2-D Disco

There wasn't enough room to give the solution to Pyrgic Puzzle No. 3 on May 7th:

'Down at the 2-D disco, 3 identical discs gang up on a slightly smaller disc and they trap it between themselves and a straight line, as shown.

      If the radius of the small disc is 1 unit, what is the radius of the bigger discs?'

Many readers became lost or bogged down in detail. Solving a problem like this one is rather like swimming from A to B except

that there are no markers to tell you in which direction you must set off and— perhaps as importantly— how far to go. So some readers went around in circles and others gave up before they could reach the conclusion.

You need to find an expression giving R in terms of r. Don't imagine that this must be a linear equation. As it happens it turns out to be a quadratic.

Let's start with the construction shown and let's label as many of these lengths as we can in terms of R and r.

Now AB = 2R; AD = R + r.

Note that the point C is a vertical distance R above the horizontal line. D is r above the horizontal line. So CD = R - r.

BC = R + r - CD = R + r - (R - r) = 2r.

To summarise:

AB = 2R.

AD = R + r.

CD = R - r.

BC = 2r.

There are two ways of finding AC: from the upper right-angled triangle, and from the lower right-angled triangle. Equating these will give us a direct link between R and r.

From the upper triangle:

AC² + BC² = AB², so that AC² + 4r² = 4R², so that:

AC² = 4R² - 4r².

From the lower triangle:
                                            AC² + (R - r)² = (R + r)².

This boils down to:
                                           AC² = 4Rr.

Equating these two expressions for AC², we find:

4R² - 4r² = 4Rr.

This can be written:
                                                    R² - Rr - r² = 0,

and since we are told that r = 1, we have:

                                                 R² - R - 1 = 0,

This may be regarded as a quadratic with R as the unknown.

Most people solve their quadratics by referring to the standard formula for the canonical quadratic, which is expressed in standard form as:

ax² + bx + c = 0.*

This has solutions:

x = [-b ± √(b² - 4ac)]/2a.

In this particular instance, a = 1, b = -1, c = -1, so the solutions are:

R = [1 ± √(5)]/2.

R is positive, so we must reject the negative root and we are left with:

R = [1 + √(5)]/2,

which is known as the golden ratio, which all goes to show that the golden ratio turns up in all sorts of unexpected places.

*If you don't remember or know the formula (if you are short on memory but long on cunning) you can certainly proceed as follows:

R² - R = 1.

Add 1/4 to both sides:

R² - R + 1/4 = 5/4.

The left-hand side is simply (R - 1/2)², so

(R -1/2)² = 5/4.

We can take square roots of both sides without affecting the balance of the equation:

R - 1/2 = ±√5/2

So that:

R = (1 ±√5)2/.

Again we discard the negative root, since it gives us a negative value for R and the Radius of a circle is a positive quantity.

R = (1 + √5)/2

as before.