# 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}

\begin{align} \vec{h} \cdot (\vec{b} - \vec{a}) - \vec{a} \cdot (\vec{b} - \vec{c}) - \vec{b} \cdot (\vec{c} - \vec{a}) &= 0 \\ \vec{h} \cdot (\vec{b} - \vec{a}) - \vec{c} \cdot (\vec{b} - \vec{a}) &= 0 \\ (\vec{h} - \vec{c}) \cdot (\vec{b} - \vec{a}) &= 0 \end{align}

### Circumcenter

The skill is similar. There’s no need to actually solve for the vector of the desired point in terms of $\vec{a}$, $\vec{b}$ and $\vec{c}$.

Let

• $O$ be the point of intersection of two perpendicular bisectors of sides $\overrightarrow{AB}$ and $\overrightarrow{BC}$.
• $\vec{v} = \overrightarrow{PO}$

Remark: I used v instead of o above to avoid confusion with the little-O notation.

\begin{align} \left(\vec{v} - \frac{\vec{a}+\vec{b}}{2}\right) \cdot (\vec{a}-\vec{b}) &= 0 \\ \left(\vec{v} - \frac{\vec{b}+\vec{c}}{2}\right) \cdot (\vec{b}-\vec{c}) &= 0 \end{align}

\begin{align} \vec{v}\cdot(\vec{a}-\vec{c}) - \frac{\normlr{\vec{a}}^2-\normlr{\vec{b}}^2}{2} - \frac{\normlr{\vec{b}}^2-\normlr{\vec{c}}^2}{2} &= 0 \\ \vec{v}\cdot(\vec{a}-\vec{c}) - \frac{\normlr{\vec{a}}^2-\normlr{\vec{c}}^2}{2} & = 0 \\ \left(\vec{v} - \frac{\vec{a}+\vec{c}}{2}\right) \cdot (\vec{a}-\vec{c}) &= 0 \end{align}

### Incenter

Let $I$ be the intersection point of two angle bisectors $\overrightarrow{AD}$ and $\overrightarrow{BE}$.

I tried computing dot products two days ago, but I got something like

$$2 \vec{v} \cdot \frac{\vec{b}-\vec{a}}{\normlr{\vec{b}-\vec{a}}} - \frac{\normlr{\vec{b}}^2-\normlr{\vec{a}}^2}{\normlr{\vec{b}-\vec{a}}} - \normlr{\vec{a} - \vec{c}} + \normlr{\vec{b} - \vec{c}},$$

where $\vec{v} = \overrightarrow{PI}$, and I didn’t know what to do next.

I have to combine two elementary formulae

1. an elementary fact about the angle bisector, which can be derived from Co-Angle Theorem (共角定理)

$$\frac{\normlr{\overrightarrow{AB}}}{\normlr{\overrightarrow{AC}}} = \frac{\normlr{\overrightarrow{BD}}}{\normlr{\overrightarrow{DC}}}$$
2. the convex combination of two vectors (定比分點公式)

$$\overrightarrow{AD} = \frac{\normlr{\overrightarrow{DC}}}{\normlr{\overrightarrow{BD}}+ \normlr{\overrightarrow{DC}}} \, \overrightarrow{AB} + \frac{\normlr{\overrightarrow{BD}}}{\normlr{\overrightarrow{BD}}+ \normlr{\overrightarrow{DC}}} \, \overrightarrow{AC}$$

so that I know precisely where the incenter is.

\begin{align} \overrightarrow{AD} &= \frac{\normlr{\overrightarrow{AC}}}{\normlr{\overrightarrow{AB}}+ \normlr{\overrightarrow{AC}}} \, \overrightarrow{AB} + \frac{\normlr{\overrightarrow{AB}}}{\normlr{\overrightarrow{AB}}+ \normlr{\overrightarrow{AC}}} \, \overrightarrow{AC} \\ \overrightarrow{BE} &= \frac{\normlr{\overrightarrow{BC}}}{\normlr{\overrightarrow{BA}}+ \normlr{\overrightarrow{BC}}} \, \overrightarrow{BA} + \frac{\normlr{\overrightarrow{BA}}}{\normlr{\overrightarrow{BA}}+ \normlr{\overrightarrow{BC}}} \, \overrightarrow{BC} \end{align}

Let $s = \normlr{\overrightarrow{AI}}/\normlr{\overrightarrow{AD}}$ and $t = \normlr{\overrightarrow{BI}}/\normlr{\overrightarrow{BE}}$. Then $\overrightarrow{AI} = s \overrightarrow{AB}$ and $\overrightarrow{AI} = t \overrightarrow{AB}$. We solve for $s$ and $t$ by considering $\overrightarrow{CA}+\overrightarrow{AI} = \overrightarrow{CB} + \overrightarrow{BI} = \overrightarrow{CI}$.

