Vietnam TST 2017 problem 3

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Vietnam TST 2017 problem 3

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Source: Vietnam TST 2017

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#1 h Mar 26, 2017, 9:27 AM • 4 Y

Y by mihajlon, artsolver, nguyendangkhoa17112003, Adventure10

Triangle $ABC$ with incircle $(I)$ touches the sides $AB, BC, AC$ at $F, D, E$ , res. $I_b, I_c$ are $B$ - and $C-$ excenters of $ABC$ . $P, Q$ are midpoints of $I_bE, I_cF$ . $(PAC)\cap AB=\{ A, R\}$ , $(QAB)\cap AC=\{ A,S\}$ .
a. Prove that $PR, QS, AI$ are concurrent.
b. $DE, DF$ cut $I_bI_c$ at $K, J$ , res. $EJ\cap FK=\{ M\}$ . $PE, QF$ cut $(PAC), (QAB)$ at $X, Y$ res. Prove that $BY, CX, AM$ are concurrent.

This post has been edited 3 times. Last edited by gausskarl, Mar 26, 2017, 1:33 PM
Reason: Typo + LaTeX

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#2 h Mar 26, 2017, 11:58 AM • 2 Y

Y by Adventure10, Mango247

gausskarl wrote:

Triangle ABC with incircle (I) touches the sides AB, BC, AC at D, E, F, res.

Are you sure that that is correct order?

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#3 h Mar 26, 2017, 12:58 PM • 1 Y

Y by Adventure10

gausskarl wrote:

$\triangle ABC$ with incircle $(I)$ touches the sides $BC,AB,AC$ at $D,E,F$ , res. $I_b, I_c$ are $B- and C- excenters$ of $\triangle ABC$ . $Q,P$ are midpoints of $I_cE , I_bF$ . Circle $(PAC)$ cuts $AB$ at $R$ ,Circle $(QAB)$ cuts $AC$ at $S$ .
$A.$ Prove that $PR, QS, AI$ are concurrent.
$B.$ $DE, DF$ cut $I_bI_c$ at $K, J$ . $EJ$ cuts $FK$ at $M$ . $PF, QE$ cut $(PAC), (QAB)$ at $X, Y$ res. Prove that $BY, CX, AM$ are concurrent.

I think you meant above

This post has been edited 1 time. Last edited by aryanna30, Mar 26, 2017, 1:00 PM

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#5 h Mar 26, 2017, 1:34 PM • 1 Y

Y by Adventure10

Again, another serious mistake in translating the problem statement. Sorry about all.

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#6 h Mar 26, 2017, 1:39 PM • 1 Y

Y by Adventure10

Here, mathlinker Livetolove212 has a solution to this problem, in Vietnamese

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#7 h Mar 26, 2017, 2:25 PM • 2 Y

Y by mihajlon, Adventure10

a)Wow this is really cool!
Notice that $B$ -midline and $A$ -midline in triangles $\triangle BDF$ and $\triangle AEF$ cut in $Q$ .Notice also that $B$ -midline is radical axis of $B$ and $\odot I$ and $A$ -midline is the radical axis of $A$ and $\odot I$ and hence $Q$ is the radical center of $B,A,\odot I$ $\implies$ $BQ=QA=t(Q,\odot I)$ $\implies$ $AF\cdot BQ+BF\cdot AQ=t(Q,\odot I)\cdot AB$ $\implies$ by Casey,there exist a circle passing thru $B,Q,A$ tangenting $\odot I$ and hence $\odot BSA$ touches $\odot I$ and analogously so does $\odot CRA$ ,however this implies that $\triangle BSA$ , $\triangle CRA$ share the mixtlinear incircle $\odot I$ $\implies$ they also share the incenter,namely the midpoint of $EF$ .The previous implies $\triangle BSA$ and $\triangle CRA$ have the same incircle $\implies$ $AS\{BS\cap CR\}R$ is tangential and hence the result follows as $SQ,RP$ are angle bisectors of $\angle BSA,\angle CRA$ . $\blacksquare$

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#8 h Mar 26, 2017, 3:42 PM • 2 Y

Y by Adventure10, Mango247

$A)$ By the result of imo shortlist $2002 G7$ we have: $I_bF$ cuts $\odot I$ at $V$ so $(AVC)$ is tangent to $\odot I$ . $VF$ cuts $(AVC)$ at $P'$ so foot of $P'$ on $AC$ be the midpoint of $AC$ so $P',P$ are coincide. so $(APC)$ is tangent to $\odot I$ .also $(AQB)$ is tangent to $\odot I$ similarly. so $\odot I$ be the the mixtlinear incircle of $(ASB),(ARC)$ . The midpoint of $EF$ be the incenter of $\triangle ASB , \triangle ARC$ so $RBSC$ be a tangential quadrilateral. so $SQ,RP,AI$ that are angle bisectors of $\angle ASB$ , $\angle ARC$ , $\angle BAC$ are concurrent.

