In a triangle,what is the ratio of the distance between a vertex and the orthocenter ...

Question

In a triangle,what is the ratio of the distance between a vertex and the orthocenter and the distance of the circumcenter from the side opposite vertex.

Answer 1

The triangle is $\Delta ABC$. Let's say that $O$ is the circumcenter, $H$ the orthocenter and $G$ is the centroid. We are going to use a very usefull result that comes from the Euler Line. The Euler Line says that $OH=3GO \quad (1)$.

If we use vectors to solve that problem the relation $(1)$ says that if we place the origin at $O$ then the coordinates of the orthocenter will be given by $H=A+B+C$ and then:

$$\vec{AH}=H-A=B+C$$

We also know that the distance from the circuncenter ($O$) to the side ($BC$) is the distance from $O$ to the midpoint (called $M$) of $BC$.

$$\vec{OM}=M=\frac{B+C}{2}$$

Then

$$\vec{AH}=2\cdot \vec{OM}\Leftrightarrow\frac{|\vec{AH}|}{|\vec{OM}|}=2$$

Answer 2

Hint: If $A$ is the vertex, $A_1$ is the midpoint of the opposite side, $O$ is the orthocenter, $C$ is the circumcenter and $M$ is the centroid, just check the similar triangles $AOM$ and $A_1CM$.