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It follows from the law of cosines that if $a,b,c$ are the lengths of the sides of a triangle with respective opposite angles $\alpha,\beta,\gamma$, then $$ a^2+b^2+c^2 = 2ab\cos\gamma + 2ac\cos\beta + 2bc\cos\alpha. $$

For a cyclic (i.e. inscribed in a circle) polygon, consider the angle "opposite" a side to be the angle between adjacent diagonals whose endpoints are those of that side (it doesn't matter which vertex of the polygon serves as the vertex of that angle because of the Inscribed Angle Theorem). Then for a cyclic quadrilateral with sides $a,b,c,d$ and opposite angles $\alpha,\beta,\gamma,\delta$, one can show that $$ \begin{align} a^2+b^2+c^2+d^2 = {} & 2ab\cos\gamma\cos\delta + 2ac\cos\beta\cos\delta + 2ad\cos\beta\cos\gamma \\ & {} +2bc\cos\alpha\cos\delta+2bd\cos\alpha\cos\gamma+2cd\cos\alpha\cos\beta \\ & {}-4\frac{abcd}{(\text{diameter})^2} \end{align} $$ And for a cyclic pentagon, with sides $a,b,c,d,e$ and respective opposite angles $\alpha,\beta,\gamma,\delta,\varepsilon$, $$ \begin{align} a^2 + \cdots + e^2 = {} & 2ab\cos\gamma\cos\delta\cos\varepsilon+\text{9 more terms} \\& {} - 4\frac{abcd}{(\text{diameter})^2}\cos\varepsilon+ \text{4 more terms} \end{align} $$ And for a cyclic $n$-gon with sides $a_i$ and opposite angles $\alpha_i$, $$ \begin{align} \sum_{i=1}^n a_i^2 = {} & \text{a sum of }\binom{n}{2}\text{ terms each with coefficient 2} \\ & {} - \text{a sum of }\binom{n}{4}\text{ terms each with coefficient 4} \\ & {} + \text{a sum of }\binom{n}{6}\text{ terms each with coefficient 6} \\ & {} - \cdots \end{align} $$ The number of terms depends on $n$ and the power of the diameter on the bottom is in each case what is needed to make the term homogeneous of degree $2$ in the side lengths ("dimensional correctness" if you like physicists' language), and the alternation of signs continues.

I showed this by induction. It should work for infinitely many sides, too, by taking limits. Each term would then have a product of infinitely many cosines.

My question is: Is there some reasonable geometric interpretation of the sum of squares of sides of a polygon inscribed in a circle?

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    $\begingroup$ This was crossposted at MO. Michael, please wait some time before posting your question in multiple fora, and when you do, provide links to the other posts (there is already an answer on the MO one by Noam Elkies). $\endgroup$ Jul 31, 2011 at 6:11
  • $\begingroup$ I mildly object to the notion that the first identity follows from the law of cosines: in fact, the usual trigonometric proof proceeds by multiplying $c = a\cos{\beta} + b\cos{\alpha}$ by $c$. Doing this for each side and writing out $a^2 + b^2 - c^2$ gives you the law of cosines, writing out $a^2 + b^2 + c^2$ gives you the identity you ask about. $\endgroup$
    – t.b.
    Jul 31, 2011 at 19:48
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    $\begingroup$ No geometric interpretation, but an observation: Assuming the polygon convex (a requirement for each $\alpha_i$ to be uniquely determined), we have $a_i = 2 r \sin{\alpha_i}$, where $r$ is the radius of the circle. Moreover, the arcs determined by the polygon's sides (obviously) comprise the full circle; hence the inscribed angles sum to $\pi$. So, dividing your formula by $4 r^2$ leaves a trig identity for the sum of squares of sines of $n$ (non-negative) angles that total $\pi$. $\endgroup$
    – Blue
    Jul 31, 2011 at 23:04
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    $\begingroup$ @Theo: Here's a picture. :) math.stackexchange.com/questions/803/… $\endgroup$
    – Blue
    Jul 31, 2011 at 23:42

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I haven't worked it out up to arbitrary $n$, but I've managed the cases $n=3$ and $n=4$ to follow your lead and it becomes tempting to write "+4 terms" and so on.

I believe geometric interpretation of that ultimately goes down to interpreting geometrically the cosine law, since it is the main tool in your arguments.

EDIT : I just thought of something : suppose the sides of the inscribed polygon are almost are equal, or even better, that they form a regular polygon. Consider the following image :

enter image description here

You can consider the squares as a prism that rose over the inscribed polygon and then got unfolded. The sum of the squares can then be thought of an approximation of the lateral area of a cylinder with height being the average of the $a_i$'s, i.e. if you let $$ \mu = \frac{\sum_{i=1}^n a_i}n, $$ we can intuitively argue that $$ \sum_{i=1}^n a_i^2 \approx 2 \pi r \mu. $$ Hope that helps,

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    $\begingroup$ Well, instead of saying "approximately", one could just put the perimeter of the polygon in place of $2\pi r$ and they'd be exactly equal. (And the approximation would be good only when all the side lengths are small.) $\endgroup$ Aug 1, 2011 at 20:06
  • $\begingroup$ @Michael Hardy : I never said I couldn't make it right, I just didn't want to, so I intuitively argued. There is obviously a way to define what approximately means here, and get a bound for the error term, but OP asks for intuition, so I didn't bother. =) $\endgroup$ Aug 2, 2011 at 4:57
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Instead of an answer to the question you asked, let me give several observations.

