0
$\begingroup$

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.

enter image description here

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.

$\endgroup$
  • $\begingroup$ the two is different, let A=90, B=C=45,you will see the difference. your is wrong. $\endgroup$ – chenbai Apr 26 '19 at 13:02
1
$\begingroup$

"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 $ABC$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.