https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=I+like+pie&feedformat=atom AoPS Wiki - User contributions [en] 2020-10-19T20:59:45Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=User_talk:I_like_pie&diff=34001 User talk:I like pie 2010-03-24T02:47:53Z <p>I like pie: Blanked the page</p> <hr /> <div></div> I like pie https://artofproblemsolving.com/wiki/index.php?title=User:I_like_pie&diff=34000 User:I like pie 2010-03-24T02:47:27Z <p>I like pie: Redirected page to User talk:I like pie</p> <hr /> <div>#REDIRECT[[User_talk:I_like_pie]]</div> I like pie https://artofproblemsolving.com/wiki/index.php?title=2006_USAMO_Problems&diff=31289 2006 USAMO Problems 2009-04-16T19:27:11Z <p>I like pie: Fixed level of Resources section, added link to subsequent USAMO in box</p> <hr /> <div>= Day 1 =<br /> == Problem 1 ==<br /> Let &lt;math&gt;p&lt;/math&gt; be a prime number and let &lt;math&gt;s&lt;/math&gt; be an integer with &lt;math&gt;0&lt;s&lt;p&lt;/math&gt;. Prove that there exist integers &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; with &lt;math&gt;0&lt;m&lt;n&lt;p&lt;/math&gt; and<br /> &lt;cmath&gt;\left\lbrace\frac{sm}{p}\right\rbrace&lt;\left\lbrace\frac{sn}{p}\right\rbrace&lt;\frac{s}{p}&lt;/cmath&gt;<br /> if and only if &lt;math&gt;s&lt;/math&gt; is not a divisor of &lt;math&gt;p-1&lt;/math&gt;.<br /> <br /> Note: For &lt;math&gt;x&lt;/math&gt; a real number, let &lt;math&gt;\lfloor x\rfloor&lt;/math&gt; denote the greatest integer less than or equal to &lt;math&gt;x&lt;/math&gt;, and let &lt;math&gt;\lbrace x\rbrace=x-\lfloor x\rfloor&lt;/math&gt; denote the fractional part of &lt;math&gt;x&lt;/math&gt;.<br /> <br /> [[2006 USAMO Problems/Problem 1 | Solution]]<br /> <br /> == Problem 2 ==<br /> For a given positive integer &lt;math&gt;k&lt;/math&gt; find, in terms of &lt;math&gt;k&lt;/math&gt;, the minimum value of &lt;math&gt;N&lt;/math&gt; for which there is a set of &lt;math&gt;2k+1&lt;/math&gt; distinct positive integers that has sum greater than &lt;math&gt;N &lt;/math&gt; but every subset of size &lt;math&gt;k&lt;/math&gt; has sum at most &lt;math&gt;N/2&lt;/math&gt;.<br /> <br /> [[2006 USAMO Problems/Problem 2 | Solution]]<br /> <br /> == Problem 3 ==<br /> For integral &lt;math&gt;m&lt;/math&gt;, let &lt;math&gt;p(m)&lt;/math&gt; be the greatest prime divisor of &lt;math&gt;m&lt;/math&gt;. By convention, we set &lt;math&gt; p(\pm1)=1&lt;/math&gt; and &lt;math&gt;p(0)=\infty&lt;/math&gt;. Find all polynomials &lt;math&gt;f&lt;/math&gt; with integer coefficients such that the sequence &lt;math&gt;\lbrace p(f(n^2))-2n\rbrace_{n\ge0}&lt;/math&gt; is bounded above. (In particular, this requires &lt;math&gt;f(n^2)\neq0&lt;/math&gt; for &lt;math&gt;n\ge0&lt;/math&gt;.)<br /> <br /> [[2006 USAMO Problems/Problem 3 | Solution]]<br /> <br /> = Day 2 =<br /> == Problem 4 ==<br /> Find all positive integers &lt;math&gt;n&lt;/math&gt; such that there are &lt;math&gt;k\ge 2&lt;/math&gt; positive rational numbers &lt;math&gt;a_1,a_2,\ldots a_k&lt;/math&gt; satisfying &lt;math&gt;a_1+a_2+\ldots+a_k=a_1\cdot a_2\cdots a_k=n&lt;/math&gt;.<br /> <br /> [[2006 USAMO Problems/Problem 4 | Solution]]<br /> <br /> == Problem 5 ==<br /> 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 &lt;math&gt;n &lt;/math&gt;, then it can jump either to &lt;math&gt;n+1&lt;/math&gt; or to &lt;math&gt;n+2^{m_n+1}&lt;/math&gt; where &lt;math&gt;2^{m_n}&lt;/math&gt; is the largest power of 2 that is a factor of &lt;math&gt;n&lt;/math&gt;. Show that if &lt;math&gt;k\ge2&lt;/math&gt; is a positive integer and &lt;math&gt;i&lt;/math&gt; is a nonnegative integer, then the minimum number of jumps needed to reach &lt;math&gt;2^ik&lt;/math&gt; is greater than the minimum number of jumps needed to reach &lt;math&gt;2^i&lt;/math&gt;.<br /> <br /> [[2006 USAMO Problems/Problem 5 | Solution]]<br /> <br /> == Problem 6 ==<br /> Let &lt;math&gt;ABCD &lt;/math&gt; be a quadrilateral, and let &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;F&lt;/math&gt; be points on sides &lt;math&gt;AD&lt;/math&gt; and &lt;math&gt;BC&lt;/math&gt;, respectively, such that &lt;math&gt;AE/ED=BF/FC&lt;/math&gt;. Ray &lt;math&gt;FE&lt;/math&gt; meets rays &lt;math&gt;BA&lt;/math&gt; and &lt;math&gt;CD&lt;/math&gt; at &lt;math&gt;S&lt;/math&gt; and &lt;math&gt;T&lt;/math&gt; respectively. Prove that the circumcircles of triangles &lt;math&gt;SAE&lt;/math&gt;, &lt;math&gt;SBF&lt;/math&gt;, &lt;math&gt;TCF&lt;/math&gt;, and &lt;math&gt;TDE&lt;/math&gt; pass through a common point.<br /> <br /> [[2006 USAMO Problems/Problem 6 | Solution]]<br /> <br /> = Resources =<br /> *[[USAMO Problems and Solutions]]<br /> *[http://www.artofproblemsolving.com/Forum/resources.php?c=182&amp;cid=27&amp;year=2006 USAMO 2006 Problems on the Resources Page]<br /> *[http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2006-ua/2006usamoQ.pdf USAMO 2006 Questions Document]<br /> *[http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2006-ua/2006usamoS.pdf USAMO 2006 Solutions Document]<br /> <br /> {{USAMO newbox|year=2006|before=[[2005 USAMO]]|after=[[2007 USAMO]]}}</div> I like pie https://artofproblemsolving.com/wiki/index.php?title=User:I_like_pie&diff=31288 User:I like pie 2009-04-16T19:24:20Z <p>I like pie: Removed random stuff</p> <hr /> <div></div> I like pie https://artofproblemsolving.com/wiki/index.php?title=2006_USAMO_Problems&diff=31287 2006 USAMO Problems 2009-04-16T19:23:18Z <p>I like pie: Removed extra parenthesis</p> <hr /> <div>= Day 1 =<br /> == Problem 1 ==<br /> Let &lt;math&gt;p&lt;/math&gt; be a prime number and let &lt;math&gt;s&lt;/math&gt; be an integer with &lt;math&gt;0&lt;s&lt;p&lt;/math&gt;. Prove that there exist integers &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; with &lt;math&gt;0&lt;m&lt;n&lt;p&lt;/math&gt; and<br /> &lt;cmath&gt;\left\lbrace\frac{sm}{p}\right\rbrace&lt;\left\lbrace\frac{sn}{p}\right\rbrace&lt;\frac{s}{p}&lt;/cmath&gt;<br /> if and only if &lt;math&gt;s&lt;/math&gt; is not a divisor of &lt;math&gt;p-1&lt;/math&gt;.<br /> <br /> Note: For &lt;math&gt;x&lt;/math&gt; a real number, let &lt;math&gt;\lfloor x\rfloor&lt;/math&gt; denote the greatest integer less than or equal to &lt;math&gt;x&lt;/math&gt;, and let &lt;math&gt;\lbrace x\rbrace=x-\lfloor x\rfloor&lt;/math&gt; denote the fractional part of &lt;math&gt;x&lt;/math&gt;.<br /> <br /> [[2006 USAMO Problems/Problem 1 | Solution]]<br /> <br /> == Problem 2 ==<br /> For a given positive integer &lt;math&gt;k&lt;/math&gt; find, in terms of &lt;math&gt;k&lt;/math&gt;, the minimum value of &lt;math&gt;N&lt;/math&gt; for which there is a set of &lt;math&gt;2k+1&lt;/math&gt; distinct positive integers that has sum greater than &lt;math&gt;N &lt;/math&gt; but every subset of size &lt;math&gt;k&lt;/math&gt; has sum at most &lt;math&gt;N/2&lt;/math&gt;.<br /> <br /> [[2006 USAMO Problems/Problem 2 | Solution]]<br /> <br /> == Problem 3 ==<br /> For integral &lt;math&gt;m&lt;/math&gt;, let &lt;math&gt;p(m)&lt;/math&gt; be the greatest prime divisor of &lt;math&gt;m&lt;/math&gt;. By convention, we set &lt;math&gt; p(\pm1)=1&lt;/math&gt; and &lt;math&gt;p(0)=\infty&lt;/math&gt;. Find all polynomials &lt;math&gt;f&lt;/math&gt; with integer coefficients such that the sequence &lt;math&gt;\lbrace p(f(n^2))-2n\rbrace_{n\ge0}&lt;/math&gt; is bounded above. (In particular, this requires &lt;math&gt;f(n^2)\neq0&lt;/math&gt; for &lt;math&gt;n\ge0&lt;/math&gt;.)<br /> <br /> [[2006 USAMO Problems/Problem 3 | Solution]]<br /> <br /> = Day 2 =<br /> == Problem 4 ==<br /> Find all positive integers &lt;math&gt;n&lt;/math&gt; such that there are &lt;math&gt;k\ge 2&lt;/math&gt; positive rational numbers &lt;math&gt;a_1,a_2,\ldots a_k&lt;/math&gt; satisfying &lt;math&gt;a_1+a_2+\ldots+a_k=a_1\cdot a_2\cdots a_k=n&lt;/math&gt;.<br /> <br /> [[2006 USAMO Problems/Problem 4 | Solution]]<br /> <br /> == Problem 5 ==<br /> 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 &lt;math&gt;n &lt;/math&gt;, then it can jump either to &lt;math&gt;n+1&lt;/math&gt; or to &lt;math&gt;n+2^{m_n+1}&lt;/math&gt; where &lt;math&gt;2^{m_n}&lt;/math&gt; is the largest power of 2 that is a factor of &lt;math&gt;n&lt;/math&gt;. Show that if &lt;math&gt;k\ge2&lt;/math&gt; is a positive integer and &lt;math&gt;i&lt;/math&gt; is a nonnegative integer, then the minimum number of jumps needed to reach &lt;math&gt;2^ik&lt;/math&gt; is greater than the minimum number of jumps needed to reach &lt;math&gt;2^i&lt;/math&gt;.<br /> <br /> [[2006 USAMO Problems/Problem 5 | Solution]]<br /> <br /> == Problem 6 ==<br /> Let &lt;math&gt;ABCD &lt;/math&gt; be a quadrilateral, and let &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;F&lt;/math&gt; be points on sides &lt;math&gt;AD&lt;/math&gt; and &lt;math&gt;BC&lt;/math&gt;, respectively, such that &lt;math&gt;AE/ED=BF/FC&lt;/math&gt;. Ray &lt;math&gt;FE&lt;/math&gt; meets rays &lt;math&gt;BA&lt;/math&gt; and &lt;math&gt;CD&lt;/math&gt; at &lt;math&gt;S&lt;/math&gt; and &lt;math&gt;T&lt;/math&gt; respectively. Prove that the circumcircles of triangles &lt;math&gt;SAE&lt;/math&gt;, &lt;math&gt;SBF&lt;/math&gt;, &lt;math&gt;TCF&lt;/math&gt;, and &lt;math&gt;TDE&lt;/math&gt; pass through a common point.<br /> <br /> [[2006 USAMO Problems/Problem 6 | Solution]]<br /> <br /> == Resources ==<br /> *[[USAMO Problems and Solutions]]<br /> *[http://www.artofproblemsolving.com/Forum/resources.php?c=182&amp;cid=27&amp;year=2006 USAMO 2006 Problems on the Resources Page]<br /> *[http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2006-ua/2006usamoQ.pdf USAMO 2006 Questions Document]<br /> *[http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2006-ua/2006usamoS.pdf USAMO 2006 Solutions Document]<br /> <br /> {{USAMO newbox|year=2006|before=[[2005 USAMO]]|after=2007 USAMO}}</div> I like pie https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10A_Problems/Problem_1&diff=30301 2009 AMC 10A Problems/Problem 1 2009-02-15T18:47:52Z <p>I like pie: Fixed year in AMC box</p> <hr /> <div>== Problem ==<br /> One can holds &lt;math&gt;12&lt;/math&gt; ounces of soda. What is the minimum number of cans needed to provide a gallon (&lt;math&gt;128&lt;/math&gt; ounces) of soda?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 7\qquad<br /> \mathrm{(B)}\ 8\qquad<br /> \mathrm{(C)}\ 9\qquad<br /> \mathrm{(D)}\ 10\qquad<br /> \mathrm{(E)}\ 11&lt;/math&gt;<br /> <br /> == Solution ==<br /> &lt;math&gt;10&lt;/math&gt; cans would hold &lt;math&gt;120&lt;/math&gt; ounces, but &lt;math&gt;128&gt;120&lt;/math&gt;, so &lt;math&gt;11&lt;/math&gt; cans are required. Thus, the answer is &lt;math&gt;\mathrm{(E)}&lt;/math&gt;.