The X+Y Files

Issue 8

The Pentagon – analysis

In our last issue we folded a piece of A4 paper to produce a pentagon which looked regular, but we would like to know if it really is.

From the way in which it was produced we can say, by symmetry, that AP = AS, PQ = SR and that P = S, Q = R, but we need more than that.

We suggested using EQUATION.

Let us call AP x, where the sides of the paper are a and ka. For a piece of A4, k is supposed
__
to be

2 , but we may want to find
out if we can change k in the hope of creating a regular pentagon if this one isn’t.

If we now look at the pentagon unfolded we can put quite a lot of information onto it.

 Figure 2

  1. In Fig. 1 PBQR, and from Fig. 2, where EQUATION, by unfolding, EQUATION (unfolded). [They are alternate angles.] So EQUATION (folded or unfolded) = ß also. So the angles of the pentagon are 2 at A and 180° – ß at P, Q, R and S.
  2. If we now look at ABC, and let BC = b, we have a right-angled triangle with sides a and b and hypotenuse AC, which is ka – b, since AC + CB is originally one of the longer sides of the paper. Applying Pythagoras’ Theorem we have

    a² + b² = (ka – b)² = k²a² – 2kab + b²
    a = k²a – 2kb, which gives
    We can now find 2 = 90° + EQUATION BAC, where tan(BAC) =

    b

    k² – 1

    		 __

    __

    		  =

    ______

    		 , with k = 2

    a

    2k
    		for A4 paper
    This will make EQUATION = 2 = 109.47°, and the remaining four equal angles each 107.63° 


    EQUATION       EQUATION

      EQUATION , with EQUATION  

     

    This is not far from the 108° required for a regular pentagon, but it’s not good enough.

    What about the sides?
  3. Look at EQUATION in Fig. 1. AP = x, BP = a – x.

    EQUATION
  4. Finding expressions for PQ (=SR) and QR is more difficult. First note that in Fig. 2 EQUATION.

    From EQUATION

    We then work with projections onto AM: EQUATION. Rearranging this, and putting it through the calculator to produce an approximate value gives PQ = SR EQUATION.

    Projecting onto QR we can see that EQUATION, so doubling gives QR = (0.550510)a.

This is the same as the value we found for AP and AS, but we’ve used the calculator, so, although intriguing, we can’t trust it.

However it is possible to prove that these three sides are indeed the same — it requires some Sixth Form trigonometry using double angles. So it would seemt aht any rectangle will generate a pentagon with four equal angles and 3 + 2 equal sides. The question is whether there is a value for k which will produce a regular pentagon.

In triangle ABC, b = atan (BAC), ka - b = a/cos(BAC), and for a regular figure EQUATION.

EQUATION adding these two results : EQUATION

This would give a length of 289mm for a width of 210mm, so we would need to take about 8mm off the length fo a standard piece of A4. That would make a pentagon with equal angles – would it be regular?

Yes, it would! Given that AP = AS, the only possible non-regular pentagons with all equal angles could be super-imposed on top of a regular pengaton of side AP. They would sit symmetrically and be either longer or shorter than the regular one. If shorter then QR > AP and if longer then QR < AP. However, we already know that QR is equal to AP, so the regular shape is the only possibility.

MLP