Irish Intervarsity Mathematics Competition 2015

Transcription

1 Irish Intervarsity Mathematics Competition 215 Trinity College Dublin Vladimir Dotsenko & Timothy Murphy Saturday, 28 March 215 9:3 12:3 1. Each point of a circle is coloured red or blue. Show that there exists an isosceles triangle with vertices on the circle which is monochromatic (ie all its vertices are the same color). Answer: Take a regular pentagon ABCDE on the circle. Two adjacent points, say A and B, must be the same colour, say blue. Then C must be red, otherwise ABC is monocrome and isosceles. Similarly E is red. If now D is blue then ABD is a monocrome and isosceles; while if D is red then CDE is a monocrome and isosceles. 2. Does 215 n end with the digits 215 for any integer n > 1? Answer: Since n if n > 1, 215 n will end in, 25, 5 or 75, and so cannot end in Does 215 n begin with the digits 215 for any integer n > 1? Answer: Suppose 215 n begins with the digits 215, and has m digits after these four. Then Taking logs to base 1, m 215 n < m. log m n log 215 < log m. Conversely, if we can find integers m, n with n > satisfying this inequality then 215 n will start with the digits

2 Writing ln x for the natural logarithm, for if < x < 1/2 then since log(216) = log(215) + log(1 + 1/215) > log(215) + ln(1 + 1/215) > log(215) + 1/215; ln(1 + x) = x x 2 /2 + x 3 /3 < x x 3 /3 x 4 /4 + < x 3 + x 4 + = x3 1 x < x2 /2. Thus it is sufficient to find integers m, n with n > 1 such that m (n 1) log 215 < m + 1/215. We will show that given any real number θ and any ɛ > we can find integers m, n with n > 1 such that nθ m < ɛ. To see this, take N > 1/ɛ and consider the fractional parts of, θ,..., Nθ. By the pigeon-hole principle, two if these must differ by < 1/N, say {rθ} < {sθ} < {rθ} + ɛ. Setting n = r s, We may have nθ m < ɛ. ɛ < nθ m <. But in that case we can find a larger n such that ɛ = nθ m < n θ m < ɛ ; and then either n or n n will give the required result. 4. Suppose A is an invertible square matrix in which each row has exactly one nonzero entry ±1. Show that A k = A T for some k >, where A T is the transpose of A. 2

3 Answer: Since the n n-matrix A has n nonzero entries, and is nonsingular, it must have one nonzero entry ±1 in each column, as in each row. Thus there is a permutation π S n { ±1 if j = π(i) a ij = otherwise It follows that the matrices of this type form a group G of order 2 n n!. If A has order d then B = A d 1 = A T, since AB = I, so each entry a ij = ±1 is multiplied by an equal entry a ji to give a diagonal entry 1 in I. 5. The sequence a n is defined by a n+1 = a 2 n a n 1 for n >, with a = and a 1 = 1. Show that 215 divides infinitely many numbers in the sequence. Answer: Consider the pairs (a n mod 215, a n+1 mod 215). Since there are at most such pairs, there must be a repeat. Suppose the first repeat is a m a n mod 215, a m+1 a m mod 215, where m < n. If m > then it follows from the recursive relation that a m 1 a n 1, contradiction the assumption that this is the first repeat. We conclude that m =, so that a n a m = mod 215, a n+1 a n = 1 mod 215. Thus 215 a n, and it follows from the argument that for t = 1, 2, 3, a tn 6. Show that there do not exist polynomials p 1 and p 2 in x 1, x 2, x 3 with real coefficients for which we have p p 2 2 = x x x

4 Answer: The polynomials p 1, p 2 must be linear; for suppose x i 1x j 2x k 3 is a term of maximal total order i + j + k with coefficients c 1, c 2 in p 1, p 2. Then x 2i 1 x 2j 2 x 2k 3 occurs with coefficient c c 2 2 > on the left, and does not occur on the right. Also p 1, p 2 have zero constant term, for the same reason. Thus p 1, p 2 are 2 homogeneous linear forms in 3 unknowns, and so have a common non-trivial zero x 1, x 2, x 3 which will give a contradiction, since x x x 2 3 >. 7. Does there exist a continuous function f : R R for which f(x) is rational whenever x is irrational, and irrational whenever x is rational? Justify your answer. Answer: Evidently f(x) is not constant. (For if it were, it would either be always rational or else always irrational). Suppose f(a) = u, f(b) = v, where a < b, then f(x) takes all values in [u, v] (or [v, u]) as x increases from a to b. But the set of rational numbers is enumerable, so the set of numbers S = {f(x) : x rational} is enumerable. But the set of irrational numbers in this range is nonenumerable, so they cannot all be included in S. 8. Three points A, B, C are chosen at random on the circumference of a circle. What is the probability that the centre of the circle lies inside ABC? Answer: Let P denote the antipodal point to P (so that P P passes through the centre O of the circle). Choose a point A on the circle, draw two random diameters UU, V V and choose points B {U, U }, C {V, V }, at random. The triangle ABC contains the centre if and only if A lies in the arc BC. Thus the probability that the triangle contains the centre O is 1/4. 9. Show that if n points P 1,..., P n lie on the surface of a sphere of radius 1 then the 2 closest points lie at a distance less that 4/ n from each other (measured along the surface of the sphere). Answer: Suppose the n points are all at distance > d from each other. Then the circles of radius d/2 around each point do not overlap. Each of these circles has area > d 2 π/4. Since the surface area of the sphere is 4π, nd 2 π 4 < 4π = d < 4 n. 4

5 1. Compute the integral 1 (e x + 1)(x 2 + 1). Answer: We have 1 I = 1 (e x + 1)(x 2 + 1) + (e x + 1)(x 2 + 1) = J + K, say. Setting x = y, Thus dy J = 1 (e y + 1)(y 2 + 1) e y dy = (1 + e y )(y 2 + 1). I = = (e x + 1) (e x + 1)(x 2 + 1) x = [arctan x] 1 = π/4. 5