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#10 h Mar 27, 2017, 5:00 AM • 2 Y

Y by Adventure10, Mango247

any solution for part b

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#11 h Mar 27, 2017, 5:30 AM • 1 Y

Y by Adventure10

Related link for part b) https://artofproblemsolving.com/community/q1h1180223

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#16 h Mar 27, 2017, 8:54 AM • 1 Y

Y by Adventure10

$B)$ $I_bF,I_cE$ cut $\odot I$ at $V,W$ res. It's easy to show that $A$ be the miquel point of $MEDF$ so $M,E,D,F$ are concyclic. $\tfrac {sin \angle MAF}{sin \angle MAE}$ = $\tfrac {sin \angle EJD}{sin \angle FKD}$ = $\tfrac {\tfrac {ED}{DJ}}{\tfrac {DF}{DK}}$
= $\tfrac {{DE}^2}{{DF}^2}$ . Lemma 1: $K,V,D,C,I_b$ are concyclic and also $B,D,W,J,I_c$ are concyclic similarly.(It's easy to prove with some angle chasing).
Lemma 2: $FW,EV$ cut $JK$ at $Z,T$ res and let $L$ be the midpoint of $EF$ so $L,M,Z,J,W$ are concyclic and also $L,M,T,K,V$ are concyclic similarly.(similar toLemma 1 we can prove this).
Lemma 3: $L,F,C,V$ are concyclic and $L,E,W,B$ are concyclic too.(proof: $L$ is incenter of $\triangle ARC$ , $\triangle ASB$ and $V,W$ be tangency points of mixtlinear incircle $\odot I$ .and rest is well-known.)
so $\angle VCD = \angle VKE$ , $\angle VCF=\angle ELV$ , $\angle WBE=\angle FLW$ , $\angle WBD=\angle FJW$ . $\tfrac {sin \angle ABW}{sin \angle WBC}$ = $\tfrac {sin \angle FLW}{sin \angle WJF}$ = $\tfrac {sin \angle ZWL=sin \angle ZJL }{LF}$ . $\tfrac {JF}{sin \angle ZWJ= sin \angle ZLJ}$ = $\tfrac {ZL}{ZJ}$ . $\tfrac {JF}{LF}$
Lemma 4: $ZL \parallel JD$ , $TL \parallel KD$ (proof : $\angle TZL= 180 - \angle JZL= 180 - \angle JML=\angle EML$ , $\tfrac {sin \angle EML}{sin \angle LMF}$ = $\tfrac {MF}{ME}$ = $\tfrac {ED}{DF}$ , $\angle EML+\angle LMF=\angle EFD+\angle FED$ so $\angle EML=\angle EFD$ so done.)
$\tfrac {sin \angle ABW}{sin \angle WBC}$ . $\tfrac {sin \angle BCV}{sin \angle VCA}$ =
$\tfrac {ZL}{ZJ}$ . $\tfrac {JF}{LF}$ . $\tfrac {TK}{TL}$ . $\tfrac{LE}{KE}$ = $\tfrac {FD}{ED}$ . $\tfrac {DF}{ED}$ = $\tfrac{{DF}^2}{{DE}^2}$ = $\tfrac{sin \angle MAE}{sin \angle MAF}$ so by $Ceva's$ theorem $BW ,CV,AM$ are concurrent. so $BY, CX ,AM$ are concurrent. so done.

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#17 h Feb 8, 2019, 10:58 AM • 2 Y

Y by Adventure10, Mango247

Here is my solution for this problem
Solution
a) Firstly, we need to prove this lemma
Lemma: Given $\triangle$ $ABC$ with incircle ( $I$ ) and its tangency points $D$ , $E$ , $F$ . Let $I_a$ be $A$ - excenter of $\triangle$ $ABC$ . The circle ( $\omega$ ) passes through $B$ , $C$ and touches ( $I$ ) at $M$ . Let $S$ be the second intersection of ( $\omega$ ) with $AC$ . Prove that:
1) ( $\omega$ ) passes through midpoint of $I_aD$
2) The internal bisector of $\widehat{BSC}$ passes through midpoint of $DE$
1)

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#18 h Feb 8, 2019, 12:17 PM • 1 Y

Y by Adventure10

Let $G$ $\equiv$ $EF$ $\cap$ $BC$ , $N$ be midpoint of $DG$ , $T$ $\equiv$ $MD$ $\cap$ ( $\omega$ ), $P$ be midpoint of $II_a$
We have: $NG^2$ = $ND^2$ = $\overline{NB}$ . $\overline{NC}$ or $P_{N / (I)}$ = $P_{N / (\omega)}$
Hence: $NM$ is common tangent of ( $I$ ) and ( $\omega$ ) or $NM$ = $ND$ = $NG$
So: $MD$ $\perp$ $MP$
But: $M(BCDG)$ = $-$ 1 then $MD$ is internal bisector of $\widehat{BMC}$
Hence: $T$ is midpoint of $\stackrel\frown{BC}$ which not contain $M$ of ( $\omega$ ) or $TP$ $\parallel$ $ID$
Let $U$ , $V$ be intersections of ( $I$ ) and ( $BICI_a$ )
Since: $ND^2$ = $\overline{NB}$ . $\overline{NC}$ or $P_{N / (I)}$ = $P_{N / (BICI_a)}$ , so: $N$ $\in$ $UV$
But: $UV$ is polar of $I_a$ with respect to ( $I$ ) then: $I_a$ lies on polar of $N$ WRT ( $I$ ) or $I_a$ $\in$ $MD$
So: $M$ , $D$ , $T$ , $I_a$ are collinear
Combine with: $TP$ $\parallel$ $ID$ , $P$ is midpoint of $II_a$ , we have: $T$ is midpoint of $I_aD$

