## Folding a Trapezium: Solution

### A Geometric Secret of A4 Paper

## Problem:

We wish to make **two** folds in a rectangular piece of paper to form a trapezium (or trapezoid for the chaps across the water). There are infnitely many ways in which this may be achieved, but there is only one way that fits the following criteria:

- The lower ends of the folds must be in the lower corners of the piece of paper
- The upper corners of the paper, once folded, must meet on the sheet’s vertical line of symmetry with no overlapping

Take a piece of A4 paper (or A5 or any rectangular or square paper) and have a go. You’ll probably want to fold the paper down the middle first so you know where you’re aiming. Chances are you’ll find it quite difficult to get it neat and accurate. You might argue that it doesn’t *need* to be neat and accurate, but this is a geometry problem and if this were part of an engineering project you’d probably require a very high degree of accuracy.

### The question is thus:

Find the distance x from the top of the line of symmetry to the top of one of the folds, as in the following diagram:

**Solution**:

It’s easy to make the mistake of thinking there isn’t enough information to calculate this, but there’s actually more than enough.

Firstly it’s important to redraw the diagram with all the information we can immediately ascertain, to make it easier to see how to progress.

Since the problem is symmetrical we can ignore half the piece of paper. There are then two things to identify:

- The longer diagonal, running from the line of symmetry to the lower-right corner, is the entire right-edge of the piece of paper so we know it is 21.0cm.
- We don’t yet know anything about angles p and q, but, because the internal angles of a triangle add up to 180º, and so do the angles on a straight line, triangles A and B must be
.*similar*

Because triangle A is right-angled, we can use Pythagoras’ Theorem to work out the length of y:

21.0^{2}-14.85^{2}=y^{2} → y~14.85cm

(That’s weird?! That appears to make triangle A isoceles.)

And if we know y, then we know the length of the vertical side of triangle B (21.0-y) and since triangles A and B are similar, x must equal 21.0-y, i.e. **x~6.15cm**.

## Easy?

Perhaps. So what’s the secret? Well if triangles A and B are both similar and isoceles, then angles p and q must be equal at 45º. Returning to the original diagram, we see that this means the complete diagonal lines which come from the lower corners, “meet” or “cross” on the line of symmetry and extend to the top edge (shown here in purple), **are *** straight lines*,

**a characteristic entirely due to the proportional dimensions of A4 paper.**

And because A5, A3, A2, A1 and A0 paper sizes are all proportionally identical, this characteristic prevails across the range.

## A note on accuracy

Because y is only *approximately *equal to 14.85cm (the real figure is the square root of 220.4775 which is 14.848484771…), p and q are actually *not exactly* equal and the purple lines are *not* *exactly* straight, but within reasonable bounds of accuracy it’s fair to declare the above result.

But, for the sake of curiosity, let’s calculate what the exact length of A4 would need to be for the angles to be *exactly* 45º.

To do this we use trigonometry:

But there’s the secret of the A-standard paper range! **You can get from a length to a width by simply dividing it by the square root of 2.** Those 18th century printers certainly knew their algebra, but it looks like they rounded all the lengths to the nearest millimetre, assumedly to make it easier to replicate (you try cutting a piece of paper to the length of 21/√2 centimetres!)

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