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Isogonal Conjugates of Nagel and Gergonne Point
SerdarBozdag   6
N 16 minutes ago by ohiorizzler1434
Source: http://math.fau.edu/yiu/Oldwebsites/Geometry2013Fall/Geometry2013Chapter12.pdf
Proposition 12.1.
(a) The isogonal conjugate of the Gergonne point is the insimilicenter of
the circumcircle and the incircle.
(b) The isogonal conjugate of the Nagel point is the exsimilicenter of the circumcircle and
the incircle.
Note: I need synthetic solution.
6 replies
SerdarBozdag
Apr 17, 2021
ohiorizzler1434
16 minutes ago
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Weird ninja points collinearity
americancheeseburger4281   1
N Apr 27, 2025 by MathLuis
Source: Someone I know
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:
[list]
[*]$O$ 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
[/list]
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
1 reply
americancheeseburger4281
Apr 26, 2025
MathLuis
Apr 27, 2025
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Weird ninja points collinearity
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americancheeseburger4281
31 posts
americancheeseburger4281
#1 h  Apr 26, 2025, 10:12 PM
Y by
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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MathLuis
1556 posts
MathLuis
#2 h  Apr 27, 2025, 3:00 AM • 1 Y
Y by Yiyj1
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:.
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