<br /> <br /> {{AMC10 box|year=2009|ab=A|before=First Question|num-a=2}}</div> I like pie https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10A_Problems/Problem_1&diff=30300 2009 AMC 10A Problems/Problem 1 2009-02-15T18:46:14Z <p>I like pie: Added solution</p> <hr /> <div>== Problem ==<br /> One can holds &lt;math&gt;12&lt;/math&gt; ounces of soda. What is the minimum number of cans needed to provide a gallon (&lt;math&gt;128&lt;/math&gt; ounces) of soda?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 7\qquad<br /> \mathrm{(B)}\ 8\qquad<br /> \mathrm{(C)}\ 9\qquad<br /> \mathrm{(D)}\ 10\qquad<br /> \mathrm{(E)}\ 11&lt;/math&gt;<br /> <br /> == Solution ==<br /> &lt;math&gt;10&lt;/math&gt; cans would hold &lt;math&gt;120&lt;/math&gt; ounces, but &lt;math&gt;128&gt;120&lt;/math&gt;, so &lt;math&gt;11&lt;/math&gt; cans are required. Thus, the answer is &lt;math&gt;\mathrm{(E)}&lt;/math&gt;.<br /> <br /> {{AMC10 box|year=2008|ab=A|before=First Question|num-a=2}}</div> I like pie https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10A_Problems/Problem_22&diff=30242 2009 AMC 10A Problems/Problem 22 2009-02-14T00:11:51Z <p>I like pie: Added Second Solution</p> <hr /> <div>== Problem ==<br /> <br /> Two cubical dice each have removable numbers &lt;math&gt;1&lt;/math&gt; through &lt;math&gt;6&lt;/math&gt;. The twelve numbers on the two dice are removed, put into a bag, then drawn one at a time and randomly reattached to the faces of the cubes, one number to each face. The dice are then rolled and the numbers on the two top faces are added. What is the probability that the sum is &lt;math&gt;7&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \mathrm{(A)}\ \frac{1}{9}<br /> \qquad<br /> \mathrm{(B)}\ \frac{1}{8}<br /> \qquad<br /> \mathrm{(C)}\ \frac{1}{6}<br /> \qquad<br /> \mathrm{(D)}\ \frac{2}{11}<br /> \qquad<br /> \mathrm{(E)}\ \frac{1}{5}<br /> &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> === Solution 1 ===<br /> <br /> At the moment when the numbers are in the bag, imagine that each of them has a different color. Clearly the situation is symmetric at this moment. Hence after we draw them, attach them and throw the dice, the probability of getting some pair of colors is the same for any two colors.<br /> <br /> There are &lt;math&gt;{12 \choose 2} = 66&lt;/math&gt; ways how to pick two of the colors. We now have to count the ways where the two chosen numbers will have sum &lt;math&gt;7&lt;/math&gt;. <br /> <br /> Sum &lt;math&gt;7&lt;/math&gt; can be obtained as &lt;math&gt;1+6&lt;/math&gt;, &lt;math&gt;2+5&lt;/math&gt;, or &lt;math&gt;3+4&lt;/math&gt;. Each number in the bag has two different colors, hence each of these three options corresponds to four pairs of colors. <br /> <br /> Out of the &lt;math&gt;66&lt;/math&gt; pairs of colors we can get when throwing the dice, &lt;math&gt;3\cdot 4=12&lt;/math&gt; will give us the sum &lt;math&gt;7&lt;/math&gt;. Hence the probability that this will happen is &lt;math&gt;\frac{12}{66} = \boxed{\frac 2{11}}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> Ignoring the numbers that do not affect the probability of the desired outcome (the ones that are not on top of the dice), say that the number on top of the first die is &lt;math&gt;n&lt;/math&gt;. For the sum of the &lt;math&gt;2&lt;/math&gt; numbers to be &lt;math&gt;7&lt;/math&gt;, the second die must have the number &lt;math&gt;7-n&lt;/math&gt; on top. There are &lt;math&gt;11&lt;/math&gt; remaining numbers that could be on top of the second die, &lt;math&gt;2&lt;/math&gt; of which are &lt;math&gt;7-n&lt;/math&gt; (since &lt;math&gt;n\ne7-n&lt;/math&gt; in all cases).<br /> <br /> Thus, the probability of the sum of the &lt;math&gt;2&lt;/math&gt; numbers being &lt;math&gt;7&lt;/math&gt; is &lt;math&gt;\frac{2}{11}&lt;/math&gt;, so the answer is &lt;math&gt;\mathrm{(D)}&lt;/math&gt;.<br /> <br /> == See Also ==<br /> <br /> {{AMC10 box|year=2009|ab=A|num-b=21|num-a=23}}</div> I like pie https://artofproblemsolving.com/wiki/index.php?title=Chemistry&diff=28798 Chemistry 2008-12-26T21:02:25Z <p>I like pie: Fixed link 'Atoms' to 'Atom' (still not written)</p> <hr /> <div>'''Chemistry''' is the study of interactions between [[atom]]s at a macroscopic level.<br /> <br /> ==See also==<br /> * [[AoPSWiki:Table of Contents|Table of Contents]]<br /> <br /> {{stub}}</div> I like pie https://artofproblemsolving.com/wiki/index.php?title=User:I_like_pie&diff=28758 User:I like pie 2008-12-25T17:59:30Z <p>I like pie: </p> <hr /> <div>I am a user on AoPS, and go by the username i_like_pie. Besides math, I enjoy mountain biking, playing soccer, programming, and many other things.</div> I like pie https://artofproblemsolving.com/wiki/index.php?title=Kelvin&diff=28629 Kelvin 2008-12-18T22:23:35Z <p>I like pie: </p> <hr /> <div>The '''Kelvin''' is the [[SI]] unit for [[temperature]]. &lt;math&gt;0\,\mathrm{K}&lt;/math&gt; is defined as [[absolute zero]], the theoretical temperature at which there is no movement.<br /> <br /> ==Conversion==<br /> *Kelvin to [[Celsius]]: &lt;math&gt;^\circ\mathrm{C}=\mathrm{K}+273.15&lt;/math&gt;<br /> *Kelvin to [[Fahrenheit]]: &lt;math&gt;^\circ\mathrm{F}=\tfrac{9}{5}\cdot\mathrm{K}-459.67&lt;/math&gt;<br /> <br /> {{stub}}<br /> [[Category:Definition]]<br /> [[Category:Physics]]<br /> [[Category:Units]]</div> I like pie https://artofproblemsolving.com/wiki/index.php?title=Temperature&diff=28628 Temperature 2008-12-18T22:12:53Z <p>I like pie: </p> <hr /> <div>'''Temperature''' is how hot or cold an object or a region is. The [[SI]] unit for temperature is the [[Kelvin]]; other commonly used scaled are [[Celsius]] and [[Fahrenheit]].<br /> <br /> {{stub}}<br /> [[Category:Definition]]<br /> [[Category:Physics]]</div> I like pie https://artofproblemsolving.com/wiki/index.php?title=Second&diff=28627 Second 2008-12-18T22:09:59Z <p>I like pie: </p> <hr /> <div>The '''second''' is the [[Système international|SI]] unit for [[time]]. It is defined as the time in which light travels approximately 299792458 [[meter]]s, or about 299792 [[kilometer]]s. There are 60 seconds in a [[minute]], 3600 seconds in an [[hour]], and 86400 seconds in a [[day]].<br /> <br /> {{stub}}<br /> [[Category:Units]]</div> I like pie https://artofproblemsolving.com/wiki/index.php?title=Atm&diff=28626 Atm 2008-12-18T22:07:20Z <p>I like pie: </p> <hr /> <div>'''Atm''', which stands for Earth atmospheres at sea level, is a unit of pressure. 1 atm is defined as the pressure at 0 [[meter]]s above sea level.<br /> <br /> ==Conversion==<br /> To convert from atm to the [[Système_international|SI]] unit of pressure, the [[pascal]], and vice versa:<br /> *&lt;math&gt;1\,\mathrm{Pa}=9.8692\times10^{-6}\,\mathrm{atm}&lt;/math&gt;<br /> *&lt;math&gt;1\,\mathrm{atm}=101325\,\mathrm{Pa}&lt;/math&gt;<br /> <br /> {{stub}}<br /> [[Category:Units]]</div> I like pie https://artofproblemsolving.com/wiki/index.php?title=Celsius&diff=28625 Celsius 2008-12-18T22:02:33Z <p>I like pie: </p> <hr /> <div>The '''Celsius''' (also '''Centigrade''') scale is a was [[measure|measuring]] [[temperature]]. Originally, &lt;math&gt;0^\circ\mathrm{C}&lt;/math&gt; was defined as the temperature at which water freezes at 1 [[atm]], and &lt;math&gt;100^\circ\mathrm{C}&lt;/math&gt; as the temperature at which water boils at 1 atm.<br /> <br /> However, since different thermometers register slightly different temperatures at these points, the [[absolute zero]] and the [[triple point]] of water are used to fix the Celsius scale. They are defined, respectively, to be &lt;math&gt;-273.15^\circ\mathrm{C}&lt;/math&gt; and &lt;math&gt;0.01^\circ\mathrm{C}&lt;/math&gt;.<br /> <br /> ==Conversion==<br /> *Celsius to [[Fahrenheit]]: &lt;math&gt;^\circ\mathrm{F}=\tfrac{9}{5}\cdot{}^\circ\mathrm{C}+32&lt;/math&gt;<br /> *Celsius to [[Kelvin]]: &lt;math&gt;\mathrm{K}={}^\circ\mathrm{C}-273.15&lt;/math&gt;<br /> <br /> {{stub}}<br /> [[Category:Definition]]<br /> [[Category:Physics]]<br /> [[Category:Units]]</div> I like pie https://artofproblemsolving.com/wiki/index.php?title=FLOPS&diff=28624 FLOPS 2008-12-18T21:51:05Z <p>I like pie: </p> <hr /> <div>'''FLOPS''' is an acronym for [[Float|'''FL'''oating]] point '''O'''perations '''P'''er [[Second|'''S'''econd]].<br /> <br /> It is a commonly used metric for measuring computational power.<br /> <br /> {{stub}}</div> I like pie https://artofproblemsolving.com/wiki/index.php?title=Composite_number&diff=28604 Composite number 2008-12-15T05:49:20Z <p>I like pie: Reference to Prime already exists in article; removed from See also section.</p> <hr /> <div>A '''composite number''' is a [[positive integer]] with at least one [[divisor]] different from 1 and itself. Some composite numbers are &lt;math&gt;4=2^2&lt;/math&gt; and &lt;math&gt;12=2\times 6=3\times 4&lt;/math&gt;. <br /> <br /> Note that the number one is neither prime nor composite. It follows that two is the only even [[prime number]], three is the only multiple of three that is prime, and so on. <br /> <br /> Every positive integer either is prime, composite, or 1.<br /> <br /> ==See also==<br /> * [[Number Theory]]<br /> <br /> {{stub}}<br /> [[Category:Definition]]<br /> [[Category:Number theory]]</div> I like pie https://artofproblemsolving.com/wiki/index.php?title=Chord&diff=28591 Chord 2008-12-13T02:45:34Z <p>I like pie: </p> <hr /> <div>A '''chord''' of a [[circle]] &lt;math&gt;O&lt;/math&gt; is a [[line segment]] joining two [[point]]s on &lt;math&gt;O&lt;/math&gt;.<br /> <br /> &lt;asy&gt;size(100);<br /> pair O=origin,A=dir(135),B=dir(30);<br /> D(unitcircle);<br /> D(A--B);<br /> MP(&quot;O&quot;,D(O),S);<br /> MP(&quot;A&quot;,D(A),W);<br /> MP(&quot;B&quot;,D(B),E);&lt;/asy&gt;<br /> <br /> The [[diameter]] of a circle is the longest chord of that circle. The diameter goes through the center of the circle.<br /> <br /> &lt;asy&gt;size(120);<br /> pair O=origin,A=dir(170),B=dir(-10);<br /> D(unitcircle);<br /> D(A--B);<br /> MP(&quot;O&quot;,D(O),N);<br /> MP(&quot;A&quot;,D(A),W);<br /> MP(&quot;B&quot;,D(B),E);&lt;/asy&gt;<br /> <br /> {{stub}}<br /> [[Category:Geometry]]</div> I like pie https://artofproblemsolving.com/wiki/index.php?title=String&diff=28590 String 2008-12-13T02:39:53Z <p>I like pie: </p> <hr /> <div>A '''string''' is a [[datatype]] that signifies an ordered sequence of [[character]]s.<br /> <br /> For example, &quot;4f38fjy7a&quot; is a string.<br /> <br /> Note that &quot;222&quot; does not equal the number 222.<br /> <br /> A special string function is [[concatenation]]: &quot;abc&quot; + &quot;789&quot; = &quot;abc789&quot;.<br /> <br /> {{stub}}<br /> [[Category:Datatypes]]</div> I like pie https://artofproblemsolving.com/wiki/index.php?title=String&diff=28589 String 2008-12-13T02:39:01Z <p>I like pie: </p> <hr /> <div>A '''string''' is a [[datatype]] that signifies an ordered sequence of [[character]]s.<br /> <br /> For example, &quot;4f38fjy7a&quot; is a string.<br /> <br /> Note that &quot;222&quot; does not equal the number 222.<br /> <br /> A special string function is [[concatenation]]: &quot;abc&quot;+&quot;789&quot;=&quot;abc789&quot;.