22nd VTRMC, 2000, Solutions

1. Let I = dq/(5 - 4cosq). Using the half angle formula cosq = 2cos2(q/2) - 1, we obtain

I = dq/(9 - 8cos2(q/2)) = (sec2(q/2) dq)/(9sec2(q/2) - 8) = (sec2(q/2) dq)/(9tan2(q/2) + 1).

Now make the substitution x = 3tan(q/2). Then dx = 3sec2(q/2) dq, consequently

3I = 2 dx/(1 + x2) = 2tan-1(3tan(a/2)).

Therefore I = (2/3)tan-1(3tan(a/2)). By using the facts that tan(p/6) = 1/ and tan(p/3) = , we see that when a = p/3,

I = (2/3)tan-1 = 2p/9.

2. Let J denote the Jordan canonical form of A. Then A and J will have the same trace, and the entries on the main diagonal of J will satisfy 4x4 + 1 = 0. This equation has roots ±1/2±i/2, so the trace of A will be a sum of such numbers. But the trace of A is real, hence the imaginary parts must cancel and we see that there must be an even number of terms in the sum. It follows that the trace of A is an integer.

3. Make the substitution y = x - t. Then the equation becomes x' = x2 - 2xt + 1. We will show that limt - > x'(t) exists and is 0, and then it will follow that limt - > y'(t) exists and is -1.

When t = 0, the initial condition tells us that x = 0, so x'(0) = 1 and we see that x(t) > 0 for small t. Suppose for some positive t we have x(t) 0. Then there is a least positive number T such that x(T) = 0. Then x'(T) = 1, which leads to a contradiction because x(t) > 0 for t < T. We deduce that x(t) > 0 for all t.

Now x' - 1 = x(x - 2t) and since x'(0) = 1, we see that x - 2t < 0 for small t. We deduce that for t sufficiently small, x(t) < t, consequently y(t) < 0 for small t. We now claim that y(t) < 0 for all positive t. If this is not the case, then there is a least positive number T such that y(T) = 0, and then we must have y'(S) = 0 for some S with 0 < S < T. But from y' = (y - t)(y + t) and (y + t) > 0, we would have to have y(S) = S, a contradiction because y(S) < 0. We deduce that x(t) < t for all positive t.

Now consider x' = x(x - 2t) + 1. Note that we cannot have x'(t) 0 for all t, because then x - > 0 as t - > which is clearly impossible, consequently x' takes on negative values. Next we have x'' = 2(xx' - tx' - x), so if x'(t) = 0, we see that x''(t) < 0. We deduce that if x'(T) < 0, then x'(t) < 0 for all t > T. Thus there is a positive number T such that x'(t) < 0 for all t > T. Now differentiate again to obtain x''' = 2(xx'' + x'x' - tx'' - 2x'). Then we see that if x''(t) = 0 and t > T, then x'''(t) > 0, consequently there is a positive number S > T such that either x''(t) < 0 for all t > S or x''(t) > 0 for all t > S. We deduce that x'(t) is monotonic increasing or decreasing for t > S and hence limt - > x'(t) exists (possibly infinite).

We now have x'(t) is monotonic and negative for t > S, yet x(t) > 0 for all t > S. We deduce that limt - > x'(t) = 0 and the result follows.

4. Set y = . Then
 l22 = (l - x)2 + y2 + 2(l - x)ycosq l12 = x2 + y2 - 2xycosq

Subtracting the second equation from the first we obtain

l22 - l12 = l2 - 2lx + 2lycosq

which yields

2ycosq = (l22 - l12 + 2lx - l2)/l.

From the second equation and the above, we obtain

l12 - x2 + x(l22 - l12 + 2lx - l2)/l = y2.

By differentiating the above with respect to x, we now get

2x + (l22 - l12 - l2)/l = 2y dy/dx

and we deduce that dy/dx = cosq. Therefore

l2 - l1 = y(l )- y(0) = cosq dx

as required.

5. Open out the cylinder so that it is an infinitely long rectangle with width 4. Then the brush paints out two ellipses (one on either side of the cylinder) which have radius in the direction of the axis of the cylinder, which we shall call the y-axis, and radius 2 in the perpendicular direction, which we shall call the x-axis. Then the equation of the ellipse is x2/4 + y2/3 = 1. By considering just one of the ellipses, we see that the area required is

4y dx = 8 dx.

By making the substitution x = 2sinq, this evaluates to 6 + 4p/3.

6. Let a = Sn = 1antn. Then

a2 = Sn = 1antn Sn = 1antn.

Consider the coefficient of tn on the right hand side of the above; it is

an - 1a1 + an - 2a2 + ... + a2an - 2 + a1an - 1

for n2 and 0 for n = 1. Using the given hypothesis, we see that this is an for all n2. We deduce that t + a2 = a. For the problem under consideration, we need to calculate a when t = 2/9, so we want to find a when a2 - a + 2/9 = 0. Since the roots of this equation are 1/3 and 2/3, we are nearly finished. However we ought to check that a2/3,.

Suppose this is not the case. Let bn = Sm = 1nam(2/9)m. If a = 2/3 or , then there exists a positive integer N such that bN < 1/3 and bN + 11/3. Then the same argument as above gives

2/9 + bN2 > bN + 1,

which is not possible, so the result follows.

7. There are three possible positions for the tiles, namely   ,    and , which we shall call A,B and C respectively. Consider An + 2. Then whatever the n + 1st tile is in the chain, we can complete it to a chain of length n + 2. Therefore An + 2 = An + 1 + x for some nonnegative integer x. However if the n + 1st tile is position A, then there is exactly one way to add a tile to get a chain of length n + 2 (namely add tile B), whereas if the tile is B or C, then there are exactly two ways to add a tile to get a chain of length n + 2 (namely add tiles A or C). Therefore x is the number of ways a chain ends of length n + 1 ends in B plus the number of ways a chain of length end in C, which is precisely An. Therefore An + 2 = An + 1 + An. This recurrence relation is valid for n1. Since A1 = 3 and A2 = 5, we get A3 = 8, A4 = 13 and A10 = 233.

Peter Linnell
2000-11-22