# Cutting polygons in half

This Riddler puzzle is about cutting polygons in half. Here is the problem:

The archvillain Laser Larry threatens to imminently zap Riddler Headquarters (which, seen from above, is shaped like a regular pentagon with no courtyard or other funny business). He plans to do it with a high-powered, vertical planar ray that will slice the building exactly in half by area, as seen from above. The building is quickly evacuated, but not before in-house mathematicians move the most sensitive riddling equipment out of the places in the building that have an extra high risk of getting zapped.

Where are those places, and how much riskier are they than the safest spots? (It’s fine to describe those places qualitatively.)

Extra credit: Get quantitative! Seen from above, how many high-risk points are there? If there are infinitely many, what is their total area?

Here is my solution:
[Show Solution]

And here is a bonus interactive graphic showing the solution

## 12 thoughts on “Cutting polygons in half”

1. Dmytro Taranovsky says:

Assuming that the direction is random–or for any other reasonable measure of the random cuts–the relative risk diverges (becomes infinite) at midpoints of the segments that bisect the pentagon in half by area. To see this, note that as the cut direction continuously changes, the infinitesimal movement is rotation around the midpoint so as to preserve equal area, and hence the density diverges there. These points form a curved pentagonal star, which is shown in blue in the picture in your post.
As we move from the center to an edge of the pentagon, the risk profile changes as follows: High initially, rises to infinity (like inverse square root) as we approach the star, falls abruptly and then rises to infinity as we approach the star again, then falls abruptly (to a fraction of the value at the center (except at the corners)) and continues to fall further as we move away from the center.

1. Laurent says:

It hadn’t occurred to me that the star could also be obtained by simply drawing the locus of midpoints of XY. Nice observation! I might edit my post to reflect this fact.

I don’t follow your argument that “relative risk diverges” at the midpoints of the segments that bisect the pentagon in half by area. How are you measuring risk?

In my solution, I’m admittedly using a simple notion of risk, which I explain is just the number of valid cuts passing through a given point. The advantage of this measure is that it is unambiguous and doesn’t depend on the random distribution of cuts. Of course, there are other ways of measuring risk.

1. Dmytro Taranovsky says:

The most natural model is that risk density at a point p is lim(ε→0, Risk(ε,p)) where Risk(ε,p) is the probability that point p is within distance ε/2 of the laser beam. Equivalent formulations include:
* Risk per width ε of the laser beam as ε→0.
* Assuming that the laser sweeps 360°, the risk density is sum(1/v_i)/(2π) where each time the beam touches the point p, v_i is the speed at which the beam moves through p. (Hence, zero speed leads to infinite risk density.)
* Using measure theory: The risk to a region is integral (over the probability distribution of the rays) of the length of the part of the ray that is in the region.

For the probability distribution of the rays, I assume that each direction is equally likely (rotational symmetry), though for divergence at the pentagonal star it is sufficient that every angle has a nonzero probability density.

1. Dmytro Taranovsky says:

Correction: lim(ε→0, Risk(ε,p)) should have been lim(ε→0, Risk(ε,p)/ε)

2. Hector says:

Very nice! I chose to calculate the star’s area instead of the inner polygon, but it looks like we agree in the case of the triangle, as we should: the ratio is .01986?

1. Laurent says:

yup, that’s what I found! Area for the triangle is $\frac14(\log(8)-2)\approx 0.01986$

3. Jim Crimmins says:

Hi Laurent – very cool stuff. I think if e.g. you express one of the hyperbolas in polar coordinates from the origin (0,0) as r^2 = a^2*b^2 / (b^2*cos^2(theta)-a^2*sin^2(theta)) with theta measured from the y-axis you can integrate the area under the hyperbola and get 1/2n of the total area of the polygon in a compact form. The answer will look something like 2*n *[a*b/2 * tanh-1[a/b * Xp/Yp] – Xp*Yc/2] , where (Xp, Yp) is the closest vertex of the polygon to the y-axis in the first quadrant and Yc is the Y coordinate of the center of the hyperbola and the polygon (not the origin). The first term is the total area under the hyperbola swept from the origin and the second term nets out the un-needed triangle in the swept area using Heron’s formula. Using this result and expressing result as area ratio of the large polygon I get {n,area ratios} of {3, 1.986e-2};{5, 3.339e-4};{7, 3.741e-5};{9, 7.763e-6};[11, 2.256e-6};{13, 8.133e-7};{15, 3.410e-7}. These look pretty close to your values! Thanks for the great blog!

1. Laurent says:

I ran my code again and I got precisely the same numerical answers as yours, so it looks like we both probably did it right! Polar coordinates do seem like a more natural choice. Maybe I’ll take another look at my derivation and try to clean it up… thanks!

4. Etienne Vouga says:

I think the numbered diagram (above “Calculating the area”) is not quite correct. The centroid (only) should be labeled 5; the other points in the inner pentagon are 3, except for the points on the line segments from the centroid to the outer pentagon’s vertices, which are 4; similarly the regions labeled “3” should be 2, except for along the line segment from the centroid to outer pentagon’s vertices, where the label is indeed 3.

1. Laurent says:

I don’t think that’s true. If you imagine the centroid and draw the five bisectors, and then place a new point close to the centroid (still inside the inner pentagon), then each of the five bisectors can be perturbed slightly in order to intersect the new point.

1. Laurent says:

I made a little applet that draws the bisectors so you can see what I’m talking about. I added it to the solution.