12th VTRMC, 1990, Solutions
- 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
|10(a + 9b)
|24(a + 18b)
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.
- 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.
N/60≈50(2 ln 2 - 1). Using
we conclude that it takes approximately 19 hours to complete the
- 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
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,....
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
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
c≤ - 2. The result follows.
- 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/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.
- 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.
- 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.
- 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
≤1. It follows that yn converges to 1.
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
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
F(x)≤u. The result is proven.
- 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