83. A metrical characterization of concurrence.

by Virgil Nicula, Aug 7, 2010, 9:27 PM

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=360757

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=38&t=367293

Let $\triangle ABC$ with area $[ABC]=1$ and lines $d_k$ , $k\in \overline{1,3}$ . Denote some intersections of given lines with sidelines of $\triangle ABC$ :

$\left\{\begin{array}{cccccccc}
d_1\equiv MN\ : & M\in AB\cap d_1 & ; & \overline {MB}=m\cdot\overline {MA} & ; & N\in AC\cap d_1 & ; & \overline {NC}=n\cdot\overline {NA}\\\
d_2\equiv PQ\ : & P\in BC\cap d_2 & ; & \overline {PC}=p\cdot\overline {PB} & ; & Q\in BA\cap d_2 & ; & \overline {QA}=q\cdot\overline {QB}\\\
d_3\equiv RS\ : & R\in CA\cap d_3 & ; & \overline {RA}=r\cdot\overline {RC} & ; & S\in CB\cap d_3 & ; & \overline {SB}=s\cdot\overline {SC}\end{array}\right|$

$\boxed{\blacktriangleright d_1\cap d_2\cap d_3\ne\emptyset\Longleftrightarrow\Delta\equiv\left|\begin{array}{ccc}
1 & m & n\\\
q & 1 & p\\\
r & s & 1\end{array}\right|=0\Longleftrightarrow 1+mpr+nqs=mq+ps+nr}$

$\boxed{\blacktriangleright\left|\begin{array}{c}
X\in d_2\cap d_3\\\
Y\in d_3\cap d_1\\\
Z\in d_1\cap d_2\end{array}\right|\implies [XYZ]=\frac {1}{\lambda}\cdot\Delta^2=\frac {1}{\lambda}\cdot \left|\begin{array}{ccc}
1 & m & n\\\
q & 1 & p\\\
r & s & 1\end{array}\right|^2}$

where $\lambda =(1+pr+qs-ps-q-r)(1+ns+mr-nr-m-s)(1+mp+nq-mq-n-p)$ .


Proof 1 (without the second point of conclusion).[/size] Choose an origin $O$ of the vectorial system and for any point $P$ denote $\overrightarrow {OX}=X$ . Thus,

$\left\|\begin{array}{ccc} 
(1-m)M=B-mA & ; & (1-n)N=C-nA\\\
(1-p)P=C-pB & ; & (1-q)Q=A-qB\\\
(1-r)R=A-rC & ; & (1-s)S=B-sC\end{array}\right\|$ . Define $U\in AC\cap PQ$ , $V\in AB\cap d_3$ and $\left\|\begin{array}{c}
Y\in d_1\cap d_3=MN\cap SR\\\
Z\in d_1\cap d_2=MN\cap PQ\end{array}\right\|$ .

$1\blacktriangleright$ Observe that $\frac {\overline{RA}}{\overline{RN}}=\frac {A-R}{N-R}=$ $\frac {A-\frac {A-rC}{1-r}}{\frac {C-nA}{1-n}-\frac {A-rC}{1-r}}$ $\implies$ $\boxed {\frac {\overline{RA}}{\overline{RN}}=\frac {r(1-n)}{1-rn}}\ (1)$ . Apply Menelaus' theorem to $\overline{VRS}/ABC$ $\implies$ $\frac {\overline{VB}}{\overline{VA}}\cdot\frac {\overline {RA}}{\overline{RC}}\cdot\frac {\overline{SC}}{\overline {SB}}=1$ $\implies$ $\frac {\overline{VB}}{\overline{VA}}=\frac sr$ $\implies$ $V=\frac {sA-rB}{s-r}$ $\implies$ $\frac {\overline{VM}}{\overline{VA}}=\frac {M-V}{A-V}=\frac {\frac {B-mA}{1-m}-\frac {sA-rB}{s-r}}{A-\frac {sA-rB}{s-r}}$ $\implies$ $\boxed {\frac {\overline{VM}}{\overline{VA}}=\frac {s-mr}{r(1-m)}}\ (2)$ . Apply Menelaus' theorem in $\overline{VRY}/AMN$ $\implies$ $\frac {\overline{YM}}{\overline{YN}}\cdot\frac {\overline {RN}}{\overline{RA}}\cdot\frac {\overline{VA}}{\overline {VM}}=1$ $\implies$ $\frac {\overline{YM}}{\overline{YN}}=\frac {\overline{RA}}{\overline{RN}}\cdot\frac{\overline{VM}}{\overline{VA}}$ $\stackrel{1 \wedge 2}{\ \implies\ }$ $\boxed {\frac {\overline{YM}}{\overline{YN}}=\frac {(1-n)(s-mr)}{(1-m)(1-nr)}}\ (3)$ .

