thinking

Here are the solutions for the crazy cut puzzles I posted last week.  Did you solve them all?

This post explains solutions for the pizza puzzle. If you haven’t read that puzzle yet, you should try it first!

There are multiple ways to solve this one. For these solutions, we’ll treat the radius of the pizza as length 1. Even if it had a different radius, we could scale all the work in a solution accordingly, and the end result would be the same, since it’s a percentage of the total area.

First, let’s draw another hexagon inside the pizza, with vertices at the center of each pepperoni circle.

The bold hexagon is the one we already see on the pizza; the dashed hexagon is drawn with vertices on the center of each pepperoni circle.  We’ve also drawn new circles at the corner of each vertex of the bold hexagon. Since each triangle in the large hexagon is equilateral, and two of its sides are radii of the pizza circle, each edge of the bold hexagon has length 1. Each outside circle (the new circles) have radius half that, which means they each have area A = π r² = π/4 (that should read “pi over 4,” in case the “pi” isn’t showing up correctly for you). The total area of the outside circles is thus six times that, which is 3π/2.

Notice that the large bold hexagon and its circles are similar to the small dashed hexagon and its circles (the pepperoni circles). If we can find the ratio of areas between the large hexagon and the small hexagon, this will also tell us the ratio of areas between the outer circles and the inner circles. This is a general rule of geometry; as another example of this rule, see the next figure. Whenever shapes are resized together, the scale factor of their areas is the same for each shape – either all are doubled, or all are halved, etc.

If area(big M-shape) = 2×area(little M-shape), then area(big circle) = 2×area(little circle).

Let’s take a closer look at one of the equilateral triangles in the big hexagon, from the first figure. The bold and dashed lines here show how this is a close-up on the bold and dashed hexagons.

There are six copies of the wavy triangle in the large equilateral triangle, and two copies of it in the smaller equilateral triangle (lower-right).  So the area of the smaller equilateral triangle is exactly 1/3rd the area of the larger. This means the area of the pepperoni circles is also a third of the area of the outer circles. Hence the pepperoni area is exactly (1/3) · (3π/2) = π/2, which means they take up half the area of pizza!

I personally enjoy this first solution because it involves no knowledge of trigonometry. However, there is another way to find the answer that looks more explicitly at the angles involved. In the next figure, we see the same wavy triangle from the last drawing.

We could find an equation between r and R through either the Pythagorean theorem, or using the fact that

r/R = sin(π/6) = 1/2.

The altitude of the large equilateral triangle is sin(π/3) = √3 /2 = r+R = 3r, so r = √3/ 6. Notice that r is the radius of a pepperoni circle, so the area of each pepperoni is π/12. The area of six pepperoni circles is then π/2, confirming it’s half the area of the pizza.

A mathematical pizzeria sells pizzas with six slices, each slice containing a maximally-sized equilateral triangle, as shown in the figure. Their pepperoni pizza features one maximally-sized circle of pepperoni per slice. What fraction of the area is taken up by the pepperoni circles?

This is the solution for the pirate-catching puzzle I posted a few days ago.

Here’s the general idea:

It’s helpful to put things into a coordinate system where the starting position of the pirate ship is (0,0), and the government ship is at (3,0).  This works because we know they start 3 km apart, so any two points at distance 3 are fine as starting points.  In the picture above, the blue line is the path of the pursuing government ship, and the red line is one possible straight line the pirates could take.

The strategy is in two parts:

1. Start by travelling straight toward the pirate ship for 2 km.  If, by luck, the pirates are headed directly toward the government ship, they’ll meet at this point, since the pirates will have travelled 1 km in the same time (1 minute).
2. After that, go in an expanding spiral around the origin.  Think in polar coordinates.  You know the pirates are going directly away from (0,0) at constant speed 1 km/min, so their distance r from the origin is the same as the time t that has passed since the start.  Your ship (the government one) just needs to keep at the same radius as the pirate ship, while sweeping its angle around in a circle.  The government ship has to meet the pirate ship by the time it completes 360 degrees.

This strategy is possible because the government ship is going faster than the pirates.  That gives us a general idea of how to solve the puzzle – but it’s still a little vague.  We still don’t know the exact spiral to take, or how long it will take at most to catch them.

That’s where the real fun starts.  Let’s find the spiral’s equation, using as the coordinates of the government ship at time (where is the instant they spotted the pirates and went for them).

We have two constraints:

• , and
• speed .

We also know that the angle and radius are increasing over time.  To put this all together, we can use this equation for the speed of the government ship as a derivative of time:

Where did that equation come from?  Well, it’s basically the derivative (with respect to t) of arc length in polar coordinates.  We know , so

We also know that and therefore , so

Almost got it!  Using that again, we see that , so solve for to get

That’s the path of the government ship, which looks like this:

How long does it take to traverse?  The highest the angle can get is arbitrarily close to 360 degrees.  Our equations use radians (for example, the speed equation assumes is in radians), so we can just plug in for to find out the answer to our puzzle.  The longest time is exactly

minutes.

As a mathematician, I think this is a great answer to such a simple-to-state problem!

This weekend I heard a good mathy puzzle from George Miller.  Here’s a version I found online, attributed to Howard Lederer:

You’re on a government ship, looking for a pirate ship.  You know that the pirate ship travels at a constant speed, and you know what that speed is.  Your ship can travel twice as fast as the pirate ship.  Moreover, you know that the pirate ship travels along a straight line, but you don’t know what that line is.  It’s very foggy, so foggy that you see nothing.  But then!  All of a sudden, and for just an instant, the fog clears enough to let you determine the exact position of the pirate ship.  Then, the fog closes in again and you remain (forever) in the thick fog.  Although you were able to determine the position of the pirate ship during that fog-free moment, you were not able to determine its direction.  How will you navigate your government ship so that you will capture the pirate ship?

If you wanted, you could give a convincing but math-lite answer to this.  But you can also do better, so I’m going to ask my own version of the question:

Using your answer to the above question, what is the longest amount of time that may pass from the instant you saw the pirate ship until you capture it?  Suppose the pirate ship travels at 1 km/minute, and you first see it 3 km away.  How many minutes, at most, until you capture it?

I’ll post my answer on Friday (in 5 days).