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 June 2nd, 2019, 10:41 AM #1 Senior Member     Joined: Jan 2012 Posts: 137 Thanks: 2 Circumradius of an especial triangle Hello, There is a triangle ABC, the orthocenter of which is the point P. How can we find the circumradius of the triangle PBC in terms of R (where R is the circumradius of the triangle ABC)? Thanks. Last edited by skipjack; June 2nd, 2019 at 11:35 AM.
 June 2nd, 2019, 12:15 PM #2 Global Moderator   Joined: Dec 2006 Posts: 20,746 Thanks: 2133 They're equal.
June 2nd, 2019, 12:34 PM   #3
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Quote:
 Originally Posted by happy21 There is a triangle ABC, the orthocenter of which is the point P. How can we find the circumradius of the triangle PBC in terms of R (where R is the circumradius of the triangle ABC)?
Two constructions attached. D is the circumcenter of ABC.

Seems like the circumradii are congruent. Maybe you can see a way to prove it?
Attached Images
 ortho1.jpg (45.2 KB, 4 views) ortho2.jpg (43.3 KB, 3 views)

 June 2nd, 2019, 02:44 PM #4 Global Moderator   Joined: Dec 2006 Posts: 20,746 Thanks: 2133 BC is a chord of both circumcircles. Consider the angles it subtends at A and P.
 June 3rd, 2019, 01:55 AM #5 Senior Member     Joined: Jan 2012 Posts: 137 Thanks: 2 I guess we can also prove using Sine rule...but I am not getting exactly how to do...Can u elaborate that? Thx.
June 3rd, 2019, 04:12 AM   #6
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Quote:
 Originally Posted by skeeter Two constructions attached. D is the circumcenter of ABC. Seems like the circumradii are congruent. Maybe you can see a way to prove it?
Can u prove it in steps also? Thx.

 June 3rd, 2019, 10:41 AM #7 Global Moderator   Joined: Dec 2006 Posts: 20,746 Thanks: 2133 The angles I mentioned are supplementary, so they have equal sine, which makes proof by the sine rule easy. Thanks from happy21
June 3rd, 2019, 11:50 AM   #8
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Quote:
 Originally Posted by skipjack The angles I mentioned are supplementary, so they have equal sine, which makes proof by the sine rule easy.
Ohk...I try that way.

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