$2\blacktriangleright$ Observe that $\frac {\overline{QM}}{\overline{QA}}=\frac {M-Q}{A-Q}=$ $\frac {\frac {B-mA}{1-m}-\frac {A-qB}{1-q}}{A-\frac {A-qB}{1-q}}$ $\implies$ $\boxed {\frac {\overline{QM}}{\overline{QA}}=\frac {1-mq}{q(1-m)}}\ (4)$ . Apply Menelaus' theorem in $\overline{UPQ}/ABC$ $\implies$ $\frac {\overline{UA}}{\overline{UC}}\cdot\frac {\overline {PC}}{\overline{PB}}\cdot\frac {\overline{QB}}{\overline {QA}}=1$ $\implies$ $\frac {\overline{UA}}{\overline{UC}}=\frac qp$ $\implies$ $U=\frac {pA-qC}{p-q}$ $\implies$ $\frac {\overline{UA}}{\overline{UN}}=\frac {A-U}{N-U}=\frac {A-\frac {pA-qC}{p-q}}{\frac {C-nA}{1-n}-\frac {pA-qC}{p-q}}$ $\implies$ $\boxed {\frac {\overline{UA}}{\overline{UN}}=\frac {q(1-n)}{p-nq}}\ (5)$ .
Apply Menelaus' theorem in $\overline{UQZ}/AMN$ $\implies$ $\frac {\overline{ZM}}{\overline{ZN}}\cdot\frac {\overline {UN}}{\overline{UA}}\cdot\frac {\overline{QA}}{\overline {QM}}=1$ $\implies$ $\frac {\overline{ZM}}{\overline{ZN}}=\frac {\overline{QM}}{\overline{QA}}\cdot\frac{\overline{UA}}{\overline{UN}}$ $\stackrel{4\wedge 5}{\ \implies\ }$ $\boxed {\frac {\overline{ZM}}{\overline{ZN}}=\frac {(1-n)(1-mq)}{(1-m)(p-nq)}}\ (6)$ .
Thus, $d_1\cap d_2\cap d_3\ne\emptyset\Longleftrightarrow$ $\frac {\overline{YM}}{\overline{YN}}=\frac {\overline{ZM}}{\overline{ZN}}$ $\stackrel{3\wedge 6}{\ \Longleftrightarrow\ }$ $\frac {(1-n)(s-mr)}{(1-m)(1-nr)}=$ $\frac {(1-n)(1-mq)}{(1-m)(p-nq)}$ $\Longleftrightarrow$ $1+mpr+nqs =mq+ps+nr$ .


Proof 2 (with the second point of conclusion). $\left|\begin{array}{ccc}
\frac {\overline{MB}}{\overline {MA}}=m & \implies & M(-m,1,0)\\\\
\frac {\overline{NC}}{\overline{NA}}=n & \implies & N(-n,0,1)\end{array}\right|\implies$ equation of $d_1$ is $\left|\begin{array}{ccc}
x & y & z\\\
-m & 1 & 0\\\
-n & 0 & 1\end{array}\right|$ , i.e. $x+my+nz=0$

$\left|\begin{array}{ccc}
\frac {\overline{PC}}{\overline {PB}}=p & \implies & P(0,-p,1)\\\\
\frac {\overline{QA}}{\overline{QB}}=q & \implies & Q(1,-q,0)\end{array}\right|\implies$ equation of $d_2$ is $\left|\begin{array}{ccc}
x & y & z\\\
0 & -p & 1\\\
 1 & -q & 0\end{array}\right|$ , i.e. $qx+y+pz=0$

$\left|\begin{array}{ccc}
\frac {\overline{RA}}{\overline {RC}}=r & \implies & R(1,0,-r)\\\\
\frac {\overline{SB}}{\overline{SC}}=s & \implies & S(0,1,-s)\end{array}\right|\implies$ equation of $d_3$ is $\left|\begin{array}{ccc}
x & y & z\\\
1 & 0 & -r\\\
0 & 1 & -s\end{array}\right|$ , i.e. $rx+sy+z=0$

$\left|\begin{array}{cccc}
X\in d_2\cap d_3\ : & \left|\begin{array}{c}
qx+y+pz=0\\\
rx+sy+z=0\end{array}\right| & \implies & X(1-ps,pr-q,qs-r)\\\\
Y\in d_3\cap d_1\ : & \left|\begin{array}{c}
rx+sy+z=0\\\
x+my+nz=0\end{array}\right| & \implies & Y(sn-m,1-nr,rm-s)\\\\
Z\in d_1\cap d_2\ : & \left|\begin{array}{c}
x+my+nz=0\\\
qx+y+pz=0\end{array}\right| & \implies & Z(mp-n,nq-p,1-mq)\end{array}\right|$ .

