123. My extension and its proof.

by Virgil Nicula, Sep 13, 2010, 3:56 AM

PP. Let the incircle $w=C(I,r)$ of $\triangle ABC$ which touches $BC$ , $CA$ , $AB$ at the points $X$ , $Y$ , $Z$ . The parallels to $AX$ , $BY$ ,

$CZ$ through $I$ meet the lines $BC$ , $CA$ , $AB$ at the points $D$ , $E$ , $F$ respectively. Prove that $I$ is the centroid of $\triangle DEF$ .


Proof (vectorial). Denote $M\in AI\cap BC$ and $\overrightarrow{OX}=X$ , where $O$ is the origin of vectorial system. Thus, $a\cdot A+b\cdot B+c\cdot C=2s\cdot I$ , where $2s=a+b+c$

From $\left\{\begin{array}{ccc}
\frac {XB}{XC}=\frac {s-b}{s-c} & \implies & a\cdot X=(s-c)\cdot B+(s-b)\cdot C\\\\
\frac {MB}{MC}=\frac cb & \implies & (b+c)\cdot M=b\cdot B+c\cdot C\\\\
\frac {DX}{DM}=\frac {b+c}{a} & \implies & 2s\cdot D=a\cdot X+(b+c)\cdot M\end{array}\right\|$ $\implies$ $2s\cdot D=[(s-c)\cdot B+(s-b)\cdot C]+[b\cdot B+c\cdot C]$ $\implies$

$2s\cdot D=(s+b-c)\cdot B+(s+c-b)\cdot C$ . Obtain so $\left\{\begin{array}{c}
2s\cdot D=(s+b-c)\cdot B+(s+c-b)\cdot C\\\\
2s\cdot E=(s+c-a)\cdot C+(s+a-c)\cdot A\\\\
2s\cdot F=(s+a-b)\cdot A+(s+b-a)\cdot B\end{array}\right\|\ \bigoplus\ \implies$

$2s\cdot (D+E+F)=3\cdot (a\cdot A+b\cdot B+c\cdot C)$ $\implies$ $2s\cdot (D+E+F)=3\cdot 2s\cdot I$ $\implies$ $D+E+F=3\cdot I$ $\implies$ $I$ is the centroid of $\triangle DEF$ .


Generalization. Let $\triangle ABC$ and $P(x, y, z)$ , $R(u, v, w)$ with normalized barycentric co-ordinates w.r.t. $\triangle ABC$ . Let $D$ , $E$ , $F$ which belong to $BC$ , $CA$ , $AB$ respectively

so that $PD\parallel AR$ , $PE\parallel BR$ , $PF\parallel CR$ . Prove that $P$ is the centroid of $\triangle DEF\ \iff$ exist relations $\boxed{\ \frac{u(v+w)}{x}=\frac{v(w+u)}{y}=\frac{w(u+v)}{z}\ }$ .

Particular case. . If $P(a,b,c)$ is the incenter and $R(r_a, r_b, r_c)$ is the Gergogne's point, then obtain the propose problem.


Proof. Denote $\overrightarrow{OX}=X$ , where $O$ is origin of vectorial system, $M\in AP\cap BC$ , $X\in AR\cap BC$ and another two pairs of analogous $\{Y,E\}\subset CA\ \wedge\ \{Z,F\}\subset AB$ . Suppose

w.l.o.g. that $xyz\ne 0$ and $uvw\ne 0$ . From relations $\left\{\begin{array}{ccc}
xA+yB+zC=P & \wedge & (y+z)M=yB+zC\\\\
uA+vB+wC=R & \wedge & (v+w)X=vB+wC\end{array}\right\|$ and $\frac {\overline{XD}}{\overline{DM}}=\frac {\overline{AP}}{\overline{PM}}=\frac {y+z}{x}$ obtain $D=xX+(y+z)M$ $\iff$

$(v+w)D=x(vB+wC)+(v+w)(yB+zC)$ $\iff$ $D=\left(y+\frac {xv}{v+w}\right)B+\left(z+\frac {xw}{v+w}\right)C$ . Obtain analogous $E$ , $F$ . Thus, $\left|\begin{array}{c}
D=\left(y+\frac {xv}{v+w}\right)B+\left(z+\frac {xw}{v+w}\right)C\\\\
E=\left(z+\frac {yw}{w+u}\right)C+\left(x+\frac {yu}{w+u}\right)A\\\\
F=\left(x+\frac {zu}{u+v}\right)A+\left(y+\frac {zv}{u+v}\right)B\end{array}\right|$

and $P$ is the centroid of $\triangle DEF$ $\iff$ $3\cdot P=D+E+F$ $\iff$ $3\cdot (xA+yB+zC)=\sum\left(y+\frac {xv}{v+w}\right)B+\left(z+\frac {xw}{v+w}\right)C$ $\iff$ $\left\{\begin{array}{c}
\frac xu=\frac {y}{u+w}+\frac {z}{u+v}\\\\
\frac yv=\frac {z}{v+u}+\frac {x}{v+w}\\\\
\frac zw=\frac {x}{w+v}+\frac {y}{w+u}\end{array}\right\|\ (*)$ . Exists

$m$ , $n$ , $p$ so that $\left\{\begin{array}{c}
x=mu(v+w)\\\
y=nv(w+u)\\\
z=pw(u+v)\end{array}\right\|$ . Relations $(*)$ becomes $\left\{\begin{array}{c}
m(v+w)=nv+pw\\\
n(w+u)=pw+mu\\\
p(u+v)=mu+nv\end{array}\right\|$ $\iff$ $\left\{\begin{array}{c}
(m-n)v=(p-m)w\\\
(n-p)w=(m-n)u\\\
(p-m)u=(n-p)v\end{array}\right\|$ $\iff$ $\frac {n-p}{u}=\frac {p-m}{v}=\frac {m-n}{w}=\frac 01\iff$

$m=n=p\ \iff$ $\boxed{\frac{u(v+w)}{x}=\frac{v(w+u)}{y}=\frac{w(u+v)}{z}=2\cdot (uv+vw+wu)}$ and $m=n=p=\frac {1}{2(uv+vw+wu)}$ . In my opinion this problem is a nice extension with a

nice proof and mostly with a nice conclusion !


Proposed problem. The incircle $w=C(I,r)$ of $\triangle ABC$ touches $BC$ , $CA$ , $AB$ at the points $D$ , $E$ , $F$ . Let $N$ be the Nagel's point of $\triangle ABC$ .

Denote the points $\left\{\begin{array}{ccc}
R\in (EF) & \wedge & \frac {RF}{RE}=\frac {s-c}{s-b}\\\\
S\in (FD) & \wedge & \frac {SD}{SF}=\frac {s-a}{s-c}\\\\
T\in (DE) & \wedge & \frac {TE}{TD}=\frac {s-b}{s-a}\end{array}\right\|$ , where $2s=a+b+c$ . Prove easily that exists $P\in AR\cap BS\cap CT$ . The parallels

lines to $AN$ , $BN$ , $CN$ through $P$ meet the lines $BC$ , $CA$ , $AB$ at the points $X$ , $Y$ , $Z$ respectively. Prove that $P$ is the centroid of $\triangle XYZ$ .


Remark. Can construct the point $R$ so : $A$-Nagel cevian meets $BC$ in $U$ ; find the point $V\in (BE)$ such that $UV\parallel AC$ $\implies$ $VR\parallel AB$ .
This post has been edited 56 times. Last edited by Virgil Nicula, Nov 26, 2015, 9:30 PM

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