do you mean perpendicular bisector?

baldeagle123 wrote: do you mean perpendicular bisector? Yes. I fixed it.

It is true for any point on the Neuberg Cubic. clair23 wrote: Let $\triangle ABC$ be a triangle with circumcenter $O$. The perpendicular bisectors of the segments $OA,OB$ and $OC$ intersect the lines $BC,CA$ and $AB$ at $D,E$ and $F$, respectively. Prove that $D,E,F$ are collinear. This is just equivalent to proving that $\Delta ABC$ and $\Delta O_AO_BO_C$ are perspective where $O_A,O_B,O_C$ are the circumcenters of $\Delta BOC,\Delta COA,\Delta AOB$ rspectively. It's well known that the perspector of $\Delta ABC,\Delta O_AO_BO_C$ is the Kosnita Point. (Apply Jacobi). Hence, by desargues the collinearity is immediate. $\blacksquare$ @2below my solution is not projective. but yeah I indeed used Desargues Theorem which is a projective theorem, but don't get confused with DIT/DDIT, that's something different.

@2above Ok thanks... 1) you might want to hide your solution 2) I'm working on a complex number solution and then I will try with the Apollonius circles although those might actually be easier in complex numbers, maybe you should try @above I wish I knew what you were talking about

amar is a pr0 at high-tech projective geo The main idea is using desargues: https://artofproblemsolving.com/community/c6h1509866p8957048 but idk how to do the rest lol btw, @amar, where did u learn such nice projective stuff?

Note that $f(P)=PA^2-PO^2$ is linear, so $$\frac{BD}{CD}=\frac{f(B)-f(D)}{f(C)-f(D)}=\frac{c^2-R^2}{b^2-R^2}$$and similar for $B,C$, where lengths are directed. Thus, $$\frac{BD}{CD}\cdot \frac{CE}{AE}\cdot \frac{AF}{BF}=1$$, so we are done by Menelaus.

Does anyone have a complex number solution?

The most direct way is using complex numbers. Assume W.L.O.G. $|a|=|b|=|c|=1$ From here: $$a \bar a=1, b \bar b =1, c \bar c = 1$$ $$\bar a = \frac{1}{a}, \bar b = \frac{1}{b}, \bar c = \frac{1}{c}$$ Using that $d$ lies on the bisector of $AO$: $$|d-a|=|d|$$ From there: $$(d-a)(\bar d - \bar a) = d \bar d$$$$d\bar d - a \bar d - \bar a d +a \bar a = d \bar d$$ The points $D,B,C$ are collinear so: $$\frac{ d - b}{\bar d - \bar b} = \frac{d-c}{\bar d - \bar c}$$$$(d-b)(\bar d - \bar c) = (d-c)(\bar d - \bar b)$$$$d\bar d + b \bar c - b \bar d - \bar c d = d \bar d + c \bar b - d \bar b - c \bar d$$$$d(\bar b - \bar c)=\bar d (b-c)+c\bar b - b \bar c$$$$d(b \bar b c - b c \bar c) = bc \bar d (b-c) + b\bar b c^2 - b^2 c \bar c$$$$d(c-b)=bc \bar d (b-c) + (c-b)(c+b)$$$$d = b+c-bc \bar d = b+c - bc \frac{a-d}{a^2}$$$$a^2d = a^2b + a^2c-abc +bcd$$$$d = \frac{a^2b+a^2c-abc}{a^2-bc}$$ $$\bar d = \frac{\bar a^2 \bar b + \bar a^2 \bar c - \bar a \bar b \bar c}{\bar a^2 - \bar b \bar c}=\frac{a^2\bar a^2 \bar bbc + a^2\bar a^2 b c\bar c - a^2 b c \bar a \bar b \bar c}{a^2\bar a^2 bc - a^2 b \bar b c \bar c} = \frac{c+b-a}{bc-a^2}$$In the same way we get: $$e = \frac{b^2c+b^2a-abc}{b^2-ac}$$$$\bar e = \frac{a+c-b}{ac-b^2}$$$$f = \frac{c^2a+c^2b-abc}{c^2-ab}$$$$\bar f = \frac{a+b-c}{ab-c^2}$$ Now it remains to show: $$\frac{d-e}{\bar d - \bar e} = \frac{d-f}{\bar d - \bar f}$$ This is a straightforward calculation.

@above what is the motivation for multiplying both sides by $bc$ in the $DBC$ collinearity simplification? is that a common trick for simplification?

baldeagle123 wrote: @above what is the motivation for multiplying both sides by $bc$ in the $DBC$ collinearity simplification? You know that $a \bar a=1, b \bar b =1, c \bar c = 1$ so this is just a way to get rid of the conjugates.

@mira74 which technique did you use? How can I learn that linearity thing?

SerdarBozdag wrote: @mira74 which technique did you use? How can I learn that linearity thing? This is basically an application of linearity of PoP to the circles with radius $0$ at $A$ and $O$. Linearity of PoP is explained in section 4 of this handout.

mira74 wrote: Note that $f(P)=PA^2-PO^2$ is linear, so $$\frac{BD}{CD}=\frac{f(B)-f(D)}{f(C)-f(D)}=\frac{c^2-R^2}{b^2-R^2}$$ Can someone explain me this move?

By a $2 \times$ dilation about $O$, we see that the result is equivalent to proving that the intersection points between each of the tangents to $A$, $B$, $C$ and the opposite sides are collinear. This is then true by Pascal's on $AABBCC$, so we are done.

sotpidot wrote: By a $2 \times$ dilation about $O$, we see that the result is equivalent to proving that the intersection points between each of the tangents to $A$, $B$, $C$ and the opposite sides are collinear. This is then true by Pascal's on $AABBCC$, so we are done. When you dilate, $BC$, $CA$, and $AB$ also move...

Here's my solution: Let $O_A$ be the circumcenter of $\triangle BOC$, and define $O_B$, $O_C$ similarly. Clearly, $O$ is the incenter of triangle $O_AO_BO_C$ and $A$, $B$, $C$ are the reflections of $O$ across $O_BO_C$, $O_CO_A$, $O_AO_B$, respectively. Hence, by a well-known lemma, $O_AA$, $O_BB$, $O_CC$ concur. Now, the result follows by Desargues' Theorem on triangles $ABC$ and $O_AO_BO_C$. $\blacksquare$

With a homogeneity of O, the sentence becomes a question that can be easily solved! https://artofproblemsolving.com/community/c6h553088p3212802

edit: wrong

CyclicISLscelesTrapezoid wrote: Take a homothety with scale factor $2$ centered at $O$, which sends the perpendicular bisectors to the tangents to the circumcircle of $\triangle ABC$ at $A$, $B$, and $C$. Then, use Pascal on $AABBCC$. sheesh

Dear Mathlinkers, http://jl.ayme.pagesperso-orange.fr/vol16.html then Perpendiculaires à trois céviennes concourantes p. 10... Sincerely Jean-Louis

Let $\triangle A'B'C'$ is the triangle formed by the three perpendicular bisectors. By the Desargues' thereom it is sufficient to prove $AA', BB', CC'$ are concurrent. Notice that $\triangle ABC$ and $\triangle A'B'C'$ are orthologic and the two orthologic centers are same which is $O$. So by Sondat's Theorem $AA', BB', CC'$ are concurrent.

By Linearity of PoP on zero-radius circles at $A$ and $O$, $BD/DC = (\lvert AB \rvert^2 - R^2) \div (\lvert AC \rvert - R^2)$ so Menelaus finishes.