## 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. [Read More]

## Dot Products

### My complicated derivation

Preface There’s a simpler derivation using the geometric defintion linearity from the definition 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: $\sum_{i=3}^6 i^2 = 3^2 + 4^2 + 5^2 + 6^2 = 86$ [Read More]