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.
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:
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:
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:
We can use a trig identity…
…and the expression for R above, to rewrite the expression as follows:
Now we will use a very useful product of sines identity……:
…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:
This simplifies to:
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:
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:
We can proceed as above in the Classic up to this point:
Interestingly this product is equal to:
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:
Making the substitution N’ = (N-1)/2:
We will use another useful product of sines identity here:
And our expression simplifies to:
Finally, the answer for S=2, N=1001 is or about 31.64 .