Let E be the midpoint of the side AD of a square ABCD.
Problem: Determine which has the greater perimeter, the square ABCD or the circle through E, B, C? (You can assume you know the value of π. Otherwise try to use elementary methods.)
The neat solution is, I think, rather satisfying, even though it doesn’t require any special ingenuity to find it. Metaproblem: Why is this solution aesthetically pleasing?
The answer — to the problem, if not the metaproblem — is below the fold: but try blowing the dust of your school geometry before looking at the cheat sheet!
OK: obviously we need to compare the diameter of the circle with the side of the square. But we haven’t yet got a diameter drawn — so put in the obvious one, through E to the opposite side of the circle at F, with EF intersecting BC at G.
The hint was “use elementary methods”, which in the first year of geometry typically means drawing some triangles, and using basic facts about similar triangles, angles of triangles inscribed in circles, etc. So the absolutely obvious next thing to do is to draw a triangle to get the augmented diagram.
Now we know that |BG| = |AE| = |EG|/2. And it is entirely elementary that EGB and BGF are similar. So |FG|/|BG| = |BG|/|EG|. From these two facts we immediately have |FG| = |EG|/4.
Set |EG| to be one unit. Then the perimeter of the square is 4, and the perimeter of the circle is π5/4 < (16/5)5/4 < 4 (for we know π < 3.2). So the square has the larger perimeter.
That is — you would have thought — entirely straightforward. But the proof is also rather satisfying, possibly easier than you were expecting, as it uses minimal input. You don’t need to know about Cartesian coordinates or anything like that.
It was therefore interesting (and to an extent puzzling) to see that when this question was posed on math.stackexchange and taken up on reddit answers were all over the place. And the most upvoted answer was initially one that depended on it just dawning on you (from your familiarity with Pythagorean triangles) that the situation must be as diagrammed on the left. So the perimeter of the square is (in those units) 32, and of the circle is 2π.5. But such guesswork doesn’t make for an elegant proof, even if in this case your very first guess might deliver the goods. We can remove the guesswork by applying brute force and the Pythagorean theorem to the diagram on the right. 
And now we just use the facts that $latex x^2 + y^2 = 1$ and $latex 1 + x = 2y$ and do the algebra. But that is still, surely, not at all such a neat proof as our original one.
What this illustrates, anyway, is perhaps that satisfying proofs don’t have to be ingenious or depend on unexpected tricks. Sometimes, the aesthetic pleasure comes from slightly surprising simplicity and economy of means.
[Diagrams snaffled from math.stackexchange!]
Here is another nice problem with an elementary solution, which I remember doing at school. I don’t know how widely known it is.
Draw two identical parabolas, touching nose to nose (eg $latex y = x^2$ and $latex y = -x^2$).
Keep one of them fixed, and roll the other one round it.
What is the locus of the focus of the rolled one?
Ah yes. That’s neat!
Back in the day, that used to be a whole geometry paper for the Cambridge scholarship examination, including a fair amount of projective geometry, which was an inexhaustible source of often rather cute problems.
I’m very rusty. How do we know that 1 + x = 2y?
Because (in that last diagram with the origin at the centre of the circle wit unit radius) the side of the square |AD| = |AB| = 1 + x; while y = |AE|, and by hypothesis E is the midpoint of |AD| …
Oh right. Thank you!
Presumably trigonometry is not considered “elementary”. What about the Pythagorean Theorem though? If that’s permissible, then I see a fairly elegant solution.