You Won’t Believe This Amazing Formula

[This is a back-issue of one of this site’s newsletters]

If you’ve been reading for a while, you might remember this puzzle: Can you find a rectangle whose area equals its perimeter?

For example, a 6 x 3 rectangle has a perimeter of 18 units, and the area is 18 square units. The same number!

In two emails last year, I solved the puzzle in two different ways. Try it yourself first, or read my solutions here and here.

As they stand, these two solutions show that for some puzzles, there’s an easy way and a hard way to solve them. However, the real magic starts to happen when you combine the two answers.

The first (harder) method says:

  • Start with a pythagorean triangle with sides P, Q and R. The hypotenuse is R.
  • Work out N = 8P/Q and S=8R/Q+8
  • Then, work out A=(S+N)/4 and B=(S-N)/4
  • The rectangle with sides A and B has area equal to the perimeter.

For example,

  • Start with a 3,4,5 triangle, P=3, Q=4, R=5.
  • This gives N=24/4=6, S=40/4+8 = 18.
  • This in turn gives A=24/4=6 and B=12/4=3. It’s our 3×6 rectangle again!

That’s great – if you happen to have a ready supply of pythagorean triplets. Most people only know a handful off the bat.

The second method says:

  • Start with a ratio T. It doesn’t have to be a whole number, fractions work fine. (In fact, T ends up being the ratio A/B of the sides of the rectangle)
  • Work out A=2+2T and B=2+2/T.
  • The rectangle with sides A and B has area equal to the perimeter.

For example, starting with T=2 gives our old friend, the 3×6 rectangle again.

Two very different methods for solving the same problem.

When you have two method for solving the same problem, there’s often something amazing lurking in the background. The amazing thing lurking here can be found when we forget about A and B, and find the relationship between P, Q and R.

  • So, we start with A=(S+N)/4=2+2T and B=(S-N)/4=2+2/T.
  • Adding A and B gives S/2 = 4+2T+2/T, or S=8+4T+4/T.
  • That means, S-8 = 4T + 4/T. However, S-8 is 8R/Q.
  • Subtracting B from A gives N/2 = 2T-2/T.
  • Now we’ve got N=4T-4/T and S-8=4T+4/T.
  • Remember, though, that N=8P/Q and S-8=8R/Q.
  • Therefore, P/Q = (T-1/T)/2 and R/Q=(T+1/T)/2.

Suddenly, we have a formula that gives us right-angled triangles. Watch this:

  • I’ll start with T=2.
  • So, 1/T is 1/2.
  • T+1/T is 5/2 and T-1/T is 3/2.
  • That means, P/Q is 3/4 and R/Q is 5/4.
  • This gives me the triangle P=3, Q=4, R=5.

Let’s try that again:

  • Start with T=3/2.
  • So, 1/T is 2/3.
  • T + 1/T is 13/6 and T – 1/T is 5/6.
  • That means P/Q is 5/12 and R/Q is 13/12.
  • This gives the triangle P=5, Q=12 and R=13.

A simple formula for generating pythagorean triplets. Ain’t that amazing?



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