30th VTRMC, 2008, Solutions

1. Write f (x) = xy³ + yz³ + zx³ - x³y - y³z - z³x. First we look for local maxima, so we need to solve ∂f /∂x = ∂f /∂y = ∂f /∂z = 0. Now ∂f /∂x = y³ +3x²z - z³ -3x²y. If y = z, then f (x, y, z) = 0 and this is not a maximum. Thus we may divide by y - z and then ∂f /∂x = 0 yields y² + yz + z² = 3x². Similarly x² + xz + z² = 3y² and x² + xy + y² = 3z². Adding these three equations, we obtain (x - y)² + (y - z)² + (z - x)² = 0, which yields x = y = z. This does not give a maximum, because f = 0 in this case, and we conclude that the maximum of f must occur on the boundary of the region, so at least one of x, y, z is 0 or 1.

   Let's look at f on the side x = 0. Here f = yz³ - y³z and 0≤y≤1, 0≤z≤1. To find local maxima, we solve ∂f /∂y = ∂f /∂z = 0. This yields y = z = 0 and f = 0, which is not a maximum, so the maximum occurs on the edges of the region considered. If y or z = 0, we get f = 0 which is not a maximum. If y = 1, then f = z³ - z≤ 0, which won't give a maximum. Finally if z = 1, then f = y - y³. Since df /dy = 1 - 3y², we see that f has a maximum at y = 1/√3. This gives that the maximum value of f on x = 0 is 1/√3 -1/√3³ = 2√3/9.

   Similarly if y or z = 0, the maximum value of f is 2√3/9. Now let's look at f on the side x = 1. Here f = y³ + yz³ + z - y - y³z - z³. Again we first look for local maxima: ∂f /∂y = 3y² + z³ -1 - 3y²z. Then ∂f /∂y = 0 yields either z = 1 or 3y² = z² + z + 1. If z = 1, then f = 0 which is not a maximum, so 3y² = z² + z + 1. Similarly 3z² = y² + y + 1. Adding these two equations, we find that y² - y/2 + z² - z/2 = 1. Thus (y - 1/2)² + (z - 1/2)² = 3/2. This has no solution in the region considered 0≤y≤1, 0≤z≤1. Thus f must have a maximum on one of the edges. If y or z is 0, then we are back in the previous case. On the other hand if y or z is 1, then f = 0, which is not a maximum.

   We conclude that the maximum value of f on 0≤x≤1, 0≤y≤1, 0≤z≤1 is 2√3/9.

2. For each positive integer n, let f (n) denote the number of sequences of 1's and 3's that sum to n. Then f (n + 3) = f (n + 2) + f (n), and we have f (1) = 1, f (2) = 1, and f (3) = 2. Thus f (4) = f (3) + f (1) = 3, f (5) = f (4) + f (2) = 4, f (6) = 6, ..., f (15) = 189, f (16) = 277. Thus the number of sequences required is 277.

3. Let R denote the specified region, i.e. {(x, y) | x⁴ + y⁴≤x² - x²y² + y²}. Then R can be described as the region inside the curve x⁴ + x²y² + y⁴ = x² + y² ( (x, y)≠(0, 0)). This can be rewritten as
   (x² + y² - xy)(x² + y² + xy) = x² + y².
   Now change to polar coordinates: write x = r cosθ, y = r sinθ; then the equation becomes (r² - r²cosθsinθ)(r² + r²cosθsinθ) = r². Since r≠ 0 and 2 cosθsinθ = sin 2θ, we now have r²(1 - (1/4)sin²2θ) = 1. Therefore the area A of R is

   ∬_Rr drdθ = ∫₀^2π∫₀^{(1-(1/4)sin22θ)-1/2}r drdθ = ∫₀^2πdθ/(2(1 - (1/4)sin²2θ)) = ∫₀^π/416 dθ/(3 + cos²2θ) = ∫₀^π/416 sec²2θ dθ/(4 + 3 tan²2θ).

   
   
   Now make the substitution 2z = √3tan 2θ, so dz = √3sec²2θ dθ and we obtain
   A = (4/√3)∫₀^∞dz/(1 + z²) = 2π/√3.

4. Ceva's theorem applied to the triangle ABC shows that (AP/PB)(BM/MC)(CN/NA) = 1. Since BM = MC, we see that AP/PB = AN/NC and we deduce that PN is parallel to BC. Therefore ∠NPX = ∠PCB = ∠NAX and we conclude that APXN is a cyclic quadrilateral. Since that opposite angles of a cyclic quadrilateral sum to 180^o, we see that ∠APX + ∠XNA = 180^o, and the result follows.

5. Let T = {a_n | n∈N} and for t a positive number, let A_t = {n∈N | a_n≥t}. Since ∑a_n = 1 and a_n≥ 0 for all n, we that if δ > 0, then there are only finitely many numbers in T greater than δ, and also A_t is finite. Thus we may label the nonzero elements of T as t₁, t₂, t₃,..., where t₁ > t₂ > t₃ > ... > 0. We shall use the notation XΔY to indicate the symmetric difference {X\Y∪Y\X} of two subsets X, Y.

   Consider the sum

   ∑_i≥1(t_i - t_i+1)| A_tiΔπ^-1A_ti|.
   Note that n∈A_t\π^-1A_t if and only if a_n≥t > a_πn, and n∈π^-1A_t\A_t if and only if a_n < t≤a_πn. Write a_n = t_p and a_πn = t_q. We have three cases to examine:
   1. a_n = a_πn. Then n does not appear in the above sum.

