Chapter 6.1, Problem 62E

Solution

a.

To prove:

  $\begin{array}{c}\overrightarrow{O{M}_1}=\frac{1}{2}\overrightarrow{OA}+\frac{1}{2}\overrightarrow{OB}\\ \overrightarrow{O{M}_2}=\frac{1}{2}\overrightarrow{OC}+\frac{1}{2}\overrightarrow{OB}\\ \overrightarrow{O{M}_3}=\frac{1}{2}\overrightarrow{OA}+\frac{1}{2}\overrightarrow{OC}\end{array}$ .

Given information:

The medians of a triangle are in the ratio $1:2$ .

  $M_1,M_2\text{ and }{M}_3$ are the midpoints of the sides of the triangle.

Proof:

The given triangle $ABC$ ,

  

From the exercise 61b the triangle $OAB$ , $M_1$ is the midpoint of $AB$

  $\begin{array}{c}x=\frac{1}{2}\\ y=1-x\\ =1-\frac{1}{2}\\ =\frac{1}{2}\end{array}$

Substituting the values,

  $\begin{array}{c}\overrightarrow{O{M}_1}=x\overrightarrow{OA}+y\overrightarrow{OB}\\ =\frac{1}{2}\overrightarrow{OA}+\frac{1}{2}\overrightarrow{OB}\end{array}$

Similarly,

From triangle $OBC$ ,

  $\begin{array}{c}\overrightarrow{O{M}_2}=x\overrightarrow{OB}+y\overrightarrow{OC}\\ =\frac{1}{2}\overrightarrow{OB}+\frac{1}{2}\overrightarrow{OC}\end{array}$

From triangle $OCA$ ,

  $\begin{array}{c}\overrightarrow{O{M}_3}=x\overrightarrow{OC}+y\overrightarrow{OA}\\ =\frac{1}{2}\overrightarrow{OC}+\frac{1}{2}\overrightarrow{OA}\end{array}$

Hence $\overrightarrow{O{M}_1}=\frac{1}{2}\overrightarrow{OA}+\frac{1}{2}\overrightarrow{OB}$ , $\overrightarrow{O{M}_2}=\frac{1}{2}\overrightarrow{OC}+\frac{1}{2}\overrightarrow{OB}$ and $\overrightarrow{O{M}_3}=\frac{1}{2}\overrightarrow{OA}+\frac{1}{2}\overrightarrow{OC}$ has been proved.

b.

To prove:

  $\begin{array}{c}2\overrightarrow{O{M}_1}+\overrightarrow{OC}=\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}\\ 2\overrightarrow{O{M}_2}+\overrightarrow{OA}=\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}\\ 2\overrightarrow{O{M}_3}+\overrightarrow{OB}=\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}\end{array}$

Given information:

The medians of a triangle are in the ratio $1:2$ .

  $M_1,M_2\text{ and }{M}_3$ are the midpoints of the sides of the triangle.

Proof:

The given triangle $ABC$ ,

  

From part (a), it is known that

  $\overrightarrow{O{M}_1}=\frac{1}{2}\overrightarrow{OA}+\frac{1}{2}\overrightarrow{OB}$

Multiply by $2$ on both sides to get,

  $2\overrightarrow{O{M}_1}=\overrightarrow{OA}+\overrightarrow{OB}$

Adding $\overrightarrow{OC}$ on both sides to get,

  $2\overrightarrow{O{M}_1}+\overrightarrow{OC}=\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}$

Similarly, need to find the other two identities

  $\begin{array}{c}\overrightarrow{O{M}_2}=\frac{1}{2}\overrightarrow{OB}+\frac{1}{2}\overrightarrow{OC}\\ 2\overrightarrow{O{M}_2}=\overrightarrow{OC}+\overrightarrow{OB}\\ 2\overrightarrow{O{M}_2}+\overrightarrow{OA}=\overrightarrow{OC}+\overrightarrow{OB}+\overrightarrow{OA}\\ \overrightarrow{O{M}_3}=\frac{1}{2}\overrightarrow{OA}+\frac{1}{2}\overrightarrow{OC}\\ 2\overrightarrow{O{M}_3}=\overrightarrow{OA}+\overrightarrow{OC}\\ 2\overrightarrow{O{M}_3}+\overrightarrow{OC}=\overrightarrow{OA}+\overrightarrow{OC}+\overrightarrow{OB}\end{array}$

Therefore, the identities $2\overrightarrow{O{M}_1}+\overrightarrow{OC}=\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}$ , $2\overrightarrow{O{M}_2}+\overrightarrow{OA}=\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}$ and $2\overrightarrow{O{M}_3}+\overrightarrow{OB}=\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}$ has been proved.

c.

To prove: Why part (b) establishes the desired outcome.

Given information:

The medians of a triangle are in the ratio $1:2$ .

  $M_1,M_2\text{ and }{M}_3$ are the midpoints of the sides of the triangle.

Proof:

The given triangle $ABC$ ,

  

Now, draw the two medians

  

From part (b), it is known that

  $\begin{array}{c}2\overrightarrow{O{M}_1}+\overrightarrow{OC}=2\overrightarrow{O{M}_3}+\overrightarrow{OB}\\ =2\overrightarrow{O{M}_2}+\overrightarrow{OA}\\ =\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}\end{array}$

Considering parallelogram $BOCD$ ,

Let’s take the vector on $AD$ ,

  $\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}=\overrightarrow{OA}+\overrightarrow{OD}$

Let’s take the vector on $CE$ ,

  $\overrightarrow{OC}+\overrightarrow{OA}+\overrightarrow{OB}=\overrightarrow{OC}+\overrightarrow{OE}$

Since it is situated on two different directions, then

  $\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}=0$

From part (b),

  $\begin{array}{c}2\overrightarrow{O{M}_1}+\overrightarrow{OC}=\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}\\ 2\overrightarrow{O{M}_1}+\overrightarrow{OC}=0\\ \overrightarrow{OC}=-2\overrightarrow{O{M}_1}\end{array}$

Similarly,

  $\begin{array}{c}\overrightarrow{OB}=-2\overrightarrow{O{M}_3}\\ \overrightarrow{OA}=-2\overrightarrow{O{M}_2}\end{array}$

Thus, the vector $\overrightarrow{OB}$ and $\overrightarrow{O{M}_3}$ passes through the same direction at point $O$ .

Therefore, the third median passes through the intersection of the other because the points $B,O,M_3$ are collinear.