Bust of Pythagoras, Museo Capitolino, Rome

A couple of nights ago I woke up at around 2 am and, as you do, immediately started thinking about Pythagoras and his famous theorem concerning right-angled triangles.  Now, you may recall that Pythagoras of Samos was an Ionian Greek mathematician and philosopher who lived some five hundred years before the Common Era started.  There is no doubt that he was a great man, a deep thinker, a traveller and a believer in reincarnation but what we also know is that he was put on this earth to torment us in the eighth grade (age 13 or thereabout) with his theorem concerning right-angled triangles.  Well can I remember chanting “In a right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the other two sides,” where the hypotenuse is the side of the triangle opposite the right angle.  Here’s the theorem:

“So what’s the problem?” you ask.  I first encountered this theorem in 1952, or maybe 1953 when I started geometry in secondary school.  I trotted it out in examination questions using logarithm tables to evaluate the square roots of a side.  In later years, I may even have used the theorem in anger to compute the length of a ladder that would enable me to reach an upstairs window of a house over an obstruction such as a bush.  But, sixty-four years later, I had forgotten how to prove it.  That was what kept me awake the other morning – how to prove Pythagoras’s Theorem!

Later that day, I turned to the Web and found a site that listed 118 proofs of this theorem – one hundred and eighteen!  Well now, I wasn’t going to wade through 118 solutions and a further search revealed a simple solution that brought it all flooding back to me.  Here is what I now recall from sixty-four years ago.

Start with a red square of size c x c.

Fig. 1

Turn the red square an arbitrary amount about its centre.

Fig. 2

Draw another square (blue) that exactly encloses the twisted red square.

Fig. 3

Note: the red square lies on top of the blue square and we see four blue identical right-angled triangles.  Let a and b be the length of the triangle sides that embrace the right angle.  The hypotenuse of each triangle is of length c.

Now it’s easy!

Fig. 4

Just looking at the blue square, the area is (a + b) x (a + b) = a² + 2ab + b²

But, looking at Fig. 3, we see that the area of the blue square is also defined by the area of the four blue right-angled triangles plus the area of the red square.  That is 4 x (ab/2) + c² = 2ab + c²

Therefore a² + 2ab + b² = 2ab + c² which tidies to a² + b² = c² which is Pythagoras’s Theorem!

That does it.  Now I can sleep easy until the next time: the Königsberg Bridge Problem might get me going … again.

Königsberg, now renamed Kaliningrad,  is a town on the Preger River in Russia. Within the town are two river islands that are connected to the river banks via six bridges plus a seventh bridge that connects the two islands together. Can you cross each of the seven bridges once and once only without retracing your steps?  If you think the answer is ‘yes’, what’s the route?  If you think the answer is ‘no’, can you prove it?

Answers on a postcard, please…

(^_^)