Let $ABC$ be a triangle, with $AB<AC$, and let $\Gamma$ be its circumcircle. Let $D$, $E$ and $F$ be the tangency points of the incircle with $BC$, $CA$ and $AB$ respectively. Let $R$ be the point in $EF$ such that $DR$ is an altitude in the triangle $DEF$, and let $S$ be the intersection of the external bisector of $\angle BAC$ with $\Gamma$. Prove that $AR$ and $SD$ intersect on $\Gamma$.
Problem
Source: Spanish MO 2022 P3
Tags: Spain, geometry, circumcircle
jbaca
05.05.2022 19:26
Solution. Let $T = \overline{EF}\cap \overline{BC}$. We have that $(T,D;B,C) = -1$. Since $\angle TRD = 90^{\circ}$, this means that $RD$ and $RT$ bisect $\angle BRC$ internally and externally, respectively. It readily follows that $\bigtriangleup BFR \sim \bigtriangleup CER$, thus $$\dfrac{FR}{RE} = \dfrac{BF}{CE} = \dfrac{BD}{DC}$$We conclude that $R$ and $D$ are corresponding points of the similar triangles $\bigtriangleup FAE$ and $\bigtriangleup BSC$; hence $\angle BAR = \angle FAR = \angle BSD$, which yields the desired assertion. $\Box$
MartinPR05
14.05.2022 13:12
Let $P=(AEFI)\cap \Gamma$, $L=AI\cap \Gamma$, $K=AI\cap BC$, $R'=AP\cap EF$, $D'=EF\cap BC$, $T_1=SD\cap\Gamma$, $T_2=AR\cap\Gamma$ and $\omega$ be the nine point circle of $DEF$. Claim 1. $P$, $R$ and $I$ are collinear.
Consider the inversion around $(DEF)$. There,$(AEFI)$ goes to $EF$ and $(ABC)$ goes to $\omega$. Therefore $P$ goes to $\omega \cap EF$, which is the feet of the altitude from $D$ to $EF$. $\square$
Claim 2. $P$, $D$ and $L$ are collinear.
Consider the inversion around $(BIC)$. There,$(ABC)$ goes to $BC$, so $A$ goes to $K$. $(AEFPI)$ goes to $(IK)$. Therefore $P$ goes to $BC \cap (IK)=D$. $\square$
Therefore, we have that $PD$ is the angle bisector of $\angle BPC$, so $P$, $S$ and $D'$ are collinear. $$-1=(B,C;D,D')\stackrel{S}{=}(B,C;T_1,P)$$$$-1=(F,E;I,A)\stackrel{P}{=}(F,E;R,R')\stackrel{A}{=}(B,C;T_2,P)$$so we conclude $T_1=T_2$. $\blacksquare$
Ru83n05
14.05.2022 15:07
I'll sketch the solution I wrote during the exam. Let $X=AR\cap DS$, $A'$ is the $A$ antipode, $T$ is the $A$ - sharkydevil point, $N$ midpoint of arc $BC$ (not including $A$). Recall the well known colinearities: $N-D-T$ and $T-R-A'$. By the inverse of Pascal's theorem on hexagon $A'SXANT$, since $AN\cap SA'$, $SX\cap NT=D$ and $XA\cap TA'=R$ are colinear (the lines $DR, AN$ and $A'S$ are all parallel), these six points must be conconic. Given that $A', S, A, N, T$ lie on $(ABC)$ and 5 points define a conic, $X$ must also lie on the circle. The end.
gambi
17.05.2022 15:57
Let $I$ be the incenter, and $K=(\overline{AI})\cap \Gamma$. Let $U=ND\cap \Gamma, V=AR\cap (AEF)$. Let $\Psi$ be the spiral similarity mapping $BC$ to $FE$. It is clearly centered at $K$, which is the Miquel point of quadrilateral $BCFE$. Since $\triangle FRD\sim \triangle IDC$, $\triangle ERD\sim \triangle IDB$, then $$ \frac{FR}{ER}=\frac{BD}{CD}, $$Therefore, $\Psi(B)=F, \Psi(C)=E, \Psi(D)=R, \Psi(\Gamma)=(AEF), \Psi(S)=A$. Hence $$ \Psi(U)=\Psi(ND\cap \Gamma)=AR\cap (AEF)=V, $$which implies $$ \measuredangle BAU=\measuredangle BKU=\measuredangle FKV=\measuredangle FAV, $$so $A-U-V$ collinear, as desired.
Mahdi_Mashayekhi
17.05.2022 19:31
Note that $AFE$ and $SBC$ are similar so we need to prove $R,D$ are corresponding points of $EF,CB$. Claim $: RD$ is angle bisector of $\angle BRC$. Proof $:$ Let $EF$ meet $BC$ at $P$ Note that $\angle DRP = \angle 90$ and Note that $\frac{BD}{BP} = \frac{BF}{BP} = \frac{\sin{BPF}}{\sin{AFE}} = \frac{\sin{CPE}}{\sin{AEF}} = \frac{CE}{CP} = \frac{CD}{CP}$. Now Note that $\angle RFB = \angle 180 - \angle AFE = \angle 180 - \angle AEF = \angle REC$ and $\angle FRB = \angle 90 - \angle BRD = \angle 90 - CRD = \angle ERC$ so $\frac{FR}{RE} = \frac{BR}{RC}$ and since $RD$ was angle bisector we have $\frac{BR}{RC} = \frac{BD}{DC}$ so $R,D$ are corresponding points of $EF,CB$. we're Done.