vectors and or barycentres

Omid Hatami wrote: $H,I,O,N$ are orthogonal center, incenter, circumcenter, and Nagelian point of triangle $ABC$. $I_{a},I_{b},I_{c}$ are excenters of $ABC$ corresponding vertices $A,B,C$. $S$ is point that $O$ is midpoint of $HS$. Prove that centroid of triangles $I_{a}I_{b}I_{c}$ and $SIN$ concide. A cute little problem! Here is my solution to it. Define $M$ to be the centroid of $ABC$, $L$ to be the centroid of $I_{a}I_{b}I_{c}$ and $P$ be such point that $O$ is the midpoint of a line segment $IP$. Because $I$ is an orthocenter and $O$ - a nine-point center in $I_{a}I_{b}I_{c}$ we infer that $P$ is a circumcenter of $I_{a}I_{b}I_{c}$ and $\vec{IL}=2\vec{LP}$. Thus note that we are done if we showed that $\vec{SP}=\vec{PN}$. But $\vec{SP}=\vec{IH}$ because of the central symmetry at $O$. Thus it remains to show that $\vec{PN}=\vec{IH}$. But it is well known that $\vec{NM}=2\vec{MI}$ and $\vec{HM}=2\vec{MO}$, thus having considered a homothety centered at $M$ with coefficient $-1/2$ we see that $\vec{HN}=2\vec{IO}=\vec{IP}$ (the last one follows from the definition of $P$). We deduce that $HNPI$ is a parallelogram and thus $\vec{PN}=\vec{IH}$. Hence the result. What's the point of your post, arline (besides showing that you know barycentic coordinates and vectors)? It's absolutely useless.

xixas,its straighforward computation using barycantres and or vectors.its true remark.and somewhat general method.

arline wrote: vectors and or barycentres barycentric give us a solution much simpler

Is there an elementary solution without vectors or barycentric co-ordinates? Thank you.

Since $N,H$ are the incenter and circumcenter of the antimedial triangle of $\triangle ABC,$ it follows that $IO \parallel HN$ and $IO=\frac{_1}{^2}HN.$ Therefore, reflection $B_e$ of $I$ about $O$ (Bevan point of ABC) is the midpoint of $\overline{NS}$ $(\star).$ $B_e$ is the circumcenter of $\triangle I_aI_bI_c$ since $B_e$ is the concurrency point of the perpendiculars from $I_a,I_b,I_c$ to the sides $BC,CA,AB$ of its orthic triangle $\triangle ABC.$ Thus, the centroid $G_0$ of $\triangle I_aI_bI_c$ lies on the segment connecting its circumcenter $B_e$ and orthocenter $I$ such that $\overline{G_0B_e}:\overline{G_0I}=-1:2.$ Together with $(\star),$ we conclude that $G_0$ is the centroid of $\triangle SIN$ as well.

I understand that you find your solution's arguments a little trivial, but can you please help me by elaborating more? Why is the Nagel point the incentre of the anti-medial triangle? and Why does it being so, and H being the circumcentre result in the parallelism with IO? and so much more of what you have said.. P.S. I have no doubts that your solution is right, It is still too complicated for me.

Ramchandran wrote: I understand that you find your solution's arguments a little trivial, but can you please help me by elaborating more? Why is the Nagel point the incentre of the anti-medial triangle? and Why does it being so, and H being the circumcentre result in the parallelism with IO? and so much more of what you have said.. I find this post a bit unnecessary. Why not try it for yourselves? It could've been a good challenge Anyways also, instead of posting it here, you could've easily got the answers in a second by searching AoPS or Mathworld. But well, Ramchandran wrote: Why is the Nagel point the incentre of the anti-medial triangle? The answer that comes most quickly to me is just a homothety centred about $G$, the centroid of $\triangle ABC$, and ratio $-2$. This takes $\triangle ABC$ to its anti-median triangle. Also, by this, it takes the Incentre of $\triangle ABC$ to the Nagel Point of $\triangle ABC$. Since it must take the Incentre of $\triangle ABC$ to the Incentre of its Anti-Medial Triangle, the result follows. Ramchandran wrote: Why does it being so, and H being the circumcentre result in the parallelism with IO? Think about that homothety discussed above. Ramchandran wrote: And so much more of what you have said.. What else?

Let $IO\cap SN={L}$, let $Be$,$Z$ be circumcenter and centroid of $\triangle I_{A}I_{B}I_{C}$. Homothety $\mathcal{H}_{G,-\frac{1}{2}}$ takes $I$$\rightarrow$$N$,(well-known) and hence $2IG=GN$.From the Menelaus on the $\triangle SNG$ and the transversal $IO$ we get : $$\frac{NI}{IG}\cdot \frac{GO}{OS}\cdot \frac{SL}{LN}=-1$$Along with the $\frac{NI}{IG}=-3$,$\frac{GO}{OS}=\frac{1}{3}$ and hence $SL=LN$ and so the centroid of $\triangle SIN$ lies $OI$,and so it suffices to prove that $SZ$ bisects $IN$.From Menalaus on $\triangle NIL$ and transversal $GO$ we have $IO=OL$ and so $Be\equiv L$.From $\triangle I_{A}I_{B}I_{C}$ we have $\frac{BeZ}{IZ}=\frac{1}{2}$ so we are done.$\clubsuit$

As i remember, Omid Hatami wrote: $H,I,O,N$ are orthogonal center, incenter, circumcenter, and Nagelian point of triangle $ABC$. $I_{a},I_{b},I_{c}$ are excenters of $ABC$ corresponding vertices $A,B,C$. $S$ is point that $O$ is midpoint of $HS$. Prove that centroid of triangles $I_{a}I_{b}I_{c}$ and $SIN$ concide. it's a property of De Longchamps point.