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Source: Bosnia and Herzegovina TST 2015 day 2 problem 3

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#1 h May 18, 2015, 8:18 AM • 2 Y

Y by Adventure10, Mango247

Let $D$ , $E$ and $F$ be points in which incircle of triangle $ABC$ touches sides $BC$ , $CA$ and $AB$ , respectively, and let $I$ be a center of that circle.Furthermore, let $P$ be a foot of perpendicular from point $I$ to line $AD$ , and let $M$ be midpoint of $DE$ . If $\{N\}=PM\cap{AC}$ , prove that $DN \parallel EF$

This post has been edited 1 time. Last edited by gobathegreat, Sep 16, 2018, 11:39 AM

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#2 h May 18, 2015, 8:41 AM • 4 Y

Y by phantranhuongth, enhanced, Adventure10, Mango247

gobathegreat wrote:

Let $D$ , $E$ and $F$ be points in which incircle of triangle $ABC$ touches sides $BC$ , $CA$ and $AB$ , respectively, and let $I$ be a center of that circle.Furthermore, let $P$ be a foot of perpendicular from point $I$ to line $AD$ , and let $M$ be midpoint of $DE$ . If $\{N\}=PM\cap{AC}$ , prove that $\color{red}{ D }\normalcolor N \parallel EF$

Typo corrected

My solution :

Let $T=AD \cap \odot (I)$ ( $T \neq D$ ) .

Easy to see $A, E, F, I, P$ are concyclic .

From $\angle ENP=\angle AET=\angle EDP$ $\Longrightarrow D, E, P, N$ are concyclic ,
so we get $\angle AND=\angle EPA=\angle EFA=\angle AEF$ $\Longrightarrow DN \parallel EF$ .

Q.E.D

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jayme

#3 h May 18, 2015, 10:47 AM • 1 Y

Y by Adventure10

Dear Mathlinkers,
the circle (PIMD) is also fruitfull...

Sincerely
Jean-Louis

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tranquanghuy7198

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tranquanghuy7198

#4 h May 18, 2015, 4:01 PM • 2 Y

Y by Adventure10, Mango247

My solution:
$AD\cap{(I)} = T$ , $PM\cap{EF} = K$
$\Rightarrow$ $DP = PT, DM = ME$
$\Rightarrow$ $PM\parallel{TE}$ (1)
Now: $-1 = (EFDT) = E(EFDT) = E(NKMT)$ (2)
(1), (2) $\Rightarrow$ $MK = MN$ $\Rightarrow$ $EF\parallel{DN}$
Q.E.D

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#5 h May 20, 2015, 8:18 AM • 2 Y

Y by Adventure10, Mango247

Here is a solution involving no extra points to be used.

Because $\angle API= \angle AEI= 90^{\circ}$ we have that $A,P,I,E$ are concyclic.
From that we have $\angle APE = \angle AIE = 90^{\circ}- \frac{\alpha}{2}$ .

It's easy to see that the points $I,M,C$ are collinear.
From $\angle IPD = \angle IMD = 90^{\circ}$ we see that $P,I,M,D$ are concyclic.
Now we have $\angle NPD = \angle MPD = \angle MID = \angle CID = 90^{\circ}-\frac{\gamma}{2}=\angle DEC= \angle DEN$ , and here we see that $P,D,N,E$ are concyclic too and $\angle DNA=\angle DNE=\angle APE= 90-\frac{\alpha}{2}=\angle FEA \Rightarrow DN \parallel EF$ , Q.E.D.

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#6 h Jun 21, 2015, 7:15 PM • 1 Y

Y by Adventure10

My solution:

$AD\cap EF\equiv R$ . Apply Menelaus theorem for $\triangle DAE$ has $\overline{P,M,N}\Rightarrow\frac{PD}{PA}=\frac{EN}{AN}$

We'll prove $\frac{PD}{PA}=\frac{DR}{DA}\Leftrightarrow PD^2=PA.PR\Leftrightarrow AP^2-AF^2=PA.PR\Leftrightarrow AF^2=PA.RA$

It's correct because $\triangle FAR$ and $\triangle PAF$ are similar $\Rightarrow AF^2=AR.AP$

So, $\frac{EN}{AN}=\frac{DR}{DA}\Rightarrow DN\parallel EF$

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#7 h Jul 10, 2019, 8:47 AM • 2 Y

Y by Adventure10, Mango247

Rename $P$ as $Q$ . Let $R$ be midpoint of $DE$ and $QR$ $\cap$ $AC$ $=$ $S$ . We need to show, $DS||EF$ . Notice, $Q$ as the midpoint of the $D-$ symmedian chord. $\angle DQS=\angle DD'E=\angle DFE=\angle DES$ Hence, $QDSE$ cyclic. Now,
$\angle AEF=\angle FDE=\angle QDE+\angle FDQ=\angle QSE + \angle QED =\angle ASD$

This post has been edited 1 time. Last edited by AlastorMoody, Jul 10, 2019, 8:47 AM

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