Euler's theorem in geometry

In geometry, Euler's theorem states that the distance d between the circumcentre and incentre of a triangle can be expressed as[1][2][3][4]

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

where R and r denote the circumradius and inradius respectively (the radii of the above two circles). The theorem is named for Leonhard Euler, who published it in 1767.[5] However, the same result was published earlier by William Chapple in 1746.[6]

From the theorem follows the Euler inequality:[2][3]

$R \ge 2r,$

which holds with equality only in the equilateral case.[7]:p. 198

Proof

Proof of Euler's theorem in geometry

Letting O be the circumcentre of triangle ABC, and I be its incentre, the extension of AI intersects the circumcircle at L. Then L is the midpoint of arc BC. Join LO and extend it so that it intersects the circumcircle at M. From I construct a perpendicular to AB, and let D be its foot, so ID = r. It is not difficult to prove that triangle ADI is similar to triangle MBL, so ID / BL = AI / ML, i.e. ID × ML = AI × BL. Therefore 2Rr = AI × BL. Join BI. Because

BIL = ∠ A / 2 + ∠ ABC / 2,
IBL = ∠ ABC / 2 + ∠ CBL = ∠ ABC / 2 + ∠ A / 2,

we have ∠ BIL = ∠ IBL, so BL = IL, and AI × IL = 2Rr. Extend OI so that it intersects the circumcircle at P and Q; then PI × QI = AI × IL = 2Rr, so (R + d)(R − d) = 2Rr, i.e. d2 = R(R − 2r).

Stronger version of the inequality

A stronger version[7]:p. 198 is

$\frac{R}{r} \geq \frac{abc+a^3+b^3+c^3}{2abc} \geq \frac{a}{b}+\frac{b}{c}+\frac{c}{a}-1 \geq \frac{2}{3} \left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a} \right) \geq 2.$