Difference between revisions of "2006 USAMO Problems"
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− | = Day 1 = | + | == Day 1 == |
− | == Problem 1 == | + | === Problem 1 === |
Let <math>p</math> be a prime number and let <math>s</math> be an integer with <math>0<s<p</math>. Prove that there exist integers <math>m</math> and <math>n</math> with <math>0<m<n<p</math> and | Let <math>p</math> be a prime number and let <math>s</math> be an integer with <math>0<s<p</math>. Prove that there exist integers <math>m</math> and <math>n</math> with <math>0<m<n<p</math> and | ||
<cmath>\left\lbrace\frac{sm}{p}\right\rbrace<\left\lbrace\frac{sn}{p}\right\rbrace<\frac{s}{p}</cmath> | <cmath>\left\lbrace\frac{sm}{p}\right\rbrace<\left\lbrace\frac{sn}{p}\right\rbrace<\frac{s}{p}</cmath> | ||
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[[2006 USAMO Problems/Problem 1 | Solution]] | [[2006 USAMO Problems/Problem 1 | Solution]] | ||
− | == Problem 2 == | + | === Problem 2 === |
For a given positive integer <math>k</math> find, in terms of <math>k</math>, the minimum value of <math>N</math> for which there is a set of <math>2k+1</math> distinct positive integers that has sum greater than <math>N </math> but every subset of size <math>k</math> has sum at most <math>N/2</math>. | For a given positive integer <math>k</math> find, in terms of <math>k</math>, the minimum value of <math>N</math> for which there is a set of <math>2k+1</math> distinct positive integers that has sum greater than <math>N </math> but every subset of size <math>k</math> has sum at most <math>N/2</math>. | ||
[[2006 USAMO Problems/Problem 2 | Solution]] | [[2006 USAMO Problems/Problem 2 | Solution]] | ||
− | == Problem 3 == | + | === Problem 3 === |
For integral <math>m</math>, let <math>p(m)</math> be the greatest prime divisor of <math>m</math>. By convention, we set <math> p(\pm1)=1</math> and <math>p(0)=\infty</math>. Find all polynomials <math>f</math> with integer coefficients such that the sequence <math>\lbrace p(f(n^2))-2n\rbrace_{n\ge0}</math> is bounded above. (In particular, this requires <math>f(n^2)\neq0</math> for <math>n\ge0</math>.) | For integral <math>m</math>, let <math>p(m)</math> be the greatest prime divisor of <math>m</math>. By convention, we set <math> p(\pm1)=1</math> and <math>p(0)=\infty</math>. Find all polynomials <math>f</math> with integer coefficients such that the sequence <math>\lbrace p(f(n^2))-2n\rbrace_{n\ge0}</math> is bounded above. (In particular, this requires <math>f(n^2)\neq0</math> for <math>n\ge0</math>.) | ||
[[2006 USAMO Problems/Problem 3 | Solution]] | [[2006 USAMO Problems/Problem 3 | Solution]] | ||
− | = Day 2 = | + | == Day 2 == |
− | == Problem 4 == | + | === Problem 4 === |
Find all positive integers <math>n</math> such that there are <math>k\ge 2</math> positive rational numbers <math>a_1,a_2,\ldots a_k</math> satisfying <math>a_1+a_2+\ldots+a_k=a_1\cdot a_2\cdots a_k=n</math>. | Find all positive integers <math>n</math> such that there are <math>k\ge 2</math> positive rational numbers <math>a_1,a_2,\ldots a_k</math> satisfying <math>a_1+a_2+\ldots+a_k=a_1\cdot a_2\cdots a_k=n</math>. | ||
[[2006 USAMO Problems/Problem 4 | Solution]] | [[2006 USAMO Problems/Problem 4 | Solution]] | ||
− | == Problem 5 == | + | === Problem 5 === |
