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.