Solution to puzzle 37: Five marbles

Consider two adjacent marbles, of radii a < b. We will show that b/a is a constant, whose value is dependent only upon the slope of the funnel wall.

The marbles are in contact with each other, and therefore the vertical distance between their centers is b + a.

The marbles are also in contact with the funnel wall. Since the slope of the funnel wall (in cross section) is a constant, the two green triangles are similar. Hence the horizontal distance from the center of each marble to the funnel wall is bc and ac, respectively, where c = sec(x) is a constant dependent upon the slope of the funnel wall. (x is the angle the funnel wall makes with the vertical.)

Cross section of five marbles in a conical funnel.

Close-up cross section of two adjacent marbles in a conical funnel, one vertically above the other. A green triangle has one vertex at the center of the larger marble. From this center, a line is produced horizontally to the funnel wall, meeting at the second vertex. A second line is produced at right angles to the (same) funnel wall, meeting at the third vertex. A similar green triangle is drawn for the smaller marble.

Let the slope of the funnel wall be m.
Then m = (b + a) / [(b − a)c].
Rearranging, b/a = (mc + 1)/(mc − 1).

Hence the ratio of the radii of adjacent marbles is a constant, dependent only upon the slope of the funnel wall. Let this constant be k.

In this case, we have 18 = 8k⁴.
So k² = 3/2.

Therefore the radius of the middle marble is 8 · (3/2) = 12mm.

Remarks

Note that 12 is the geometric mean of 8 and 18. For any odd number of marbles in such a configuration, the radius of the middle marble is the geometric mean of the radii of the smallest and largest marbles.

Source: Mark Ganson