Irish Intervarsity Mathematics Competition

University College Cork 1990

1. If x, y and z are real numbers solve the equations

x + y + z = \sqrt[3]{2}
x^{2} + y^{2} + z^{2} = \sqrt[3]{4}
x^{3} + y^{3} + z^{3} = 2

Answer: We can simplify the equations by setting

x = ^{3}\sqrt{2}X, y = ^{3}\sqrt{2}Y, z = ^{3}\sqrt{2}Z.

The equations become

X + Y + Z = 1
X^{2} + Y^{2} + Z^{2} = 1
X^{3} + Y^{3} + Z^{3} = 1

From the power-sums on the left-hand side of these equations we can compute the symmetric products

c_{1} = X + Y + Z, c_{2} = Y Z + ZX + XY, c_{3} = XY Z. The solutions to the equations are then given by

{X, Y, Z} = {α, β, γ}, where α, β, γ are the roots of a cubic

t^{3} − c_{1}t^{2} + c_{2}t^{1} − c_{3} = 0.

But it is clear from the equations that one solution is

(X, Y, Z) = (1, 0, 0).

It follows that there are just 3 solutions:

(X, Y, Z) = (1, 0, 0), (0, 1, 0), (0, 0, 1),

2. A hand of 13 cards is dealt from a pack of 52 cards containing 13 spades, 13 diamonds, 13 clubs and 13 hearts. If the hand has 3 clubs, 4 spades, 2 diamonds and 4 hearts, it is siad to have the distribution type 4 − 4 − 3 − 2, while if it has 6 hearts and 7 spades it is said to have distribution type 7 − 6. How many distribution types are there? List them all.

Answer: We are asked, in effect, for the number p_{4}(13) of partitions of 13 into 4 or fewer parts. It is well-known that there is no simple formula for p_{r}(n), although there are various recurrence formulae, expressing p_{r}(n) in terms of partitions of smaller numbers.

However, we are asked to list the partitions, so there seems little point in looking for a nice method. The partitions are:

13, 12 − 1, 11 − 2, 11 − 1 − 1, 10 − 3, 10 − 2 − 1, 10 − 1 − 1 − 1, 9 − 4, 9 − 3 − 1, 9 − 2 − 2, 9 − 2 − 1 − 1, 8 − 5, 8 − 4 − 1, 8 − 3 − 2, 8 − 3 − 1 − 1, 8 − 2 − 2 − 1, 7 − 6, 7 − 5 − 1, 7 − 4 − 2, 7 − 4 − 1 − 1, 7 − 3 − 3, 7 − 3 − 2 − 1, 7 − 2 − 2 − 2, 6 − 6 − 1, 6 − 5 − 2, 6 − 5 − 1 − 1, 6 − 4 − 3, 6 − 4 − 2 − 1, 6 − 3 − 3 − 1, 6 − 3 − 2 − 2, 5 − 5 − 3, 5 − 5 − 2 − 1, 5 − 4 − 4, 5 − 4 − 3 − 1, 5 − 4 − 2 − 2, 5 − 3 − 3 − 2, 4 − 4 − 4 − 1, 4 − 4 − 3 − 2, 4 − 3 − 3 − 3.

Thus

p_{4}(13) = 39.

3. Evaluate

∑_{n=1}^{∞} \frac{n^{3}}{3^{n}} .

Answer: We have

∑_{n=0}^{∞} \frac{x^{n}}{3^{n}} = (1 − \frac{x}{3})^{−1}
= \frac{3}{3 − x}.

Differentiating,

∑_{n=1}^{∞} \frac{nx^{n−1}}{3^{n}} = \frac{3}{(3 − x)^{2}}

Multiplying by x,

∑_{n=1}^{∞} \frac{nx^{n}}{3^{n}} = \frac{3x}{(3 − x)^{2}}

Differentiating again,

∑_{n=1}^{∞} \frac{n^{2}x^{n−1}}{3^{n}} = \frac{3}{(3 − x)^{2}} + \frac{6x}{(3 − x)^{3}} .

Multiplying by x,

∑_{n=1}^{∞} \frac{n^{2}x^{n}}{3^{n}} = \frac{3x}{(3 − x)^{2}} + \frac{6x^{2}}{(3 − x)^{3}} .

Differentiating a third time,

∑_{n=1}^{∞} \frac{n^{3}x^{n−1}}{3^{n}} = \frac{3}{(3 − x)^{2}} + \frac{18x}{(3 − x)^{3}} + \frac{18x^{2}}{(3 − x)^{4}} .

Subsituting x = 1,

∑_{n=1}^{∞} \frac{n^{3}}{3^{n}} = \frac{3}{4} + \frac{18}{8} + \frac{18}{16}
= \frac{33}{8}.

4. Find the minimum value of the real function

f(x) = x^{x}, for x > 0.

Answer: We have

log f(x) = x log x.

