16th Vietnamese Mathematical Olympiad 1978 Problems

16th Vietnamese Mathematical Olympiad 1978 Problems

A1. Find all three digit numbers abc such that 2(abc) = bca + cab.

A2. Find all values of the parameter m such that the equations x2 = 2|x| + |x| - y - m = 1 - y2 have only one root.

A3. The triangle ABC has angle A = 30o and AB = 3/4 AC. Find the point P inside the triangle which minimises 5 PA + 4 PB + 3 PC.

B1. Find three rational numbers a/d, b/d, c/d in their lowest terms such that they form an arithmetic progression and b/a = (a + 1)/(d + 1), c/b = (b + 1)/(d + 1).

B2. A river has a right-angle bend. Except at the bend, its banks are parallel lines a distance a apart. At the bend the river forms a square with the river flowing in across one side and out across an adjacent side. What is the longest boat of length c and negligible width which can pass through the bend?

B3. ABCDA'B'C'D' is a rectangular parallelepiped (so that ABCD and A'B'C'D' are faces and AA', BB', CC', DD' are edges). We have AB = a, AD = b, AA' = c. The perpendicular distances of A, A', D from the line BD' are p, q, r. Show that there is a triangle with sides p, q, r. Find a relation between a, b, c and p, q, r.

Solution

16th VMO 1978

Problem A1

Find all three digit numbers abc such that 2(abc) = bca + cab.

Answer

111, 222, 333, 370, 407, 444, 481, 518, 555, 592, 629, 666, 777, 888, 999.

Solution

We have 200a + 20b + 2c = 100b + 10c + a + 100c + 10a + b, so 7a = 3b + 4c (*). There are the obvious solutions a = b = c. If any two of a, b, c are equal, then (*) implies that they are all equal. So we can assume they are all distinct. Note that we must have a = b mod 4. It is now a question of looking in turn at a = 1, 2, ... , 9. For example a = 1, so b = 5 or 9. But in both cases 3b > 7a.

Thanks to Suat Namli

16th VMO 1978

Problem A2

Find all values of the parameter m such that the equations x2 = 2|x| + |x| - y - m = 1 - y2 have only one root.

Answer

0, 2

Solution

If x is a non-zero solution, then -x is another, so we must have x = 0. Hence 0 = 1 - y - m = 1 - y2. So y = ±1, so m = 0 or 2, and each of these gives just one root.

Thanks to Suat Namli