Triangle ABC is constructed on vectors overrightarrowBC=a…

Triangle $ABC$ is constructed on vectors $\overrightarrow{BC}=\vec{a}$ and $\overrightarrow{CA}=\vec{b}$. a) Express the median vectors $\overrightarrow{AA_1}$, $\overrightarrow{BB_1}$ and $\overrightarrow{CC_1}$ in terms of $\vec{a}$ and $\vec{b}$. b) Show that the three medians can form a triangle.

Explanation of Method

Medians of a triangle expressed via side vectors; the triangle of medians.

Step 1: Setup

The sides form a closed loop, so

\[ \overrightarrow{BC}+\overrightarrow{CA}+\overrightarrow{AB} = \vec{0} \quad\Longrightarrow\quad \overrightarrow{AB} = -\vec{a}-\vec{b}. \]

$A_1$, $B_1$, $C_1$ are the midpoints of $BC$, $CA$ and $AB$ respectively.

Step 2: Part a): the median vectors

Median $\overrightarrow{AA_1}$

\[ \overrightarrow{AA_1} = \overrightarrow{AB} + \tfrac{1}{2}\overrightarrow{BC} = (-\vec{a}-\vec{b}) + \tfrac{1}{2}\vec{a} \]

\[\overrightarrow{AA_1} = -\frac{1}{2}\vec{a} - \vec{b}\]

Median $\overrightarrow{BB_1}$

\[ \overrightarrow{BB_1} = \overrightarrow{BC} + \tfrac{1}{2}\overrightarrow{CA} \]

\[\overrightarrow{BB_1} = \vec{a} + \frac{1}{2}\vec{b}\]

Median $\overrightarrow{CC_1}$

\[ \overrightarrow{CC_1} = \overrightarrow{CA} + \tfrac{1}{2}\overrightarrow{AB} = \vec{b} + \tfrac{1}{2}(-\vec{a}-\vec{b}) \]

\[\overrightarrow{CC_1} = -\frac{1}{2}\vec{a} + \frac{1}{2}\vec{b}\]

$A$ $B$ $C$ $A_1$ $B_1$ $C_1$ $G$ $\vec a$ $\vec b$

$\vec a=\overrightarrow{BC}$
$\vec b=\overrightarrow{CA}$
$\overrightarrow{AA_1}=-\tfrac12\vec a-\vec b$
$\overrightarrow{BB_1}=\vec a+\tfrac12\vec b$
$\overrightarrow{CC_1}=-\tfrac12\vec a+\tfrac12\vec b$

Triangle $ABC$ with its three medians meeting at the centroid $G$.

Step 3: Part b): the medians form a triangle

Three vectors form a triangle when placed head-to-tail exactly when their sum is $\vec 0$ (and they are not all collinear). Adding the medians: \[\left(-\tfrac{1}{2}\vec{a}-\vec{b}\right)+\left(\vec{a}+\tfrac{1}{2}\vec{b}\right)+\left(-\tfrac{1}{2}\vec{a}+\tfrac{1}{2}\vec{b}\right) = \left(-\tfrac12+1-\tfrac12\right)\vec a+\left(-1+\tfrac12+\tfrac12\right)\vec b = \vec{0}.\]

\[\overrightarrow{AA_1}+\overrightarrow{BB_1}+\overrightarrow{CC_1} = \vec{0}\]

Why this holds in general: the sum of the medians is

\[ \overrightarrow{AA_1}+\overrightarrow{BB_1}+\overrightarrow{CC_1} = (A_1+B_1+C_1)-(A+B+C), \] and the three midpoints have the same sum as the three vertices: $A_1+B_1+C_1=\tfrac{B+C}{2}+\tfrac{C+A}{2}+\tfrac{A+B}{2}=A+B+C$.

Construction of the median triangle

Denote $\vec{m}_a = \overrightarrow{AA_1}$, $\vec{m}_b = \overrightarrow{BB_1}$, $\vec{m}_c = \overrightarrow{CC_1}$. Start at a point $P$, go along $\vec{m}_a$ to $Q$, then along $\vec{m}_b$ to $R$:

\[ \overrightarrow{PQ} = \vec{m}_a, \quad \overrightarrow{QR} = \vec{m}_b \quad\Longrightarrow\quad \overrightarrow{RP} = -(\vec m_a+\vec m_b) = \vec{m}_c. \]

So the third median closes the path, and $PQR$ is a triangle whose sides are the three medians. Since $\vec a$ and $\vec b$ are not collinear, neither are $\vec m_a$ and $\vec m_b$, so the triangle is non-degenerate. $\blacksquare$

Example: side lengths

For $\vec{a}\perp\vec{b}$ with $|\vec{a}|=|\vec{b}|=2$: \[|\vec{m}_a| = \sqrt{1+4} = \sqrt{5}, \qquad |\vec{m}_b| = \sqrt{4+1} = \sqrt{5}, \qquad |\vec{m}_c| = \sqrt{1+1} = \sqrt{2},\] which satisfy the triangle inequality ($\sqrt5+\sqrt2\gt\sqrt5$, $\sqrt5+\sqrt5\gt\sqrt2$).

$P$ $Q$ $R$ $\vec m_a$ $\vec m_b$ $\vec m_c$

The same three median vectors placed head-to-tail close up into triangle $PQR$.

Result

Median	Via $\vec{a}$, $\vec{b}$
$\overrightarrow{AA_1}$	$-\dfrac{1}{2}\vec{a} - \vec{b}$
$\overrightarrow{BB_1}$	$\vec{a} + \dfrac{1}{2}\vec{b}$
$\overrightarrow{CC_1}$	$-\dfrac{1}{2}\vec{a} + \dfrac{1}{2}\vec{b}$
Sum	$\vec 0$ ⇒ the medians form a triangle