386. Barycentrical coordinates.

by Virgil Nicula, Sep 1, 2013, 2:55 AM

Barycentrical coordinates in Olympiad Geometry.


Here are barycentrical coordinates w.r.t. $\triangle ABC$ for: centroid $G$ , orthocenter $H$, circumcenter $O$, incenter $I$ , $A$-excenter $I_a$ and the points Nagel's $N$ , Gergogne's $\Gamma$ and Lemoine's $L$ .

$\left\{\begin{array}{ccccc}
G(1,1,1) & ; & I(a,b,c) & ; & I_a(-a,b,c)\\\\
N(s-a,s-b,s-c) & ; & \Gamma\left(\frac 1{s-a},\frac 1{s-b},\frac 1{s-c}\right) & ; &  H(\tan A,\tan B,\tan C)\\\\
O(\sin 2A,\sin 2B,\sin 2C) & ; & L\left(a^2,b^2,c^2\right) & ; & \end{array}\right\|\  \mathrm{where}\ \left\{\begin{array}{ccc}
\sum(s-b)(s-c)=r(4R+r) &  ; & \sum\cot B\cot C=1\\\\
\sum \sin 2A=\frac {2S}{R^2} & ; & \sum a^2\left(b^2+c^2-a^2\right)=16S^2\end{array}\right\|$

Theorem 1. If $M(\alpha ,\beta ,\gamma )$ , where $\alpha +\beta +\gamma \ne 0$ , then $\alpha \cdot\overrightarrow{MA}+\beta \cdot\overrightarrow{MB}+\gamma\cdot\overrightarrow{MC}=0$ and $(\forall )\ X$ have $\boxed{\alpha \cdot\overrightarrow {XA}+\beta\cdot\overrightarrow{XB}+\gamma\cdot\overrightarrow{XC}=(\alpha +\beta +\gamma )\cdot \overrightarrow{XM}}\ (1^*)$ .


Remark. Observe that $D\in AM\cap BC\implies$ $\beta\cdot\overline{DB}+\gamma\cdot\overline{DC}=0\ ,\ (\beta +\gamma )\overline{MD}$ $+\alpha\cdot\overline{MA}=0$ and $(\beta +\gamma )\cdot\overrightarrow{AD}=\beta \cdot\overrightarrow{AB}+\gamma\cdot\overrightarrow {AC}=$ $\overrightarrow{AM}$ .

Theorem 2. The power $p_w(M)$ of $M(\alpha ,\beta ,\gamma )\ ,\ \alpha +\beta +\gamma =1$ w.r.t. $w=C(O,R)$ of $\triangle ABC\ :$

$\boxed{-\mathrm{p}_w(M)=\beta\gamma a^2+\gamma\alpha b^2+\alpha\beta c^2=\alpha\cdot MA^2+\beta\cdot MB^2+\gamma\cdot MC^2=OP^2-R^2}\ (2^*)$ .


Examples: $\ \mathrm{p}_w(G)=-\frac {a^2+b^2+c^2}{9}\implies OG^2=R^2-\frac {a^2+b^2+c^2}{9}\implies a^2+b^2+c^2\le 9R^2\ ;$ $\mathrm{p}_w(I)=-2Rr\implies OI^2=R^2-2Rr\implies  2r\le R\ ;$

$\mathrm{p}_w(I_a)=2Rr_a \implies OI_a^2=R^2+2Rr_a\ ;$ $\mathrm{p}_w(N)=-4r(R-r)\implies ON^2=R^2-4r(R-r)\implies ON=R-2r\ ;$

$\mathrm{p}_w(H)=-8R^2\cos A\cos B\cos C\implies OH^2=R^2\left(1-8\cos A\cos B\cos C\right) \implies \cos A\cos B\cos C\le\frac 18\ ;$

$\mathrm{p}_w(\Gamma )=-r(R+r)\left(\frac {2s}{4R+r}\right)^2 \implies O\Gamma^2=R^2-r(R+r)\left(\frac {2s}{4R+r}\right)^2\implies\frac {R(4R+r)}{2s}\ge\sqrt{r(R+r)}\ ;$

$\mathrm{p}_w(L)=-3\left(\frac {abc}{a^2+b^2+c^2}\right)^2\implies OL^2=R^2-3\left(\frac {abc}{a^2+b^2+c^2}\right)^2\implies a^2+b^2+c^2\ge4S\sqrt 3\ .$

Remark. $\overline{OH}=3\cdot\overline {OG}\implies$ $OH^2=9R^2-\left(a^2+b^2+c^2\right)$ .