This post has been edited 1 time. Last edited by khanhnx, Apr 26, 2021, 2:26 AM

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#19 h Feb 8, 2019, 1:28 PM • 2 Y

Y by Adventure10, Mango247

Here is part 2 of the lemma
2)

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#20 h Feb 8, 2019, 1:40 PM • 1 Y

Y by Adventure10

Let $W$ be midpoint of $DE$ , $X$ $\equiv$ $MW$ $\cap$ ( $I$ ), $Y$ $\equiv$ $MC$ $\cap$ ( $I$ )
We have: $\widehat{WME}$ = $\widehat{CMD}$ = $\widehat{BMD}$ , $\widehat{MED}$ = $\widehat{BDM}$
Hence: $\widehat{MWE}$ = $\widehat{MBD}$ or $M$ , $B$ , $D$ , $W$ lie on a circle
So: $\widehat{BWC}$ = $\widehat{BWD}$ + $\widehat{CWD}$ = $\widehat{BMD}$ + $90^o$ or $W$ $\in$ ( $T$ , $TB$ )
Then: $TW^2$ = $TB^2$ = $TC^2$ = $\overline{TD}$ . $\overline{TM}$
Therefore: $\widehat{DEI_a}$ = $\widehat{DWT}$ = $\widehat{DMW}$ = $\widehat{DEX}$ or $E$ , $X$ , $I_a$ are collinear
So: $\widehat{CEX}$ = $\widehat{EMX}$ = $\widehat{DMC}$ = $\widehat{TSC}$
Hence: $ST$ $\parallel$ $EI_a$
In other words, $S$ , $W$ , $T$ are collinear

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#21 h Feb 8, 2019, 1:47 PM • 1 Y

Y by Adventure10

Back to the main problem:
Let $T$ be midpoint of $EF$
Then by the above lemma, we have: $S$ , $T$ , $Q$ are collinear; $P$ , $T$ , $R$ are collinear
It means that: $AI$ , $QS$ , $PR$ concurrent at point $T$

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#22 h Feb 9, 2019, 1:37 PM • 2 Y

Y by Adventure10, Mango247

b) We consider these architectures:
Architecture 1:

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#23 h Feb 9, 2019, 2:35 PM • 1 Y

Y by Adventure10

Let $D$ , $E$ , $F$ be tangency points of incircle ( $I$ ) with $BC$ , $CA$ , $AB$ ; $M$ $\equiv$ $EF$ $\cap$ $BC$ ; $N$ , $P$ , $Q$ be midpoints of $DM$ , $DE$ , $DF$
From the proof of the above lemma, we have: $NX$ is common tangent of ( $I$ ) and ( $BXC$ )
So: $NX^2$ = $ND^2$ = $\overline{NP}$ . $\overline{NQ}$ or ( $XPQ$ ) tangent ( $I$ ) at $X$

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#24 h Feb 9, 2019, 3:04 PM • 1 Y

Y by Adventure10

Architecture 2:

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#25 h Feb 9, 2019, 3:05 PM • 1 Y

Y by Adventure10

Let $G$ be centroid of $\triangle$ $ABC$ ; $E$ , $F$ be orthogonal projections of $B$ , $C$ on $CA$ , $AB$ ; $Y$ , $Z$ be intersections of rays $GE$ , $GF$ with ( $O$ )
It's easy to see that: $\overline{GE}$ . $\overline{GY}$ = $\overline{GF}$ . $\overline{GZ}$ or $E$ , $F$ , $Z$ , $Y$ lie on a circle
We have: $YZ$ , $EF$ , $BC$ are radical axises of (( $O$ ); ( $EFZY$ )), (( $EFZY$ ), ( $BCEF$ )), (( $BCEF$ ); ( $O$ ))
So: $YZ$ , $EF$ , $BC$ concurrent or the intersections of these pair of lines: ( $YZ$ ; $BC$ ), ( $GY$ ; $AC$ ), ( $GZ$ ; $AB$ ) are collinear
Then applying Desargues theorem for $\triangle$ $ABC$ and $\triangle$ $GYZ$ , we have: $AG$ , $BZ$ , $CY$ concurrent

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#26 h Feb 9, 2019, 3:18 PM • 2 Y

Y by Adventure10, Mango247

From the lemma in part a, we have: $X$ , $Y$ are tangency points of ( $PAC$ ), ( $QAB$ ) with ( $I$ )
It's easy to see that: $M$ $\in$ ( $I$ ) and $AM$ passes through midpoint of $EF$
So by these 2 above architectures, we have: $DM$ , $EY$ , $FX$ concurrent
Then by Steinbart theorem, we have: $AM$ , $BY$ , $CX$ concurrent

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