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?

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

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

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

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 $\angle BSA,\angle CRA$.$\blacksquare$

$A)$ 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.

any solution for part b

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

$B)$ $I_bF,I_cE$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.

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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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 $I_aD$

Here is part 2 of the lemma 2)

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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

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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b) We consider these architectures: Architecture 1:

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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 $X$

Architecture 2:

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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

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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