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 + xdxN.

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, dZ. 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=11/np 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, limn-> ∞| 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 < yn < 1. Then yn+1/yn = 2 - yn > 1, so yn+1 > yn. Also yn+1 = 1 - (1 - yn)2, so yn+1 < 1. We deduce that yn is a monotonic positive increasing function that is bounded above by 1, in particular yn converges to some positive number ≤1. It follows that yn 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 max0≤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 max0≤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.

Peter Linnell 2010-06-02