13th Mexican Mathematical Olympiad Problems 1999

13th Mexican Mathematical Olympiad Problems 1999

A1.  1999 cards are lying on a table. Each card has a red side and a black side and can be either side up. Two players play alternately. Each player can remove any number of cards showing the same color from the table or turn over any number of cards of the same color. The winner is the player who removes the last card. Does the first or second player have a winning strategy?

A2.  Show that there is no arithmetic progression of 1999 distinct positive primes all less than 12345.

A3.  P is a point inside the triangle ABC. D, E, F are the midpoints of AP, BP, CP. The lines BF, CE meet at L; the lines CD, AF meet at M; and the lines AE, BD meet at N. Show that area DNELFM = (1/3) area ABC. Show that DL, EM, FN are concurrent.

B1.  10 squares of a chessboard are chosen arbitrarily and the center of each chosen square is marked. The side of a square of the board is 1. Show that either two of the marked points are a distance ≤ √2 apart or that one of the marked points is a distance 1/2 from the edge of the board.

B2.  ABCD has AB parallel to CD. The exterior bisectors of ∠B and ∠C meet at P, and the exterior bisectors of ∠A and ∠D meet at Q. Show that PQ is half the perimeter of ABCD.

B3.  A polygon has each side integral and each pair of adjacent sides perpendicular (it is not necessarily convex). Show that if it can be covered by non-overlapping 2 x 1 dominos, then at least one of its sides has even length.

Solutions

Problem A1

1999 cards are lying on a table. Each card has a red side and a black side and can be either side up. Two players play alternately. Each player can remove any number of cards showing the same color from the table or turn over any number of cards of the same color. The winner is the player who removes the last card. Does the first or second player have a winning strategy?

Solution

We show that if an equal number of red and black cards are on the table, then the next player loses, and in all other cases the next player wins.

If there are an equal number, then whatever move the next player makes he must leave unequal numbers of red and black cards. In particular, he cannot immediately take the last card. On the other hand, if there are an unequal number, then the next player can take enough of the larger number to equalise. The game terminates because one player is always taking.

Since 1999 is odd, there must be unequal numbers, so the first player wins by always taking cards (not turning cards) and always leaving equal numbers of red and black cards.

Problem A2

Show that there is no arithmetic progression of 1999 distinct positive primes all less than 12345.

Solution

Let the progression be a, a+d, a+2d, ... , a+1998d. Note that a+ad is composite, so we must have a > 1998. 1999 + 6·1998 > 12345, so the difference d of the progression must be less than 6. It must be even (or alternate terms will be even). So d = 2 or 4. But a = 1 or 2 mod 3, so either a+4 or a+8 is a multiple of 3. Contradiction.

Problem A3

P is a point inside the triangle ABC. D, E, F are the midpoints of AP, BP, CP. The lines BF, CE meet at L; the lines CD, AF meet at M; and the lines AE, BD meet at N. Show that area DNELFM = (1/3) area ABC. Show that DL, EM, FN are concurrent.

Solution

We use vectors. Take P to be the origin. Let the vectors PA, PB, PC be a, b, c. Then PD is a/2. L is the centroid of the triangle PBC, so PL is (2/3)(b+c)/2 = (b+c)/3. Take Q to be the point (a+b+c)/5. Then since (a+b+c)/5 = (2/5)a/2 + (3/5)(b+c)/3, Q lies on DL. Similarly, Q lies on EM and FN.

Note that area ANB = (1/3) area APB, area ADB = area BEA = (1/2) area APB. Hence area PDNE = (1 - 1/2 - 1/2 + 1/3) area APB = (1/3) area APB. Similarly for PELF and PFMD. Hence area DNELFM = (1/3) area ABC.

Problem B2
ABCD has AB parallel to CD. The exterior bisectors of ∠B and ∠C meet at P, and the exterior bisectors of ∠A and ∠D meet at Q. Show that PQ is half the perimeter of ABCD.
Solution

Let the internal bisectors at B and C meet at P'. Let PP' and BC meet at M. The exterior angle at B equals the interior angle at C, so BP is parallel to CP'. Similarly BP' is parallel to CP, so BPCP' is a parallelogram. But CP' and CP are perpendicular, and similarly BP and BP', so BPCP' is a rectangle. Hence M is the midpoint of BC and ∠DCP' = ∠MCP' = ∠MP'C, so MP is parallel to CD. Similarly, if N is the midpoint of AD, then QN is parallel to BC, and so P and Q both lie on MN.

But MN = (AB+CD)/2. Since ∠BPC = 90^o and M is the midpoint of BC, M is the the circumcenter of BCP and so MP = BC/2. Similarly, NQ = AD/2. Hence PQ = perimeter/2.