1960 IMO Problems

Problems of the 2nd IMO 1960 Romania.

Day I

Problem 1

Determine all three-digit numbers $N$ having the property that $N$ is divisible by 11, and $\dfrac{N}{11}$ is equal to the sum of the squares of the digits of $N$ .

Solution

Problem 2

For what values of the variable $x$ does the following inequality hold:

$\dfrac{4x^2}{(1 - \sqrt {2x + 1})^2} < 2x + 9 \ ?$

Solution

Problem 3

In a given right triangle $ABC$ , the hypotenuse $BC$ , of length $a$ , is divided into $n$ equal parts ( $n$ and odd integer). Let $\alpha$ be the acute angle subtending, from $A$ , that segment which contains the midpoint of the hypotenuse. Let $h$ be the length of the altitude to the hypotenuse of the triangle. Prove that:

$\tan{\alpha}=\frac{4nh}{(n^2-1)a}.$

Solution

Day II

Problem 4

Construct triangle $ABC$ , given $h_a$ , $h_b$ (the altitudes from $A$ and $B$ ), and $m_a$ , the median from vertex $A$ .

Solution

Problem 5

Consider the cube $ABCDA'B'C'D'$ (with face $ABCD$ directly above face $A'B'C'D'$ ).

a) Find the locus of the midpoints of the segments $XY$ , where $X$ is any point of $AC$ and $Y$ is any point of $B'D'$ ;

b) Find the locus of points $Z$ which lie on the segment $XY$ of part a) with $ZY = 2XZ$ .

Solution

Problem 6

Consider a cone of revolution with an inscribed sphere tangent to the base of the cone. A cylinder is circumscribed about this sphere so that one of its bases lies in the base of the cone. Let $V_1$ be the volume of the cone and $V_2$ be the volume of the cylinder.

a) Prove that $V_1 \neq V_2$ ;

b) Find the smallest number $k$ for which $V_1 = kV_2$ ; for this case, construct the angle subtended by a diamter of the base of the cone at the vertex of the cone.

Solution

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1960 IMO Problems

Contents

Day I

Problem 1

Problem 2

Problem 3

Day II

Problem 4

Problem 5

Problem 6

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