Let $ABC$ be a triangle with incentre $I$ and circumcircle $\omega$. The incircle of the triangle $ABC$ touches the sides $BC$, $CA$ and $AB$ at $D$, $E$ and $F$, respectively. The circumcircle of triangle $ADI$ crosses $\omega$ again at $P$, and the lines $PE$ and $PF$ cross $\omega$ again at $X$and $Y$, respectively. Prove that the lines $AI$, $BX$ and $CY$ are concurrent.
Problem
Source: RMM Shortlist 2023 G1
Tags: trigonometry, Angle Chasing, concurrency, geometry, RMM Shortlist, circumcircle
01.03.2024 00:08
Let: $M$ be the second intersection of $AI$ with $\omega$ $N$ be the antipode of $M$ wrt $\omega$ $T = \overline{EF} \cap \overline{BC}$, which satisfies $(T, D; B, C) = -1$ since $AD$, $BE$, $CF$ concur by Ceva's Theorem Note that because $\overline{MN} \parallel \overline{DI}$, by Reim's we find that $P$, $D$, and $N$ are collinear. However $$-1 = (M, N; B, C) \overset{P}{=} (\overline{MP} \cap \overline{BC}, D; B, C),$$so $PM$ passes through $T$ too. Thus, Pascal's on $AMPXBC$ and $AMPYCB$, we find that both $\overline{BX} \cap \overline{AIM}$ and $\overline{CY} \cap \overline{AIM}$ lie on $EF$, so they must be the same point.
01.03.2024 12:13
Here is a solution with trig bash which is long but motivated. First we need a lemma: Lemma: $PD$ is the bisector of $\angle BPC$ Proof: We need $\frac{BP}{CP}=\frac{BD}{CD}$. For this we do inversion around incircle. $\newline$ Let $X\to X'$ under this inversion $\newline$ Then the lemma becomes $\newline$ Triangle $D'E'F'$, $I$ is circumcenter $A',B',C'$ are the midpoints of the sides, $P'=A'D'\cap(A'B'C')$ and we want to prove $\frac{B'D'}{C'D'}=\frac{B'P'}{C'P'}$ $\newline$ For this just use $\frac{B'P'}{C'P'}=\frac{sin \angle B'A'D'}{sin \angle C'A'D'}=\frac{sin \angle A'D'C'}{sin \angle C'A'D'}=\frac{A'C'}{C'D'}=\frac{B'D'}{C'D'}$ which is what we wanted so the lemma is proved $\newline$ Now coming back to our problem let $L=BX \cap CY$ by pascal on $BXPYCA$ we have $L,E,F$ are collinear so it suffices to prove $FL=LE$ $\newline$ Deonte $\alpha=\angle FLY=\angle ELC$ and by law of sines we have $\newline$ $\frac{FL}{sin \angle PAC}=\frac{YF}{sin \alpha}$ and $\frac{EL}{sin \angle APF}=\frac{CE}{sin \alpha}$ $\newline$ Dividing the relations it suffices to prove $\frac{sin \angle APF}{sin \angle CAP}=\frac{YF}{CE}$ $\newline$ But $YF\cdot FP=FA\cdot FB$ so we want $\frac{sin \angle APF}{sin \angle CAP}=\frac{FA\cdot FB}{FP\cdot CE}$ (*) $\newline$ By law of sines we have $\frac{AF}{sin \angle APF}=\frac{FP}{sin \angle BAP}$ $\newline$ Replacing in (*) we want $\frac{sin \angle BAP}{sin \angle CAP}=\frac{FB}{CE}$. But this is trivial since $\newline$ $\frac{sin \angle BAP}{sin \angle CAP}=\frac{BP}{CP}=\frac{BD}{CD}=\frac{FB}{CE}$ by the lemma. The problem is solved
02.03.2024 00:02
Assassino9931 wrote: Let $ABC$ be a triangle with incentre $I$ and circumcircle $\omega$. The incircle of the triangle $ABC$ touches the sides $BC$, $CA$ and $AB$ at $D$, $E$ and $F$, respectively. The circumcircle of triangle $ADI$ crosses $\omega$ again at $P$, and the lines $PE$ and $PF$ cross $\omega$ again at $X$and $Y$, respectively. Prove that the lines $AI$, $BX$ and $CY$ are concurrent. Let $L = AI \cap BC$ and $S$ be a mid of arc $BAC$. Claim: $PD$ is angle bisector of $\angle BPC. $ Proof: note that $\angle APD=\angle LID = \pi /2 -(\angle A/2+\angle C) = \angle B/2 -\angle C/2=\angle ABS = \angle APS$, so $D \in PS$. $\square$ Now note that condition is equivalent with $CX/XA \cdot AY/YB =1$. But since ratio lemma (law of sines) it's easy to rewrite it how $BF/CE=BP/CP=BD/CD$, which is true, so we are done!