Hence

\frac{f′(x)}{f(x)} = log x + 1.

Thus

f′(x) = 0 ⟺ log x = −1 ⟺ x = e^{−1}.

As x → 0+, log f(x) → 0 and so f(x) → 1. As x → ∞, log f(x) → ∞ and so f(x) → ∞. Thus the minimum value is

f(e^{−1}) = e^{−e^{−1}} = \frac{1}{e^{1/e}} .

5. Prove that

tan^{2} (\frac{π}{7} ) + tan^{2} (\frac{2π}{7} ) + tan^{2} (\frac{3π}{7} ) = 21.

Answer: We have

tan 7θ = \frac{P (tan θ)}{Q(tan θ)},

where P (t) and Q(t) are polynomials of degree ≤ 7 in t. Now

tan 7θ = 0 ⟺ θ = \frac{nπ}{7}

for some integer n. It follows that the roots of P (t) = 0 are the 7 numbers

tan \frac{nπ}{7} , (n = 0, 1, …, 6).

In fact, since

tan \frac{(7 − n)π}{7} = − tan \frac{nπ}{7} ,

the roots are

0, ± tan \frac{π}{7} , ± tan \frac{2π}{7} , ± tan \frac{3π}{7} .

Hence

P (t) = x(at^{6} + bt^{4} + ct^{2} + d).

Thus

tan^{2} (\frac{π}{7} ) , tan^{2} (\frac{2π}{7} ) , tan^{2} (\frac{3π}{7} )

are the roots of

ax^{3} + bx^{2} + cx + d.

In particular

tan^{2} (\frac{π}{7} ) + tan^{2} (\frac{2π}{7} ) + tan^{2} (\frac{3π}{7} ) = −\frac{b}{a}.

It only remains to compute the polynomial P (t). Writing tan θ = t, we have

tan 2θ = \frac{2t}{1 − t^{2}} ,
tan 3θ = \frac{t + \frac{2t}{1−t^{2}}}{1 − \frac{2t^{2}}{1−t^{2}}}
= \frac{3t − t^{3}}{1 − 3t^{2}} ,
tan 4θ = \frac{2\frac{2t}{1−t^{2}}}{1 − \frac{4t^{2}}{(1−t^{2})^{2}}}

= \frac{4t(1 − t^{2})}{1 − 6t^{2} + t^{4}} ,
tan 7θ = \frac{\frac{3t−t^{3}}{1−3t^{2}} + \frac{4t(1−t^{2})}{1−6t^{2}+t^{4}}}{1 − \frac{3t−t^{3}}{1−3t^{2}} \frac{4t(1−t^{2})}{1−6t^{2}+t^{4}}}
= \frac{P (t)}{Q(t)},

where

P (t) = (3t − t^{3})(1 − 6t^{2} + t^{4}) + 4t(1 − t^{2})(1 − 3t^{2})
= −t [t^{6} − 21t^{4} + 34t^{2} − 7] .

Hence

tan^{2} (\frac{π}{7} ) + tan^{2} (\frac{2π}{7} ) + tan^{2} (\frac{3π}{7} ) = 21.

6. A ladder of constant length 2L extends from a horizontal floor to a wall which is inclined at 85^{∘} to the horizontal. If the ladder slips while remaining in contact with both the floor and the wall, find the locus of the mid-point of the ladder.

Answer: Taking cartesian coordinates centred on the point where the floor and wall meet, the coordinate of the point where the ladder meets the wall is

(Y sin 5^{∘}, Y cos 5^{∘})

for some Y. If the point where the ladder meets the floor is (X, 0) then

(X − sin 5^{∘}Y)^{2} + (cos 5^{∘}Y)^{2} = 4L^{2}.

In other words

X^{2} + Y^{2} − 2 sin 5^{∘}XY = 4L^{2}.

The mid-point of the ladder has coordinates

(x, y) = (\frac{X + sin 5^{∘}Y}{2} , \frac{cos 5^{∘}Y}{2} ) .

Thus

Y = 2 sec 5^{∘}y, X = 2x − 2 tan 5^{∘}y.

Thus the point (x, y) lies on the locus

(x − tan 5^{∘}y)^{2} + sec^{2} 5^{∘}y^{2} − 2 sin 5^{∘}(x − tan 5^{∘}y) sec 5^{∘}y = L^{2}.

Thus the locus of the mid-point of the ladder is the ellipse

x^{2} + (tan^{2} 5^{∘} + sec^{2} 5^{∘})y^{2} − 2 tan 5^{∘}xy = L^{2};

or rather, that part of the ellipse between the wall and the floor.

7. Prove that 3, 5 and 7 are the only 3 consecutive odd numbers all of which are prime.

Answer: One of the 3 numbers n, n + 2, n + 4 is divisible by 3. So if all 3 are prime, one of the numbers must be 3. Thus the 3 numbers must be 3, 5, 7.