<br /> <br /> {{stub}}<br /> [[Category:Datatypes]]</div> I like pie https://artofproblemsolving.com/wiki/index.php?title=Fixed&diff=28588 Fixed 2008-12-13T02:20:29Z <p>I like pie: </p> <hr /> <div>The '''fixed''' [[datatype]] is a very accurate datatype, but is inefficient in memory storage. Since it does not utilize [[float]]ing points, each bit must be separately encoded.<br /> <br /> {{stub}}<br /> [[Category:Datatypes]]</div> I like pie https://artofproblemsolving.com/wiki/index.php?title=Float&diff=28587 Float 2008-12-13T02:03:18Z <p>I like pie: </p> <hr /> <div>The '''float''' [[datatype]] is a type of computing [[variable]]. It has three parts: the [[mantissa]], the [[exponent]], and the [[sign]].<br /> <br /> The mantissa is the number itself; it is multiplied by &lt;math&gt;10\textsuperscript{exponent}&lt;/math&gt; to get the actual number represented by this float. The sign determines whether the number is positive or negative.<br /> <br /> == Advantages and Disadvantages with [[Fixed]] Datatype ==<br /> Advantages:<br /> <br /> *Float has a significantly larger range of numbers than fixed. For example, 123450000000 and 0.00012345 take up the same space.<br /> <br /> *It is an important metric in measuring computational power (in [[FLOPS]]).<br /> <br /> Disadvantages:<br /> <br /> *Float must take up a slightly larger amount of space than fixed in order to exercise its full potential.<br /> <br /> *With floats, there is a possibility of low-precision rounding.<br /> <br /> {{stub}}<br /> [[Category:Datatypes]]</div> I like pie https://artofproblemsolving.com/wiki/index.php?title=Fraction&diff=28585 Fraction 2008-12-13T01:46:02Z <p>I like pie: </p> <hr /> <div>A '''fraction''' is the [[ratio]] of two [[number]]s. Most commonly, we consider [[rational number]]s, those fractions which are the ratio of two [[integer]]s. For instance, &lt;math&gt;\frac 34&lt;/math&gt; is such a fraction.<br /> <br /> {{stub}}<br /> [[Category:Definition]]</div> I like pie https://artofproblemsolving.com/wiki/index.php?title=Remainder&diff=28347 Remainder 2008-11-07T21:01:45Z <p>I like pie: </p> <hr /> <div>{{stub}}<br /> <br /> The '''remainder''' of a division of two integers &lt;math&gt;\frac {a}{b},\ b \neq 0&lt;/math&gt; is the integer &lt;math&gt;r &lt; b&lt;/math&gt; such that &lt;math&gt;a = qb + r&lt;/math&gt;, where &lt;math&gt;q&lt;/math&gt; is the [[Division|quotient]]; in other words, &lt;math&gt;r&lt;/math&gt; is the part of &lt;math&gt;a&lt;/math&gt; that is not [[Divisibility|divisible]] by &lt;math&gt;b&lt;/math&gt;. If &lt;math&gt;a = 4&lt;/math&gt;, and &lt;math&gt;b = 3&lt;/math&gt;, for example, the division &lt;math&gt;\frac {4}{3}&lt;/math&gt; would have remainder &lt;math&gt;1&lt;/math&gt;, since &lt;math&gt;4 = (1)3 + 1&lt;/math&gt; (notice that the quotient, in this case, is one). If &lt;math&gt;b&lt;/math&gt; is a [[divisor]] of &lt;math&gt;a&lt;/math&gt;, the remainder is said to be zero.<br /> <br /> <br /> The concept of a remainder is related to [[modular arithmetic]]: &lt;math&gt;r&lt;/math&gt; is said to be the [[residue class]] of &lt;math&gt;a&lt;/math&gt; in modulo &lt;math&gt;b&lt;/math&gt; [[iff]] &lt;math&gt;a = qb + r&lt;/math&gt; (an equivalent statement would be &lt;math&gt;a \equiv r \mod b&lt;/math&gt;).<br /> <br /> <br /> It is important to notice that the remainder is most useful when an integer quotient is desired, as we can always say that &lt;math&gt;a = qb&lt;/math&gt; for any [[real number]] &lt;math&gt;q&lt;/math&gt; (in the example provided earlier, &lt;math&gt;q = 1.\overline{3}&lt;/math&gt;).</div> I like pie https://artofproblemsolving.com/wiki/index.php?title=Template:Stub&diff=28346 Template:Stub 2008-11-07T21:00:53Z <p>I like pie: Removed space at top</p> <hr /> <div>''This article is a stub. Help us out by &lt;span class=&quot;plainlinks&quot;&gt;[{{fullurl:{{FULLPAGENAME}}|action=edit}} expanding it]&lt;/span&gt;.''<br /> &lt;noinclude&gt;<br /> ==Template Documentation==<br /> &lt;noinclude&gt;This template will categorize articles that include it into [[:Category:Stubs]]. Add it with &lt;nowiki&gt;{{stub}}&lt;/nowiki&gt; at the bottom of an article. <br /> <br /> [[Category:Stubs|*]]&lt;/noinclude&gt;&lt;includeonly&gt;[[Category:Stubs]]&lt;/includeonly&gt;</div> I like pie https://artofproblemsolving.com/wiki/index.php?title=MIST_Academy&diff=26051 MIST Academy 2008-05-19T06:21:46Z <p>I like pie: Fixed MIST Academy Link</p> <hr /> <div>'''MIST Academy''' is a school for motivated math and science students in the Birmingham, Alabama area run by former [[AoPS]] teacher, and class and textbook author [[Mathew Crawford]]. MIST stands for Mathematics, Informatics, Science, and Technology.<br /> <br /> Started in 2007, MIST Academy is currently focused on developing its math and problem solving programs.<br /> <br /> <br /> == External Links ==<br /> * [http://www.mistacademy.com/index.htm MIST Academy homepage]<br /> * [http://www.mistacademy.com/blog/ Mathew Crawford's blog]</div> I like pie https://artofproblemsolving.com/wiki/index.php?title=Center_(geometry)&diff=25980 Center (geometry) 2008-05-12T06:50:40Z <p>I like pie: List, minor edits</p> <hr /> <div>The '''center''' of a [[circle]] or [[sphere]] is a [[point]] inside the circle which is [[equidistant]] from all points on the circle.<br /> <br /> ==Triangle centers==<br /> *The [[centroid]] is where the three [[median]]s of the triangle meet.<br /> <br /> *The [[incenter]] of the triangle is where the three [[angle bisector]]s meet. It is also the center of the [[incircle]].<br /> <br /> *The [[circumcenter]] is where the [[perpendicular bisector]]s of the triangles sides meet. It is also the center of the [[circumcircle]].<br /> <br /> *The [[orthocenter]] Is where the [[altitude]]s of the triangle meet.<br /> <br /> {{stub}}<br /> [[Category:Geometry]]</div> I like pie https://artofproblemsolving.com/wiki/index.php?title=Center_(algebra)&diff=25979 Center (algebra) 2008-05-12T06:48:36Z <p>I like pie: Links, minor edits</p> <hr /> <div>In general, the '''center''' of an algebraic structure is the [[set]] of [[element]]s which commute with every of the structure. With [[magma]]s (such as [[group]]s), this definition is straightforward; for [[ring]]s and [[field]]s, the commutativity in question is multiplicative commutativity.<br /> <br /> The center of a group is never empty, as the identity commutes with every element of a group. The center of a group is a [[subgroup]] of the group&amp;mdash;a [[normal subgroup]], in fact; it is also stable under any [[endomorphism]] on the group.<br /> <br /> == See also ==<br /> *[[Centralizer]]<br /> <br /> {{stub}}<br /> [[Category:Abstract algebra]]</div> I like pie https://artofproblemsolving.com/wiki/index.php?title=Surjection&diff=25924 Surjection 2008-05-09T06:23:13Z <p>I like pie: </p> <hr /> <div>A '''surjection''' is a [[function]] which takes each value in its [[codomain]] at some value in its [[domain]]. That is, the [[range]] (or [[image]]) of the function is equal to its codomain. (For every function, the range is a subset of the codomain.) In adjectival form, we say that a function is ''surjective'' or ''onto''.<br /> <br /> For instance, the function &lt;math&gt;f: \mathbb Z \to \mathbb Z&lt;/math&gt; defined by &lt;math&gt;f(x) = x+1&lt;/math&gt; is surjective because every [[integer]] is one more than some other integer, but the function &lt;math&gt;f: \mathbb N \to\mathbb N&lt;/math&gt; defined by &lt;math&gt;f(x) = x+1&lt;/math&gt; is not surjective because there exists a [[natural number]] which is not one more than any other natural number.<br /> <br /> ==See also ==<br /> * [[Bijection]]<br /> * [[Injection]]<br /> <br /> {{stub}}</div> I like pie https://artofproblemsolving.com/wiki/index.php?title=2005_USAMO_Problems&diff=25816 2005 USAMO Problems 2008-05-03T17:40:28Z <p>I like pie: Problems 4, 5, 6, authors needed</p> <hr /> <div>= Day 1 =<br /> == Problem 1 ==<br /> (''Zuming Feng'') Determine all composite positive integers &lt;math&gt;n&lt;/math&gt; for which it is possible to arrange all divisors of &lt;math&gt;n&lt;/math&gt; that are greater than 1 in a circle so that no two adjacent divisors are relatively prime.<br /> <br /> [[2005 USAMO Problems/Problem 1 | Solution]]<br /> <br /> == Problem 2 ==<br /> (''Răzvan Gelca'') Prove that the<br /> system<br /> &lt;cmath&gt;<br /> \begin{align*}<br /> x^6+x^3+x^3y+y &amp; = 147^{157} \\<br /> x^3+x^3y+y^2+y+z^9 &amp; = 157^{147}<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> has no solutions in integers &lt;math&gt;x&lt;/math&gt;, &lt;math&gt;y&lt;/math&gt;, and &lt;math&gt;z&lt;/math&gt;.<br /> <br /> [[2005 USAMO Problems/Problem 2 | Solution]]<br /> <br /> == Problem 3 ==<br /> (''Zuming Feng'') Let &lt;math&gt;ABC&lt;/math&gt; be an acute-angled triangle, and let &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt; be two points on side &lt;math&gt;BC&lt;/math&gt;. Construct point &lt;math&gt;C_1 &lt;/math&gt; in such a way that convex quadrilateral &lt;math&gt;APBC_1&lt;/math&gt; is cyclic, &lt;math&gt;QC_1 \parallel CA&lt;/math&gt;, and &lt;math&gt;C_1&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt; lie on opposite sides of line &lt;math&gt;AB&lt;/math&gt;. Construct point &lt;math&gt;B_1&lt;/math&gt; in such a way that convex quadrilateral &lt;math&gt;APCB_1&lt;/math&gt; is cyclic, &lt;math&gt;QB_1 \parallel BA &lt;/math&gt;, and &lt;math&gt;B_1 &lt;/math&gt; and &lt;math&gt;Q &lt;/math&gt; lie on opposite sides of line &lt;math&gt;AC&lt;/math&gt;. Prove that points &lt;math&gt;B_1, C_1,P&lt;/math&gt;, and &lt;math&gt;Q&lt;/math&gt; lie on a circle.<br /> <br /> [[2005 USAMO Problems/Problem 3 | Solution]]<br /> <br /> = Day 2 =<br /> == Problem 4 ==<br /> Legs &lt;math&gt;L_1,L_2,L_3,L_4&lt;/math&gt; of a square table each have length &lt;math&gt;n&lt;/math&gt;, where &lt;math&gt;n&lt;/math&gt; is a positive integer. For how many ordered 4-tuples &lt;math&gt;\left(k_1,k_2,k_3,k_4\right)&lt;/math&gt; of nonnegative integers can we cut a piece of length &lt;math&gt;k_i&lt;/math&gt; from the end of leg &lt;math&gt;L_i\ (i=1,2,3,4)&lt;/math&gt; and still have a stable table?<br /> <br /> (The table is stable if it can be placed so that all four of the leg ends touch the floor. Note that a cut leg of length 0 is permitted.)<br /> <br /> [[2005 USAMO Problems/Problem 4 | Solution]]<br /> <br /> == Problem 5 ==<br /> Let &lt;math&gt;n&lt;/math&gt; be an integer greater than 1. Suppose &lt;math&gt;2n&lt;/math&gt; points are given in the plane, no three of which are collinear. Suppose &lt;math&gt;n&lt;/math&gt; of the given &lt;math&gt;2n&lt;/math&gt; points are colored blue and the other &lt;math&gt;n&lt;/math&gt; colored red. A line in the plane is called a ''balancing line'' if it passes through one blue and one red point and, for each side of the line, the number of blue points on that side is equal to the number of red points on the same side.