Here's another nice geometrical theorem. I hadn't seen it before today but it seems it's usually called Thébault's problem II and is credited to Victor Thébault (1882–1960), a mathematician who published over 1000 problems
Here's how it goes (this is a slight extension of the original problem, which had a square where the following requires only a parallelogram). Given a parallelogram ABCD as shown, erect equilateral triangles on two adjacent sides. Then the two free vertices of those triangles together with the vertex of the parallelogram opposite form an equilateral triangle.
The following is my proof. I haven't looked at any published ones, but I'm sure this is one of the standard ones and I imagine we could use complex numbers quite nicely as well, but this uses only ideas that would have been familiar to Euclid.
In this diagram there are two different lengths, marked with | and ||. Call these L1 and L2 and suppose the angle BCD is α. Then using the properties of equilateral triangles (all angles are 60) and parallelograms (adjacent angles add up to 180) we can mark in a series of angles.
The angle EDF is 60 + α by subtracting the other three angles at D from 360.
Now if we consider triangles EFD, FCD and BAE, we see they all have an angle of 60 + α between sides of length L1 and L2, which means these are all congruent and it follows immediately that the three dotted distances are equal and BEF is equilateral as claimed.
Thébault's theorem has attracted some deal of interest, because it is so neat and quite simple, but seemingly was not spotted for over 2,000 years. It has ben generalised at least twice, most recently in 2019, see The Mathematical Gazette, Vol. 103, No. 557 (July 2019), pp. 343-346.
More Beautiful Geometry
Here's another nice geometrical theorem. I hadn't seen it before today but it seems it's usually called Thébault's problem II and is credited to Victor Thébault (1882–1960), a mathematician who published over 1000 problems
Here's how it goes (this is a slight extension of the original problem, which had a square where the following requires only a parallelogram). Given a parallelogram ABCD as shown, erect equilateral triangles on two adjacent sides. Then the two free vertices of those triangles together with the vertex of the parallelogram opposite form an equilateral triangle.
The following is my proof. I haven't looked at any published ones, but I'm sure this is one of the standard ones and I imagine we could use complex numbers quite nicely as well, but this uses only ideas that would have been familiar to Euclid.
In this diagram there are two different lengths, marked with | and ||. Call these L1 and L2 and suppose the angle BCD is α. Then using the properties of equilateral triangles (all angles are 60) and parallelograms (adjacent angles add up to 180) we can mark in a series of angles.
The angle EDF is 60 + α by subtracting the other three angles at D from 360.
Now if we consider triangles EFD, FCD and BAE, we see they all have an angle of 60 + α between sides of length L1 and L2, which means these are all congruent and it follows immediately that the three dotted distances are equal and BEF is equilateral as claimed.
Thébault's theorem has attracted some deal of interest, because it is so neat and quite simple, but seemingly was not spotted for over 2,000 years. It has ben generalised at least twice, most recently in 2019, see The Mathematical Gazette, Vol. 103, No. 557 (July 2019), pp. 343-346.