https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Not+Trig&feedformat=atomAoPS Wiki - User contributions [en]2024-03-28T13:26:07ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=USAMO_historical_results&diff=68846USAMO historical results2015-03-13T23:27:05Z<p>Not trig: /* USAMO Honorable Mentions */ https://web.archive.org/web/20100201140542/http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2009-ua/2009usamo-highest.shtml</p>
<hr />
<div>This is the '''USAMO historical results''' page. <!-- Post results here so the [[USAMO]] page does not become cluttered. --><br />
<br />
==USAMO Perfect Scorers==<br />
<br />
Since 1996, a perfect score on the [[USAMO]] has been 42 points for 6 problems. Prior to 1996 a perfect score was 100 points across 5 problems.<br />
*2014: <br />
**Joshua Brakensiek<br />
*2013: None<br />
*2012: <br />
**Alex Zhu<br />
**Thomas Swayze<br />
**Xiaoyu He<br />
**Mitchell Lee<br />
**Samuel Zbarsky<br />
*2011:<br />
**David Yang<br />
**Evan O'Dorney<br />
*2010: None<br />
*2008: None<br />
*2007: None<br />
*2006:<br />
**Brian Lawrence<br />
*2005: None<br />
*2004:<br />
**Tiankai Liu<br />
*2003:<br />
**Tiankai Liu<br />
**Po-Ru Loh<br />
*2002: <br />
**Daniel Kane<br />
**Ricky Liu<br />
**Tiankai Liu<br />
**Po-Ru Loh<br />
**Inna Zakharevich<br />
*1996:<br />
**Chris Chang<br />
*1992:<br />
**Kiran Kedlaya<br />
**Lenny Ng<br />
<br />
==USAMO Winners==<br />
<br />
The top 12 scorers on the [[USAMO]] are currently designated as USAMO winners. In the past, this number was smaller.<br />
*2014:<br />
**Joshua Brakensiek<br />
**Evan Chen<br />
**Ravi Jagadeesan<br />
**Allen Liu<br />
**Nipun Pitimanaaree<br />
**Mark Sellke<br />
**Zhuoqun Song<br />
**David Stoner<br />
**Kevin Sun<br />
**James Tao<br />
**Alexander Whatley<br />
**Scott Wu<br />
*2013<br />
**Calvin Deng<br />
**Andrew He<br />
**Ravi Jagadeesan<br />
**Pakawut Jiradilok<br />
**Ray Li<br />
**Kevin Li<br />
**Mark Sellke<br />
**Bobby Shen<br />
**Zhuoqun Song<br />
**David Stoner<br />
**Thomas Swayze<br />
**Victor Wang<br />
*2012:<br />
**Andre Arslan<br />
**Joshua Brakensiek<br />
**Calvin Deng<br />
**Xiaoyu He<br />
**Ravi Jagadeesan<br />
**Mitchell Lee<br />
**Alex Song<br />
**Thomas Swayze<br />
**Victor Wang<br />
**David Yang<br />
**Samuel Zbarsky<br />
**Alex Zhu<br />
*2011:<br />
**Wenyu Cao<br />
**Zijing Gao<br />
**Benjamin Gunby<br />
**Xiaoyu He<br />
**Ravi Jagadeesan<br />
**Spencer Kwon<br />
**Mitchell Lee<br />
**Ray Li<br />
**Evan O'Dorney<br />
**Mark Sellke<br />
**David Yang<br />
**Joy Zheng<br />
*2010:<br />
**Timothy Chu<br />
**Calvin Deng<br />
**Michael Druggan<br />
**Brian Hamrick<br />
**Travis Hance<br />
**Xiaoyu He<br />
**Mitchell Lee<br />
**In Sung Na<br />
**Evan O'Dorney<br />
**Toan Phan<br />
**Hunter Spink<br />
**Allen Yuan<br />
*2009:<br />
**John Berman<br />
**Sergei Bernstein<br />
**Wenyu Cao<br />
**Robin Cheng<br />
**Vlad Firoiu<br />
**Eric Larson<br />
**Delong Meng<br />
**Qinxuan Pan<br />
**Panupang Pasupat<br />
**Toan Phan<br />
**David Rush<br />
**David Yang<br />
*2008:<br />
**David Benjamin<br />
**TaoRan Chen<br />
**Paul Christiano<br />
**Sam Elder<br />
**Shaunak Kishore<br />
**Delong Meng<br />
**Evan O'Dorney<br />
**Qinxuan Pan<br />
**David Rolnick<br />
**Colin Sandon<br />
**Krishanu Sankar<br />
**Alex Zhai<br />
*2007:<br />