<br /> <br /> Prove that there exist at least two balancing lines.<br /> <br /> [[2005 USAMO Problems/Problem 5 | Solution]]<br /> <br /> == Problem 6 ==<br /> For &lt;math&gt;m&lt;/math&gt; a positive integer, let &lt;math&gt;s(m)&lt;/math&gt; be the sum of the digits of &lt;math&gt;m&lt;/math&gt;. For &lt;math&gt;n\ge 2&lt;/math&gt;, let &lt;math&gt;f(n)&lt;/math&gt; be the minimal &lt;math&gt;k&lt;/math&gt; for which there exists a set &lt;math&gt;S&lt;/math&gt; of &lt;math&gt;n&lt;/math&gt; positive integers such that &lt;math&gt;s\left(\sum_{x\in X} x\right) = k&lt;/math&gt; for any nonempty subset &lt;math&gt;X\subset S&lt;/math&gt;. Prove that there are constants &lt;math&gt;0 &lt; C_1 &lt; C_2&lt;/math&gt; with<br /> &lt;cmath&gt;<br /> C_1 \log_{10} n \le f(n) \le C_2 \log_{10} n.<br /> &lt;/cmath&gt;<br /> <br /> [[2005 USAMO Problems/Problem 6 | Solution]]<br /> <br /> = Resources =<br /> * [[USAMO Problems and Solutions]]<br /> * [http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2005-ua/questions/Day1/05USAMOday1.pdf 2005 USAMO Day 1 Problems]<br /> * [http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2005-ua/questions/Day2/05USAMOday2.pdf 2005 USAMO Day 2 Problems]<br /> * [http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2005-ua/Solutions/05SOL.pdf 2005 USAMO Solutions]<br /> * [http://www.artofproblemsolving.com/Forum/resources.php?c=182&amp;cid=27&amp;year=2005 USAMO Problems on the Resources page]<br /> {{USAMO newbox|year=2005|before=[[2004 USAMO]]|after=2006 USAMO}}</div> I like pie https://artofproblemsolving.com/wiki/index.php?title=2005_USAMO_Problems&diff=25815 2005 USAMO Problems 2008-05-03T17:28:55Z <p>I like pie: Authors</p> <hr /> <div>= Day 1 =<br /> == Problem 1 ==<br /> (''Zuming Feng'') Determine all composite positive integers &lt;math&gt;n&lt;/math&gt; for which it is possible to arrange all divisors of &lt;math&gt;n&lt;/math&gt; that are greater than 1 in a circle so that no two adjacent divisors are relatively prime.<br /> <br /> [[2005 USAMO Problems/Problem 1 | Solution]]<br /> <br /> == Problem 2 ==<br /> (''Răzvan Gelca'') Prove that the<br /> system<br /> &lt;cmath&gt;<br /> \begin{align*}<br /> x^6+x^3+x^3y+y &amp; = 147^{157} \\<br /> x^3+x^3y+y^2+y+z^9 &amp; = 157^{147}<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> has no solutions in integers &lt;math&gt;x&lt;/math&gt;, &lt;math&gt;y&lt;/math&gt;, and &lt;math&gt;z&lt;/math&gt;.<br /> <br /> [[2005 USAMO Problems/Problem 2 | Solution]]<br /> <br /> == Problem 3 ==<br /> (''Zuming Feng'') Let &lt;math&gt;ABC&lt;/math&gt; be an acute-angled triangle, and let &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt; be two points on side &lt;math&gt;BC&lt;/math&gt;. Construct point &lt;math&gt;C_1 &lt;/math&gt; in such a way that convex quadrilateral &lt;math&gt;APBC_1&lt;/math&gt; is cyclic, &lt;math&gt;QC_1 \parallel CA&lt;/math&gt;, and &lt;math&gt;C_1&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt; lie on opposite sides of line &lt;math&gt;AB&lt;/math&gt;. Construct point &lt;math&gt;B_1&lt;/math&gt; in such a way that convex quadrilateral &lt;math&gt;APCB_1&lt;/math&gt; is cyclic, &lt;math&gt;QB_1 \parallel BA &lt;/math&gt;, and &lt;math&gt;B_1 &lt;/math&gt; and &lt;math&gt;Q &lt;/math&gt; lie on opposite sides of line &lt;math&gt;AC&lt;/math&gt;. Prove that points &lt;math&gt;B_1, C_1,P&lt;/math&gt;, and &lt;math&gt;Q&lt;/math&gt; lie on a circle.<br /> <br /> [[2005 USAMO Problems/Problem 3 | Solution]]<br /> <br /> = Day 2 =<br /> == Problem 4 ==<br /> {{solution}}<br /> <br /> [[2005 USAMO Problems/Problem 4 | Solution]]<br /> <br /> == Problem 5 ==<br /> {{solution}}<br /> <br /> [[2005 USAMO Problems/Problem 5 | Solution]]<br /> <br /> == Problem 6 ==<br /> {{solution}}<br /> <br /> [[2005 USAMO Problems/Problem 6 | Solution]]<br /> <br /> = Resources =<br /> * [[USAMO Problems and Solutions]]<br /> * [http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2006-ua/2006usamoQ.pdf 2005 USAMO Problems]<br /> * [http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2006-ua/2006usamoS.pdf 2005 USAMO Solutions]<br /> * [http://www.artofproblemsolving.com/Forum/resources.php?c=182&amp;cid=27&amp;year=2005 USAMO Problems on the Resources page]<br /> {{USAMO newbox|year=2005|before=[[2004 USAMO]]|after=2006 USAMO}}</div> I like pie https://artofproblemsolving.com/wiki/index.php?title=2005_USAMO_Problems&diff=25814 2005 USAMO Problems 2008-05-03T17:27:02Z <p>I like pie: Wrong year...</p> <hr /> <div>= Day 1 =<br /> == Problem 1 ==<br /> Determine all composite positive integers &lt;math&gt;n&lt;/math&gt; for which it is possible to arrange all divisors of &lt;math&gt;n&lt;/math&gt; that are greater than 1 in a circle so that no two adjacent divisors are relatively prime.<br /> <br /> [[2005 USAMO Problems/Problem 1 | Solution]]<br /> <br /> == Problem 2 ==<br /> Prove that the<br /> system<br /> &lt;cmath&gt;<br /> \begin{align*}<br /> x^6+x^3+x^3y+y &amp; = 147^{157} \\<br /> x^3+x^3y+y^2+y+z^9 &amp; = 157^{147}<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> has no solutions in integers &lt;math&gt;x&lt;/math&gt;, &lt;math&gt;y&lt;/math&gt;, and &lt;math&gt;z&lt;/math&gt;.<br /> <br /> [[2005 USAMO Problems/Problem 2 | Solution]]<br /> <br /> == Problem 3 ==<br /> Let &lt;math&gt;ABC&lt;/math&gt; be an acute-angled triangle, and let &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt; be two points on side &lt;math&gt;BC&lt;/math&gt;. Construct point &lt;math&gt;C_1 &lt;/math&gt; in such a way that convex quadrilateral &lt;math&gt;APBC_1&lt;/math&gt; is cyclic, &lt;math&gt;QC_1 \parallel CA&lt;/math&gt;, and &lt;math&gt;C_1&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt; lie on opposite sides of line &lt;math&gt;AB&lt;/math&gt;. Construct point &lt;math&gt;B_1&lt;/math&gt; in such a way that convex quadrilateral &lt;math&gt;APCB_1&lt;/math&gt; is cyclic, &lt;math&gt;QB_1 \parallel BA &lt;/math&gt;, and &lt;math&gt;B_1 &lt;/math&gt; and &lt;math&gt;Q &lt;/math&gt; lie on opposite sides of line &lt;math&gt;AC&lt;/math&gt;. Prove that points &lt;math&gt;B_1, C_1,P&lt;/math&gt;, and &lt;math&gt;Q&lt;/math&gt; lie on a circle.<br /> <br /> [[2005 USAMO Problems/Problem 3 | Solution]]<br /> <br /> = Day 2 =<br /> == Problem 4 ==<br /> {{solution}}<br /> <br /> [[2005 USAMO Problems/Problem 4 | Solution]]<br /> <br /> == Problem 5 ==<br /> {{solution}}<br /> <br /> [[2005 USAMO Problems/Problem 5 | Solution]]<br /> <br /> == Problem 6 ==<br /> {{solution}}<br /> <br /> [[2005 USAMO Problems/Problem 6 | Solution]]<br /> <br /> = Resources =<br /> * [[USAMO Problems and Solutions]]<br /> * [http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2006-ua/2006usamoQ.pdf 2005 USAMO Problems]<br /> * [http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2006-ua/2006usamoS.pdf 2005 USAMO Solutions]<br /> * [http://www.artofproblemsolving.com/Forum/resources.php?c=182&amp;cid=27&amp;year=2005 USAMO Problems on the Resources page]<br /> {{USAMO newbox|year=2005|before=[[2004 USAMO]]|after=2006 USAMO}}</div> I like pie https://artofproblemsolving.com/wiki/index.php?title=Template:Alternate_solutions&diff=25813 Template:Alternate solutions 2008-05-03T17:25:57Z <p>I like pie: </p> <hr /> <div>''Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please [{{fullurl:{{FULLPAGENAME}}|action=edit}} add it] to this page.''&lt;noinclude&gt;<br /> <br /> This template is to be placed on problem pages, to encourage users to post alternate solutions.&lt;/noinclude&gt;</div> I like pie https://artofproblemsolving.com/wiki/index.php?title=2005_USAMO_Problems/Problem_1&diff=25812 2005 USAMO Problems/Problem 1 2008-05-03T17:24:32Z <p>I like pie: Dscussion on forum, alternate solutions, author</p> <hr /> <div>== Problem ==<br /> (''Zuming Feng'') Determine all composite positive integers &lt;math&gt;n&lt;/math&gt; for which it is possible to arrange all divisors of &lt;math&gt;n&lt;/math&gt; that are greater than 1 in a circle so that no two adjacent divisors are relatively prime.<br /> <br /> == Solution ==<br /> === Solution 1 (official solution) ===<br /> No such circular arrangement exists for &lt;math&gt;n=pq&lt;/math&gt;, where &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;q&lt;/math&gt; are distinct primes. In that case, the numbers to be arranged are &lt;math&gt;p&lt;/math&gt;; &lt;math&gt;q&lt;/math&gt; and &lt;math&gt;pq&lt;/math&gt;, and in any circular arrangement, &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;q&lt;/math&gt; will be adjacent. We claim that the desired circular arrangement exists in all other cases. If &lt;math&gt;n=p^e&lt;/math&gt; where &lt;math&gt;e\ge2&lt;/math&gt;, an arbitrary circular arrangement works. Henceforth we assume that &lt;math&gt;n&lt;/math&gt; has prime factorization &lt;math&gt;p^{e_1}_{1}p^{e_2}_{2}\cdots p^{e_k}_k&lt;/math&gt;, where &lt;math&gt;p_1&lt;p_2&lt;\cdots&lt;p_k&lt;/math&gt; and either &lt;math&gt;k&gt;2&lt;/math&gt; or else &lt;math&gt;\max(e1,e2)&gt;1&lt;/math&gt;. To construct the desired circular arrangement of &lt;math&gt;D_n:=\lbrace d:d|n\ \text{and}\ d&gt;1\rbrace&lt;/math&gt;, start with the circular arrangement of &lt;math&gt;n,p_{1}p_{2},p_{2}p_{3}\ldots,p_{k-1}p_{k}&lt;/math&gt; as shown.<br /> <br /> {{image}}<br /> <br /> Then between &lt;math&gt;n&lt;/math&gt; and &lt;math&gt;p_{1}p_{2}&lt;/math&gt;, place (in arbitrary order) all other members of &lt;math&gt;D_n&lt;/math&gt; that have &lt;math&gt;p_1&lt;/math&gt; as their smallest prime factor. Between &lt;math&gt;p_{1}p_{2}&lt;/math&gt; and &lt;math&gt;p_{2}p_{3}&lt;/math&gt;, place all members of &lt;math&gt;D_n&lt;/math&gt; other than &lt;math&gt;p_{2}p_{3}&lt;/math&gt; that have &lt;math&gt;p_2&lt;/math&gt; as their smallest prime factor. Continue in this way, ending by placing &lt;math&gt;p_k,p^{2}_{k},\ldots,p^{e_k}_{k}&lt;/math&gt; between &lt;math&gt;p_{k-1}p_k&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt;. It is easy to see that each element of &lt;math&gt;D_n&lt;/math&gt; is placed exactly one time, and any two adjacent elements have a common prime factor. Hence this arrangement has the desired property.<br /> <br /> ''Note.'' In graph theory terms, this construction yields a Hamiltonian cycle in the graph with vertex set &lt;math&gt;D_n&lt;/math&gt; in which two vertices form an edge if the two corresponding numbers have a common prime factor. The graphs below illustrate the construction for the special cases &lt;math&gt;n=p^{2}q&lt;/math&gt; and &lt;math&gt;n=pqr&lt;/math&gt;.