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GGPiku

#1 h Apr 25, 2016, 3:49 PM • 1 Y

Y by Adventure10

Let $ABC$ be an acute triangle with $AB<AC$ and $D,E,F$ be the contact points of the incircle $(I)$ with $BC,AC,AB$ . Let $M,N$ be on $EF$ such that $MB \perp BC$ and $NC \perp BC$ . $MD$ and $ND$ intersect the $(I)$ in $D$ and $Q$ . Prove that $DP=DQ$ .

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#2 h Apr 25, 2016, 4:45 PM • 1 Y

Y by Adventure10

Let $T\in MN$ such that $BM=BT$ $\Longrightarrow$ $\measuredangle TMB=\measuredangle ETB=\measuredangle FNC$ $\Longrightarrow$ $\triangle TEB\sim \triangle NFC$ $\Longrightarrow$ $\tfrac{MB}{BE}=\tfrac{TB}{BE}=\tfrac{NC}{FC}$ , since $BE=BD$ and $FC=CD$ we get $\tfrac{BM}{BD}=\tfrac{CN}{CD}$ then $\triangle DMB\sim \triangle DNC$ $\Longrightarrow$ $\measuredangle MDC=\measuredangle NDC$ $\Longrightarrow$ $\measuredangle PQD=\measuredangle QPD$ hence $DP=DQ$ .

This post has been edited 1 time. Last edited by FabrizioFelen, Apr 25, 2016, 4:46 PM

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#3 h Apr 25, 2016, 5:07 PM • 2 Y

Y by Adventure10, Mango247

If we prove that $\angle PDI= \angle QDI$ we are done, because that implies $\triangle PIB$ and $\triangle QIB$ are congruent.

In order to do that, we will prove that $\angle PDB = \angle QDC$ .

Notice that $\angle AFE= \angle AEF$ and since $MB \parallel NC$ we have $\angle BMF= 180 - \angle ENC$ .

Hence by law of sines in $\triangle ENC$ and $\triangle BFM$ we get:

$\frac{NC}{EC} = \frac{ \sin \angle NEC}{ \sin 180 - \angle BMF} = \frac{ \sin \angle MFB}{ \sin \angle BMF} = \frac{MB}{BF}$

Since $BF=BD$ and $CD=CE$ we get:

$\frac{NC}{CD} = \frac{NC}{CE} = \frac{MB}{BF} = \frac{MB}{BD}$

And since $\angle MBD= \angle DCN=90$ , that implies:

$\triangle MBD \sim \triangle NCD$ . Hence $\angle MDB= \angle NDC$ and $\angle PDI= \angle QDI$

Done!

This post has been edited 1 time. Last edited by Math_CYCR, Apr 25, 2016, 5:08 PM

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#4 h Apr 25, 2016, 5:41 PM • 1 Y

Y by Adventure10

Let's prove $\triangle DMB\sim\triangle DNC$ , thus being done. let the $A-$ altitude of $\triangle ABC$ intersect $EF$ at $R$ . Then $\triangle MBF\sim\triangle RAF$ , implying $\frac{MB}{AR}=\frac{BF}{AF}\ (\ 1\ )$ and $\triangle NCE\sim\triangle RAE$ , implying $\frac{NC}{AR}=\frac{CE}{AE}\ (\ 2\ )$ . Dividing (1) and (2) side by side we get $\frac{MB}{NC}=\frac{BF}{CE}$ , but $BF=BD, CE=CD$ hence $\triangle DMB\sim\triangle DNC$ , so we are done.

Best regards,
sunken rock

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PROF65

#5 h Apr 25, 2016, 6:15 PM • 1 Y

Y by Adventure10

Let $G$ the intersection of $EF$ and $BC$ ; $H$ the intersection of the perpendicular of $BC$ through $D$ and $EF$ then $(M,N;G,H)=-1$ but $GD \perp DH$ so $DH$ is the angle bisector of $\widehat{PDQ}$ then $\widehat{PDB}=\widehat{QDC}$ therefore $PD=DQ$ .
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#6 h May 13, 2016, 6:20 PM • 1 Y

Y by Adventure10

Hi dude can you send me all jbmo 2016 Tst problems of Romania please?

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#7 h May 14, 2016, 12:44 PM • 2 Y

Y by Adventure10, Mango247

Suffices prove $DI$ bisects $\angle PDQ$ . Let $DI\cap EF=K$ and $EF\cap BC=L$ , then we want $(M, N;K, L)=-1$ , but $(M, N;K, L)=(B, C;D, L)=-1$ , and last equality is obvious.

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#8 h Aug 17, 2023, 6:58 PM • 1 Y

Y by erkosfobiladol

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#9 h Oct 24, 2024, 6:34 AM

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What is definition for P?.

This post has been edited 1 time. Last edited by zaidova, Oct 24, 2024, 6:53 PM

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#10 h Oct 24, 2024, 10:46 AM

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What is definition for P?

GGPiku wrote:

Let $ABC$ be an acute triangle with $AB<AC$ and $D,E,F$ be the contact points of the incircle $(I)$ with $BC,AC,AB$ . Let $M,N$ be on $EF$ such that $MB \perp BC$ and $NC \perp BC$ . $MD$ and $ND$ intersect the $(I)$ in $D$ and $Q$ . Prove that $DP=DQ$ .

This post has been edited 2 times. Last edited by zaidova, Oct 24, 2024, 10:47 AM

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#11 h Oct 24, 2024, 1:59 PM • 1 Y

Y by zaidova

Quote:

What is definition for P?

Quote:

Let $ABC$ be an acute triangle with $AB<AC$ and $D,E,F$ be the contact points of the incircle $(I)$ with $BC,AC,AB$ . Let $M,N$ be on $EF$ such that $MB \perp BC$ and $NC \perp BC$ . $MD$ and $ND$ intersect the $(I)$ in $D$ and $Q$ . Prove that $DP=DQ$ .

I am quite sure about $P$ is intersection point of incircle and $MD$ . There should be typo.
My solution:
Let $MN \cap BC$ at $X$ , $MB \cap ND=T$ , $MD \cap NC=Y$ . $(X,D;B,C)=-1$ take pencil from $N$ $\implies$ $(NX,ND;NB,NC)=-1$ . $MB$ is parallel to $NC$ so, $MB=BT$ which solves problem.

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