Carnot's Theorem (Lote)

Carnot's theorem for plumb bobs on triangle sides:
blue area = red area

The set of Carnot (after Lazare Nicolas Marguerite Carnot ) provides a necessary and sufficient condition for whether three straight line on the three (extended) sides of a triangle are perpendicular, intersect at a point. In addition, it can also be understood as a generalization of the Pythagorean theorem .

statement

For a triangle with sides , three straight lines are given, each of which is perpendicular to an (extended) side of the triangle and which intersect at a common point . Denoting the base points on the (extended) triangle sides with , then the following equation holds: ${\displaystyle \triangle ABC}$ ${\displaystyle a,b,c}$ ${\displaystyle F}$ ${\displaystyle a,b,c}$ ${\displaystyle P_{a},P_{b},P_{a}}$

${\displaystyle |AP_{c}|^{2}+|BP_{a}|^{2}+|CP_{b}|^{2}=|BP_{c}|^{2}+|CP_{a}|^{2}+|AP_{b}|^{2}}$.

The reverse of this proposition also applies, that is: If the base points of three perpendiculars meet the above equation, then they intersect at a common point.

Special cases

If the triangle has a right angle and the point of intersection lies on one of the two corner points or , one obtains the Pythagorean theorem. Is, for example, on , then applies , , , , and , and supplies the above equation . ${\displaystyle \triangle ABC}$ ${\displaystyle C}$ ${\displaystyle F}$ ${\displaystyle A}$ ${\displaystyle B}$ ${\displaystyle F}$ ${\displaystyle A}$ ${\displaystyle |AP_{b}|=0}$ ${\displaystyle |AP_{c}|=0}$ ${\displaystyle |CP_{a}|=0}$ ${\displaystyle |CP_{b}|=b}$ ${\displaystyle |BP_{a}|=a}$ ${\displaystyle |BP_{c}|=c}$ ${\displaystyle a^{2}+b^{2}=c^{2}}$

If the three straight lines are perpendicular to the center , then , and . Hence the above equation exists and as a special case we get the theorem that the perpendiculars of a triangle intersect at a point. ${\displaystyle |AP_{c}|=|BP_{c}|}$ ${\displaystyle |BP_{a}|=|CP_{a}|}$ ${\displaystyle |CP_{b}|=|AP_{b}|}$

If the three straight lines are the extensions of the triangular heights , the straight lines run through the corner points. The height divides the triangle into two right-angled triangles, for which the Pythagorean theorem provides the equations and , and follows by taking the difference . The relationships and follow in exactly the same way or by mentally rotating the triangle . If you add these three relationships, you get ${\displaystyle CP_{c}}$ ${\displaystyle |AP_{c}|^{2}+|CP_{c}|^{2}=b^{2}}$ ${\displaystyle |BP_{c}|^{2}+|CP_{c}|^{2}=a^{2}}$ ${\displaystyle |AP_{c}|^{2}-|BP_{c}|^{2}=b^{2}-a^{2}}$ ${\displaystyle |BP_{a}|^{2}-|CP_{a}|^{2}=c^{2}-b^{2}}$ ${\displaystyle |CP_{b}|^{2}-|AP_{b}|^{2}=a^{2}-c^{2}}$

${\displaystyle |AP_{c}|^{2}-|BP_{c}|^{2}+|BP_{a}|^{2}-|CP_{a}|^{2}+|CP_{b}|^{2}-|AP_{b}|^{2}=b^{2}-a^{2}+c^{2}-b^{2}+a^{2}-c^{2}=0}$,

that is, the equation consists of the above sentence. So we also get the theorem of the vertical intersection as a special case of Carnot's theorem.

literature

• Martin Wohlgemuth (Ed.): Mathematical for advanced beginners. More popular posts from Matroids Matheplanet. Springer, 2010, ISBN, 9783827426079, pp. 273-276.

Web links

• Florian Modler: Forgotten sentences on the triangle - Carnot's sentence. On: Matroids Math Planet.