5
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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 $\alpha,\beta,\gamma$, at the next, $\beta,\gamma,\delta$, at the next $\gamma,\delta,\varepsilon$, then $\delta,\varepsilon,\alpha$, and finally $\varepsilon,\alpha,\beta$. Note that $$ \alpha+\beta+\gamma+\delta+\varepsilon=\pi. \tag{constraint} $$ Later note: (Lest anything be misunderstood, notice that what I wrote above is true of all pentagons inscribed in a circle. Angles with vertices on the circle have the same measure if they're subtended by the same arc. Consequently if the three angles between adjacent diagonals at one vertex are $\alpha,\beta,\gamma$, in that order, then two of the angles between adjacent diagonals at one of the neighboring vertices must be $\alpha$ and $\beta$, and two of those at the other neighboring vertex must be $\beta$ and $\gamma$. And regardless of the shape of the pentagon, the sum of the five angles must be a half-circle. That's a general proposition about polygons inscribed in a circle, which, when applied to triangles, says the sum of the three angles is a half-circle.) End of later note

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 $$ \frac 1 2 \left(\overbrace{\sin\alpha\sin\beta\sin\gamma}^\text{sines}\ \overbrace{\cos\delta\cos\varepsilon}^\text{cosines} +\ \cdots\text{nine more terms }\cdots\ - \overbrace{2\sin\alpha\sin\beta\sin\gamma\sin\delta\sin\varepsilon}^\text{all sines} \right). $$ (It should be obvious what the nine more terms are: 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 here at least once before.)

Can the eleven terms be interpreted as areas?

LATER EDIT: Even for quadrilaterals it seems mysterious. If the angles between adjacent diagonals are $\alpha+\beta+\gamma+\delta=\pi$, and two of them occur at each vertex, and each occurs at two of the four vertices, the the area is $$ \frac{\sin(2\alpha)+\sin(2\beta)+\sin(2\gamma)+\sin(2\delta)}{8} = \frac 1 2\Big(\overbrace{\sin\alpha\sin\beta\sin\gamma}^\text{sines}\ \overbrace{\cos\delta}^\text{cosine}+\cdots\text{ three more terms }\cdots\Big) $$

You might think that the four terms are areas of the four triangles into which the polygon is divided by the diagonals. But guess what?? They're not! Similarly, the pentagram divides the pentagon into $11$ triangles, and there are $11$ terms on the right side, but they don't correspond to the areas.

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    Unless your circle has radius $\frac{1}{2}$, the formula (1) seems off by a factor of $4$.2012-07-05
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    @ThomasAndrews : I've added the words "of unit diamter". That's how you do it if you want the lengths of the three sides of a triangle to be equal to the sines of the angles.2012-07-05
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    @ThomasAndrews : A small elaboration: If you state the law of sines by saying the lengths of the sides are proportional to the sines, then the constant of proportionality is the _diameter_ of the circumscribed circle. So that's what we want to be equal to $1$ if we want the lengths to be the sines.2012-07-05
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    A picture would be very nice here. Can you deduce all the internal angles in the pentagram formed by the pentagon and its diagonals? (I believe that drawing the picture reveals additional constraints that force all five of your angles to be $\pi/5$.)2012-07-06
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    @mjqxxxx : Certainly it's easy to deduce those angles. But I'm not assuming any regularity, so I don't know how you get $\pi/5$.2012-07-06
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    It's clear that the trigonometric terms (11 for the pentagon, or 4 for the quadrilateral) cannot correspond directly to the areas of the (11 or 4) apparent sub-triangles: There's nothing keeping one of the angles from being obtuse, which would make terms involving the cosine of that angle negative, even though the total area remains the sum of *positive* sub-triangle areas. Whether the terms can correspond to "signed areas" of some cleverly-related auxiliary figures is another matter.2012-07-06
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    However: when you got to hexagons, you end up with 26 terms in the sum and only 25 sub-regions (and the sub-regions are not always triangles, even with the pentagon), and with heptagons, you have 57 terms in the sum and 50 subregions.2012-07-06
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    @MichaelHardy: That the term-counts don't match the region-counts in more-sided polygons is just extra evidence that there's no correspondence between the two things, isn't it?2012-07-06
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    @DayLateDon : There is not a one-to-one correspondence between them. But the question remains: Do the terms have a geometric meaning?2012-07-06
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    http://mathoverflow.net/q/101534/123572014-02-26

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