# Challenge of the Week 05/11/2010

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 $\triangle ABC$, M is a point on BC. Let N be the point not on BA such that $\triangle BMN$ is equilateral. P, Q, and R are midpoints of AB, BN, and CM respectively. Prove that $\triangle PQR$ 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: $B = (0,0)$ $C = (4,0)$ $M = (4x,0)$

Since we know B and C, we can calculate A and N: $A = (2,2 \sqrt{3})$ $N = (2x, -2x \sqrt{3})$

It becomes easy to calculate P, Q, and R: $P = (1, \sqrt{3})$ $Q = (x, -x \sqrt{3})$ $R = (2x+2, 0)$

We now calculate the distances of PQ, QR, and PR using the distance formula: $\begin{array}{rcl} PQ &=& \sqrt{(x-1)^2 + (x \sqrt{3} + \sqrt{3})^2} \\ &=& 2 \sqrt{x^2 + x + 1} \end{array}$ $\begin{array}{rcl} QR &=& \sqrt{(2x+2-x)^2 + (x \sqrt{3})^2} \\ &=& 2 \sqrt{x^2 + x + 1)} \end{array}$ $\begin{array}{rcl} PR &=& \sqrt{(2x+2-1)^2 + \sqrt{3}^2} \\ &=& 2 \sqrt{x^2 + x + 1)} \end{array}$

We find that the three simplify to the same thing. That means $PQ = QR = PR$ and $\triangle PQR$ is equilateral.

### Solution via trigonometry

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

Using the law of cosines, with the fact that $\cos 60^\circ = \frac{1}{2}$ and $\cos 120^\circ = -\frac{1}{2}$, we can get these equations: $PR^2 = (\frac{a}{2})^2 + (\frac{a+b}{2})^2 - (\frac{a}{2})(\frac{a+b}{2})$ $QR^2 = (\frac{b}{2})^2 + (\frac{a+b}{2})^2 - (\frac{b}{2})(\frac{a+b}{2})$ $PQ^2 = (\frac{a}{2})^2 + (\frac{b}{2})^2 + (\frac{a}{2})(\frac{b}{2})$

If we just simplify these three expressions, we find that they are exactly the same: $PR^2 = QR^2 = PQ^2 = \frac{a^2+ab+b^2}{4}$

which proves the result.

### Solution via homothetic triangles

A more elegant and less ‘bashy’ solution exists via homothety. We expand triangle $PQR$ by a factor of 2 about the point $B$, giving the new triangle $ANK$. That is, $BA$ is twice of $BP$, and $BK$ is twice of $BR$, and so on.

It is sufficient to prove that $\triangle ANK$ is equilateral, as it is homothetic to $\triangle PQR$.

Since $BK = 2BR$, it follows that $BM=CK$. It is obvious that $\angle ABN = \angle ACK = 120^\circ$, and also that $AB = AC$, thus triangles $ABN$ and $ACK$ are congruent. Thus, $AN = AK$.

As $\angle BAC = 60^\circ$, it is also true that $\angle NAK = 60^\circ$, thus proving triangle $ANK$ to be equilateral by SAS.