21st VTRMC, 1999, Solutions

1. Since the value of f (x, y) is unchanged when we swap x with y,

2f (x + y) dydx = f (x + y) dydx.

Also

f (x + y) dy = f (z) dz = f (z) dz

because f (z) = f (1 + z) for all z. Since f (z) dz = 1999, we conclude that

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/, we wee that f'(0) = 0. Set c = f''(0). By Taylor's theorem, f (y) = 1 + cy2/2 + e(y2), where limy - > 0e(y2)/y2 = 0. By taking a = b = 1/ again, we see that f (x/)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 mN, 2-m(| c| + d)x2 < 1, and

(1 + 2-m - 1(c - d)x2)2mf (x)(1 + 2-m - 1(c + d)x2)2m.

Now let m - > . We obtain

e(c - d)x2/2f (x)e(c + d)/x2.

Since d was arbitrary, f (x) = ecx2/2. Using the condition f (1) = 2, we conclude that f (x) = 2x2.

3. Note that any eigenvalue of An has absolute value at most M, because the sum of the absolute values of the entries in any row of An is at most M. We may assume that M > 1. By considering the characteristic polynomial, we see that the product of nonzero eigenvalues of An is a nonzero integer. Write d = en(d). Then we have Mndd > 1. This can be written as

en(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 limx - > 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)a is required, otherwise f (x) = x would be a solution.

6. (i)
Obviously n > 4. Next, n5 because 4 divides 3 + 5. Also n6 because 3 divides 6 and n7 because 7 divides 3 + 4. Finally n8 since 4 divides 8, and n9 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.

Peter Linnell
2000-09-17