<br /> <br /> {{alternate solutions}}<br /> <br /> == See also ==<br /> * &lt;url&gt;viewtopic.php?t=34314 Discussion on AoPS/MathLinks&lt;/url&gt;<br /> <br /> {{USAMO newbox|year=2005|before=First Question|num-a=2}}<br /> <br /> [[Category:Olympiad Number Theory Problems]]</div> I like pie https://artofproblemsolving.com/wiki/index.php?title=2005_USAMO_Problems/Problem_1&diff=25811 2005 USAMO Problems/Problem 1 2008-05-03T17:21:37Z <p>I like pie: Official solution (image needed)</p> <hr /> <div>== Problem ==<br /> Determine all composite positive integers &lt;math&gt;n&lt;/math&gt; for which it is possible to arrange all divisors of &lt;math&gt;n&lt;/math&gt; that are greater than 1 in a circle so that no two adjacent divisors are relatively prime.<br /> <br /> == Solution ==<br /> === Solution 1 (official solution) ===<br /> No such circular arrangement exists for &lt;math&gt;n=pq&lt;/math&gt;, where &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;q&lt;/math&gt; are distinct primes. In that case, the numbers to be arranged are &lt;math&gt;p&lt;/math&gt;; &lt;math&gt;q&lt;/math&gt; and &lt;math&gt;pq&lt;/math&gt;, and in any circular arrangement, &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;q&lt;/math&gt; will be adjacent. We claim that the desired circular arrangement exists in all other cases. If &lt;math&gt;n=p^e&lt;/math&gt; where &lt;math&gt;e\ge2&lt;/math&gt;, an arbitrary circular arrangement works. Henceforth we assume that &lt;math&gt;n&lt;/math&gt; has prime factorization &lt;math&gt;p^{e_1}_{1}p^{e_2}_{2}\cdots p^{e_k}_k&lt;/math&gt;, where &lt;math&gt;p_1&lt;p_2&lt;\cdots&lt;p_k&lt;/math&gt; and either &lt;math&gt;k&gt;2&lt;/math&gt; or else &lt;math&gt;\max(e1,e2)&gt;1&lt;/math&gt;. To construct the desired circular arrangement of &lt;math&gt;D_n:=\lbrace d:d|n\ \text{and}\ d&gt;1\rbrace&lt;/math&gt;, start with the circular arrangement of &lt;math&gt;n,p_{1}p_{2},p_{2}p_{3}\ldots,p_{k-1}p_{k}&lt;/math&gt; as shown.<br /> <br /> {{image}}<br /> <br /> Then between &lt;math&gt;n&lt;/math&gt; and &lt;math&gt;p_{1}p_{2}&lt;/math&gt;, place (in arbitrary order) all other members of &lt;math&gt;D_n&lt;/math&gt; that have &lt;math&gt;p_1&lt;/math&gt; as their smallest prime factor. Between &lt;math&gt;p_{1}p_{2}&lt;/math&gt; and &lt;math&gt;p_{2}p_{3}&lt;/math&gt;, place all members of &lt;math&gt;D_n&lt;/math&gt; other than &lt;math&gt;p_{2}p_{3}&lt;/math&gt; that have &lt;math&gt;p_2&lt;/math&gt; as their smallest prime factor. Continue in this way, ending by placing &lt;math&gt;p_k,p^{2}_{k},\ldots,p^{e_k}_{k}&lt;/math&gt; between &lt;math&gt;p_{k-1}p_k&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt;. It is easy to see that each element of &lt;math&gt;D_n&lt;/math&gt; is placed exactly one time, and any two adjacent elements have a common prime factor. Hence this arrangement has the desired property.<br /> <br /> ''Note.'' In graph theory terms, this construction yields a Hamiltonian cycle in the graph with vertex set &lt;math&gt;D_n&lt;/math&gt; in which two vertices form an edge if the two corresponding numbers have a common prime factor. The graphs below illustrate the construction for the special cases &lt;math&gt;n=p^{2}q&lt;/math&gt; and &lt;math&gt;n=pqr&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{USAMO newbox|year=2005|before=First Question|num-a=2}}<br /> <br /> [[Category:Olympiad Number Theory Problems]]</div> I like pie https://artofproblemsolving.com/wiki/index.php?title=2005_USAMO_Problems/Problem_3&diff=25810 2005 USAMO Problems/Problem 3 2008-05-03T17:06:09Z <p>I like pie: </p> <hr /> <div>== Problem ==<br /> (''Zuming Feng'') Let &lt;math&gt;ABC&lt;/math&gt; be an acute-angled triangle, and let &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt; be two points on side &lt;math&gt;BC&lt;/math&gt;. Construct point &lt;math&gt;C_1 &lt;/math&gt; in such a way that convex quadrilateral &lt;math&gt;APBC_1&lt;/math&gt; is cyclic, &lt;math&gt;QC_1 \parallel CA&lt;/math&gt;, and &lt;math&gt;C_1&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt; lie on opposite sides of line &lt;math&gt;AB&lt;/math&gt;. Construct point &lt;math&gt;B_1&lt;/math&gt; in such a way that convex quadrilateral &lt;math&gt;APCB_1&lt;/math&gt; is cyclic, &lt;math&gt;QB_1 \parallel BA &lt;/math&gt;, and &lt;math&gt;B_1 &lt;/math&gt; and &lt;math&gt;Q &lt;/math&gt; lie on opposite sides of line &lt;math&gt;AC&lt;/math&gt;. Prove that points &lt;math&gt;B_1, C_1,P&lt;/math&gt;, and &lt;math&gt;Q&lt;/math&gt; lie on a circle.<br /> <br /> == Solution ==<br /> Let &lt;math&gt;B_1'&lt;/math&gt; be the second intersection of the line &lt;math&gt;C_1A&lt;/math&gt; with the circumcircle of &lt;math&gt;APC&lt;/math&gt;, and let &lt;math&gt;Q'&lt;/math&gt; be the second intersection of the circumcircle of &lt;math&gt;B_1' C_1P&lt;/math&gt; and line &lt;math&gt;BC&lt;/math&gt;. It is enough to show that &lt;math&gt;B_1'=B_1&lt;/math&gt; and &lt;math&gt;Q' =Q&lt;/math&gt;. All our angles will be directed, and measured mod &lt;math&gt;\pi&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> size(300);<br /> defaultpen(1);<br /> <br /> pair A=(2,5), B=(-1,0), C=(5,0);<br /> pair C1=(.5,5.7);<br /> path O1=circumcircle(A,B,C1);<br /> pair P=IntersectionPoint(O1,B--C,1);<br /> path O2=circumcircle(A,P,C);<br /> pair B1=IntersectionPoint(O2,C1--5A-4C1,0);<br /> path O=circumcircle(B1,C1,P);<br /> pair Q=IntersectionPoint(O,B--C,1);<br /> <br /> draw(C1--P--A--B--C--A);<br /> draw(P--B1--C1--Q--B1);<br /> draw(O1,dashed+linewidth(.7));<br /> draw(O2,dashed+linewidth(.7));<br /> draw(O,dotted);<br /> <br /> label(&quot;$A$&quot;,A,N);<br /> label(&quot;$B$&quot;,B,SW);<br /> label(&quot;$C$&quot;,C,SE);<br /> label(&quot;$P$&quot;,P,S);<br /> label(&quot;$Q'$&quot;,Q,S);<br /> label(&quot;$C_1$&quot;,C1,N);<br /> label(&quot;$B_1'$&quot;,B1,E);<br /> &lt;/asy&gt;<br /> <br /> Since points &lt;math&gt;C_1, P, Q', B_1'&lt;/math&gt; are concyclic and points &lt;math&gt;C_1, A,B_1'&lt;/math&gt; are collinear, it follows that<br /> &lt;cmath&gt; \angle C_1 Q' P \equiv \angle C_1 B_1' P \equiv \angle A B_1' P . &lt;/cmath&gt;<br /> But since points &lt;math&gt;A, B_1', P, C&lt;/math&gt; are concyclic,<br /> &lt;cmath&gt; \angle AB_1'P \equiv \angle ACP . &lt;/cmath&gt;<br /> It follows that lines &lt;math&gt;AC&lt;/math&gt; and &lt;math&gt;C_1 Q'&lt;/math&gt; are parallel. If we exchange &lt;math&gt;C&lt;/math&gt; with &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;C_1&lt;/math&gt; with &lt;math&gt;B_1'&lt;/math&gt; in this argument, we see that lines &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;B_1' Q'&lt;/math&gt; are likewise parallel.<br /> <br /> It follows that &lt;math&gt;Q'&lt;/math&gt; is the intersection of &lt;math&gt;BC&lt;/math&gt; and the line parallel to &lt;math&gt;AC&lt;/math&gt; and passing through &lt;math&gt;C_1&lt;/math&gt;. Hence &lt;math&gt;Q' = Q&lt;/math&gt;. Then &lt;math&gt;B_1'&lt;/math&gt; is the second intersection of the circumcircle of &lt;math&gt;APC&lt;/math&gt; and the line parallel to &lt;math&gt;AB&lt;/math&gt; passing through &lt;math&gt;Q&lt;/math&gt;. Hence &lt;math&gt;B_1' = B_1&lt;/math&gt;, as desired. &lt;math&gt;\blacksquare&lt;/math&gt;<br /> <br /> {{alternate solutions}}<br /> <br /> == See also ==<br /> * &lt;url&gt;Forum/viewtopic.php?p=213011#213011 Discussion on AoPS/MathLinks&lt;/url&gt;<br /> <br /> {{USAMO newbox|year=2005|num-b=2|num-a=4}}<br /> <br /> [[Category:Olympiad Geometry Problems]]</div> I like pie https://artofproblemsolving.com/wiki/index.php?title=2005_USAMO_Problems/Problem_2&diff=25809 2005 USAMO Problems/Problem 2 2008-05-03T17:05:22Z <p>I like pie: </p> <hr /> <div>== Problem ==<br /> (''Răzvan Gelca'') Prove that the system<br /> &lt;cmath&gt;<br /> \begin{align*}x^6 + x^3 + x^3y + y &amp;= 147^{157} \\<br /> x^3 + x^3y + y^2 + y + z^9 &amp;= 157^{147}\end{align*}<br /> &lt;/cmath&gt;<br /> has no solutions in integers &lt;math&gt;x&lt;/math&gt;, &lt;math&gt;y&lt;/math&gt;, and &lt;math&gt;z&lt;/math&gt;.<br /> <br /> == Solution ==<br /> It suffices to show that there are no solutions to this system in the integers mod 19. We note that &lt;math&gt;152 = 8 \cdot 19&lt;/math&gt;, so &lt;math&gt;157 \equiv -147 \equiv 5 \pmod{19}&lt;/math&gt;. For reference, we construct a table of powers of five:<br /> &lt;cmath&gt; \begin{array}{c|c||c|c}<br /> n&amp; 5^n &amp;n &amp; 5^n \\\hline<br /> 1 &amp; 5 &amp; 6 &amp; 7 \\<br /> 2 &amp; 6 &amp; 7 &amp; -3 \\<br /> 3 &amp; -8 &amp; 8 &amp; 4 \\<br /> 4 &amp; -2 &amp; 9 &amp; 1 \\<br /> 5 &amp; 9 &amp;&amp;<br /> \end{array} &lt;/cmath&gt;<br /> Evidently, then the order of 5 is 9. Hence 5 is the square of a multiplicative generator of the nonzero integers mod 19, so this table shows all nonzero squares mod 19, as well.<br /> <br /> It follows that &lt;math&gt;147^{157} \equiv (-5)^{13} \equiv -5^4 \equiv 2&lt;/math&gt;, and &lt;math&gt;157^{147} \equiv 5^3 \equiv -8&lt;/math&gt;. Thus we rewrite our system thus:<br /> &lt;cmath&gt; \begin{align*}<br /> (x^3+y)(x^3+1) &amp;\equiv 2 \\<br /> (x^3+y)(y+1) + z^9 &amp;\equiv -8 .<br /> \end{align*} &lt;/cmath&gt;<br /> Adding these, we have<br /> &lt;cmath&gt; (x^3+y+1)^2 - 1 + z^9 &amp;\equiv -6, &lt;/cmath&gt;<br /> or<br /> &lt;cmath&gt; (x^3+y+1)^2 \equiv -z^9 - 5 . &lt;/cmath&gt;<br /> By [[Fermat's Little Theorem]], the only possible values of &lt;math&gt;z^9&lt;/math&gt; are &lt;math&gt;\pm 1&lt;/math&gt; and 0, so the only possible values of &lt;math&gt;(x^3+y+1)^2&lt;/math&gt; are &lt;math&gt;-4,-5&lt;/math&gt;, and &lt;math&gt;-6&lt;/math&gt;. But none of these are squares mod 19, a contradiction. Therefore the system has no solutions in the integers mod 19. Therefore the solution has no equation in the integers. &lt;math&gt;\blacksquare&lt;/math&gt;<br /> <br /> {{alternate solutions}}<br /> <br /> == See also ==<br /> * &lt;url&gt;Forum/viewtopic.php?p=213009#213009 Discussion on AoPS/MathLinks&lt;/url&gt;<br /> <br /> {{USAMO newbox|year=2005|num-b=1|num-a=3}}<br /> <br /> [[Category:Olympiad Number Theory Problems]]</div> I like pie https://artofproblemsolving.com/wiki/index.php?title=2005_USAMO_Problems/Problem_1&diff=25808 2005 USAMO Problems/Problem 1 2008-05-03T17:04:38Z <p>I like pie: </p> <hr /> <div>== Problem ==<br /> Determine all composite positive integers &lt;math&gt;n&lt;/math&gt; for which it is possible to arrange all divisors of &lt;math&gt;n&lt;/math&gt; that are greater than 1 in a circle so that no two adjacent divisors are relatively prime.<br /> <br /> == Solution ==<br /> {{solution}}<br /> <br /> == See also ==<br /> {{USAMO newbox|year=2005|before=First Question|num-a=2}}<br /> <br /> [[Category:Olympiad Number Theory Problems]]</div> I like pie https://artofproblemsolving.com/wiki/index.php?title=2005_USAMO_Problems&diff=25807 2005 USAMO Problems 2008-05-03T17:03:35Z <p>I like pie: USAMO box, {{problem}} tags, minor edits</p> <hr /> <div>= Day 1 =<br /> == Problem 1 ==<br /> Determine all composite positive integers &lt;math&gt;n&lt;/math&gt; for which it is possible to arrange all divisors of &lt;math&gt;n&lt;/math&gt; that are greater than 1 in a circle so that no two adjacent divisors are relatively prime.