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).


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.

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