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It is well known (?) that if $\alpha+\beta+\gamma=\pi$ then $4\sin\alpha\sin\beta\sin\gamma = \sin(2\alpha)+\sin(2\beta)+\sin(2\gamma)$ (I think I've seen it in some late-19th-century books, and I read somewhere on the internet (therefore it's true!! right?) that it has repeatedly appeared on the joint entrance examination of the Indian Institutes of Technology).

It seems very probable that this similar identity is in the literature somewhere, and I wonder where: $$ \begin{align} & {}\qquad \text{If }\alpha+\beta+\gamma=\pi\text{ then }4\sin^2\alpha\;\sin^2\beta\;\sin^2\gamma \\ & = (\sin\alpha+\sin\beta+\sin\gamma)(\sin\alpha+\sin\beta-\sin\gamma)(\sin\alpha-\sin\beta+\sin\gamma)(-\sin\alpha+\sin\beta+\sin\gamma). \end{align} $$

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I have tagged it as reference-request. – Aryabhata Aug 24 '11 at 18:22
Heron's formula + the law of sines? – Hans Lundmark Aug 24 '11 at 18:24
@Hans: Looking at this now, I think there may be essentially nothing more involved than the two identities you mention. Every few months I look at something I scrawled months ago about things like this and go on from where I left off without stopping to recall how I got there. About three years ago I actually found the sum of the distances from the vertices of a triangle to the Fermat point a.k.a. the Steiner point. Surprisingly hairy and I don't remember much about how I did it. Everybody knows that point minimizes the sum; few if any know the value of the sum. – Michael Hardy Aug 24 '11 at 20:53
@Michael: I get $2m^2=a^2+b^2+c^2+4\sqrt{3s(s-a)(s-b)(s-c)}$ where $s=(a+b+c)/2$. – robjohn Aug 24 '11 at 23:25
@Ragib: It would be, but, as shown, these are all simple consequences of the Law of Sines, the Law of Cosines, and Heron's Formula. Why invoke a CA package? I find proof by computer a little less satisfying. – robjohn Aug 25 '11 at 2:52
up vote 4 down vote accepted

It can be done using Heron's formula (*) as stated in the comments, but notice that this is simply the constraint (or the Zariski closure of the constraint) on the sines of three angles to be sines of a triangle. I don't think there is a lower-degree polynomial expressing the same condition. This raises the question of whether a conceptual solution exists, avoiding the details of classical geometry.

[(*) Start from Heron's formula (squared), replace Area by $abc/4R$, replace $a,b,c$ by $2R\sin \alpha, 2R \sin \beta, 2R \sin \gamma$. Factors of $R$ will disappear from the final result, leaving the formula on sines.

The "well known" formula at the top of the question is a similar restatement of the fact that the area of a triangle is the sum of the small triangles formed by joining vertices to the center of the circumscribed circle.]

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The areas of the small triangles are signed areas. – Michael Hardy Aug 25 '11 at 5:31

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