# Line joining the orthocenter to the circumcenter of a triangle ABC is inclined to BC at an angle $\tan^{-1}(\frac{3-\tan B\tan C}{\tan B-\tan C})$

Show that the line joining the orthocenter to the circumscribed center of a triangle ABC is inclined to BC at an angle $\tan^{-1}\left(\frac{3-\tan B\tan C}{\tan B-\tan C}\right)$

I let the foot of perpendicular from A,B,C to opposite sides is D,E,F.Then

$$\tan B=\frac{AD}{BD},\tan C=\frac{AD}{CD}$$

$$\frac{3-\tan B\tan C}{\tan B-\tan C}=\frac{3-\frac{AD}{BD}\frac{AD}{CD}}{\frac{AD}{BD}-\frac{AD}{CD}}$$

Suppose $D=(0,0), A=(0,1),B=(b,0),C=(c,0)$.

Then the orthocenter $H=(0,-bc)$ and the circumcenter $O=(\frac{b+c}2,\frac{bc+1}2)$ after simple computation. The slope of $OH$ is $\frac{3bc+1}{b+c}$.

Note that $\tan B=-\frac1b$ and $\tan C=\frac1c$. The proof is completed.

• You have taken all the three points of ABC on $y$ axis,how is this possible?
– diya
Aug 6, 2015 at 15:14
• @diya That's a typo. Thanks. Aug 6, 2015 at 15:16
• Sir,i understood your method,but i want to ask if there is some other geometrical method to prove this formula.Can we do it without supposing the coordinate of the vertices?Thanks.
– diya
Aug 6, 2015 at 15:39
• You can just note that by the property of scaling of axis, rotation of axis, and translation of axis. This is just good enough to work with any other case. Aug 6, 2015 at 16:35
• @diya You can translate it into trigonometric language. But the equality to be proved is so complicated that without coordinates the computation would be difficult. Aug 7, 2015 at 0:49

Let $a$ be the length of side BC coinciding with the x-axis such that vertex B is at origin $(0, 0)$ & vertex C is at $(a, 0)$ then vertex A will at $\left(\frac{a\tan C}{\tan B+\tan C}, \frac{a\tan B\tan C}{\tan B+\tan C}\right)$ Hence, the ortho-center will be $$H\equiv \left(\frac{0+a+\frac{a\tan C}{\tan B+\tan C}}{3}, \frac{0+0+\frac{a\tan B\tan C}{\tan B+\tan C}}{3} \right)\equiv \left(\frac{a(\tan B+2\tan C)}{3(\tan B+\tan C)}, \frac{a\tan B\tan C}{3(\tan B+\tan C)} \right)$$ & the circumscribed center say $D$ can be calculated as $$D\equiv\left(\frac{a}{2}, \frac{a(\tan B\tan C-1)}{2(\tan B+\tan C)} \right)$$

Hence, the slope of the line HD joining $H$ & $D$ with the BC (x-axis) is given as $$m=\frac{y_2-y_1}{x_2-x_1}$$ $$=\frac{\frac{a(\tan B\tan C-1)}{2(\tan B+\tan C)}-\frac{a\tan B\tan C}{3(\tan B+\tan C)}}{\frac{a}{2}-\frac{a(\tan B+2\tan C)}{3(\tan B+\tan C)}}$$ $$=\frac{3\tan B\tan C-3-2\tan B\tan C}{3\tan B+3\tan C-2\tan B-4\tan C}$$ $$=\frac{\tan B\tan C-3}{\tan B-\tan C}$$ Hence the angle of the line joining H & D with side BC is given as $$\tan \theta=|m|$$ $$\implies \theta=\tan^{-1}\left|\frac{\tan B\tan C-3}{\tan B-\tan C}\right|$$ or $$\bbox [5px, border:2px solid #C0A000]{\color{red}{\theta=\tan^{-1}\left|\frac{3-\tan B\tan C}{\tan B-\tan C}\right|}}$$