8. If x, y and z are positive integers, all greater than 1, find any solution

of the equation

x^{x}y^{y} = z^{z}.

Answer: [Solution by James Ward, of UCG] Any solution must have

z < x + y,

since

(x + y)^{x+y} > x^{x}y^{y}

(as one of the terms in the binomial expansion of the left-hand side will be a multiple of the right-hand side).

Suppose

d = hcf(x, y).

Then

d^{x+y}|z^{z} ⟹ d|z,

since z < x + y. Let

x = dX, y = dY, z = dZ.

Then (taking the dth root on each side)

d^{e}X^{X}Y^{Y} = Z^{Z},

where e = X + Y − Z.

In particular

X^{X}Y^{Y} |Z^{Z}.

Moreover, if we can find X, Y, Z satisfying this relation and also satisfying

X + Y − Z = 1

then dX, dY, dZ will satisfy the original equation, with

d = \frac{Z^{Z}}{X^{X}Y^{Y}} .

Suppose

u − v = 1.

Squaring,

u^{2} + v^{2} − 2uv = 1.

Let us therefore take

X = u^{2}, Y = v^{2}, Z = 2uv.

Then

\frac{Z^{Z}}{X^{X}Y^{Y}} = 2^{2uv}u^{2uv−2u^{2}}v^{2uv−2v^{2}}
= \frac{2^{2uv}v^{2v}}{u^{2u}} .

We can make this to be an integer—as we want—by taking u = 2^{a}. In this case 2 occurs to the power 2uv on the top, and 2au on the bottom. Thus the condition will be satisfied if

v = 2^{a} − 1 ≥ a.

This holds for all a. However, we only get a non-trivial solution if a ≥ 2.

So we have the family of solutions

X = 2^{2a}, Y = (2^{a} − 1)^{2} , Z = 2^{a+1} (2^{a} − 1)

for a = 2, 3, . . ..

Taking a = 2 this gives

X = 2^{4} = 16, Y = 3^{2} = 9, Z = 2^{3}3 = 24.

In this case d = 2^{8}3^{6}, and so

x = 2^{12}3^{6}, y = 2^{8}3^{8}, z = 2^{11}3^{7}.

9. If a, b, c are the lengths of the side of a triangle prove that

3(ab + bc + ca) ≤ (a + b + c)^{2} ≤ 4(ab + bc + ca),

and in each case find the necessary and sufficient condition.

Answer: Taking the inequality on the left first, we have

(a + b + c)^{2} − 3(ab + bc + ca) = a^{2} + b^{2} + c^{2} − (ab + bc + ca)
= \frac{1}{2}(a − b)^{2} + \frac{1}{2}(b − c)^{2} + \frac{1}{2}(c − a)^{2}.

Thus the inequality holds, with equality if and only if a = b = c, ie the triangle is equiangular.

The inequality on the right can be written

2(ab + bc + ca) − (a^{2} + b^{2} + c^{2}) ≥ 0.

Now

ab + bc = b(a + c) ≥ b^{2},

since a + c ≥ b. Similarly

bc + ca ≥ c^{2}, ca + ab ≥ a^{2}.

Adding the 3 inequalities,

2(ab + bc + ca) ≥ a^{2} + b^{2} + c^{2}.

Now a = b+c only if the 3 vertices A, B, C lie on a line with A between B and C. Thus the 3 inequalities cannot all be equalities simultaneously unless the 3 vertices coincide, ie a = b = c = 0.

10. Let S be the set of all natural numbers, which can be written in the form x^{3} + y^{3} + z^{3} − 3xyz for some x, y, z ∈ N, where N is the set of natural numbers. if a ∈ S and b ∈ S prove that ab ∈ S.

Answer: We have

x^{3} + y^{3} + z^{3} − 3xyz = (x + y + z)(x + ωy + ω^{2}z)(x + ω^{2}y + ωz),

where ω = e^{2πi/3}.

Now if a, b, c are integers then

a + bω + cω^{2} = 0 ⟺ a = b = c ⟺ a + bω^{2} + cω = 0.

Suppose

m = x^{3} + y^{3} + z^{3} − 3xyz, n = X^{3} + Y^{3} + Z^{3} − 3XY Z.

We can certainly satisfy the equation

u + ωv + ω^{2}w = (x + ωy + ω^{2}z)(X + ωY + ω^{2}Z)

by taking

u = xX + yZ + zY, v = xY + yX + zZ, w = xZ + yY + zX.

Moreover, by the argument above this will also imply that

u + ω^{2}v + ωw = (x + ω^{2}y + ωz)(X + ω^{2}Y + ωZ)

And it is easy to see that

u + v + w = (x + y + z)(X + Y + Z).

Multiplying these factors together,

mn = u^{3} + v^{3} + w^{3} − 3uvw.

Thus mn ∈ S.