19th VTRMC, 1997, Solutions

1. We change to polar coordinates. Thus x = rcosq, y = rsinq, and dA = r drdq. The circle (x - 1)2 + y2 = 1 becomes r2 - 2rcosq = 0, which simplifies to r = 2cosq. Also as one moves from (2, 0) to (0, 0) on the semicircle C (see diagram below), q moves from 0 to p/2. Therefore

 x3/(x2 + y2) dA = (r3cos3q/r2)rdrdq = r2cos3q drdq = (8/3)cos6q dq = (1/3)(1 + cos 2q)3 dq = (1/3)(1 + 3cos22q) dq = 5p/12.

2. Since r1r2 = 2, the roots r1, r2 will satisfy a quadratic equation of the form x2 + px + 2 = 0, where p C. Therefore we may factor

x4 - x3 + ax2 - 8x - 8 = (x2 + px + 2)(x2 + qx - 4)

where q C. Equating the coefficients of x3 and x, we obtain p + q = - 1 and 2q - 4p = - 8. Therefore p = 1 and q = - 2. We conclude that a = - 4 and r1, r2 are the roots of x2 + x + 2, so r1 and r2 are (- 1±i)/2.

3. The number of different combinations of possible flavors is the same as the coefficient of x100 in

(1 + x + x2 + ...)4

This is the coefficient of x100 in (1 - x)-4, that is 103!/(3!100!) = 176851.

4. We can represent the possible itineraries with a matrix. Thus we let

A = (
 1 3 1 1
)

and let aij indicate the (i, j)th entry of A. Then for a one day period, a11 is the number of itineraries from New York to New York, a12 is the number of itineraries from New York to Los Angeles, a21 is the number of itineraries from Los Angeles to New York, and a22 is the number of itineraries from Los Angeles to Los Angeles. The number of itineraries for an n day period will be given by An; in particular the (1, 1) entry of A100 will be the number of itineraries starting and finishing at New York for a 100 day period.

To calculate A100, we diagonalize it. Then the eigenvalues of A are and the corresponding eigenvectors (vectors u satisfying Au = lu where l = 1±) are , 1). Therefore if

P = (
 - 1 1
)

then

P-1AP = (
 1 + 0 0 1 -
)

Thus

A100 = P (
 (1 + )100 0 0 (1 - )100
) P-1

We conclude that the (1, 1) entry of A100 is ((1 + )100 + (1 - )100)/2, which is the number of itineraries required.

5. For each city x in S, let Gx S denote all the cities which you can travel from x (this includes x). Clearly Gx is well served and | Gx|>3 (where | Gx| is the number of cities in Gx). Choose x so that | Gx| is minimal. We need to show that if y, z Gx, then one can travel from y to z stopping only at cities in Gx; clearly we need only prove this in the case z = x. So suppose by way of contradiction y Gx and we cannot travel from y to x stopping only at cities in Gx. Since Gy Gx and xGy, we have | Gy| < | Gx|, contradicting the minimality of | Gx| and the result follows.

6. Let O denote the center of the circle with radius 2cm., let C denote the center of the disk with radius 1cm., and let H denote the hole in the center of the disk. Choose axes so that the origin is at O, and then let the initial position have C and H on the positive x-axis with H furthest from O. The diagram below is in general position (i.e. after the disk has been moved round the inside of the circle). Let P be the point of contact of the circle and the disk, (so OCP will be a straight line), let Q be where CH meets the circumference of the disk (on the x-axis, though we need to prove that), and let R be where the circle meets the positive x-axis. Since the arc lengths PQ and PR are equal and the circle has twice the radius of that of the disk, we see that PCQ = 2POR and it follows that Q does indeed lie on the x-axis.

Let (a, b) be the coordinates of C. Then a2 + b2 = 1 because the disk has radius 1, and the coordinates of H are (3a/2, b/2). It follows that the curve H traces out is the ellipse 4x2 + 36y2 = 9. We now use the formula that the area of an ellipse with axes of length 2p and 2q is ppq. Here p = 3/2, q = 1/2, and we deduce that the area enclosed is 3p/4.

7. Let x = {x0, x1,..., xn} J. Then
 Tx = LA({x0, x0 + x1, x0 + x1 + x2,...}) = L({1 + x0, 1 + x0 + x1, 1 + x0 + x1 + x2,...}) = {1, 1 + x0, 1 + x0 + x1, 1 + x0 + x1 + x2,...}.

Therefore T2y = T({1, 2, 3,...}) = {1, 1 + 1, 1 + 1 + 2, 1 + 1 + 2 + 3,...}. We deduce that T2y = {1, 2, 4, 7, 11, 16, 22, 29,...} and in general (T2y)n = n(n + 1)/2 + 1.

Suppose z = limi - > Tiy exists. Then Tz = z, so 1 = z0, 1 + z0 = z1, 1 + z0 + z1 = z2, 1 + z0 + z1 + z2 = z3, etc. We now see that zn = 2n. To verify this, we use induction on n, the case n = 0 already having been established. Assume true for n; then

zn + 1 = 1 + z0 + z1 + ... + zn = 1 + 1 + 2 + ... + 2n = 2n + 1,

so the induction step is complete and the result is proven.

Peter Linnell
2002-06-30