I posted this question to stackexchange, where it's generated some comments but no progress toward answering it. I'm going to say somewhat more here than I did there.

At one vertex of a pentagon inscribed in a circle of unit diameter (unit diameter, not unit radius) let the angles between adjacent diagonals be $α,β,γ$. It follows that at one of the two vertices adjacent to that one, two of the angles, since they are subtended by the same arcs, must be $\beta,\gamma$; call the next one $\delta$, so we have $β,γ,δ$. At the next vertex we have $γ,δ,ε$, then $δ,ε,α$, and finally $ε,α,β$. It necessarily follows that $$α+β+γ+δ+ε=π.\tag{constraint}$$

It's not hard to show that the area of the pentagon is $$ \frac{\sin(2\alpha)+\sin(2\beta)+\sin(2\gamma)+\sin(2\delta)+\sin(2\varepsilon)}{8}.\tag{1} $$ It's somewhat more work than that to show that if the "constraint" above holds, then $(1)$ is equal to $$ \frac 1 2 \left(\overbrace{\sin\alpha\sin\beta\sin\gamma}^\text{sines}\ \overbrace{\cos\delta\cos\varepsilon}^\text{cosines} +\ \cdots\text{nine more terms }\cdots\ - \overbrace{2\sin\alpha\sin\beta\sin\gamma\sin\delta\sin\varepsilon}^\text{all sines} \right). $$ (For the nine more terms: choose three factors in each term to be sines and then the other two are cosines.)

(As far as I know, this is my own. I've mentioned it on stackexchange at least once before.)

QUESTION: Can the eleven terms be interpreted as areas, or do they otherwise have a geometric meaning?

One temptation is to think that the eleven terms correspond to the eleven regions into which the diagonals divide the interior of the pentagon. But the numbers don't correspond to their areas, even if we try to make sense of using "signed" areas in some cases.

With a cyclic quadrilateral we have $\alpha+\beta+\gamma+\delta=\pi$ and $$ \begin{align} & \frac{\sin(2\alpha)+\sin(2\beta)+\sin(2\gamma)+\sin(2\delta)}{8} \\ = {} & \frac 1 2 \left(\overbrace{\sin\alpha\sin\beta\sin\gamma}^\text{sines}\ \overbrace{\cos\delta}^\text{cosine}+ \cdots\text{ three more terms }\cdots\right). \end{align} $$ Again, the sum of the four terms equals the sum of the areas of the four smaller polygons into which the diagonals divide the polygon, but again, the areas don't coincide at all.

With hexagons we have $\alpha+\beta+\gamma+\delta+\varepsilon+\zeta=\pi$ and $$ \begin{align} & \frac{\sin(2\alpha)+\sin(2\beta)+\sin(2\gamma)+\sin(2\delta)+\sin(2\varepsilon)+\sin(2\zeta)}{8} \\ = {} & \frac 1 2 \left(\overbrace{\sin\alpha\sin\beta\sin\gamma}^\text{sines}\ \overbrace{\cos\delta\cos\varepsilon\cos\zeta}^\text{cosines}+ \cdots\text{ 19 more terms }\cdots\right) \\ & {} - \frac 1 2 \left(2\overbrace{\sin\alpha\sin\beta\sin\gamma\sin\delta\sin\varepsilon}\ \overbrace{\cos\zeta}+ \cdots\text{ five more terms }\cdots\right) \end{align} $$ This time there are $26$ terms and only $25$ regions into which the polygon gets divided, so the tempting false conjecture is no longer there.

With heptagons, we have $\alpha+\beta+\gamma+\delta+\varepsilon+\zeta+\eta=\pi$ and $$ \begin{align} & \frac{\sin(2\alpha)+\sin(2\beta)+\sin(2\gamma)+\sin(2\delta)+\sin(2\varepsilon)+\sin(2\zeta)+\sin(2\eta)}{8} \\ = {} & \frac 1 2 \left(\overbrace{\sin\alpha\sin\beta\sin\gamma}^\text{sines}\ \overbrace{\cos\delta\cos\varepsilon\cos\zeta\cos\eta}^\text{cosines}+ \cdots\text{ 34 more terms }\cdots\right) \\ & {} - \frac 1 2 \left(2\overbrace{\sin\alpha\sin\beta\sin\gamma\sin\delta\sin\varepsilon}\ \overbrace{\cos\zeta\cos\eta} + \cdots\text{ 20 more terms }\cdots\right) \\ & {} + \frac 1 2 \left( 3 \,\overbrace{ \sin\alpha \sin\beta \sin\gamma \sin\delta \sin\varepsilon \sin\zeta \sin\eta}^\text{all sines} \right) \end{align} $$ There are $57$ terms. The number of sub-polygons is $50$.

The pattern goes on, with the alternation of signs and the arithmetically growing coefficients $1,2,3,\ldots$.

QUESTION: Do the terms on the right sides of these identities have a geometric meaning?

  • $\begingroup$ All of the denominators under the sums of double angles should be $8$. Some of them were not. I've fixed all (?) of the typos, I hope. $\endgroup$ Jul 7, 2012 at 0:20
  • $\begingroup$ BTW, the reason I used unit diameter rather than unit radius is that that's when the constant of proportionality in the law of sines is $1$, i.e. the length of each side is equal to the sine of the opposite angle. The length of each side is generally the diameter of the circumscribed circle times the sine of the opposite angle. $\endgroup$ Jul 7, 2012 at 0:22
  • 2
    $\begingroup$ math.stackexchange.com/questions/167203/… $\endgroup$
    – JRN
    Jul 7, 2012 at 9:23

2 Answers 2


To help visualize Michael's setup:
      Circular Pentagon


This won't answer the question, but Joseph O'Rourke got four up-votes for doing some nice graphics to clarify the question.

In the case of the quadrilateral, we have $$ \sin\alpha\sin\beta\sin\gamma\cos\delta + \cdots\text{ three more terms } \cdots $$ And then for the first of the four terms: $$ \begin{align} & \sin\alpha\sin\beta\sin\gamma\cos\delta \\[10pt] = {} & -\sin\alpha\sin\beta\sin\gamma\cos(\alpha+\beta+\gamma)\qquad\text{ (since }\alpha+\beta+\gamma+\delta=\pi) \\[10pt] = {} & \frac 1 8 {\Big(} \sin(2\alpha+2\beta+2\gamma) - \sin(2\alpha+2\beta) - \sin(2\alpha+2\gamma) - \sin(2\beta+2\gamma) \\[6pt] & {} + \sin(2\alpha)+\sin(2\beta) + \sin(2\gamma) - 0{\Big)} \end{align} $$ The last term is the sine of the empty sum (so if these were cosines, we'd see a $1$ there). Clearly this is reminiscent of usual inclusion-exclusion principle.


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