In conclusion, $d_1\cap d_2\cap d_3\ne\emptyset\Longleftrightarrow$ $\Delta\equiv\left|\begin{array}{ccc}
1 & m & n\\\
q & 1 & p\\\
r & s & 1\end{array}\right|=0$ $\Longleftrightarrow$ $1+mpr+nqs=mq+ps+nr$ and

$[XYZ]=\frac {1}{\lambda}\cdot \left|\begin{array}{ccc}
1-ps & pr-q & qs-r\\\\
ns-m & 1-nr & mr-s\\\\
mp-n & nq-p & 1-mq\end{array}\right|\stackrel{(*)}{=}\frac {1}{\lambda}\left|\begin{array}{ccc}
1 & m & n\\\\
q & 1 & p\\\\
r & s & 1\end{array}\right|^2$

where $\lambda =(1+pr+qs-ps-q-r)(1+ns+mr-nr-m-s)(1+mp+nq-mq-n-p)$ .

$(*)$ Indeed, observe that $\left|\begin{array}{ccc}
1-ps & pr-q & qs-r\\\\
ns-m & 1-nr & mr-s\\\\
mp-n & nq-p & 1-mq\end{array}\right|\ \begin{array}{c}
C_1:=q\cdot C_1\ \Downarrow\\\\
C_1:=C_1+C_2+p\cdot C_3\end{array}$

$=\ \frac {1}{q}$ $\cdot \left|\begin{array}{ccc}
0 & pr-q & qs-r\\\\
\Delta & 1-nr & rm-s\\\\
0 & nq-p & 1-mq\end{array}\right|=$ $-\frac {\Delta}{q}\cdot\left|\begin{array}{cc}
pr-q & qs-r\\\\
nq-p & 1-mq\end{array}\right|=$ $\frac {-\Delta}{q}\cdot \left(-q\Delta\right)=$ $\Delta ^2$ .

Remark. $\begin{array}{ccc}\left|\begin{array}{ccc}1-ps & pr-q & qs-r\\\\ ns-m & 1-nr & mr-s\\\\ mp-n & nq-p & 1-mq\end{array}\right|=\left|\begin{array}{ccc}1 & m & n\\\\ q & 1 & p\\\\ r & s & 1\end{array}\right|^{2}=\left|\begin{array}{cccc} 1+m^2+n^2 & m+q+np & n+r+ms \\\\ m+q+np & 1+p^2+q^2 & p+s+q r \\\\ n+r+ms & p+s+qr & 1+r^2+s^2\end{array}\right|\end{array}$ .


Quote:
Let $D\in BC$ , $E\in CA$ , $F\in AB$ be the middlepoints of the sides in the triangle $ABC\ .$ For $P$ define $X\in AP\cap BC$ , $Y\in BP\cap CA$ , $Z\in CP\cap AB\ .$

Prove that the lines $DZ$ , $EX$ , $FY$ are concurrently if and only if $\boxed{\ \frac{XB}{XC}+\frac{YC}{YA}+\frac{ZA}{ZB}=1\ }\ .$ See http://www.artofproblemsolving.com/Forum/viewtopic.php?t=114378

Two remarkable particular cases.

$1\blacktriangleright$ $R: =A\ ,$ $M: =B$ and $P: =C\ ,$ i.e. $d_{1}=BN\ ,$ $d_{2}=CQ$ and $d_{3}=AS$ $\Longrightarrow$ $m=p=r=0$ $\Longrightarrow$ $1+nqs=0$ $\Longrightarrow$ $\frac{NC}{NA}\cdot \frac{QA}{QB}\cdot\frac{SB}{SC}=-1$ .

$2\blacktriangleright$ $P: =C$ and $S: =B\ ,$ i.e. $d_{2}=CQ$ and $d_{3}=BR$ $\Longrightarrow$ $p=s=0$ $\Longrightarrow$ $mq+rn=1$ $\Longrightarrow$ $\boxed{\ \frac{MB}{MA}\cdot \frac{QA}{QB}+\frac{RA}{RC}\cdot \frac{NC}{NA}=1\ }\ .$

Apply the Menelaus' theorem to $\overline{CXQ}/ABY$ and $\overline{BXR}/ACY$ : $\left\{\begin{array}{c}\frac{CY}{CB}\cdot \frac{QB}{QA}\cdot \frac{XA}{XY}=1\Longrightarrow\frac{QA}{QB}=\frac{CY}{CB}\cdot\frac{XA}{XY}\\\\ \frac{BY}{BC}\cdot\frac{RC}{RA}\cdot\frac{XA}{XY}=1\Longrightarrow\frac{RA}{RC}=\frac{BY}{BC}\cdot\frac{XA}{XY}\end{array}\right\|$ $\Longrightarrow$

$\frac{MB}{MA}\cdot\frac{CY}{CB}\cdot\frac{XA}{XY}+\frac{NC}{NA}\cdot\frac{BY}{BC}\cdot\frac{XA}{XY}=1\Longrightarrow$ $\boxed{\ \frac{MB}{MA}\cdot YC+\frac{NC}{NA}\cdot BY=\frac{XY}{XA}\cdot BC\ }\ .$
This post has been edited 60 times. Last edited by Virgil Nicula, Nov 23, 2015, 3:23 PM

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