2nd British Mathematical Olympiad 1966 Problems

2nd British Mathematical Olympiad 1966 Problems

1.  Find the greatest and least values of f(x) = (x⁴ + x² + 5)/(x² + 1)² for real x.

2.  For which distinct, real a, b, c are all the roots of ±√(x - a) ±√(x - b) ±√(x - c) = 0 real?

3.  Sketch y² = x²(x + 1)/(x - 1). Find all stationary values and describe the behaviour for large x.

4.  A₁, A₂, A₃, A₄ are consecutive vertices of a regular n-gon. 1/A₁A₂ = 1/A₁A₃ + 1/A₁A₄. What are the possible values of n?

5.  A spanner has an enclosed hole which is a regular hexagon side 1. For what values of s can it turn a square nut side s?

6.  Find the largest interval over which f(x) = √(x - 1) + √(x + 24 - 10√(x - 1) ) is real and constant.

7.  Prove that √2, √3 and √5 cannot be terms in an arithmetic progression.

8.  Given 6 different colours, how many ways can we colour a cube so that each face has a different colour? Show that given 8 different colours, we can colour a regular octahedron in 1680 ways so that each face has a different colour.

9.  The angles of a triangle are A, B, C. Find the smallest possible value of tan A/2 + tan B/2 + tan C/2 and the largest possible value of tan A/2 tan B/2 tan C/2.

10.  One hundred people of different heights are arranged in a 10 x 10 array. X, the shortest of the 10 people who are the tallest in their row, is a different height from Y, the tallest of the 10 people who are the shortest in their column. Is X taller or shorter than Y?

11.  (a) Show that given any 52 integers we can always find two whose sum or difference is a multiple of 100. (b) Show that given any set 100 integers, we can find a non-empty subset whose sum is a multiple of 100. 

Problem 1

Find the greatest and least values of f(x) = (x⁴ + x² + 5)/(x² + 1)² for real x.

Solution

Answer: 5, 0.95.

f(x) = 5 - (4x⁴ + 9x²)/(x⁴ + 2x² + 1) <= 5, with equality iff x = 0.

Put z = (1 + x²), then f(x) = (z(z - 1) + 5)/z². Put w = 1/z = 1/(1 + x²), then f(x) = 5w² - w + 1 = 5(w - 1/10)² + 95/100 ≥ 0.95 with equality iff w = 1/10 or x = ±3.

[Of course, the question could be done less elegantly but much faster using calculus, but these problems are aimed at pre-university students, who are assumed not to know calculus.]

Problem 2

For which distinct, real a, b, c are all the roots of ±√(x - a) ±√(x - b) ±√(x - c) = 0 real?

Solution

Answer: always true.

This is slightly tricky.

Suppose x, a, b, c satisfy the equation for some combination of signs. We show that they must also satisfy a quadratic, which is independent of the choice of sign.

±√(x - a) = ±√(x - b) ±√(x - c). Squaring: x - a = 2x - b - c ±2 √(x - b)√(x - c). So x + a - b - c = ±2 √(x - b)√(x - c). Squaring: x² + 2(a - b - c)x + a² + b² + c² - 2(ab - bc + ca) = 4x² - 4(b + c)x + 4bc, so 3x² - 2(a + b + c)x - (a² + b²+ c²) + 2(ab + bc + ca) = 0. Set f(x) = 3x² - 2(a + b + c)x - (a² + b²+ c²) + 2(ab + bc + ca). Clearly f(x) > 0 for |x| large. But f(a) = 3a² - 2a² - 2ab - 2ca - a² - b² - c² + 2ab + 2bc + 2ca = - (b - c)² < 0. So f(x) = 0 always has two distinct real roots, one < a and one > a. The same is true for b and c. So if a > b > c, we have two distinct real roots, one > a and one < c.

Reading the question strictly, that gives us the answer: if x is a root of the original equations, then it is real. But that does not tell us whether the original equations have 0, 1 or 2 roots. The status of the root < c is slightly problematical. It is in fact a real root of the original equation, but we need complex numbers to see it. For example, let a = 8, b = 5, c = 0. The roots are 9 (giving +3 - 2 - 1 = - 3 + 2 + 1 = 0), which is straightforward, and -1/3 (giving i/√3 (1 + 4 - 5) = i/√3 (-1 - 4 + 5) = 0). But we still have to prove this is true in general.