**Sergei Bernstein<br />
**Sherry Gong<br />
**Adam Hesterberg<br />
**Eric Larson<br />
**Brian Lawrence<br />
**Tedrick Leung<br />
**Haitao Mao<br />
**Delong Meng<br />
**Krishanu Sankar<br />
**Jacob Steinhardt<br />
**Arnav Tripathy<br />
**Alex Zhai<br />
*2006:<br />
**Brian Lawrence<br />
**Alex Zhai<br />
**Yufei Zhao<br />
**Peng Shi<br />
**Sherry Gong<br />
**Richard McCutchen<br />
**Yi Sun<br />
**Arnav Tripathy<br />
**Taehyeon (Ryan) Ko<br />
**Yi Han<br />
**Yakov Berchenko-Kogan<br />
**Tedrick Leung<br />
*2005:<br />
**Robert Cordwell<br />
**Zhou Fan<br />
**Sherry Gong<br />
**Rishi Gupta<br />
**Hyun Soo Kim<br />
**Brian Lawrence<br />
**Albert Ni<br />
**Natee Pitiwan<br />
**Eric Price<br />
**Peng Shi<br />
**Yi Sun<br />
**Yufei Zhao<br />
*2004:<br />
**Tiankai Liu<br />
**Jae Bae<br />
**Jongmin Baek<br />
**Oleg Golberg<br />
**Matt Ince<br />
**Janos Kramar<br />
**Alison Miller<br />
**Aaron Pixton<br />
**Brian Rice<br />
**Jacob Tsimerman<br />
**Ameya Velingker<br />
**Tony Zhang<br />
*1996:<br />
**Chris Chang<br />
**Alex Saltman<br />
**Josh Nichols-Barrer<br />
**Carl Bosley<br />
**Michael Korn<br />
**Carl Miller<br />
**Nathan Curtis<br />
**Daniel Stronger<br />
*1995:<br />
**Aleksander Khazanov<br />
**Jacob Lurie<br />
**Chris Chang<br />
**Jay Chyung<br />
**Andrei Gnepp<br />
**Josh Nichols-Barrer<br />
**Samit Dasgupta<br />
**Craig Helfgott<br />
*1994:<br />
**Jeremy Bem<br />
**Aleksander Khazanov<br />
**Jonathan Weinstein<br />
**Jacob Lurie<br />
**Noam Shazeer<br />
**Stephen Wang<br />
**Chris Chang<br />
**Jacob Rasmussen<br />
*1989:<br />
**Jordan Ellenberg<br />
**Andrew Kresh<br />
**Jonathan Higa<br />
**[[Richard Rusczyk]]<br />
**Jeff Vanderkam<br />
**[[Sam Vandervelde]]<br />
**David Carlton<br />
<br />
==USAMO Honorable Mentions==<br />
<br />
Other students who finish (or tie to finish) in the top 24 of the [[USAMO]] receive Honorable Mention (often abbreviated HM).<br />
<br />
*2014<br />
**Lewis Chen<br />
**Ernest Chiu<br />
**Samuel Korsky<br />
**Yang Liu<br />
**Sammy Luo<br />
**Alexander Meiburg<br />
**Shyam Narayanan<br />
**Ashwin Sah<br />
**Nat Sothanaphan<br />
**Danielle Wang<br />
**Kevin Yang<br />
*2013:<br />
**Joshua Brakensiek<br />
**Lewis Chen<br />
**Sohail Farhangi<br />
**Kevin Huang<br />
**Allen Liu<br />
**Sammy Luo<br />
**Eric Schneider<br />
**Rachit Singh<br />
**Vikram Sundar<br />
**James Tao<br />
**David Yang<br />
**Kevin Yang<br />
*2011:<br />
**Kevin Chen<br />
**Calvin Deng<br />
**Michael Druggan<br />
**Albert Gu<br />
**Bobby Shen<br />
**Alex Song<br />
**Thomas Swayze<br />
**Tianqi Wu<br />
**Dai Yang<br />
**Kevin Yin<br />
**Allen Yuan<br />
**Samuel Zbarsky<br />
**Brian Zhang<br />
*2010:<br />
**Thomas Swayze<br />
**Kerry Xing<br />
**Tian-Yi Jiang<br />
**Brian Zhang<br />
**Archit Kulkarni<br />
**Patrick Hulin<br />
**Carl Lian<br />
**George Xing<br />
**Potcharapol Suteparuk<br />
**Jason Wu<br />
**Shijie Zheng<br />
**Robin Cheng<br />
**Bowei Liu<br />
**Wenyu Cao<br />
**Kevin Yin<br />
**Albert Gu<br />
*2009:<br />
**Timothy Chu<br />
**Brian Hamrick<br />
**Travis Hance<br />
**Jason Hoch<br />
**Tian-Yi Jiang<br />
**Sam Keller<br />
**Holden Lee<br />
**In-Sung Na<br />
**Ofir Nachum<br />
**Jarno Sun <br />
**Matthew Superdock<br />
**Tong Zhan<br />
**Mark Zhang<br />
*2008:<br />
