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.

Lemma

 * Incenter Circumcenter Distance.png

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 $(1)$: