The University holds an entrance exam for its new students. It's a simple practical test. They are put one at a time in a small, bare, room, in which there is an ordinary sink with cold water tap, a gas ring connected to gas in the ordinary way, an empty kettle on the floor, and a box of matches. They are asked to boil water. Those who fail are sent to the chemistry department, but anyone who manages to take the kettle to the sink, fill it with water, put it on the gas ring, turn on the gas, strike a match, light the gas, and wait for the water to boil goes through to stage two of the test. In stage two, the prospective student is put in a very similar bare room, containing sink, gas ring, kettle, and matches, but the kettle is already on the gas ring, and full of water. The task is the same. Students who simply turn on the gas, strike a match, light the gas and wait for the water to boil go to the physics department. Only those who take the kettle off the gas ring, empty it down the sink, and put it on the floor, thus reducing the problem to one they have already solved, go to the maths department.

Here's a boring-looking problem to which I can present a proof which can cause even a non-mathematician at a dinner table to go "Cor! Neat, eh!". In a plane (i.e. on a table napkin) draw three circles all of different sizes (they may overlap, but may not be completely inside each other). Now for each pair of circles, draw the two tangents that touch both on the same side (i.e. place two chopsticks to sandwich each pair of circles). Each of these pairs of lines (chopsticks) meets on a point (because the circles aren't the same size). Prove that the three points at which the three pairs of chopsticks intersect all lie on a straight line.

OK, imagine that on each circle on the napkin stands a hemispherical bubble of the same radius (and of rather strong material as bubbles go). Also imagine that the table has a mirror surface, so you appear to see three spheres floating in space with the knives and forks. Now imagine that for each pair of spheres, the pair of touching chopsticks is replaced by a carefully cut piece of paper from the inside of the menu, curved around the bubbles. Now you see a floating cone, and obviously the vertex (er, "point") of the cone is at the same place where the chopsticks used to intersect. Almost there: we have to show that the vertices of the three cones lie on a straight line. So we take the stiff (flat) outer cover of the menu, and lay it on top of the three bubbles. It isn't hard to see that it therefore lies neatly on top of the three cones, and therefore intersects the table at each of the three vertices. But both the menu and the table are planes, and two planes intersect in a straight line. Therefore the three vertices lie on a straight line. Therefore the two-dimensional problem follows trivially, as a subset of the three-dimensional one.

Now that's Elegant. (And no, I didn't think of it, but I claim responsibility for reducing it to ASCII.)