Irish Intervarsity Mathematics Competition 1993

University College Galway

Answer all ten questions

1. Evaluate ∑_{j=2}^{∞} (∑_{i=2}^{∞} \frac{1}{i^{j}} ) .

Answer: Since the terms are all positive, the order of summation does not matter:

∑_{j=2}^{∞} ∑_{i=2}^{∞} \frac{1}{i^{j}} = ∑_{i=2}^{∞} ∑_{j=2}^{∞} \frac{1}{i^{j}}

But

∑_{j=2}^{∞} \frac{1}{i^{j}} = \frac{1}{i^{2}} + \frac{1}{i^{3}} + ⋯
= \frac{1}{i^{2}} (1 − \frac{1}{i})^{−1}
= \frac{1}{i(i − 1)}.

Thus

∑_{i=2}^{∞} ∑_{j=2}^{∞} \frac{1}{i^{j}} = ∑_{i=2}^{∞} \frac{1}{i(i − 1)}

= ∑_{i=2}^{∞} (\frac{1}{i − 1} − \frac{1}{i})
= 1.

2. abcd is a square and e is a point inside the square such that |ae| = |be| and eb̂a = 15^{∘}. Show that dec is an equilateral triangle.

[figure: a; b; d; c; e]

Answer: Let the square have side 1. If dec is an equilateral triangle then

|de| = |dc| = |ad|.

Thus ade is an isosceles triangle. This gives 2 ways of computing |ae|. Firstly, from the isosceles triangle aeb,

|ae| = \frac{1}{2 cos 15^{∘}} .

Secondly, from the isosceles triangle ade, since dâe = 75^{∘},

|ae| = 2 cos 75^{∘} = 2 sin 15^{∘}.

Hence

4 sin 15^{∘} cos 15^{∘} = 1.

This is true, since 2 sin 15^{∘} cos 15^{∘} = sin 30^{∘} = \frac{1}{2}.

Conversely, working backwards we deduce that the triangle ade is isosceles, since the perpendicular from d to ae bisects ae; and so |de| = |ab| = |dc|. Similarly |ce| = |dc|. Thus the triangle dec is equilateral.

3. If x, y and n are all natural numbers and t = x^{4n} + y^{4n} + x^{2n}y^{2n} is a prime number, find all possible values of t.

Answer: Clearly x ≠ 0, y ≠ 0. We have

t = (x^{2n} + y^{2n})^{2} − (x^{n}y^{n})^{2}
= (x^{2n} + y^{2n} − x^{n}y^{n}) (x^{2n} + y^{2n} + x^{n}y^{n})

If t is prime, the first of these factors must be 1, giving

(x^{n} − y^{n})^{2} + x^{n}y^{n} = 1,

which is only possible if

x^{n} − y^{n} = 0, x^{n}y^{n} = 1,

in which case

t = (x^{2n} + y^{2n} + x^{n}y^{n})
= (x^{n} − y^{n})^{2} + 3x^{n}y^{n}
= 3.

4. Construct, with proof, a non–constant arithmetic progression of positive integers which contains no squares, cubes or higher powers of integers.

Answer: Suppose p is a prime. Consider any arithmetic progression of the form

p + kp^{2}n (n = 0, 1, …)

Each term in the progression is divisible by p, but not by p^{2}. Hence it cannot be a square or higher power.

The simplest sequence of this form, with p = 2 is

2, 6, 10, 14, …

Each term is divisible by 2, but not by 4.

5. If ^{n}C_{r} is the number of combinations of n objects r at a time, find the

value of ∑_{r=0}^{n} (\frac{1}{r + 1})^{n}C_{r}.

Answer: By the binomial theorem,

∑_{r=0}^{n}^{n}C_{r}x^{r} = (1 + x)^{n}.

Integrating from 0 to 1,

∑_{r=0}^{n}^{n}C_{r} ∫_{0}^{1} x^{r} dx = ∫_{0}^{1}(1 + x)^{n} dx,

giving

∑_{r=0}^{n} (\frac{1}{r + 1}) ^{n}C_{r} = [\frac{(1 + x)^{n+1}}{n + 1} ]_{0}^{1}
= \frac{2^{n+1} − 1}{n + 1}.

6. If x and y are real numbers, find all solutions of the equation

(x^{4} + 1) = (x^{2})(2^{1−y^{2}}).

Answer: [There was an error in tn earlier version of this: x^{2} for x^{4} on the left.] We have to find x, y such that

x^{2} + \frac{1}{x^{2}} = 2^{1−y^{2}}.

Since the arithmetic mean is greater than the geometric,

x^{2} + \frac{1}{x^{2}} ≥ 2,

with equality only if x^{2} = 1, ie x = ±1. On the other hand,

2^{1−y^{2}} ≤ 2,

with equality only if y = 0. So the only solutions to the given equation are: (x, y) = (±1, 0).

7. Find a formula for the nth derivative of the function f(x) = \frac{q}{x^{2} + x^{2}} where a is a constant.

Answer: There is obviously a misprint in this question.

8. If abc is an obtuse angled triangle with bâc > \frac{π}{2} prove that

54|ac|^{3}|ab|^{3}| cos bâc| ≤ |bc|^{6}

and find conditions under which equality holds.

Answer: By the cosine law,

|bc|^{2} = |ab|^{2} + |ac|^{2} + 2|ab||ac|(− cos bâc).

(Note that cos bâc < 0.) Since the arithmetic mean of the 3 terms on the right is less than their geometric mean,

\frac{|ab|^{2} + |ac|^{2} + 2|ab||ac|(− cos bâc)}{3} ≥ (2|ab|^{3}|ac|^{3}| cos bâc|)^{1/3} .

Cubing,

|bc|^{6} ≥ 2 · 27|ab|^{3}|ac|^{3}| cos bâc|.

The arithmetic and geometric means are equal only when all the terms are equal. So there is equality in this case if and only if

|ab|^{2} = |ac|^{2} = 2|ab||ac|| cos bâc|;

in other words,

|ab| = |ac|, | cos bâc| = \frac{1}{2}.

Thus there is equality if and only if bac is an isosceles triangle with angle 2π/3 = 120^{∘}.

9. Show that there do not exist polynomials f(x) and g(x) such that

e^{x} = \frac{f(x)}{g(x)} for all x.

Answer: Well, e^{x} increases faster than any polynomial …

10. Find the real factors of x^{4} + y^{4} + (x + y)^{4}.

Answer: We may observe that if y = ωx, where ω = e^{2πi/3}, then

x + y = (1 + ω)x = −ω^{2}x.

Thus

x^{4} + y^{4} + (x + y)^{4} = (1 + ω^{4} + (−ω^{2})^{4}) x^{4}
= (1 + ω + ω^{2})x^{4}
= 0.

Thus x − ωy is a factor, and similarly so is x − ω^{2}. Hence one real factor is

(x − ωy)(x − ω^{2}y) = x^{2} + xy + y^{2}.

Dividing

x^{4} + y^{4} + (x + y)^{4} = 2 (x^{4} + 2x^{3}y + 3x^{2}y^{2} + 2xy^{3} + y^{4})

by x^{2} + xy + y^{2}, we see that this factor occurs twice:

x^{4} + y^{4} + (x + y)^{4} = 2 (x^{2} + xy + y^{2})^{2} .