The post I was planning on putting up today got delayed by completely foreseeable events, so here’s a cool nerd snipe I ran across recently.

Start with a shape that looks like a right triangle except that the hypotenuse has been replaced with part of the arc of a circle. (Convex inward, and intersects the triangle at only those two points.)

Like this but less blurry.

**Prove that the curve is shorter than the sum of the two sides.**

I like this question because it’s immediately obvious just from looking at it, but you’ll probably have to think to get a proof. It seems like there should be a completely obvious proof that you could explain to a six-year-old, and there probably is, but if so I don’t know what. Even the best proof I’ve seen involved a limit and wasn’t all that satisfying. Can you find a good one?

(As incentive for trying, a proof here will also work as an explanation of the troll piÂ phenomenon. Because there’s only a countably infinite number of points where the shape with all the right angles intersects the circles, and there’s an *un*countably infinite number of points on the circle, there must be some points where no matter how many iterations you go, that point will never be on one of the corners. So zoom in on one of those points, and you get this shape. Proving this will prove that the shape with the right angles has a larger perimeter than the circle, and Archimedes will praise you as the savior of mathematics.)

Anyway, the post that was meant for today will probably go up in a day or two.

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Stuart ArmstrongProof that works because the “hypotenuse” is exactly an arc of circle:

Let O be the centre of the circle. Let p(t) be a point moving along the arc of the circle at 1m/s (ie 1 in SI units – or any other units, but SI works here). Extend the radius Op(t) until it hits the two sides; call this intersection point q(t).

When we increase t by a very small amount dt, q(t) moves a distance Op(t) times dt. Except at the two ends, Op(t) < Oq(t). Furthermore, the q(t) does not move orthogonally to Oq(t), but at an angle. Both of these facts contribute to the fact that q(t) must move a greater distance during dt than does p(t).

So if q(t) moves more for every small increase in time, it must move a greater distance overall.

Eli DupreeA way that a six year old could understand? You’re dealing with path length, and that’s defined using limits (specifically integrals), so a rigorous proof will involve those, either in itself or by referring to an abstraction that’s based on integrals (Stuart Armstrong’s post refers to integrals, although using an intuitive explanation rather than a rigorous one). If you use them directly, you needn’t restrict it to an arc of a circle – you might as well prove it for any differentiable, nondecreasing function.

I guess you can take a shortcut by starting with the circle and using some facts about trig functions (because we can assume we already know the length of the path they trace out), but it’s even uglier:

WLOG take the arc endpoints to be (sin(\theta_1), 1 – cos(\theta_1)) and (sin(\theta_2), 1 – cos(\theta_2)), for 0 < \theta_1 < \theta_2 (\theta_2 – \theta_1), by the fact that the (sin-cos) function always has derivative > 1 in the range (0, \tau/4).

Eli DupreeHuh, it ate some of the lines of my proof. Let’s try that again:

WLOG take the arc endpoints to be (sin(\theta_1), 1 – cos(\theta_1)) and (sin(\theta_2), 1 – cos(\theta_2)), for 0 < \theta_1 < \theta_2 (\theta_2 – \theta_1), by the fact that the (sin-cos) function always has derivative > 1 in the range (0, \tau/4).

Eli DupreeNope, still not working, so I didn’t just accidentally delete part of it – it must be a text substitution from the site, but I don’t know what. Here it is in a pastebin. http://pastebin.com/BknTkxcH

Stuart ArmstrongHow about this (works whenever the curve is convex):

Plant a green ring at each ends of the hypotenuse. Then put a red nail at the right angle of the triangle.

Then put blue nails at each point of the curve. You need infinitely many. but it’s a thought experiment, so just make them really thin.

Then thread a cord through the green rings, round all the nails, and pull hard on both ends. This cord will catch on the red nail, forming the triangle green ring-red nail-green ring (the red lines in your completely non-blurry drawing).

Now remove the red nail, and continue pulling. You’ve removed an obstacle the cord was caught on, so you must be able to pull more of the cord out now. The cord will now be caught on the blue nails instead.

Since the curve is convex, being caught on the blue nails means it will perfectly follow the curved blue hypotenuse. Since we’ve been able to pull out more of the cord, the length of the curve must be shorter now than it was before, when it lay on the red lines.

Nate GabrielPost authorI like those last two proofs. Especially the algebraic one; it reached how-did-I-miss-that levels of simplicity.

The argument I ended up making was that you can just draw a diagonal line cutting off part of the corner without intersecting the curve. If you do keep doing that forever, you approach the curve. Since each line made the total perimeter smaller (triangle inequality), the thing being approached must be smaller than the starting length. Like the one with the cord but less intuitive.