A mathematical frog jumps along the number line. The frog starts at 1, and jumps according to the following rule: if the frog is at integer <math>n </math>, then it can jump either to <math>n+1</math> or to <math>n+2^{m_n+1}</math> where <math>2^{m_n}</math> is the largest power of 2 that is a factor of <math>n</math>. Show that if <math>k\ge2</math> is a positive integer and <math>i</math> is a nonnegative integer, then the minimum number of jumps needed to reach <math>2^ik</math> is greater than the minimum number of jumps needed to reach <math>2^i</math>. | A mathematical frog jumps along the number line. The frog starts at 1, and jumps according to the following rule: if the frog is at integer <math>n </math>, then it can jump either to <math>n+1</math> or to <math>n+2^{m_n+1}</math> where <math>2^{m_n}</math> is the largest power of 2 that is a factor of <math>n</math>. Show that if <math>k\ge2</math> is a positive integer and <math>i</math> is a nonnegative integer, then the minimum number of jumps needed to reach <math>2^ik</math> is greater than the minimum number of jumps needed to reach <math>2^i</math>. | ||
[[2006 USAMO Problems/Problem 5 | Solution]] | [[2006 USAMO Problems/Problem 5 | Solution]] | ||
− | == Problem 6 == | + | === Problem 6 === |
Let <math>ABCD </math> be a quadrilateral, and let <math>E</math> and <math>F</math> be points on sides <math>AD</math> and <math>BC</math>, respectively, such that <math>AE/ED=BF/FC</math>. Ray <math>FE</math> meets rays <math>BA</math> and <math>CD</math> at <math>S</math> and <math>T</math> respectively. Prove that the circumcircles of triangles <math>SAE</math>, <math>SBF</math>, <math>TCF</math>, and <math>TDE</math> pass through a common point. | Let <math>ABCD </math> be a quadrilateral, and let <math>E</math> and <math>F</math> be points on sides <math>AD</math> and <math>BC</math>, respectively, such that <math>AE/ED=BF/FC</math>. Ray <math>FE</math> meets rays <math>BA</math> and <math>CD</math> at <math>S</math> and <math>T</math> respectively. Prove that the circumcircles of triangles <math>SAE</math>, <math>SBF</math>, <math>TCF</math>, and <math>TDE</math> pass through a common point. | ||
[[2006 USAMO Problems/Problem 6 | Solution]] | [[2006 USAMO Problems/Problem 6 | Solution]] | ||
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{{USAMO newbox|year=2006|before=[[2005 USAMO]]|after=[[2007 USAMO]]}} | {{USAMO newbox|year=2006|before=[[2005 USAMO]]|after=[[2007 USAMO]]}} | ||
+ | {{MAA Notice}} |
Latest revision as of 08:24, 14 May 2021
Contents
[hide]Day 1
Problem 1
Let be a prime number and let be an integer with . Prove that there exist integers and with and if and only if is not a divisor of .
Note: For a real number, let denote the greatest integer less than or equal to , and let denote the fractional part of .
Problem 2
For a given positive integer find, in terms of , the minimum value of for which there is a set of distinct positive integers that has sum greater than but every subset of size has sum at most .
Problem 3
For integral , let be the greatest prime divisor of . By convention, we set and . Find all polynomials with integer coefficients such that the sequence is bounded above. (In particular, this requires for .)
Day 2
Problem 4
Find all positive integers such that there are positive rational numbers satisfying .
Problem 5
A mathematical frog jumps along the number line. The frog starts at 1, and jumps according to the following rule: if the frog is at integer , then it can jump either to or to where is the largest power of 2 that is a factor of . Show that if is a positive integer and is a nonnegative integer, then the minimum number of jumps needed to reach is greater than the minimum number of jumps needed to reach .
Problem 6
Let be a quadrilateral, and let and be points on sides and , respectively, such that . Ray meets rays and at and respectively. Prove that the circumcircles of triangles , , , and pass through a common point.
2006 USAMO (Problems • Resources) | ||
Preceded by 2005 USAMO |
Followed by 2007 USAMO | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.