17th VTRMC, 1995, Solutions

1. Let A = {(x, y) | 0≤x≤2, 0≤y≤3, 3x≤2y} and B = {(x, y) | 0≤x≤2, 0≤y≤3, 3x≤2y}. Let I = ∫03021/(1 + max(3x, 2y))2 dxdy. Then max(3x, 2y) = 2y for (x, y)∈A and max(3x, 2y) = 3x for xB.

Therefore

 I = ∬A1/(1 + 2y)2 dA + ∬B1/(1 + 3x)2 dA = ∫03∫02y/31/(1 + 2y)2 dxdy + ∫02∫03x/21/(1 + 3x)2 dydx = ∫032y/(3(1 + 2y)2) dy + ∫023x/(2(1 + 3x)2) dx = ∫031/(3(1 + 2y)) - 1/(3(1 + 2y)2) dy + ∫021/(2(1 + 3x)) - 1/(2(1 + 3x)2) dx = [(ln(1 + 2y))/6 + 1/(6(1 + 2y))]03 + [(ln(1 + 3x))/6 + 1/(6(1 + 3x))]02 = (ln 7)/6 + 1/42 - 1/6 + (ln 7)/6 + 1/42 - 1/6 - (7 ln 7 - 6)/21.

2. Let

A = (
 4 -3 2 -1
)

We want to calculate powers of A, and to do this it is useful to find the Jordan canonical form of A. The characteristic polynomial of A is det(xI - A) = (x - 4)(x + 1) + 6 = x2 - 3x + 2 which has roots 1,2. Set

u = (
 1 1
)

An eigenvector corresponding to 1 is u and an eigenvector corresponding to 2 is

(
 3 2
)

Set
P = (
 1 3 1 2
)

and D = diag(1, 2) (diagonal matrix with 1,2 on the main diagonal). Then

P-1 = (
 -2 3 1 -1
)

and P-1AP = D. Let

v = (
 1 0
)

and let T denote transpose. Since A = PDP-1, we find that

 (θ100v)T = A100v + (A99 + A98 + ... + A + A0)u = PD100P-1v + P(D99 + D98 + ... + D + D0)P-1u

= P (
 1 0 0 2100
) P-1v + P (
 100 0 0 2100 - 1
) P-1u = (
 98 + 3·2100 98 + 2·2100
)

Thus θ100(1, 0) = (98 + 3·2100, 98 + 2·2100).

3. Let g(x) denote the power series in x

1 - (x + x2 + ... + xn) + (x + x2 + ... + xn)2 - ... + (- 1)n(x + x2 + ... + xn)n + ....

Then for 2≤rn, the coefficient of xr in f (x) is the same as the coefficient of xr in g(x). Since x + x2 + ... + xn = x(1 - xn)/(1 - x), we see that g(x) is a geometric series with ratio between successive terms - x(1 - xn)/(1 - x), hence its sum is

1/(1 + x(1 - xn)/(1 - x)) = (1 - x)/(1 - xn+1) = (1 - x)(1 + xn+1 + x2n+2 + ...).

clearly the coefficient of xr in the above is 0 for 2≤rn, which proves the result.

4. Write n] = p. Then p is the unique integer satisfying p < τn < p + 1 because p≠τn (otherwise τ = p/n, a rational number), that is p/τ < n < p/τ + 1. Since 1/τ = τ - 1, we see that pτ - p < n < pτ - p + 1 and we deduce that n + p - 1 < pτ < n + p. Therefore [pτ] = n + p - 1 and hence [τ[τn] + 1] = n + p. But τ2n = τn + n, consequently 2n] = p + n and the result follows.

5. Suppose xR and θ(x)≤ - 1. Fix yR with y < x. Then if n is a positive integer and x > p1 > ... > pn > y, we have for 1≤in

 θ(x) ≥θ(x)3 > θ(p1), θ(pi) > θ(pi)3 > θ(pi+1), θ(pn) > θ(pn)3 > θ(y),

and we deduce that θ(x)θ(p1)2n-2 > θ(y), for all n. this is not possible, so θ(x) > - 1 for all xR. The same argument works if 0≤θ(y) < θ(x)≤1.

6. We will concentrate on the bottom left hand corner of the square and determine the area A of that portion of the square that can be painted by the brush, and then multiply that by 4. We make the bottom of the square the x-axis and the left hand side of the square the y-axis. The equation of a line of length 4 from (a, 0) to the y-axis is x/a + y/√(16 - a2) = 1, that is y = (1 - x/a)√(16 - a2). For fixed x, we want to know the maximum value y can take by varying a. To do this, we differentiate y with respect to a and then set the resulting expression to 0. Thus we need to solve

(x/a2)√(16 - a2) - a(1 - x/a)/√(16 - a2) = 0.

On multiplying by √(16 - a2) and simplifying, we obtain 16x = a3 and hence dx/da = 3a2/16. Therefore

 A = ∫x=0x=4(1 - x/a)√(16 - a2) dx = ∫a=0a=4(1 - x/a)√(16 - a2) dx/da da = ∫a=0a=43a2(1 - a2/16)√(16 - a2)/16 da = ∫043a2(16 - a2)3/2/256 da.

This ia a standard integral which can be evaluated by a trigonometric substitution. Specifically we set a = 4 sin t, so da/dt = 4 cos t and we find that

 A = ∫0π/248 cos4t sin2t dt = ∫0π/26 sin22t(1 + cos 2t) dt = ∫0π/23(1 - cos 4t) dt = 3π/2.

We conclude that the total area that can be painted by the brush is 6πin2.

7. Note that if p is a prime, then f (p) = p. Thus f (100) = f (22·52) = 4 + 10 = 14, f (2·7) = 2 + 7 = 9, f (32) = 3·2 = 6. Therefore g(100) = 6. Next f (1010) = f (210·510) = 20 + 50 = 70, f (2·5·7) = 14, f (2·7) = 2 + 7 = 9, f (32) = 3·2 = 6. Therefore g(1010) = 6.

Since f (p) = p if p is prime, we see that g(p) = p also and thus primes cannot have property H. Note that if r, s are coprime, then g(rs)≤f (r)s. Suppose n has property H and let p be a prime such that p2 divides n, so n = pkr where k≥2 and r is prime to p. It is easy to check that if pk > 9, then pk > 2pk, that is f (pk) < pk/2, thus f (n) < n/2 and we see that n cannot have property H. Also if p, q are distinct odd primes and pq > 15, then f (pq) < pq/2 and so if n = pqr with r prime to pq, then we see again that n cannot have property H.

The only cases to be considered now are n = 9, 15, 45. By direct calculation, 9 has property H, but 15 and 45 do not. So the only positive odd integer larger than 1 that has property H is 9.

Peter Linnell 2007-06-05