Welcome to The Riddler. Every week, I offer up problems related to the things we hold dear around here: math, logic and probability. There are two types: Riddler Express for those of you who want something bite-size and Riddler Classic for those of you in the slow-puzzle movement. Submit a correct answer for either, and you may get a shoutout in next week’s column. If you need a hint, or if you have a favorite puzzle collecting dust in your attic, find me on Twitter.

This week, we’ve got two puzzles from the forthcoming puzzle book “The Original Area Mazes,” by Alex Bellos, Naoki Inaba and Ryoichi Murakami. The goal of these puzzles, which are also known by the Japanese term “menseki meiro,” is to figure out what the “?” equals. The only math you’ll need to know is that length times width equals area. Keep in mind that the diagrams aren’t necessarily to scale — this is about logic, not measuring.

Riddler Express

Submit your answer

Riddler Classic

Submit your answer

Solution to last week’s Riddler Express

Congratulations to 👏 Kristian Hougaard 👏 of Copenhagen, Denmark, winner of last week’s Express puzzle!

Twenty ghostbusters are on their annual camping retreat. Two of them, Abe and Betty, have discovered that another pair, Candace and Dan, are in fact ghosts posing as ghostbusters. Abe and Betty hatch a plan: When all 20 campers are sitting in a circle around the campfire, Abe will fire his proton pack at Candace, and Betty will simultaneously fire her proton pack at Dan, annihilating the ghosts. However, if two proton streams cross, it means the end of all life on Earth. If the ghostbusters are arranged randomly around the fire, what are the chances that Abe and Betty will cross the streams?

The chances are 1/3.

There are 20 ghostbusters, but we only really care about four of them — Abe, Betty, Candace and Dan. The position of the other 16 won’t affect the possible crossing of the streams, so let’s ignore them. (Sorry, you 16 irrelevant ghostbusters.)

Fix Abe’s spot at the campfire — say he’s on the north side. There are then three places his co-ghostbuster Betty can sit — east, west or south. The ghosts, Candace and Dan, will sit in the other two seats. There are 3*2*1 or six possibilities for the seating in the east, west and south seats. In exactly two of these arrangements — the two where Candace occupies the southern seat, forcing Abe to fire across the circle — the ghostbusters will cross their proton streams. Each of these six arrangements is equally likely, so the chances of stream-crossing disaster are 2/6 = 1/3.

Solution to last week’s Riddler Classic

Congratulations to 👏 Carl Ober 👏 of New Britain, Connecticut, winner of last week’s Classic puzzle!

Last week, you faced four sticky questions:

If you break a stick in two places at random, forming three pieces, what is the probability of being able to form a triangle with the pieces? If you select three sticks, each of random length (between 0 and 1), what is the probability of being able to form a triangle with them? If you break a stick in two places at random, what is the probability of being able to form an acute triangle — where each angle is less than 90 degrees — with the pieces? If you select three sticks, each of random length (between 0 and 1), what is the probability of being able to form an acute triangle with the sticks?

The probabilities are, respectively, 1/4, 1/2, \(\ln(8)-2\) and \(1-\pi/4\).

To solve these geometry problems, let’s draw some pictures! We’ll take the questions one by one.

For No. 1: Call the lengths of the three pieces \(x\), \(y\) and \(z\). They form a triangle if the sum of any two sides are larger than the third side (\(x+y>z\), \(y+z>x\) and \(x+z>y\)). This is called the triangle inequality!

Now to our stick, which we’ll assume is one unit long. Call the points where we broke it \(a\) and \(b\), both chosen at random. That gives us pieces of length \(a\), \(b\) minus \(a\) and 1 minus \(b\). Substituting those lengths in for \(x\), \(y\) and \(z\) above simplifies those triangle equalities:

\(x+y>z \Rightarrow a+(b-a)>1-b \Rightarrow b>1-b \Rightarrow 2b>1 \Rightarrow b>1/2\)

\(y+z>x \Rightarrow (b-a)+(1-b)>a \Rightarrow 1-a>a \Rightarrow a<1/2\)

\(x+z>y \Rightarrow a+(1-b)>b-a \Rightarrow 2a+1 > 2b \Rightarrow b-a<1/2\)

So, armed with the lengths of those three pieces, we can visualize the answer to the problem. As long as our randomly selected points on the stick (\(a\) and \(b\)) satisfy those inequalities we just made, we know we can make a triangle. We can plot values of \((a,b)\) as coordinates in a plane, as Laurent Lessard did:

The shaded areas are where the inequalities are satisfied. (There are two triangles because one corresponds to when \(a>b\) and the other when \(b>a\).) Those areas take up ¼ of the total square, which gives us our answer: 25 percent.

