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:


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 .

A Case of Unmistaken Identities

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