Nagelian point

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Source: Iranian National Olympiad (3rd Round) 2002

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#1 h Oct 1, 2006, 4:45 PM • 1 Y

Y by Adventure10

$H,I,O,N$ 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.

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arline

#2 h Oct 10, 2006, 3:50 PM • 1 Y

Y by Adventure10

vectors and or barycentres

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Xixas

#3 h Oct 10, 2006, 6:42 PM • 3 Y

Y by horizon, Adventure10, Mango247

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.

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arline

#4 h Oct 11, 2006, 2:34 AM • 2 Y

Y by Adventure10, Mango247

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

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hucht

#5 h Oct 13, 2006, 10:12 PM • 1 Y

Y by Adventure10

arline wrote:

vectors and or barycentres

barycentric give us a solution much simpler

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#6 h Jan 27, 2011, 3:02 AM • 2 Y

Y by Adventure10, Mango247

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

Thank you.

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#7 h Jan 27, 2011, 5:32 AM • 2 Y

Y by Adventure10, Mango247

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.

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#8 h Jan 27, 2011, 6:42 AM • 2 Y

Y by Adventure10, Mango247

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.

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904 posts

#9 h Jan 27, 2011, 9:52 AM • 2 Y

Y by Adventure10, Mango247

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?

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1246 posts

#10 h Sep 8, 2016, 6:48 PM • 2 Y

Y by Adventure10, Mango247

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. $\clubsuit$

This post has been edited 3 times. Last edited by nikolapavlovic, Sep 9, 2016, 7:31 AM

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#11 h Sep 15, 2016, 2:12 AM • 2 Y

Y by Adventure10, Mango247

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.

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