Triangle Point Distance Formula
Let $ G $ and $ H $ denote the centroid and orthocenter of triangle $ ABC, $ respectively. Let $ F $ be the midpoint of $ \overline{GH} $. Express $ AF^2 + BF^2 + CF^2 $ in terms of the side lengths $ a, $ $ b, $ $ c $ and circumradius $ R $ of triangle $ ABC $.
- 1
- 2
- 3
- +
- 4
- 5
- 6
- -
- 7
- 8
- 9
- $\frac{a}{b}$
- .
- 0
- =
- %
- $a^n$
- $a^{\circ}$
- $a_n$
- $\sqrt{}$
- $\sqrt[n]{}$
- $\pi$
- $\ln{}$
- $\log$
- $\theta$
- $\sin{}$
- $\cos{}$
- $\tan{}$
- $($
- $)$
- $[$
- $]$
- $\cap$
- $\cup$
- $,$
- $\infty$
Solution
Let the circumcenter $ O $ of triangle $ ABC $ be the origin. Then
\[\overrightarrow{G} = \frac{\overrightarrow{A} + \overrightarrow{B} + \overrightarrow{C}}{3}\]and $ \overrightarrow{H} = \overrightarrow{A} + \overrightarrow{B} + \overrightarrow{C}, $ so
\[\overrightarrow{F} = \frac{2}{3} (\overrightarrow{A} + \overrightarrow{B} + \overrightarrow{C}).\]Then
\begin{align*}
AF^2 &= \|\overrightarrow{A} - \overrightarrow{F}\|^2 \\
&= \left\| \overrightarrow{A} - \frac{2}{3} (\overrightarrow{A} + \overrightarrow{B} + \overrightarrow{C}) \right\|^2 \\
&= \left\| \frac{1}{3} \overrightarrow{A} - \frac{2}{3} \overrightarrow{B} - \frac{2}{3} \overrightarrow{C} \right\|^2 \\
&= \frac{1}{9} \|\overrightarrow{A} - 2 \overrightarrow{B} - 2 \overrightarrow{C}\|^2 \\
&= \frac{1}{9} (\overrightarrow{A} - 2 \overrightarrow{B} - 2 \overrightarrow{C}) \cdot (\overrightarrow{A} - 2 \overrightarrow{B} - 2 \overrightarrow{C}) \\
&= \frac{1}{9} (\overrightarrow{A} \cdot \overrightarrow{A} + 4 \overrightarrow{B} \cdot \overrightarrow{B} + 4 \overrightarrow{C} \cdot \overrightarrow{C} - 4 \overrightarrow{A} \cdot \overrightarrow{B} - 4 \overrightarrow{A} \cdot \overrightarrow{C} + 8 \overrightarrow{B} \cdot \overrightarrow{C}) \\
&= \frac{1}{9} (9R^2 - 4 \overrightarrow{A} \cdot \overrightarrow{B} - 4 \overrightarrow{A} \cdot \overrightarrow{C} + 8 \overrightarrow{B} \cdot \overrightarrow{C}).\end{align*}
Similarly,
\begin{align*}
BF^2 &= \frac{1}{9} (9R^2 - 4 \overrightarrow{A} \cdot \overrightarrow{B} + 8 \overrightarrow{A} \cdot \overrightarrow{C} - 4 \overrightarrow{B} \cdot \overrightarrow{C}), \\
CF^2 &= \frac{1}{9} (9R^2 + 8 \overrightarrow{A} \cdot \overrightarrow{B} - 4 \overrightarrow{A} \cdot \overrightarrow{C} - 4 \overrightarrow{B} \cdot \overrightarrow{C}).\end{align*}
Thus, $ AF^2 + BF^2 + CF^2 = \boxed{3R^2} $.
Freeflows