<br /> <br /> [[2005 USAMO Problems/Problem 1 | Solution]]<br /> <br /> == Problem 2 ==<br /> Prove that the<br /> system<br /> &lt;cmath&gt;<br /> \begin{align*}<br /> x^6+x^3+x^3y+y &amp; = 147^{157} \\<br /> x^3+x^3y+y^2+y+z^9 &amp; = 157^{147}<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> has no solutions in integers &lt;math&gt;x&lt;/math&gt;, &lt;math&gt;y&lt;/math&gt;, and &lt;math&gt;z&lt;/math&gt;.<br /> <br /> [[2005 USAMO Problems/Problem 2 | Solution]]<br /> <br /> == Problem 3 ==<br /> Let &lt;math&gt;ABC&lt;/math&gt; be an acute-angled triangle, and let &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt; be two points on side &lt;math&gt;BC&lt;/math&gt;. Construct point &lt;math&gt;C_1 &lt;/math&gt; in such a way that convex quadrilateral &lt;math&gt;APBC_1&lt;/math&gt; is cyclic, &lt;math&gt;QC_1 \parallel CA&lt;/math&gt;, and &lt;math&gt;C_1&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt; lie on opposite sides of line &lt;math&gt;AB&lt;/math&gt;. Construct point &lt;math&gt;B_1&lt;/math&gt; in such a way that convex quadrilateral &lt;math&gt;APCB_1&lt;/math&gt; is cyclic, &lt;math&gt;QB_1 \parallel BA &lt;/math&gt;, and &lt;math&gt;B_1 &lt;/math&gt; and &lt;math&gt;Q &lt;/math&gt; lie on opposite sides of line &lt;math&gt;AC&lt;/math&gt;. Prove that points &lt;math&gt;B_1, C_1,P&lt;/math&gt;, and &lt;math&gt;Q&lt;/math&gt; lie on a circle.<br /> <br /> [[2005 USAMO Problems/Problem 3 | Solution]]<br /> <br /> = Day 2 =<br /> == Problem 4 ==<br /> {{solution}}<br /> <br /> [[2005 USAMO Problems/Problem 4 | Solution]]<br /> <br /> == Problem 5 ==<br /> {{solution}}<br /> <br /> [[2005 USAMO Problems/Problem 5 | Solution]]<br /> <br /> == Problem 6 ==<br /> {{solution}}<br /> <br /> [[2005 USAMO Problems/Problem 6 | Solution]]<br /> <br /> = Resources =<br /> * [[USAMO Problems and Solutions]]<br /> * [http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2006-ua/2006usamoQ.pdf 2005 USAMO Problems]<br /> * [http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2006-ua/2006usamoS.pdf 2005 USAMO Solutions]<br /> * [http://www.artofproblemsolving.com/Forum/resources.php?c=182&amp;cid=27&amp;year=2005 USAMO Problems on the Resources page]<br /> {{USAMO newbox|year=2006|before=[[2005 USAMO]]|after=2007 USAMO}}</div> I like pie https://artofproblemsolving.com/wiki/index.php?title=2006_USAMO_Problems&diff=25804 2006 USAMO Problems 2008-05-03T05:51:05Z <p>I like pie: Added USAMO box, minor edits</p> <hr /> <div>= Day 1 =<br /> == Problem 1 ==<br /> Let &lt;math&gt;p&lt;/math&gt; be a prime number and let &lt;math&gt;s&lt;/math&gt; be an integer with &lt;math&gt;0&lt;s&lt;p&lt;/math&gt;. Prove that there exist integers &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; with &lt;math&gt;0&lt;m&lt;n&lt;p&lt;/math&gt; and<br /> &lt;cmath&gt;\left\lbrace\frac{sm}{p}\right\rbrace&lt;\left\lbrace\frac{sn}{p}\right\rbrace&lt;\frac{s}{p}&lt;/cmath&gt;<br /> if and only if &lt;math&gt;s&lt;/math&gt; is not a divisor of &lt;math&gt;p-1&lt;/math&gt;.<br /> <br /> Note: For &lt;math&gt;x&lt;/math&gt; a real number, let &lt;math&gt;\lfloor x\rfloor&lt;/math&gt; denote the greatest integer less than or equal to &lt;math&gt;x&lt;/math&gt;, and let &lt;math&gt;\lbrace x\rbrace=x-\lfloor x\rfloor&lt;/math&gt; denote the fractional part of &lt;math&gt;x&lt;/math&gt;.<br /> <br /> [[2006 USAMO Problems/Problem 1 | Solution]]<br /> <br /> == Problem 2 ==<br /> For a given positive integer &lt;math&gt;k&lt;/math&gt; find, in terms of &lt;math&gt;k&lt;/math&gt;, the minimum value of &lt;math&gt;N&lt;/math&gt; for which there is a set of &lt;math&gt;2k+1&lt;/math&gt; distinct positive integers that has sum greater than &lt;math&gt;N &lt;/math&gt; but every subset of size &lt;math&gt;k&lt;/math&gt; has sum at most &lt;math&gt;N/2&lt;/math&gt;.<br /> <br /> [[2006 USAMO Problems/Problem 2 | Solution]]<br /> <br /> == Problem 3 ==<br /> For integral &lt;math&gt;m&lt;/math&gt;, let &lt;math&gt;p(m)&lt;/math&gt; be the greatest prime divisor of &lt;math&gt;m&lt;/math&gt;. By convention, we set &lt;math&gt; p(\pm1)=1&lt;/math&gt; and &lt;math&gt;p(0)=\infty&lt;/math&gt;. Find all polynomials &lt;math&gt;f&lt;/math&gt; with integer coefficients such that the sequence &lt;math&gt;\lbrace p(f(n^2))-2n)\rbrace_{n\ge0}&lt;/math&gt; is bounded above. (In particular, this requires &lt;math&gt;f(n^2)\neq0&lt;/math&gt; for &lt;math&gt;n\ge0&lt;/math&gt;.)<br /> <br /> [[2006 USAMO Problems/Problem 3 | Solution]]<br /> <br /> = Day 2 =<br /> == Problem 4 ==<br /> Find all positive integers &lt;math&gt;n&lt;/math&gt; such that there are &lt;math&gt;k\ge 2&lt;/math&gt; positive rational numbers &lt;math&gt;a_1,a_2,\ldots a_k&lt;/math&gt; satisfying &lt;math&gt;a_1+a_2+\ldots+a_k=a_1\cdot a_2\cdots a_k=n&lt;/math&gt;.<br /> <br /> [[2006 USAMO Problems/Problem 4 | Solution]]<br /> <br /> == Problem 5 ==<br /> 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 &lt;math&gt;n &lt;/math&gt;, then it can jump either to &lt;math&gt;n+1&lt;/math&gt; or to &lt;math&gt;n+2^{m_n+1}&lt;/math&gt; where &lt;math&gt;2^{m_n}&lt;/math&gt; is the largest power of 2 that is a factor of &lt;math&gt;n&lt;/math&gt;. Show that if &lt;math&gt;k\ge2&lt;/math&gt; is a positive integer and &lt;math&gt;i&lt;/math&gt; is a nonnegative integer, then the minimum number of jumps needed to reach &lt;math&gt;2^ik&lt;/math&gt; is greater than the minimum number of jumps needed to reach &lt;math&gt;2^i&lt;/math&gt;.<br /> <br /> [[2006 USAMO Problems/Problem 5 | Solution]]<br /> <br /> == Problem 6 ==<br /> Let &lt;math&gt;ABCD &lt;/math&gt; be a quadrilateral, and let &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;F&lt;/math&gt; be points on sides &lt;math&gt;AD&lt;/math&gt; and &lt;math&gt;BC&lt;/math&gt;, respectively, such that &lt;math&gt;AE/ED=BF/FC&lt;/math&gt;. Ray &lt;math&gt;FE&lt;/math&gt; meets rays &lt;math&gt;BA&lt;/math&gt; and &lt;math&gt;CD&lt;/math&gt; at &lt;math&gt;S&lt;/math&gt; and &lt;math&gt;T&lt;/math&gt; respectively. Prove that the circumcircles of triangles &lt;math&gt;SAE&lt;/math&gt;, &lt;math&gt;SBF&lt;/math&gt;, &lt;math&gt;TCF&lt;/math&gt;, and &lt;math&gt;TDE&lt;/math&gt; pass through a common point.<br /> <br /> [[2006 USAMO Problems/Problem 6 | Solution]]<br /> <br /> == Resources ==<br /> *[[USAMO Problems and Solutions]]<br /> *[http://www.artofproblemsolving.com/Forum/resources.php?c=182&amp;cid=27&amp;year=2006 USAMO 2006 Problems on the Resources Page]<br /> *[http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2006-ua/2006usamoQ.pdf USAMO 2006 Questions Document]<br /> *[http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2006-ua/2006usamoS.pdf USAMO 2006 Solutions Document]<br /> <br /> {{USAMO newbox|year=2006|before=[[2005 USAMO]]|after=2007 USAMO}}</div> I like pie https://artofproblemsolving.com/wiki/index.php?title=2007_USAMO_Problems&diff=25803 2007 USAMO Problems 2008-05-03T05:45:37Z <p>I like pie: </p> <hr /> <div>=Day 1=<br /> ==Problem 1==<br /> Let &lt;math&gt;n&lt;/math&gt; be a positive integer. Define a sequence by setting &lt;math&gt;a_1=n&lt;/math&gt; and, for each &lt;math&gt;k&gt;1&lt;/math&gt;, letting &lt;math&gt;a_k&lt;/math&gt; be the unique integer in the range &lt;math&gt;0\le a_k\le k-1&lt;/math&gt; for which &lt;math&gt;a_1+a_2+\cdots+a_k&lt;/math&gt; is divisible by &lt;math&gt;k&lt;/math&gt;. For instance, when &lt;math&gt;n=9&lt;/math&gt; the obtained sequence is &lt;math&gt;9, 1, 2, 0, 3, 3, 3, \ldots&lt;/math&gt;. Prove that for any &lt;math&gt;n&lt;/math&gt; the sequence &lt;math&gt;a_1, a_2, a_3, \ldots&lt;/math&gt; eventually becomes constant.<br /> <br /> [[2007 USAMO Problems/Problem 1 | Solution]]<br /> <br /> ==Problem 2==<br /> A square grid on the Euclidean plane consists of all points &lt;math&gt;(m,n)&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are integers. Is it possible to cover all grid points by an infinite family of discs with non-overlapping interiors if each disc in the family has radius at least 5?<br /> <br /> [[2007 USAMO Problems/Problem 2 | Solution]]<br /> <br /> ==Problem 3==<br /> Let &lt;math&gt;S&lt;/math&gt; be a set containing &lt;math&gt;n^2+n-1&lt;/math&gt; elements, for some positive integer &lt;math&gt;n&lt;/math&gt;. Suppose that the &lt;math&gt;n&lt;/math&gt;-element subsets of &lt;math&gt;S&lt;/math&gt; are partitioned into two classes. Prove that there are at least &lt;math&gt;n&lt;/math&gt; pairwise disjoint sets in the same class.<br /> <br /> [[2007 USAMO Problems/Problem 3 | Solution]]<br /> <br /> =Day 2=<br /> ==Problem 4==<br /> An ''animal'' with &lt;math&gt;n&lt;/math&gt; ''cells'' is a connected figure consisting of &lt;math&gt;n&lt;/math&gt; equal-sized cells.&lt;math&gt;{}^1&lt;/math&gt; The figure below shows an 8-cell animal.<br /> <br /> [[Image:2007 USAMO-4.PNG|300px|center]]<br /> <br /> A ''dinosaur'' is an animal with at least 2007 cells. It is said to be ''primitive'' it its cells cannot be partitioned into two or more dinosaurs. Find with proof the maximum number of cells in a primitive dinosaur.<br /> <br /> &lt;small&gt;&lt;math&gt;{}^1&lt;/math&gt;Animals are also called ''polyominoes''. They can be defined inductively. Two cells are ''adjacent'' if they share a complete edge. A single cell is an animal, and given an animal with &lt;math&gt;n&lt;/math&gt; cells, one with &lt;math&gt;n+1&lt;/math&gt; cells is obtained by adjoining a new cell by making it adjacent to one or more existing cells.&lt;/small&gt;<br /> <br /> [[2007 USAMO Problems/Problem 4 | Solution]]<br /> <br /> ==Problem 5==<br /> Prove that for every nonnegative integer &lt;math&gt;n&lt;/math&gt;, the number &lt;math&gt;7^{7^n}+1&lt;/math&gt; is the product of at least &lt;math&gt;2n+3&lt;/math&gt; (not necessarily distinct) primes.<br /> <br /> [[2007 USAMO Problems/Problem 5 | Solution]]<br /> <br /> ==Problem 6==<br /> Let &lt;math&gt;ABC&lt;/math&gt; be an acute triangle with &lt;math&gt;\omega&lt;/math&gt;, &lt;math&gt;\Omega&lt;/math&gt;, and &lt;math&gt;R&lt;/math&gt; being its incircle, circumcircle, and circumradius, respectively. Circle &lt;math&gt;\omega_A&lt;/math&gt; is tangent internally to &lt;math&gt;\Omega&lt;/math&gt; at &lt;math&gt;A&lt;/math&gt; and tangent externally to &lt;math&gt;\omega&lt;/math&gt;. Circle &lt;math&gt;\Omega_A&lt;/math&gt; is tangent internally to &lt;math&gt;\Omega&lt;/math&gt; at &lt;math&gt;A&lt;/math&gt; and tangent internally to &lt;math&gt;\omega&lt;/math&gt;. Let &lt;math&gt;P_A&lt;/math&gt; and &lt;math&gt;Q_A&lt;/math&gt; denote the centers of &lt;math&gt;\omega_A&lt;/math&gt; and &lt;math&gt;\Omega_A&lt;/math&gt;, respectively. Define points &lt;math&gt;P_B&lt;/math&gt;, &lt;math&gt;Q_B&lt;/math&gt;, &lt;math&gt;P_C&lt;/math&gt;, &lt;math&gt;Q_C&lt;/math&gt; analogously. Prove that<br /> &lt;cmath&gt;8P_AQ_A \cdot P_BQ_B \cdot P_CQ_C \le R^3,&lt;/cmath&gt;<br /> with equality if and only if triangle &lt;math&gt;ABC&lt;/math&gt; is equilateral.