AoPS - Problem

Problem

Source: IMO ShortList 1990, Problem 5 (FRA 1)

Tags: geometry, incenter, vector, Euler, trigonometry, orthocenter, IMO Shortlist

payman_pm

31.05.2005 13:00

Given a triangle $ABC$ . Let $G$ , $I$ , $H$ be the centroid, the incenter and the orthocenter of triangle $ABC$ , respectively. Prove that $\angle GIH > 90^{\circ}$ .

Sailor

31.05.2005 20:00

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.

darij grinberg

31.05.2005 20:26

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

pbornsztein

01.06.2005 00:11

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

Virgil Nicula

23.09.2006 02:36

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. $m(\widehat{HIO})\ge 90^{\circ}\ .$

sunchips

23.09.2006 21:46

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

Virgil Nicula

23.09.2006 21:57

$P\equiv\overrightarrow{AB}\odot\overrightarrow{CD}=AB\cdot CD\cdot\cos\phi$ , 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.

sunchips

23.09.2006 22:01

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

lomos_lupin

07.11.2006 08:38

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 )

Little Gauss

22.12.2006 18:42

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.

sayantanchakraborty

12.05.2014 12:42

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...

DottedCaculator

15.11.2021 17:03

It suffices to show

$(G-I)\cdot(H-I)<0$ . Let

$O$ be the origin. Then,

$H=A+B+C$ ,

$G=\frac13H$ , and

$I=\frac1{a+b+c}(aA+bB+cC)$ . Therefore, we have

$\begin{align*} (G-I)\cdot(H-I)&=(G-I)\cdot(3G-I)\\ &=3G\cdot G-4I\cdot G+I\cdot I\\ &=\frac13(A\cdot A+B\cdot B+C\cdot C+2A\cdot B+2B\cdot C+2C\cdot A)-4\cdot I\cdot G+I\cdot I\\ &=\frac13(9R^2-a^2-b^2-c^2)-\frac4{3a+3b+3c}(aA+bB+cC)\cdot(A+B+C)+R^2-\frac{abc}{a+b+c}\\ &=3R^2-\frac13(a^2+b^2+c^2)-\frac4{3(a+b+c)}(3(a+b+c)R^2-\frac12((a+b)c^2+(b+c)a^2+(a+c)b^2))+R^2-\frac{abc}{a+b+c}\\ &=3R^2-\frac13(a^2+b^2+c^2)-4R^2+\frac2{3(a+b+c)}(ab^2+ac^2+ba^2+bc^2+ca^2+cb^2)+R^2-\frac{abc}{a+b+c}\\ &=-\frac13(a^2+b^2+c^2)+\frac2{3(a+b+c)}(ab^2+ac^2+ba^2+bc^2+ca^2+cb^2)-\frac{abc}{a+b+c}\\ &=\frac1{3(a+b+c)}(-a^3-a^2b-a^2c-b^2a-b^3-b^2c-c^2a-c^2b-c^3+2ab^2+2ac^2+2ba^2+2bc^2+2ca^2+2cb^2-3abc)\\ &=\frac1{3(a+b+c)}{-a^3-b^3-c^3-3abc+a^2b+a^2c+b^2a+b^2c+c^2a+c^2b}\\ &<0, \end{align*}$ where the last inequality follows from Schur's Inequality.

pinkpig

01.02.2022 21:44

Note that

$\overrightarrow{IG}\cdot\overrightarrow{IH}=|\overrightarrow{IG}|\cdot|\overrightarrow{IH}|\cdot\cos{\angle{GIH}}.$ Thus, if

$\angle{GIH}>90^{\circ},$ then

$\cos{\angle{GIH}}<0,$ which means that

$\overrightarrow{IG}\cdot\overrightarrow{IH}<0.$ Now,

$\overrightarrow{IG}\cdot\overrightarrow{IH}=(\overrightarrow{G}-\overrightarrow{I})(\overrightarrow{H}-\overrightarrow{I})=\overrightarrow{G}\cdot\overrightarrow{H}-\overrightarrow{I}\cdot(\overrightarrow{H}+\overrightarrow{G})+\overrightarrow{I}\cdot\overrightarrow{I}.$ Let the circumcenter of triangle

$ABC$ be the origin. This means that

$\overrightarrow{G}=\frac{\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C}}{3},\overrightarrow{H}=\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C},$ and

$\overrightarrow{I}=\frac{a\cdot\overrightarrow{A}+b\cdot\overrightarrow{B}+c\cdot\overrightarrow{C}}{a+b+c}.$ Setting the circumradius to

$R,$ we find that

$\overrightarrow{G}\cdot\overrightarrow{H}=\left(\frac{\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C}}{3}\right)\cdot(\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C})=$

