Problem 1 of the IberoAmerican Mathematical Olympiad 1988

The measures of the angles of a triangle is an arithmetic progression and its altitudes is also another arithmetic progression. Prove that the triangle is equilateral.

Solution by Steve Dinh, a.k.a. Vo Duc Dien

Let I, J, and K be the feet of A, B and C to BC, AC and AB, respectively. Now let BC = a, AC = b, AB = c, AI = d, BJ = e, CK = f, BI = g, CJ = i and BK = h. Also let ∠BAC = α, ∠ABC = β and ∠ACB = γ.

Assume α is the smallest angle of the triangle and ε is the angle of common difference. We have

β = α + ε, γ = α + 2ε but the sum of the angles is 180°, we then have 3(α + ε) = 180° or β = α + ε = 60° and α = 120° - γ. Now we need to prove a = c for the triangle ABC to be equilateral.

Since β = 60°, we have a = 2h, c = 2g and f˛ = a˛ - h˛ = 3h˛ or f = h \sqrt{3} = a \sqrt{3}/2.

Similarly d = c \sqrt{3}/2 and since d, e, and f form another arithmetic progression, we have

	e = (f + d)/2 = (a + c) \sqrt{3}/4						(*) 

We also have sinα = e/c and sinγ = e/a or

	sinα = sin(120° - γ) =  \sqrt{3}/2 cosγ + ˝ e/a = e/c			(**)

but cosγ = i/a, (**) becomes \sqrt{3}/2 i/a + ˝ e/a = e/c (***)

Apply Pythagorean’s theorem to right triangle BJC, we have i = \sqrt{a˛ - e˛}

Now substitute i and e from (*) to (***), we have

	a^4 + c^4 + a^3c + ac^3 – 4a^2c^2 = 0   	or
	(a –c)^2 (a^2 + 3ac + c^2) = 0 	or  a = c