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| [[2014 IMO Problems/Problem 6|Solution]] | | [[2014 IMO Problems/Problem 6|Solution]] |
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− | ==Problem 1==
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− | ''Let <math>\mathbb{Z}</math> be the set of integers. Determine all functions <math>f : \mathbb{Z} \to \mathbb{Z}</math> such that, for all
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− | ''integers <math>a</math> and <math>b</math>, <cmath>f(2a) + 2f(b) = f(f(a + b)).</cmath>''
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− | [[2019 IMO Problems/Problem 1|Solution]]
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− | ==Problem 2==
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− | In triangle <math>ABC</math>, point <math>A_1</math> lies on side <math>BC</math> and point <math>B_1</math> lies on side <math>AC</math>. Let <math>P</math> and <math>Q</math> be points on segments <math>AA_1</math> and <math>BB_1</math>, respectively, such that <math>PQ</math> is parallel to <math>AB</math>. Let <math>P_1</math> be a point on line <math>PB_1</math>, such that <math>B_1</math> lies strictly between <math>P</math> and <math>P_1</math>, and <math>\angle PP_1C=\angle BAC</math>. Similarly, let <math>Q_1</math> be the point on line <math>QA_1</math>, such that <math>A_1</math> lies strictly between <math>Q</math> and <math>Q_1</math>, and <math>\angle CQ_1Q=\angle CBA</math>.
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− | Prove that points <math>P,Q,P_1</math>, and <math>Q_1</math> are concyclic.
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− | [[2019 IMO Problems/Problem 2|Solution]]
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− | ==Problem 3==
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− | A social network has <math>2019</math> users, some pairs of whom are friends. Whenever user <math>A</math> is friends with user <math>B</math>, user <math>B</math> is also friends with user <math>A</math>. Events of the following kind may happen repeatedly, one at a time:
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− | Three users <math>A</math>, <math>B</math>, and <math>C</math> such that <math>A</math> is friends with both <math>B</math> and <math>C</math>, but <math>B</math> and <math>C</math> are not friends, change their friendship statuses such that <math>B</math> and <math>C</math> are now friends, but <math>A</math> is no longer friends with <math>B</math>, and no longer friends with <math>C</math>. All other friendship statuses are unchanged.
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− | Initially, <math>1010</math> users have <math>1009</math> friends each, and <math>1009</math> users have <math>1010</math> friends each. Prove that there exists a sequence of such events after which each user is friends with at most one other user.
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− | [[2019 IMO Problems/Problem 3|Solution]]
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− | ==Problem 4==
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− | Find all pairs <math>(k,n)</math> of positive integers such that
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− | <cmath>k!=(2^n-1)(2^n-2)(2^n-4)\dots(2^n-2^{n-1}).</cmath>
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− | [[2019 IMO Problems/Problem 4|Solution]]
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− | ==Problem 5==
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− | The Bank of Bath issues coins with an <math>H</math> on one side and a <math>T</math> on the other. Harry has <math>n</math> of these coins arranged in a line from left to right. He repeatedly performs the following operation:
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− | If there are exactly <math>k > 0</math> coins showing <math>H</math>, then he turns over the <math>k^{th}</math> coin from the left; otherwise, all coins show <math>T</math> and he stops. For example, if <math>n = 3</math> the process starting with the configuration <math>THT</math> would be <math>THT \rightarrow HHT \rightarrow HTT \rightarrow TTT</math>, which stops after three operations.
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− | (a) Show that, for each initial configuration, Harry stops after a finite number of operations.
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− | (b) For each initial configuration <math>C</math>, let <math>L(C)</math> be the number of operations before Harry stops. For
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− | example, <math>L(THT) = 3</math> and <math>L(TTT) = 0</math>. Determine the average value of <math>L(C)</math> over all <math>2^n</math>
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− | possible initial configurations <math>C</math>.
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− | [[2019 IMO Problems/Problem 5|Solution]]
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− | ==Problem 6==
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− | Let <math>I</math> be the incenter of acute triangle <math>ABC</math> with <math>AB \neq AC</math>. The incircle <math>\omega</math> of <math>ABC</math> is tangent to sides <math>BC</math>, <math>CA</math>, and <math>AB</math> at <math>D</math>, <math>E</math>, and <math>F</math>, respectively. The line through <math>D</math> perpendicular to <math>EF</math> meets ω again at <math>R</math>. Line <math>AR</math> meets ω again at <math>P</math>. The circumcircles of triangles <math>PCE</math> and <math>PBF</math> meet again at <math>Q</math>.
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− | Prove that lines <math>DI</math> and <math>PQ</math> meet on the line through <math>A</math> perpendicular to <math>AI</math>.
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− | [[2019 IMO Problems/Problem 6|Solution]]
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| {{IMO box|year=2014|before=[[2013 IMO Problems]]|after=[[2015 IMO Problems]]}} | | {{IMO box|year=2014|before=[[2013 IMO Problems]]|after=[[2015 IMO Problems]]}} |
Problem 1
Let be an infinite sequence of positive integers, Prove that there exists a unique integer such that
Solution
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.
Solution
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
Solution
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 .
Solution
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 .
Solution
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.
Solution