Difference between revisions of "2012 EGMO Problems"
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Let <math>ABC</math> be a triangle with circumcentre <math>O</math>. The points <math>D,E,F</math> lie in the interiors of the sides <math>BC,CA,AB</math> respectively, such that <math>DE</math> is perpendicular to <math>CO</math> and <math>DF</math> is perpendicular to <math>BO</math>. (By interior we mean, for example, that the point <math>D</math> lies on the line <math>BC</math> and <math>D</math> is between <math>B</math> and <math>C</math> on that line.) | Let <math>ABC</math> be a triangle with circumcentre <math>O</math>. The points <math>D,E,F</math> lie in the interiors of the sides <math>BC,CA,AB</math> respectively, such that <math>DE</math> is perpendicular to <math>CO</math> and <math>DF</math> is perpendicular to <math>BO</math>. (By interior we mean, for example, that the point <math>D</math> lies on the line <math>BC</math> and <math>D</math> is between <math>B</math> and <math>C</math> on that line.) | ||
Let <math>K</math> be the circumcentre of triangle <math>AFE</math>. Prove that the lines <math>DK</math> and <math>BC</math> are perpendicular. | Let <math>K</math> be the circumcentre of triangle <math>AFE</math>. Prove that the lines <math>DK</math> and <math>BC</math> are perpendicular. | ||
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[[2012 EGMO Problems/Problem 1|Solution]] | [[2012 EGMO Problems/Problem 1|Solution]] | ||
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[[2012 EGMO Problems/Problem 4|Solution]] | [[2012 EGMO Problems/Problem 4|Solution]] | ||
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==Day 2== | ==Day 2== | ||
===Problem 5=== | ===Problem 5=== | ||
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There are infinitely many people registered on the social network Mugbook. Some pairs of (different) users are registered as friends, but each person has only finitely many friends. Every user has at least one friend. (Friendship is symmetric; that is, if <math>A</math> is a friend of <math>B</math>, then <math>B</math> is a friend of <math>A</math>.) | There are infinitely many people registered on the social network Mugbook. Some pairs of (different) users are registered as friends, but each person has only finitely many friends. Every user has at least one friend. (Friendship is symmetric; that is, if <math>A</math> is a friend of <math>B</math>, then <math>B</math> is a friend of <math>A</math>.) | ||
Each person is required to designate one of their friends as their best friend. If <math>A</math> designates <math>B</math> as her best friend, then (unfortunately) it does not follow that <math>B</math> necessarily designates <math>A</math> as her best friend. Someone designated as a best friend is called a <math>1</math>-best friend. More generally, if <math>n> 1</math> is a positive integer, then a user is an <math>n</math>-best friend provided that they have been designated the best friend of someone who is an <math>(n-1)</math>-best friend. Someone who is a <math>k</math>-best friend for every positive integer <math>k</math> is called popular. | Each person is required to designate one of their friends as their best friend. If <math>A</math> designates <math>B</math> as her best friend, then (unfortunately) it does not follow that <math>B</math> necessarily designates <math>A</math> as her best friend. Someone designated as a best friend is called a <math>1</math>-best friend. More generally, if <math>n> 1</math> is a positive integer, then a user is an <math>n</math>-best friend provided that they have been designated the best friend of someone who is an <math>(n-1)</math>-best friend. Someone who is a <math>k</math>-best friend for every positive integer <math>k</math> is called popular. | ||
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(a) Prove that every popular person is the best friend of a popular person. | (a) Prove that every popular person is the best friend of a popular person. | ||
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(b) Show that if people can have infinitely many friends, then it is possible that a popular person is not the best friend of a popular person. | (b) Show that if people can have infinitely many friends, then it is possible that a popular person is not the best friend of a popular person. | ||
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[[2012 EGMO Problems/Problem 6|Solution]] | [[2012 EGMO Problems/Problem 6|Solution]] | ||
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===Problem 8=== | ===Problem 8=== | ||
A word is a finite sequence of letters from some alphabet. A word is repetitive if it is a concatenation of at least two identical subwords (for example, <math>ababab</math> and <math>abcabc</math> are repetitive, but <math>ababa</math> and <math>aabb</math> are not). Prove that if a word has the property that swapping any two adjacent letters makes the word repetitive, then all its letters are identical. (Note that one may swap two adjacent identical letters, leaving a word unchanged.) | A word is a finite sequence of letters from some alphabet. A word is repetitive if it is a concatenation of at least two identical subwords (for example, <math>ababab</math> and <math>abcabc</math> are repetitive, but <math>ababa</math> and <math>aabb</math> are not). Prove that if a word has the property that swapping any two adjacent letters makes the word repetitive, then all its letters are identical. (Note that one may swap two adjacent identical letters, leaving a word unchanged.) | ||
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[[2012 EGMO Problems/Problem 8|Solution]] | [[2012 EGMO Problems/Problem 8|Solution]] |
Latest revision as of 10:15, 23 December 2022
Contents
Day 1
Problem 1
Let be a triangle with circumcentre
. The points
lie in the interiors of the sides
respectively, such that
is perpendicular to
and
is perpendicular to
. (By interior we mean, for example, that the point
lies on the line
and
is between
and
on that line.)
Let
be the circumcentre of triangle
. Prove that the lines
and
are perpendicular.
Problem 2
Let be a positive integer. Find the greatest possible integer
, in terms of
, with the following property: a table with
rows and
columns can be filled with real numbers in such a manner that for any two different rows
and
the following holds:
Solution
Problem 3
Find all functions such that
for all
.
Problem 4
A set of integers is called sum-full if
, i.e. each element
is the sum of some pair of (not necessarily different) elements
. A set
of integers is said to be zero-sum-free if
is the only integer that cannot be expressed as the sum of the elements of a finite nonempty subset of
.
Does there exist a sum-full zero-sum-free set of integers?
Day 2
Problem 5
The numbers and
are prime and satisfy
for some positive integer
. Find all possible values of
.
Problem 6
There are infinitely many people registered on the social network Mugbook. Some pairs of (different) users are registered as friends, but each person has only finitely many friends. Every user has at least one friend. (Friendship is symmetric; that is, if is a friend of
, then
is a friend of
.)
Each person is required to designate one of their friends as their best friend. If
designates
as her best friend, then (unfortunately) it does not follow that
necessarily designates
as her best friend. Someone designated as a best friend is called a
-best friend. More generally, if
is a positive integer, then a user is an
-best friend provided that they have been designated the best friend of someone who is an
-best friend. Someone who is a
-best friend for every positive integer
is called popular.
(a) Prove that every popular person is the best friend of a popular person.
(b) Show that if people can have infinitely many friends, then it is possible that a popular person is not the best friend of a popular person.
Problem 7
Let be an acute-angled triangle with circumcircle
and orthocentre
. Let
be a point of
on the other side of
from
. Let
be the reflection of
in the line
, and let
be the reflection of
in the line
. Let
be the second point of intersection of
with the circumcircle of triangle
.
Show that the lines
,
and
are concurrent. (The orthocentre of a triangle is the point on all three of its altitudes.)
Problem 8
A word is a finite sequence of letters from some alphabet. A word is repetitive if it is a concatenation of at least two identical subwords (for example, and
are repetitive, but
and
are not). Prove that if a word has the property that swapping any two adjacent letters makes the word repetitive, then all its letters are identical. (Note that one may swap two adjacent identical letters, leaving a word unchanged.)