Euler Triangle Formula/Proof 2
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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.
Proof
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 |
$\blacksquare$
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\(\ds \dfrac 1 \rho\) | \(=\) | \(\ds \dfrac {2 R} {\paren {R + d} \paren {R - d} }\) | from $(1)$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {R + d + R - d} {\paren {R + d} \paren {R - d} }\) | add and subtract $d$ in right hand side numerator | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 1 {\paren {R - d} } + \dfrac 1 {\paren {R + d} }\) | separating by partial fractions |