Can You Calculate the Product of Perpendicular Diagonals?

This week’s Riddler Classic asks:

Lately, Rushabh has been thinking about very large regular polygons — that is, a polygon all of whose sides and angles are congruent. His latest construction is a particular regular 1,000-gon, which has 1,000 sides of length 2. Rushabh picks one of its longest diagonals, which connects two opposite vertices.

Now, this 1,000-gon has many diagonals, but only some are perpendicular to that first diagonal Rushabh picked. If you were to slice the polygon along all these perpendicular diagonals, you’d break the first diagonal into 500 distinct pieces. Rushabh is curious — what is the product of the lengths of all these pieces?

Extra credit: Now suppose you have a regular 1,001-gon, each of whose sides has length 2. You pick a vertex and draw an altitude to the opposite side of the polygon. Again, you slice the polygon along all the perpendicular diagonals, breaking the altitude into 500 distinct pieces. What’s the product of the lengths of all these pieces this time?


To approach this problem, what I did first was to try to define the mathematical expression we would have to use to calculate the product of all the perpendicular diagonals.

Thinking about a smaller example first—say, 6 sides of length 2 instead of 1,000—I went about constructing the the component parts we want to multiply together to get the answer.

We need to figure out how far apart the vertices are from each other. If we can imagine an inscribed circle around this polygon, the radius would be \(\csc(\pi/n)\). We can then calculate the length of the first (moving from 0 radians clockwise; i.e., the right-most) red line as \[2\cdot\csc(\pi/n)\cdot\sin(2\pi/n)\]

To calculate the product of every diagonal perpendicular to the longest (blue) one, we can generate a product of the following form: \[\prod_{i=1}^{\frac{n}{2}-1}2\cdot\csc(\pi/n)\cdot\sin(i\cdot2\pi/n)\]

To solve the original question–what is that value for a 1000-gon?–the number is so astronomically large, we need to articulate a closed-form expression for it, rather than waste digital ink typing out 1251 digits of a number nobody could possible process. As such, I set out to identify a pattern in the results as we escalate from smaller, more manageable \(n\)s to larger ones.

The following is the table of results for the first handful:


With this pattern, we can identify that the answer to the first question is \(500\csc^{499}(\pi/1000)\) or approximately 1.835*101251.

To answer the extra credit question, we’ll adapt our product expression and lean on ever-helpful WolframAlpha to do the heavy lifting for us. The result is \[\prod_{i=1}^{500}2\cdot\csc(\pi/1001)\cdot\sin(i\cdot2\pi/1001)\approx 1.3889*10^{1253}\]