Let $ABC$ be a scalene triangle and $P$ be an interior point such that $\angle PBA=\angle PCA$. The lines $PB$ and $PC$ intersect the internal and external bisectors of $\angle BAC$ at $Q$ and $R$, respectively. Let $S$ be the point such that $CS$ is parallel to $AQ$ and $BS$ is parallel to $AR$. Prove that $Q$, $R$ and $S$ are colinear.
Problem
Source: Spain MO 2024 P3
Tags: geometry, Spain
v4913
16.03.2024 16:09
Let $R' = BP \cap AR$; then $\triangle{ABR'} \sim \triangle{ACR}$, so $\frac{AR}{AR'} = \frac{AC}{AB}$. If $X = AQ \cap BS$ then $\frac{SX}{BX} = \frac{AC}{AB}$ by the Angle Bisector theorem, so $AX, SR, BR'$ concur which implies $Q, R, S$ are collinear. $\square$
Davsch
16.03.2024 16:19
Note that $P$ lies on the isogonal conjugate of the perpendicular bisector of $BC$. $S$ is a fixed point and $Q,R$ move on fixed lines, so if we move $P$ projectively, then by suitable projections from $B,C$, so do $Q$ and $R$. It therefore suffices to check the claim for three choices of $P$. If $P=A$, then $Q=R=A$ and the claim holds.
Next, we will show the statement for $P=B$. By symmetry, it will also holds for $P=C$, finishing the problem. For this, let $D$ be the intersection of $BC$ and the $A$-angle bisector. Then since $AD\perp AR$, we have $(B,C;D,R)=-1$. By angle-chasing, we find that $BS$ bisects angle $\angle CBQ$ and also that $\angle QDB=\angle BQD$, i.e. $|BD|=|BQ|$. Let $S'=BS\cap QR$.
Then $BS$ also bisects $\angle QS'D$. Since $(B,C;R,D)$ is harmonic, this implies that $CS'\perp BS'=BS$, so $S=S'$, as desired.
MathSaiyan
16.03.2024 20:19
Define $X = AQ \cap BS$, $Y = AR\cap CS$. It's clear that $AYSX$ is a rectangle. Let $K$ be the point on $CS$ such that $RK\perp AB$. By Thales/homothecy with center $Q$, it's enough to show that
\[
\frac{AQ}{QX} = \frac{AR}{AY} = \frac{AR}{XS}.
\]The second equality is obvious, so we're done if we prove the second one.
Notice that triangles $RYK$ and $AXB$ are similar (pairwise perpendicular sides). Now notice that $RAKC$ is cyclic as
\[
\angle ARK = \angle BAX = \angle XAC = \angle ACK.
\]But now we get $\angle AKR = \angle ACR = \angle ABQ$, implying that shapes $ABXQ$ and $RKYA$ are similar with those points on the sides included. But this similarity gives us exactly the ratios we need.
Attachments:
asdf334
16.03.2024 21:05
Let $BP\cap AR=X$. Project $B$ and $C$ onto $AR$; call these points $U$ and $V$. Project $A$ onto $BS$; call this point $L$. Reflect $C$ across $AQ$; by homothety this proves \[\frac{AR}{AV}=\frac{AX}{AU}\implies \frac{AR}{LS}=\frac{AX}{LB}\]or that $R$, $Q$, and $S$ are collinear. Done.
MathSaiyan
17.03.2024 00:40
Also, forgot to mention: probably the coolest solution to this problem (sadly not mine though). Recall that the locus of $P$ is a circumrectangular hyperbola (i.e. goes through $A$,$B$,$C$ and $H$). Claim. Both points of infinity along the hyperbola coincide with the infinity points of the $A$-bisectors. Proof. Just take the infinity cases. I.e. suppose $P$ lies at infinity, and now a simple anglechase shows that $P$ must be at infinity along the bisectors. $\square$ Now apply Pascal's theorem with $BPC\infty_1 A\infty_2$ where $\infty_1$ and $\infty_2$ are the infinities along the internal and external bisectors, respectively. We get exactly what we wanted.
axolotlx7
11.04.2024 18:40
Consider pole-polar duality at $A$, with $X_{\ell}$ denoting the pole of line $\ell$. The diagram becomes (details left to the reader, extra points added for convenience): And we want to show that $X_{SB}, X_{QR}, X_{SC}$ are collinear. However, noting that $\Delta AX_1X_{SB} \cup X_{PB}$ is similar to $\Delta AX_{SC}X_2 \cup X_{PC}$, we have \[ \frac{X_1X_{PB}}{X_{PB}X_{SB}} = \frac{X_{SC}X_{PC}}{X_{PC}X_{2}} \]which proves the desired result.
sami1618
21.06.2024 16:26
Let the foots of $A$ onto $BS$ and $CS$ be $E$ and $F$. Notice that right triangles $ABE$ and $CAF$ are similar. Right triangles $RFC$ and $BEQ$ are similar as $\angle FRC=\angle FAC-\angle PCA=\angle ABF-\angle PBA=\angle QBE$. It is sufficient to show that right triangles $RFS$ and $SEQ$ are similar which follows from $$RF\cdot EQ=FC\cdot EB=AF\cdot AE=SE\cdot FS$$
Attachments:
ezpotd
21.06.2024 19:47
We use untethered moving points. Consider $BP \cap AC = E, CP \cap AB = F$, clearly then $BEFC$ is cyclic. Then we know line $EF$ is at a fixed orientation, so animate $E$ with degree $1$ and $F$ consequently also is the intersection of $E\infty_{EF}$ and has degree $1$. Then $BE,CF$ also have degree $1$, and their intersections with the bisectors which have degree $0$ is just degree $1$ each, so $Q,R$ have degree $1$ each as well. Finally, $S$ is fixed since $AQ$ and $AR$ have fixed orientations, so it has degree $0$. Now the statement of $Q,R,S$ being collinear has degree $1 + 1 + 0 - 1$, since when $E = A$, $Q,R$ become the same point. Thus it suffices to check $2$ cases. In the first case, let $E$ lie on $BS$, then it suffices to prove that $CF$ is parallel to the external bisector, which would result in $R$ being the point at infinity along the external bisector, $Q$ being point $E$, and everything working. Observe that $BAE$ is isosceles by angle chasing, so $CAF$ must also be isosceles by angle chasing, thus the statement is true. In the second case, we can virtually do the same thing except let $F = CS \cap AB$.