5th Mexican Mathematical Olympiad Problems 1991

5th Mexican Mathematical Olympiad Problems 1991

A1.  Find the sum of all positive irreducible fractions less than 1 whose denominator is 1991.

A2.  n is palindromic (so it reads the same backwards as forwards, eg 15651) and n = 2 mod 3, n = 3 mod 4, n = 0 mod 5. Find the smallest such positive integer. Show that there are infinitely many such positive integers.

A3.  4 spheres of radius 1 are placed so that each touches the other three. What is the radius of the smallest sphere that contains all 4 spheres?

B1.  ABCD is a convex quadrilateral with AC perpendicular to BD. M, N, R, S are the midpoints of AB, BC, CD, DA. The feet of the perpendiculars from M, N, R, S to CD, DA, AB, BC are W, X, Y, Z. Show that M, N, R, S, W, X, Y, Z lie on the same circle. 

B2.  The sum of the squares of two consecutive positive integers can be a square, for example 3² + 4² = 5². Show that the sum of the squares of 3 or 6 consecutive positive integers cannot be a square. Give an example of the sum of the squares of 11 consecutive positive integers which is a square.

B3.  Let T be a set of triangles whose vertices are all vertices of an n-gon. Any two triangles in T have either 0 or 2 common vertices. Show that T has at most n members. 

Solutions

Problem A1

Find the sum of all positive irreducible fractions less than 1 whose denominator is 1991.

Answer

900

Solution

1991 = 11·181 (and 181 is prime). So the 180 fractions 11/1991, 22/1991, ... , 11·180/1991 are reducible, and the 10 fractions 181/1991, 363/1991, ... , 1810/1991 are reducible. All the others are irreducible. Thus their sum is (1/1991)(1 + 2 + ... + 1990 - 11(1 + 2 + ... + 180) - 181(1 + 2 + ... + 10)) = (1990·1991/2 - 11·180·181/2 - 181·10·11/2)/1991 = 995 - 90 - 5 = 900. 

Problem A2

n is palindromic (so it reads the same backwards as forwards, eg 15651) and n = 2 mod 3, n = 3 mod 4, n = 0 mod 5. Find the smallest such positive integer. Show that there are infinitely many such positive integers.

Answer

515

Solution

To meet the congruence conditions, n must be 35 mod 60. So it has last digit 5. Hence it must have first digit 5 and hence must be at least 35 + 8·60 = 515, which works. So 515 is the smallest.

Consider 51...15, where there are 3k+1 1s. It is obviously 0 mod 5 and 3 mod 4. The sum of the digits is 2 mod 3 and hence the number is 2 mod 3. It is obviously palindromic. 

Problem A3

4 spheres of radius 1 are placed so that each touches the other three. What is the radius of the smallest sphere that contains all 4 spheres?

Answer

√(3/2) + 1

Solution

The centers of the 4 spheres form a regular tetrahedron side 2. The center of the fifth sphere must obviously be at the center of the tetrahedron. An altitude of the tetrahedron will have one end at the centroid of a face, so its length is √(2² - ((2/3)√3)²) = √(4 - 4/3) = 2√(2/3). So the distance from the center of the fifth sphere to one of the other centers is (3/4)2√(2/3) = √(3/2). Hence the radius of the fifth sphere is √(3/2) + 1. 

Problem B1

ABCD is a convex quadrilateral with AC perpendicular to BD. M, N, R, S are the midpoints of AB, BC, CD, DA. The feet of the perpendiculars from M, N, R, S to CD, DA, AB, BC are W, X, Y, Z. Show that M, N, R, S, W, X, Y, Z lie on the same circle.

Solution

MN is a midline of the triangle ABC, so MN is parallel to AC. Similarly, NR is parallel to BD. But AC and BD are perpendicular, so ∠MNR = 90^o. Similarly, ∠NRS = ∠RSM = ∠SMN = 90^o, so MNRS is a rectangle and hence cyclic. Moreover, MR and NS are diameters of the circle. But ∠MWR = 90^o, so W also lies on the circle. Similarly for X, Y, Z. 

Problem B2

The sum of the squares of two consecutive positive integers can be a square, for example 3² + 4² = 5². Show that the sum of the squares of 3 or 6 consecutive positive integers cannot be a square. Give an example of the sum of the squares of 11 consecutive positive integers which is a square.

Solution

(n-1)² + n² + (n+1)² = 3n² + 2 = 2 mod 3, but a square must be 0 or 1 mod 3.

Given 6 consecutive positive integers, 3 must be even and 3 must be odd, hence the sum of their squares must be 3 mod 4, so it cannot be a square. (An odd square is 1 mod 4 and an even square is 0 mod 4).

18² + 19² + ... + 28² = 77².