Firstly, by homogeneity, we can consider the case where the diameter of the circle is 2, or the radius being 1. In this case, then, basic trigonometry gives you that $a = 2 \sin \alpha$, $b = 2\sin \beta$, etc.

Thus the right hand side consists of terms of the form $$\sin \times \sin \times \cos \times \cos\times \cos + \sin \times\sin \times\sin\times \sin \times \cos + \dots $$ which should be reminiscent of the angle addition rule for cosine.

Indeed, we have that

$$ \cos \sum \theta_k = \Re \exp \sum i\theta_k = \Re \prod \left( \cos \theta_k + i \sin\theta_k\right) $$

which is similar to the RHS that you've written down, except that on your RHS you do not have the term $\prod \cos\theta_k$. This term can be recovered by what you have on the LHS.

Note that if you have an inscribed polygon, $\sum \theta_k = \pi$. So $\cos \sum\theta_k = -1$. Notice that the LHS of your equation is proportional to $\sum \sin^2\theta_k$, and observe the following:

$$\begin{align} \sum 2\sin^2\theta_k & = \sum (1 - \cos(2\theta_k))\\ & = N - \sum_{k\neq 1} \cos(2\theta_k) - \cos 2\theta_1 \end{align}$$

Now using that $\sum \theta_k = \pi$, you get that $\cos2\theta_j = \cos(2\pi - 2\theta_j) = \cos(\sum_{k\neq j} 2\theta_k)$. And we observe that

$$ \cos(\sum_{k\neq j} 2\theta_k) = 2\cos(\sum_{k\neq j}\theta_j)\cos(\pi - \theta_k) - 1$$

where for $\cos(\sum_{k\neq j}\theta_j)$ we can inductively use the polynomial expression we have already derived. Doing this for all terms you end up with

$$ \sum \sin^2 \theta_k = 2N(1+\prod_k\cos\theta_k) + P \quad\quad (\sharp)$$

where $P$ is some polynomial in $\cos\theta_k$ and $\sin\theta_k$ that contains terms with positive, even number of factors in $\sin$.

So perhaps a better interpretation of your identity is actually as the identity for the angle addition formula for cosine, coupled with identity (#) above, which holds for a list of angles $\theta_k$ that sums to $\pi$.

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  • $\begingroup$ "Coupled with". (To be continued.....) $\endgroup$ Aug 3, 2011 at 23:19
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As @t.b. noted all those years ago, the identity for the triangle doesn't follow from the Law of Cosines, it proves the Law of Cosines. Here's a(nother) link to my answer depicting that.

I can get close to illustrating the identity for the cyclic quadrilateral using a comparable strategy of dropping myriad perpendiculars. For simplicity, I'll just show perpendiculars dropped to side $a$:

enter image description here

Simple right triangle trigonometry shows that we can decompose side $a$ into lengths $b \cos\gamma \cos\delta$, $d \cos\gamma \cos\beta$, and $c \cos\delta \cos\beta - c\sin\beta \sin\delta$ (where we take on faith that all the trig signs do the correct things for obtuse angles). Thus, the area of the square on side $a$ is the sum of the areas of rectangles with side $a$ and those other lengths:

$$a^2 = a b \cos\gamma \cos\delta + a d \cos\gamma \cos\beta + a ( c \cos \beta \cos \delta - c \sin\beta \sin\delta)$$

Likewise for other sides, so that we can throw everything together and write

$$\begin{align} a^2 + b^2 + c^2 + d^2 &= 2 a b \cos\gamma\cos\delta + 2 a c \cos\beta \cos\delta + 2 a d \cos\beta \cos\gamma \\ &+2b c\cos\alpha\cos\delta + 2 b d\cos\alpha\cos\gamma+2cd\cos\alpha\cos\beta \\ &-2a c \sin\beta\sin\delta - 2 b d\sin\alpha \sin \gamma \end{align} \tag{$\star$}$$

What isn't clear from the diagram is that, with $k$ denoting the circumdiameter, $$a c \sin\beta\sin\delta = b d\sin\alpha \sin\gamma = \frac{abcd}{k^2} \tag{$\star\star$}$$ which gives the desired result. Of course, we know $(\star\star)$ is true, because the chord subtended by an inscribed angle $\theta$ has length $k \sin\theta$ ---this is effectively one proof of the Law of Sines--- but an ideal diagram would incorporate the relation.


I'll note that, by dropping only two perpendiculars from the endpoints of side $c$, one can derive the decomposition $$a = b \cos(\gamma+\delta) + c\cos(\beta-\delta)+d\cos(\beta+\gamma)$$ This may be easier to see, but having to expand the compound trig factors defeats the purpose of illustrating the target relation, since too much symbol manipulation is required.

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