**John Berman<br />
**Sergei Bernstein<br />
**Gregory Gauthier<br />
**Brian Hamrick<br />
**Eric Larson<br />
**Gaku Liu<br />
**Jeffrey Manning<br />
**Palmer Mebane<br />
**Danny Shi<br />
**Jacob Steinhardt<br />
**Matthew Superdock<br />
**Nicholas Triantafillou<br />
*2007:<br />
**David Benjamin<br />
**Gregory Brockman<br />
**Yingyu Gao<br />
**Yan Li<br />
**Gaku Liu<br />
**Jeffrey Manning<br />
**Palmer Mebane<br />
**Evan O'Dorney<br />
**Alexander Remorov<br />
**Max Rosett<br />
**Qiaochu Yuan<br />
**Bohua Zhan<br />
*2006:<br />
**Joseph Chu<br />
**Zachary Abel<br />
**Zarathustra Brady<br />
**Peter Diao<br />
**Richar Peng<br />
**Adam Hesterberg<br />
**Bohua Zhan<br />
**Kevin Modzelewski<br />
**Daniel Poore<br />
**Lin Fei<br />
**Jason Trigg<br />
**Tony Liu<br />
**Viktoriya Krakovna<br />
*2005:<br />
**Zachary Abel<br />
**Charles Chen<br />
**Evan Dummit<br />
**Adam Hesterberg<br />
**Richard Ho<br />
**Alexander Marcus<br />
**Richard McCutchen<br />
**Thomas Mildorf<br />
**Charles Nathanson<br />
**Keenan Pepper<br />
**Timothy Pollio<br />
*2004:<br />
**Boris Alexeev<br />
**Joshua Batson<br />
**Thomas Belulovich<br />
**Rishi Gupta<br />
**Anders Kaseorg<br />
**Hyun Soo Kim<br />
**John Kim<br />
**Po Ling Loh<br />
**Sira Sriswasdi<br />
**Yi Sun<br />
**Elena Udovina<br />
**Yufei Zhao<br />
*1995<br />
** Mathew Crawford<br />
*1993<br />
** Mathew Crawford<br />
<br />
==See also==<br />
*[http://www.unl.edu/amc/e-exams/e8-usamo/archiveusamo.html USAMO Archive]<br />
<br />
[[Category:Historical results]]</div>Not trighttps://artofproblemsolving.com/wiki/index.php?title=North_Carolina_ARML&diff=56897North Carolina ARML2013-08-12T06:28:32Z<p>Not trig: /* Historical Results */</p>
<hr />
<div>North Carolina has historically sent two teams of 15 to ARML each year, along with a handful of alternates. These two teams were previously called, and continue to be colloquially referred to as, the "A" and "B" teams. Due to the B team's 4th place finish in Division B in 2007, however, it has been required to compete in the A division, so the teams have been renamed to "A1" and "A2," respectively.<br />
<br />
Team selection is done primarily via the [http://www.ncssm.edu/smc/ North Carolina State Math Contest], with consideration given to results on the [[AMC]] series of tests and performance in local meets, notably the [[Duke Math Meet]]. One to three official practices are held each year at [http://www.ncssm.edu NCSSM].<br />
<br />
== Historical Results ==<br />
{| class="wikitable" style="text-align:center; width:200px; height:200px;"<br />
! Year<br />
! A/A1<br />
! B/A2<br />
|-<br />
<br />
! 2013 <br />
| 2nd <br />
| 33rd<br />
|-<br />
! 2012 <br />
| 1st <br />
| 25th<br />
|-<br />
! 2011 <br />
| 3rd <br />
| 26th<br />
|-<br />
! 2010 <br />
| 4th <br />
| 40th<br />
|-<br />
! 2009 <br />
| 7th <br />
| 38th<br />
|-<br />
! 2008 <br />
| 3rd <br />
| 39rd<br />
|-<br />
! 2007 <br />
| 4th <br />
| 4th*<br />
|-<br />
! 2006 <br />
| 1st <br />
| 14th*<br />
|-<br />
! 2005 <br />
| 7th <br />
| 25th*<br />
|-<br />
! 2004 <br />
| 15th <br />
| 13th*<br />
|}<br />
''B division performances marked by *''<br />
<br />
Earlier results of the A team may be found [http://www.ncssm.edu/smc/Inform/ncarmlhist.htm here].<br />
<br />
== Resources ==<br />
* [http://www.ncssm.edu/smc/ North Carolina State Math Contest homepage]<br />
* [[North Carolina mathematics competitions]]<br />
* [[ARML historical results]]<br />
* [[Mathematics competition resources]]<br />
