198. Generalized Carnot's lemma.

by Virgil Nicula, Dec 26, 2010, 8:12 AM

Generalized Carnot's lemma. Let $ABC\ ,\ MNP$ be two triangles. Prove that the following sentencies are equivalently :

$\blacktriangleright$ The perpendicular lines from the points $A\ ,\ B\ ,\ C$ on the lines $NP\ ,\ PM\ ,\ MN$ respectively are concurrently.

$\blacktriangleright$ The perpendicular lines from the points $M\ ,\ N\ ,\ P$ on the lines $BC\ ,\ CA\ ,\ AB$ respectively are concurrently.

$\blacktriangleright$ Exists the relation $\boxed {\ AN^2+BP^2+CM^2=AP^2+BM^2+CN^2\ }$ .

Proof. Denote $\left\|\begin{array}{ccc} X\in NP & ; & AX\perp NP\\\ Y\in PM & ; & BY\perp PM\\\ Z\in MN & ; & CZ\perp MN\end{array}\right\|$ and $\left\|\begin{array}{ccc} U\in BC & ; & MU\perp BC\\\ V\in CA & ; & NV\perp CA\\\ W\in AB & ; & PW\perp AB\end{array}\right\|$ . Observe that

$\left\|\begin{array}{ccc} AX\perp PN & \iff & XN^2-XP^2=AN^2-AP^2\\\ BY\perp MP & \iff & YP^2-YM^2=BP^2-BM^2\\\ CZ\perp NM & \iff & ZM^2-ZN^2=CM^2-CN^2\end{array}\right\|$ and $\left\|\begin{array}{ccc} MU\perp BC & \iff & MC^2-MB^2=UC^2-UB\\\ NV\perp CA & \iff & NA^2-NC^2=VA^2-VC^2\\\ PW\perp AB & \iff & PB^2-PA^2=WB^2-WA^2\end{array}\right\|$ .

Therefore, $\boxed{AX\cap BY\cap CZ\ne\emptyset}$ $\stackrel{\mathrm{(Carnot's\ lemma\ MNP)}}{\iff}$ $\left(XN^2-XP^2\right)+\left(YP^2-YM^2\right)+\left(ZM^2-ZN^2\right)=0$ $\iff$

$\left(AN^2-AP^2\right)+\left(BP^2-BM^2\right)+\left(CM^2-CN^2\right)=0$ $\iff$ $\boxed {AN^2+BP^2+CM^2=AP^2+BM^2+CN^2}$ $\iff$

$\left(NA^2-NC^2\right)+\left(PB^2-PA^2\right)+\left(MC^2-MB^2\right)=0$ $\iff$ $\left(VA^2-VC^2\right)+\left(WB^2-WA^2\right)+\left(UC^2-UB^2\right)=0$ $\stackrel{\mathrm{(Carnot's \ lemma\ ABC)}}{\iff}$

$\boxed{MU\cap NV\cap PW\ne\emptyset}$ . In this case the triangles $ABC$ and $MNP$ are orthologically and write $ABC\ \boxed *\ MNP$ . The points $K\in AX\cap BY\cap CZ$ ,

$L\in MU\cap NV\cap PW$ are named the orthology centers for given triangles. Prove easily that $\boxed *$ is an equivalence relation (reflexive, symmetric and transitive)

and $\boxed{\ ABC\ \boxed *\ MNP\ \iff\ (\forall )\ S\ ,\ \overline{SM}\cdot\overline{BC}+\overline{SN}\cdot\overline{CA}+\overline{SP}\cdot\overline{AB}=\overline 0\ \ \vee\ \ \overline{SA}\cdot\overline{NP}+\overline{SB}\cdot\overline{PM}+\overline{SC}\cdot\overline{MN}=\overline 0\ }$ .
This post has been edited 19 times. Last edited by Virgil Nicula, Nov 22, 2015, 5:28 PM

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