Imagine a point a little over 10 km from the south pole. If it is just the right distance away, you can go south for 10 km, east 10 km circling exactly once around the pole, then north 10 km back you your starting point. The set of all such points forms a circle just over 10 km from the south pole.

Starting a little closer, you can go twice around the pole before heading back to your starting point. These points form a circle too. In fact there are an infinite number of such concentric circles, closer and closer to 10 km from the south pole.

So what are the actual values?

Assume the planet Capra is perfectly spherical, with the same radius as earth (6371 km).

The circle around the pole that you travel along (heading east) has a circumference of 10 / n, where n is the number of times you repeat it. Let a be the radius of that circle. Then 10 / n = 2πa, so a = 5 / nπ.

Now we can grab some formulas from Wikipedia. The circle forms the base of a spherical cap with the following values:

• a: circle radius = 5 / nπ
• r: sphere radius = 6371
• h: distance from the centre of the circle to the pole.
• θ: angle at the centre of the sphere between the pole and the edge of the circle.

One of the formulas is:

r = (a² + h²) / 2h

Solving for h gives:

h² - 2rh + a² = 0
h = (2r ± sqrt(4r² - 4a²)) / 2    (by the quadratic formula)
h = r ± sqrt(r² - a²)

There are two solutions, one for each pole (imagine placing a ring onto a sphere; it fits over either end). We want the solution where h < r, so:

h = r - sqrt(r² - a²)
h = 6371 - sqrt(6371² - (5 / nπ)²)

Now find θ by equating two different formulas for the surface area of the spherical cap:

π(a² + h²) = 2πr²(1 - cos θ)
1 - cos θ = (a² + h²) / 2r²
θ = cos^-1(1 - (a² + h²) / 2r²)

The distance overland from the pole to the circle is rθ (the fraction of a whole circle θ / 2π times the circumference 2πr).

Here are some results:

n	a	h	θ	Pole to circle	Pole to start point
1	1.6 km	2 m	0.00025	1.6 km	11.6 km
2	796 m	5 cm	0.000125	0.80 km	10.80 km
3	531 m	2.2 cm	0.0000833	0.53 km	10.53 km
1000	1.6 m	0.0002 mm	0.00000025	1.6 m	10.0016 km

The values of a and the pole to circle distance rθ are almost identical, since the surface of the planet is virtually flat at these distances.

Since you know there is a unique solution, the ring’s area must be the same no matter what size the circles are. So you can set the radius of the inner circle to 0, making the radius of the outer circle (1/2 AC). The ring’s area is now the same as the outer circle: π * (1/2 AC)².

Problem 2

Here’s similar problem mentioned in the notes for the same puzzle.

A hole 6 inches long has been drilled through a wooden sphere. What is the volume remaining in the sphere? There is only one possible answer.

Show answer 2

If you try the “zero radius” approach from the previous problems and set the radii to 0 and 2 cm, the total area is π * 2² = 4π. Intuitively the minimum area might be when the circles have an equal radius of 1 cm, which gives a total area of 2 * (π * 1¹) = 2π. So far so good, but we need to confirm.

Call the two radii x and y, so x + y = 2 and the total area of the circles is:

πx² + πy² = π(x² + y²).

To find the minimum and maximum area, first express it in terms of just one variable. y = 2 - x, so the area is:

π(x² + (2 - x)²) = π(x² + 4 - 4x + x²) = 2πx² - 4πx + 4π

This is an upward parabola (since the x² coefficient is positive), so the minimum value is at x = -b/2a (using the standard quadratic form ax² + bx + c):

x = -(-4π)/(2*2π) = 1
y = 2 - x = 1

So the values for the minimum area we originally guessed are correct (x = y = 1). But what about the maximum value? Being an upward parabola means there is no maximum. How is that possible if both radii are between 0 and 2?

The problem has a hidden assumption. The way the circles were drawn implies they are non-overlapping. But one circle could be inside the other and still satisfy the condition of touching at a single point:

Now the relationship between the radii is different. If x is the radius of the larger circle, y = x - 2, which allows both radii to have values higher than 2. (The formula for the total area still works out to be the same).

Both cases can be combined using the absolute value: y = | x - 2 |.

The horizontal motion of the pen is identical to gear B, and the vertical motion is identical to gear A (at the point where the rod is attached). However, gear B completes m revolutions for each revolution of gear A.

Ignoring scale and translation, and assuming the rods attached to A and B both start at angle 0° and gear A completes one revolution in time 2π, then at time t the pen is at location:

x = cos(mt)    (gear B)
y = sin(t)       (gear A)

Which looks like this (courtesy of GraphSketch.com):

These are known as Lissajous curves. Here are a few more:

Question 2

Does it make a difference if gears A and B don’t start at the same angle?

Show answer 2

You can find the possible arrangements as follows:

• For one or two beads, there is only one arrangement.
• For n >= 3 beads, add one bead to each arrangement of n-1 beads at all possible positions to create new arrangements.
• When adding to an arrangement, existing clusters of beads are never changed, but beads can be moved in a way that maintains their existing relationships with the other beads. This allows newly added beads to touch two beads that do not already touch, forming a loop of 4 or more.
• Exclude any duplicate arrangements.

This procedure gives 2 arrangements of 3, 5 arrangements of 4 and 13 arrangements of 5 beads.

If you draw a line between the centres of touching beads, the arrangements are a little easier to see.

Question 2

If you think of these diagrams as connected graphs, what constraints are there for a bead arrangement to be valid? Can you think of some example graphs that cannot be drawn without violating a constraint?

Show answer 2

Each student selected the chord a different way.

Sue’s method was to chose two random points on the circle and draw a chord between them. In the diagram below, if the first point is A, then choosing any point along the arc BC creates a chord longer than the side of the triangle. The length of arc BC is one third of the circumference, making the probability 1/3.

Mark’s approach started the same way by picking a point A somewhere on the circle. Then he chose a point at a random distance between A and the centre C and drew a chord through that point perpendicular to AC.

In the diagram below, B is the intersection of AC and a perpendicular edge of the triangle. If the randomly selected point is anywhere on BC, the chord is longer than the triangle edge. B is the midpoint of AC (since angle CXB is 30°, BC/XC = sin(30°) = 1/2, so BC = XC/2 = AC/2), so the probability of selecting a point on BC is 1/2.

Alex decided to select a point anywhere in the circle (with equal probability) and draw a chord with that as its midpoint.

If a smaller circle is inscribed inside the triangle, then selecting any point inside that circle results in a chord longer than the side of the triangle.

Call the radii of the smaller and larger circle r and R respectively. Then r = R/2 (as in Mark’s approach). So the probability of selecting a point in the smaller circle is (πR²/4) / πR² = 1/4.

This problem is known as Betrand’s Paradox. The issue is that the method of randomly selecting a chord is never clearly stated.

The reason for the three different answers is that each method involves a different distribution. If you reduce each of the methods to a single variable, it is possible to visualise the distributions as 2D graphs of chord length vs the variable value.

What variable might you use for each method?

Show answer 2