Napoleon's Theorem/Proof 2

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Theorem

Let $\triangle ABC$ be an arbitrary triangle.

Let $\triangle ABF$, $\triangle BCD$ and $\triangle ACE$ be equilateral triangles constructed on $AB$, $BC$ and $AC$ respectively on the exterior of $\triangle ABC$.

Let $O_1$, $O_2$ and $O_3$ be the incenters of $\triangle ABF$, $\triangle BCD$ and $\triangle ACE$.

Then $\triangle O_1 O_2 O_3$ is an equilateral triangle.

Proof

Lemma $1$

Let $T = \triangle ABC$ be an equilateral triangle in the Cartesian plane $\CC$.

Let the sides of $\triangle ABC$ be the vectors $\mathbf u$, $\mathbf v$, and $\mathbf w$.

Let the interior of $T$ lie to the left of the vector path.

Let $\mathbf v$ be rotated by an angle of $60 \degrees$ anticlockwise.

Let the rotated vector be $\mathbf v'$.

Then:

$\mathbf u + \mathbf v' = \mathbf 0$

where $\mathbf 0$ denotes the zero vector.

$\Box$

Lemma $2$

Let $T = \triangle ABC$ be an equilateral triangle in the plane $\CC$.

Let $\mathbf{v}$ be a vector in $\CC$ with magnitude $\dfrac 1 3 \norm {AB}$ and direction $\vec {BA}$.

Let a unit rotation be anticlockwise by $60^{\circ}$, and denote vector $\mathbf{v}$ after this rotation as $\mathbf{v}'$.

Then the vector path from $B$ to the incenter $O$ of $T$ is $\mathbf{v} + -\mathbf{v}' '$ and the vector path from the incenter $O$ to $A$ is $\mathbf{v}' + \mathbf{v}$.

$\Box$

Vectors

Side $a = BC$ lies opposite vertex $A$ of $\triangle ABC$.

Let the vector $\mathbf{a}$ have magnitude $\dfrac 1 3 a$ in the direction of $\vec {BC}$.

$O_1$ is the incenter of the $\triangle BCD$ with side $a$ from $\triangle ABC$.

Side $b = CA$ lies opposite vertex $B$ of $\triangle ABC$.

Let the vector $\mathbf{b}$ have magnitude $\dfrac 1 3 b$ in the direction of $\vec {CA}$.

$O_2$ is the incenter of the $\triangle CAE$ with side $b$ from $\triangle ABC$.

Side $c = AB$ lies opposite vertex $C$ of $\triangle ABC$.

Let the vector $\mathbf{c}$ have magnitude $\dfrac 1 3 c$ in the direction of $\vec {AB}$.

$O_3$ is the incenter of the $\triangle ABF$ with side $c$ from of $\triangle ABC$.

Construction of Vector Paths

Let $\mathbf{p}$ be the vector path from $O_1$ to $O_2$ through vertex $C$.

By Lemma $2$:

The first part of $\mathbf{p}$ from $O_1$ to $C$ is $\mathbf{a}' + \mathbf{a}$.

Also by Lemma $2$:

The second part of $\mathbf{p}$ from $C$ to $O_2$ is $\mathbf{b} + - \mathbf{b}' '$.

By addition:

$\mathbf{p} = \mathbf{a}' + \mathbf{a} + \mathbf{b} + - \mathbf{b}' '$

Let $\mathbf{q}$ be the vector path from $O_2$ to $O_3$ through vertex $A$.

By Lemma $2$:

The first part of $\mathbf{q}$ from $O_2$ to $A$ is $\mathbf{b}' + \mathbf{b}$.

By Lemma $2$:

The second part of $\mathbf{q}$ from $A$ to $O_3$ is $\mathbf{c} + - \mathbf{c}' '$.

By addition:

$\mathbf{q} = \mathbf{b}' + \mathbf{b} + \mathbf{c} + - \mathbf{c}' '$

Test for Equilateral Triangle

By the test from Lemma $1$, we construct $\mathbf{p} + \mathbf{q}'$:

If the result is $\mathbf{p} + \mathbf{q}' = \mathbf{0}$, the two vectors $\mathbf{p}$ and $\mathbf{q}$ are sides of an equilateral triangle.

$\mathbf{q}' = \mathbf{b}' ' + \mathbf{b}' + \mathbf{c}' - \mathbf{c}' ' '$

By definition of plane rotation and that $\theta = \frac 1 6$ of a complete rotation:

$- \mathbf{c}' ' ' = \mathbf{c}$

Substituting:

$\mathbf{q}' = \mathbf{b}' ' + \mathbf{b}' + \mathbf{c}' + \mathbf{c}$

Then:

\(\ds \mathbf{p} + \mathbf{q}'\)

\(=\)

\(\ds \mathbf{a}' + \mathbf{a} + \mathbf{b} - \mathbf{b}' ' + \mathbf{b}' ' + \mathbf{b}' + \mathbf{c}' + \mathbf{c}\)

Addition

\(\ds \)

\(=\)

\(\ds \mathbf{a}' + \mathbf{a} + \mathbf{b} + \mathbf{b}' + \mathbf{c}' + \mathbf{c}\)

Cancel terms

\(\ds \)

\(=\)

\(\ds \mathbf{a} + \mathbf{b} + \mathbf{c}\)

Vector Sum of Rotated Triangle is Zero

\(\ds \)

\(=\)

\(\ds \mathbf{0}\)

Vector Sum of Triangle is Zero

The result follows.

$\blacksquare$