03.03.2024 05:53
Let $M$ and $N$ be the midpoints of arcs $BC$ and $BAC$. Note that $\measuredangle APD=\measuredangle AID=\measuredangle (\overline{AM},\perp_{\overline{BC}})=\measuredangle AMN=\measuredangle APN$, hence $P,D,N$ are collinear. By trig Ceva it suffices to show that $$\frac{\sin \angle ABX}{\sin \angle CBX}=\frac{AX}{XC}=\frac{AY}{YB}=\frac{\sin \angle ACY}{\sin \angle BCY} \iff (A,C;X,M)=\frac{AX}{XC} \div \frac{AM}{MC}=\frac{AY}{YB} \div \frac{AM}{MB}=(A,B;Y,M).$$Note that we have $(A,C;X,M)\stackrel{P}{=}(A,C;E,\overline{PM} \cap \overline{AC})$ and $(A,B;Y,M)\stackrel{P}{=}(A,B;F,\overline{PM} \cap \overline{AB})$. Hence by prism lemma it suffices to show that $\overline{EF},\overline{BC},\overline{PM}$ concur. Let $T=\overline{EF} \cap \overline{BC}$; then $$-1=(B,C;D,T)\stackrel{=}{P}(B,C;N,\overline{TP} \cap \omega) \implies \overline{TP} \cap \omega=M,$$which finishes. $\blacksquare$
04.03.2024 03:46
For me the most natural solution is the trig ceva + angle bisector angle chase approach. However, here is another natural start, which eliminates a lot of points from the diagram and it is interesting if it can be finished in a reasonable way. By Pascal's theorem we have that $BA \cap YP = F$, $AC \cap PX = E$ and $BX \cap CY$ are collinear, hence if $N = AI \cap EF$, then it suffices to show that $N$ lies on $BX$ (it would follow analogously that $N$ lies on $CY$). This is equivalent to proving $\angle ABN = \angle APE$ and here we do not need $X$, $Y$, $F$, which simplifies the diagram a lot. Any idea how to continue?
04.03.2024 13:25
Yes, it's possible to finish that approach by spiral similarity, here are the details.
09.03.2024 15:04
appreciable, love the config grind
09.03.2024 16:36
hmm well-known config well known that if $EF \cap BC = K $ then $KM_A \cap (ABC)=P$, let $S$ be the sharky devil point, let $M_{EF},M$ denote the midpoints of $EF,BC$. Redefine $Y$ to be $PF \cap (ABC)$, $(SP;BC)=-1$ follows from projecting from $K$, so notice $YLFM_{EF}$ is cyclic so done.
22.05.2024 18:51
Let $M$ be the midpoint of $EF$ and $N$ be the midpoint of the arc $BC$ which is not containng $A$. $EF\cap BC=K,\ KN\cap (ABC)=\{N,R\}, \ RD\cap (ABC)=\{R,Q\}, \ RE\cap (ABC)=\{R,L\}$ $\textbf{Claim 1: } \ A,I,D,R$ are cyclic.
. $\angle ARD=\angle ARQ=\frac{\angle B-\angle C}{2}=\angle NID$ which gives that $(AIDR)$ is cyclic. $\textbf{Claim 2:} \ BM$ and $RE$ intersect on $(ABC)$. Pascal at $BLRNAC$ gives the desired result. Claim $1$ gives that $P=R$. Also $L=RE\cap (ABC)=PE\cap (ABC)=X$ and $B,L,M$ are collinear. Hence $BX$ passes through $M$. Similarly $CY$ passes through $M$. Thus $BX,CY,AI$ pass through $M$ as desired.$\blacksquare$