Answer to Question #95767 in Geometry for Max

Question

The base BC of triangle ABC is trisected at X and Y .
Show that |AX|
2 + |AY|
2 + 4|XY|
2 = |AB|
2 + |AC|
2

Answer 1

Let $BX=\frac{1}{3}BC$ and $BY=\frac{2}{3}BC$ .

Consider $\overrightarrow{AB}$ and $\overrightarrow{AC}$ as a basis of a vector space.

We have

1)$\overrightarrow{BC}=\overrightarrow{AC}-\overrightarrow{AB}$ , so $|\overrightarrow{BC}|^2=|\overrightarrow{AC}|^2+|\overrightarrow{AB}|^2-2(\overrightarrow{AC},\overrightarrow{AB}).$ That is $(\overrightarrow{AC},\overrightarrow{AB})=\frac{1}{2}(AC^2+AB^2-BC^2)$

2)$\overrightarrow{AX}=\overrightarrow{AB}+\overrightarrow{BX}=\overrightarrow{AB}+\frac{1}{3}\overrightarrow{BC}=\overrightarrow{AB}+\frac{1}{3}(\overrightarrow{AC}-\overrightarrow{AB})=$

$=\frac{2}{3}\overrightarrow{AB}+\frac{1}{3}\overrightarrow{AC}$, so $|\overrightarrow{AX}|^2=\frac{4}{9}|\overrightarrow{AB}|^2+\frac{1}{9}|\overrightarrow{AC}|^2+\frac{4}{9}(\overrightarrow{AB},\overrightarrow{AC})=$

$=\frac{4}{9}AB^2+\frac{1}{9}AC^2+\frac{2}{9}(AC^2+AB^2-BC^2)=$

$=\frac{2}{3}AB^2+\frac{1}{3}AC^2-\frac{2}{9}BC^2$

3)$\overrightarrow{AY}=\overrightarrow{AB}+\overrightarrow{BY}=\overrightarrow{AB}+\frac{2}{3}\overrightarrow{BC}=\overrightarrow{AB}+\frac{2}{3}(\overrightarrow{AC}-\overrightarrow{AB})=$

$=\frac{1}{3}\overrightarrow{AB}+\frac{2}{3}\overrightarrow{AC}$, so $|\overrightarrow{AY}|^2=\frac{1}{9}|\overrightarrow{AB}|^2+\frac{4}{9}|\overrightarrow{AC}|^2+\frac{4}{9}(\overrightarrow{AB},\overrightarrow{AC})=$

$\frac{1}{9}AB^2+\frac{4}{9}AC^2+\frac{2}{9}(AC^2+AB^2-BC^2)=$

$=\frac{1}{3}AB^2+\frac{2}{3}AC^2-\frac{2}{9}BC^2$

4)From 2 and 3 we obtain $AX^2+AY^2=\bigl(\frac{2}{3}AB^2+\frac{1}{3}AC^2-\frac{2}{9}BC^2\bigr)+$

$+\bigl(\frac{1}{3}AB^2+\frac{2}{3}AC^2-\frac{2}{9}BC^2\bigr)=AB^2+AC^2-\frac{4}{9}BC^2$

Since $XY=\frac{1}{3}BC$, we have $XY^2=\frac{1}{9}BC^2$ , so $AX^2+AY^2=AB^2+AC^2-\frac{4}{9}BC^2=AB^2+AC^2-4XY^2$ . Moving $4XY^2$ to the left side, we obtain $AX^2+AY^2+4XY^2=AB^2+AC^2$ .