21st VTRMC, 1999, Solutions

1. Since the value of f (x, y) is unchanged when we swap x with y,
   2$$\int_{0}^{1}$$$$\int_{0}^{x}$$f (x + y) dydx = $$\int_{0}^{1}$$$$\int_{0}^{1}$$f (x + y) dydx.
   Also
   $$\int_{0}^{1}$$f (x + y) dy = $$\int_{x}^{1+x}$$f (z) dz = $$\int_{0}^{1}$$f (z) dz
   because f (z) = f (1 + z) for all z. Since $\int_{0}^{1}$f (z) dz = 1999, we conclude that
   $$\int_{0}^{1}$$$$\int_{0}^{1}$$f (x + y) dydx = 1999/2.

2. For a = 1, b = 0 and x = 1, we have f (1)f (0) = f (1). Therefore f (0) = 1. By differentiation
   af'(ax)f (bx) + bf (ax)f'(bx) = f'(x)
   holds for all x, and for all a,b satisfying a² + b² = 1. Hence (a + b)f'(0) = f'(0) holds. By taking a = b = 1/$\sqrt{2}$, we wee that f'(0) = 0. Set c = f''(0). By Taylor's theorem, f (y) = 1 + cy²/2 + e(y²), where lim_{y - > 0}e(y²)/y² = 0. By taking a = b = 1/$\sqrt{2}$ again, we see that f (x/$\sqrt{2}$)² = f (x) for all x. By repetition, for every positive integer m,
   f (x) = (f (2^-m/2x))^2m.
   Now fix any x, and d > 0. There is a positive integer N such that for all m$\ge$N, 2^-m(| c| + d)x² < 1, and
   (1 + 2^{-m - 1}(c - d)x²)^2m$$\le$$f (x)$$\le$$(1 + 2^{-m - 1}(c + d)x²)^2m.
   Now let m - > $\infty$. We obtain
   e^{(c - d)x2/2}$$\le$$f (x)$$\le$$e^{(c + d)/x2}.
   Since d was arbitrary, f (x) = e^cx2/2. Using the condition f (1) = 2, we conclude that f (x) = 2^x2.

3. Note that any eigenvalue of A_n has absolute value at most M, because the sum of the absolute values of the entries in any row of A_n is at most M. We may assume that M > 1. By considering the characteristic polynomial, we see that the product of nonzero eigenvalues of A_n is a nonzero integer. Write d = e_n(d). Then we have Mⁿd^d > 1. This can be written as
   e_n(d)/n < ln(M)(ln(1/d)).
   The result follows.

4. The points inside the box which are distance at least 1 from all of the sides form a rectangular box with sides 1,2,3, which has volume 6. The volume of the original box is 60. The points outside the box which are distance at most 1 from one of the sides have volume

   3 X 4 + 3 X 4 + 3 X 5 + 3 X 5 + 4 X 5 + 4 X 5 = 94

   plus the points at the corners, which form eight 1/8th spheres of radius 1, plus the points which form 12 1/4th cylinders whose heights are 3,4,5. It follows that the volume required is

   60 - 6 + 94 + 4p/3 + 12p = 148 + 40p/3.

5. By differentiating f (f (x)) = x, we obtain f'(f (x))f'(x) = 1. Since f is continuous, f'(x) can never cross zero. This means that either f'(x) > 0 for all x of f'(x) < 0 for all x. If f'(x) > 0 for all x, then x > y implies f (x) > f (y), and we get a contradiction by considering f (f (a)) = a. We deduce that f is monotonically decreasing, and since f is bounded below by 0, we see that lim_{x - > $\infty$}f (x) exists and is some nonnegative number, which we shall call L. If L > 0, then we obtain a contradiction by considering f (f (L/2)) = L/2. The result follows. Remark The condition f (a)$\ne$a is required, otherwise f (x) = x would be a solution.

6. (i)

   Obviously n > 4. Next, n$\ne$5 because 4 divides 3 + 5. Also n$\ne$6 because 3 divides 6 and n$\ne$7 because 7 divides 3 + 4. Finally n$\ne$8 since 4 divides 8, and n$\ne$9 since 3 divides 9. On the other hand n = 10 because 3 does not divide 4, 10 and 14. Furthermore 4 does not divide 3, 10 and 13, and 10 does not divide 3, 4 and 7.

   (ii)

   Suppose {3, 4, 10, m} is contained in a set which has property ND. Then 3 should not divide m, so m is not of the form 3k. Also 3 should not divide 10 + m, so m is not of the form 3k + 2. Furthermore 33 should not divide m + 4 + 10, so m is not of the form 3k + 1. Here k denotes some integer. If s has property ND and contains 3,4 10 and m, then m cannot be of the form 3k, 3k + 1, 3k + 2. This is impossible and the statement is proven.