# Two mysteries cleared up in one

Since the dawn of time, two mysteries have plagued mathematicians:

a) How do you find a centre of a 90º rotation? and b) What’s the 45º set square for?

Imagine my surprise when I discovered that each is the answer to the other!

### Some facts about the centre of rotation

Imagine you’ve got two right-angled triangles, one of which is rotated by 90º with respect to the other. The centre of rotation, $X$, needs to be the same distance from each of the right angles - let’s call them $A$ and $P$.

So you could, if you wanted, draw the perpendicular bisector of those two points and say “it lies somewhere on that line.” However, we can do better.

Wherever the centre is on that line, the angle $A\hat XP$ is a right angle. We also know that $AX = XP$, so the triangle is isosceles - which means that the angles $X\hat P A$ and $X\hat A P$ need to be the same - and in fact, need to be 45º.

### Which is where the set square comes in

Rather than drawing the bisector, you can simply reach for your set square. Here’s what to do:

- Draw a line between the two right angles
- Use your set square to draw a 45º angle to the line at each of the two points
- These should form a right-angled isosceles triangle
- The centre of rotation is at the right angle

### But wait…

There are two possible ways to draw the triangle. How do you pick between them?

That’s simple enough: one of them will look badly wrong. Once you’ve drawn your triangle, imagine rotating the page 90º around that point. Will one shape be where the other one was? If not, draw the triangle on the other side and try again.

## A selection of other posts

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