\begin{align} & \overrightarrow{CA} + \overrightarrow{AI} \\ =& \vec{a} - \vec{c} + s \left[\frac{\normlr{\vec{c} - \vec{a}}}{\normlr{\vec{b} - \vec{a}} + \normlr{\vec{c} - \vec{a}}} \left(\vec{b} - \vec{a}\right) + \frac{\normlr{\vec{b} - \vec{a}}}{\normlr{\vec{b} - \vec{a}} + \normlr{\vec{c} - \vec{a}}} \left(\vec{c} - \vec{a}\right)\right] \\ =& (1-s) \vec{a} + s \, \frac{\normlr{\vec{c} - \vec{a}}}{\normlr{\vec{b} - \vec{a}} + \normlr{\vec{c} - \vec{a}}} \, \vec{b} - \left(1 - s \, \frac{\normlr{\vec{b} - \vec{a}}}{\normlr{\vec{b} - \vec{a}} + \normlr{\vec{c} - \vec{a}}}\right) \vec{c} \\ & \overrightarrow{CB} + \overrightarrow{BI} \\ =& \vec{b} - \vec{c} + t \left[\frac{\normlr{\vec{c} - \vec{b}}}{\normlr{\vec{b} - \vec{a}} + \normlr{\vec{c} - \vec{b}}} \left(\vec{a} - \vec{b}\right) \\ + \frac{\normlr{\vec{a} - \vec{b}}}{\normlr{\vec{c} - \vec{b}} + \normlr{\vec{a} - \vec{b}}} \left(\vec{c} - \vec{b}\right)\right] \\ =& t \, \frac{\normlr{\vec{c} - \vec{b}}}{\normlr{\vec{b} - \vec{a}} + \normlr{\vec{c} - \vec{b}}} \, \vec{a} + (1-t) \, \vec{b} - \left(1 - t \, \frac{\normlr{\vec{b} - \vec{a}}}{\normlr{\vec{b} - \vec{a}} + \normlr{\vec{c} - \vec{b}}}\right) \, \vec{c} \end{align}

From the $\vec{c}$-component of $\overrightarrow{CI}$, we have

$$\frac{s}{\normlr{\vec{a} - \vec{b}} + \normlr{\vec{c} - \vec{a}}} = \frac{t}{\normlr{\vec{a} - \vec{b}} + \normlr{\vec{b} - \vec{c}}}.$$

Substitute the above equation into the $\vec{b}$- component of $\overrightarrow{CI}$.

\begin{align} 1 - t &= \normlr{\vec{c} - \vec{a}} \, \frac{t}{\normlr{\vec{a} - \vec{b}} + \normlr{\vec{b} - \vec{c}}} \\ 1 &= \left(1 + \frac{\normlr{\vec{c} - \vec{a}}}{\normlr{\vec{a} - \vec{b}} + \normlr{\vec{b} - \vec{c}}}\right) t \\ t &= 1 - \frac{\normlr{\vec{c} - \vec{a}}}{\normlr{\vec{a} - \vec{b}} + \normlr{\vec{b} - \vec{c}} + \normlr{\vec{c} - \vec{a}}} \end{align}

Substitute the above equation into the $\vec{a}$-component of $\overrightarrow{CI}$.

$$s = 1 - \frac{\normlr{\vec{b} - \vec{c}}}{\normlr{\vec{a} - \vec{b}} + \normlr{\vec{b} - \vec{c}} + \normlr{\vec{c} - \vec{a}}}$$

Substitute the above two equations into $\overrightarrow{CI}$.

\begin{align} & \overrightarrow{CI} \\ =& \frac{\normlr{\vec{b} - \vec{c}}}{\normlr{\vec{a} - \vec{b}} + \normlr{\vec{b} - \vec{c}} + \normlr{\vec{c} - \vec{a}}} \, \vec{a} + \frac{\normlr{\vec{c} - \vec{a}}}{\normlr{\vec{a} - \vec{b}} + \normlr{\vec{b} - \vec{c}} + \normlr{\vec{c} - \vec{a}}} \, \vec{b} \\ & -\frac{\normlr{\vec{b} - \vec{c}} + \normlr{\vec{c} - \vec{a}}}{\normlr{\vec{a} - \vec{b}} + \normlr{\vec{b} - \vec{c}} + \normlr{\vec{c} - \vec{a}}} \, \vec{c} \\ =& \frac{\normlr{\vec{b} - \vec{c}}}{\normlr{\vec{a} - \vec{b}} + \normlr{\vec{b} - \vec{c}} + \normlr{\vec{c} - \vec{a}}} \left(\vec{a} - \vec{c}\right) + \frac{\normlr{\vec{c} - \vec{a}}}{\normlr{\vec{a} - \vec{b}} + \normlr{\vec{b} - \vec{c}} + \normlr{\vec{c} - \vec{a}}} \left(\vec{b} - \vec{c}\right) \end{align}

From the last equality, we see that $\overrightarrow{CI} \parallel \overrightarrow{CF}$.