23rd VTRMC, 2001, Solutions

1. 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.

2. Let the circle with radius 1 have center P, the circle with radius 2 have center Q, and let R be the center of the third circle, as shown below. Let = a, = b, and let the radius of the third circle be r. By Pythagoras on the triangles PAR and QRB, we obtain

(1 - r)2 + a2 = (1 + r)2,    (2 - r)2 + b2 = (2 + r)2.

Therefore a2 = 4r and b2 = 8r. Also (a + b)2 + 1 = 9, so 2 + 2 = and we deduce that r = 6 - 4.

3. Let m, n be a positive integers where m < n. For each m X m square in an n X n grid, replace it with the (m - 1) X (m - 1) square obtained by deleting the first row and column; this means that the 1 X 1 squares become nothing. Then these new squares (don't include the squares which are nothing) are in a one-to-one correspondence with the squares of the (n - 1) X (n - 1) grid obtained by deleting the first row and column of the n X n square. Therefore Sn - 1 is the number of squares in an n X n grid which have size at least 2 X 2. Since there are n2 1 X 1 squares in an n X n grid, we deduce that Sn = Sn - 1 + n2.

Thus

S8 = 12 + 22 + ... + 82 = 204.

4. If p < q < (p + 1)2, then p divides q if and only if q = p2, p2 + p or p2 + 2p. Therefore if ak = p2, then ak + 3 = (p + 1)2. Since a1 = 12, we see that

a10000 = (1 + 9999/3)2 = 33342 = 11115556.

5. Let an = nnxn/n!. First we use the ratio test:

an + 1/an = ((n + 1)n + 1xn + 1n!)/(nnxn(n + 1)!) = (1 + 1/n)nx.

Since limn - > (1 + 1/n)n = e, we see that the interval of convergence is of the form { - 1/e, 1/e}, where we need to decide whether the interval is open or closed at its two endpoints. By considering dx/x, we see that

1/(n + 1) < ln(1 + 1/n) < (1/n + 1/(n + 10))/2 < 3/(3n + 1)

because 1/n is a concave function, consequently (1 + 1/n)n + 1/3 < e < (1 + 1/n)n + 1. Therefore when | x| = 1/e, we see that | an| is a decreasing sequence. Furthermore by induction on n and the left hand side of the last inequality, we see that | an| < 1/. Thus when x = - 1/e, we see that limn - > an = 0 and it follows that the given series is convergent, by the alternating series test. On the other hand when x = 1/e, the series is Snn/(enn!). By induction on n and the above inequality, we see that nn/(en/n!) > 1/(en) for all n > 1. Since S1/n is divergent, we deduce that the given series is divergent when x = 1/e, consequently the interval of convergence is [- 1/e, 1/e).

6. Let

A = (
 3 1 1 3
).

If we can find a matrix

B = (
 a b c d
)

such that B2 = A or even 4A and set f (x) = (ax + b)/(cx + d ), then f (f (x)) = (3x + 1)/(x + 3). So we want to find a square root of A. The eigenvalues of A are 2 and 4, and the corresponding eigenvectors are (1,-1) and (1,1) respectively. Thus if

X = (
 1 -1 1 1
),

then

XAX-1 = (
 2 0 0 4
).

Set

C = (
 0 0 2
).

Then C2 = XAX-1, so if we let B = (X-1CX)2, then B2 = A. Since

2B = (
 2 + 2- 2- 2+
),

we may define (multiply top and bottom by 1 + /2)

f (x) = ((3 + 2)x + 1)/(x + 3 + 2).

Finally we should remark that f still maps R+ to R+. Of course there are many other solutions and answers.

7. Choose x A and y B so that f (xy) is as large as possible. Suppose we can write xy in another way as ab with a A and b B (so a =/= x). Set g = ax-1 and note that I =/= g G. Therefore either f (gxy) > f (xy) or f (g-1xy) > f (xy). We deduce that either f (ay) > f (xy) or f (xb) > f (xy), a contradiction and the result follows.

Peter Linnell
2001-11-24