**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