FIGO is perpendicular

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FIGO is perpendicular

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Source: 2018 Taiwan TST Round 3

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#1 h Apr 2, 2020, 6:57 AM • 4 Y

Y by kiyoras_2001, Bassiskicking, Mango247, GeoKing

Let $I,G,O$ be the incenter, centroid and the circumcenter of triangle $ABC$ , respectively. Let $X,Y,Z$ be on the rays $BC, CA, AB$ respectively so that $BX=CY=AZ$ . Let $F$ be the centroid of $XYZ$ .

Show that $FG$ is perpendicular to $IO$ .

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#2 h Apr 3, 2020, 4:42 AM • 1 Y

Y by Tamako22

Lemma 1: As $X$ varies on ray $BC$ , $F$ lie on a line passing through $G$ .
Proof: Set up barycentric co-ordinate system with $A=(1,0,0),B=(0,1,0)$ and $C=(0,0,1)$ . Let $X=(0,x,1-x),Y=(1-y,0,y)$ and $Z=(z,1-z,0)$ . Then $BX^2=CY^2=AZ^2\implies a^2(x-1)^2=b^2(y-1)^2=c^2(z-1)^2\implies y,z$ are of the form $ax+b$ for some $a,b\in\mathbb{R}$ .
$\therefore$ Centroid of $\triangle XYZ$ lie on a line as $X$ varies on ray $BC$ . For $X=B$ , $F=G$ . This proves lemma 1.

Lemma 2:Let WLOG, $BC$ be smallest side of $\triangle ABC$ . Let $D,E$ lie on segments $CA$ and $AB$ respectively s.t. $BC=CD=BE$ . Then $OI\perp DE$
Proof: Let $M$ be the second intersection of $BI$ and $(O)$ . Then $CDE\sim MIO$ because $\angle DCE = \angle IMO$ and $\frac{CD}{CE} = \frac{CD}{CB} = \frac{MA}{MO} = \frac{MI}{MO}.$ Furthermore, $CD \perp MI$ and $CE \perp MO$ , so $DE \perp IO$ as well. This proves lemma 2.
(Note: Proof of lemma 2 is due to huajd)

Lemma 3: Let $X=C$ . Then $Y=D$ . Let $Z=K$ . Then line joining centroid $F$ of $\triangle CDK$ and $G$ is parallel to $DE$ ,ie $FG$ is parallel to $DE$ .
Proof: Set up the barycentric co-ordinate system with $A=(1,0,0),B=(0,1,0)$ and $C=(0,0,1)$ . Let $BC=a,CA=b$ and $AB=c$ . Let $\frac{1}{a}=u,\frac{1}{b}=v$ and $\frac{1}{c}=w$ . Then $D=(\frac{a}{b},0,\frac{b-a}{b})$ , $E=(\frac{a}{c},\frac{c-a}{c},0)$ .
The equation of line $DE$ is $\frac{x}{a}+\frac{ya}{a-c}+\frac{za}{a-c}=0$ , ie

$x+\frac{yw}{w-u}+\frac{zv}{v-u}=0$ .
Also we obtain $Z=K=(\frac{c-a}{c},\frac{a}{c},0)$ .
So centroid $F$ of $XYZ=(\frac{c-a}{c}+\frac{a}{b}:\frac{a}{c}:2-\frac{a}{b})$ . Also $G=(1:1:1)$ .
Equation of line $FG$ becomes $(2v+2w-u)x+(2w+2u-v)y+(2u+2v-w)z=0$ .

$\begin{vmatrix} 2(u+v+w) & 2(u+v+w) & 2(u+v+w) \\ 1 & \frac{w}{w-u} & \frac{v}{v-u} \\ 1 & 1 & 1 \end{vmatrix}=0$
$\begin{vmatrix} -3u & -3v & -3w \\ 1 & \frac{w}{w-u} & \frac{v}{v-u} \\ 1 & 1 & 1 \end{vmatrix}=0$
$\therefore$ $\begin{vmatrix} 2v+2w-u & 2w+2u-v & 2u+2v-w \\ 1 & \frac{w}{w-u} & \frac{v}{v-u} \\ 1 & 1 & 1 \end{vmatrix}=0$
$\therefore FG\parallel DE$ as desired.

Now let $X,Y,Z$ be the on ray $BC,CA,AB$ respectively s.t. $BX=CY=AZ$ . Then centroid of $\triangle XYZ$ lies on line joining centroid of $\triangle CDK$ and $G$ . This line is perpendicular to $OI$ from lemma 1 and lemma 2. This completes the proof.
Q.E.D.

This post has been edited 6 times. Last edited by Euler365, Apr 3, 2020, 4:51 AM

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#3 h Apr 10, 2020, 2:10 AM • 3 Y

Y by YS_PARKGONG, cmjun1109, JH_K2IMO

$BS=CT=BC$
S lies on AB, T lies on AC, ST and IO are perpendicular. (Pythagorean)
And prove FG and ST are parellel by vector (it's trivial)

This post has been edited 1 time. Last edited by jeusson, Apr 10, 2020, 2:10 AM

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#4 h Sep 13, 2020, 7:34 PM

Y by

An extension?!

In $\triangle ABC$ , we select three points $X,Y,Z$ on the sides $BC,CA,AB$ respectively such that $BX=CY=AZ$ . Define the points $S,T$ in the lines $AB,AC$ such that $BS=BC=CT$ , the centroid of $\triangle ABC$ is $G$ , the centroid of $\triangle XYZ$ is $G_1$ . Prove that the lines $SG$ and $TG_1$ meets on the $B$ -median of $\triangle BZC$ .

