What is wrong with the solution of 'Find that the distance between the circumcenter and the orthocenter of triangle ABC'.

This is NOT a duplicate of - Distance between orthocenter and circumcenter.

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As O is circumcenter $ \angle BOD = \angle A$

$\angle OBD = 90$° - A

$\angle ABL = 90°- A$ ( $ \angle ALB is 90 degrees)

$B = -2A + \pi + \angle OBL$

$\angle OBL = B + 2A - \pi$ now using sine rule in triangle OHB

$OH = - sin(\pi - (2A+B) 2R$

$OH = - 2Rsin(2A+B)$

But the answer is R $ \sqrt{1-8\ cosA \ cosB \ cosC} $

maybe my answer is correct but just in the correct form.

  • $\begingroup$ the two is different, let A=90, B=C=45,you will see the difference. your is wrong. $\endgroup$
    – chenbai
    Commented Apr 26, 2019 at 13:02

1 Answer 1


"now using sine rule in triangle OHB"

How do you know the circumradius of $OHB$?

In the answer $R \sqrt{1−8 cosA cosB cosC}$ the letter $R$ probably refers to the circumradius of $ABC$.


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