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The solution to Putnam 2000 A5 uses this formula, for which the following proof is given: (source: https://mks.mff.cuni.cz/kalva/putnam/psoln/psol005.html)

Let the sides (of triangle $ABC$) have lengths $a, b, c$ as usual. The question suggests that we use some relationship of the form $abc = constant \times R$. ...

To prove the relation, let $O$ be the centre of the circumcircle. Project $AO$ to meet the circle again at $K$. Let $AH$ be the altitude. Then angle $ABC = \angle AKC$, so triangles $ABH$ and $AKC$ are similar. Hence $\frac{AB}{AH} = \frac{AK}{AC}$ or $\frac{c}{AH} = \frac{2R}b$. Hence $abc = 2R·a·AH = 4\Delta R$.

The bold part confuses me. How does $ABC = AKC$? $K$ is dependent wholly on $O$ and $A$ whereas $B$ is independent of both. And if $ABC = AKC$, how does that lead to $ABH \sim AKC$?

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  • $\begingroup$ Inscribed Angle Theorem? $\endgroup$
    – Blue
    Jul 30, 2014 at 23:08
  • $\begingroup$ The bold part is in fact the result of a theorem called "angles in the same segment". Or equivalently, "equal arcs, then equal angles". $\endgroup$
    – Mick
    Jul 31, 2014 at 5:28
  • $\begingroup$ Note also that angle ABH = 90 degrees (AH is the altitude from A to BC) and angle ACK = 90 degrees (angles in semi-circle). Therefore, the two triangles are similar (AAA). $\endgroup$
    – Mick
    Jul 31, 2014 at 8:25
  • $\begingroup$ @Mick Thanks for the comment. But I am still confused. Don't you mean AHB not ABH? AHB = ACK = 90 and ABC = AKC. But how does that make ABH ~ AKC? In order to prove similarity we need ABH = AKC but we have ABC = AKC instead. imgur.com/2oUBqPr $\endgroup$
    – user1299784
    Jul 31, 2014 at 21:28
  • $\begingroup$ @user1299784 The confusion comes from you and the author (and also me) are drawing different pictures. It will be quite clear if you re-draw your picture and let your angle B be acute (instead of obtuse). Then, H will lie inside the circle (instead of outside). The next question is then “why an obtuse triangle won’t work?” The answer is “it will also work" but the proof will run differently. $\endgroup$
    – Mick
    Aug 1, 2014 at 2:24

2 Answers 2

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Angles intercepting the same arc are equal. Both angles are intercepting $\widehat{AC}$.

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Area of the inscribed $ \triangle = \frac 12 a.b\sin C$

Try to draw the plane and we will get $\sin C = \frac{c}{2R}$

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