Isogonal conjugates

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Isogonal conjugates

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#1 h Mar 10, 2025, 10:26 PM • 2 Y

Y by hectorleo123, Geometrylife

Let $ABC$ a triangle with circumcircle $\Gamma$ and isogonal conjugates $P$ and $Q$ . Take $X$ and $Y$ points on $AB$ and $BC$ respectively such that $\angle PXA=\angle CYQ$ . If $\odot(PXA)$ cut $\Gamma$ again at $R$ . Prove that $RY$ and $QA$ cut at $\Gamma$

If $P$ and $Q$ are external just read it by directed angles
Btw i found this using well-known lemma so it might not be mine

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#2 h Mar 11, 2025, 12:21 AM • 1 Y

Y by drmzjoseph

$A$ -center the problem, such that the condition becomes $\angle PXB = \angle CYQ$ . Fix $P,Q$ , we animate $X$ along $AB$ . Let $AQ$ meet $\Gamma$ at $G$ . Let $R'$ be the intersection of $GY$ and $\Gamma$ . Our goal is to show $R' = R$ , which implies the desired result.

Claim 1: The map from $X$ to $R'$ is projective.
Since $\angle AXP = \angle AYQ, \angle XAP = \angle YAQ$ by isogonality, we know $\triangle AXP$ and $\triangle AYQ$ are similar. Thus the map from $X$ to $Y$ can be written as the composition of reflecting $X$ over the angle bisector of $\angle A$ , followed by a homothety with fixed ratio $\frac{AQ}{AP}$ , both of which are obviously projective. The map from $Y$ to $R'$ is just projecting from $G$ from line $AC$ to the circle $\Gamma$ , which is also well known to be projective. Composing these maps gives the result.

Claim 2: The map from $X$ to $R$ is projective.
This is well known, trivial by inversion at $B$ .

To verify $R' = R$ , it suffices to verify that these maps are the same map, so it suffices to check that $R' = R$ for $3$ different values of $X$ .

When $X = A$ , we get $Y = A$ and $R' = A$ , as well as $R = A$ .

When $X$ is such that $R = B$ , we have $\angle PXB = \angle PBC + \angle A$ . Let $BQ$ meet $AC$ at $Z$ , we can compute $\angle BZC= 180 - \angle C - \angle QBC = \angle PBC+ \angle A$ , so $Z = Y$ . Thus $B, Y, Q, G$ are collinear, so $R' = B$ as desired.

When $X$ is such that $R = C$ , we have $\angle QYC = \angle PXB = 180 - \angle PCB$ , so $Y = C$ , and $R' = C$ , as desired.

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#3 h Mar 11, 2025, 2:09 AM • 1 Y

Y by Geometrylife

Thanks @above for your solution using MMP
I guess this configuration exist, coz it's a well-known problem when both angles are equal a $90^{\circ}$
I found this problem with this well-known lema:
Let $ABC$ a triangle with $P$ and $Q$ isogonal conjugates, take $Y$ on $BC$ so $X$ spiral similarity center sendind $YQ$ to $PA$ lies on $\odot(ABC)$

See diagram (just cut $\odot (XYQ)$ again at $BC$ do angle chasing and finish this conf)

This lema is same thing that:
Let $ABC$ be a triangle with $P,Q$ arbitrary points, and $P',Q'$ isogonal conjugates resp. Then the spiral similarity center sending $PQ$ to $Q'P'$ lies on $\odot (ABC)$

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#4 h Mar 12, 2025, 9:15 PM • 1 Y

Y by drmzjoseph

It took me a long time but here is my solution :-D

:-D

Well-known lema:
Let $ABC$ a triangle with $P$ and $Q$ isogonal conjugates, take $Y$ on $BC$ so $X$ spiral similarity center sendind $YQ$ to $PA$ lies on $(ABC)$
Let $S$ be the spiral similarity center that sends $AP$ to $QY$ , by the lemma $S \in (ABC)$
Let $T\equiv RP \cap (ABC), L\equiv AQ \cap (ABC), N\equiv BC\cap SL. \Rightarrow \angle SNY=\angle LAC-\angle BAS=\angle SAP=\angle SQY$
$\Rightarrow NSYQ$ is cyclic $\Rightarrow \angle QYC=\angle QSL=\angle AXP=\angle ARP=\angle ART=\angle ALT\Rightarrow \angle AQS=\angle TLS$ , and since $\angle SAQ=\angle STL$
$\Rightarrow \triangle SAQ\sim \triangle STL$ Then the spiral center that sends $YQ$ to $PA$ sends $A\to T, Q\to L$
$\Rightarrow TP\cap LY\in (ABC)\Rightarrow R,Y,L$ are collinear $_\blacksquare$

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