3rd USA Mathematical Olympiad 1974 Problems

3rd USA Mathematical Olympiad 1974 Problems

1.  p(x) is a polynomial with integral coefficients. Show that there are no solutions to the equations p(a) = b, p(b) = c, p(c) = a, with a, b, c distinct integers.

2.  Show that for any positive reals x, y, z we have x^xy^yz^z ≥ (xyz)^a, where a is the arithmetic mean of x, y, z.

3.  Two points in a thin spherical shell are joined by a curve shorter than the diameter of the shell. Show that the curve lies entirely in one hemisphere.

4.  A, B, C play a series of games. Each game is between two players, The next game is between the winner and the person who was not playing. The series continues until one player has won two games. He wins the series. A is the weakest player, C the strongest. Each player has a fixed probability of winning against a given opponent. A chooses who plays the first game. Show that he should choose to play himself against B.

5.  A point inside an equilateral triangle with side 1 is a distance a, b, c from the vertices. The triangle ABC has BC = a, CA = b, AB = c. The sides subtend equal angles at a point inside it. Show that sum of the distances of the point from the vertices is 1. 

Solution

3rd USA Mathematical Olympiad 1974

Problem 1

p(x) is a polynomial with integral coefficients. Show that there are no solutions to the equations p(a) = b, p(b) = c, p(c) = a, with a, b, c distinct integers.

Solution

Suppose there a, b, c satisfy the equations. Then p(x) = (x - a)q(x) + b = (x - b)r(x) + c = (x - c)s(x) + a for some polynomials q(x), r(x), s(x) with integer coefficients. Hence (b - a)q(b) + b = p(b) = c, so (b - a) divides (c - b). Similarly, (c - b) divides (a - c), and (a - c) divides (b - a). But (b - a) divides (c - b) divides (a - c) implies that (b - a) divides (a - c). So we have (b - a) and (a - c) dividing each other. Hence (b - a) = ±(a - c).

If b - a = a - c, then b - c = (b - a) + (a - c) = 2(a - c). But (b - a) divides 2(a - c) (and both are non-zero since a, b, c are distinct), so that is impossible. If b - a = -(a - c), the b = c, contradicting the fact that they are distinct. So there are no solutions. Problem 2

Show that for any positive reals x, y, z we have x^xy^yz^z ≥ (xyz)^a, where a is the arithmetic mean of x, y, z.

Solution

Without loss of generality x ≥ y ≥ z. We have x^xy^y ≥ x^yy^x, because that is equivalent to (x/y)^x ≥ (x/y)^y which is obviously true. Similarly y^yz^z ≥ y^zz^y and z^zx^x ≥ z^xx^z. Multiplying these three together we get (x^xy^yz^z)^x ≥ x^y+zy^z+xz^x+y. Multiplying both sides by x^xy^yz^z gives (x^xy^yz^z)³ ≥ (xyz)^3a. Taking cube roots gives the required result. 

Problem 3

Two points in a thin spherical shell are joined by a curve shorter than the diameter of the shell. Show that the curve lies entirely in one hemisphere.

Solution

Suppose the shell has diameter 2. Let M be the midpoint of the curve. Let O be the center of the shell and X the midpoint of MO. Let S be the circle center X radius √3)/2 in the plane normal to OM. Then S lies in the shell and every point of S is a distance (in space) of 1 from M. Hence the curve cannot cross S (because if it crossed at Y, we would have SY ≥ 1 along the curve, but the curve has length < 2, so SY < 1. So we have a stronger result than required. 

Problem 4

A, B, C play a series of games. Each game is between two players, The next game is between the winner and the person who was not playing. The series continues until one player has won two games. He wins the series. A is the weakest player, C the strongest. Each player has a fixed probability of winning against a given opponent. A chooses who plays the first game. Show that he should choose to play himself against B.

Solution

It must be wrong to choose B against C, for then after the first game (whatever its outcome) A would be playing one of the other players (X), and that player would already have won a game. That is a worse position than playing that person as the first game, because if he loses the game then X has won the series, whereas if he lost to X on the first game, there is still a chance A could win the series. [If he wins the game as the second game, then he is certainly no better off than he would be after winning the match as the first game.]

Use XbY to denote that X beats Y. If A chooses to play B in the first game, then he wins the series if either (1) AbB, AbC, (2) AbB, CbA, BbC, AbB, or (3) BbA, CbB, AbC, AbB. If A plays C in the first game, then he wins the series if (1') AbC, AbB, (2') AbC, BbA, CbB, AbC, (3') CbA, BbC, AbB, AbC. Evidently the probability of (1) is the same as the probability of (1'). If we compare (2) and (3') they are the same except that in (2) A must beat B and in (3') A must beat C. Similarly, if we compare (3) and (2') they are the same except that in (3) A must beat B and in (2') A must beat C. We assume that, since C is a stronger player than B, A is more likely to beat B than C. Hence prob(2) > prob(3') and prob(3) > prob(2'). Thus A should choose to play B in the first game. 

Problem 5

A point inside an equilateral triangle with side 1 is a distance a, b, c from the vertices. The triangle ABC has BC = a, CA = b, AB = c. The sides subtend equal angles at a point inside it. Show that sum of the distances of the point from the vertices is 1.

Solution

Let D be the point inside ABC, so that ∠ADB = ∠BDC = 120^o. The key is to start from ABC and to rotate the triangle BDC through 60^o away from the triangle ADB. After that everything is routine.

Suppose D goes to D' and C to C'. Then BD = BD' and ∠DBD' = 60^o, so BDD' is equilateral. Hence ∠D'DB = 60^o. ∠BDA = 120^o, so ADD' is a straight line. Also ∠DD'B = 60^o and ∠C'D'B = 120^o, so DD'C' is a straight line. Thus AC' has length DA + DB + DC.

Note that BC = BC' and ∠CBC' = 60^o, so CBC' is equilateral. Hence ∠CC'B = 60^o. Now take Y such that AC'Y is equilateral, Y is on the opposite side of AC' to C. Then ∠BC'Y = 60^o - ∠AC'B = ∠CC'A. Also BC' = CC' and YC' = AC', so triangles BC'Y and CC'A are congruent. Hence BY = CA = b. Also BC' = BC = a and BA = c. Thus B is a point inside an equilateral triangle and distances a, b, c from the vertices. Hence the triangle must have side 1. So DA + DB + DC = AC' = 1.