Difference between revisions of "2014 IMO Problems"
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==Problem 3== | ==Problem 3== | ||
− | + | Convex quadrilateral <math>ABCD</math> has <math>\angle{ABC}=\angle{CDA}=90^{\circ}</math>. Point <math>H</math> is the foot of the perpendicular from <math>A</math> to <math>BD</math>. Points <math>S</math> and <math>T</math> lie on sides <math>AB</math> and <math>AD</math>, respectively, such that <math>H</math> lies inside <math>\triangle{SCT}</math> and | |
+ | <cmath>\angle{CHS}-\angle{CSB}=90^{\circ},\quad \angle{THC}-\angle{DTC} = 90^{\circ}.</cmath> | ||
+ | |||
+ | Prove that line <math>BD</math> is tangent to the circumcircle of <math>\triangle{TSH}.</math> | ||
[[2014 IMO Problems/Problem 3|Solution]] | [[2014 IMO Problems/Problem 3|Solution]] | ||
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[[2014 IMO Problems/Problem 6|Solution]] | [[2014 IMO Problems/Problem 6|Solution]] | ||
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+ | {{IMO box|year=2014|before=[[2013 IMO Problems]]|after=[[2015 IMO Problems]]}} |
Latest revision as of 18:35, 14 March 2021
Problem 1
Let be an infinite sequence of positive integers, Prove that there exists a unique integer
such that
Problem 2
Let be an integer. Consider an
chessboard consisting of
unit squares. A configuration of
rooks on this board is
if every row and every column contains exactly one rook. Find the greatest positive integer
such that, for each peaceful configuration of
rooks, there is a
square which does not contain a rook on any of its
squares.
Problem 3
Convex quadrilateral has
. Point
is the foot of the perpendicular from
to
. Points
and
lie on sides
and
, respectively, such that
lies inside
and
Prove that line is tangent to the circumcircle of
Problem 4
Points and
lie on side
of acute-angled
so that
and
. Points
and
lie on lines
and
, respectively, such that
is the midpoint of
, and
is the midpoint of
. Prove that lines
and
intersect on the circumcircle of
.
Problem 5
For each positive integer , the Bank of Cape Town issues coins of denomination
. Given a finite collection of such coins (of not necessarily different denominations) with total value at most
, prove that it is possible to split this collection into
or fewer groups, such that each group has total value at most
.
Problem 6
A set of lines in the plane is in if no two are parallel and no three pass through the same point. A set of lines in general position cuts the plane into regions, some of which have finite are; we call these its
. Prove that for all sufficiently large
, in any set of
lines in general position it is possible to colour at least
of the lines blue in such a way that none of its finite regions has a completely blue boundary.
2014 IMO (Problems) • Resources | ||
Preceded by 2013 IMO Problems |
1 • 2 • 3 • 4 • 5 • 6 | Followed by 2015 IMO Problems |
All IMO Problems and Solutions |