[[Category:ARML teams]]</div>Not trighttps://artofproblemsolving.com/wiki/index.php?title=North_Carolina_ARML&diff=56896North Carolina ARML2013-08-12T06:22:05Z<p>Not trig: </p>
<hr />
<div>North Carolina has historically sent two teams of 15 to ARML each year, along with a handful of alternates. These two teams were previously called, and continue to be colloquially referred to as, the "A" and "B" teams. Due to the B team's 4th place finish in Division B in 2007, however, it has been required to compete in the A division, so the teams have been renamed to "A1" and "A2," respectively.<br />
<br />
Team selection is done primarily via the [http://www.ncssm.edu/smc/ North Carolina State Math Contest], with consideration given to results on the [[AMC]] series of tests and performance in local meets, notably the [[Duke Math Meet]]. One to three official practices are held each year at [http://www.ncssm.edu NCSSM].<br />
<br />
== Historical Results ==<br />
{| class="wikitable" style="text-align:center; width:200px; height:200px;"<br />
! Year<br />
! A/A1<br />
! B/A2<br />
|-<br />
<br />
! 2013 <br />
| 2nd <br />
| 33rd<br />
|-<br />
! 2012 <br />
| 1st <br />
| 25th<br />
|-<br />
! 2011 <br />
| 3rd <br />
| 26th<br />
|-<br />
! 2010 <br />
| 4th <br />
| 40th<br />
|-<br />
! 2009 <br />
| 7th <br />
| 38th<br />
|-<br />
! 2008 <br />
| 3rd <br />
| 39rd<br />
|-<br />
! 2007 <br />
| 4th <br />
| 4th*<br />
|-<br />
! 2006 <br />
| 1st <br />
| 14th*<br />
|-<br />
! 2005 <br />
| 7th <br />
| 25th*<br />
|-<br />
! 2004 <br />
| 14th <br />
| 13th*<br />
|}<br />
''B division performances marked by *''<br />
<br />
Earlier results of the A team may be found [http://www.ncssm.edu/smc/Inform/ncarmlhist.htm here].<br />
<br />
== Resources ==<br />
* [http://www.ncssm.edu/smc/ North Carolina State Math Contest homepage]<br />
* [[North Carolina mathematics competitions]]<br />
* [[ARML historical results]]<br />
* [[Mathematics competition resources]]<br />
[[Category:ARML teams]]</div>Not trighttps://artofproblemsolving.com/wiki/index.php?title=North_Carolina_mathematics_competitions&diff=56895North Carolina mathematics competitions2013-08-12T06:20:28Z<p>Not trig: /* North Carolina high school mathematics competitions */</p>
<hr />
<div>== North Carolina middle school mathematics competitions==<br />
<br />
[[AoPS]] hosts [http://www.artofproblemsolving.com/Forum/index.php?f=298 middle school math forums] where students can discuss contest problems and mathematics.<br />
<br />
* [[North Carolina MathCounts]] is part of the national [[MathCounts]] competition.<br />
<br />
<br />
== North Carolina high school mathematics competitions==<br />
<br />
[[AoPS]] hosts [http://www.artofproblemsolving.com/Forum/index.php?f=214 high school math forums] where students can discuss contest problems and mathematics<br />
<br />
*[[CMSTE Super Competition at the UNC Charlotte]] [http://www.cmste.uncc.edu/supercompetition/supercomp.htm website] <br />
*[[Duke Math Meet]] [http://www.math.duke.edu/dumu/contests/ website] <br />
*[[North Carolina State Mathematics Contest]] -- Cash prizes for top students. [http://ncssm.edu/smc/ website]<br />
*[[UNC Charlotte High School Math Contest]] [http://www.math.uncc.edu/%7Ehbreiter/problems/UNCC.html website]<br />
*[[UNC Wilmington High School Math Contest]] [http://people.uncw.edu/karlof/MathContest/ website]<br />
*[[North Carolina ARML]]<br />
<br />
== National math contests ==<br />
