126. Routh's theorem.

by Virgil Nicula, Sep 16, 2010, 3:21 AM

PP1. In $\triangle ABC\ \wedge\ \left\{\begin{array}{ccc}
M\in (BC) & , & \frac {MB}{m}=\frac {MC}{1}=\frac {a}{m+1}\\\\
N\in (CA) & , & \frac {NC}{n}=\frac {NA}{1}=\frac {b}{n+1}\\\\
P\in (AB) & , & \frac {PA}{p}=\frac {PB}{1}=\frac {c}{p+1}\end{array}\right\|\ \ \wedge\ \ \left\{\begin{array}{ccc}
X\in BN\cap CP\\\
Y\in CP\cap AM\\\
Z\in AM\cap BN\end{array}\right\|$ .

Prove that there are the relations $[MNP]=\frac {1+mnp}{(1+m)(1+n)(1+p)}\cdot [ABC]$ and $[XYZ]=\frac {(1-mnp)^2}{(1+m+mn)(1+n+np)(1+p+pm)}\cdot [ABC]$ .


Proof. Prove easily the first relation. For second relation apply the Menelaus' theorem to transversals

$\left\{\begin{array}{cccc}
\overline{BZN}/ACM\ : & \frac {BM}{BC}\cdot \frac {NC}{NA}\cdot \frac {ZA}{ZM}=1 & \implies & \frac {ZA}{m+1}=\frac {ZM}{mn}=\frac {AM}{1+m+mn}\\\\
\overline{CXP}/BAN\ : & \frac {CN}{CA}\cdot\frac {PA}{PB}\cdot \frac {XB}{XN}=1 & \implies & \frac {XB}{n+1}=\frac {XN}{np}=\frac {BN}{1+n+np}\\\\
\overline{AYM}/CBP\ : & \frac {AP}{AB}\cdot\frac {MB}{MC}\cdot \frac {YC}{YP}=1 & \implies & \frac {YC}{p+1}=\frac {YP}{pm}=\frac {CP}{1+p+pm}\end{array}\right\|$ .

From the relation $[XYZ]=[ABC]-[ABM]-[BCN]-[CAP]+[CXN]+[AYP]+[BZM]$ obtain

$\frac {[XYZ]}{[ABC]}=\frac {[ABC]}{[ABC]}-\frac {[ABM]}{[ABC]}-\frac {[BCN]}{[ABC]}-$ $\frac {[CAP]}{[ABC]}+\frac {[CXN]}{[ABC]}+$ $\frac {[AYP]}{[ABC]}+\frac {[BZM]}{[ABC]}$ $\implies$

$\frac {[XYZ]}{[ABC]}=1-\frac {[BM]}{[BC]}-\frac {[CN]}{[CA]}-\frac {[AP]}{[AP]}+$ $\frac {[NC]}{[AC]}\cdot\frac {[XN]}{[BN]}+$ $\frac {[PA]}{[BA]}\cdot\frac {[YP]}{[CP]}+\frac {[MB]}{[BC]}\cdot\frac {[ZM]}{[AM]}\implies$

$\frac {[XYZ]}{[ABC]}=1-\frac {m}{m+1}-\frac {n}{n+1}-\frac {p}{p+1}+$ $\frac {n}{n+1}\cdot \frac {np}{1+n+np}$ $+\frac {p}{p+1}\cdot\frac {pm}{1+p+pm}$ $+\frac {m}{m+1}\cdot\frac {mn}{1+m+mn}\implies$

$\frac {[XYZ]}{[ABC]}=1-\frac {m}{1+m+mn}-\frac {n}{1+n+np}-\frac {p}{1+p+pm}$ $\implies$ $\frac {[XYZ]}{[ABC]}=\frac {(1-mnp)^2}{(1+m+mn)(1+n+np)(1+p+pm)}\cdot [ABC]$ .

