20th Vietnamese Mathematical Olympiad 1982 Problems

20th Vietnamese Mathematical Olympiad 1982 Problems

A1.  Find a quadratic with integer coefficients whose roots are cos 72^o and cos 144^o.

A2.  Find all real solutions to x(x + 1)(x + 2)(x + 3) = m - 1.

A3.  ABC is a triangle. A' is on the same side of BC as A, and A" is on the opposite side of BC. A'BC and A"BC are equilateral. B', B", C', C" are defined similarly. Show that area ABC + area A'B'C' = area A"B"C".  

B1.  Find all positive integer solutions to 2^a + 2^b + 2^c = 2336. 

B2.  n is a positive integer. x and y are reals such that 0 ≤ x ≤ 1 and x ⁿ⁺¹ ≤ y ≤ 1. Show that the absolute value of (y - x)(y - x²)(y - x³) ... (y - xⁿ )(1 + x)(1 + x²) ... (1 + xⁿ ) is at most (y + x)(y + x²) ... (y + xⁿ )(1 - x)(1 - x²) ... (1 - xⁿ ). 

B3.  ABCDA'B'C'D' is a cube (ABCD and A'B'C'D' are faces and AA', BB', CC', DD' are edges). L is the line joining the midpoints of BB' and DD'. Show that there is no line which meets L and the lines AA', BC and C'D'. 

Solution

20th VMO 1982

Problem A2

Find all real solutions to x(x + 1)(x + 2)(x + 3) = m - 1.

Answer

m < 0, no solutions
m = 0, two solutions, x = -3±√5)/2
0 < m < 25/16, four solutions, x = -3/2 ±√(5/4 ±√m)
m = 25/16, three solutions, x = -3/2, -3/2 ±√(5/2)
m > 25/16, two solutions, x = -3/2 ±√(5/4 + √m)

Solution

We have m = (x²+3x+1)² = (x + 3/2)² - 5/4. So if m is negative there are no solutions. For non-negative m we have solutions as above

Thanks to Suat Namli 

20th VMO 1982

Problem B1

Find all positive integer solutions to 2^a + 2^b + 2^c = 2336.

Answer

{a,b,c} = {5,8,11}

Solution

Assume a ≥ b ≥ c (then we get other solutions by permuting a, b, c). 3·512 < 2336 < 4096, so a = 10 or 11. If a = 10, then since 512 + 512 < 2336 - 1024 = 1312, we must have b = 10. But that leaves 288, which is not a power of 2. Hence a = 11. So 2^b + 2^c = 288. But 128 + 128 < 288 < 512, so b = 8. Then c = 5.

Thanks to Suat Namli