Let $$ABC$$ be a triangle whose circumcentre is at $$P$$. If the position vectors of $$A$$, $$B$$, $$C$$ and $$P$$ are $$\vec{a}$$, $$\vec{b}$$, $$\vec{c}$$ and $$\frac{\vec{a}+\vec{b}+\vec{c}}{4}$$ respectively, then the position vector of the orthocentre of this triangle, is:

We are given that the position vectors of the vertices of the triangle are $$\vec{a},\; \vec{b},\; \vec{c}$$ and the position vector of its circum-centre is

$$\vec{P}= \frac{\vec{a}+\vec{b}+\vec{c}}{4}\;.$$

First recall the standard results concerning a triangle whose circum-centre is taken as the origin.

1. If we shift the origin to the circum-centre, the position vectors of the vertices become $$\vec{a}\,'=\vec{a}-\vec{P},\; \vec{b}\,'=\vec{b}-\vec{P},\; \vec{c}\,'=\vec{c}-\vec{P}.$$

2. In that coordinate system (origin at circum-centre) the orthocentre has the well-known vector form

$$\vec{H}\,'=\vec{a}\,'+\vec{b}\,'+\vec{c}\,'$$

because each altitude is perpendicular to a side chord through the centre of the circle.

3. Also, in any triangle the centroid is the average of the vertex vectors. Hence, with origin at the circum-centre, the centroid vector is

$$\vec{G}\,'=\frac{\vec{a}\,'+\vec{b}\,'+\vec{c}\,'}{3}\;.$$

From point 2 we immediately have the Euler line relation in that system:

$$\vec{H}\,'=3\vec{G}\,'\;.$$

Now we translate every quantity back to the original origin, where the circum-centre has the (non-zero) vector $$\vec{P}.$$

• The centroid in the original system is obtained by adding $$\vec{P}$$ to $$\vec{G}\,'.$$ Hence

$$\vec{G}= \vec{P}+ \vec{G}\,'= \vec{P}+ \frac{\vec{a}\,'+\vec{b}\,'+\vec{c}\,'}{3}.$$

Since $$\vec{a}\,'+\vec{b}\,'+\vec{c}\,' = (\vec{a}-\vec{P})+(\vec{b}-\vec{P})+(\vec{c}-\vec{P}) = \vec{a}+\vec{b}+\vec{c}-3\vec{P},$$ we have

$$\vec{G}= \vec{P}+ \frac{\vec{a}+\vec{b}+\vec{c}-3\vec{P}}{3}

= \frac{\vec{a}+\vec{b}+\vec{c}}{3}.$$

(This is just the familiar centroid formula, so the translation checks out.)

• The orthocentre vector in the original system is obtained by adding $$\vec{P}$$ to $$\vec{H}\,'.$$ Using $$\vec{H}\,'= \vec{a}\,'+\vec{b}\,'+\vec{c}\,'$$ gives

$$\vec{H}= \vec{P}+ (\vec{a}\,'+\vec{b}\,'+\vec{c}\,')

= \vec{P}+ \bigl(\vec{a}+\vec{b}+\vec{c}-3\vec{P}\bigr)

= \bigl(\vec{a}+\vec{b}+\vec{c}\bigr)-2\vec{P}.$$

But $$\vec{P}= \dfrac{\vec{a}+\vec{b}+\vec{c}}{4},$$ so substituting this value yields

$$\vec{H}= \bigl(\vec{a}+\vec{b}+\vec{c}\bigr)-2\left(\frac{\vec{a}+\vec{b}+\vec{c}}{4}\right)

= \bigl(\vec{a}+\vec{b}+\vec{c}\bigr)-\frac{\vec{a}+\vec{b}+\vec{c}}{2}

= \frac{\vec{a}+\vec{b}+\vec{c}}{2}\;.$$

Thus the position vector of the orthocentre of the triangle is

$$\boxed{\dfrac{\vec{a}+\vec{b}+\vec{c}}{2}}.$$

Hence, the correct answer is Option C.