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) = x^n-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)! + ... + a_n-1/1! + (- 1)ⁿ/(n!0!) = 0
   and the result follows.

7. Suppose 2/3 < a_n, b_n < 7/6. Then 2/3 < a_n+2, b_n+2 < 7/6. Now if c = 1.26, then 2/3 < a₃, b₃ < 1, so if x_n = a_2n+1 or b_2n+1, then x_n+1 = x_n/4 + 1/2 for all n≥1. This has the general solution of the form x_n = C(1/4)ⁿ + 2/3. We deduce that as n -> ∞, a_2n+1, b_2n+1 decrease monotonically with limit 2/3, and a_2n, b_2n decrease monotonically with limit 4/3.

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