# cyclic polygons & trigonometry

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 $4$. –  Thomas Andrews Jul 5 '12 at 20:21
@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. –  Michael Hardy Jul 5 '12 at 20:45
@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. –  Michael Hardy Jul 5 '12 at 21:06
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 $\pi/5$.) –  mjqxxxx Jul 6 '12 at 0:20
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. –  Blue Jul 6 '12 at 20:52

The figure below shows the pentagon, its diagonals, and the labeling of a few of the angles in accordance with the problem definition as I understand it. The important point to note is that there are additional constraints among the five angles. For example, consider the grey triangle. Its unlabeled angles, since they are exterior angles to the small triangles to the left and right, have to be $\alpha + \beta$ and $\beta + \gamma$. For its interior angles to sum to $\pi$, then, we must have $\alpha + 3\beta + \gamma = \pi$. Each of the other four "star point" triangles gives a similar (cyclically permuted) constraint, for a total of five linear relations among the five angles. One can check that the relations are non-singular: they have a unique joint solution, which is $\alpha=\beta=\gamma=\delta=\varepsilon=\pi/5$. In other words, if the diagonal-to-diagonal angles are to be as described in the problem, then the pentagon must be regular.
These angles are not correctly labeled. If the three angles at the uppermost vertex in this picture are as labeled, then the one you've labeld $\gamma$ on the right should be $\alpha$, and the lowest one at that vertex, which you've left unlabeled, should be $\beta$. Similar objections apply to the one on the left, where the lowest one, left unlabeled, should be labeled $\beta$, and the one you've called $\alpha$ should be $\gamma$. A simple theorem of geometry says that if the same arc of the circle subtends two angles with vertices on the circle, then both have the same measure. –  Michael Hardy Jul 6 '12 at 18:47
@MichaelHardy: This image matches my (apparent mis-)reading of the first paragraph of your question; I believe I got confused by the way you stated the order of the angles. Perhaps you could re-phrase in an order-agnostic way, such as: "Label the sides of the pentagon $a$, $b$, $c$, $d$, $e$, and at the vertices, let the angles sub-tending these chords be $\alpha$, $\beta$, $\gamma$, $\delta$, $\epsilon$, respectively." (If that's consistent with your intent.) Or, maybe you could just post your own image. –  Blue Jul 6 '12 at 19:56
@MichaelHardy: You're right, of course. The way you stated the problem made me think the angles appeared in the specified order at each vertex; but when you recall that the angles subtending a common arc have common measure, the order doesn't need to be specified at all. The "star point" triangles now all give the same constraint, namely, $\alpha+\beta+\gamma+\delta+\varepsilon=\pi$. –  mjqxxxx Jul 6 '12 at 23:04