I have a minor personal triumph to relate today. So, I try to have several cute, quick math puzzles/facts on hand to bust out at parties, because I am cool like that. The most popular of these is a participation-based trick that goes as follows (feel free to play along at home):
Take a pen and paper and draw a quadrilateral. There are no restrictions (it can be concave or self-intersecting), but don’t make it too close to the sides of the paper. Now, for each edge, draw the square containing that edge that is outside the quadrilateral. Put a dot in the center of each of the four squares, and draw a line connecting opposite dots, ie, those that came from opposite edges.
The Punchline: The lines you just drew are the same length, and perpendicular.
(If you lack pen and paper, theres an applet here)
It works pretty well because it isn’t very sensitive to sloppy geometry on the part of the artist; so you will pretty consistantly get ‘perpendicular-looking’ lines. Also, I often bend the truth a bit and attribute it to Napolean Bonaparte, even though he proved something different but closely related. The usual attribution of this result is to van Aubel, who to my knowledge conquered very little of Europe.
I’ve been using this for a couple of years now, and periodically I attempt to find a nice geometric proof without passing to coordinates. Such a proof eluded me up to last night, when I came up with a reasonably nice vector-based proof. Its a little cheap since using vectors isn’t totally different from passing to coordinates, but in my mind a geometric proof is one which can be done only with pictures (though I will use words for laziness’ sake). For the record, I was unaware of van Aubel’s proof until this morning, which is a more traditional geometric proof, but a little bit more indirect.
The first step is to notice that the vector connecting the midpoints of opposite sides is the average of the vectors composing the other two sides:
So, in the picture, the red vector is the average of the two purple vectors. Of course, instead of averaging them, we can cut each of them in half and then add them.
Now, the red vector is the sum of the two purple ones.
Next, lets rotate the purple vectors 90 degrees to the right, and position them so they go from the midpoint of the squares to the midpoints of the edges:
We haven’t changed the length of the vectors at all, and we’ve rotated them both the same amount, so their sum is still the same length as the red vector but perpedicular.
We can apply identical logic to the other vector connecting opposite midpoints:
We have that the sum of the purple vectors is the same length and perpendicular to the red vector, and that the sum of the pink vectors is the same length and perpendicular to the blue vector. Hence, the purple vectors plus the blue vector is the same length and perpendicular to the pink vectors plus the red vector. QED
Its not hard to see the coordinate-based version of this proof beneath the surface, especially if one use complex coordinates, where rotation by 90 degrees is multiplication by . In fact, I think this theorem is a pretty good illustration of the power and usefulness of cartesian coordinates.
(Note: The images above are based on a picture I stole from Mathworld)