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+zx>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=0y=0, z=0z=0, x=1x=1 and x=y+zx=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

From this we get that \(y and \((1-x)^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:

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:

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?

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 cc 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 lookscomplicated, 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.

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