Irish Intervarsity Mathematics Competition 2001

Trinity College Dublin

9.30–12.30 Saturday 3^{rd} March 2001

Answer as many questions as you can; all carry the same mark. Give

reasons in all cases. Tables and calculators are not allowed.

1. For which integers a does the equation

x^{3} − 3x + a

have three integer roots? Answer: Suppose the roots are u, v, w. Then

u + v + w = 0, uv + uw + vw = −3, uvw = −a.

Setting

w = −(u + v),

the second equation becomes

uv + uw + vw = −3,

ie

uv − (u + v)^{2} = −3,

ie

u^{2} + uv + v^{2} = 3.

We have to find solutions u, v ∈ ℤ of this equation. We can write the equation in the form

(u + v/2)^{2} + 3v^{2}/4 = 3.

Thus

3v^{2}/4 ≤ 3,

ie

v^{2} ≤ 4,

ie

|v| ≤ 2.

Similarly

|u| ≤ 2.

On going through the possibilities

u, v ∈ {−2, −1, 0, 1, 2}

we see that the only integer solutions are

(u, v) = (−2, 1), (−1, −1), (1, 1), (2, −1).

Substituting for w, the complete solutions are

(u, v, w) = (−2, 1, 1), (−1, −1, 2), (1, 1, −2), (2, −1, −1).

Thus there are really just two solutions (up to order) namely

±(1, 1, −2)

giving

a = −uvw = ±2.

2. Find all solutions in integers a, b, c of

a^{2} + b^{2} + c^{2} = a^{2}b^{2}.

Answer: One solution is evidently

a = b = c = 0.

Suppose we have a different solution. If x ∈ ℤ then

x^{2} ≡ 0 or 1 mod 4

according as x is even or odd. Suppose a, b are both odd. Then

a^{2}b^{2} ≡ 1 mod 4,

while

a^{2} + b^{2} + c^{2} ≡ 2 or 3 mod 4.

Thus one (at least) of a, b is even. Hence

a^{2}b^{2} ≡ 0 mod 4,

and so

a^{2} + b^{2} + c^{2} ≡ 0 mod 4,

which is only possible if a, b, c are all even. Let 2^{r} be the highest power of 2 dividing a, b, c. Then r ≥ 1 from above. Suppose

a = 2^{r}a′, b = 2^{r}b′, c = 2^{r}c′.

Then

a′^{2} + b′^{2} + c′^{2} = 2^{2r}a′^{2}b′^{2}.

Hence

a′^{2} + b′^{2} + c′^{2} ≡ 0 mod 4,

which as we have seen implies that a′, b′, c′ are all even, contrary to our choice of r as the highest power of 2 dividing a, b, c. We conclude that

a = b = c = 0

is the only solution.

3. Find all polynomials P (x) satisfying

P (x^{2}) = P (x)P (x − 1).

Answer: Suppose the roots of P (x) are

α_{1}, … α_{n}.

Then the roots of P (x^{2}) are

±\sqrt{α_{1}}, ⋯ ± \sqrt{α_{n}},

while the roots of P (x)P (x − 1) are

α_{1}, α_{1} − 1, …, α_{n}, α_{n} − 1.

These must be the same, up to order. Let

max|α_{i}| = M.

Then

max|\sqrt{α_{i}}| = \sqrt{M}.

Thus if M > 1 there is no way of matching the roots. Similarly, let

min_{α_{i}≠0} |α_{i}| = m.

Then

min_{α_{i}≠0} |\sqrt{α_{i}}| = \sqrt{m}.

Thus if m < 1 there is no way of matching the roots. We conclude that the non-zero roots must all lie on the unit circle:

α_{i} ≠ 0 ⟹ |α_{i}| = 1.

Now suppose α ≠ 0 is a root of P (x). Then α + 1 is a root of P (x − 1). It follows that either α = −1 or else

|α| = 1, |α + 1| = 1.

These two equations represent the two circles of radius 1 centred on 0 and −1, meeting in the points ω, ω^{2} (where ω = exp(2πi/3)). We conclude that if α is a roots of P (x) then

α ∈ {0, −1, ω, ω^{2}}.

If 0 is a root of P (x) then −1 is a root of P (x − 1), and so ±i are roots of P (x^{2}). Hence i (for example) is a root of P (x) or P (x − 1), ie i or i + 1 is a root of P (x), both of which we have already excluded.

Similarly if −1 is a root of P (x) then ±i are roots of P (x^{2}), which we have just seen is impossible.

It follows that the only possible roots of P (x) are

α = ω or ω^{2}.

Moreover these roots must occur equally often. For

ω + 1 = −ω^{2}, ω^{2} + 1 = −ω;

and

\sqrt{ω} = ±ω^{2}, \sqrt{ω^{2}} = ±ω.

Thus if ω occurs r times and ω^{2} occurs s times as roots of P (x) then ω, −ω, ω^{2}, −ω^{2} occur with multiplicities s, s, r, r in P (x^{2}), and with multiplicies r, s, r, s in P (x)P (x − 1).

