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Difference between revisions of "1960 IMO Problems"

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[[1960 IMO Problems/Problem 3 | Solution]]
 
[[1960 IMO Problems/Problem 3 | Solution]]
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== Day II ==
 
== Day II ==
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Consider the cube <math>ABCDA'B'C'D'</math> (with face <math>ABCD</math> directly above face <math>A'B'C'D'</math>).  
 
Consider the cube <math>ABCDA'B'C'D'</math> (with face <math>ABCD</math> directly above face <math>A'B'C'D'</math>).  
  
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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 piont of <math>B'D'</math>;
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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>.
 
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>.
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=== Problem 6 ===
 
=== Problem 6 ===
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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.
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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>;
 
a) Prove that <math>V_1 \neq V_2</math>;
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[[1960 IMO Problems/Problem 6 | Solution]]
 
[[1960 IMO Problems/Problem 6 | Solution]]
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=== Problem 7 ===
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An isosceles trapezoid with bases <math>a</math> and <math>c</math> and altitude <math>h</math> is given.
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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>;
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b) Calculate the distance of <math>p</math> from either base;
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c) Determine under what conditions such points <math>P</math> actually exist. Discuss various cases that might arise.
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[[1960 IMO Problems/Problem 7 | Solution]]
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== Resources ==
 
== Resources ==
 
* [[1960 IMO]]
 
* [[1960 IMO]]
 
* [http://www.artofproblemsolving.com/Forum/resources.php?c=1&cid=16&year=1960 IMO 1960 Problems on the Resources page]
 
* [http://www.artofproblemsolving.com/Forum/resources.php?c=1&cid=16&year=1960 IMO 1960 Problems on the Resources page]
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* [[IMO Problems and Solutions, with authors]]
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* [[Mathematics competition resources]]
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 +
{{IMO box|year=1960|before=[[1959 IMO]]|after=[[1961 IMO]]}}

Latest revision as of 21:21, 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
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