$\frac{1}{3}\left(\overrightarrow{A}\cdot\overrightarrow{A}+\overrightarrow{B}\cdot\overrightarrow{B}+\overrightarrow{C}\cdot\overrightarrow{C}+2\overrightarrow{A}\cdot\overrightarrow{B}+2\overrightarrow{A}\cdot\overrightarrow{C}+2\overrightarrow{B}\cdot\overrightarrow{C}\right)=$

$\frac{1}{3}\left(3R^2+2R^2\left(\frac{2R^2-a^2}{2R^2}+\frac{2R^2-b^2}{2R^2}+\frac{2R^2-c^2}{2R^2}\right)\right)=$

$\frac{1}{3}\left(9R^2-(a^2+b^2+c^2)\right).$ Also,

$\overrightarrow{I}\cdot(\overrightarrow{H}+\overrightarrow{G})=\frac{4}{3}\left(\frac{a\cdot\overrightarrow{A}+b\cdot\overrightarrow{B}+c\cdot\overrightarrow{C}}{a+b+c}\right)\cdot(\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C})=$

$\frac{4}{3}\left(\frac{a\cdot\overrightarrow{A}\cdot\overrightarrow{A}+b\cdot\overrightarrow{B}\cdot\overrightarrow{B}+c\cdot\overrightarrow{C}\cdot\overrightarrow{C}+(a+b)\overrightarrow{A}\cdot\overrightarrow{B}+(a+c)\overrightarrow{A}\cdot\overrightarrow{C}+(b+c)\overrightarrow{B}\cdot\overrightarrow{C}}{a+b+c}\right)=$

$\frac{4}{3}\left(\frac{(a+b+c)R^2+R^2\left((a+b)\left(\frac{2R^2-c^2}{2R^2}\right)+(a+c)\left(\frac{2R^2-b^2}{2R^2}\right)+(b+c)\left(\frac{2R^2-a^2}{2R^2}\right)\right)}{a+b+c}\right)=$

$\frac{4}{3}\left(3R^2-\frac{a^2b+ab^2+ac^2+a^2c+b^2c+bc^2}{2(a+b+c)}\right).$ In addition,

$\overrightarrow{I}\cdot\overrightarrow{I}=\left(\frac{a\cdot\overrightarrow{A}+b\cdot\overrightarrow{B}+c\cdot\overrightarrow{C}}{a+b+c}\right)\left(\frac{a\cdot\overrightarrow{A}+b\cdot\overrightarrow{B}+c\cdot\overrightarrow{C}}{a+b+c}\right)=$

$\frac{a^2\cdot\overrightarrow{A}\cdot\overrightarrow{A}+b^2\cdot\overrightarrow{B}\cdot\overrightarrow{B}+c^2\cdot\overrightarrow{C}\cdot\overrightarrow{C}+2ab\cdot\overrightarrow{A}\cdot\overrightarrow{B}+2ac\cdot\overrightarrow{A}\cdot\overrightarrow{C}+2bc\cdot\overrightarrow{B}\cdot\overrightarrow{C}}{(a+b+c)^2}=$

$\frac{R^2(a^2+b^2+c^2)+2R^2\left(ab\cdot\frac{2R^2-c^2}{2R^2}+ac\cdot\frac{2R^2-b^2}{2R^2}+bc\cdot\frac{2R^2-a^2}{2R^2}\right)}{(a+b+c)^2}=$

$\frac{R^2(a^2+b^2+c^2+2ab+2ac+2bc)-(abc^2+ab^2c+a^2bc)}{(a+b+c)^2}=$

$R^2-\frac{abc}{a+b+c}.$ This means that

$\frac{1}{3}(9R^2-(a^2+b^2+c^2))-\frac{4}{3}\left(3R^2-\frac{a^2b+ab^2+ac^2+a^2c+b^2c+bc^2}{2(a+b+c)}\right)+\left(R^2-\frac{abc}{a+b+c}\right)=$

$-\frac{a^2+b^2+c^2}{3}+\frac{2(a^2b+ab^2+ac^2+a^2c+b^2c+bc^2)}{3(a+b+c)}-\frac{abc}{a+b+c}=$

$-\frac{a^3+b^3+c^3+3abc-a^2b-ab^2-a^2c-ac^2-b^2c-bc^2}{3(a+b+c)}.$ From Schur's inequality,

$a^3+b^3+c^3+3abc-a^2b-ab^2-a^2c-ac^2-b^2c-bc^2>0.$ Now, since

$a,b,c$ are the sidelengths of a triangle,

$a,b,c>0\Leftrightarrow3(a+b+c)>0.$ Thus,

$-\frac{a^3+b^3+c^3+3abc-a^2b-ab^2-a^2c-ac^2-b^2c-bc^2}{3(a+b+c)}<0,$ which implies that

$\angle{GIH}>90^{\circ},$ as desired.