Theorem 3. $(\forall )\ X$ and $M\left(\alpha , \beta , \gamma \right)$ for which $\alpha +\beta +\gamma =1\ ,\ \boxed{\alpha\cdot XA^2+\beta\cdot XB^2+\gamma\cdot XC^2=XM^2-p_w(M)}\ (3^*)$ . For $X:=M$ obtain the second part of $\left(3^*\right)$ .

Theorem 4. If $M_k\left(\alpha_k,\beta_k,\gamma_k\right)\ ,\ \alpha_k+\beta_k+\gamma_k=1$ where $k\in\overline{1,2}$ , then $\boxed{M_1M_2^2=-\left[(\beta_1-\beta_2)(\gamma_1-\gamma_2)a^2+(\gamma_1-\gamma_2)(\alpha_1-\alpha_2)b^2+(\alpha_1-\alpha_2)(\beta_1-\beta_2)c^2\right]}\ (4^*)$ and

$m\left(\widehat{M_1OM_2}\right)$ is given by $\overrightarrow{OM_1}\cdot\overrightarrow{OM_2}=OM_1\cdot OM_2\cdot\cos \widehat{M_1OM_2}$ , where $\boxed{\overrightarrow{OM_1}\cdot\overrightarrow{OM_2}=R^2-\frac 12\cdot\left[\left(\beta_1\gamma_2+\beta_2\gamma_1\right)a^2+\left(\gamma_1\alpha_2+\gamma_2\alpha_1\right)b^2+\left(\alpha_1\beta_2+\alpha_2\beta_1\right)c^2\right]}\ (5^*)$ .

Remark. For $M(\alpha ,\beta ,\gamma )\ ,\ \alpha +\beta +\gamma =1$ obtain that $OM^2=\overrightarrow{OM}\cdot\overrightarrow{OM}=R^2-\left(\beta\gamma a^2+\gamma\alpha b^2+\alpha \beta c^2\right)=R^2+p_w(M)\ ,$ where $p_w(M)=-\left(\beta\gamma a^2+\gamma\alpha b^2+\alpha \beta c^2\right)$ .


Applications.

AP1 (Miguel Ochoa Sanchez). Let $P$ be an interior point of an equilateral $\triangle ABC$ with $AB=1$ so that $\frac {[BPC]}2=\frac{[CPA]}4=\frac {[APB]}1$ . Prove that $PA^2+PB^2+PC^2=\frac 87$ .

Proof 1. Oberve that $P\left(\frac 27,\frac 47,\frac 17\right)$ . From the relation $\left(2^*\right)$ obtain that $\boxed{p_w(P)=-\frac 27}\ (1)$ . Observe that $GA=\frac {\sqrt 3}{3}$ . From the relation $\left(3^*\right)$

for $M:=P$ and $X:=G$ obtain that $2GA^2+4GB^2+GC^2=$ $7GP^2+2\implies$ $\frac 73=7GP^2+2\implies$ $\boxed{PG^2=\frac 1{21}}\ (2)$ . Apply again $\left(3^*\right)$

for $M:=G$ and $X:=P$ and obtain that $PA^2+PB^2+PC^2=3PG^2+AB^2=\frac 17+1\implies$ $\boxed{PA^2+PB^2+PC^2=\frac 87}$ .

Proof 2 (elementary). Observe that $\left\{\begin{array}{ccc}
\frac {DB}{DC}=\frac {[APB]}{[APC]}=\frac 14 & \implies & \frac{DB}1=\frac {DC}4=\frac 15\\\\
\frac {EC}{EA}=\frac {[BPC]}{[BPA]}=\frac 21 & \implies & \frac{EC}2=\frac {EA}1=\frac 13\\\\
\frac {FA}{FB}=\frac {[CPA]}{[CPB]}=\frac 42 & \implies & \frac{FA}4=\frac {BF}2=\frac 16\end{array}\right\|$ . Apply the Stewart's relation in the particular case $(*)$ when

$\triangle ABC$ is equilateral to the cevians: $\left\{\begin{array}{ccc}
AD^2=AB^2-DB\cdot DC & \implies & AD^2=1-\frac 15\cdot\frac 45=\frac {21}{25}\\\\
BE^2=BC^2-EC\cdot EA & \implies & BE^2=1-\frac 23\cdot\frac 13=\frac {7}{9}\\\\   
CF^2=CA^2-FA\cdot FB & \implies & CF^2=1-\frac 23\cdot\frac 13=\frac {7}{9}\end{array}\right\|$ . Apply the Van Aubel relation:

