7th Mexican Mathematical Olympiad Problems 1993

7th Mexican Mathematical Olympiad Problems 1993

A1.  ABC is a triangle with ∠A = 90^o. Take E such that the triangle AEC is outside ABC and AE = CE and ∠AEC = 90^o. Similarly, take D so that ADB is outside ABC and similar to AEC. O is the midpoint of BC. Let the lines OD and EC meet at D', and the lines OE and BD meet at E'. Find area DED'E' in terms of the sides of ABC.

A2.  Find all numbers between 100 and 999 which equal the sum of the cubes of their digits.

A3.  Given a pentagon of area 1993 and 995 points inside the pentagon, let S be the set containing the vertices of the pentagon and the 995 points. Show that we can find three points of S which form a triangle of area ≤ 1.

B1.  f(n,k) is defined by (1) f(n,0) = f(n,n) = 1 and (2) f(n,k) = f(n-1,k-1) + f(n-1,k) for 0 < k < n. How many times do we need to use (2) to find f(3991,1993)?

B2.  OA, OB, OC are three chords of a circle. The circles with diameters OA, OB meet again at Z, the circles with diameters OB, OC meet again at X, and the circles with diameters OC, OA meet again at Y. Show that X, Y, Z are collinear.

B3.  p is an odd prime. Show that p divides n(n+1)(n+2)(n+3) + 1 for some integer n iff p divides m² - 5 for some integer m. 

Solutions

Problem A2

Find all numbers between 100 and 999 which equal the sum of the cubes of their digits.

Answer

153, 407,

Solution

This seems to require one to look at a large number of cases. There are two basic approaches: to deal with each value of a in turn (for example a = 9, so 171 < b3+c3 < 271, so one of b,c is 5 or 6. If one is 5 etc); or to deal with each value of c in turn. The latter requires more cases but less thought:

The idea is that if we fix c and one other digit, then the last digit is fixed by the requirement that a³+b³+c³ = c mod 10. Thus, if c = 0 or 1, then the other two digits must be 1,9 giving 73x, or 2,8 giving 52x, or 3,7 giving 40x, or 4,6 giving 28x, or 5,5 giving 25x, none of which work.

If c = 2, then the other two digits must be 0,4 giving 72, or 1,7 giving 353, or 2,6 giving 232, or 3,3 giving 62, or 5,9 giving 862, or 8,8 giving 1032, none of which work.

If c = 3, then the other two digits must be 0,6 giving 243, or 1,5 giving 153, or 2,2 giving 43, or 3,9 giving 783, or 4,8 giving 603, or 7,7 giving 713. Only 153 works.

If c = 4, then 1,9 →794, 2,8 →584, 3,7 →464, 4,6 →344, 5,5 →314, none of which work.

If c = 5, then 1,9 →855, 2,8 → 645, 3,7 → 525, 4,6 → 405, 5,5 →375, none of which work.

If c = 6, then 1,9 → 946, 2,8 → 736, 3,7 → 616, 4,6 →496, 5,5 → 466, none of which work.

If c = 7, then 1,7 → 687, 2,6 → 567, 3,3 → 397, 4,0 → 407, 5,9 → 1197, 8,8 → 1367, and only 407 works.

If c = 8, then 0,6 → 728, 1,5 → 638, 2,2 → 528, 3,9 → 1268, 4,8 → 1088, 7,7 → 1198, none of which work.

If c = 9, then 1,9 and 2,8, and 3,7, and 4,6 give results which are too big and 5,5 → 979, which does not work.

Problem A3

Given a pentagon of area 1993 and 995 points inside the pentagon, let S be the set containing the vertices of the pentagon and the 995 points. Show that we can find three points of S which form a triangle of area ≤ 1.

Solution

If any three points are collinear, then we have a triangle of area nil. So assume no three points are collinear. We show that there are 1993 non-overlapping triangles. By drawing two diagonals we divide the pentagon into 3 non-overlapping triangles. Now take each of the inside points in turn. Each point P must lie inside one of the existing triangles ABC. So we can join P to A, B, C, thus increasing the number of triangles by 2. Repeating for each of the inside points gives a total of 3 + 2·995 = 1993 non-overlapping triangles. Hence at least one of these triangles must have area ≤ 1993/1993 = 1.

Thanks to Vikram Nayak

Problem B2

OA, OB, OC are three chords of a circle. The circles with diameters OA, OB meet again at Z, the circles with diameters OB, OC meet again at X, and the circles with diameters OC, OA meet again at Y. Show that X, Y, Z are collinear.

Solution

This is just the Simson line.

Note that X, Y, Z are the feet of the perpendiculars from O to BC, CA, AB (because ∠OXC = ∠OBX = 90^o etc). Now OXCY is cyclic, so ∠OYX = ∠OCX. Similarly, OYZA is cyclic, so ∠OYZ = 180^o - ∠OAB. But OABC is cyclic, so ∠OAB = ∠OCX. Hence ∠OYX + ∠OYZ = 180^o, and so X,Y,Z are collinear.

The Simson line result is that the feet of the perpendiculars from K to the three sides of a triangle are collinear iff K lies on the circumcircle.

Problem B3

p is an odd prime. Show that p divides n(n+1)(n+2)(n+3) + 1 for some integer n iff p divides m² - 5 for some integer m.

Solution

n(n+1)(n+2)(n+3) + 1 = (n²+3n+1)², so if p divides n(n+1)(n+2)(n+3) + 1 then it must divide n²+3n+1 and hence 4(n²+3n+1) = (2n+3)² - 5.

Conversely, suppose an odd prime p divides m² - 5. Note that m cannot be 1, 2, 3. If m = 0, then we can take n = 1 (because p must be 5 which divides 4! + 1). Now if m is odd and ≥ 5, then we can write m = 2n+3 with n ≥ 1 and hence p divides n(n+1)(n+2)(n+3) + 1. If m is even, then p must also divide k² - 5, where k = |p-m|, which is odd.

Thanks to Suat Namli