# Orthocentre distance to triangle vertices as a function of triangle angles and side lengths. While solving one of Brilliant problems I came across an interesting property of an orthocentre which I have not thought of before, so I decided to share it with Brilliant community. Let's consider a triangle $ABC$ with circumcentre $O$, orthocentre $H$, radius $R$, $AB=c, AC = b, BC = a$. Let's create circumcircle with diameter $AF$ and connect points $BF$. $BF \perp AB$ (from Thales' Theorem) and $CH \perp AB$ ($CH$ is part of the altitude of the triangle) $\Rightarrow BF \parallel CH$. For the same reasons we can state that $CF \parallel BH$. This means that quadrilateral $BFCH$ is a parallelogram and $BF=CH$ and $CF= BH$. It is also follows from Inscribed Angle Theorem that $\angle BFA = \angle ACB$ and $\angle AFC = \angle ABC$. $BF = 2R \cos(\angle BFA ) = \dfrac{AB}{\tan(\angle BFA)} \Rightarrow$ $CH = 2R \cos(C) = \dfrac{c}{\tan(C)}$

For the same reasons the following holds true: $AH = 2R \cos(A) = \dfrac{a}{\tan(A)}$ $BH = 2R \cos(B) = \dfrac{b}{\tan(B)}$

Because of special cases for right angles, It is better rewritten as: $AH = 2R \cos(A) = a \times \cot(A)$ $BH = 2R \cos(B) = b \times \cot(B)$ Note by Maria Kozlowska
1 year, 5 months ago

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Some criticisms:

1. You will need to expand on your proof. Specifically, say that $BF \perp AB$ because of Thales' theorem and that $CH \perp AB$, since the line joining CH is an altitude of the triangle $\overline{ABC}$ through $C$.

2. What is $R$? Please define your terms precisely. I will also make the same argument with the upper-case and lower-case $a$'s, $b$'s and $c$'s... are they points? Angles? Distances?

All in all, an unclear proof to an interesting but probably well-known result (I do not think this is new, though; this would be something that was well-known to probably Euler or Monge, but I would have no clue on the historical context). I would be interested to see if this result generalises to arbitrary dot product geometries.

P.S. If I find a result in the rational analog for not only the triangle but also the tetrahedron, I'll be sure to credit you!

- 1 year, 3 months ago

The symbols are commonly used as a standard, but I added the information. The very reason I wrote that note is that I have not seen that information before. I am not claiming it is new. Most likely it is not. It is for the information for other members of Brilliant. It is nothing advanced, rather simple, but could be useful in some instances. I was not going for a strict proof, rather sketchy one. Just sharing information with others. For more information on orthocentre properties you can also check my solution for problem

- 1 year, 3 months ago

No worries. I am currently finishing my thesis in three-dimensional rational geometry, but when I’m done I have two papers to write on tetrahedron centres in rational geometry. Will have a look at your stuff when I’m onto it.

- 1 year, 3 months ago

Wow ! That's pretty cool . Please keep up the good work 😀

- 6 months, 3 weeks ago