Euler's Formula for the Distance OI

A geometric proof of Euler's formula for the distance between the circumcenter and incenter, using a derived chord product, an isosceles-triangle lemma, and a pair of similar right triangles.

Keywords: Euler's formula, circumcenter, incenter, chord product, Euler's inequality

Prerequisites: incenter-proof, circumcenter-proof, inscribed-angle-proof, thales-theorem-proof · Difficulty: advanced

Euler's formula relates the distance between the circumcenter $O$ and the incenter $I$ to the circumradius $R$ and the inradius $r$:

Since $|OI{|}^2\ge 0$, this forces $R\ge 2r$ — Euler's inequality, with equality only for the equilateral triangle.

Strategy: extend the bisector of angle $A$ to meet the circumcircle at $M$ and derive the chord product $R^2-|OI{|}^2=IA\cdot IM$. Then show $IM=MB$ via an isosceles triangle, and use a pair of similar right triangles to prove $IA\cdot MB=2Rr$ without trigonometry.

Step 1: Circumcircle and Incircle

We start with the circumcircle (center $O$, radius $R$) and the incircle (center $I$, radius $r$). Euler's formula will pin down the distance $|OI|$ between their centers.

Step 2: $|OI{|}^2=R^2-2Rr$

We will prove

The proof has three moves. First, derive the chord product $R^2-|OI{|}^2=IA\cdot IM$. Second, show $IM=MB$. Third, use two similar right triangles to prove $IA\cdot MB=2Rr$ without trigonometry.

Step 3: Extend the Bisector to $M$

Extend the bisector of angle $A$ beyond $I$ until it meets the circumcircle again at $M$. Since $\angle BAM=\angle MAC$, the intercepted arcs $BM$ and $MC$ are equal. Thus $M$ is the midpoint of arc $BC$, and $MB=MC$.

Step 4: Derive the Chord Product

Let $F$ be the foot from $O$ to the chord $AM$. Since a perpendicular from the center to a chord bisects the chord, $AF=FM$.

Write $x=AF=FM$ and $y=IF$. Then the two pieces from $I$ to the chord endpoints have lengths $x-y$ and $x+y$, in either order:

Now use the right triangles $OAF$ and $OIF$:

Subtracting cancels $O{F}^2$ and leaves

Step 5: $\angle MBI=A/2+B/2$

Look at triangle $MIB$ and compare its base angles. Since $M$ is the midpoint of arc $BC$, $\angle MBC=A/2$; since $BI$ bisects angle $B$, $\angle CBI=B/2$. Adding them, $\angle MBI=A/2+B/2$.

Step 6: The Exterior Angle $\angle MIB$

At $I$, the angle $\angle MIB$ is the exterior angle of triangle $ABI$, so

which matches $\angle MBI$.

Step 7: $IM=MB$

The base angles of triangle $MIB$ are equal, so it is isosceles, giving $IM=MB$.

Step 8: A Similar-Triangle Product

Let $N$ be the point opposite $M$ on the circumcircle. Then $MN$ is a diameter, so $MN=2R$, and Thales' theorem makes $\angle MBN$ right.

The inradius $IZ$ is perpendicular to $AB$, so $\angle AZI$ is right. Also $\angle IAZ=\angle BAM$, and $\angle BNM$ equals $\angle BAM$ because both inscribed angles subtend chord $BM$. Therefore $△AZI\sim △NBM$.

From corresponding sides,

and hence

Step 9: Euler's Formula

Using $IM=MB$ and the similar-triangle product,

Recalling the chord product $R^2-|OI{|}^2=IA\cdot IM$, this gives $R^2-|OI{|}^2=2Rr$, that is

In particular $R\ge 2r$, Euler's inequality.

The chord product derived along the bisector chord $AM$ gives $R^2-|OI{|}^2=IA\cdot IM=2Rr$, so

This is the metric fact behind Feuerbach's theorem, and its non-negativity is Euler's inequality $R\ge 2r$. ∎

Notes

The full written argument behind the animation, in four steps.

Derive the chord product

Extend the bisector of angle $A$ to meet the circumcircle again at $M$, the midpoint of arc $BC$. Let $F$ be the foot from $O$ to chord $AM$. A perpendicular from the center to a chord bisects the chord, so $AF=FM$. Write $x=AF=FM$ and $y=IF$. The two pieces from $I$ to the endpoints of the chord have lengths $x-y$ and $x+y$ (in either order), so

The right triangles $OAF$ and $OIF$ give

Subtracting,

Replace $IM$ by $MB$

Because $AI$ bisects angle $A$, the point $M$ is the midpoint of the arc $BC$ not containing $A$. Hence $\angle MBC=A/2$. Also $\angle CBI=B/2$, so $\angle MBI=A/2+B/2$. On the other side, $\angle MIB$ is the exterior angle of triangle $ABI$, hence $\angle MIB=\angle IAB+\angle IBA=A/2+B/2$. Therefore triangle $MIB$ is isosceles and

Find $IA\cdot MB$ without trigonometry

Let $N$ be the point opposite $M$ on the circumcircle, so $MN$ is a diameter and $MN=2R$. Drop the inradius $IZ$ to $AB$, so $IZ=r$ and $\angle AZI$ is right. By Thales' theorem, $\angle MBN$ is also right.

The remaining acute angles match: $\angle IAZ=\angle BAM$, and $\angle BNM$ subtends the same chord $BM$ as $\angle BAM$. Thus $\angle IAZ=\angle BNM$, and

Corresponding sides give

so

Putting it together

hence $R^2-|OI{|}^2=2Rr$, i.e. $|OI{|}^2=R^2-2Rr$.