Euler's theorem in geometry

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

In geometry, Euler's theorem states that the distance d between the circumcenter and incenter of a triangle is given byTemplate:R $$d^2=R(R-2r)$$ or equivalently $$\frac{1}{R-d}+\frac{1}{R+d}=\frac{1}{r},$$ where $R$ and $r$ denote the circumradius and inradius respectively (the radii of the circumscribed circle and inscribed circle respectively). The theorem is named for Leonhard Euler, who published it in 1765.Template:R However, the same result was published earlier by William Chapple in 1746.Template:R

From the theorem follows the Euler inequality:Template:R $$R\ge 2r,$$ which holds with equality only in the equilateral case.Template:R

Proof

Proof of Euler's theorem

Let $O$ be the center of the circumcircle of triangle $ABC$, and let $I$ be the center of its incircle.

If the ray $AI$ intersects the circumcircle at the point $L$, then $L$ is the midpoint of arc $BC$. Draw the ray $LO$ and denote its intersection with the circumcircle by $M$.

Then $LM$ is a diameter of the circumcircle. Drop the perpendicular $ID$ from the point $I$ to $AB$. Then $ID=r$. Rewrite Euler's formula in the following form:

$d^2-R^2=-2rR$

Notice that the left-hand side is the power of the point $I$ with respect to the circumcircle.

Therefore, it suffices to prove the equality $LI\cdot IA=2Rr$.

By the trident lemma, $LI=LB,$ so it suffices to prove that $LB\cdot IA=2Rr$.

Now observe that $2R=LM$ and $r=ID,$ that is, the required equality can be rewritten as $LB\cdot IA=LM\cdot ID.$ Rewriting it once more, we obtain $LB/LM=ID/IA$.

This equality follows from the similarity of triangles $\mathrm{△}AID$ and $\mathrm{△}MLB$.

Indeed, the angles at $B$ and $D$ in these triangles are right angles, while the angles at $A$ and $M$ are equal because they both subtend the arc $BL$ (moreover, the ratio $LB/LM=ID/IA$ is equal to the sine of the angle $\mathrm{\angle }BAL$).

Stronger version of the inequality

A stronger versionTemplate:R is $$\frac{R}{r}\ge \frac{abc+a^3+b^3+c^3}{2abc}\ge \frac{a}{b}+\frac{b}{c}+\frac{c}{a}-1\ge \frac{2}{3}(\frac{a}{b}+\frac{b}{c}+\frac{c}{a})\ge 2,$$ where $a$, $b$, and $c$ are the side lengths of the triangle.

Euler's theorem for the escribed circle

If $r_a$ and $d_a$ denote respectively the radius of the escribed circle opposite to the vertex $A$ and the distance between its center and the center of the circumscribed circle, then $d_a^2=R(R+2r_a)$.

Euler's inequality in absolute geometry

Euler's inequality, in the form stating that, for all triangles inscribed in a given circle, the maximum of the radius of the inscribed circle is reached for the equilateral triangle and only for it, is valid in absolute geometry.Template:R

See also

• Fuss' theorem for the relation among the same three variables in bicentric quadrilaterals
• Poncelet's closure theorem, showing that there is an infinity of triangles with the same two circles (and therefore the same R, r, and d)
• Egan conjecture, generalization to higher dimensions
• List of triangle inequalities

External links

• Template:Mathworld