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Source: IDK

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#1 h Apr 25, 2025, 4:15 AM • 1 Y

Y by Funcshun840

Let $ABC$ be a triangle and let $O$ be its circumcenter and $I$ its incenter.
Let $P$ be the radical center of its three mixtilinears and let $Q$ be the isogonal conjugate of $P$ .
Let $G$ be the Gergonne point of the triangle $ABC$ .
Prove that line $QG$ is parallel with line $OI$ .

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#2 h May 2, 2025, 2:00 AM

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#4 h May 12, 2025, 2:02 AM

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#5 h May 14, 2025, 7:39 PM • 1 Y

Y by Lufin

very cool!!!
Lemma 1: The point $P$ lies on ray $OI$ with $IP:OP = r:2R$ , where $r$ , $R$ are the inradius and circumradius respectively.
Proof:Take a homothety with ratio $0.5$ at $I$ of the intouch triangle. Then it suffices to prove that $P$ is the exsimilicenter of this homothetied triangle and the medial triangle of the excentral triangle. Consider the $B$ - and $C$ - mixtilinear incircles $\omega _B$ and $\omega _ C$ . We have that the south pole lies on the radical axis of $\omega _B$ and $\omega _ C$ by the shooting lemma. Additionally, the midpoint of $ID$ lies on the radical axis by a simple application of linpop and Pythagoras, hence the radical axis of $\omega _B$ and $\omega _ C$ passes through the desired exsimilicenter. By symmetry, we find that $P$ is the desired point.
Lemma 2:Let T be the insimilicenter of the incircle and the circumcircle and $J$ the inverse of $J$ wrt the circumcircle, then we have $(J, I; P, T) = -1$ .
Proof: This is a length chase, recall that $OI = \sqrt{R^2 - 2rR}$ , and $OJ= \frac{R^2}{OI}$ . We also have $\frac{OT}{TI} = \frac{R}{r}$ , so $OT = \frac{R}{r+R} OI$ . We also have $OP = \frac{2R}{2R-r} OI$ . Hence
$\frac{JP}{IP} = \frac{JO - PO}{PO - IO} = \frac{\frac{R^2}{R^2 - 2rR} - \frac{2R}{2R-r}}{\frac{2R}{2R-r}-1} = \frac{2R^3 - rR^2 - 2R^3 + 4r R^2}{rR(R-2r)} = \frac{3R}{(R-2r)}.$ and $\frac{JT}{IT} = \frac{JO - TO}{IO - TO} = \frac{\frac{R^2}{R^2-2rR} - \frac{R}{r+R}}{1- \frac{R}{r+R}} = \frac{R^2r + R^3 - R^3 + 2rR^2 }{rR(R-2r)}=\frac{3R}{R-2r}.$
Now we are ready for the problem. Consider an isogonal conjugation, which preserves cross ratios. Then $I$ is fixed and $J$ is sent to the antipode $K$ of $I$ on the Feuerbach hyperbola (proof: let $OI$ intersect the circumcircle at $X$ , $Y$ Then $-1=(X,Y;I,J)=(X', Y';I, K)$ and $X'$ , $Y'$ are the points at infinity of the hyperbola). The point $T$ is sent to the Gergonne point $G$ (proof: Monge + mixtilinear properties). Finally $P$ is sent to $Q$ by definition. Now since $J$ , $I$ , $P$ , $T$ lie on $OI$ with $(J, I; P, T) = -1$ , we have $K$ , $G$ , $Q$ , $I$ lie on the Feuerbach hyperbola with $(K, I; Q, G) = -1$ . However, as the tangents to the hyperbola are parallel to $OI$ (they are antipodes and $IO$ is tangent at $I$ ), we find that $QG$ is parallel to $OI$ .
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