For No. 2: What if, instead of pieces from one stick, we pick up three sticks of random length somewhere between zero and one? This is a little trickier. When we plotted the first problem, it could be collapsed to two dimensions, because we were only really worried about two sticks — the length of our third piece was automatically determined by the length of our first two pieces. But this problem is in three dimensions, so the solution needs to be plotted not in a one-by-one square but rather in a one-by-one-by-one cube.

To help solve the problem, consider what wouldn’t solve it: a violation of our triangle inequalities. Suppose, for example, that \(x>y+z\), which makes it impossible to build a triangle. In that formulation, those points are contained in a pyramid bounded by the planes \(y=0\), \(z=0\), \(x=1\) and \(x=y+z\). There are three such pyramids in this cube, one for each of the ways the triangle inequality can be violated.

Each of those pyramids has a volume of ⅙. (A pyramid has a volume equal to the area of its base times its height, all divided by three. Our pyramid in question has height 1 and a triangular base with area ½.). Therefore, there is a ½ chance we can’t make a triangle, and a ½ chance we can. And so we have the answer to the second problem.

No. 3: Getting tougher still (as though that were possible)! Now it’s time for calculus. Guy D. Moore explains this one for us:

The problem asks us to ensure that three pieces form an acute triangle. Consider three pieces with lengths \(x>y>z\).

First, think about a right triangle. The formula for that, as our middle school teachers drilled into our heads, is \(x^2=y^2+z^2\). (Otherwise known as the Pythagorean Theorem.) To have an acute triangle, all angles must be less than 90 degrees, so we tweak that formula: \(x^2<y^2+z^2\).

From this we get that y<x and \((1-x)^2<y^2+(x-y)^2\), which is the same as

\begin{equation*}x<\frac{1-2y^2}{2(1-y)}\end{equation*}

Since we’re dealing with pieces of the same stick and not three separate sticks, we can return to plotting in two dimensions, not three. And our mirrored-triangle plot is useful again since our answer lies within those two original triangles. This time, though, we need to draw two new three-pointed shapes within those two triangles. The area of those shapes will be our answer — the probability of an acute triangle.

So to calculate our new shapes, we need to cut pieces out of our original triangles. The area of one of those pieces is expressed in an integral (which is the calculus part of the solution). That integral is:

\begin{equation*}\int_0^{1/2}\frac{1-2y}{2-2y}dy=\frac{1-\ln(2)}{2}\end{equation*}

There are six shapes, each with the same area, cut out of our one-by-one square, leaving:

\begin{equation*}1-3(1-\ln(2))=\ln(8)-2\end{equation*}

In that equation, “ln” is the natural log, which equals an implied probability of acute triangle-formation of about 7.9 percent. (Who knew that natural logs are a great way to solve stick problems?) Guy also provided this illustration of the curvy areas we calculated:

No. 4: We’re back to three dimensions again for the final question. This solution furthers the solution from problem No. 2, the way that solution No. 3 furthers solution No. 1. Laurent explained his solution this way:

We’ll solve this problem the same way we solved No. 2, but we’ll replace the triangle inequalities with the acute triangle inequalities. As in No. 2, we end up with a three-dimensional volume rather than a two-dimensional area. For simplicity again, we’ll assume that \(c\) is the largest length, which accounts for one-third of all possibilities.

Laurent provided a lovely illustration of this volume:

Our answer will ultimately be three times the area of this shape (this shape only accounts for stick c being longest, and two identical shapes will be generated for stick b being longest and stick a being longest).

Our solution lies in the filled-in parts of that shape. While this looks complicated, the curved surface inside that area has the equation \(c^2 = a^2 + b^2\), which is, conveniently, the equation of a right circular cone! So we can calculate the volume of the region of interest by subtraction. It’s ⅓ of the volume of the cube minus ¼ of the volume of the cone. (One-third because we’re considering only one out of three scenarios, the one where c is longest. And ¼ because the cone’s base is ¼ of a circle.) The total probability is three times this volume, because we must account for the remaining identical pieces. The final answer is \(3(1/3 − 1/4 ( \pi/3 ) ) = 1 − \pi/4\) or about 0.2146. So the probability of forming an acute triangle with three randomly chosen lengths is about 21.5 percent.

Want to submit a riddle?

Email me at oliver.roeder@fivethirtyeight.com.