39th VTRMC, 2017, Solutions

  1. Set f (x) = 2x6 -6x4 -6x3 +12x2 + 1 = 0 and g(x) = 2x6 -6x4 -4√2x3 +12x2. By raising to the sixth power, we see that a solution to the given equation also satisfies f. Furthermore to have a real solution, we need x≤√2. Therefore if we can show that f (x) has no solutions with x≤√2, then it will follow that the original equation has no solutions. Now g(x) = 2x2(x - √2)2(x2 +2√2x + 3). Thus g has zeros at 0 and √2 (of multiplicity 2), and is positive otherwise, because x2 +2√2x + 3 > 0 for all x∈ℝ. Now f (x) -g(x) = (4√2 -6)x3 + 1 which is positive for x≤√2, because the function is decreasing and (4√2 -6)√23 +1 > 0. To see this, we need to show that 17 - 12√2 > 0. However multiplying by 17 + 12√2, we see that we need to show 172 -144·2 > 0, which is true. It follows that the given equation has no real solutions.

  2. Write t = tan(x/2). Then cos2(x/2) = 1/(1 + t2), so

    cos x = cos2(x/2) - sin2(x/2) = (1 -t2)/(1 + t2)

    and since tan x = 2t/(1 -t2),

    sin x = cos x tan x = 2t/(1 + t2).

    Write I = ∫0adx/(1 + cos x + sin x). Since dt/dx = (sec2(x/2))/2 = (1 + t2)/2, we see that

    I = ∫0tan(a/2)2dt/((1 + t2) + (1 -t2) + 2t) = ∫0tan(a/2)dt/(1 + t).

    Therefore I = ln(1 + tan(a/2)). (An alternative answer is ½ln(1 + sin a)/(1 + cos a) + ½ln 2.) When a = π/2, we have tan(a/2) = 1 and we deduce that I = ln 2 as required.

  3. We may assume that AB = 1. Since ∠APB = 150, the sine rule yields, sin 150/AB = sin 20/AP = sin 10/BP and sin 30/AP = sin 40/CP. Therefore PC = 4 sin 20 sin 40 = 2 cos 20 - 1. Write ∠PBC = θ. Since ∠BPC = 100, we see that ∠PCB = 80 - θ, and then the sine rule for triangle BPC yields

    (2 cos 20 - 1)/(sinθ) = 2 sin 10/(sin(80 - θ)) = (2 sin 10)/(cos(θ + 10)).

    Therefore

    2 cos 20 cos(θ +10) = 2 sin 10 sinθ + cos(θ +10) = cos(θ - 10).

    We deduce that cos(30 + θ) + cos(10 - θ) = cos(θ - 10) and hence cos(30 + θ) = 0. We conclude that θ = 60.

  4. Denote the vertices of the triangle by A, B and C (counterclockwise). Let P be an interior point of the triangle and draw lines parallel to the three sides, partitioning the triangle into three triangles and three parallelograms. Let EH be the segment parallel to AC, let FI be the segment parallel to BC, and let JG be the segment parallel AB. Here the points E, F lie on the edge AB; the points G, H lie on the edge BC, and the points I, J lie on the edge AC. Suppose that the area of the triangle EFP is a, the area of the triangle PGH is b, and the area of the triangle JPI is c. Note that the triangles EFP, PGH, JPI and ABC are similar. Therefore EF/PG = √a/√b and JP/PG = √c/√b. Thus (EF + JP)/PG = (√a + √c)/√b and hence 1 + (EF + JP)/PG = 1 + (√a + √c)/√b, i.e.

    (PG + EF + JP)/PG = (√a + √b + √c)/√b.

    Since PG = FB and JP = AE, because FBGP and AEJP are parallelograms, AB/PG = (√a + √b + √c)/√b. Because ABC is similar to PGH, we have AB/PG = √T/√b. Therefore √T = √a + √b + √c.