Let g(x) = √(x - c) - √(x - b) - √(x - a). For x > a, g(x) is clearly real and continuous. But g(a) = √(a - c) - √(a - b) > 0. Whereas for x large, g(x) = √x < 0, so g(x) has at least one real root > a. Similarly, if x < c, then (with a suitable choice of signs) g(x) = i ( √(a - x) - √(b - x) - √(c - x) ). So g(c) > 0, but g(x) < 0 for x large and negative, so g(x) has at least one real root < c. But we already know that the equations have at most two real roots, so they have exactly two. If we choose to regard only values of x >= a as admissible, then they have one real root (and no imaginary roots).  

 Problem 3

Sketch y² = x²(x + 1)/(x - 1). Find all stationary values and describe the behaviour for large x.

Solution

It is not easy to incorporate sketches on this web page, and anyway this is rather trivial bookwork, so I will just give a brief description.

The obvious point is that x²(x + 1)/(x - 1) is negative for -1 < x < 1, so there are no points on the graph for these values of x. Also the graph is symmetrical about the x-axis. x = 1 is an asymptote. Focussing on the positive y-part, y tends to infinity as x tends to 1 from above. It has a minimum at (1 + √5)/2 and tends to infinity as x² as x tends to infinity. Obviously, the reflection in the x-axis has a maximum at the same value of x. Differentiating, the gradient is infinite at x = -1. So for large negative x, the curve is similar to y = ± x². It there are two symmetrically placed inflections and the curve is vertical as it cuts the x-axis.

Problem 4

A₁, A₂, A₃, A₄ are consecutive vertices of a regular n-gon. 1/A₁A₂ = 1/A₁A₃ + 1/A₁A₄. What are the possible values of n?

Solution

Answer: n = 7.

It is a nice question whether one has to show that n = 7 is a solution! [Does the question give as a fact that there exists a solution?] Clearly, there are no other possible solutions. Suppose the n-gon has side 1. Then A₁A₃ and A₁A₄ are obviously strictly increasing functions of n. So 1/A₁A₃ + 1/A₁A₄ is a strictly decreasing function of n, whereas 1/A₁A₂ is always 1. So there can be at most one solution. Moreover, for n = 6, the lhs is 1/√3 + 1/2 > 1, whereas for n = 8, 1/√(2 + √2) + (2 - 1) < 1. So if there is a solution it is 7.

If we are not allowed to assume that there is a solution, then we have to prove equality for n = 7.

A kludgy (but easy) proof is as follows. A₁A₃ = 2 cos π/7, A₁A₄ = (1 + 2 cos 2π/7). Put c = cos π/7. The cos 2π/7 = 2c² - 1. So the result is true if 8c³ - 4c² - 4c + 1 = 0. Put s = sin π/7. Then (c + is)⁷ = - 1. Expanding and comparing real parts, we get: c⁷ - 21c⁵(1 - c²) + 35c³(1 - c²)² - 7c(1 - c²)³ + 1 = 0, or 64c⁷ - 112c⁵ + 56c³ - 7c + 1 = 0. But we know that c = -1 is a solution (not the one we want), so we can factor that out to get: 64c⁶ - 64c⁵ - 48c⁴ + 48c³ + 8c² - 8c + 1 = 0. But we know that this has three repeated roots (cos π/7 = cos (-π/7) etc), so it should factorize as the square of a cubic and indeed it does: (8c³ - 4c² - 4c + 1)² = 0, so 8c³ - 4c² - 4c + 1 = 0 as required.  

Problem 5

A spanner has an enclosed hole which is a regular hexagon side 1. For what values of s can it turn a square nut side s?

Solution

Answer: 3 - √3 >= s > √(3/2). Note that 3 - √3 = 1.27, √(3/2) = 1.22, so the range is fairly narrow.

The square must be small enough to fit into the hole and big enough to stick.

The sticking condition is easy. Inscribe a circle in the hexagon. If the square will fit into the circle, then it obviously does not stick. So we require the diagonal of the square to be > √3, or s > √(3/2). On the other hand if the diagonal exceeds √3, then it obviously does stick because the parallel opposite sides of the hexagon are only a distance √3 apart.

It seems fairly clear that the maximum square has sides parallel to a pair of opposite sides and each vertex a distance x from on endpoint of those sides. For the square to have equal sides, we then need √3 (1 - x) = 1 + x, or x = 2 - √3, and hence s = 3 - √3. But how do we prove it?

Problem 6

Find the largest interval over which f(x) = √(x - 1) + √(x + 24 - 10√(x - 1) ) is real and constant.

Solution

Answer: [1, 26].

f(x) is not real for x < 1. For 1 <=x <= 26, we have 0 < √(x - 1) <= 5. Now (5 - √(x - 1) )² = x + 24 - 10√(x - 1). So for 1 <= x <= 26, f(x) = 5. For x > 26, f(x) = 2√(x - 1) - 5.