<br /> <br /> [[2007 USAMO Problems/Problem 6 | Solution]]<br /> <br /> = See also =<br /> *[[USAMO Problems and Solutions]]<br /> <br /> {{USAMO newbox|year=2007|before=[[2006 USAMO]]|after=2008 USAMO}}</div> I like pie https://artofproblemsolving.com/wiki/index.php?title=2008_USAMO_Problems&diff=25802 2008 USAMO Problems 2008-05-03T05:42:52Z <p>I like pie: Problems, links to solutions, USAMO box</p> <hr /> <div>=Day 1=<br /> ==Problem 1==<br /> (''Titu Andreescu'') Prove that for each positive integer &lt;math&gt;n&lt;/math&gt;, there are pairwise relatively prime integers &lt;math&gt;k_0,k_1\ldots,k_n&lt;/math&gt;, all strictly greater than 1, such that &lt;math&gt;k_0k_1\cdots k_n-1&lt;/math&gt; is the product of two consecutive integers.<br /> <br /> [[2008 USAMO Problems/Problem 1|Solution]]<br /> <br /> ==Problem 2==<br /> (''Zuming Feng'') Let &lt;math&gt;ABC&lt;/math&gt; be an acute, [[scalene]] triangle, and let &lt;math&gt;M&lt;/math&gt;, &lt;math&gt;N&lt;/math&gt;, and &lt;math&gt;P&lt;/math&gt; be the midpoints of &lt;math&gt;\overline{BC}&lt;/math&gt;, &lt;math&gt;\overline{CA}&lt;/math&gt;, and &lt;math&gt;\overline{AB}&lt;/math&gt;, respectively. Let the [[perpendicular]] [[bisect]]ors of &lt;math&gt;\overline{AB}&lt;/math&gt; and &lt;math&gt;\overline{AC}&lt;/math&gt; intersect ray &lt;math&gt;AM&lt;/math&gt; in points &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; respectively, and let lines &lt;math&gt;BD&lt;/math&gt; and &lt;math&gt;CE&lt;/math&gt; intersect in point &lt;math&gt;F&lt;/math&gt;, inside of triangle &lt;math&gt;ABC&lt;/math&gt;. Prove that points &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;N&lt;/math&gt;, &lt;math&gt;F&lt;/math&gt;, and &lt;math&gt;P&lt;/math&gt; all lie on one circle.<br /> <br /> [[2008 USAMO Problems/Problem 2|Solution]]<br /> <br /> ==Problem 3==<br /> (''Gabriel Carroll'') Let &lt;math&gt;n&lt;/math&gt; be a positive integer. Denote by &lt;math&gt;S_n&lt;/math&gt; the set of points &lt;math&gt;(x, y)&lt;/math&gt; with integer coordinates such that<br /> &lt;cmath&gt;\left|x\right|+\left|y+\frac{1}{2}\right|&lt;n&lt;/cmath&gt;<br /> A path is a sequence of distinct points &lt;math&gt;(x_1,y_1),(x_2,y_2),\ldots,(x_\ell,y_\ell)&lt;/math&gt; in &lt;math&gt;S_n&lt;/math&gt; such that, for &lt;math&gt;i=2,\ldots,\ell&lt;/math&gt;, the distance between &lt;math&gt;(x_i,y_i)&lt;/math&gt; and &lt;math&gt;(x_{i-1},y_{i-1})&lt;/math&gt; is &lt;math&gt;1&lt;/math&gt; (in other words, the points &lt;math&gt;(x_i,y_i)&lt;/math&gt; and &lt;math&gt;(x_{i-1},y_{i-1})&lt;/math&gt; are neighbors in the lattice of points with integer coordinates). Prove that the points in &lt;math&gt;S_n&lt;/math&gt; cannot be partitioned into fewer than &lt;math&gt;n&lt;/math&gt; paths (a partition of &lt;math&gt;S_n&lt;/math&gt; into &lt;math&gt;m&lt;/math&gt; paths is a set &lt;math&gt;\mathcal{P}&lt;/math&gt; of &lt;math&gt;m&lt;/math&gt; nonempty paths such that each point in &lt;math&gt;S_n&lt;/math&gt; appears in exactly one of the &lt;math&gt;m&lt;/math&gt; paths in &lt;math&gt;\mathcal{P}&lt;/math&gt;).<br /> <br /> [[2008 USAMO Problems/Problem 3|Solution]]<br /> <br /> =Day 2=<br /> ==Problem 4==<br /> (''Gregory Galparin'') Let &lt;math&gt;\mathcal{P}&lt;/math&gt; be a [[convex polygon]] with &lt;math&gt;n&lt;/math&gt; sides, &lt;math&gt;n\ge3&lt;/math&gt;. Any set of &lt;math&gt;n-3&lt;/math&gt; diagonals of &lt;math&gt;\mathcal{P}&lt;/math&gt; that do not intersect in the interior of the polygon determine a ''triangulation'' of &lt;math&gt;\mathcal{P}&lt;/math&gt; into &lt;math&gt;n - 2&lt;/math&gt; triangles. If &lt;math&gt;\mathcal{P}&lt;/math&gt; is regular and there is a triangulation of &lt;math&gt;\mathcal{P}&lt;/math&gt; consisting of only isosceles triangles, find all the possible values of &lt;math&gt;n&lt;/math&gt;.<br /> <br /> [[2008 USAMO Problems/Problem 4|Solution]]<br /> <br /> ==Problem 5==<br /> (''Kiran Kedlaya'') Three nonnegative real numbers &lt;math&gt;r_1&lt;/math&gt;, &lt;math&gt;r_2&lt;/math&gt;, &lt;math&gt;r_3&lt;/math&gt; are written on a blackboard. These numbers have the property that there exist integers &lt;math&gt;a_1&lt;/math&gt;, &lt;math&gt;a_2&lt;/math&gt;, &lt;math&gt;a_3&lt;/math&gt;, not all zero, satisfying &lt;math&gt;a_1r_1+a_2r_2+a_3r_3=0&lt;/math&gt;. We are permitted to perform the following operation: find two numbers &lt;math&gt;x&lt;/math&gt;, &lt;math&gt;y&lt;/math&gt; on the blackboard with &lt;math&gt;x\le y&lt;/math&gt;, then erase &lt;math&gt;y&lt;/math&gt; and write &lt;math&gt;y-x&lt;/math&gt; in its place. Prove that after a finite number of such operations, we can end up with at least one &lt;math&gt;0&lt;/math&gt; on the blackboard.<br /> <br /> [[2008 USAMO Problems/Problem 5|Solution]]<br /> <br /> ==Problem 6==<br /> (''[[Sam Vandervelde]]'') At a certain mathematical conference, every pair of mathematicians are either friends or strangers. At mealtime, every participant eats in one of two large dining rooms. Each mathematician insists upon eating in a room which contains an even number of his or her friends. Prove that the number of ways that the mathematicians may be split between the two rooms is a power of two (i.e., is of the form &lt;math&gt;2^k&lt;/math&gt; for some positive integer &lt;math&gt;k&lt;/math&gt;).<br /> <br /> [[2008 USAMO Problems/Problem 6|Solution]]<br /> <br /> = See also =<br /> *[[USAMO Problems and Solutions]]<br /> <br /> {{USAMO newbox|year=2008|before=[[2007 USAMO]]|after=2009 USAMO}}</div> I like pie https://artofproblemsolving.com/wiki/index.php?title=2008_USAMO_Problems/Problem_3&diff=25788 2008 USAMO Problems/Problem 3 2008-05-01T23:59:16Z <p>I like pie: Fixed link</p> <hr /> <div>== Problem ==<br /> (''Gabriel Carroll'') Let &lt;math&gt;n&lt;/math&gt; be a positive integer. Denote by &lt;math&gt;S_n&lt;/math&gt; the set of points &lt;math&gt;(x, y)&lt;/math&gt; with integer coordinates such that<br /> &lt;cmath&gt;\left|x\right| + \left|y + \frac {1}{2}\right| &lt; n&lt;/cmath&gt;<br /> A path is a sequence of distinct points &lt;math&gt;(x_1 , y_1 ), (x_2 , y_2 ), \ldots , (x_\ell, y_\ell)&lt;/math&gt; in &lt;math&gt;S_n&lt;/math&gt; such that, for &lt;math&gt;i = 2, \ldots , \ell&lt;/math&gt;, the distance between &lt;math&gt;(x_i , y_i )&lt;/math&gt; and &lt;math&gt;(x_{i - 1} , y_{i - 1} )&lt;/math&gt; is &lt;math&gt;1&lt;/math&gt; (in other words, the points &lt;math&gt;(x_i , y_i )&lt;/math&gt; and &lt;math&gt;(x_{i - 1} , y_{i - 1} )&lt;/math&gt; are neighbors in the lattice of points with integer coordinates). Prove that the points in &lt;math&gt;S_n&lt;/math&gt; cannot be partitioned into fewer than &lt;math&gt;n&lt;/math&gt; paths (a partition of &lt;math&gt;S_n&lt;/math&gt; into &lt;math&gt;m&lt;/math&gt; paths is a set &lt;math&gt;\mathcal{P}&lt;/math&gt; of &lt;math&gt;m&lt;/math&gt; nonempty paths such that each point in &lt;math&gt;S_n&lt;/math&gt; appears in exactly one of the &lt;math&gt;m&lt;/math&gt; paths in &lt;math&gt;\mathcal{P}&lt;/math&gt;).<br /> <br /> __TOC__<br /> == Solution ==<br /> === Solution 1 ===<br /> === Solution 2 ===<br /> <br /> {{alternate solutions}}<br /> <br /> == Resources ==<br /> {{USAMO newbox|year=2008|num-b=2|num-a=4}}<br /> <br /> * &lt;url&gt;viewtopic.php?t=202936 Discussion on AoPS/MathLinks&lt;/url&gt;<br /> <br /> [[Category:Olympiad Combinatorics Problems]]</div> I like pie https://artofproblemsolving.com/wiki/index.php?title=2008_USAMO_Problems/Problem_6&diff=25787 2008 USAMO Problems/Problem 6 2008-05-01T23:58:46Z <p>I like pie: Fixed link</p> <hr /> <div>== Problem ==<br /> (''Sam Vandervelde'') At a certain mathematical conference, every pair of mathematicians are either friends or strangers. At mealtime, every participant eats in one of two large dining rooms. Each mathematician insists upon eating in a room which contains an even number of his or her friends. Prove that the number of ways that the mathematicians may be split between the two rooms is a power of two (i.e., is of the form &lt;math&gt;2^k&lt;/math&gt; for some positive integer &lt;math&gt;k&lt;/math&gt;).<br /> <br /> __TOC__<br /> == Solution ==<br /> === Solution 1 ===<br /> === Solution 2 ===<br /> <br /> {{alternate solutions}}<br /> <br /> == Resources ==<br /> {{USAMO newbox|year=2008|num-b=5|after=Last Problem}}<br /> <br /> * &lt;url&gt;viewtopic.php?t=202908 Discussion on AoPS/MathLinks&lt;/url&gt;<br /> <br /> [[Category:Olympiad Combinatorics Problems]]</div> I like pie https://artofproblemsolving.com/wiki/index.php?title=2008_USAMO_Problems/Problem_5&diff=25786 2008 USAMO Problems/Problem 5 2008-05-01T23:58:25Z <p>I like pie: Fixed link</p> <hr /> <div>== Problem ==<br /> (''Kiran Kedlaya'') Three nonnegative real numbers &lt;math&gt;r_1&lt;/math&gt;, &lt;math&gt;r_2&lt;/math&gt;, &lt;math&gt;r_3&lt;/math&gt; are written on a blackboard. These numbers have the property that there exist integers &lt;math&gt;a_1&lt;/math&gt;, &lt;math&gt;a_2&lt;/math&gt;, &lt;math&gt;a_3&lt;/math&gt;, not all zero, satisfying &lt;math&gt;a_1r_1 + a_2r_2 + a_3r_3 = 0&lt;/math&gt;. We are permitted to perform the following operation: find two numbers &lt;math&gt;x&lt;/math&gt;, &lt;math&gt;y&lt;/math&gt; on the blackboard with &lt;math&gt;x \le y&lt;/math&gt;, then erase &lt;math&gt;y&lt;/math&gt; and write &lt;math&gt;y - x&lt;/math&gt; in its place. Prove that after a finite number of such operations, we can end up with at least one &lt;math&gt;0&lt;/math&gt; on the blackboard.<br /> <br /> __TOC__<br /> == Solution ==<br /> === Solution 1 ===<br /> === Solution 2 ===<br /> <br /> {{alternate solutions}}<br /> <br /> == Resources ==<br /> {{USAMO newbox|year=2008|num-b=4|num-a=6}}<br /> <br /> * &lt;url&gt;viewtopic.php?t=202910 Discussion on AoPS/MathLinks&lt;/url&gt;<br /> <br /> [[Category:Olympiad Number Theory Problems]]</div> I like pie https://artofproblemsolving.com/wiki/index.php?title=2008_USAMO_Problems/Problem_2&diff=25785 2008 USAMO Problems/Problem 2 2008-05-01T23:57:55Z <p>I like pie: Fixed link</p> <hr /> <div>== Problem ==<br /> (''Zuming Feng'') Let &lt;math&gt;ABC&lt;/math&gt; be an acute, scalene triangle, and let &lt;math&gt;M&lt;/math&gt;, &lt;math&gt;N&lt;/math&gt;, and &lt;math&gt;P&lt;/math&gt; be the midpoints of &lt;math&gt;\overline{BC}&lt;/math&gt;, &lt;math&gt;\overline{CA}&lt;/math&gt;, and &lt;math&gt;\overline{AB}&lt;/math&gt;, respectively. Let the perpendicular bisectors of &lt;math&gt;\overline{AB}&lt;/math&gt; and &lt;math&gt;\overline{AC}&lt;/math&gt; intersect ray &lt;math&gt;AM&lt;/math&gt; in points &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; respectively, and let lines &lt;math&gt;BD&lt;/math&gt; and &lt;math&gt;CE&lt;/math&gt; intersect in point &lt;math&gt;F&lt;/math&gt;, inside of triangle &lt;math&gt;ABC&lt;/math&gt;. Prove that points &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;N&lt;/math&gt;, &lt;math&gt;F&lt;/math&gt;, and &lt;math&gt;P&lt;/math&gt; all lie on one circle.