Let the position vectors of the vertices of a triangle be such that the orthocenter is p+ …

Question

Let the position vectors of the vertices of a triangle be such that the orthocenter is $\dfrac{\vec p+\vec q+\vec r}4$ and the circumcenter is $\alpha\vec p+\beta\vec q+\gamma\vec r$. Then $\alpha+2\beta+5\gamma$ is equal to:

Accepted Answer

Using the Euler line relation $O + 2\cdot(\text{circumcenter}) = 3\cdot(\text{centroid})$, with centroid $=\dfrac{\vec p+\vec q+\vec r}3$:
$$2(\alpha\vec p+\beta\vec q+\gamma\vec r) + \frac{\vec p+\vec q+\vec r}4 = 3\cdot\frac{\vec p+\vec q+\vec r}3$$
$$8(\alpha\vec p+\beta\vec q+\gamma\vec r) = 3(\vec p+\vec q+\vec r) \implies \alpha=\beta=\gamma=\frac38$$

$$\alpha+2\beta+5\gamma = \frac38+\frac68+\frac{15}8 = \frac{24}8=3$$

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