1960 IMO Problems: Difference between revisions

Latest revision as of 20:21, 20 August 2020

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

Problem 7

An isosceles trapezoid with bases $a$ and $c$ and altitude $h$ is given.

a) On the axis of symmetry of this trapezoid, find all points $P$ such that both legs of the trapezoid subtend right angles at $P$ ;

b) Calculate the distance of $p$ from either base;

c) Determine under what conditions such points $P$ actually exist. Discuss various cases that might arise.

Solution

Resources

1960 IMO (Problems) • Resources
Preceded by 1959 IMO	1 • 2 • 3 • 4 • 5 • 6	Followed by 1961 IMO
All IMO Problems and Solutions

@@ Line 4: / Line 4: @@
 === Problem 1 ===
+Determine all three-digit numbers <math>N</math> having the property that <math>N</math> is divisible by 11, and <math>\dfrac{N}{11}</math> is equal to the sum of the squares of the digits of <math>N</math>.
 [[1960 IMO Problems/Problem 1 | Solution]]
 === Problem 2 ===
+For what values of the variable <math>x</math> does the following inequality hold:
+<cmath>\dfrac{4x^2}{(1 - \sqrt {2x + 1})^2} < 2x + 9 \ ?</cmath>
@@ Line 17: / Line 19: @@
 In a given right triangle <math>ABC</math>, the hypotenuse <math>BC</math>, of length <math>a</math>, is divided into <math>n</math> equal parts (<math>n</math> and odd integer). Let <math>\alpha</math> be the acute angle subtending, from <math>A</math>, that segment which contains the midpoint of the hypotenuse. Let <math>h</math> be the length of the altitude to the hypotenuse of the triangle. Prove that:
 <center><math>
-\displaystyle\tan{\alpha}=\frac{4nh}{(n^2-1)a}.
+\tan{\alpha}=\frac{4nh}{(n^2-1)a}.
 </math>
 </center>
@@ Line 25: / Line 27: @@
 [[1960 IMO Problems/Problem 3 | Solution]]
 == Day II ==
 === Problem 4 ===
+Construct triangle <math>ABC</math>, given <math>h_a</math>, <math>h_b</math> (the altitudes from <math>A</math> and <math>B</math>), and <math>m_a</math>, the median from vertex <math>A</math>.
 [[1960 IMO Problems/Problem 4 | Solution]]
 === Problem 5 ===
+Consider the cube <math>ABCDA'B'C'D'</math> (with face <math>ABCD</math> directly above face <math>A'B'C'D'</math>).
+a) Find the locus of the midpoints of the segments <math>XY</math>, where <math>X</math> is any point of <math>AC</math> and <math>Y</math> is any point of <math>B'D'</math>;
+b) Find the locus of points <math>Z</math> which lie on the segment <math>XY</math> of part a) with <math>ZY = 2XZ</math>.
 [[1960 IMO Problems/Problem 5 | 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 <math>V_1</math> be the volume of the cone and <math>V_2</math> be the volume of the cylinder.
+a) Prove that <math>V_1 \neq V_2</math>;
+b) Find the smallest number <math>k</math> for which <math>V_1 = kV_2</math>; for this case, construct the angle subtended by a diamter of the base of the cone at the vertex of the cone.
 [[1960 IMO Problems/Problem 6 | Solution]]
+=== Problem 7 ===
+An isosceles trapezoid with bases <math>a</math> and <math>c</math> and altitude <math>h</math> is given.
+a) On the axis of symmetry of this trapezoid, find all points <math>P</math> such that both legs of the trapezoid subtend right angles at <math>P</math>;
+b) Calculate the distance of <math>p</math> from either base;
+c) Determine under what conditions such points <math>P</math> actually exist. Discuss various cases that might arise.
+[[1960 IMO Problems/Problem 7 | Solution]]
 == Resources ==
 * [[1960 IMO]]
 * [http://www.artofproblemsolving.com/Forum/resources.php?c=1&cid=16&year=1960 IMO 1960 Problems on the Resources page]
+* [[IMO Problems and Solutions, with authors]]
+* [[Mathematics competition resources]]
+{{IMO box|year=1960|before=[[1959 IMO]]|after=[[1961 IMO]]}}

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1960 IMO Problems: Difference between revisions

Latest revision as of 20:21, 20 August 2020

Contents

Day I

Problem 1

Problem 2

Problem 3

Day II

Problem 4

Problem 5

Problem 6

Problem 7

Resources