Weird ninja points collinearity

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Weird ninja points collinearity

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americancheeseburger4281

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americancheeseburger4281

#1 h Apr 26, 2025, 10:12 PM

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For some triangle, define its Ninja Point as the point on its circumcircle such that its Steiner line coincides with the Euler line of the triangle. For an triangle $ABC$ , define:

$O$ as its circumcentre, $H$ as its orthocentre and $N_9$ as its nine-point centre.
$M_a$ , $M_b$ and $M_c$ to be the midpoint of the smaller arcs.
$G$ as the isogonal conjugate of the Nagel point (i.e. the exsimillicenter of the incircle and circumcircle)
$S$ as the ninja point of $\Delta M_aM_bM_c$
$K$ as the ninja point of the contact triangle

Prove that:
$(a)$ Points $K$ , $N_9$ and $I$ are collinear, that is $K$ is the Feuerbach point.
$(b)$ Points $H$ , $G$ and $S$ are collinear

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1546 posts

#2 h Apr 27, 2025, 3:00 AM • 1 Y

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Denote $\mathcal H$ as the feuerbach hyperbola, then redefine $S$ as the antigonal conjugate of $H$ in $\triangle ABC$ , clearly its isogonal conjugate in $\triangle ABC$ is $\infty_{OI}$ which therefore shows that (notice $OI$ is by homothety euler line of $\triangle M_AM_BM_C, \triangle DEF$ where incircle of $\triangle ABC$ touches $BC,CA,AB$ at $D,E,F$ respectively) in fact $S$ is our desired ninja point.
Now redefine $K$ as the center of $\mathcal H$ then let $L,J$ orthocenters of $\triangle AIB, \triangle AIC$ respectively, then they also lie on $\mathcal H$ but also let $ID \cap EF=M$ and $N$ midpoint of $BC$ , either from ratio lemma or homothety we can see that $A,M,N$ are colinear but also recall that the poles of $L,J$ w.r.t. $(DEF)$ are the $B,C$ -midbases of $\triangle ABC$ respectively therefore $L,D,J$ are colinear and lie on the polar of $N$ w.r.t. $(DEF)$ so by homothety and letting $LJ \cap EF=V$ we have to notice from La'Hire twice that $M$ lies on the polar of $V$ w.r.t. $(DEF)$ and therefore $-1=(E, F; M, V)$ , we will use this later.
Now recall from Iran Lemma twice that $AL, DE, BI$ meet at a point $P$ and $AJ, DF, CI$ meet at a point $Q$ and however $P,Q$ lie on $(AI)$ and the $A$ -midbase on $\triangle ABC$ which is sufficient to tell that $I$ is the orthocenter of $\triangle PDQ$ , now let $IN \cap LJ=R$ then $\angle IRD=90$ but we will prove $R$ is the $D$ -queue point of $\triangle PDQ$ because $-1=(E, F; M, V) \overset{D}{=} (DP, DQ; DI, DR)$ which is enough by the radax that gives an harmonic showing it lies on line $DR$ however by the angle condition mentioned we force $R$ to be our desired point.
And this is so useful because now notice on $\triangle ALJ$ that since $A,L,J \in \mathcal H$ we have that by invariant of pedal circles through the center of $\mathcal H$ that $QKPRD$ is cyclic but also reflecting $K$ over $DE,DF$ let them be $K_1,K_2$ then you also reflect this recently found circle to see that $K_2QID, K_1PID$ are cyclic and thus note $\measuredangle K_2ID=\measuredangle K_2QD=\measuredangle DQK=\measuredangle DPK=\measuredangle K_1PD=\measuredangle K_1ID$ which shows that $K_1,I,K_2$ are colinear so we have that $K$ is indeed anti-steiner of $IO$ as desired.
And now we are basically done because $S$ is just $H$ reflected over $K$ so the colinearity $H,G,K,S$ is trivial by considering homothety at $G$ sending $\triangle DEF$ to $\triangle M_AM_BM_C$ thus we are done :cool:

:cool:

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