# Euler Triangle Formula

## Theorem

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

Then:

$d^2 = R \left({R - 2 \rho}\right)$

where:

$R$ is the circumradius
$\rho$ is the inradius.

## Proof

### Lemma

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:

$AP = BP = IP$

$\Box$

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 $F$ 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 $R$.

Thus:

$OI = d$
$OG = OJ = R$

Therefore:

$IJ = R + d$
$GI = R - d$

By the Intersecting Chord Theorem:

$GI \cdot IJ = IP \cdot CI$

By the lemma:

$IP = PB$

and so:

$GI \cdot IJ = PB \cdot CI$

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

$\dfrac {PB} {\sin \left({\angle PCB}\right)} = 2 R$

and so:

$GI \cdot IJ = 2 R \sin \left({\angle PCB}\right) \cdot CI$
$\angle PCB = \angle ICF$

and so:

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

We have that:

$IF = \rho$
$\angle IFC$ is a right angle.

By the definition of sine:

$\sin \left({\angle ICF}\right) = \dfrac {\rho} {CI}$

and so:

$\sin \left({\angle ICF}\right) \cdot CI = \rho$

Substituting in $(1)$:

 $\ds GI \cdot IJ$ $=$ $\ds 2 R \rho$ $\ds \implies \ \$ $\ds \left({R + d}\right) \left({R - d}\right)$ $=$ $\ds 2 R \rho$ $\ds \implies \ \$ $\ds R^2 - d^2$ $=$ $\ds 2 R \rho$ Difference of Two Squares $\ds \implies \ \$ $\ds d^2$ $=$ $\ds R^2 - 2 R \rho$ $\ds$ $=$ $\ds R \left({R - 2 \rho}\right)$

$\blacksquare$

## Source of Name

This entry was named for Leonhard Paul Euler.