9th VTRMC, 1987, Solutions

1. The length of $\overline{{P_0P_1}}$ is √2/2. For n≥1, the horizontal distance from P_n to P_n+1 is 2^-n-1, while the vertical distance is 3·2^-n-1. Therefore the length of $\overline{{P_n,P_{n+1}}}$ 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 a_r is odd, and then a_r - r is even. It follows that (a₁ -1)(a₂ -2)...(a_n - 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 lim_{x-> 0}p(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 n_i = 31 for all i≥2.

   (b)

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

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

7. Let D = {d_ij} be the diagonal matrix with d_nn = t, d_ii = 1 for i≠n, and d_ij = 0 if i≠j. 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.