This post has been edited 1 time. Last edited by Bassiskicking, Sep 14, 2020, 3:02 AM
Reason: Mistake

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#5 h Jan 1, 2023, 3:51 PM • 2 Y

Y by Ru83n05, euclides05

Let $D$ and $R$ lie on $AC$ and $E$ and $S$ lie on $AB$ such that $D$ and $E$ are intouch points and $SB = BC = CR$ .

Lemma: $RS \perp OI$ .

Proof: If $AC = AB$ , the result is trivial. Without loss of generality assume $AC > AB$ . Then,
$\begin{align*} OS^2 - OR^2 & = (OS^2-OA^2)+(OA^2-OR^2) \\ & = -SA\cdot SB + RA \cdot RC = a(RA - SA) = a(b - c). \end{align*}$ We also find
$\begin{align*} IS^2 - IR^2 & = ES^2 - DR^2 \\ & = ((a - s + b)^2 - (a - s + c)^2) = a(b - c). \end{align*}$ Hence, by Perpendicularity Lemma we obtain the desired. $\square$

Let $BX = CY = AZ = k$ . We can determine the direction of $FG$ :
$\begin{align*} \overrightarrow{FG} & = \frac{A + B + C}{3} - \frac{X + Y + Z}{3} = \frac{1}{3}(\overrightarrow{ZA} + \overrightarrow{XB} + \overrightarrow{YC}) \\ & = \frac{k}{3}(\vec u_{BA} + \vec u_{CB} + \vec u_{AC}) = \frac{k}{3a}(\overrightarrow{BS} + \overrightarrow{CB} + \overrightarrow{RC}) = \frac{k}{3a}\overrightarrow{RS}. \end{align*}$ We conclude that $FG \parallel RS$ . Hence $FG \perp OI$ .

This post has been edited 1 time. Last edited by DongerLi, Jan 3, 2023, 9:47 PM

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#6 h Jun 5, 2024, 7:47 PM • 1 Y

Y by GeoKing

Too easy with linpop. Let $BX = CY = AZ = s$ and $BC = y+z, CA = z+x, AB = x+y$ . Let $f(\bullet) = \text{Pow}(\bullet, \omega) - \text{Pow}(\bullet, (ABC))$ where $\omega$ is the incircle, then
$f(F) = \frac 13 \sum_{\text{cyc}} f(X) = \frac 13 \sum_{\text{cyc}} ((y-s)^2 + s(y+z-s)) = \frac 13 \sum_{\text{cyc}} y^2$ which does not depend on $s$ , so $F$ moves along a line perpendicular to $OI$ . Note that $F=G$ when $s=0$ , hence proved.

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#7 h Apr 8, 2025, 9:52 AM

Y by

The $OI$ vector trick instantly solve this problem.

First, we recall a well known lemma which state that $\overrightarrow{OI}$ is perpendicular to the sum of unit vectors along $CB,AC,BA$
Next, we’ll show that $\overrightarrow{FG}$ is parallel to the sum of units vectors along $CB,AC,BA$
Note that $\begin{align*} \overrightarrow{FG} &= \overrightarrow{FY} + \overrightarrow{YA} + \overrightarrow{AG} \\ &= \frac{2}{3} \left( \frac{\overrightarrow{XZ}}{2} + \overrightarrow{ZY} \right) + \overrightarrow{YA} + \frac{2}{3} \left( \overrightarrow{AB} + \frac{\overrightarrow{BC}}{2} \right) \\ &= \frac{2}{3} \cdot \frac{\overrightarrow{XZ}}{2} + \frac{2}{3} \overrightarrow{ZY} + \overrightarrow{YA} + \frac{2}{3} \overrightarrow{AB} + \frac{1}{3} \overrightarrow{BC} \\ &= \frac{\overrightarrow{XB} + \overrightarrow{BZ}}{3} + \frac{2}{3} (\overrightarrow{ZA}) + \frac{\overrightarrow{YA}}{3} + \frac{2}{3} \overrightarrow{AB} + \frac{1}{3} \overrightarrow{BC} \\ &= \frac{\overrightarrow{XB}}{3} + \frac{\overrightarrow{BA} + \overrightarrow{AZ}}{3} + \frac{2}{3} \overrightarrow{ZA} + \frac{\overrightarrow{YA}}{3} + \frac{\overrightarrow{AC}}{3} + \frac{\overrightarrow{AB}}{3} \\ &= \frac{\overrightarrow{XB}}{3} + \frac{\overrightarrow{YC}}{3} + \frac{\overrightarrow{ZA}}{3} \end{align*}$ which is clearly in the same direction as the sum of units vector along vectors along $CB,AC,BA$

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#8 h Apr 19, 2025, 12:26 PM

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Let $l_a$ be the line passing through $X$ and parallel to $BI$ . Define $l_b,l_c$ similarly. Let $D= l_a \cap l_b$ and define $E,F$ similarly. let $G_1,G_2$ be the centroids of $\triangle XYZ, \triangle DEF$ , respectively. From $X(BC \cap l_b)=BC$ , we know that $DI = BX$ and $DI \parallel BC$ , so from its symmetry, we know that the $GG_1G_2I$ is parallelogram and $I$ is the circumcenter of $\triangle DEF$ .
On the other hand, obviously $\triangle DEF$ is orthogonally similar to the intouch triangle of $\triangle ABC$ and $OI$ is the Euler line of the intouch triangle of $\triangle ABC$ .
Therefore we get that $GG_1 \perp OI$ .

This post has been edited 1 time. Last edited by kaede_Arcadia, Apr 21, 2025, 5:33 AM

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