$\left\{\begin{array}{ccccccc}
\frac {PA}{PD}=\frac {EA}{EC}+\frac {FA}{FB}=\frac 12+\frac 42=\frac 52 & \implies & \frac {PA}5=\frac {PD}2=\frac {AD}7 & \implies & PA=\frac 57\cdot AD & \implies & PA^2=\frac 37\\\\
\frac {PB}{PE}=\frac {FB}{FA}+\frac {DB}{DC}=\frac 24+\frac 14=\frac 34 & \implies & \frac {PB}3=\frac {PE}4=\frac {BE}7 & \implies & PB=\frac 37\cdot BE & \implies & PB^2=\frac 17\\\\
\frac {PC}{PF}=\frac {DC}{DB}+\frac {EC}{EA}=\frac 41+\frac 21=\frac 61 & \implies & \frac {PC}6=\frac {PF}1=\frac {CF}7 & \implies & PC=\frac 67\cdot CF & \implies & PC^2=\frac 47\end{array}\right\|$ $\implies$ $PA^2+PB^2+PC^2=\frac 87$ .

Remark. Consider an $A$-isosceles triangle $ABC$ and a point $D\in (BC)$ . Denote the midpoint $M$ od $[BC]$ and suppose w.l.o.g. $D\in (BM)$ . Then $AM\perp BD\iff$

$AB^2-AD^2=MB^2-MD^2=$ $(MB-MD)(MB+MD)=DB(MC+MD)=$ $DB\cdot DC\iff$ $\boxed{AD^2=AB^2-DB\cdot DC}\ (*)$ .

Otherwise. Apply the Pythagoras's theorem: $AD^2=BA^2+BD^2-2\cdot BD\cdot BM=$ $AB^2-BD\cdot (BC-BD)\implies$ $AD^2=AB^2-DB\cdot DC$ .



AP2. Let $DEF$ be the cevian triangle of the point $P$ w.r.t. $\triangle ABC$ and let $L$ , $M$ , $N$ be the midpoints of $[BC]$ , $[CA]$ , $[AB]$ respectively. Denote the midpoints $X$ , $Y$ , $Z$

of $[AD]$ , $[BE]$ , $[CF]$ . Prove that $LX$ , $MY$ , $NZ$ are concurrently. If $T\in LX\cap MY\cap NZ$ and $P(x,y,z)$ , then $\boxed{T\left(\ \frac 1y+\frac 1z\ ,\ \frac 1z+\frac 1x\ ,\ \frac 1x+\frac 1y\ \right)}$ .


Proof. Observe that $D(0,y,z)$ , $X(y+z,y,z)$ and $L(0,1,1)$ . Thus, $T(\alpha , \beta , \gamma )\in XL\iff$ $\left|\begin{array}{ccc}
\alpha  & \beta & \gamma\\\\
y+z & y & z\\\\
0 & 1 & 1\end{array}\right|=0\iff$ $(y-z)\alpha -(y+z)\beta +(y+z)\gamma =0$ . Therefore,

$T\in LX\cap MY\cap NZ\iff$ the homogeneous system with $\Delta =\left|\begin{array}{ccc}
y-z & -(y+z) & y+z\\\\
z+x & z-x & -(z+x)\\\\
-(x+y) & x+y & x-y\end{array}\right|$ has at least solution, i.e. $\Delta =0$ what is evidently (the sum of its lines is zero).

The system $\left\{\begin{array}{ccc}
(y-z)\alpha -(y+z)\beta +(y+z)\gamma & = & 0\\\\
(z+x)\alpha +(z-x)\beta -(z+x)\gamma & = & 0\end{array}\right\|=0$ has the solution $\frac {\alpha}{\left|\begin{array}{cc}
-(y+z) & y+z\\\\
z-x & -(z+x)\end{array}\right|}=\frac {\beta}{\left|\begin{array}{cc}
y+z & y-z\\\\
-(z+x) & z+x\end{array}\right|}=\frac{\gamma}{\left|\begin{array}{cc}
y-z & -(y+z)\\\\
z+x & z-x\end{array}\right|}$ ,

i.e. $T\left(\ \frac 1y+\frac 1z\ ,\ \frac 1z+\frac 1x\ ,\ \frac 1x+\frac 1y\ \right)$



AP3. Let $DEF$ be the cevian triangle of the point $P$ w.r.t. $\triangle ABC$ and let $L$ , $M$ , $N$ be midpoints of $[BC]$ , $[CA]$ , $[AB]$ respectively. Denote midpoints

$X$ , $Y$ , $Z$ of $[EF]$ , $[DF]$ , $[DE]$ . Prove that $LX$ , $MY$ , $NZ$ are concurrently. If $T\in LX\cap MY\cap NZ$ and $P(x,y,z)$ , then $\boxed{T\left(1+x,1+y,1+z\right)}$ .


