Euler Triangle Formula

Theorem

Let $d$ be the distance between the incenter and the circumcenter of a triangle.

Then:

$d^2 = R \paren {R - 2 \rho}$

where:

$R$ is the circumradius

$\rho$ is the inradius.

Corollary

Let $\rho$ be the inradius of a triangle.

Then:

$\dfrac 1 \rho = \dfrac 1 {R - d} + \dfrac 1 {R + d}$

where:

$R$ is the circumradius

$d$ is the distance between the incenter and the circumcenter.

Proof 1

Lemma 1

Let the incenter of $\triangle ABC$ be $I$.

Let the circumcenter of $\triangle ABC$ be $O$.

Let $OI$ be produced to the circumcircle at $G$ and $J$.

Let $CI$ be produced to the circumcircle at $P$.

Let $F$ be the point where the incircle of $\triangle ABC$ meets $BC$.

We are given that:

the distance between the incenter and the circumcenter is $d$

the inradius is $\rho$

the circumradius is $R$.

Then

$IP \cdot CI = \paren {R + d} \paren {R - d}$

$\Box$

Lemma 2

Let the bisector of angle $C$ of triangle $\triangle ABC$ be produced to the circumcircle at $P$.

Let $I$ be the incenter of $\triangle ABC$.

Then:

$AP = BP = IP$

$\Box$

By Lemma $1$:

$GI \cdot IJ = IP \cdot CI$

substituting:

$IP \cdot CI = \paren {R + d} \paren {R - d}$

By Lemma $2$:

$IP = PB$

and so:

$GI \cdot IJ = PB \cdot CI$

Now using the Extension of Law of Sines in $\triangle CPB$:

$\dfrac {PB} {\map \sin {\angle PCB} } = 2 R$

and so:

$GI \cdot IJ = 2 R \map \sin {\angle PCB} \cdot CI$

By the $4$th of Euclid's common notions:

$\angle PCB = \angle ICF$

and so:

$(1): \quad GI \cdot IJ = 2 R \map \sin {\angle ICF} \cdot CI$

We have that:

$IF = \rho$

and by Radius at Right Angle to Tangent:

$\angle IFC$ is a right angle.

By the definition of sine:

$\map \sin {\angle ICF} = \dfrac {\rho} {CI}$

and so:

$\map \sin {\angle ICF} \cdot CI = \rho$

Substituting in $(1)$:

\(\ds GI \cdot IJ\)

\(=\)

\(\ds 2 R \rho\)

\(\ds \leadsto \ \ \)

\(\ds \paren {R + d} \paren {R - d}\)

\(=\)

\(\ds 2 R \rho\)

\(\ds \leadsto \ \ \)

\(\ds R^2 - d^2\)

\(=\)

\(\ds 2 R \rho\)

Difference of Two Squares

\(\ds \leadsto \ \ \)

\(\ds d^2\)

\(=\)

\(\ds R^2 - 2 R \rho\)

\(\ds \)

\(=\)

\(\ds R \paren {R - 2 \rho}\)

$\blacksquare$

Proof 2

Lemma 1

Let the incenter of $\triangle ABC$ be $I$.

Let the circumcenter of $\triangle ABC$ be $O$.

Let $OI$ be produced to the circumcircle at $G$ and $J$.

Let $CI$ be produced to the circumcircle at $P$.

Let $F$ be the point where the incircle of $\triangle ABC$ meets $BC$.

We are given that:

the distance between the incenter and the circumcenter is $d$

the inradius is $\rho$

the circumradius is $R$.

Then

$IP \cdot CI = \paren {R + d} \paren {R - d}$

$\Box$

Lemma 2

Let the bisector of angle $C$ of triangle $\triangle ABC$ be produced to the circumcircle at $P$.

Let $I$ be the incenter of $\triangle ABC$.

Then:

$AP = BP = IP$

$\Box$

\(\ds \leadsto \ \ \)

\(\ds IP \cdot CI\)

\(=\)

\(\ds \paren {R + d} \cdot \paren {R - d}\)

Lemma $1$

\(\ds IP\)

\(=\)

\(\ds PB\)

Lemma $2$

\(\ds CI \cdot PB\)

\(=\)

\(\ds \paren {R + d} \cdot \paren {R - d}\)

Common Notion $1$

Draw diameter $POQ$.

\(\ds \angle PBQ\)

\(=\)

\(\ds 90 \degrees\)

Thales' Theorem

\(\ds \angle PCB\)

\(=\)

\(\ds \angle BQP\)

Equal Angles in Equal Circles

\(\ds \triangle IFC\)

\(\sim\)

\(\ds \triangle PBQ\)

Equiangular Right Triangles are Similar

\(\ds \leadsto \ \ \)

\(\ds \dfrac {2R} {PB}\)

\(=\)

\(\ds \dfrac {CI} \rho\)

\(\ds CI \cdot PB\)

\(=\)

\(\ds 2 R \rho\)

rearranging

\(\text {(1)}: \quad\)

\(\ds \paren {R + d} \paren {R - d}\)

\(=\)

\(\ds 2 R \rho\)

substitution from above

\(\ds R^2 - d^2\)

\(=\)

\(\ds 2 R \rho\)

Difference of Two Squares

\(\ds d^2\)

\(=\)

\(\ds R \paren {R - 2 \rho}\)

rearranging

$\blacksquare$

Also presented as

Some writers prefer the following form:

$d^2 + 2 \rho R = R^2$

Source of Name

This entry was named for Leonhard Paul Euler.