9th VTRMC, 1987, Solutions

1. The length of is √2/2. For n≥1, the horizontal distance from Pn to Pn+1 is 2-n-1, while the vertical distance is 3·2-n-1. Therefore the length of is 2-n-1√10 for n≥1, and it follows that the distance of the path is (√2 + √10)/2.

2. We want to solve in positive integers a2 = x2 + d2, b2 = d2 + y2, a2 + b2 = (x + y)2. These equations yield xy = d2, so we want to find positive integers x, y such that x(x + y) and y(x + y) are perfect squares. One way to do this is to choose positive integers x, y such that x, y, x + y are perfect squares, so one possibility is x = 9 and y = 16. Thus we could have a = 15, b = 20 and c = 25.

3. If n is odd, then there are precisely (n + 1)/2 odd integers in {1, 2,...}. Since (n + 1)/2 > n/2, there exists an odd integer r such that ar is odd, and then ar - r is even. It follows that (a1 -1)(a2 -2)...(an - n) is even.

4. (a)
Since | a0| = | p(0)|≤| 0|, we see that a0 = 0.

(b)
We have | p(x)/x|≤1 for all x≠ 0 and limx-> 0p(x)/x = a1. Therefore | a1|≤1.

5. (a)
Set n1 = 231, which has binary representation 1 followed by 31 0's. Since 31 has binary representation 11111, we see that ni = 31 for all i≥2.

(b)
It is clear that {ni} is monotonic increasing for i≥2, so we need to prove that {ni} is bounded. Suppose 2kni < 2k+1 where i≥2. If ℓ is the number of zeros in the binary representation of ni, then ni < 2k+1 - ℓ and we see that ni+1 < 2k+1. We deduce that ni < 2k+1 for all i≥2 and the result follows.

6. Of course, {an} is the Fibonacci sequence (so in particular an > 0 for all n), and it is obvious that x = - 1 is a root of pn(x) for all n (because an+2 - an+1 - an = 0). Since the roots of pn(x) are real and their product is - an/an+2, we see that limn-> ∞rn = - 1. Finally sn = limn-> ∞an/an+2. If f = limn-> ∞an+1/an, then f2 = f + 1, so f = (1 + √5)/2 because f > 0. Thus f2 = (3 + √5)/2 and we deduce that limn-> ∞sn = (3 - √5)/2.

7. Let D = {dij} be the diagonal matrix with dnn = t, dii = 1 for in, and dij = 0 if ij. Then A(t) = DA and B(t) = DB. Therefore

A(t)-1B(t) = (DA)-1DB = A-1D-1DB = A-1B

as required.

8. (a)
x'(t) = u(t) - x(t), y'(t) = v(t) - y(t), u'(t) = - x(t) - u(t), v'(t) = y(t) - v(t).

(b)
Set

Y = (
 u(t) x(t)
)

and

A = (
 -1 -1 1 -1
)

Then we want to solve Y' = AY. The eigenvalues of A are -1±i, and the corresponding eigenvectors are

(
 ±i 1
)

Therefore u(t) = e-t(- A sin t + B cos t) and x(t) = e-t(A cos t + B sin t), where A, B are constants to be determined. However when t = 0, we have u(t) = x(t) = 10, so A = B = 10. Therefore u(t) = 10e-t(cos t - sin t) and x(t) = 10e-t(cos t + sin t). Finally the cat will hit the mirror when u(t) = 0, that is when t = π/4.

Peter Linnell 2011-06-28