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Graphically, it's obvious that given two different points $a$ and $b$ on a circle of radius $r$, the linear distance (chord length) from $a$ to $b$ is less than the arc length from $a$ to $b$. How do you prove this?

I know (because I've proven it before) that if the angle between $a$ and $b$ is $\theta$ in radians, then the arc length $s$ is $s = r \theta$, and the chord length $d$ is $d = r * \sqrt{2 - 2\cos(\theta)}$.

The proof should be simple, but somehow it's not apparent to me how to prove $d < s$.

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There is not going to be any generic answer to this question, because it depends completely on what set of axioms and definitions you're using. It's fairly common to define a line as a curve of extremal length, in which case the thing you want to prove is simply proved by pointing to the definition. –  Ben Crowell Feb 9 at 4:28
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3 Answers

up vote 4 down vote accepted

You wish to prove that (note the missing $r$): $$r\theta >r\sqrt{2-2\cos\theta}$$ Now use the half angle identity, to get: $$r\sqrt{2-2\cos\theta}=2r \sin(\theta/2)$$ And since for $x>0$ we have $\sin x < x$, $$2r \sin(\theta/2)<r\theta$$

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Thanks, I missed r in the formula for chord length. I'm curious about theta being between pi/2 and pi. –  user8473 Feb 8 at 21:25
    
@user8473 - my bad. I fixed the answer. –  nbubis Feb 8 at 21:26
    
Thanks. It's so obvious now –  user8473 Feb 8 at 21:29
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OK, but why is $\sin x<x$? –  John Bentin Feb 8 at 22:04
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$\sin x < x$ is essentially a more specialized form of exactly what you're trying to prove, I don't think it's valid to assume it unless you know of some very clever proof. –  Jack M Feb 8 at 22:07
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The question is: What is arc length anyway? One possible definition is: For $\epsilon>0$, consider all point sequences $a=x_0,x_1,\ldots, x_n=b$ such that all $x_i$ are on the circle (or in any other set $S$ with respect to which we want to measure an arc length) and the distance $d(x_i,x_{i+1})$ between $x_i$ and $x_{i+1}$ is $<\epsilon$. Let $d_\epsilon(a,b)$ be the infimum of $d(x_0,x_1)+d(x_1,x_2)+\ldots +d(x_{n-1},x_n)$ over all such sequences. Then the arc length from $a$ to $b$ can be defined as $\lim_{\epsilon\to 0} d_\epsilon(a,b)$.

By the triangle inequality, $$d(x_0,x_1)+d(x_1,x_2)+\ldots +d(x_{n-1},x_n)\ge d(x_0,x_n)=d(a,b)$$ for all considered point sequences. Hence $d_\epsilon(a,b)\ge d(a,b)$ for all $\epsilon$, and hence the same inequality holds for the limit. To show that the arc length is in fact strictly greater, pick any point $c$ on the arc between $a$ and $b$, and note that all point sequences with step width $<\epsilon$ have an intermediate point $x_i$ close to $c$ (that is: with $d(x_i,c)<\epsilon$). Hence in the limit we obtain that the arc length from $a$ to $b$ is $\ge d(a,c)+d(c,b)>d(a,b)$, where the last strict inequality follows from the fact that $c$ is not on the straight line segment $ab$.

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Briefly: It's true for triangles. By induction, it's true for polygonal paths. Taking the limit, it's true for curves. –  Jack M Feb 8 at 23:42
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One short answer(I have failed to comment), the line segment has the shortest distance among all other curves joining your points.

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