20th British Mathematical Olympiad 1984 Problems

1.  In the triangle ABC, ∠C = 90^o. Find all points D such that AD·BC = AC·BD = AB·CD/√2.

2.  ABCD is a tetrahedron such that DA = DB = DC = d and AB = BC = CA = e. M and N are the midpoints of AB and CD. A variable plane through MN meets AD at P and BC at Q. Show that AP/AD = BQ/BC. Find the value of this ratio in terms of d and e which minimises the area of MQNP.

3.  Find the maximum and minimum values of cos x + cos y + cos z, where x, y, z are non-negative reals with sum 4π/3.

4.  Let b_n be the number of partitions of n into non-negative powers of 2. For example b⁴ = 4: 1 + 1 + 1 + 1, 1 + 1 + 2, 2 + 2, 4. Let c_n be the number of partitions which include at least one of every power of 2 from 1 up to the highest in the partition. For example, c₄ = 2: 1 + 1 + 1 + 1, 1 + 1 + 2. Show that b_n+1 = 2c_n.

5.  Show that for any positive integers m, n we can find a polynomial p(x) with integer coefficients such that | p(x) - m/n | ≤ 1/n² for all x in some interval of length 1/n. 

Problem 1

In the triangle ABC, ∠C = 90^o. Find all points D such that AD·BC = AC·BD = AB·CD/√2.

Solution

Take rectangular coordinates with C as (0,0), A as (a,0), B as (0,b). Take D as (x,y). Then ((x-a)²+y²)b² = a²(x²+(y-b)²) = (a²+b²)(x²+y²)/2.

Hence (a²-b²)(x²+y²) + 2ab(bx-ay) = 0 and (a²-b²)(x²+y²) - 4a²by + 2a²b² = 0. Hence bx + ay = ab. So D lies on the line AB. Substituting back, we find (x,y) = (ab/(a+b),ab/(a+b)) or (ab/(a-b),ab/(a-b)).

Geometrically, the two points are the intersections of the internal and external angle bisectors of angle C with the line AB. 

Problem 2

ABCD is a tetrahedron such that DA = DB = DC = d and AB = BC = CA = e. M and N are the midpoints of AB and CD. A variable plane through MN meets AD at P and BC at Q. Show that AP/AD = BQ/BC. Find the value of this ratio in terms of d and e which minimises the area of MQNP.

Answer

(d²+e²)/4d²

Solution

 

Call the plane Π. If Π is parallel to BD, then P must be the midpoint of AD, and Q must be the midpoint of BC, so AP/AD = BQ/BC = 1/2. So assume it is not parallel and let the point of intersection be X. The line PM lies in the plane ABD and in Π so it must be their intersection. But X must lie in the intersection of ABD and Π, so it must lie on PM. Similarly it must lie on QN. Now applying Menelaus to the line XNQ and triangle BCD we get (BQ/QC)(CN/ND)(DX/XB) = 1. Applying it to line MPX and triangle ABD we get (AP/PD)(DX/XB)(BM/MA) = 1. Hence AP/PD = BQ/QC and hence AP/AD = BQ/BC.

It follows that the midpoint of PQ must lie on MN. That is obvious if we use vectors, because P = λA + (1-λ)D, Q = λB + (1-λ)C, so ½P + ½Q = λM + (1-λ)N. Hence area PNM = area QNM and it is sufficient to minimise area PNM. 

Take G to be the midpoint of MN and take P to be the foot of the perpendicular from G to AD. Then G is the centroid of ABCD. Since ABC is equilateral and DA = DB = DC, the plane ADG is perpendicular to BC. Hence GP is perpendicular to BC. Let L be the midpoint of AC, then LM is parallel to BC and hence perpendicular to GP. Also NL is parallel to AD which is perpendicular to GP, so NL is perpendicular to GP. So the plane LMN is perpendicular to GP and hence MN is perpendicular to GP. So GP is the shortest distance between the lines AD and MN and hence this choice of P minimises area MNP.

