateral $APMQ$ is cyclic.
$\Rightarrow \angle PAQ=\angle PMF=\angle PEF+ \angle MPE$
Addition , we have $PQ$ is the symedian . So $\angle FED=\angle FPD=\angle MPE$
$\Rightarrow \angle PAQ=\angle PEF+ \angle FED=\angle PED=\angle BDF$
$\Rightarrow AQ // BC \Rightarrow AX=AF=AE=AY$
Q.E.D

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#6 h Mar 18, 2009, 12:12 AM • 3 Y

Y by ImSh95, Adventure10, Mango247

well know lemma.
$A^{*}$ is inversed point with respect to circle $C$ with center $O$ .
A line through $A$ meets $C$ at two points $X, Y$ .
Then, $O,X,Y,A^{*}$ are on the same circle.

using this lemma,
$AI\perp EF$ and , $AI \cap EF=Y$
Then, it is easy to elicit conclusion .

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guldam

#7 h Jan 27, 2015, 6:07 PM • 3 Y

Y by ImSh95, Adventure10, Mango247

I hope that my solution is not duplicate.

Here I first invoke a lemma

$\textbf{Lemma}$ $\triangle{ABC}$ is inscribed in circle $O$ . Let $P$ be the intersection of tangent lines of circle $O$ at $B,C$ , i.e, the pole of $BC$ with respect to circle $O$ . It is well-known that $AD$ is symmedian of $\triangle ABC$ . $AP$ meets with circle $O$ at $D(\ne A)$ . Line which is perpendicular to $AP$ and pass through $D$ meets with circle $O$ at point $Y$ and with line $BC$ at $Q$ . $AY$ , which is a diameter of circle $O$ , meets with $BC$ at $X$ . Then $B,X,C,Q$ are harmonic.

$\textbf{proof}$ Let $M$ be the midpoint of $BC$ . It is sufficient to prove $PC \times PB = PX \times PM$ . In turn it is sufficient to show $PX \times PM = PY \times PD$ , which mean that $M,X,D,Y$ are cyclic.
Let's denote the intersection of $AM$ with circle $O$ as $Z$ .
Since $AD$ is the symmedian line, it follows that $\angle{DMB}= \angle{ZMC} = \angle{AMB} = \angle{ACD} = \angle{AYD}$ . Therfore $M,X,Y,D$ are cyclic and we're done.

Let' move to the original problem.

$AD$ meets $EF$ at $Z$ . $PQ$ meets $DX$ at $T$ and $DY$ at $S$ and with the incircle at $N$ again. $DN$ meets $FE$ at $W$ .

It is well observed that if $\triangle{DEF}$ and $\triangle{DYX}$ are similar, then $DA$ becomes median of $\triangle{DXY}$ so the proof becomes done. Therefore it is suffice to prove $X,E,F,Y$ are cyclic.

It is well known that $A,P,Z,D$ are harmonic. Then it is followed that $X,T,E,D$ and $Y,S,F,D$ are also harmonic since they are $Q-pencil$ .

If $N'$ denotes $XF \cap YE$ , then it is well known that $QN'$ pass through $S,T$ .

$F,W,E,Q$ are harmonic from the lemma.
Here $S,N',T,Q$ are also harmonic.
Therefore $N'W$ passes through $D$ , which means $N'=N$ , in turn, $XF,EY,DW,PQ$ are all concurrent at $N$ .

Since $\angle{NFD}=\angle{NED}=90^\circ$ , $N$ is orthocenter of $\triangle{DXY}$ .

Therefore $E,X,Y,F$ are cyclic and we are done.

Attachments:

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#8 h Mar 21, 2016, 10:11 AM • 3 Y

Y by ImSh95, Adventure10, Mango247

Let $R$ be the antipode of $D$ wrt the inscribed circle and let the inersection of $AD$ , $EF$ be $S$ .
Claim: $AQ$ is parallel to $BC$
Proof:
$AD$ is the symmedian $\implies$ $(P,D;F,E)=-1=R(P,D;F,E)=(Q,S;F,E)=(F,E;S,Q)=A(F,E;S,Q)=(B,C;M,Q^{'})$ where $M$ is the intersection of $AS$ , $BC$ and $Q^{'}$ is the intersection of $AQ$ , $BC$ .
Now it is well known $M$ is the midpoint of $BC$ so $Q^{'}$ is at infinity so $AQ$ is parallel to $BC$ .

Now $YFEX$ is cyclic and so $FE$ and $XY$ are antiparallel but $AD$ is the symmedian in $\triangle YDX$ so we are done

This post has been edited 3 times. Last edited by nikolapavlovic, Mar 21, 2016, 10:12 AM

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#9 h Jan 11, 2017, 4:26 PM • 4 Y

Y by Horseradish, ImSh95, Adventure10, Mango247

I found a solution using only angle chasing and power of point.