Proof. Observe that $L(0,1,1)$ , $E(x,0,z)$ , $F(x,y,0)$ and $X\left[x(1+x), y(1-y) , z(1-z)\right]$ . Thus, $T(\alpha , \beta , \gamma )\in XL\iff$ $\left|\begin{array}{ccc}
\alpha  & \beta & \gamma\\\\
x(1+x) & y(1-y) & z(1-z)\\\\
0 & 1 & 1\end{array}\right|=0\iff$

$(y-z)\alpha -(1+x)\beta +(1+x)\gamma =0$ . Hence $T\in LX\cap MY\cap NZ\iff$ the homogenous system with $\Delta=\left|\begin{array}{ccc}
y-z & -(1+x) & 1+x\\\\
1+y & z-x & -(1+y)\\\\
-(1+z) & 1+z & x-y\end{array}\right|$

has at least solution, i.e. $\Delta =0$ what prove easily. The system $\left\{\begin{array}{ccc}
(y-z)\alpha -(1+x)\beta +(1+x)\gamma & = & 0\\\\
(1+y)\alpha +(z-x)\beta -(1+y)\gamma & = & 0\end{array}\right\|=0$ has the solution

$\frac {\alpha}{\left|\begin{array}{cc}
-(1+x) & 1+x\\\\
z-x & -(1+y)\end{array}\right|}=\frac {\beta}{\left|\begin{array}{cc}
1+x & y-z\\\\
-(1+y) & 1+y\end{array}\right|}=\frac{\gamma}{\left|\begin{array}{cc}
y-z & -(1+x)\\\\
1+y & z-x\end{array}\right|}$ , i.e. $T\left(1+x,1+y,1+z\right)$



AP4. Ascertain the area $[KIG]=f(S,a,b,c)$ , where $K\ ,\ I\ ,\ G$ are the Lemoine's (symmedian) point, the incenter and the centroid of $\triangle ABC$ respectively.

Proof. $\left\{\begin{array}{c}
G(1,1,1)\\\\
I(a,b,c)\\\\
K\left(a^2,b^2,c^2\right)\end{array}\right\|\implies$ $(1+1+1)(a+b+c)\left(a^2+b^2+c^2\right)\cdot [KIG]=\mod\left|\begin{array}{ccc}
1 & 1 & 1\\\\
a & b & c\\\\
a^2 & b^2 & c^2\end{array}\right|\cdot [ABC]\implies$ $\boxed{[KIG]=\frac {|(a-b)(b-c)(c-a)|S}{3(a+b+c)\left(a^2+b^2+c^2\right)}}$ .


AP5 (Miguel Ochoa Sanchez). Let $\triangle ABC$ with the circumcircle $w=\mathbb C(O,R)$ and a point $P(\alpha ,\beta ,\gamma )_w$ , where $\alpha +\beta +\gamma =1$ . Prove that $(\forall )\ S$

there is the identity $\alpha\cdot p_sA+\beta\cdot p_sB+\gamma\cdot p_sC+p_wP=0$ , where $p_sX$ is the power of the point $X$ w.r.t. the circle $s=\mathbb (S,\rho )$ and $SP=\rho$ .


Proof. $\alpha\cdot p_sA+\beta\cdot p_sB+\gamma\cdot p_sC=$ $\sum\alpha\cdot \left(AS^2-SP^2\right)=$ $\sum\alpha\cdot SA^2-SP^2=-p_wP\implies$ $\alpha\cdot p_sA+\beta\cdot p_sB+\gamma\cdot p_sC+p_wP=0\ .$

Remark. Let $\triangle ABC$ with the circumcircle $w$ and two fixed points $F$ and $P(\alpha ,\beta ,\gamma )_w$ , where $\alpha +\beta +\gamma =1$ . Consider a variable circle

$s=\mathbb C(M,\rho )$ such that $F\in s$ and $M$ belongs to a fixed line $d\perp PF$ . Prove that the sum $S\equiv \alpha\cdot p_sA+\beta\cdot p_sB+\gamma\cdot p_sC$ is constant
This post has been edited 90 times. Last edited by Virgil Nicula, Nov 24, 2016, 3:35 PM

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