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 \[\prod_{i=1}^{500}2\cdot\csc(\pi/1001)\cdot\sin(i\cdot2\pi/1001)\approx 1.3889*10^{1253}\]

Can You Decipher The Secret Message?

This week’s Riddler Classic asks:

This week’s Classic comes courtesy of Alexander Zhang of Lynbrook High School, California. Alexander won first place in the mathematics category at this year’s International Science and Engineering Fair for his work at the intersection of topology and medicine. He developed his own highly efficient algorithms to detect and remove defects (like “handles” or “tunnels”) from three-dimensional scans (e.g., MRI). Alexander has long had an interest in topology, which just might be related to his submitted puzzle.

Consider the following image showing a particular uppercase sans serif font:


Alexander thinks many of these letters are equivalent, but he leaves it to you to figure out how and why. He also has a message for you:


It may not look like much, but Alexander assures me that it is equivalent to exactly one word in the English language.

What is Alexander’s message?


There are two important hints embedded in this question — one obvious and one more subtle. The obvious one, Alexander has long had an interest in topology, which just might be related to his submitted puzzle, tells us to consider the shape of the letters rather than any lexicographical characteristics. The more subtle one, Consider the following image showing a particular uppercase sans serif font: (emphasis mine), is important because of the topographical nuances on which this question hinge.

Conventional topography rules would divide the letters provided into three groups: A’s, which have one hole (A, D, O, P, Q, R); B’s which have two (just B), and C’s which have no holes (C, E, F, G, H, I, J, K, L, M, N, S, T, U, V, W, X, Y, Z). The variations in the lines can be stretched or moved into or out of existence, but the holes must be preserved.

Looking for words (from the Scrabble dictionary) that match the ‘YIRTHA’ pattern (CCACCA) yielded 413 results, including ‘CHOKED’, ‘FIASCO’, and ‘WHACKO’.

In search of a more restrictive topographical definition, I recognized the hint in the phrase sans serif. Serifed fonts (like the one I use on this site) have ornamentation which give letters extra lines. If we impose the requirement that for shapes to belong to the same topographical class they must be equivalent only through the stretching and rearranging of lines, but that neither lines nor holes can be created nor destroyed, then we get seven distinct classes: A’s, which have one hole and two additional lines (A, R); B’s, which have two holes (B); C’s, which have no holes and only one undivided line (C, G, I, J, L, M, N, S, U, V, W, Z); D’s, which have one whole and no additional lines (D, O); E’s, which have one two-way fork (E, F, T, Y); H’s, which have two two-way forks (H, K, X); and P’s, which have one hole and one additional line (P, Q).

Searching for the new pattern — EIAEHA — we find only a single matching word: EUREKA!

Can You Bake The Biggest π?

This week’s Riddler Classic asks:

This Sunday, March 14, is Pi Day! To celebrate, you are planning to bake a pie. You have a sheet of crust laid out in front of you. After baking, your pie crust will be a cylinder of uniform thickness (or rather, thinness) with delicious filling inside.

To maximize the volume of your pie, what fraction of your crust should you use to make the circular base (i.e., the bottom) of the pie?

Note: If you solve this riddle by baking an optimal pie, you automatically win.


First, happy Pi Day! Now on to new business. Setting up this problem, we are trying to maximize the volume of pie filling that can fit inside the cylindrical pie crust. The volume of filling will be:
\[V=\pi r^2\cdot h\]
with \(V\) being the volume of pie filling, \(pi r^2\) is the surface area of the base of the pie crust, and \(h\) the height of the walls. We are going to maximize \(V\) with respect to \(r\), but first we have to constrain \(h\) in terms of \(r\). The sheet of pie crust, before being shaped into a pie, will have some area \(A\), and from it we must extract two circular sections with area \(A_b=\pi r^2\) and also the rectangular wall which will have area \(A_h=2\pi r\cdot h\). This gives
\[A=2\pi r^2+2\pi r\cdot h.\]
From this, we can extract that
\[h=\frac{A-2\pi r^2}{2\pi r}\]
which will be substituted into our volume equation:
\[V=\pi r^2\cdot\left(\frac{A-2\pi r^2}{2\pi r}\right).\]
We need to maximize \(V\) with respect to \(r\), which we will do by taking the partial derivative and setting it equal to 0, to find the maximum
\[\frac{\partial V}{\partial r}=\frac{A}{2}-3\pi r^2=0.\]
Now, we just need to find the value of \(\pi r^2\) (the area of the circle) in terms of \(A\) to find the fraction of \(A\) used to make the base. With some simple arithmetic, we get that
\[\pi r^2=\frac{A}{6},\]
and that the fraction is 1/6.

Can You Skillfully Ski The Slopes?

This week’s Riddler Classic asks:

Congratulations, you’ve made it to the finals of the Riddler Ski Federation’s winter championship! There’s just one opponent left to beat, and then the gold medal will be yours.

Each of you will complete two runs down the mountain, and the times of your runs will be added together. Whoever skis in the least overall time is the winner. Also, this being the Riddler Ski Federation, you have been presented detailed data on both you and your opponent. You are evenly matched, and both have the same normal probability distribution of finishing times for each run. And for both of you, your time on the first run is completely independent of your time on the second run.

For the first runs, your opponent goes first. Then, it’s your turn. As you cross the finish line, your coach excitedly signals to you that you were faster than your opponent. Without knowing either exact time, what’s the probability that you will still be ahead after the second run and earn your gold medal?

Extra credit: Over in the snowboarding championship, there are 30 finalists, including you (apparently, you’re a dual-sport threat!). Again, you are the last one to complete the first run, and your coach signals that you are in the lead. What is the probability that you’ll win gold in snowboarding?


Right away, since you have the same exact underlying distribution, each of you is equally likely to win the second trial. That means you have at least 50% chance of prevailing, since your opponent’s chances of being fastest overall after losing each trial are quite slim.

The next part is a little… abstract. Since we know nothing about your margin of victory in the first trial, the likelihood that your opponent’s margin of victory in the second trial, assuming that you don’t win the second trial, is the same in expectation. This means that you have a 50% chance to prevail overall despite losing the second trial (by having the faster overall time). This gives us an overall probability of 75%!

Just to be sure, we’ll simulate:

# r
skiRace <- function(){
    times <- rnorm(4)
    if(times[1] > times[2]){
      return(times[1] + times[3] > times[2] + times[4])

mean(replicate(100000, skiRace()))

Extra Credit: The second answer we arrive at by simulation:

# r
snowboardRace <- function(){
    firstTrial <- rnorm(30)
    if(which.max(firstTrial) == 30){
      secondTrial <- nrorm(30)
      times <- firstTrial + secondTrial
      return(which.max(times) == 30)

mean(replicate(10000000, snowboardRace()))

We get 31.5%.

Finally, a visulazation of how the probability of victory changes over number of competitors.