Euler's theorem (geometry)

Euler's theorem:
${\displaystyle d=|UI|={\sqrt {R(R-2r)}}}$

In geometry , Euler's theorem , named after Leonhard Euler , describes a formula for the distance between the centers of the circumference and the incircle of a triangle . ${\displaystyle d}$

${\displaystyle d^{2}=R(R-2r)\,}$

This relationship is also often represented using fractions in the following equivalent equation:

${\displaystyle {\frac {1}{R+d}}+{\frac {1}{R-d}}={\frac {1}{r}}}$

It denotes the circumferential radius and the incircle radius. ${\displaystyle R}$ ${\displaystyle r}$

Euler's inequality follows directly from the theorem :

${\displaystyle R\geq 2r}$

proof

Sketch for proof

Let the center of the circle and the center of the circle of the triangle be . As an angle bisector, the straight line intersects the circumference according to the south pole law at a point that is also on the associated perpendicular . The second point of intersection of this perpendicular line ( ) with the circumference is . Designating the low end of of of precipitated solder to with , then applies . ${\displaystyle U}$ ${\displaystyle I}$ ${\displaystyle ABC}$ ${\displaystyle AI}$ ${\displaystyle L}$ ${\displaystyle UL}$ ${\displaystyle M}$ ${\displaystyle I}$ ${\displaystyle AB}$ ${\displaystyle D}$ ${\displaystyle {\overline {ID}}=r}$

Because of the correspondence in two angles ( ( theorem of circumferential angles ) and (the perpendicular and theorem of Thales )) the triangles and are similar to each other . Therefore, and on . This shows: ${\displaystyle \angle DAI=\angle BML}$ ${\displaystyle \angle ADI=\angle MBL=90^{\circ }}$ ${\displaystyle ADI}$ ${\displaystyle MBL}$ ${\displaystyle {\overline {ID}}/{\overline {LB}}={\overline {AI}}/{\overline {ML}}}$ ${\displaystyle {\overline {ML}}\cdot {\overline {ID}}={\overline {AI}}\cdot {\overline {LB}}}$

${\displaystyle 2Rr={\overline {AI}}\cdot {\overline {LB}}}$

If you connect with , you can use the set of external angles , according to which an external angle ( ) of a triangle ( ) is as large as the two non-adjacent internal angles : ${\displaystyle B}$ ${\displaystyle I}$ ${\displaystyle \angle BIL}$ ${\displaystyle ABI}$

${\displaystyle \angle BIL={\frac {\alpha }{2}}+{\frac {\beta }{2}}}$

In addition, with the help of the circumferential angle set

${\displaystyle \angle LBI={\frac {\beta }{2}}+\angle LBC={\frac {\beta }{2}}+\angle LAC={\frac {\beta }{2}}+{\frac {\alpha }{2}}}$,

from which it follows. So triangle is isosceles ; it applies . From what has already been proven one obtains ${\displaystyle \angle BIL=\angle LBI}$ ${\displaystyle IBL}$ ${\displaystyle {\overline {LI}}={\overline {LB}}}$

${\displaystyle {\overline {AI}}\cdot {\overline {LI}}=2Rr}$.

Now let and be the intersection of the straight line with the circumference Applying the chord theorem results ${\displaystyle P}$ ${\displaystyle Q}$ ${\displaystyle UI}$

${\displaystyle {\overline {PI}}\cdot {\overline {QI}}={\overline {AI}}\cdot {\overline {LI}}=2Rr}$.

The route lengths on the left can be expressed using the circumferential radius and the distance : ${\displaystyle R}$ ${\displaystyle d={\overline {UI}}}$

${\displaystyle (R+d)\cdot (R-d)=2Rr}$

A short transformation gives the claim:

${\displaystyle R^{2}-d^{2}=2Rr\,}$

${\displaystyle d^{2}=R^{2}-2Rr=R(R-2r)\,}$

Related statements

Is the radius of the side associated Ankreises , then for the distance between the center of this Ankreises and the circumcenter: ${\displaystyle r_{a}}$ ${\displaystyle a}$ ${\displaystyle d_{a}}$

${\displaystyle {d_{a}}^{2}=R(R+2r_{a})}$

The same applies to the other two arrivals.

The set of foot provides a set of analog Euler statement for tendons tangential quadrilateral.

history

The set is named after Euler, who published it in 1765. The English surveyor William Chapple had published the same result in an English magazine as early as 1746.

Euler's inequality in absolute geometry

Euler's inequality, in the form that asserts that the maximum of the incircular radii of all triangles inscribed in a given circle is only reached for an equilateral triangle, is valid in absolute geometry .

literature

• Günter Aumann : Circular Geometry: An Elementary Introduction . Springer, 2015, ISBN 978-3-662-45306-3 , pp. 137-140 ( excerpt (Google) )
• Gerry Leversha, GC Smith: Euler and Triangle Geometry . In: The Mathematical Gazette , Vol. 91, No. 522, Nov., 2007, pp. 436-452 ( JSTOR 40378417 )
• Roger B. Nelsen: Euler's Triangle Inequality via Proofs without Words . In: Mathematics Magazine , Vol. 81, No. 1, Feb., 2008, pp. 58-61 ( JSTOR 27643082 )
• Victor Pambuccian, Celia Schacht: Euler's inequality in absolute geometry . In: Journal of Geometry , Vol. 109, Art. 8, 2018, pp. 1–11

Individual evidence

1. ↑ Victor Pambuccian, Celia Schacht: Euler's inequality in absolute geometry. In: Journal of Geometry Vol. 109, 2018, Art. 8, pp. 1–11.

Web link

• Eric W. Weisstein : Euler Triangle Formula . In: MathWorld (English).