Bài thi toán châu Mỹ thuộc địa 1988

Saturday, October 23rd, 2021

[ Sunday, 25 April, 2010 ]

Have a little spare time. Let’s do math. This problem has not been solved since 1988.

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 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 f² = a² – h² = 3h² 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 + ½ e/a = e/c (**)
but cosγ = i/a, (**) becomes sqrt(3)/2 i/a + ½ e/a = e/c (***)
Apply Pythagorean’s theorem to right triangle BJC, we have i = sqrt(a² – e²)

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

a4 + c4 + a3c + ac3 – 4a²c² = 0 or
(a –c)² (a² + 3ac + c²) = 0 or a = c
and the problem is solved.