Problem 7

Prove that √2, √3 and √5 cannot be terms in an arithmetic progression.

Solution

If they are, then for some non-zero rational R, (√5 - √2) = R(√3 - √2). Squaring and rearranging: √10 = (7 - 5R²)/2 + R²√6. If 7 - 5R² = 0, then squaring we have an obvious contradiction: 10 = (49/25) 6. So (7 - 5R²)/2 is non-zero and squaring again gives that √6 is rational. Contradiction by the standard argument. [If √6 = m/n in lowest terms, then m² = 6 n². 2 divides 6, so it must divide m, hence also n. ]

Problem 8

Given 6 different colours, how many ways can we colour a cube so that each face has a different colour? Show that given 8 different colours, we can colour a regular octahedron in 1680 ways so that each face has a different colour.

Solution

Answer: 30.

Take the colours as C₁, C₂, ... , C₆. We may always rotate the cube so that C₁ is on the top face. There are then 5 choices for the opposite face. The lowest remaining colour may then be rotated into a fixed position (say north). The three remaining colours may then be distributed over the three remaining faces in 6 ways. Total 5 x 6 = 30.

Similarly, for the octahedron. Take C₁ on the top face. There are 7 possibilities for the bottom face. There are now 20 (3 chosen from 6) choices for the colours of the three sides adjacent to the top face. Rotate the lowest colour into a fixed position. There are then 2 choices for the other two faces. Finally, the remaining 3 colours may be assigned to the remaining 3 faces in 6 ways. Total: 7 x 20 x 2 x 6 = 1680.  

Problem 9

The angles of a triangle are A, B, C. Find the smallest possible value of tan A/2 + tan B/2 + tan C/2 and the largest possible value of tan A/2 tan B/2 tan C/2.

Solution

Answer: equilateral in both cases, √3, 1/(3√3).

Using the familiar tan(a + x) = (tan a + tan x)/(1 - tan a tan x), we have that for a and x in the interval [0, π/4], tan(a + x) + tan(a - x) =2 tan a (1 + tan²x)/(1 - tan²a tan²x) >= 2 tan a, with equality iff x = 0. Hence the minimum value of tan A/2 + tan B/2 + tan C/2 occurs when A = B = C (if any pair of A, B, C are unequal then we can lower the value by replacing the pair by two angles equal to their mean).

Similarly, we show that for a and x in the interval (0, π/4), tan(a - x) tan(a + x) < tan²a. The lhs = (tan²a - tan²x)/(1 - tan²a tan²x) = tan²a - tan²x (1 - tan⁴a)/(1 - tan²a tan²x) < tan²a.

This proves that the maximum value of tan A/2 tan B/2 tan C/2 occurs for A = B = C. For if A ≠ B, then we can increase the value by replacing A and B by their mean (and similarly for any other unequal pair). This does not deal with the degenerate case of a flat triangle. But if the angles are A = 2δ, B = 2ε, C = π - (2δ + 2ε), where δ and ε are small, then tan A/2 tan B/2 tan C/2 = δ ε/(δ + ε) which tends to zero as δ and ε tend to zero. So the flat triangle gives a smaller value.

Problem 10

One hundred people of different heights are arranged in a 10 x 10 array. X, the shortest of the 10 people who are the tallest in their row, is a different height from Y, the tallest of the 10 people who are the shortest in their column. Is X taller or shorter than Y?

Solution

Answer: taller.

If X and Y are in the same row or column, then the result is immediate (by definition, X is taller than others in the row, and Y is shorter than others in the column). So suppose they are not. Take Z in the same row as X and the same column as Y. Then Z is shorter than X and taller than Y.

Problem 11

(a) Show that given any 52 integers we can always find two whose sum or difference is a multiple of 100.

(b) Show that given any set 100 integers, we can find a non-empty subset whose sum is a multiple of 100.

Solution

(a)   Consider residues mod 100. We can divide the residues into 51 subsets: two singletons {0} and {50}, and 49 pairs: {1, 99}, {2, 98}, ... {49, 51}. If we have two nubers from the same subset, then either their sum or their difference is divisible by 100. But any collection of 52 numbers must have two in the same subset.

(b)   Denote the integers as a₁, a₂, ... , a₁₀₀. Consider the 100 numbers: a₁, a₁ + a₂, a₁ + a₂ + a₃, ... , a₁ + ... + a₁₀₀. Either one of them is divisible by 100, in which case we are done, or two of them must be congruent mod 100 (because there are only 99 non-zero residues mod 100). But in that case their difference, which is the sum a_i + a_i+1 + ... + a_j for some i ≤ j, is zero mod 100, as required.