*Update (7/31/2010): This article has been rewritten, with several more elegant solutions added, rather than the coordinate solution presented originally.*

This is the second time I’m attempting a Challenge of the Week problem, after solving this one and winning coupons for Baskin Robbins’ Ice Cream (which by the way I can’t use because I’m in Canada).

Well my solution here is pretty inelegant and is a sort of ‘coordinate bash’ geometric solution. Pretty sure a better solution exists, but this is the best I could do for now.

The problem is originally from here (although I expect it will be somewhere else by the end of this week).

### Problem

In equilateral triangle , *M* is a point on *BC*. Let *N* be the point not on *BA* such that is equilateral. *P*, *Q*, and *R* are midpoints of *AB*, *BN*, and *CM* respectively. Prove that is equilateral.

### Solution by coordinate geometry

As with coordinate geometry, I put the figure on a cartesian grid with *B* as the origin.

Without losing generality, we can let *BC* be on the x-axis and *C* be *(4,0)*. Also let *M* be *(4x,0)* for some value *x* between 0 and 1. So *4x* is between 0 and 4 and is any point on *BC*.

Doing this doesn’t lose generality because any instance of this problem can be rotated and scaled so that *B* is the origin and *C* is *(4,0)*.

I choose these values for the initial points in order to end up with easier to work with points for *PQR*.

Noting the coordinates of the points:

Since we know *B* and *C*, we can calculate *A* and *N*:

It becomes easy to calculate *P*, *Q*, and *R*:

We now calculate the distances of *PQ*, *QR*, and *PR *using the distance formula:

We find that the three simplify to the same thing. That means and is equilateral.

### Solution via trigonometry

Rather than a coordinate bash, we can use a similar trigonometry bash. We let be the larger equilateral side (), and be the smaller equilateral side ().

Using the law of cosines, with the fact that and , we can get these equations:

If we just simplify these three expressions, we find that they are exactly the same:

which proves the result.

### Solution via homothetic triangles

A more elegant and less ‘bashy’ solution exists via *homothety*.

We expand triangle by a factor of 2 about the point , giving the new triangle . That is, is twice of , and is twice of , and so on.

It is sufficient to prove that is equilateral, as it is homothetic to .

Since , it follows that . It is obvious that , and also that , thus triangles and are congruent. Thus, .

As , it is also true that , thus proving triangle to be equilateral by SAS.