20th VTRMC, 1998, Solutions

1. Set r = x2 + y2. Then f (x, y) = ln(1 - r) - 1/(2r - (x + y)2), so for given r, we see that f is maximized when x + y = 0. Therefore we need to maximize ln(1 - r) - 1/(2r) where 0 < r < 1. The derivative of this function is

1/(r - 1) + 1/(2r2) = (2r2 + r - 1)/(2(r - 1)r2)

which is positive when r < 1/2, 0 when r=1/2, and negative when r > 1/2. It follows that the maximum value of this function occurs when r = 1/2 and we deduce that M = - 1 - ln 2.

2. We cut the cone along PV and then open it out flat, so in the picture below P and P1 are the same point. We want to find Q on VP1 so that the length of MQP is minimal. To do this we reflect in VP1 so P2 is the image of P under this reflection, and then MQP2 will be a straight line and the problem is to find the length of MP2.

Since the radius of the base of the cone is 1, we see that the length from P to P1 along the circular arc is 2p, hence the angle PVP1 is p/3 because VP = 6. We deduce that PVP2 = 2p/3, and since VM = 3 and VP2 = VP = 6, we conclude that MP2 = = 3.

3. We calculate the volume of the region which is in the first octant and above {(x, y, 0) | x>y}; this is 1/16 of the required volume. The volume is above R, where R is the region in the xy-plane and bounded by y = 0, y = x and y = , and below z = . This volume is

 dzdydx + dzdydx =  dydx +  dydx = x dx + (1 - x2) dx = [- (1 - x2)3/2/3]01/ + [x - x3/3]1/1 = 1/3 - 1/(6) + 2/3 + 1/(6) - 1/ = 1 - 1/.

Therefore the required volume is 16 - 8.

4. We shall prove that AB = BC. Using the cosine rule applied to the triangle ABC, we see that BC2 = AB2 + AC2 - 2(AB)(AC)cos 70. Therefore we need to prove AC = 2ABcos 70. By the sine rule applied to the triangle APC, we find that AP = 2sin 50, so we need to prove = 2(1 + 2sin 50)cos 70. However sin(50 + 70) + sin(50 - 70) = 2sin 50cos 70, sin 120 = /2 and sin(50 - 70) = - cos 70. The result follows.

5. Since S1/an is a convergent series of positive terms, we see that given M > 0, there are only finitely many positive integers n such that an < M. Also rearranging a series with positive terms does not affect its convergence, hence we may assume that {an} is a monotonic increasing sequence. Then b2n + 1>b2n>an/2, so the terms of the sequence {1/bn} are at most the corresponding terms of the sequence

2/a1, 2/a1, 2/a2, 2/a2, 2/a3, 2/a3,...

Since S1/an is convergent, so is the sum of the above sequence and the result now follows from the comparison test for positive term series.

6. We shall assume the theory of writing permutations as a product of disjoint cycles, though this is not necessary. Rule 1 corresponds to the permutation (1 2 3 4 5 6 7 8 9 10) and Rule 2 corresponds to the permutation (2 6)(3 4)(5 9)(7 8). Since

(2 6)(3 4)(5 9)(7 8)(1 2 3 4 5 6 7 8 9 10) = (1 6 8 5 2 4 9 10)

(where we have written mappings on the left) has order 8, we see the position of the cats repeats once every 16 jumps. Now 10p.m. occurs after 900 jumps, hence the cats are in the same position then as after 4 jumps and we conclude that the white cat is on post 8 at 10p.m.

Peter Linnell
2002-06-30