* [[List of United States elementary school mathematics competitions]] for national contests.<br />
* [[List of United States middle school mathematics competitions]] for national contests.<br />
* [[List of United States high school mathematics competitions]] for national contests.<br />
<br />
<br />
== See also ==<br />
* [[List of mathematics competitions]]<br />
* [[Mathematics competitions resources]]</div>Not trighttps://artofproblemsolving.com/wiki/index.php?title=North_Carolina_mathematics_competitions&diff=56894North Carolina mathematics competitions2013-08-12T06:19:49Z<p>Not trig: /* North Carolina high school mathematics competitions */</p>
<hr />
<div>== North Carolina middle school mathematics competitions==<br />
<br />
[[AoPS]] hosts [http://www.artofproblemsolving.com/Forum/index.php?f=298 middle school math forums] where students can discuss contest problems and mathematics.<br />
<br />
* [[North Carolina MathCounts]] is part of the national [[MathCounts]] competition.<br />
<br />
<br />
== North Carolina high school mathematics competitions==<br />
<br />
[[AoPS]] hosts [http://www.artofproblemsolving.com/Forum/index.php?f=214 high school math forums] where students can discuss contest problems and mathematics<br />
<br />
*[[CMSTE Super Competition at the UNC Charlotte]] [http://www.cmste.uncc.edu/supercompetition/supercomp.htm website] <br />
*[[Duke Math Meet]] [http://www.math.duke.edu/dumu/contests/ website] <br />
*[[North Carolina State Mathematics Contest]] -- Cash prizes for top students. [http://ncssm.edu/smc/ website]<br />
*[[UNC Charlotte High School Math Contest]] [http://www.math.uncc.edu/%7Ehbreiter/problems/UNCC.html website]<br />
*[[UNC Wilmington High School Math Contest]] [http://people.uncw.edu/karlof/MathContest/ website]<br />
<br />
== National math contests ==<br />
* [[List of United States elementary school mathematics competitions]] for national contests.<br />
* [[List of United States middle school mathematics competitions]] for national contests.<br />
* [[List of United States high school mathematics competitions]] for national contests.<br />
<br />
<br />
== See also ==<br />
* [[List of mathematics competitions]]<br />
* [[Mathematics competitions resources]]</div>Not trighttps://artofproblemsolving.com/wiki/index.php?title=North_Carolina_ARML&diff=56893North Carolina ARML2013-08-12T06:16:13Z<p>Not trig: </p>
<hr />
<div>North Carolina has historically sent two teams of 15 to ARML each year, along with a handful of alternates. These two teams were previously called, and continue to be colloquially referred to as, the "A" and "B" teams. Due to the B team's 4th place finish in Division B in 2007, however, it has been required to compete in the A division, so the teams have been renamed to "A1" and "A2," respectively.<br />
<br />
Team selection is done primarily via the [http://www.ncssm.edu/smc/ North Carolina State Math Contest], with consideration given to results on the [[AMC]] series of tests and performance in local meets, notably the [[Duke Math Meet]].<br />
<br />
== Historical Results ==<br />
{| class="wikitable" style="text-align:center; width:200px; height:200px;"<br />
! Year<br />
! A/A1<br />
! B/A2<br />
|-<br />
<br />
! 2013 <br />
| 2nd <br />
| 33rd<br />
|-<br />
! 2012 <br />
| 1st <br />
| 25th<br />
|-<br />
! 2011 <br />
| 3rd <br />
| 26th<br />
|-<br />
! 2010 <br />
| 4th <br />
| 40th<br />
|-<br />
! 2009 <br />