Since

(x − ω)(x − ω^{2}) = x^{2} + x + 1,

we have shown that the only solutions of the given relation are

P (x) = (x^{2} + x + 1)^{n}

for n = 0, 1, 2, … (to which should be added the trivial solution P (x) = 0).

4. If the plane is partitioned into two disjoint subsets, show that one of the subsets contains three points forming the vertices of an equilateral triangle.

Answer: There are probably many ways of doing this.

Let us call the two sets A and B. If A and B are empty the result is trivial; so we may assume that there are points A ∈ A, B ∈ B.

Consider the line AB, and equidistant points

…, P_{−2}, P_{−1}, A, B, P_{1}, P_{2}, …

on the line. There are two possibilities. Either these points are alternately in A and B,

…, A_{−1}, B_{−1}, A_{0}, B_{0}, A_{1}, B_{1}, …;

or we can find three equidistant points X, Y, Z with the central point X in the same set as just one of the end-points X, Z, eg

A_{1}A_{2}B_{1}.

In the first case, consider the equilateral triangles

A_{i}X_{i}A_{i+1} and A_{i}Y_{i}A_{i+1},

with all the points X_{i} on the same side of the line. Then the points X_{i}, Y_{i} must all lie in B; and so

X_{1}Y_{2}X_{3}

is an equilateral triangle in B.

In the second case we may assume the three equidistant points are

A_{1}A_{2}B_{1}.

Consider the hexagon

A_{1}P_{1}P_{2}B_{1}P_{3}P_{4}

with centre A_{2} and diagonal A_{1}B_{1}. Then P_{1} ∈ B, or A_{1}P_{1}A_{2} would be an equilateral triangle in A. Similarly P_{4} ∈ B. But then P_{1}B_{1}P_{4} is an equilateral triangle in B.

More questions overleaf!

5. For which real numbers x is

sin(cos x) = cos(sin x)?

Answer: If sin(cos x) = cos(sin x),

then sin(cos x) = sin(π/2 − sin x).

But

sin A = sin B ⟺ A = B + nπ,

where n ∈ ℤ. Thus

cos x = nπ + π/2 − sin x.

Since

|cos x|, |sin x| ≤ 1,

this is only possible if

n = 0 or − 1.

Thus

cos x + sin x = ±\frac{π}{2} .

Squaring both sides,

cos^{2} x + 2 cos x sin x + sin^{2} x = \frac{π^{2}}{4} .

Since

cos^{2} x + sin^{2} x = 1, 2 cos x sin x = sin 2x,

our equation gives

sin 2x = \frac{π^{2}}{4} − 1.

But π^{2} > 9, and so

\frac{π^{2}}{4} − 1 > \frac{9}{4} − 1 > 1.

Since |sin 2x| ≤ 1 for all x, we conclude that the original equation has no solution.

6. For which real numbers c ≠ 0 does the sequence defined by

a_{0} = c, a_{n+1} = \frac{1}{2} (a_{n} + \frac{1}{a_{n}} )

converge? Answer: Suppose the sequence converges to ℓ. Then

ℓ = \frac{1}{2} (ℓ + \frac{1}{ℓ})

Hence

ℓ = \frac{1}{ℓ},

ie

ℓ = ±1.

If c is replaced by −c then a_{n} is replaced by −a_{n} for all n. Thus it is sufficient to consider the case c > 0; if a_{n} → ℓ when a_{0} = c > 0 then a_{n} → −ℓ when a_{0} = −c. If c > 0 then a_{n} > 0 for all n. Thus if a_{n} is convergent then a_{n} → 1. By the inequality between the arithmetic and geometric means (or by a simple calculation) if a_{n} > 0 then

\frac{1}{2} (a_{n} + \frac{1}{a_{n}} ) ≥ 1.

Thus

a_{n} ≥ 1

for n ≥ 1. Also

a_{n+1} ≤ a_{n} ⟺ \frac{1}{a_{n}} ≤ a_{n}
⟺ a_{n} ≥ 1

Hence the sequence a_{n} is decreasing for n ≥ 1, and so must converge to a limit, which we have seen must be 1. We conclude that the sequence always converges if c ≠ 0:

a_{n} → \begin{cases}1 & if c > 0, \\ −1 & if c < 0. \end{cases}

7. Show that for any sequence a_{n} of strictly positive real numbers one (or both) of the series

∑ \frac{a_{n}}{n^{2}} and ∑ \frac{1}{a_{n}}

must diverge.

Answer: Suppose the two series both converge. Then

\frac{1}{a_{n}} → 0 ⟹ a_{n} → ∞.

Since a_{n} > 0 we can re-arrange the series without affecting their convergence or divergence. Thus we may assume that the a_{n} are increasing:

a_{n+1} ≥ a_{n}.

Let π(m) denote the number of a_{n} which are ≤ m:

π(m) = |{n : a_{n} ≤ m}|.

