30th VTRMC, 2008, Solutions

1. Write f (x) = xy3 + yz3 + zx3 - x3y - y3z - z3x. First we look for local maxima, so we need to solve f /∂x = ∂f /∂y = ∂f /∂z = 0. Now f /∂x = y3 +3x2z - z3 -3x2y. If y = z, then f (x, y, z) = 0 and this is not a maximum. Thus we may divide by y - z and then f /∂x = 0 yields y2 + yz + z2 = 3x2. Similarly x2 + xz + z2 = 3y2 and x2 + xy + y2 = 3z2. Adding these three equations, we obtain (x - y)2 + (y - z)2 + (z - x)2 = 0, which yields x = y = z. This does not give a maximum, because f = 0 in this case, and we conclude that the maximum of f must occur on the boundary of the region, so at least one of x, y, z is 0 or 1.

Let's look at f on the side x = 0. Here f = yz3 - y3z and 0≤y≤1, 0≤z≤1. To find local maxima, we solve f /∂y = ∂f /∂z = 0. This yields y = z = 0 and f = 0, which is not a maximum, so the maximum occurs on the edges of the region considered. If y or z = 0, we get f = 0 which is not a maximum. If y = 1, then f = z3 - z≤ 0, which won't give a maximum. Finally if z = 1, then f = y - y3. Since df /dy = 1 - 3y2, we see that f has a maximum at y = 1/√3. This gives that the maximum value of f on x = 0 is 1/√3 -1/√33 = 2√3/9.

Similarly if y or z = 0, the maximum value of f is 2√3/9. Now let's look at f on the side x = 1. Here f = y3 + yz3 + z - y - y3z - z3. Again we first look for local maxima: f /∂y = 3y2 + z3 -1 - 3y2z. Then f /∂y = 0 yields either z = 1 or 3y2 = z2 + z + 1. If z = 1, then f = 0 which is not a maximum, so 3y2 = z2 + z + 1. Similarly 3z2 = y2 + y + 1. Adding these two equations, we find that y2 - y/2 + z2 - z/2 = 1. Thus (y - 1/2)2 + (z - 1/2)2 = 3/2. This has no solution in the region considered 0≤y≤1, 0≤z≤1. Thus f must have a maximum on one of the edges. If y or z is 0, then we are back in the previous case. On the other hand if y or z is 1, then f = 0, which is not a maximum.

We conclude that the maximum value of f on 0≤x≤1, 0≤y≤1, 0≤z≤1 is 2√3/9.

2. For each positive integer n, let f (n) denote the number of sequences of 1's and 3's that sum to n. Then f (n + 3) = f (n + 2) + f (n), and we have f (1) = 1, f (2) = 1, and f (3) = 2. Thus f (4) = f (3) + f (1) = 3, f (5) = f (4) + f (2) = 4, f (6) = 6, ..., f (15) = 189, f (16) = 277. Thus the number of sequences required is 277.

3. Let R denote the specified region, i.e. {(x, y) | x4 + y4x2 - x2y2 + y2}. Then R can be described as the region inside the curve x4 + x2y2 + y4 = x2 + y2 ( (x, y)≠(0, 0)). This can be rewritten as

(x2 + y2 - xy)(x2 + y2 + xy) = x2 + y2.

Now change to polar coordinates: write x = r cosθ, y = r sinθ; then the equation becomes (r2 - r2cosθsinθ)(r2 + r2cosθsinθ) = r2. Since r≠ 0 and 2 cosθsinθ = sin 2θ, we now have r2(1 - (1/4)sin22θ) = 1. Therefore the area A of R is

 ∬Rr drdθ = ∫02π∫0(1-(1/4)sin22θ)-1/2r drdθ = ∫02πdθ/(2(1 - (1/4)sin22θ)) = ∫0π/416 dθ/(3 + cos22θ) = ∫0π/416 sec22θ dθ/(4 + 3 tan22θ).

Now make the substitution 2z = √3tan 2θ, so dz = √3sec22θ dθ and we obtain

A = (4/√3)∫0dz/(1 + z2) = 2π/√3.

4. Ceva's theorem applied to the triangle ABC shows that (AP/PB)(BM/MC)(CN/NA) = 1. Since BM = MC, we see that AP/PB = AN/NC and we deduce that PN is parallel to BC. Therefore NPX = ∠PCB = ∠NAX and we conclude that APXN is a cyclic quadrilateral. Since that opposite angles of a cyclic quadrilateral sum to 180o, we see that APX + ∠XNA = 180o, and the result follows.

5. Let T = {an | nN} and for t a positive number, let At = {nN | ant}. Since an = 1 and an≥ 0 for all n, we that if δ > 0, then there are only finitely many numbers in T greater than δ, and also At is finite. Thus we may label the nonzero elements of T as t1, t2, t3,..., where t1 > t2 > t3 > ... > 0. We shall use the notation XΔY to indicate the symmetric difference {X\YY\X} of two subsets X, Y.

