Euler's theorem: d = | I O | = R ( R − 2 r ) {\displaystyle 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 by d 2 = R ( R − 2 r ) {\displaystyle d^{2}=R(R-2r)} or equivalently 1 R − d + 1 R + d = 1 r , {\displaystyle {\frac {1}{R-d}}+{\frac {1}{R+d}}={\frac {1}{r}},} where R {\displaystyle R} and r {\displaystyle 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. However, the same result was published earlier by William Chapple in 1746.

From the theorem follows the Euler inequality: R ≥ 2 r , {\displaystyle R\geq 2r,} which holds with equality only in the equilateral case.

Proof

Proof of Euler's theorem

Let O {\displaystyle O} be the center of the circumcircle of triangle A B C {\displaystyle ABC}, and let I {\displaystyle I} be the center of its incircle.

If the ray A I {\displaystyle AI} intersects the circumcircle at the point L {\displaystyle L}, then L {\displaystyle L} is the midpoint of arc B C {\displaystyle BC}. Draw the ray L O {\displaystyle LO} and denote its intersection with the circumcircle by M {\displaystyle M}.

Then L M {\displaystyle LM} is a diameter of the circumcircle. Drop the perpendicular I D {\displaystyle ID} from the point I {\displaystyle I} to A B {\displaystyle AB}. Then I D = r {\displaystyle ID=r}. Rewrite Euler's formula in the following form:

d 2 − R 2 = − 2 r R {\displaystyle d^{2}-R^{2}=-2rR}

Notice that the left-hand side is the power of the point I {\displaystyle I} with respect to the circumcircle.

Therefore, it suffices to prove the equality L I ⋅ I A = 2 R r {\displaystyle LI\cdot IA=2Rr}.

By the trident lemma, L I = L B , {\displaystyle LI=LB,} so it suffices to prove that L B ⋅ I A = 2 R r {\displaystyle LB\cdot IA=2Rr}.

Now observe that 2 R = L M {\displaystyle 2R=LM} and r = I D , {\displaystyle r=ID,} that is, the required equality can be rewritten as L B ⋅ I A = L M ⋅ I D . {\displaystyle LB\cdot IA=LM\cdot ID.} Rewriting it once more, we obtain L B / L M = I D / I A {\displaystyle LB/LM=ID/IA}.

This equality follows from the similarity of triangles △ A I D {\displaystyle \triangle AID} and △ M L B {\displaystyle \triangle MLB}.

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

Stronger version of the inequality

A stronger version is R r ≥ a b c + a 3 + b 3 + c 3 2 a b c ≥ a b + b c + c a − 1 ≥ 2 3 ( a b + b c + c a ) ≥ 2 , {\displaystyle {\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,} where a {\displaystyle a}, b {\displaystyle b}, and c {\displaystyle c} are the side lengths of the triangle.

Euler's theorem for the escribed circle

If r a {\displaystyle r_{a}} and d a {\displaystyle d_{a}} denote respectively the radius of the escribed circle opposite to the vertex A {\displaystyle A} and the distance between its center and the center of the circumscribed circle, then d a 2 = R ( R + 2 r a ) {\displaystyle 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.

See also

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