12th VTRMC, 1990, Solutions

1. Let a be the initial thickness of the grass, let b the rate of growth of the grass, and let c be the rate at which the cows eat the grass (in the appropriate units). Let n denote the number of cows that will eat the third field bare in 18 weeks. Then we have

   10(a + 4b)/3 = 12*4c

   10(a + 9b) = 21*9c

   24(a + 18b) = n18c

   
   
   If we multiply the first equation by -27/5 and the second equation by 14/5, we obtain 10(a + 18b) = 270c, so (a + 18b)/c = 27. We conclude that n = 36, so the answer is 36 happy cows.

2. The exact number N of minutes to complete the puzzle is ∑_x=0⁹⁹⁹3(1000 - x)/(1000 + x). Since 3(1000 - x)/(1000 + x) is a non-negative monotonic decreasing function for 0≤x < 1000, we see that
   N - 3≤∫₀¹⁰⁰⁰ -3 + 6000/(1000 + x) dx≤N.
   Therefore N/60≈50(2 ln 2 - 1). Using ln 2≈.69, we conclude that it takes approximately 19 hours to complete the puzzle.

3. One can quickly check that f (2) = 2 and f (3) = 3, so it seems reasonable that f (n) = n, so let us try to prove this. Certainly if f (n) = n, then f (1) = 1, so we will prove the result by induction on n; we assume that the result is true for all integers ≤n. Then
   f (n + 1) = f (f (n)) + f (n + 1 - f (n)) = n + f (1) = n + 1
   as required and it follows that f (n) = n for n = 1, 2,....

4. Write P(x) = ax³ + bx² + cx + d, where a, b, c, d∈Z. Let us suppose by way of contradiction that a, b, c, d≥ - 1. From P(2) = 0, we get 8a + 4b + 2c + d = 0, in particular d is even and hence d≥ 0. Since 4b + 2c + d≥ - 7, we see that a≤ 0. Also a≠ 0 because P(x) has degree 3, so a = - 1. We now have 4b + 2c + d = 8 and b + c + d = 1 from P(1) = 0. Thus -2c - 3d = 4, so -2c = 4 + 3d≥4 and we conclude that c≤ - 2. The result follows.

5. (a)

   For small positive x, we have x/2 < sin x < x, so for positive integers n, we have 1/(2n) < sin(1/n) < 1/n. Since ∑_n=1^∞1/n^p is convergent if and only if p > 1, it follows from the basic comparison test that ∑_n=1^∞(sin 1/n)^p is convergent if and only if p > 1.

   (b)

   It is not difficult to show that any real number x, there exists an integer n > x such that | sin n| > 1/2. Thus whatever p is, lim_{n-> ∞}| sin n|^p≠ 0. Therefore ∑_n=1^∞| sin n|^p is divergent for all p.

6. (a)

   If y^* is a steady-state solution, then y^* = y^*(2 - y^*), so y^* = 0 or 1 = 2 - y^*. Therefore the steady-state solutions are y^* = 0 or 1.

   (b)

   Suppose 0 < y_n < 1. Then y_n+1/y_n = 2 - y_n > 1, so y_n+1 > y_n. Also y_n+1 = 1 - (1 - y_n)², so y_n+1 < 1. We deduce that y_n is a monotonic positive increasing function that is bounded above by 1, in particular y_n converges to some positive number ≤1. It follows that y_n converges to 1.

7. Let y∈[0, 1] be such that (g(y) + uf (y)) = u. Let us suppose we do have constants A and B such that F(x) = Ag(x)/(f (x) + B) is a continuous function on [0.1] with max_0≤x≤1F(x) = u. We will guess that the maximum occurs when x = y, so u = Ag(y)/(f (y) + B). Then A = B = -1 satisfies these equations, so F(x) = g(x)/(1 - f (x)).

   So let us prove that F(x) = g(x)/(1 - f (x)) has the required properties. Certainly F(x) is continuous because f (x) < 1 for all x∈[0, 1], and F(y) = u from above. Finally max_0≤x≤1(g(x) + uf (x)) = u, so g(x)≤u(1 - f (x)) for all x and we conclude that F(x)≤u. The result is proven.

8. Suppose we can disconnect F by removing only 8 points. Then the resulting framework will consist of two nonempty frameworks A, B such that there is no segment joining a point of A to a point of B. Let a be the number of points in A. Then there are 9 - a points in B, at most a(a - 1)/2 line segments joining the points of A, and at most (10 - a)(10 - a - 1)/2 line segments joining the points of B. It follows that the resulting framework has at most 45 - 10a + a². Since 10a - a² > 8 for 1≤a≤9, the result follows.