# A Case of Unmistaken Identities

This week’s Riddler Classic from Rushabh Mehta asks about the product of the lengths of line segments of a diagonal or altitude of regular polygons formed by the intersections with all the perpendicular diagonals.

### Riddler Classic

Here we will solve the problem in general and then show that for a side length of 2 and an even number of sides the product is independent of the number of sides.

The length of the diagonal connecting two opposite vertices for an even N-sided polygon is twice the major radius R the circumscribed circle.

[The length of the diagonal connecting a vertex to the midpoint of the opposite side for odd N is r+R (odd case shown here)]

The length of the major and minor radii for an N-sided polygon of side length S are:

$r = \frac{S}{2}*cot(\frac{\pi}{N}), \quad R = \frac{S}{2}*csc(\frac{\pi}{N})$

For the case of even number of sides, we will put a vertex on the point (R,0) and the coordinates of the N vertices are:

$x_{k}=R*cos(\frac{2k\pi}{N}), \quad y_{k}=R*sin(\frac{2k\pi}{N}), \quad k = [0,N-1]$

The lengths of the diagonal segments created by the perpendicular diagonals is the difference in the x-coordinates of N/2 pairs of adjacent vertices. And the product is:

$P = \prod\limits_{k=1}^{N/2}R*(cos(\frac{2\pi*(k-1)}{N})-cos(\frac{2\pi*k}{N}))$

We can use a trig identity…

$cos(A)-cos(B) = -2sin(\frac{A+B}{2})* sin(\frac{A-B}{2})$

…and the expression for R above, to rewrite the expression as follows:

$P = S^\frac{N}{2}\prod\limits_{k=1}^{N/2}sin(\frac{(2k-1)\pi}{N})$

Substituting N’=N/2:

$P = S^\frac{N}{2}\prod\limits_{k=1}^{N'}sin(\frac{(2k-1)\pi}{2N'})$

Now we will use a very useful product of sines identity……:

$I(n)= \prod\limits_{k=1}^{n-1}sin(\frac{k\pi}{n}) = \frac{n}{2^{n-1}}$

…in a way described here:

https://math.stackexchange.com/questions/1632023/evaluation-of-prodn-r-1-sin-left-frac2r-1-pi2n-right?noredirect=1&lq=1

To use this identity we rewrite the the last espression for P this way, where the appropriate terms on the top and the bottom cancel to make the result identical:

$P = S^\frac{N}{2}\frac{\prod\limits_{k=1}^{2N'-1}sin(\frac{k\pi}{N'})}{\prod\limits_{j=1}^{N'-1}sin(\frac{2j\pi}{2N'})}$

This simplifies to:

$P = (\frac{S}{2})^\frac{N-1}{2} *2.0$

And for the given case of S=2 we see that the product of the line segments along the diagonal = 2.0 and is independent of N:

$P = 2.0$

### Extra Credit

For the extra credit we have a 1001 sided (odd number) polygon. We will solve for general odd N. The lengths of the altitude (R+r) segments created by the perpendicular diagonals is the difference in the x-coordinates of (N-1)/2 pairs of adjacent vertices. And the product is:

$P = \prod\limits_{k=1}^{(N-1)/2}R*(cos(\frac{2\pi*(k-1)}{N})-cos(\frac{2\pi*k}{N}))$

We can proceed as above in the Classic up to this point:

$P = S^\frac{N-1}{2}\prod\limits_{k=1}^\frac{N-1}{2}sin(\frac{(2k-1)\pi}{N})$

Interestingly this product is equal to:

$P = S^\frac{N-1}{2}\prod\limits_{k=1}^\frac{N-1}{2}sin(\frac{k\pi}{N})$

To see why the last two expressions are equal, let’s look at the case where N= 17. We notice that both expressions contain k and (2k-1) values of [1,3,5,7] so those terms are the same. The remaining indexed (2k-1) values in Expression 1 are [9,11,13,15] and the remaining indexed k values in Expression 2 are [2,4,6,8]. When we pair these terms such that k+ (2k-1) = N, they are equal. IE [(9,8), (11,6), (13,4), (15,2)]. To see why, it is easy to show that:

$sin(\frac{n\pi}{N}) = sin(\frac{m\pi}{N}) \quad if \quad n+m = N$

Making the substitution N’ = (N-1)/2:

$P = S^{N'}\prod\limits_{k=1}^{N'}sin(\frac{k\pi}{2N'+1})$

We will use another useful product of sines identity here:

$\prod\limits_{k=1}^{n}sin(\frac{k\pi}{2n+1}) = \frac{\sqrt{2n+1}}{2^{n}}$

And our expression simplifies to:

$P = (\frac{S}{2})^\frac{N-1}{2}*\sqrt{N}$

For S=2:

$P = \sqrt{N}$

Finally, the answer for S=2, N=1001 is $\sqrt{1001}$ or about 31.64 .