7th International Mathematical Olympiad 1965 Problems & Solutions

A1.  Find all x in the interval [0, 2p] which satisfy:         2 cos x = |v(1 + sin 2x) - v(1 - sin 2x)| = v2.

A2.  The coefficients a_ij of the following equations         a₁₁x₁ + a₁₂ x₂+ a₁₃ x₃ = 0

        a₂₁x₁ + a₂₂x₂ + a₂₃x₃ = 0

        a₃₁x₁ + a₃₂x₂ + a₃₃x₃ = 0

satisfy the following: (a) a₁₁, a₂₂, a₃₃ are positive, (b) other a_ij are negative, (c) the sum of the coefficients in each equation is positive. Prove that the only solution is x₁ = x₂ = x₃ = 0.

A3.  The tetrahedron ABCD is divided into two parts by a plane parallel to AB and CD. The distance of the plane from AB is k times its distance from CD. Find the ratio of the volumes of the two parts.

B1.  Find all sets of four real numbers such that the sum of any one and the product of the other three is 2.

B2.  The triangle OAB has ?O acute. M is an arbitrary point on AB. P and Q are the feet of the perpendiculars from M to OA and OB respectively. What is the locus of H, the orthocenter of the triangle OPQ (the point where its altitudes meet)? What is the locus if M is allowed to vary over the interior of OAB?

B3.  Given n > 2 points in the plane, prove that at most n pairs of points are the maximum distance apart (of any two points in the set). 

Solutions

Problem A1

Find all x in the interval [0, 2π] which satisfy:

        2 cos x ≤ |√(1 + sin 2x) - √(1 - sin 2x)| ≤ √2.

Solution

Let y = |√(1 + sin 2x) - √(1 - sin 2x)|. Then y² = 2 - 2|cos 2x|. If x belongs to [0, π/4] or [3π/4, 5π/4] or [7π/4], then cos 2x is non-negative, so y² = 2 - 2 cos 2x = 4 sin² x, so y = 2|sin x|. We have cos x <= |sin x| except for x in [0, π/4] and [7π/4, 2π]. So that leaves [3π/4, 5π/4] in which we certainly have |sin x| ≤ 1/√2.

If x belongs (π/4, 3π/4) or (5π/5, 7π/4), then cos 2x is negative, so y² = 2 + 2 cos 2x = 4 cos²x. So y = 2 |cos x|. So the first inequality certainly holds. The second also holds.

Thus the inequalities hold for all x in [π/4, 7π/4]. 

Problem A2

The coefficients a_ij of the following equations

        a₁₁x₁ + a₁₂ x₂+ a₁₃ x₃ = 0
        a₂₁x₁ + a₂₂x₂ + a₂₃x₃ = 0
        a₃₁x₁ + a₃₂x₂ + a₃₃x₃ = 0

satisfy the following: (a) a₁₁, a₂₂, a₃₃ are positive, (b) other a_ij are negative, (c) the sum of the coefficients in each equation is positive. Prove that the only solution is x₁ = x₂ = x₃ = 0.

Solution

The slog solution is to multiply out the determinant and show it is non-zero. A slicker solution is to take the x_i with the largest absolute value. Say |x₁| ≥ |x₂|, |x₃|. Then looking at the first equation we have an immediate contradiction, since the first term has larger absolute value than the sum of the absolute values of the second two terms. 

Problem A3

The tetrahedron ABCD is divided into two parts by a plane parallel to AB and CD. The distance of the plane from AB is k times its distance from CD. Find the ratio of the volumes of the two parts.

Solution

Let the plane meet AD at X, BD at Y, BC at Z and AC at W. Take plane parallel to BCD through WX and let it meet AB in P.

Since the distance of AB from WXYZ is k times the distance of CD, we have that AX = k·XD and hence that AX/AD = k/(k+1). Similarly AP/AB = AW/AC = AX/AD. XY is parallel to AB, so also AX/AD = BY/BD = BZ/BC.

vol ABWXYZ = vol APWX + vol WXPBYZ. APWX is similar to the tetrahedron ABCD. The sides are k/(k+1) times smaller, so vol APWX = k³(k+1)³ vol ABCD. The base of the prism WXPBYZ is BYZ which is similar to BCD with sides k/(k+1) times smaller and hence area k²(k+1)² times smaller. Its height is 1/(k+1) times the height of A above ABCD, so vol prism = 3 k²(k+1)³ vol ABCD. Thus vol ABWXYZ = (k³ + 3k²)/(k+1)³ vol ABCD. We get the vol of the other piece as vol ABCD - vol ABWXYZ and hence the ratio is (after a little manipulation) k²(k+3)/(3k+1). 

Problem B1

Find all sets of four real numbers such that the sum of any one and the product of the other three is 2.

Answer

1,1,1,1 or 3,-1,-1,-1.

Solution

Let the numbers be x₁, ... , x₄. Let t = x₁x₂x₃x₄. Then x₁ + t/x₁ = 2. So all the x_i are roots of the quadratic x² - 2x + t = 0. This has two roots, whose product is t.

If all x_i are equal to x, then x³ + x = 2, and we must have x = 1. If not, then if x₁ and x₂ are unequal roots, we have x₁x₂ = t and x₁x₂x₃x₄ = t, so x₃x₄ = 1. But x₃ and x₄ are still roots of x² - 2x + t = 0. They cannot be unequal, otherwise x₃x₄ = t, which gives t = 1 and hence all x_i = 1. Hence they are equal, and hence both 1 or both -1. Both 1 gives t = 1 and all x_i = 1. Both -1 gives t = -3 and hence x_i = 3, -1, -1, -1 (in some order). 

Problem B2

The triangle OAB has ∠O acute. M is an arbitrary point on AB. P and Q are the feet of the perpendiculars from M to OA and OB respectively. What is the locus of H, the orthocenter of the triangle OPQ (the point where its altitudes meet)? What is the locus if M is allowed to vary of the interior of OAB?

Solution

Let X be the foot of the perpendicular from B to OA, and Y the foot of the perpendicular from A to OB. We show that the orthocenter of OPQ lies on XY.

MP is parallel to BX, so AM/MB = AP/PX. Let H be the intersection of XY and the perpendicular from P to OB. PH is parallel to AY, so AP/PX = YH/HX. MQ is parallel to AY, so AM/MB = YQ/BQ. Hence YQ/BQ = YH/HX and so QH is parallel to BX and hence perpendicular to AO, so H is the orthocenter of OPQ as claimed.

If we restrict M to lie on a line A'B' parallel to AB (with A' on OA, B' on OB) then the locus is a line X'Y' parallel to XY, so as M moves over the whole interior, the locus is the interior of the triangle OXY.

Problem B3

Given n > 2 points in the plane, prove that at most n pairs of points are the maximum distance apart (of any two points in the set).

Solution

The key is that if two segments length d do not intersect then we can find an endpoint of one which is a distance > d from an endpoint of the other.

Given this, the result follows easily by induction. If false for n, then there is a point A in three pairs AB, AC and AD of length d (the maximum distance). Take AC to lie between AB and AD. Now C cannot be in another pair. Suppose it was in CX. Then CX would have to cut both AB and AD, which is impossible.

To prove the result about the segments, suppose they are PQ and RS. We must have angle PQR less than 90^o, otherwise PR > PQ = d. Similarly, the other angles of the quadrilateral must all be less than 90^o. Contradiction.

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.