Fun fact: Take any three points, connect them with lines to form a triangle,
build equilateral triangles off of each side,
then find the centers of those equilateral triangles.
When you connect those centers, you'll get another equilateral triangle,
no matter which three points you start with.
This is Napoleon's Theorem, named after THE Napoleon Bonapart,
who, aside from being a military commander and Emperor of France,
was also an amateur mathematician with special interest in geometry.
Although, it's not at all clear whether he really had anything to do with this particular theorem.
Now, I was asked by a viewer whether Napoleon's Theorem could be proved
using vector sums like the proof in my video on Van Aubel's theorem.
These theorems are similar, so I figured, why not? Challenge accepted.
The first thing to do is decide how to prove that a triangle is equilateral.
One way is to show that all three sides have the same length. But it is also sufficient to
show that two of the sides are equal, and that the angle between them is 60 degrees.
This works well for a vector proof, since a vector encodes both a length and a direction.
So for this proof, I'll construct vectors along each of these sides, like this, and then rotate one
by 60 degrees counter-clockwise. If this makes two vectors that sum to zero, then that means they are
of equal length and direction, and were therefore part of an equilateral triangle to start with.
Next, we'll need to look at what it means to be the center of an equilateral triangle.
Triangles have many "centers", and for a random triangle, they're all different.
The four most commonly used are the circumcenter,
the incenter,
the orthocenter,
and the centroid. Fortunately, for an equilateral triangle, these are all the same point.
So we can choose whichever definition is the easiest to work with, which is the centroid.
If we put the triangle in a coordinate system, the centroid can be found simply by averaging the
three x-coordinates and the three y-coordinates of the vertices. In vector terms, that means
adding the three vectors from the origin to the vertices and dividing the result by three.
Or, if we choose one of the vertices to be the origin, then one of the vectors disappears and the
other two coincide with the sides of the triangle. Again, we can add the vectors and divide by three
to find the center. Or better yet, divide those two vectors by three before adding
them together. And that is how we will define the vectors that we'll use for this proof.
Consider vectors 'a', 'b' and 'c', which take you from each of the vertices
to a point one-third of the way along the corresponding side.
Note that these vectors added together form a closed loop, which means they sum to zero.
Rotating the vector 'a' 60 degrees counter-clockwise creates a vector that I'll call 'a'-prime.
As we saw moments ago, adding that to 'a' gets you to the center of the triangle built on that
side. Rotating it by another 60 degrees gives us 'a'-double-prime, and that can bring us to
the same center from the other direction, but we'll want to continue in the same direction,
so we'll be using negative-'a'-double-prime. Rotating that by 60 degrees counter-clockwise
gives us negative-'a'-triple-prime, which, it's not hard to see, is the same as 'a'.
Using this system of vectors and rotations, we can express the vector between the first pair of
centers as negative-'a'-double-prime, plus 'a', plus 'b', plus 'b'-prime. And the vector for the second
pair as negative-'b'-double-prime, plus 'b', plus 'c', plus 'c'-prime. Following the recipe we discussed
earlier, we'll rotate the first one 60 degrees counter-clockwise, which adds a "prime" to each of
the constituent vectors. Notice that this one is now negative-'a'-triple-prime, which is the same as 'a'.
So, if we can show that these two vector sums added together will always result in zero,
that would prove they started as two sides of an equilateral triangle. Ready?
Ta-da! The sum of all these vectors is nothing more than zero, plus zero, plus zero.
Q.E.D. and how cool is that?
Before you go, I should point out that Napoleon's and Van Aubel's Theorems are both special cases
of a theorem that works for polygons with any number of sides, called the Petr-Douglass-Neumann
Theorem. This is a marvelous theorem that lends itself to some really nice animations,
and maybe someday I'll figure out how to explain its proof clearly enough for another video.