The line DG meets the plane ABC at the center O of the triangle ABC. ∠AOG = ∠APG = 90^o, so AOGP is cyclic and hence DP·DA = DG·DO = (3/4)DO² = (3/4)(AD² - AO²) = (3/4)(d² - e²/3). Hence DP = (3d²-e²)/(4d) and AP = (d²+e²)/4d, so the required ratio AP/AD = (d²+e²)/4d².

Problem 3

Find the maximum and minimum values of cos x + cos y + cos z, where x, y, z are non-negative reals with sum 4π/3.

Answer

max 3/2 at 0,0,4π/3
min 0 at 0,2π/3,2π/3

Solution

wlog x ≤ y ≤ z. So 4π/9 ≤ z ≤ 4π/3. Put z = 4π/3 - k, so 0 ≤ k ≤ 8π/9. Then cos x + cos y + cos z = cos(4π/3 - k) + 2 cos k/2 cos(y-x)/2. Since cos(k/2) ≥ 0, the expression is maximised by taking x = y so that cos(y-x)/2 = 1. That gives cos x + cos y + cos z = cos(4π/3 - k) + 2 cos k/2 = -1/2 cos k - (√3)/2 sin k + 2 cos k/2 = - cos2k/2 + 1/2 + 2 cos k/2 - (√3)/2 sin k = 3/2 - (cos k/2 - 1)2 - (√3)/2 sin k. Since k < π, sin k ≥ 0, so we maximise this expression by taking k = 0 which gives (cos k/2 - 1) = 0 and sin k = 0.

We minimise cos(4π/3 - k) + 2 cos k/2 cos(y-x)/2 by taking cos(y-x)/2 as small as possible, in other words by taking x = 0, y = k, giving cos x + cos y + cos z = cos(4π/3 - k) + 2 cos²k/2 = cos(4π/3 - k) + cos k + 1 = 1 + 2 cos 2π/3 cos(2π/3 - k) = 1 - cos(2π/3 - k), which is minimised by taking k = 2π/3 to give 0. 

Problem 4

Let b_n be the number of partitions of n into non-negative powers of 2. For example b⁴ = 4: 1 + 1 + 1 + 1, 1 + 1 + 2, 2 + 2, 4. Let c_n be the number of partitions which include at least one of every power of 2 from 1 up to the highest in the partition. For example, c₄ = 2: 1 + 1 + 1 + 1, 1 + 1 + 2. Show that b_n+1 = 2c_n.

Solution

Let C_n be the set of partitions of n into non-negative powers of 2, where there is at least one of every power of 2 from 1 up to the highest in the partition, so |C_n| = c_n. Let D_n be the set of partitions of n into non-negative powers of 2, where there is just one of the highest power of 2 (for example 4 = 4 belongs to D₄, but 4 = 2 + 2 does not). Let E_n be the set of partitions of n into non-negative power of 2, where there is more than one of the highest power of 2. Obviously b_n+1 = |D_n+1| + |E_n+1|. We show that |D_n+1| = |E_n+1| = c_n.

Given a partition n = ∑₀^k a_i2ⁱ in C_n, we see that n+1 = ∑₀^k (a_i-1)2ⁱ + 2^k+1 is a partition in D_n+1. Moreover this is obviously a bijection between the two sets. Hence |D_n+1| = |C_n|. Similarly, given a partition n = ∑₀^k a_i2ⁱ in C_n, we see that n+1 = ∑₀^k-1 (a_i-1)2ⁱ + (a_k+1)2^k is a partition in E_n+1, and again this is obviously a bijection between the two sets. Hence |E_n+1| = |C_n|. 

Problem 5

Show that for any positive integers m, n we can find a polynomial p(x) with integer coefficients such that | p(x) - m/n | ≤ 1/n² for all x in some interval of length 1/n.

Solution

Take p(x) = (m/n) (nx - 1)^2k+1 + m/n. Exanding by the binomial theorem, it is obvious that p(x) has integer coefficients. Take the interval to be 1/n ± 1/2n. Then we have |p(x) - m/n| ≤ (m/n) 1/2^2k+1, which can be made arbitrarily small (and certainly < 1/n²) for sufficiently large k.