<br /> <br /> __TOC__<br /> == Solution ==<br /> === Solution 1 ===<br /> === Solution 2 ===<br /> <br /> {{alternate solutions}}<br /> <br /> == Resources ==<br /> {{USAMO newbox|year=2008|num-b=1|num-a=3}}<br /> <br /> * &lt;url&gt;viewtopic.php?t=202907 Discussion on AoPS/MathLinks&lt;/url&gt;<br /> <br /> [[Category:Olympiad Geometry Problems]]</div> I like pie https://artofproblemsolving.com/wiki/index.php?title=2008_USAMO_Problems/Problem_1&diff=25784 2008 USAMO Problems/Problem 1 2008-05-01T23:57:41Z <p>I like pie: Fixed link</p> <hr /> <div>== Problem ==<br /> (''Titu Andreescu'')<br /> Prove that for each positive integer &lt;math&gt;n&lt;/math&gt;, there are pairwise relatively prime integers &lt;math&gt;k_0, k_1 \dotsc, k_n&lt;/math&gt;, all strictly greater than 1, such that &lt;math&gt;k_0 k_1 \dotsm k_n -1&lt;/math&gt; is the product of two consecutive integers.<br /> <br /> __TOC__<br /> == Solutions ==<br /> === Solution 1 ===<br /> We will prove the problem for each nonnegative integer &lt;math&gt;n&lt;/math&gt;. We wish to show that<br /> &lt;cmath&gt; k_0 k_1 \dotsc k_n = a^2+a+1, &lt;/cmath&gt;<br /> for some integer &lt;math&gt;a&lt;/math&gt;. We induct on &lt;math&gt;n&lt;/math&gt;. For our base case, &lt;math&gt;n=0&lt;/math&gt;, we may let &lt;math&gt;a&lt;/math&gt; be positive integer.<br /> <br /> For the inductive step, suppose that &lt;math&gt;k_0, \dotsc, k_{n-1}&lt;/math&gt; are pairwise relatively prime integers such that<br /> &lt;cmath&gt; k_0 \dotsm k_{n-1} = a^2+a+1, &lt;/cmath&gt;<br /> for some integer &lt;math&gt;a&lt;/math&gt;. Let &lt;math&gt;k_n = a^2 + 3a+3&lt;/math&gt;. Evidently, &lt;math&gt;k_n &gt;1&lt;/math&gt;. Also,<br /> &lt;cmath&gt; \gcd(a^2+a+1, k_n) = \gcd\bigl(a^2+a+1, k_n - (a^2+a+1) \bigr) = \gcd\bigl(a^2 + a+1, 2(a+1) \bigr) . &lt;/cmath&gt;<br /> Since &lt;math&gt;a^2+a+1 = a(a+1)+1&lt;/math&gt; is odd and relatively prime to &lt;math&gt;a+1&lt;/math&gt;, it follows that &lt;math&gt;a^2+a+1&lt;/math&gt; and &lt;math&gt;k_n&lt;/math&gt; are relatively prime, so &lt;math&gt;k_n&lt;/math&gt; is relatively prime to each of &lt;math&gt;k_0, \dotsc, k_{n-1}&lt;/math&gt;. Finally,<br /> &lt;cmath&gt;\begin{align*}<br /> k_0 k_1 \dotsm k_n &amp;= (a^2+a+1)(a^2+3a+3)\\<br /> &amp;= \bigl[ (a+1)^2-a \bigr] \cdot \bigl[ (a+1)^2 + a+2 \bigr] \\<br /> &amp;= (a+1)^4 + 2(a+1)^2 - a^2-2a \\<br /> &amp;= (a+1)^4 + (a+1)^2 + 1 .<br /> \end{align*} &lt;/cmath&gt;<br /> This completes the induction. &lt;math&gt;\blacksquare&lt;/math&gt;<br /> <br /> === Solution 2 ===<br /> '''Lemma.''' If &lt;math&gt;p&lt;/math&gt; is prime such that &lt;math&gt;p\equiv 1 \pmod{3}&lt;/math&gt;, there exists a residue &lt;math&gt;r&lt;/math&gt; such that &lt;math&gt;p \mid r^2+r+1&lt;/math&gt;.<br /> <br /> ''Proof.'' Let &lt;math&gt;g&lt;/math&gt; be a multiplicative generator of the nonzero integers mod 3. Set &lt;math&gt;r \equiv g^{(p-1)/3}&lt;/math&gt;. Then &lt;math&gt;r-1 \not\equiv 0 \pmod{p}&lt;/math&gt;, but &lt;math&gt;r^3-1 \equiv 0 \pmod{p}&lt;/math&gt;, so &lt;math&gt;\frac{r^3-1}{r-1} \equiv r^2 + r+1 \equiv 0 \pmod{p}&lt;/math&gt;. &lt;math&gt;\blacksquare&lt;/math&gt;<br /> <br /> By [[Dirichlet's Theorem]], there are infinitely many primes congruent to 1 (mod 3). Let &lt;math&gt;p_0, \dotsc, p_n&lt;/math&gt; be &lt;math&gt;(n+1)&lt;/math&gt; such primes, and let &lt;math&gt;r_0, \dotsc, r_n&lt;/math&gt; be respective residues as described in the lemma. By the [[Chinese Remainder Theorem]], there is a positive integer &lt;math&gt;a&lt;/math&gt; that satisfies the relation<br /> &lt;cmath&gt; a \equiv r_i \pmod{p_i} &lt;/cmath&gt;<br /> for each integer &lt;math&gt;i \in [0,n]&lt;/math&gt;. Then<br /> &lt;cmath&gt; p_0 \dotsc p_n \mid a^2+a+1 . &lt;/cmath&gt;<br /> Now, for &lt;math&gt;1 \le i \le n&lt;/math&gt;, take &lt;math&gt;k_i&lt;/math&gt; to be the greatest power of &lt;math&gt;p_i&lt;/math&gt; that divides &lt;math&gt;a^2 + a +1&lt;/math&gt;, and let &lt;math&gt;k_0 = (a^2+a+1)/(k_1 \dotsm k_n)&lt;/math&gt;. Since all the &lt;math&gt;k_i&lt;/math&gt; are pairwise relatively prime and are greater than 1, we are done. &lt;math&gt;\blacksquare&lt;/math&gt;<br /> <br /> {{alternate solutions}}<br /> <br /> == Resources ==<br /> {{USAMO newbox|year=2008|beforetext=|before=First Problem|num-a=2}}<br /> <br /> * &lt;url&gt;viewtopic.php?t=202909 Discussion on AoPS/MathLinks&lt;/url&gt;<br /> <br /> [[Category:Olympiad Number Theory Problems]]</div> I like pie https://artofproblemsolving.com/wiki/index.php?title=2008_USAMO_Problems/Problem_4&diff=25783 2008 USAMO Problems/Problem 4 2008-05-01T23:57:18Z <p>I like pie: Fixed link</p> <hr /> <div>== Problem ==<br /> (''Gregory Galparin'') Let &lt;math&gt;\mathcal{P}&lt;/math&gt; be a convex polygon with &lt;math&gt;n&lt;/math&gt; sides, &lt;math&gt;n\ge3&lt;/math&gt;. Any set of &lt;math&gt;n - 3&lt;/math&gt; diagonals of &lt;math&gt;\mathcal{P}&lt;/math&gt; that do not intersect in the interior of the polygon determine a [i]triangulation[/i] of &lt;math&gt;\mathcal{P}&lt;/math&gt; into &lt;math&gt;n - 2&lt;/math&gt; triangles. If &lt;math&gt;\mathcal{P}&lt;/math&gt; is regular and there is a triangulation of &lt;math&gt;\mathcal{P}&lt;/math&gt; consisting of only isosceles triangles, find all the possible values of &lt;math&gt;n&lt;/math&gt;.<br /> <br /> __TOC__<br /> == Solution ==<br /> === Solution 1 ===<br /> === Solution 2 ===<br /> <br /> {{alternate solutions}}<br /> <br /> == Resources ==<br /> {{USAMO newbox|year=2008|num-b=3|num-a=5}}<br /> <br /> * &lt;url&gt;viewtopic.php?t=202905 Discussion on AoPS/MathLinks&lt;/url&gt;</div> I like pie https://artofproblemsolving.com/wiki/index.php?title=2008_USAMO_Problems/Problem_1&diff=25776 2008 USAMO Problems/Problem 1 2008-05-01T22:26:16Z <p>I like pie: Moved TOC + minor edits</p> <hr /> <div>== Problem ==<br /> (''Titu Andreescu'')<br /> Prove that for each positive integer &lt;math&gt;n&lt;/math&gt;, there are pairwise relatively prime integers &lt;math&gt;k_0, k_1 \dotsc, k_n&lt;/math&gt;, all strictly greater than 1, such that &lt;math&gt;k_0 k_1 \dotsm k_n -1&lt;/math&gt; is the product of two consecutive integers.<br /> <br /> __TOC__<br /> == Solutions ==<br /> === Solution 1 ===<br /> We will prove the problem for each nonnegative integer &lt;math&gt;n&lt;/math&gt;. We wish to show that<br /> &lt;cmath&gt; k_0 k_1 \dotsc k_n = a^2+a+1, &lt;/cmath&gt;<br /> for some integer &lt;math&gt;a&lt;/math&gt;. We induct on &lt;math&gt;n&lt;/math&gt;. For our base case, &lt;math&gt;n=0&lt;/math&gt;, we may let &lt;math&gt;a&lt;/math&gt; be positive integer.<br /> <br /> For the inductive step, suppose that &lt;math&gt;k_0, \dotsc, k_{n-1}&lt;/math&gt; are pairwise relatively prime integers such that<br /> &lt;cmath&gt; k_0 \dotsm k_{n-1} = a^2+a+1, &lt;/cmath&gt;<br /> for some integer &lt;math&gt;a&lt;/math&gt;. Let &lt;math&gt;k_n = a^2 + 3a+3&lt;/math&gt;. Evidently, &lt;math&gt;k_n &gt;1&lt;/math&gt;. Also,<br /> &lt;cmath&gt; \gcd(a^2+a+1, k_n) = \gcd\bigl(a^2+a+1, k_n - (a^2+a+1) \bigr) = \gcd\bigl(a^2 + a+1, 2(a+1) \bigr) . &lt;/cmath&gt;<br /> Since &lt;math&gt;a^2+a+1 = a(a+1)+1&lt;/math&gt; is odd and relatively prime to &lt;math&gt;a+1&lt;/math&gt;, it follows that &lt;math&gt;a^2+a+1&lt;/math&gt; and &lt;math&gt;k_n&lt;/math&gt; are relatively prime, so &lt;math&gt;k_n&lt;/math&gt; is relatively prime to each of &lt;math&gt;k_0, \dotsc, k_{n-1}&lt;/math&gt;. Finally,<br /> &lt;cmath&gt;\begin{align*}<br /> k_0 k_1 \dotsm k_n &amp;= (a^2+a+1)(a^2+3a+3)\\<br /> &amp;= \bigl[ (a+1)^2-a \bigr] \cdot \bigl[ (a+1)^2 + a+2 \bigr] \\<br /> &amp;= (a+1)^4 + 2(a+1)^2 - a^2-2a \\<br /> &amp;= (a+1)^4 + (a+1)^2 + 1 .<br /> \end{align*} &lt;/cmath&gt;<br /> This completes the induction. &lt;math&gt;\blacksquare&lt;/math&gt;<br /> <br /> === Solution 2 ===<br /> '''Lemma.''' If &lt;math&gt;p&lt;/math&gt; is prime such that &lt;math&gt;p\equiv 1 \pmod{3}&lt;/math&gt;, there exists a residue &lt;math&gt;r&lt;/math&gt; such that &lt;math&gt;p \mid r^2+r+1&lt;/math&gt;.<br /> <br /> ''Proof.'' Let &lt;math&gt;g&lt;/math&gt; be a multiplicative generator of the nonzero integers mod 3. Set &lt;math&gt;r \equiv g^{(p-1)/3}&lt;/math&gt;. Then &lt;math&gt;r-1 \not\equiv 0 \pmod{p}&lt;/math&gt;, but &lt;math&gt;r^3-1 \equiv 0 \pmod{p}&lt;/math&gt;, so &lt;math&gt;\frac{r^3-1}{r-1} \equiv r^2 + r+1 \equiv 0 \pmod{p}&lt;/math&gt;. &lt;math&gt;\blacksquare&lt;/math&gt;<br /> <br /> By [[Dirichlet's Theorem]], there are infinitely many primes congruent to 1 (mod 3). Let &lt;math&gt;p_0, \dotsc, p_n&lt;/math&gt; be &lt;math&gt;(n+1)&lt;/math&gt; such primes, and let &lt;math&gt;r_0, \dotsc, r_n&lt;/math&gt; be respective residues as described in the lemma. By the [[Chinese Remainder Theorem]], there is a positive integer &lt;math&gt;a&lt;/math&gt; that satisfies the relation<br /> &lt;cmath&gt; a \equiv r_i \pmod{p_i} &lt;/cmath&gt;<br /> for each integer &lt;math&gt;i \in [0,n]&lt;/math&gt;. Then<br /> &lt;cmath&gt; p_0 \dotsc p_n \mid a^2+a+1 . &lt;/cmath&gt;<br /> Now, for &lt;math&gt;1 \le i \le n&lt;/math&gt;, take &lt;math&gt;k_i&lt;/math&gt; to be the greatest power of &lt;math&gt;p_i&lt;/math&gt; that divides &lt;math&gt;a^2 + a +1&lt;/math&gt;, and let &lt;math&gt;k_0 = (a^2+a+1)/(k_1 \dotsm k_n)&lt;/math&gt;. Since all the &lt;math&gt;k_i&lt;/math&gt; are pairwise relatively prime and are greater than 1, we are done. &lt;math&gt;\blacksquare&lt;/math&gt;<br /> <br /> {{alternate solutions}}<br /> <br /> == Resources ==<br /> {{USAMO newbox|year=2008|beforetext=|before=First Problem|num-a=2}}<br /> <br /> * &lt;url&gt;Forum/viewtopic.php?t=202909 Discussion on AoPS/MathLinks&lt;/url&gt;<br /> <br /> [[Category:Olympiad Number Theory Problems]]</div> I like pie https://artofproblemsolving.com/wiki/index.php?title=2008_USAMO&diff=25775 2008 USAMO 2008-05-01T22:21:37Z <p>I like pie: Added links</p> <hr /> <div>'''2008 [[USAMO]]''' problems and solutions. The first link contains the full set of test problems. The rest contain each individual problem and its solution.<br /> <br /> *[[2008 USAMO Problems]]<br /> *[[2008 USAMO Problems/Problem 1]]<br /> *[[2008 USAMO Problems/Problem 2]]<br /> *[[2008 USAMO Problems/Problem 3]]<br /> *[[2008 USAMO Problems/Problem 4]]<br /> *[[2008 USAMO Problems/Problem 5]]<br /> *[[2008 USAMO Problems/Problem 6]]<br /> <br /> {{USAMO newbox|year=2008|before=[[2007 USAMO]]|after=2009 USAMO}}</div> I like pie