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).
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.