Consider the sum

i≥1(ti - ti+1)| AtiΔπ-1Ati|.

Note that nAt-1At if and only if ant > aπn, and n∈π-1At\At if and only if an < taπn. Write an = tp and aπn = tq. We have three cases to examine:
1. an = aπn. Then n does not appear in the above sum.

2. an > aπn. Then p < q and n is in Atr-1Atr whenever tptr > tq, that is q > rp and we get a contribution (tp - tp+1) + (tp+1 - tp+2) + ... + (tq-1 - tq) = tp - tq = an - aπn = | an - aπn|.

3. an < aπn. Then p > q and n is in π-1Atr\Atr whenever tqtr > tp, that is p > rq and we get a contribution (tq - tq+1) + (tq+1 - tq+2) + ... + (tp-1 - tp) = tq - tp = aπn - an = | an - aπn|.
We conclude that

n=1| an - aπn| = ∑i≥1(ti - ti+1)| AtiΔπAti|,

because | AtiΔπ-1Ati| = | AtiΔπAti|. Similarly

n=1| an - aρn| = ∑i≥1(ti - ti+1)| AtiΔρAti|,

and we deduce that

n=1(| an - aπn| + ∑n=1| an - aρn|) = ∑i≥1(ti - ti+1)(| AtiΔπAti| + | AtiΔρAti|).

Therefore i≥1(ti - ti+1)(| AtiΔπAti| + | AtiΔρAti|) < ε. We also have i≥1(ti - ti+1)| Ati| = 1. Therefore for some i, we must have | AtiΔπAti| + | AtiΔρAti| < ε| Ati| and the result follows.

6. Multiply a4 -3a2 + 1 by b and subtract (a3 - 3a)(ab - 1) to obtain a3 - 3a + b. Now multiply by b and subtract a2(ab - 1) to obtain a2 -3ab + b2. Thus we want to know when ab - 1 divides (a - b)2 - 1, where a, b are positive integers. We cannot have a = b, because a2 - 1 does not divide -1. We now assume that a > b.

Suppose ab - 1 does divide (a - b)2 - 1 where a, b are positive integers. Write (a - b)2 - 1 = k(ab - 1), where k is an integer. Since (a - b)2 -1≥ 0, we see that k is nonnegative. If k = 0, then we have (a - b)2 = 1, so a - b = ±1. In this case, ab - 1 does divide a4 -3a2 + 1, because a4 -3a2 +1 = (a2 + a - 1)(a2 - a - 1). We now assume that k≥1.

Now fix k and choose a, b with b as small as possible. Then we have a2 + a(- 2b - kb) + b2 + k - 1 = 0. Consider the quadratic equation x2 + x(- 2b - kb) + b2 + k - 1 = 0. This has an integer root x = a. Let v be its other root. Since the sum of the roots is 2b + kb, we see that v is also an integer. Also av = b2 + k - 1. Since b, k≥1, we see that v is also positive. We want to show that v < b; if this was not the case, then we would have b2 + k - 1≥ab, that is kab - b2 + 1. We now obtain

(a - b)2 -1≥(ab - b2 + 1)(ab - 1).

This simplifies to a2 - ab≥(ab - b2 + 1)ab, that is a - b≥(ab - b2 + 1)b and we obtain (a - b)(b2 -1) + b≤ 0, which is not the case. Thus v < b and we have v2 + v(- 2b - k) + b2 + k - 1 = 0. Set u = b. Then we have (u - v)2 - 1 = k(uv - 1), where u, v are positive and v < b. By minimality of b, we conclude that there are no a, b such that (a - b)2 - 1 = k(ab - 1).

Putting this altogether, the positive integers required are all a, b such that b = a±1.

7. Note that for fixed x > 1, the sequence 1/fn(x) is decreasing with respect to n and positive, so the given limit exists which means that g is well-defined. Next we show that g(e1/e)≥1/e, equivalently limn--> ∞fn(e1/e)≤e. To do this, we show by induction that fn(e1/e)≤e for all positive integers n. Certainly f1(e1/e) = e1/ee. Now if fn(e1/e)≤e, then

fn+1(e1/e) = (e1/e)fn(e1/e)≤(e1/e)e = e,

so the induction step passes and we have proven that g(e1/e)≥1/e.

We now prove that g(x) = 0 for all x > e1/e; this will show that g is discontinuous at x = e1/e. We need to prove that limn--> ∞fn(x) = ∞. If this is not the case, then we may write limn--> ∞fn(x) = y where y is a positive number > 1. We now have

y = limn--> ∞fn(x) = limn--> ∞fn+1(x) = xlimn--> ∞fn(x) = xy.

Therefore ln y = y ln x and x = y1/y. Since (dx/dy)/x = (1 - ln y)/y2, we see by considering the graph of y1/y that it reaches its maximum when y = e, and we deduce that xe1/e. This is a contradiction and we conclude that limn--> ∞fn(x) = 0. Thus we have shown that g(x) is discontinuous at x = e1/e.

Peter Linnell 2008-11-06