I used the following relations : $\frac {[\underline ABM]}{[\underline ABC]}=\frac {BM}{BC}\ \ \wedge\ \ \frac {[CXN]}{[ABC]}=\frac {[\underline CXN]}{[\underline CBN]}\cdot\frac {[\underline BNC]}{[\underline BAC]}=\frac {XN}{BN}\cdot\frac {NC}{AC}$ a.s.o.

and $|A\cup B\cup C|=|A|+|B|+|C|-|A\cap B|-|B\cap C|-|C\cap A|+|A\cap B\cap C|$ , where $|X|$ is the number of elements (cardinal) from $X$ .



PP2. $ \triangle\ ABC\ \wedge\ \ \boxed {\ \begin{array}{ccc} D\in (BC) & ; & \frac {DB}{DC} = m \\
 \\
E\in (CA) & ; & \frac {EC}{EA} = n \\
 \\
F\in (AB) & ; & \frac {FA}{FB} = p\end{array}\ \ \wedge\ \ \begin{array}{ccc} X\in (BC) & ; & \frac {XB}{XC} = x \\
 \\
Y\in (CA) & ; & \frac {YC}{YA} = y \\
 \\
Z\in (AB) & ; & \frac {ZA}{ZB} = z\end{array}\ \ \wedge\ \ \begin{array}{c} M\in AD\cap YZ \\
 \\
N\in BE\cap ZX \\
 \\
P\in CF\cap XY\end{array}\ }$ $ \implies$

$ \boxed {\ [MNP] =\frac {(1 + xyz)(1 + mnpxyz)}{[mz(1 + y) + (1 + z)]\cdot [nx(1 + z) + (1 + x)]\cdot [py(1 + x) + (1 + y)]}\cdot [ABC]\ }$ .


Proof. Apply the Routh's relation :

$ \left\|\begin{array}{ccccc} M\in YZ\cap AD & \implies & \frac {MY}{MZ} = \frac {DC}{DB}\cdot\frac {AY}{AZ}\cdot \frac {AB}{AC} & \implies & u = \frac {MY}{MZ} = \frac {1 + z}{mz(1 + y)} \\
 \\
N\in ZX\cap BE & \implies & \frac {NZ}{NX} = \frac {EA}{EC}\cdot\frac {BZ}{BX}\cdot \frac {BC}{BA} & \implies & v = \frac {NZ}{NX} = \frac {1 + x}{nx(1 + z)} \\
 \\
P\in XY\cap CF & \implies & \frac {PX}{PY} = \frac {FB}{FA}\cdot\frac {CX}{CY}\cdot \frac {CA}{CB} & \implies & w = \frac {PX}{PZ} = \frac {1 + y}{py(1 + x)}\end{array}\right\|$ $ \implies\ uvw = \frac {1}{mnpxyz}$ and

$ [MNP] = \frac {1 + uvw}{(1 + u)(1 + v)(1 + w)}\cdot [XYZ] =$ $ \frac {1 + uvw}{(1 + u)(1 + v)(1 + w)}\cdot\frac {1 + xyz}{(1 + x)(1 + y)(1 + z)}\cdot [ABC] \ \implies$

$ [MNP] =$ $ \frac {1 + \frac {1}{mnpxyz}}{\frac {[mz(1 + y) + (1 + z)]\cdot [nx(1 + z) + (1 + x)]\cdot [py(1 + x) + (1 + y)]}{mnpxyz(1 + x)(1 + y)(1 + z)}}\cdot\frac {1 + xyz}{(1 + x)(1 + y)(1 + z)}\ \implies$

$ [MNP] =$ $ \frac {(1 + xyz)(1 + mnpxyz)}{[mz(1 + y) + (1 + z)]\cdot [nx(1 + z) + (1 + x)]\cdot [py(1 + x) + (1 + y)]}\cdot [ABC]$ .
This post has been edited 20 times. Last edited by Virgil Nicula, Nov 23, 2015, 7:33 AM

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