3rd British Mathematical Olympiad 1967 Problems

3rd British Mathematical Olympiad 1967 Problems

1.  a, b are the roots of x² + Ax + 1 = 0, and c, d are the roots of x² + Bx + 1 = 0. Prove that (a - c)(b - c)(a + d)(b + d) = B² - A².

2.  Graph x⁸ + xy + y⁸ = 0, showing stationary values and behaviour for large values. [Hint: put z = y/x.]

3.  (a)   The triangle ABC has altitudes AP, BQ, CR and AB > BC. Prove that AB + CR ≥ BC + AP. When do we have equality? (b)   Prove that if the inscribed and circumscribed circles have the same centre, then the triangle is equilateral.

4.  We are given two distinct points A, B and a line l in the plane. Can we find points (in the plane) equidistant from A, B and l? How do we construct them?

5.  Show that (x - sin x)(π - x - sin x) is increasing in the interval (0, π/2).

6.  Find all x in [0, 2π] for which 2 cos x ≤ |√(1 + sin 2x) - √(1 - sin 2x)| ≤ √2.

7.  Find all reals a, b, c, d such that abc + d = bcd + a = cda + b = dab + c = 2.

8.  For which positive integers n does 61 divide 5ⁿ - 4ⁿ?

9.  None of the angles in the triangle ABC are zero. Find the greatest and least values of cos²A + cos²B + cos²C and the values of A, B, C for which they occur.

10.  A collects pre-1900 British stamps and foreign stamps. B collects post-1900 British stamps and foreign special issues. C collects pre-1900 foreign stamps and British special issues. D collects post-1900 foreign stamps and British special issues. What stamps are collected by (1) no one, (2) everyone, (3) A and D, but not B?

11.  The streets for a rectangular grid. B is h blocks north and k blocks east of A. How many shortest paths are there from A to B? 

Problem 1

a, b are the roots of x² + Ax + 1 = 0, and c, d are the roots of x² + Bx + 1 = 0. Prove that (a - c)(b - c)(a + d)(b + d) = B² - A².
Solution

We use ab = cd = 1, a + b = -A, c + d = -B repeatedly.
(a - c)(b - c) = 1 + Ac + c², (a + d)(b + d) = 1 - Ad + d². So (a - c)(b - c)(a + d)(b + d) = 1 - Ad + d² + Ac - A² + Ad + c² - Ac + 1 = 2 + (c² + d²) - A² = 2 + (c + d)² - 2 - A² = B² - A².
Problem 2

Graph x⁸ + xy + y⁸ = 0, showing stationary values and behaviour for large values. [Hint: put z = y/x.]
Solution

Answer: a figure of 8, with the crossing point at the origin, touching both axes at the origin, extending to x, y = ±0.93, with y = ±x as axes of symmetry.

Problem 3

(a)   The triangle ABC has altitudes AP, BQ, CR and AB > BC. Prove that AB + CR >= BC + AP. When do we have equality?
(b)   Prove that if the inscribed and circumscribed circles have the same centre, then the triangle is equilateral.
Solution

Problem 4

We are given two distinct points A, B and a line l in the plane. Can we find points (in the plane) equidistant from A, B and l? How do we construct them?
Solution

Provided A does not lie on l, the locus of points equidistant from A and l is (one branch of) a hyperbola, on the same side of l as A. So if A and B are on opposite sides of l, then there are no equidistant points.
Assume A and B are on the same side of l. The locus of points equidistant from A and B is the perpendicular bisector m of the line joining them. This will intersect the other locus in two points unless m is parallel to l, in which case it only intersects the other locus in one point.
The only remaining possibility is if A (or B) lies on l. In this case the locus of points equidistant from A and l is the line perpendicular to l at A. So there is just one point equidistant from A, B and l, unless both A and B lie on l, in which case there are none.

Problem 7

Find all reals a, b, c, d such that abc + d = bcd + a = cda + b = dab + c = 2.
Solution

Answer: 1, 1, 1, 1 (one solution); 3, -1, -1, -1 (four solutions).
Either all a, b, c, d are equal, which gives the first solution, or there is an unequal pair. Suppose a ≠ b. But bcd + a = cda + b, so (a - b)(cd - 1) = 0, and hence cd = 1. Hence a + b = 2. It follows that ab ≠ 1 (because the only solution to a + b = 2, ab = 1 is a = b = 1). Hence c = d (since (c - d)(ab - 1) = 0). So either c = d = 1, or c = d = -1. But the first case gives ab = 1, a + b = 2 (contradiction, since the only possibility then is a = b = 1, whereas we are assuming a ≠ b). So c = d = -1. Hence ab = -3, a + b = 2, so a = 3, b = -1 (or vice versa).
Problem 8

For which positive integers n does 61 divide 5ⁿ - 4ⁿ?
Solution

Answer: multiples of 3.
Clearly 5 - 4 = 1 (mod 61), 5² - 4² = 9 (mod 61) and 5³ - 4³ = 0 (mod 61). So it follows that 5^3m = 4^3m (mod 61). In other words 61 does divide 5ⁿ - 4ⁿ if n is a multiple of 3. It remains to show that it does not divide 5ⁿ - 4ⁿ if n = 3m + 1 or 3m + 2.
5^3m+1 = 5 5^3m = 5 4^3m = 4^3m+1 + 4^3m (mod 61). But 61 does not divide 4^3m, and hence it does not divide 5^3m+1 - 4^3m+1. Similarly, 5^3m+2 = 25 5^3m = 25 4^3m = 4^3m+2 + 9 4^3m (mod 61). But 61 does not divide 9 or 4^3m, so it does not divide 5^3m+2 - 4^3m+2.

Problem 10

A collects pre-1900 British stamps and foreign stamps. B collects post-1900 British stamps and foreign special issues. C collects pre-1900 foreign stamps and British special issues. D collects post-1900 foreign stamps and British special issues. What stamps are collected by (1) no one, (2) everyone, (3) A and D, but not B?
Solution

Answer: (1) 1900, non-special, British; (2) none; (3) pre-1900, special, British, and post-1900, non-special, foreign.
Trivial.

Problem 11

The streets for a rectangular grid. B is h blocks north and k blocks east of A. How many shortest paths are there from A to B?
Solution

Answer: (h + k)!/( h! k! ).
A shortest path must involve h moves north and k moves east, a total of h + k moves. So there is a (1, 1) correspondence between shortests paths and subsets size h taken from the set 1, 2, 3, ... , (h + k). [Move i is north if i is in the subset and east otherwise.] But there are (h + k) C h = (h + k)! /( h! k! ) such subsets.