  5. Let (a, b)∈S and let d = gcd(a, b). Then a = dm and b = dn with gcd(m, n) = 1. Since g(a, b)∈ℕ, we see that ab = d2mn is a perfect square and hence mn is a perfect square. Therefore m and n are both perfect squares, because gcd(m, n) = 1. Thus we may write a = ds2 and b = dt2 with gcd(s, t) = 1.

    By assumption, h(a, b) = 2ds2t2/(s2 + t2)∈ℕ. Since gcd(s2 + t2, s2) = gcd(s2 + t2, t2) = gcd(s2, t2) = 1, it follows that s2 + t2 divides 2d. Thus a = k(s2 + t2)s2/2 and b = k(s2 + t2)t2/2 for some k∈ℕ.

    Now a≠b because s≠±1. Also f (a, b) = k(s2 + t2)2/4∈ℕ. We have two cases to consider.

    We conclude that f (a, b)≥25. However f (5, 45) = f (10, 40) = 25, so the minimum of f over S is 25.

  6. Set g(x) = f (x) -x2 + 4x - 2. Then g(1) = g(4) = g(8) = 0. Therefore we may write g(x) = (x - 1)(x - 4)(x - 8)q(x) where q(x)∈ℤ[x]. Since f (n) = n2 - 4n - 18, we see that g(n) = -20 and hence (n - 1)(n - 4)(n - 8)q(n) = -20. By inspection, n = 3 or 6. We note that both of these values of n can be obtained, by taking (for example) q(x) = -2 and 1 respectively, and then f (x) = -2(x - 1)(x - 4)(x - 8) + x2 - 4x + 2 and (x - 1)(x - 4)(x - 8) + x2 - 4x + 2 respectively.

  7. First we look at small values of n: the given equation is a quadratic in m. If n∈{0, 1, 2, 4}, there are no solutions. If n = 3, then m = 6 or 9. If n = 5, then m = 9 or 54. We now proceed by contradiction to show that there is no solution if n≥6. So suppose (m, n) is a solution with n≥6. Then m divides 2·3n and so either m = 3a for some 0≤a≤n, or m = 2·3b for some 0≤b≤n. If m = 3a, then

    2n+1 -1 = m + 2·3n/3a = 3a +2·3n-a.

    On the other hand if m = 2·3b, then

    2n+1 -1 = m + 2·3n/m = 2·3b +3n-b.

    Therefore there must be nonnegative integers a, b such that

    2n+1 -1 = 3a +2·3b,    a + b = n.

    Note that 3a < 2n+1 < 32(n+1)/3 and 2·3b < 2n+1 < 2·32(n+1)/3, because 32/3 > 2. Thus a, b < 2(n + 1)/3. Since a + b = n, we deduce that

    (n - 2)/3 < a < 2(n + 1)/3    and    (n - 2)/3 < b < 2(n + 1)/3.

    Now let t = min(a, b). Then t > (n - 2)/3 and since n≥6, it follows that t > 1. Because 3t divides 3a and 2·3b, we see that 3t divides 2n+1 - 1. Since t≥2, we deduce that 2n+1≡1 mod 9. Now 2n+1≡1 mod 9 if and only if 6 divides n + 1, so n + 1 = 6r for some r∈ℕ. Therefore

    2n+1 -1 = 43r -1 = (42r +4r +1)(4r -1) = (42r +4r +1)(2r -1)(2r + 1).

    Since 3t divides 2n+1 - 1, we see that 3t divides (42r +4r +1)(2r -1)(2r + 1). Note that 9 does not divide 42r +4r + 1, hence 3t-1 divides (2r -1)(2r + 1). Since gcd(2r -1, 2r + 1) = 1, either 3t-1 | 2r - 1 or 2r + 1. In any case, 3t-1≤2r + 1. Then 3t-1≤2r +1≤3r = 3(n+1)/6. Therefore (n - 2)/3 - 1 < t - 1≤(n + 1)/6. This yields n < 11, which is a contradiction, because n≥6 and we proved that 6 | n + 1.



Peter Linnell 2017-10-28