2nd International Mathematical Olympiad 1960 Problems & Solutions

A1.  Determine all 3 digit numbers N which are divisible by 11 and where N/11 is equal to the sum of the squares of the digits of N.

A2.  For what real values of x does the following inequality hold:

        4x²/(1 - √(1 + 2x))²  <  2x + 9 ?

A3.  In a given right triangle ABC, the hypoteneuse BC, length a, is divided into n equal parts with n an odd integer. The central part subtends an angle α at A. h is the perpendicular distance from A to BC. Prove that:

        tan α = 4nh/(an² - a).

B1.  Construct a triangle ABC given the lengths of the altitudes from A and B and the length of the median from A.

B2.  The cube ABCDA'B'C'D' has A above A', B above B' and so on. X is any point of the face diagonal AC and Y is any point of B'D'.

(a) find the locus of the midpoint of XY;

(b) find the locus of the point Z which lies one-third of the way along XY, so that ZY=2·XZ.

B3.  A cone of revolution has an inscribed sphere tangent to the base of the cone (and to the sloping surface of the cone). A cylinder is circumscribed about the sphere so that its base lies in the base of the cone. The volume of the cone is V₁ and the volume of the cylinder is V₂.

(a)  Prove that V₁ ≠ V₂;

(b)  Find the smallest possible value of V₁/V₂. For this case construct the half angle of the cone.

B4.  In the isosceles trapezoid ABCD (AB parallel to DC, and BC = AD), let AB = a, CD = c and let the perpendicular distance from A to CD be h. Show how to construct all points X on the axis of symmetry such that ∠BXC = ∠AXD = 90^o. Find the distance of each such X from AB and from CD. What is the condition for such points to exist? 

Solutions

Problem A1

Determine all 3 digit numbers N which are divisible by 11 and where N/11 is equal to the sum of the squares of the digits of N.

Answer

550, 803.

Solution

So, put N/11 = 10a + b. If a + b ≤ 9, we have 2a² + 2ab + 2b² = 10a + b (*), so b is even. Put b = 2B, then B = a(a-5) + 2aB + 4B², which is even. So b must be a multiple of 4, so b = 0, 4 or 8. If b = 0, then (*) gives a = 5 and we get the solution 550. If b = 4, then (*) gives a² - a + 14 = 0, which has no integral solutions. If b = 8, then (since a + b ≤ 9 and a > 0) a must be 1, but that does not satisfy (*).

If a + b > 9, we have (a+1)² + (a+b-10)² + b² = 10a + b, or 2a² + 2ab + 2b² - 28a - 21b + 101 = 0 (**), so b is odd. Put b = 2B+1. Then a² + 2aB + 4B² - 13a - 17B + 41 = 0. But a(a-13) is even, so B is odd. Hence b = 3 or 7. If b = 3, then (**) gives a² - 11a + 28 = 0, so a = 4 or 7. But a + b > 9, so a = 7. That gives the solution 803. If b = 7, then (**) gives a² - 7a + 26 = 0, which has no integral solutions.

Comment

Personally, I hate this type of question. The fastest way to solve it is almost certainly to scan the 81 multiples of 11 from 110 to 990. 

Problem A2

For what real values of x does the following inequality hold:

        4x²/(1 - √(1 + 2x))²  <  2x + 9 ?

Answer

- 1/2 ≤ x < 45/8.

Solution

We require the first inequality to avoid imaginary numbers. Hence we may set x = -1/2 + a²/2, where a ≥ 0. The inequality now gives immediately a < 7/2 and hence x < 45/8. It is a matter of taste whether to avoid x = 0. I would allow it because the limit as x tends to 0 of the lhs is 4, and the inequality holds.

Thanks to Dave Arthur for the idea of putting x = -1/2 + a, which saves some tedious algebra.

Problem A3

In a given right triangle ABC, the hypoteneuse BC, length a, is divided into n equal parts with n an odd integer. The central part subtends an angle α at A. h is the perpendicular distance from A to BC. Prove that:

        tan α = 4nh/(an² - a).

Solution

Let M be the midpoint of BC, and P and Q the two points a/2n either side of it, with P nearer B. Then α = ∠PAQ = ∠QAH - ∠PAH (taking angles as negative if P (or Q) lies to the left of H). So tan α = (QH - PH)/(AH² + QH·PH) = AH·PQ/(AH² + (MH - a/2n)(MH + a/2n)) = (ah/n)/(a²/4 - a²/(4n²)) = 4nh/(an² - a). 

Problem B1

Construct a triangle ABC given the lengths of the altitudes from A and B and the length of the median from A.

Solution

Let M be the midpoint of BC, AH the altitude from A, and BI the altitude from B. Start by constructing AHM. Take X on the circle diameter AM with MX = BI/2. Let the lines AX, HM meet at C and take B so that BM = MC. [This works because CMX and CBI are similar with MX = BI/2 and hence CM = CB/2.]

Problem B2

The cube ABCDA'B'C'D' has A above A', B above B' and so on. X is any point of the face diagonal AC and Y is any point of B'D'.
(a) find the locus of the midpoint of XY;
(b) find the locus of the point Z which lies one-third of the way along XY, so that ZY=2·XZ.

Solution

The key idea is that the midpoint must lie in the plane half-way between ABCD and A'B'C'D'. Similarly, Z must lie in the plane one-third of the way from ABCD to A'B'C'D'.

(a)  Regard ABCD as horizontal. Then the locus is the square with vertices the midpoints of the vertical faces (shown shaded in the diagram).

Take Y at B' and let X vary, then we trace out MN. Similarly, we can get the other sides. Now with Y at B', take X in general position, so the midpoint of XY is on MN. Now move Y to D', the midpoint traces out a line parallel to the other two sides of the square, so we can get any point inside the square. But equally, it is clear that any point inside the triangle LMN corresponds to a point Y on the ray D'B' not between B' and D', so it does not lie in the locus. Similarly for the other three triangles. So the locus is the square.

(b)  A similar argument shows that the locus is the rectangle shown in the diagram below which is √2/3 x 2√2/3.

 

Problem B3

A cone of revolution has an inscribed sphere tangent to the base of the cone (and to the sloping surface of the cone). A cylinder is circumscribed about the sphere so that its base lies in the base of the cone. The volume of the cone is V₁ and the volume of the cylinder is V₂.
(a)  Prove that V₁ ≠ V₂;
(b)  Find the smallest possible value of V₁/V₂. For this case construct the half angle of the cone.

 

Solution

Let the vertex of the cone be V, the center of the sphere be O and the center of the base be X. Let the radius of the sphere be r and the half-angle of the cone θ.

Then the the cone's height is VO + OX = r(1 + 1/sin θ), and the radius of its base is r(1 + 1/sin θ) tan θ. Hence V₁/V₂ = (1/6) (1 + 1/sin θ)³ tan²θ = (1 + s)³(6s(1 - s²)), where s = sin θ.

We claim that (1 + s)³(6s(1 - s²)) ≥ 4/3. This is equivalent to 1 + 3s + 3s² + s³ ≥ 8s - 3s³ or 1 - 5s + 3s² + 9s³ >= 0. But we can factorise the cubic as (1 - 3s)²(1 + s). So we have V₁/V₂ ≥ 4/3 with equality iff s = 1/3.

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.