The number of a_{n} in the range [m, 2m) is

π(2m) − π(m).

The a_{n} in this range contribute

≥ \frac{π(2m) − π(m)}{2m}

to the second series ∑ 1/a_{n}.

It follows that

\frac{π(2m) − π(m)}{2m} → 0

as m → ∞. In particular there is an N > 0 such that

\frac{π(2m) − π(m)}{2m} < \frac{1}{8}

ie

π(2m) − π(m) < \frac{1}{4}m

for m ≥ N.

Thus

π(2N) − π(N) ≤ \frac{1}{4}N,

π(4N) − π(2N) ≤ \frac{1}{4}2N,

…

π(2^{r+1}N) − π(2^{r}2N) ≤ \frac{1}{4}2^{r}N.

Adding,

π(2^{r+1}N) − π(N) ≤ \frac{1}{4} (1 + 2 + 2^{2} + ⋯ + 2^{r}) N

≤ \frac{1}{4}2^{r+1}N

Thus if

2^{r}N ≤ n < 2^{r+1}N

then

π(n) ≤ π(2^{r+1}N)

≤ π(N) + \frac{1}{2}2^{r}N

≤ π(N) + \frac{1}{2}n.

We conclude that, for some N_{1} ≥ N,

π(n) ≤ n

for n ≥ N_{1}. But

π(n) ≤ n ⟹ a_{n} ≥ n.

Thus

\frac{a_{n}}{n^{2}} ≥ \frac{1}{n}

for n ≥ N_{1}; and so the first series

∑ \frac{a_{n}}{n^{2}}

diverges by comparison with

∑ \frac{1}{n}.

8. The point A lies inside a circle centre O. At what point P on the circumference of the circle is the angle OP A maximised?

Answer: It is evident that

OP̂A < π/2,

eg by drawing the diameter D_{1}OAD_{2} through A, and comparing OP̂A with D_{1}P D_{2} = π/2.

Applying the sine law to the triangle OP A,

\frac{sin OP̂A}{OA} = \frac{sin OÂP}{OP}

ie

sin OP̂A = \frac{OA}{OP} sin OÂP.

Now OP̂A is maximised when sin(OP̂A) is maximised. Since OA and OP are fixed (with OA < AP) this will occur when sin(OÂP) is maximised, ie when OÂP = π/2.

We conclude that OP̂A is maximised when P is one of the two points where the line through A perpendicular to OA meets the circle.

9. What is the maximum volume of a cylinder contained within a sphere of radius 1?

Answer: This is straightforward. Clearly the largest cylinder will have its centre at the centre of the sphere. Suppose the radius of the cylinder is r, and the height h. Then

r^{2} + h^{2} = 1.

The volume of the cylinder is

V = πr^{2}h
= πh(1 − h^{2}).

We have to maximise this function, with the restriction that 0 < h < 1.

Let

f(x) = x(1 − x^{2}) = x − x^{3}.

Then f(0) = f(1) = 0 and

f′(x) = 1 − 3x^{2}.

Thus f(x) attains its maximum when

x = \frac{1}{\sqrt{3}};

and

V_{max} = \frac{π}{\sqrt{3}}.

10. The function f(x) is twice-differentiable for all x, and both f(x) and f′′(x) are bounded. Show that f′(x) is also bounded.

Answer: Suppose

|f(x)| ≤ C, |f′′(x)| ≤ C.

Suppose f′(x) is large at some point, say x_{0}. On taking −f(x) in place of f(x) if necessary, we may suppose that

f′(x_{0}) = X > 0.

Since f′′(x) is bounded, f′(x) must remain large for a considerable interval beyond x_{0}. More precisely, suppose

f′(x) < X/2

for some x > x_{0}. Then

f′(x_{0}) − f′(x) > X/2.

But we know, from Rolle’s Theorem, that

\frac{f′(x) − f′(x_{0})}{x − x_{0}} = f′′(ξ)

for some ξ ∈ (x_{0}, x). Thus

x − x_{0} = \frac{f′(x) − f′(x_{0})}{f′′(ξ)}

≥ \frac{X}{2C} .

In other words,

f′(x) ≥ X/2

for x ∈ [x_{0}, x_{1}], where

x_{1} = x_{0} + \frac{X}{2C} .

But this in turn means that f(x) is increasing steadily over this interval. More precisely, for some η ∈ (x_{0}, x_{1}),

f(x_{1}) − f(x_{0}) = (x_{1} − x_{0})f′(η)

≥ (x_{1} − x_{0})\frac{X}{2} .

Thus

f(x_{1}) − f(x_{0}) ≥ \frac{X^{2}}{4C} .

But since

|f(x) ≤ C

it follows that

|f(x_{1} − f(x_{0})| ≤ |f(x_{1}|| + |f(x_{0})| ≤ 2C.

Hence

\frac{X^{2}}{4C} ≤ 2C,

ie

X < 2^{3/2}C.

So we have shown that

|f′(x)| ≤ 2^{3/2}C

for all x.