Let line $DI$ intersect the incircle at $R$ and the line parallel to $BC$ passing through $A$ at $S$ . $DR$ is the diameter so $\angle DPR = \angle ASR = 90^{\circ}$ which means $APRS$ is cyclic and and $S$ is on the circumcircle $AFIE$ . That means that the parallel to $BC$ through $A$ , $PR$ and $EF$ all pass through point $Q$ being radical axes of three circles $APRS$ , $AFIES$ and the incircle.

Because $AQ \parallel BC$ we have $\angle AXE = \angle CDE = \angle DPE$ so $APEX$ is cyclic. Similarly $AYFP$ is cyclic. Now the circumcircles of $DPY$ and $DPX$ are tangent to $AQ$ because $\angle AYP = \angle AFP = \angle PDY$ and similarly $\angle AXP = \angle PDX$ , so $AY^{2} = AP \cdot AD = AX^{2}$

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#10 h Jan 12, 2017, 4:52 AM • 3 Y

Y by ImSh95, Adventure10, Mango247

Dear MLs,

This makes A the midpoint of XY.

M.T.

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jayme

#11 h Jan 14, 2017, 5:57 AM • 3 Y

Y by ImSh95, Adventure10, Mango247

Dear Mathlinkers,
this nice problem is the result of

1. http://www.artofproblemsolving.com/community/c6t48f6h1310999_parallel

2. http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=113152

Sincerely
Jean-Louis

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#12 h Feb 1, 2018, 9:35 AM • 3 Y

Y by ImSh95, Adventure10, Mango247

It's enough to prove that $(AQ)\parallel (BC)$ and then project on $(AQ)$ from $D$ .
To prove this, just remark that $APMQ$ is cyclic (where M denote the midpoint of $EF$ ) , and that $(PD)$ is the symmedian. And the rest is only Angle chasing.

This post has been edited 1 time. Last edited by omarius, Feb 1, 2018, 9:52 AM

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#13 h May 17, 2022, 1:23 PM • 2 Y

Y by ImSh95, motannoir

Let $PQ \cap \odot(DEF) = \{ P, S \}$ . We apply Pascal for $DPSFFE$ . We have that $DP \cap FF = \{ A \}$ ; $PS \cap EF = \{ Q \}$ ; $DE \cap SF = \{ X' \}$ . We have that $A,Q, X'$ are collinear but $DE \cap AQ = X$ .

So $X = X'$ . We have that $F,S,X$ are collinear. But $\angle SFD = \angle FPD = 90$ so $XF \perp DY$ . In analog way $YE \perp XD$ .

So $YFEX$ is a cyclic quadrilateral.

We have that $DA$ is symedian in $DEF$ and because $EF$ and $XY$ are antiparalel we have that $DA$ is median in $DXY$ so $A$ is midpoint of $XY$ .

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#14 h Feb 7, 2024, 3:55 AM

Y by

We first prove a claim.

Claim: In $\triangle ABC$ with altitudes $\overline{BE}$ and $\overline{CF}$ , $\overline{BC}$ -midpoint $M$ and orthocenter $H$ , if $Q$ is the circumcenter of $AEF$ —hence also the midpoint of $\overline{AH}$ —then $\overline{MQ} \perp \overline{EF}$ .
Proof: By a $\times 2$ homothety at $H$ this is equivalent to $\overline{AO} \perp \overline{EF}$ where $O$ is the circumcenter of $ABC$ . This follows since $O$ and $H$ are isogonal conjugates, and $\overline{AH} \perp \overline{BC}$ . $\blacksquare$

Let $T$ be the foot of the altitude from $A$ to $\overline{DI}$ and let $D'$ be the $D$ -antipode. By radical axis on $(EFPD')$ , $(AEFIT)$ , $(APD'T)$ , it follows that $\overline{AQ} \parallel \overline{BC}$ (note: this is also a Sharygin 2018 problem).

Now, $\measuredangle AXD=\measuredangle CDE=\measuredangle DFE$ and $\measuredangle AYD=\measuredangle DEF$ , so $\triangle DEF \sim \triangle AYX$ . Furthermore, since $\overline{AI} \perp \overline{EF}$ , it follows from the claim (and a straightforward phantom points argument) that $E$ and $F$ are in fact the feet of the altitudes from $Y$ and $X$ to the opposite sides of $\triangle AYX$ . Hence $(AEFT)$ is the nine-point circle of $\triangle AXY$ , so $A$ is the midpoint of $\overline{XY}$ . $\blacksquare$

This post has been edited 1 time. Last edited by IAmTheHazard, Feb 8, 2024, 7:49 PM

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