We work with vectors; we fix an origin in the plane, and for every point $ P$, we denote by $ P$ the vector from this origin to $ P$. Denote by $ X$ vector $ IX$, and by $ K$ the Euler center (i. e. the center of the nine-point circle of triangle $ ABC$). Then: $ H = 2K - O$ and $ G = \frac {2K + O}{3}$ hence: Using $ IK = \frac {R}{2} - r$ $ H\cdot{G} = \frac {1}{3}\cdot{(4IK^{2} - IO^{2})} = - \frac {2}{3}\cdot{r(R - 2r)} < 0$ and the result follows.

payman_pm wrote: Given a triangle $ ABC$. Let $ G$, $ I$, $ H$ be the centroid, the incenter and the orthocenter of triangle $ ABC$, respectively. Prove that $ \angle GIH > 90^{\circ}$. This is equivalent to stating that the incenter I of triangle ABC lies inside the circle with diameter GH. And this is an old result of Euler; see http://www.mathlinks.ro/Forum/viewtopic.php?t=16552 . darij

Oh, really? I thought it was a not-so-old result due to Guinand... And, it was on the ISL 1990. Pierre.

Sunchips asked to explain Sailor's solution more precisely. Some explanations for Sunchips : $\overrightarrow{HG}=2\cdot\overrightarrow{GO}\ \ ;\ \ \overrightarrow{NG}=2\cdot \overrightarrow{GI}\ .$ $X\odot I=0\ \ ;\ \ X\odot X=\overrightarrow{IX}\odot \overrightarrow{IX}=IX^{2}\ .$ $OI^{2}=R^{2}-2Rr\ .$ $IH^{2}=4R(R+r)+3r^{2}-p^{2}\ .$ $HO^{2}=9GO^{2}=9R^{2}-(a^{2}+b^{2}+c^{2})=9R^{2}-(2p^{2}-2r^{2}-8Rr)\ .$ The theorem of $I$- median in the triangle $IHO$ $\Longrightarrow$ $4\cdot IK^{2}=2(IH^{2}+IO^{2})-HO^{2}$ $\Longrightarrow$ $4\cdot IK^{2}=(R-2r)^{2}\ .$ $H=2K-O\ ,\ G=\frac{1}{3}\cdot (2K+O)$ $\Longrightarrow$ $H\odot G=\frac{1}{3}\cdot (2K-O)\odot (2K+O)=$ $\frac{1}{3}\cdot (4\cdot K\odot K-O\odot O)=$ $\frac{1}{3}\cdot (4\cdot IK^{2}-IO^{2})\ .$ Thus, $H\odot G=\frac{1}{3}\cdot (4\cdot IK^{2}-IO^{2})=$ $\frac{1}{3}\left[(R-2r)^{2}-(R^{2}-2Rr)\right]=$ $\frac{1}{3}\cdot (4r^{2}-2Rr)=$ $-\frac{2}{3}\cdot r(R-2r)\le 0\ .$ Therefore, $H\odot G\le 0$ $\Longrightarrow$ $\overrightarrow{IH}\odot \overrightarrow{IG}\le 0$, i.e. $m(\widehat{HIO})\ge 90^{\circ}\ .$

Thanks for your help Virgil Nicula. P.S. what is $\odot$: is it dot product? Thanks!

$P\equiv\overrightarrow{AB}\odot\overrightarrow{CD}=AB\cdot CD\cdot\cos\phi$, where $\phi$ is the value of the acute angle between the lines $AB$ and $CD$. If $AB\perp CD$, then $P=0\ .$ The $P$ means the scalar product between the vectors $\overrightarrow{AB}$ and $\overrightarrow{CD}\ .$ Example. Let $ABC$ be a triangle. Denote the points $T\in AB$, $S\in AC$ for which $CT\perp AB$, $BS\perp AC$. Then $P\equiv\overrightarrow{AT}\odot\overrightarrow{AB}=\overrightarrow{AS}\odot\overrightarrow{AC}$ and $BC^{2}=AB^{2}+AC^{2}-2P$ (the generalized Phytagoras' theorem). Remark. Sometimes notation for the scalar product is $A\bullet B$ or $A\cdot B$ or mere $AB$ when use a only letter for the notation of a vector.

rem wrote: This is the problem for current Olymon issue. I don't think it should be discussed. Another topic about an olymon problem was locked. see here: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=110555 Mods please lock this I just saw the Theorem of Guinand in a book, very complex. Later browsing an unrelated Geometry topic, I saw the topic of Sailor, and I just wanted the proof clarified. I never meant it to get as out of hand as this, to give other Olymon people a solution. I myself had solved the Olymon problem using a different method than this already. But Rem is right, topic should be closed

rem wrote: This is the problem for current Olymon issue. I don't think it should be discussed. Another topic about an olymon problem was locked. see here: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=110555 Mods please lock this No Its not. That problem asks to prove $HIO > 90$ not $HIG>90$. Clearly we can conclude the olymon problem from this one, but there exist a solution at Olymon level for $HIO>90$.( Which is the one you were suppose to find , just kidding )

How about this solution? It doesn't use the length of IN, IO, OG... When ABC is acute, WLOG $\angle B>\angle A >\angle C$ AI, BC meet at D, define E, F similar. Then median of A is in $\angle CAD$ B is in $\angle CBE$ C is in $\angle ACF$, so triangle $CIE$ containing G. similary, triangle $BIF$ containing H. Therefore, $\angle GIH \ge min(\angle BIC, \angle EIF)=90+\frac{A}{2}>90$. When ABC is obtuse, solution is very similar but little bit differance.

Too easy for IMO Shortlist....I calculated the squares of the lengths and then showed that $GI^2+IH^2-GH^2 \le 0$ from which the result follows...