13th VTRMC, 1991, Solutions

  1. Let P denote the center of the circle. Then ACP = ∠ABP = π/2 and BAP = α/2. Therefore BP = a tan(α/2) and we see that ABPC has area a2tan(α/2). Since BPC = π - α, we find that the area of the sector BPC is (π/2 - α/2)a2tan2(α/2). Therefore the area of the curvilinear triangle is

    a2(1 + α/2 - π/2)tan2(α/2).

  2. If we differentiate both sides with respect to x, we obtain 3f (x)2f'(x) = f (x)2. Therefore f (x) = 0 or f'(x) = 1/3. In the latter case, f (x) = x/3 + C where C is a constant. However f (0)3 = 0 and we see that C = 0. We conclude that f (x) = 0 and f (x) = x/3 are the functions required.

  3. We are given that α satisfies (1 + x)xn+1 = 1, and we want to show that α satisfies (1 + x)xn+2 = x. This is clear, by multiplying the first equation by x.

  4. Set f (x) = xn/(x + 1)n+1, the left hand side of the inequality. Then

    f'(x) = xn-1(n - x)/(x + 1)n+2.

    This shows, for x > 0, that f (x) has its maximum value when x = n and we deduce that f (x)≤nn/(n + 1)n+1 for all x > 0.

  5. Clearly there exists c such that f (x) - c has a root of multiplicity 1, e.g. x = c = 0. Suppose f (x) - c has a multiple root r. Then r will also be a root of (f (x) - c)' = 5x4 -15x2 + 4. Also if r is a triple root of f (x) - c, then it will be a double root of this polynomial. But the roots of 5x4 -15x2 + 4 are ±((15±√145)/10)1/2, and we conclude that f (x) - c can have double roots, but neither triple nor quadruple roots.

  6. Expand (1 - 1)n by the binomial theorem and divide by n!. We obtain for n > 0

    1/(0!n!) - 1/(1!(n - 1)!) + 1/(2!(n - 2)!) - ... + (- 1)n/(n!0!) = 0.

    Clearly the result is true for n = 0. We can now proceed by induction; we assume that the result is true for positive integers < n and plug into the above formula. We find that

    a0/n! + a1/(n - 1)! + a2/(n - 2)! + ... + an-1/1! + (- 1)n/(n!0!) = 0

    and the result follows.

  7. Suppose 2/3 < an, bn < 7/6. Then 2/3 < an+2, bn+2 < 7/6. Now if c = 1.26, then 2/3 < a3, b3 < 1, so if xn = a2n+1 or b2n+1, then xn+1 = xn/4 + 1/2 for all n≥1. This has the general solution of the form xn = C(1/4)n + 2/3. We deduce that as n -> ∞, a2n+1, b2n+1 decrease monotonically with limit 2/3, and a2n, b2n decrease monotonically with limit 4/3.

    On the other hand suppose an > 3/2 and bn < 1/2. Then an+1 > 3/2 and bn+1 < 1/2. Now if c = 1.24, then a3 > 3/2 and b3 < 1/2. We deduce that an+1 = an/2 + 1 and bn+1 = bn/2. This has general solution an = C(1/2)n + 2, bn = D(1/2)n. We conclude that as n -> ∞, an increases monotonically to 2 and bn decreases monotonically to 0.

  8. Let A be a base campsite and let h be a hike starting and finishing at A which covers each segment exactly once. Let B be the first campsite which h visits twice (i.e. B is the earliest campsite that h reaches a second time). This could be A after all segments have been covered, and then we are finished (just choose C = {h}). Otherwise let h1 be the hike which is the part of h which starts with the first visit to B and ends with the second visit to B (so B is the base campsite for h1). Let h' be the hike obtained from h by omitting h1 (so h' doesn't visit all segments). Now do the same with h'; let C be the first campsite on h' (starting from A) that is visited twice and let h2 be the hike which is the part of h' that starts with the first visit to C and ends with the second visit to C. Then C can be chosen to be the collection of hikes {h1, h2,...} to do what is required.

Peter Linnell 2009-06-24