38th British Mathematical Olympiad 2002 Problems

38th British Mathematical Olympiad 2002 Problems

1.  From the foot of an altitude in an acute-angled triangle perpendiculars are drawn to the other two sides. Show that the distance between their feet is independent of the choice of altitude.

2.  n people wish to sit at a round table which has n chairs. The first person takes an seat. The second person sits one place to the right of the first person, the third person sits two places to the right of the second person, the fourth person sits three places to the right of the third person and so on. For which n is this possible?

3.  The real sequence x₁, x₁, x₂, ... is defined by x₀ = 1, x_n+1 = (3x_n + √(5x_n² - 4) )/2. Show that all the terms are integers.

4.  S₁, S₂, ... , S_n are spheres of radius 1 arranged so that each touches exactly two others. P is a point outside all the spheres. Let x₁, x₂, ... , x_n be the distances from P to the n points of contact between two spheres and y₁, y₂, ... , y_n be the lengths of the tangents from P to the spheres. Show that x₁x₂ ... x_n ≥ y₁y₂ ... y_n. 

Solutions

Problem 1

From the foot of an altitude in an acute-angled triangle perpendiculars are drawn to the other two sides. Show that the distance between their feet is independent of the choice of altitude.

Solution

Let the triangle be ABC and the altitude AD. Let M be the midpoint of AD, and E, F the feet of the perpendiculars from D to AB and AC respectively. Angles AED and AFD are 90 deg, so AEDF is cyclic and M is the center of the circle. Hence angle EMF = 2 angle A and EF = 2 ME sin A = AD sin A = 2 (area ABC) (sin A)/BC. Similarly the distances obtained starting with the other two altitudes are 2 (area ABC) (sin B)/AC and 2 (area ABC)(sin C)/AB. But these are all equal by the sine rule. 

Problem 2

n people wish to sit at a round table which has n chairs. The first person takes an seat. The second person sits one place to the right of the first person, the third person sits two places to the right of the second person, the fourth person sits three places to the right of the third person and so on. For which n is this possible?

Solution

Answer: n = 2^m for m a non-negative integer.

Label the seats 0, 1, 2, ... , n-1. The kth person sits in seat k(k - 1)/2 for k = 1, 2, ... , n. So the seating plan is possible iff the n numbers k(k - 1)/2 are incongruent mod n for k = 1, 2, ... , n. Suppose first that n = 2^m. We have h(h - 1)/2 - k(k - 1)/2 = (h - k)(h + k - 1)/2. If h and k are unequal, then h - k and (h + k - 1) are non-zero. But they have opposite parity, and both are less than 2n, so one is odd and the other is not divisible by a higher power of 2 than 2^m. Thus (h - k)(h + k - 1)/2 cannot be divisible by 2^m. So the seating plan works for n a power of 2.

Conversely, suppose n = 2^m(2a + 1). If a <= 2^m, take h = 2^m + a + 1, k = 2^m - a. Then h - k = 2a + 1, h + k - 1 = 2^m+1, so h(h - 1)/2 and k(k - 1)/2 are equal mod n. If a > 2^m, then take h = 2^m + a + 1, k = a - 2^m + 1. Then h - k = 2^m+1, h + k - 1 = 2a + 1. So again h(h - 1)/2 and k(k - 1)/2 are equal mod n. So no seating plan is possible if n is not a power of 2. 

Problem 3

The real sequence x₁, x₁, x₂, ... is defined by x₀ = 1, x_n+1 = (3x_n + √(5x_n² - 4) )/2. Show that all the terms are integers.

Solution

(2x_n+1 - 3x_n)² = 5x_n² - 4, so x_n+1² - 3x_n+1x_n + x_n² = -1. Subtracting the corresponding equation for n-1 gives: x_n+1² - 3x_n(x_n+1 - x_n-1) - x_n-1² = 0. So (x_n+1 - x_n)(x_n+1 - 3x_n + x_n-1) = 0. But it is obvious from the definition that x_n+1 >= 3/2 x_n, so x_n+1 - x_n is non-zero. Hence x_n+1 = 3x_n - x_n-1. But x₀ = 1, x₁ = 2, so by a trivial induction all x_n are integral. 

Problem 4

S₁, S₂, ... , S_n are spheres of radius 1 arranged so that each touches exactly two others. P is a point outside all the spheres. Let x₁, x₂, ... , x_n be the distances from P to the n points of contact between two spheres and y₁, y₂, ... , y_n be the lengths of the tangents from P to the spheres. Show that x₁x₂ ... x_n ≥ y₁y₂ ... y_n.

Solution

It is sufficient to show that if two circles centers O and O' both with radius 1 touch at X and P is a point outside the circles with tangents PY, PZ to the two circles, then PY·PZ ≤ PX². Let the angle between PX and the line of centers be θ and the distances PX, PY, PZ be x, y, z respectively. By the cosine formula, PO² = x² + 1 - 2x cos θ, so y² = x² - 2x cos θ. Similarly, z² = x² + 2x cos θ. Hence y²z² = x⁴ - 4x²cos²θ ≤ x⁴.