Generalized Carnot's theorem

Carnot's theorem is a direct application of the theorem of Pythagoras. It's a nice little theorem that is unlikely to be rediscovered with dynamic geometry software. However, Miguel Ochoa Sanchez (Peru) has come up with an interesting generalization.

Miguel Ochoa Sanchez

19 December 2015

Point $O$ is within $\Delta ABC.\;$ Let points $D,$ $E,$ and $F$ be located on the sides $BC,$ $AC,$ and, respectively, $AB$ of the triangle such that the three angles $AEO,$ $BFO,$ and $CDO$ are all equal to, say, $\phi.$

Generalized Carnot's theorem

Then

$AF^{2} - BF^{2} + BD^{2} - CD^{2} + CE^{2} - AE^{2} = 4[\Delta ABC]\cdot\cot\phi,$

where $[\Delta ABC]$ is the area of $\Delta ABC].$

The proof depends on the following

Lemma

Let $D$ be on the side of $BC$ of $\Delta ABC;$ $\angle ADC=\phi.$

Generalized Carnot's theorem - lemma

Then

$CD^{2} - BD^{2} = AC^{2} - AB^{2} + 4[\Delta ABC]\cdot\cot\phi,$

Proof of Lemma

Drop the perpendicular $AH$ to $BC.$

Generalized Carnot's theorem - proof of lemma

Then $DH=AH\cdot\cot\phi,$ $CH=CD-AH\cdot\cot\phi.$ By the Pythagorean theorem,

$\begin{align} AC^{2}-AB^{2}&=CH^{2}-BH^{2}\\ &=(CD-AH\cdot\cot\phi)^{2}-(BD+AH\cdot\cot\phi)^{2}\\ &=CD^{2}-BD^{2}-2\cdot CD\cdot AH\cdot\cot\phi-2\cdot BD\cdot AH\cdot\cot\phi\\ &=CD^{2}-BD^{2}-2\cdot BC\cdot AH\cdot\cot\phi\\ &=CD^{2}-BD^{2}-4\cdot[\Delta AB]\cdot\cot\phi. \end{align}$

Proof of Theorem

We'll use the notations as in the diagram below:

Generalized Carnot's theorem - proof of theorem

Apply the lemma to triangles $BOC,$ $AOB,$ $AOC$ to obtain

$\begin{align} CD^{2}-BD^{2}&=m^{2}-n^{2}+4[\Delta BOC]\cot\phi\\ BF^{2}-AF^{2}&=n^{2}-l^{2}+4[\Delta BOC]\cot\phi\\ AE^{2}-CE^{2}&=l^{2}-m^{2}+4[\Delta BOC]\cot\phi. \end{align}$

Add the three up:

$\begin{align} CD^{2}-BD^{2}&+BF^{2}-AF^{2}+AE^{2}-CE^{2}\\ &=4[\Delta BOC]\cot\phi+4[\Delta BOC]\cot\phi+4[\Delta BOC]\cot\phi\\ &=4[\Delta ABC]\cot\phi. \end{align}$

There is a natural generalization to the case of an $n\text{-gon},$ with $n\ge 3:$

With $P_{n+1}=P_1,$

$\displaystyle\sum_{k=1}^{n}P_iQ_i^2-\sum_{k=1}^{n}Q_iP_{i+1}^2=4[P_1P_2\ldots P_n]\cot\phi.$

Generalized Carnot's theorem - generalization

The proof is analoguous to the case where $n=3.$

|Contact| |Front page| |Contents| |Geometry| |Store|

Copyright © 1996-2017 Alexander Bogomolny

 61657348

Search by google: