1st International Mathematical Olympiad 1959 Problems & Solutions

A1.  Prove that (21n+4)/(14n+3) is irreducible for every natural number n.

A2.  For what real values of x is √(x + √(2x-1)) + √(x - √(2x-1)) = A given (a) A = √2, (b) A = 1, (c) A = 2, where only non-negative real numbers are allowed in square roots and the root always denotes the non-negative root?

A3.  Let a, b, c be real numbers. Given the equation for cos x:

        a cos²x + b cos x + c = 0,

form a quadratic equation in cos 2x whose roots are the same values of x. Compare the equations in cos x and cos 2x for a=4, b=2, c=-1.

B1.  Given the length |AC|, construct a triangle ABC with ∠ABC = 90^o, and the median BM satisfying BM² = AB·BC.

B2.  An arbitrary point M is taken in the interior of the segment AB. Squares AMCD and MBEF are constructed on the same side of AB. The circles circumscribed about these squares, with centers P and Q, intersect at M and N.

  (a) prove that AF and BC intersect at N;

  (b) prove that the lines MN pass through a fixed point S (independent of M);

  (c) find the locus of the midpoints of the segments PQ as M varies.

B3.  The planes P and Q are not parallel. The point A lies in P but not Q, and the point C lies in Q but not P. Construct points B in P and D in Q such that the quadrilateral ABCD satisfies the following conditions: (1) it lies in a plane, (2) the vertices are in the order A, B, C, D, (3) it is an isosceles trapezoid with AB parallel to CD (meaning that AD = BC, but AD is not parallel to BC unless it is a square), and (4) a circle can be inscribed in ABCD touching the sides. 

Solutions

Problem A1

Prove that (21n+4)/(14n+3) is irreducible for every natural number n.

Solution

3(14n+3) - 2(21n+4) = 1.

Problem A2

For what real values of x is √(x + √(2x-1)) + √(x - √(2x-1)) = A, given (a) A = √2, (b) A = 1, (c) A = 2, where only non-negative real numbers are allowed in square roots and the root always denotes the non-negative root?

Answer

(a) any x in the interval [1/2,1]; (b) no solutions; (c) x=3/2.

Solution

Note that we require x ≥ 1/2 to avoid a negative sign under the inner square roots. Since (x-1)² ≥ 0, we have x ≥ √(2x-1), so there is no difficulty with √(x - √(2x-1)), provided that x ≥ 1/2.

Squaring gives 2x + 2√(x²-2x+1) = A². Note that the square root is |x-1|, not simply (x-1). So we get finally 2x + 2|x-1| = A². It is now easy to see that we get the solutions above. 

Problem A3

Let a, b, c be real numbers. Given the equation for cos x:

        a cos²x + b cos x + c = 0,

form a quadratic equation in cos 2x whose roots are the same values of x. Compare the equations in cos x and cos 2x for a=4, b=2, c=-1.

Solution

You need that cos 2x = 2 cos²x - 1. Some easy manipulation then gives:

a²cos²2x + (2a² + 4ac - 2b²) cos 2x + (4c² + 4ac - 2b² + a²) = 0.

The equations are the same for the values of a, b, c given. The angles are 2π/5 (or 8π/5) and 4π/5 (or 6π/5). 

Problem B1

Given the length |AC|, construct a triangle ABC with ∠ABC = 90^o, and the median BM satisfying BM² = AB·BC.

Solution

Area = AB·BC/2 (because ∠ABC = 90^o= BM²/2 (required) = AC²/8 (because BM = AM = MC), so B lies a distance AC/4 from AC. Take B as the intersection of a circle diameter AC with a line parallel to AC distance AC/4.

 

 Problem B2

An arbitrary point M is taken in the interior of the segment AB. Squares AMCD and MBEF are constructed on the same side of AB. The circles circumscribed about these squares, with centers P and Q, intersect at M and N.
  (a) prove that AF and BC intersect at N;
  (b) prove that the lines MN pass through a fixed point S (independent of M);
  (c) find the locus of the midpoints of the segments PQ as M varies.

Solution

(a)   ∠ANM = ∠ACM = 45^o. But ∠FNM = ∠FEM = 45^o, so A, F, N are collinear. Similarly, ∠BNM = ∠BEM = 45^o and ∠CNM = 180^o - ∠CAM = 135^o, so B, N, C are collinear.

(b)   Since ∠ANM = ∠BNM = 45^o, ∠ANB = 90^o, so N lies on the semicircle diameter AB. Let NM meet the circle diameter AB again at S. ∠ANS = ∠BNS implies AS = BS and hence S is a fixed point.

(c)   Clearly the distance of the midpoint of PQ from AB is AB/4. Since it varies continuously with M, it must be the interval between the two extreme positions, so the locus is a segment length AB/2 centered over AB.

 Problem B3

The planes P and Q are not parallel. The point A lies in P but not Q, and the point C lies in Q but not P. Construct points B in P and D in Q such that the quadrilateral ABCD satisfies the following conditions: (1) it lies in a plane, (2) the vertices are in the order A, B, C, D, (3) it is an isosceles trapezeoid with AB is parallel to CD (meaning that AD = BC, but AD is not parallel to BC unless it is a square), and (4) a circle can be inscribed in ABCD touching the sides.

Solution

Let the planes meet in the line L. Then AB and CD must be parallel to L. Let H be the foot of the perpendicular from C to AB. The fact that a circle can be inscribed implies AB + CD = BC + AD (equal tangents from A, B, C, D to the circle). Also CD = AB ± 2BH. This leads to AH = AD = BC.

The construction is now easy. First construct the point H. Then using the circle center C radius AH, construct B. Using the circle center A radius AH construct D.

Note that if CH > AH then no construction is possible. If CH < AH, then there are two solutions, one with AB > CD, the other with AB < CD. If CH = AH, then there is a single solution, which is a square.

Solutions are also available in:   Samuel L Greitzer, International Mathematical Olympiads 1959-1977, MAA 1978, and in   István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48-X.