   2. a_n > a_πn. Then p < q and n is in A_tr\π^-1A_tr whenever t_p≥t_r > t_q, that is q > r≥p and we get a contribution (t_p - t_p+1) + (t_p+1 - t_p+2) + ... + (t_q-1 - t_q) = t_p - t_q = a_n - a_πn = | a_n - a_πn|.

   3. a_n < a_πn. Then p > q and n is in π^-1A_tr\A_tr whenever t_q≥t_r > t_p, that is p > r≥q and we get a contribution (t_q - t_q+1) + (t_q+1 - t_q+2) + ... + (t_p-1 - t_p) = t_q - t_p = a_πn - a_n = | a_n - a_πn|.
   We conclude that
   ∑_n=1^∞| a_n - a_πn| = ∑_i≥1(t_i - t_i+1)| A_tiΔπA_ti|,
   because | A_tiΔπ^-1A_ti| = | A_tiΔπA_ti|. Similarly
   ∑_n=1^∞| a_n - a_ρn| = ∑_i≥1(t_i - t_i+1)| A_tiΔρA_ti|,
   and we deduce that
   ∑_n=1^∞(| a_n - a_πn| + ∑_n=1^∞| a_n - a_ρn|) = ∑_i≥1(t_i - t_i+1)(| A_tiΔπA_ti| + | A_tiΔρA_ti|).
   Therefore ∑_i≥1(t_i - t_i+1)(| A_tiΔπA_ti| + | A_tiΔρA_ti|) < ε. We also have ∑_i≥1(t_i - t_i+1)| A_ti| = 1. Therefore for some i, we must have | A_tiΔπA_ti| + | A_tiΔρA_ti| < ε| A_ti| and the result follows.

6. Multiply a⁴ -3a² + 1 by b and subtract (a³ - 3a)(ab - 1) to obtain a³ - 3a + b. Now multiply by b and subtract a²(ab - 1) to obtain a² -3ab + b². Thus we want to know when ab - 1 divides (a - b)² - 1, where a, b are positive integers. We cannot have a = b, because a² - 1 does not divide -1. We now assume that a > b.

   Suppose ab - 1 does divide (a - b)² - 1 where a, b are positive integers. Write (a - b)² - 1 = k(ab - 1), where k is an integer. Since (a - b)² -1≥ 0, we see that k is nonnegative. If k = 0, then we have (a - b)² = 1, so a - b = ±1. In this case, ab - 1 does divide a⁴ -3a² + 1, because a⁴ -3a² +1 = (a² + a - 1)(a² - a - 1). We now assume that k≥1.

   Now fix k and choose a, b with b as small as possible. Then we have a² + a(- 2b - kb) + b² + k - 1 = 0. Consider the quadratic equation x² + x(- 2b - kb) + b² + k - 1 = 0. This has an integer root x = a. Let v be its other root. Since the sum of the roots is 2b + kb, we see that v is also an integer. Also av = b² + k - 1. Since b, k≥1, we see that v is also positive. We want to show that v < b; if this was not the case, then we would have b² + k - 1≥ab, that is k≥ab - b² + 1. We now obtain

   (a - b)² -1≥(ab - b² + 1)(ab - 1).
   This simplifies to a² - ab≥(ab - b² + 1)ab, that is a - b≥(ab - b² + 1)b and we obtain (a - b)(b² -1) + b≤ 0, which is not the case. Thus v < b and we have v² + v(- 2b - k) + b² + k - 1 = 0. Set u = b. Then we have (u - v)² - 1 = k(uv - 1), where u, v are positive and v < b. By minimality of b, we conclude that there are no a, b such that (a - b)² - 1 = k(ab - 1).

   Putting this altogether, the positive integers required are all a, b such that b = a±1.

7. Note that for fixed x > 1, the sequence 1/f_n(x) is decreasing with respect to n and positive, so the given limit exists which means that g is well-defined. Next we show that g(e^1/e)≥1/e, equivalently lim_{n--> ∞}f_n(e^1/e)≤e. To do this, we show by induction that f_n(e^1/e)≤e for all positive integers n. Certainly f₁(e^1/e) = e^1/e≤e. Now if f_n(e^1/e)≤e, then
   f_n+1(e^1/e) = (e^1/e)^f_n(e1/e)≤(e^1/e)^e = e,
   so the induction step passes and we have proven that g(e^1/e)≥1/e.

   We now prove that g(x) = 0 for all x > e^1/e; this will show that g is discontinuous at x = e^1/e. We need to prove that lim_{n--> ∞}f_n(x) = ∞. If this is not the case, then we may write lim_{n--> ∞}f_n(x) = y where y is a positive number > 1. We now have

   y = lim_{n--> ∞}f_n(x) = lim_{n--> ∞}f_n+1(x) = x^{lim_{n--> ∞}f_n(x)} = x^y.
   Therefore ln y = y ln x and x = y^1/y. Since (dx/dy)/x = (1 - ln y)/y², we see by considering the graph of y^1/y that it reaches its maximum when y = e, and we deduce that x≤e^1/e. This is a contradiction and we conclude that lim_{n--> ∞}f_n(x) = 0. Thus we have shown that g(x) is discontinuous at x = e^1/e.