Euler Triangle Formula

Theorem

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

Then $d^2 = R \left({R-2 \rho}\right)$, where $R$ is the circumradius and $\rho$ is the inradius.

Lemma

Let the bisector of one angle of the triangle be produced to one point of the circumcircle $P$.

Then the distance of any of the vertices, of the triangle, left are equal to $IP$, where $I$ is the incenter of the triangle.

Proof (Lemma)

In the figure we have that $CP$ is the bisector of $\angle ACB$.

Now how $\angle ABP$ and $\angle ACP$ subtend the same arc then:

$\angle ACP = \angle ABP = \dfrac{\angle C}{2}$

As $IB$ is the bisector of $B$ then:

$\angle IBA = \dfrac{\angle B}{2} \implies \angle IBP = \dfrac{\angle B + \angle C}{2}$

and in $\triangle CIB$ the supplement of $\angle CIB = \dfrac{\angle C}{2} + \dfrac{\angle B}{2}$.

Therefore $\angle PIB = \angle IBP \implies IP = BP$.

Proof (Theorem)

$OI = d$, $OG = OJ = R$.

Therefore $IJ = R + d$ and $GI = R-d$.

By the Intersecting Chord Theorem we have $GI \cdot IJ =IP \cdot CI$.

Now by the lemma we have $GI \cdot IJ = PB \cdot CI$.

Now using the Extension of Law of Sines in $\triangle CPB$ we have $\dfrac{PB}{\sin(\angle PCB)}=2R$.

$GI \cdot IJ = 2R \sin(\angle PCB) \cdot CI$.

Note that $\angle PCB = \angle ICF$ by the fourth of Euclid's common notions.

Now working with $\sin(\angle ICF) \cdot CI$ in $\triangle CFI$.

By the definition of sine, $ \sin(\angle ICF)= \dfrac{\rho}{CI}$, thus $\sin(\angle ICF) \cdot CI = \rho$.

Plugging this in to the earlier equation yields $(R+d) \cdot (R-d) = 2R \cdot \rho \iff d^2=R(R-2 \rho) \,\!$.

$\blacksquare$

Source of Name

This entry was named for Leonhard Paul Euler.