Centres of rotation

Ever since I’ve been teaching GCSE maths, I’ve struggled to explain one topic more than any other: how to find a centre of rotation.

There are several ways students approach this problem. There’s the “I dunno” way, sometimes disguised as “I dunno where to start.” This is the kind of comment that makes me wonder whether the student is making an effort at all.

It really doesn’t take that much effort to be in the second group, which says things like “Is it sort of twisted?” That’s not going to pick up many marks, but it shows a little bit of thought about the question.

The third group uses the right vocabulary, and says “It’s a rotation!”, confidently picking up one of the three marks. Some will even say “90º1 clockwise” (or something similarly correct) for a second.

It’s the final mark that’s the problem: how do you find the centre of rotation? Obviously that’s straightforward if it’s a 180º rotation, but 90º presents a difficulty.

There’s a fourth group that calls for tracing paper and sets off on a dull process of stabby trial and improvement to find the centre. I have sympathy for this group; they generally pick up the marks, and that’s a Good Thing.

But it’s a bit unsatisfying for me, as a tutor, to recommend “tracing paper stabbity” as a method. Luckily, there’s a better way — and it involves one of the least-used tools in the geometry set: the 45º set square.

Here’s what you do

  1. Find two corresponding points on the original shape and the shape that’s been rotated — typically, the pointy end of the triangle, or a convenient right angle. Draw a line between them.
  2. At each of the points, draw a line at 45º towards where you thing the centre of rotation ought to be2 . Where these lines cross is the centre of rotation.
  3. Check you’ve gone the right way: measure the distance from your centre to two other corresponding points and check they’re the same. Otherwise, you need to draw your 45º lines on the other side of your line.
  4. Two triangles and their centre of rotationTwo triangles with a line between themTwo triangles

Why does it work?

The centre of rotation is a point that’s the same distance from any pair of corresponding points. If you draw a triangle using the two corresponding points and the centre, you get an isosceles triangle, with 90º at the centre of rotation. That means the other angles both have to be 45º.

Generalising to other angles

In likelihood, you’ll never need to find a centre of rotation outside of a GCSE exam, and by that standard 90º should be enough for anyone. But, just for the sake of argument, what if you had to find the centre of a 32º rotation? You can use the same principle: you need to build an isosceles triangle with 32º at the tip. You’ve got 148º left in your triangle, which need to be split equally between the other two angles: you’d draw lines at 74º to your original line.

In general, to find a centre of rotation with an $x$º angle, you’d need to draw angles of $\frac{180-x}{2}$ (or $90- \frac{x}{2}$) off of your original line.

* Edited 2017-07-21 to fix an unfinished sentence.


Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove.

  1. Degrees are rubbish, but they’re what the syllabus asks for []
  2. I’d use a set square, but constructing 45º lines would make examiners swoon. []


17 comments on “Centres of rotation

  • AK

    I think it would be better to generalise this idea to the case where you know rotation happened, but you neither know the centre of rotation nor the angle of rotation. You just need to convince your students that the centre of rotation, say O, would have to lie on the perpendicular bisector of the line segment connecting two corresponding points (say A and A’) in the original and rotated shape, which shouldn’t be difficult; it’s just taking the “AOA’ is isoceles” argument one step further.

    Now construct the perpendicular bisectors for two pairs of corresponding points, and their intersection gives you the centre of rotation. It’s straightforward to find the angle of rotation now.

  • ntjamba augustines

    What if you are not given an isosceles triangle?

    • Colin

      Good question — the 45º triangle is part of a standard geometry kit, but if you don’t have one, you can construct a 45º angle through a point with straight-edge and compasses by bisecting a right-angle. If you don’t have a pair of compasses, you can also do it by folding the paper to create a right-angle, and folding again to make 45º.

  • Mrs. Abbas

    Many thanks for this Post. Very helpful indeed. Please advise if we turn a shape by 180 degrees, how the centre of rotation could be constructed.

    • Colin

      When a shape has been rotated 180º, the centre of rotation is the midpoint of any two corresponding points.

  • Oduc Emma

    How can I find the center of rotation by calculation given the coordinates?

    • Colin

      Can you show me the question, please?



    • Colin

      Your caps-lock key appears to be stuck on. Also, the whole post is about finding the 90º centre of rotation, so I’m not really sure what to tell you.

  • a happi boi

    thanks man you saved an American eighth graders life

  • Abdul Basit

    What if we are given two coordinates of image and and the object and we have to find the centre of rotation as well as angle of rotation? P.s i do not want to draw angle bisectors so plz suggest a more convenient way. Thank You.

    • Colin

      I don’t think there’s a much simpler way than drawing perpendicular bisectors (not angle bisectors) between each point and its image. There are algebraic methods, but they’re much more work.

  • Abdul Basit

    Thank u very much for ur help.My cies exams are just around the corner and your advice will really help me.

  • Abdul Basit

    Another thing i wanted to ask was wht if dont know the angle of rotation ,how can we apply the isosceles triangle method?

    • Colin

      If you draw the perpendicular bisector of A and A’, and the perpendicular bisector of B and B’ — assuming it really is a rotation, the bisectors will cross at the centre.

      • Abdul Basit

        Thank You!

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