A Generalization of Simson Line

19 April 2015, Created with GeoGebra

Let $X=\ell \cap AC$ and $M$ be the midpoint of $AC$. Let $Y,Z$ be the projection of $P,B$ on $AC$, respectively. Let $H_A, H_B, H_C, H_P$ be the projection of $A, B, C, P$ on $\ell$, respectively. Let $A', B', C', P'$ be the orthopole of $\ell$ with respect to $\Delta BCP, \Delta CAP, \Delta ABP, \Delta ABC$, respectively. Let $R$ be the Poncelet point of $\{ A, B, C, P \}$ (It's well-known that $R$ is the midway between $P$ and the orthocenter $H$ of $\Delta ABC).$

From a lemma at post #2 at the artofproblemsolving forum, we get $A', B', C', P'$ lie on a line $\tau$. Since $\odot (A_PA_0C )$ is the pedal circle of $A_P$ with respect to $\Delta PAC$, so from Fontene theorem we get $B' \in \odot (A_PA_0C ),$ implying $A_0, A_P, C, H_C, B'$ are concyclic.

Since $H_CA' \perp PB, H_CB' \perp PA, H_PA' \perp BC, H_PB' \perp AC$, so $\angle B'H_CA'=\angle APB=\angle ACB=\angle B'H_PA',$ from which $A', B', H_C, H_P$ are also concyclic.

From the Reim theorem and $H_PA' \parallel A_PA_0$ we get $A_0 \in A'B' \equiv \tau$ (similar discussion for $B_0, C_0).$ It's well-known that $R$ lie on the 9-point circle of $\Delta ABC$, so $P', M, Z, R$ are concyclic at the 9-point circle of $\Delta ABC$. Similarly $B', M, Y, R$ are concyclic at the 9-point circle of $\Delta ABC$. Since $B, H_B, P', X, Z$ are concyclic at the pedal circle of $X$ with respect to $\Delta ABC$, so $\angle AZP'=\angle XH_BP'=90^{\circ}-\angle (AC, \tau)$ (note that $H_BP' \perp AC).$ Similarly we can prove $\angle CYB'=90^{\circ}-\angle (AC, \tau),$ implying $ZP' \parallel YB'$, thus, from Reim theorem, we get $P', R, B'$ are collinear, i.e., $R \in \tau \equiv \overline{A_0B_0C_0}.$

Fix the line $\ell$ and animate $P.$ The pencils $PA,PB,PC$ are projective inducing a proyectivity on $\ell,$ i.e. the series $A_p,B_p,C_p$ are projective $\Longrightarrow$ series $A_0,B_0,C_0$ are projective.

Let $D,F$ be the antipodes of $A,C$ on the circumcircle $(O)$ and consider the case when $A_p \in BF.$ If $BC_p$ cuts $(O)$ again at $D',$ then by Pascal theorem for $APCFBD',$ it follows that $A_p,C_p,CF \cap AD'$ are collinear $\Longrightarrow$ $D \equiv D'$ $\Longrightarrow$ $\angle C_pBA=90^{\circ}$ $\Longrightarrow$ $B \equiv A_0 \equiv C_0$ $\Longrightarrow$ $A_0 \mapsto C_0$ is a perspectivity $\Longrightarrow$ $A_0C_0$ goes through a fixed point. When $P$ coincides with $\{X,Y \} \equiv \ell \cap (O),$ then $A_0C_0$ becomes Simson lines of $X,Y$ meeting at the orthopole $T$ of $\ell$ $\Longrightarrow$ $T \in A_0C_0$ and similarly $T \in B_0C_0$ $\Longrightarrow$ $A_0,B_0,C_0$ are collinear on a line $\tau$ passing through $T.$

Let $H$ be the orthocenter of $\triangle ABC$ and let $X$ be the midpoint of $HP$ lying on 9-point circle $(N).$ It's known that $T \in (N)$ when $O \in \ell.$ Now since $X \ \overline{\wedge} \ P \ \overline{\wedge} \ A_0$ with fixed points at $(N) \cap BC,$ then it follows that $X \mapsto A_0$ is a stereographic projection of $(N)$ onto $BC$ $\Longrightarrow$ $X \in TA_0 \equiv \tau.$

Oai Thanh Dào posted the problem in September 2014 at the Advanced Plane Geometry group (message 1781) and subsequently at the CutTheKnotMath facebook page alnog with Telv Cohl's solution (Solution 1). Solution 2 is by Luis Gonzalez.