Existence of Four Triangle Centers

A vector proof

Settings

Let

  • $P$ be an arbitrary reference point.
  • $\triangle ABC$ be a triangle.
  • $\vec{a} = \overrightarrow{PA}$, $\vec{b} = \overrightarrow{PB}$, $\vec{c} = \overrightarrow{PC}$

Remark: $O$ is reserved for circumcenter.

symbol name meaning
$G$ centroid center of gravity
$H$ orthocenter three “heights” are concurrent
$I$ incenter center of inscribed circle
$O$ circumcenter center of circumscribed circle

Centroid

Verify that $(\vec{a} + \vec{b} + \vec{c})/3$ satisfy the constraints.

Orthocenter

Let

  • $H$ be the point of intersection of two altitudes $AA_H$ and $BB_H$
  • $\vec{h} = \overrightarrow{PH}$
\begin{align} (\vec{h} - \vec{a}) \cdot (\vec{b} - \vec{c}) &= 0 \\ (\vec{h} - \vec{b}) \cdot (\vec{c} - \vec{a}) &= 0 \end{align}

Add these two equations together.

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Dot Products

My complicated derivation

Preface

There’s a simpler derivation using

  1. the geometric defintion
  2. linearity from the definition
  3. orthogonality of the canonical basic vectors $\vec{i}$, $\vec{j}$ and $\vec{k}$.

If I had read that, I wouldn’t have type this document in LaTeX. The following content was transcribed from a PDF file that I made three days ago.

Content

Recall: $\sum_{i=m}^{n} a_{i}=a_{m}+a_{m+1}+a_{m+2}+\cdots +a_{n-1}+a_{n},$ where $m$, $n$ are integers.

Example: $\sum_{i=3}^6 i^2 = 3^2 + 4^2 + 5^2 + 6^2 = 86$

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