Regular Pentagon is the Unique Largest Two-Distance Set in the Plane A two-distance set is a collection of points for which only two distinct distances appear among pairs of points. (That is, the distance between any pair of points is either $x$ or $y$, and these values may be whatever you want.)
The unique (up to similarity) largest two-distance set in the plane is the regular pentagon. The result seems to have become folklore, and I cannot find a proof anywhere online. I am primarily interested in the bound of five on the size of the two-distance set, but a proof of uniqueness would be a welcome bonus.
 A: Suppose there is a two distance set of size $n=6$. Label each of the edges according to whether they have length $d_1 \neq d_2$. From Ramsey Theory, we know that a monochromatic triangle exists. 
Let the 3 vertices of the monochromatic triangle be $A, B, C$, WLOG distance $d_1$. Now, consider the intersections of circles with radius $d_1$ and $d_2$ about centers $A$ and $B$, which have to give us the 3 other points.   
If circles of radius $d_1$ intersect at a vertex that is not $C$, then we have $d_2 = \sqrt{3} d_1$.
If circles of $d_2$ intersect at a vertex, there are 2 possibilities for vertices. By considering the distance to $C$, we must have $ d_2 = \frac{\sqrt{3}}{3} d_1$ or $ d_2 = 2 \frac{ \sqrt{3}}{3} d_1$.
Notice that all of these $d_2$ are different, and hence we can have at most 1 of these points as a vertex.
If circles of radius $d_1$ and $d_2$ intersect at a vertex that is on the different side of $AB$ as $C$, there are 4 possibilities of points, with 2 on either side of $AB$. However, the points on either side of $AB$ will result in different values of $AB$, hence we can have at most 2 of these points as a vertex.
Hence, we must have 1 from the first set of points, and 2 from the second set of points. However, it is easy to see that non of the distances from the first set, will result in valid points in the second set.      

Uniqueness:
Suppose that we have a set of 5 points that are two-distance. Color each edge as before.
Claim: It doesn't contain a monochromatic triangle.
Suppose it does. Let the vertices be $A, B, C$, and the other 2 points be $D, E$. Take the circles with centers $A, B$. Then from the above, we know that the other 2 points must be from the 3 possible intersection points, so $C, D, E $ are collinear. The only 2-distance set of collinear points is where one of them is the midpoint of the other. By considering the possible positions, we must have (WLOG) $D$ as the other intersection of the cirlces that $C$ lies on. Then, (WLOG) $ED=DC =\sqrt{3} CA$ But $EA \neq CA, EA \neq CE$ hence we have a contradiction.
Claim: No monochromatic 4-cycle exists.
Suppose it does. Then, we must have a rhombus. Consider the lengths of the diagonals.
If they are distinct, then one of them is the side length of the rhombus, so we have a monochromatic triangle, contradicting the previous claim.
If they are equal and distinct from the side length of the rhombus, we have a square. It's quick to argue that no 5th point works.  
From the above claims, the complete graph on 5 vertices must be formed from 2 5-cycles. Hence, we must have the regular pentagon.
