# Proof: Invariant angle measure - same result for any circle drawn.

Below I have quoted Wikipedia.

I am particular interested in the statement:

The value of $\theta$ thus defined is independent of the size of the circle: if the length of the radius is changed then the arc length changes in the same proportion, so the ratio $\frac{s}{r}$ is unaltered.

That is, for any circle drawn with a pair of compasses, centered at the vertex, the arc extending from the start ray to the end ray has length $s=r\theta$, satisfying the equation $\theta=\frac{s}{r}$.

This statement is fundamental, since it states that no matter how we draw a circle to measure the angle, we always get the exact same answer $\theta$.

It is not enough to prove this by saying: the radian is defined $\theta=\frac{s}{r}$. This does not show that for any circle drawn the ratio is as stated.

Please check the image below:

Quote Wikipedia:

In order to measure an angle $\theta$, a circular arc centered at the vertex of the angle is drawn, e.g. with a pair of compasses. The length of the arc $s$ is then divided by the radius of the arc $r$, and possibly multiplied by a scaling constant k (which depends on the units of measurement that are chosen):

$\theta = k \frac{s}{r}$.

The value of $\theta$ thus defined is independent of the size of the circle: if the length of the radius is changed then the arc length changes in the same proportion, so the ratio $\frac{s}{r}$ is unaltered.

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can you please be specific about what it is you are asking then? – Ittay Weiss Jul 16 '13 at 8:21
Is this specific enough? Proof: Invariant angle measure - same result for any circle drawn. – Nicolas Lykke Iversen Jul 16 '13 at 8:36
Those grappling with this question may benefit from the earlier discussion of it at math.stackexchange.com/questions/444013/…? – Gerry Myerson Jul 16 '13 at 8:57

The proof relies on a careful definition of length is. The notion of length is extremely subtle since it is sensitive to small local changes (two curves can be very close to each other but have radically different lengths). In a general manifold, the notion of length depends on a metric structure. The situation you are describing takes place in $\mathbb R^2$ viewed as a Euclidean space. Thus, the length of a curve $\gamma:[a,b]\to \mathbb R^2$ is given by $\int_a^b\sqrt vdt$, where $v(t)=\gamma_1'(t)^2+\gamma_2'(t)^2$.

Now, a parametrization of a circle is given by $\gamma:[0,2\pi]\to \mathbb R^2$, with $\gamma(t)=(p_1+r\cdot \cos(t),p_2+r\cdot \sin(t))$, $r>0$. This traces out a circle of radius $r$ with center the point $(p_1,p_2)$. It now becomes a simple matter to use the formula above for any arc of such a circle and see that changing $r$ leads to a proportional change in the length of the arc.

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Is this the easiest way to prove this ? Back in ancient time when the angle was defined people could not rely on calculus ? – Nicolas Lykke Iversen Jul 16 '13 at 10:50
Easiest way I know of. I don't exactly know how the ancient greeks argued about length and how rigorous their approach was. – Ittay Weiss Jul 16 '13 at 11:41
Do you know other ways of proving it ? If so, can you give some hints ? Otherwise I'm about to accept your answer. – Nicolas Lykke Iversen Jul 16 '13 at 20:59
Any proof will have to first define what length is. What I described is the standard definition of length in $\mathbb R^2$, and the only one I know of. – Ittay Weiss Jul 16 '13 at 21:34
I will interprete that as you know one way of proving it. I appreciate your time and help - thanks! – Nicolas Lykke Iversen Jul 17 '13 at 7:16

Since this whole concept is so elementary, it is hard to see what assumptions, definitions, axioms or intuitions you want accept up front, on which an argument can be based. I'll try two approaches nevertheless.

Both radius and arc length are measurements of length. Since Euclidean geometry only has relative length measurements, you can either compare them to one another, or compare them to some fixed length. The former gives rise to your radian, whereas the latter gives rise to units of length. If you measure the same circle using centimeters instead of inches as your unit of length, you can be sure that the quotient between the two lengths will still be the same, since the units must cancel.

Alternatively, you could imagine your situation scaled uniformly by a fixed factor. This is a similarity transformation. All angles would be preserved, all lengths be scaled by the same factor, and for fractions of length the scaling factor would again cancel out. Hence the ratio has to remain the same.

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the question is not about changing the unit of measurement. It is an absolutely non-trivial fact of Euclidean geometry that the ratio of the circumference of a circle and its diameter is a constant (know as $\pi$) which is independent of the cirlce. That it is independent of the unit of measurement is clear, but independence of the circle is not clear and requires proof. In particular, in the hyperbolic plane for instance, said ratio is still (of course) independent of the unit of measurement but not of the circle. – Ittay Weiss Jul 16 '13 at 11:40
@Ittay: Hyperbolic geometry, in contrast to Euclidean, has an absolute measure of length: you can reasonably talk about the length of a segment without comparing it to another segment. Instead the length is measured with relation to the (square root of the Gaussian) curvature. So in this sense, changing the unit of measurement in hyperbolic geometry is not the same as in Euclidean. The non-trivial fact here is the degenerate (in the sense of a Cayley-Klein geometry) distance metric of Euclidean geometry, which only allows for relative length measurements. There are no absolute lengths. – MvG Jul 16 '13 at 11:58
I don't see how this addresses OP's question about independence from the circle in Euclidean geometry. Am I missing something? – Ittay Weiss Jul 16 '13 at 15:18