Find the value | p+q |

Question

OPQ is an equilateral triangle where R is a midpoint of PQ. Given OP = p , OQ = q , and | p | = 2, find the value of | p+q | (https://i.stack.imgur.com/IATYl.jpg)

Answer 1

Given the $$ OP = p, \ \mbox{ and } \ OQ = q, $$ we find that $$ \vec{PQ} = \vec{PO} + \vec{OQ} = \vec{OQ} + \vec{PO} = \vec{OQ} - \vec{OP} = \mathbf{q} - \mathbf{p}. $$ So $$ \vec{OR} = \vec{OP} + \vec{PR} = \vec{OP} + \frac{1}{2} \vec{PQ} = \mathbf{p} + \frac{1}{2} \left( \mathbf{q} - \mathbf{p} \right) = \frac{1}{2} \left( \mathbf{p} + \mathbf{q} \right). $$ Now extend the line segment $\overline{OR}$ to twice its original length beyond the point $R$ to arrive at the point $S$. Now join the points $P$ and $S$ by a line segment. Then $$ \vec{PS} = \vec{OQ}, $$ and also $$ \vec{OS} = \mathbf{p} + \mathbf{q} = \vec{OP} + \vec{OQ}. $$

Thus, in the $\triangle OPS$, we find that $$ \left\lvert \overline{OP} \right\rvert = \lvert \mathbf{p} \rvert = 2 = \lvert \mathbf{q} \rvert = \left\lvert \overline{OQ} \right\rvert = \left\lvert \overline{PS} \right\rvert. $$ Thus the $\triangel OPS$ is an isosceles triangle.

Moreover as $$ m \angle POQ = 60^\circ, $$ so we must have $$ m \angle OPS = 120^\circ, $$ by the fact that these two angles are the so-called alternating angles when the line $OP$ cuts the parallel lines $OQ$ and $PS$.

Finally, using the law of cosines in $\triangle OPS$, we obtain $$ \begin{align} \lvert \mathbf{p} + \mathbf{q} \rvert = \left\lvert \overline{OS} \right\rvert &= \sqrt{ \left\lvert \overline{OP} \right\rvert^2 + \left\lvert \overline{PS} \right\rvert^2 - 2 \left\lvert \overline{OP} \right\rvert \cdot \left\lvert \overline{PS} \right\rvert \cdot \cos 120^\circ } \\ &= \sqrt{ 2^2 + 2^2 - 2 \times 2 \times 2 \times \frac{1}{2} } \\ &= \sqrt{4}\\ &= 2. \end{align} $$

Answer 2

Hint: since the triangle is equilateral, we have $|q|=|p|=2$.