# 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:

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

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

- $\angle PCB = \angle ICF$

and so:

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

We have that:

- $IF = \rho$

and by Radius at Right Angle to Tangent:

- $\angle IFC$ 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.