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 ABCABC with circumcentre OO, orthocentre HH, radius RR, AB=c,AC=b,BC=aAB=c, AC = b, BC = a. Let's create circumcircle with diameter AFAF and connect points BFBF. BFABBF \perp AB (from Thales' Theorem) and CHABCH \perp AB (CHCH is part of the altitude of the triangle) BFCH\Rightarrow BF \parallel CH. For the same reasons we can state that CFBHCF \parallel BH. This means that quadrilateral BFCHBFCH is a parallelogram and BF=CHBF=CH and CF=BHCF= BH. It is also follows from Inscribed Angle Theorem that BFA=ACB \angle BFA = \angle ACB and AFC=ABC\angle AFC = \angle ABC. BF=2Rcos(BFA)=ABtan(BFA)BF = 2R \cos(\angle BFA ) = \dfrac{AB}{\tan(\angle BFA)} \Rightarrow CH=2Rcos(C)=ctan(C) CH = 2R \cos(C) = \dfrac{c}{\tan(C)}

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

Because of special cases for right angles, It is better rewritten as: AH=2Rcos(A)=a×cot(A) AH = 2R \cos(A) = a \times \cot(A) BH=2Rcos(B)=b×cot(B) BH = 2R \cos(B) = b \times \cot(B)

Note by Maria Kozlowska
3 years, 1 month ago

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Some criticisms:

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

  2. What is R R ? Please define your terms precisely. I will also make the same argument with the upper-case and lower-case aa's, bb's and cc'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!

A Former Brilliant Member - 2 years, 11 months ago

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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

Maria Kozlowska - 2 years, 11 months ago

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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.

A Former Brilliant Member - 2 years, 11 months ago

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

MEGHNA SINGH - 2 years, 2 months ago

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