Equal segments in an incircle configuration

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a_507_bc

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a_507_bc

#1 h Aug 9, 2022, 7:10 AM

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Given is a triangle $ABC$ with incircle $\omega$ , tangent to $BC, CA, AB$ at $D, E, F$ . The perpendicular from $B$ to $BC$ meets $EF$ at $M$ , and the perpendicular from $C$ to $BC$ meets $EF$ at $N$ . Let $DM$ and $DN$ meet $\omega$ at $P$ and $Q$ . Prove that $DP=DQ$ .

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onlyvipers

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onlyvipers

#2 h Aug 9, 2022, 8:36 AM • 2 Y

Y by Mango247, Mango247

Let $EF$ intersect $BC$ at $T$ , and let $P$ be the foot of the perpendicular from $D$ onto $EF$ .

It is known that $(T,D;B,C)=-1$ by the Gergonne concurrency. Since $\angle TPD = 90^{\circ}$ , by properties of harmonics we have $PD$ is the bisector of $\angle BPC$ , so $\angle MPB=\angle NPC$ .

Quadrilaterals $MPDB$ and $NPDC$ are cyclic, so this implies $\angle MDB=\angle NDC$ .

Finally, since $BDC$ is tangent to the incircle this implies that arcs $DP$ and $DQ$ subtend the same angle, so they must be equal, implying $DP=DQ$ , as desired.

This post has been edited 1 time. Last edited by onlyvipers, Aug 9, 2022, 8:36 AM

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ike.chen

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ike.chen

#3 h Aug 9, 2022, 9:06 AM

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I’m pretty sure this exact problem has appeared on a past Balkan shortlist or something. I’ll post a link here if I find it.

Edit: See JBMO SL 2015/G5.

This post has been edited 1 time. Last edited by ike.chen, Aug 14, 2022, 6:16 AM

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VicKmath7

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VicKmath7

#4 h Aug 9, 2022, 9:15 AM

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Alternatively, we want $\angle BDM= \angle CDN$ , so proving the similarity of $MBD$ and $CDN$ is sufficient. We want $\frac {MB} {BF}= \frac {CN} {CE}$ which is obvious by LoS in $MBF$ and $CEN$ .

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onlyvipers

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onlyvipers

#5 h Aug 9, 2022, 9:32 AM

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ike.chen wrote:

I’m pretty sure this exact problem has appeared on a past Balkan shortlist or something. I’ll post a link here if I find it.

Yes, it did feel somewhat familiar. I think I saw it in passing, but can't remember exactly where either.

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StefanSebez

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StefanSebez

#6 h Aug 9, 2022, 10:43 AM • 4 Y

Y by ike.chen, Mango247, Mango247, Mango247

ike.chen wrote:

I’m pretty sure this exact problem has appeared on a past Balkan shortlist or something. I’ll post a link here if I find it.

https://artofproblemsolving.com/community/c6h1525171p9137012

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MathLuis

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MathLuis

#7 h Aug 10, 2022, 4:12 PM • 2 Y

Y by UI_MathZ_25, Ru83n05

When i seen this problem on the test i had no doubt about what to do, its just insane how PSC couldnt see this. (Btw this took me 10 mins on test)
Let $EF \cap BC=G$ , $I$ incenter of $\triangle ABC$ , $ID \cap EF=K$ and $D'$ the reflection of $D$ over $I$ ( $DD'$ diameter in $\omega$ )
$-1=(G, D; B, C) \overset{\infty_{\perp BC}}{=} (G, K; M, N) \overset{D}{=} (D, D'; P, Q)$ Thus we are done

This post has been edited 1 time. Last edited by MathLuis, Aug 10, 2022, 4:12 PM

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Joalro178

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Joalro178

#8 h Aug 21, 2022, 7:01 PM • 1 Y

Y by diegoca1

Let $R$ in $EF$ such that $AR$ is parallel to $MB$ and $NC$

It is known that $\frac {MB} {AR}= \frac {BF} {AF}$ and $\frac {AR} {NC}= \frac {AE} {EC}$ .

Then we multiply these last 2 expressions and obtain : $\frac {MB} {NC}= \frac {BF} {CE}$

and how $BF$ = $BD$ and $EC$ = $DC$ , then $\frac {MB} {NC}= \frac {BD} {CD}$ so that $\triangle MBD$ and $\triangle NCD$ are similar for side-angle-side

Then $\angle BDM$ = $\angle CDN$ so that $DP$ = $DQ$ . FINISH :blush:

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Joalro178

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Joalro178

#9 h Aug 21, 2022, 7:04 PM • 1 Y

Y by Mango247

MathLuis wrote:

When i seen this problem on the test i had no doubt about what to do, its just insane how PSC couldnt see this. (Btw this took me 10 mins on test)
Let $EF \cap BC=G$ , $I$ incenter of $\triangle ABC$ , $ID \cap EF=K$ and $D'$ the reflection of $D$ over $I$ ( $DD'$ diameter in $\omega$ )
$-1=(G, D; B, C) \overset{\infty_{\perp BC}}{=} (G, K; M, N) \overset{D}{=} (D, D'; P, Q)$ Thus we are done

GOOD SOLUTION , BUT IN THE TEST I COULD NOT NOTICE IT

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