| 7th <br />
| 38th<br />
|-<br />
! 2008 <br />
| 3rd <br />
| 39rd<br />
|-<br />
! 2007 <br />
| 4th <br />
| 4th*<br />
|-<br />
! 2006 <br />
| 1st <br />
| 14th*<br />
|-<br />
! 2005 <br />
| 7th <br />
| 25th*<br />
|-<br />
! 2004 <br />
| 14th <br />
| 13th*<br />
|}<br />
''B division performances marked by *''<br />
<br />
Earlier results of the A team may be found [http://www.ncssm.edu/smc/Inform/ncarmlhist.htm here].<br />
<br />
== Resources ==<br />
* [http://www.ncssm.edu/smc/ North Carolina State Math Contest homepage]<br />
* [[North Carolina mathematics competitions]]<br />
* [[ARML historical results]]<br />
* [[Mathematics competition resources]]<br />
[[Category:ARML teams]]</div>Not trighttps://artofproblemsolving.com/wiki/index.php?title=North_Carolina_ARML&diff=56892North Carolina ARML2013-08-12T06:14:33Z<p>Not trig: </p>
<hr />
<div>North Carolina has historically sent two teams to ARML each year, along with a handful of alternates. These two teams were previously called, and continue to be colloquially referred to as, the "A" and "B" teams. Due to the B team's 4th place finish in Division B in 2007, however, it has been required to compete in the A division, so the teams have been renamed to "A1" and "A2," respectively.<br />
<br />
Team selection is done primarily via the [http://www.ncssm.edu/smc/ North Carolina State Math Contest], with consideration given to results on the [[AMC]] series of tests and performance in local meets, notably the [[Duke Math Meet]].<br />
<br />
== Historical Results ==<br />
{| class="wikitable" style="text-align:center; width:200px; height:200px;"<br />
! Year<br />
! A/A1<br />
! B/A2<br />
|-<br />
<br />
! 2013 <br />
| 2nd <br />
| 33rd<br />
|-<br />
! 2012 <br />
| 1st <br />
| 25th<br />
|-<br />
! 2011 <br />
| 3rd <br />
| 26th<br />
|-<br />
! 2010 <br />
| 4th <br />
| 40th<br />
|-<br />
! 2009 <br />
| 7th <br />
| 38th<br />
|-<br />
! 2008 <br />
| 3rd <br />
| 39rd<br />
|-<br />
! 2007 <br />
| 4th <br />
| 4th*<br />
|-<br />
! 2006 <br />
| 1st <br />
| 14th*<br />
|-<br />
! 2005 <br />
| 7th <br />
| 25th*<br />
|-<br />
! 2004 <br />
| 14th <br />
| 13th*<br />
|}<br />
''B division performances marked by *''<br />
<br />
Earlier results of the A team may be found [http://www.ncssm.edu/smc/Inform/ncarmlhist.htm here].<br />
<br />
== Resources ==<br />
* [http://www.ncssm.edu/smc/ North Carolina State Math Contest homepage]<br />
* [[North Carolina mathematics competitions]]<br />
* [[ARML historical results]]<br />
* [[Mathematics competition resources]]<br />
[[Category:ARML teams]]</div>Not trighttps://artofproblemsolving.com/wiki/index.php?title=North_Carolina_ARML&diff=56891North Carolina ARML2013-08-12T06:12:52Z<p>Not trig: Created page with "North Carolina has historically sent two teams to ARML each year, along with a handful of alternates. These two teams were previously called, and continue to be colloquially refe..."</p>
<hr />
<div>North Carolina has historically sent two teams to ARML each year, along with a handful of alternates. These two teams were previously called, and continue to be colloquially referred to as, the "A" and "B" teams. Due to the B team's 4th place finish in Division B in 2007, however, it has been required to compete in the A division, so the teams have been renamed to "A1" and "A2," respectively.<br />
<br />
Team selection is done primarily via the [http://www.ncssm.edu/smc/ North Carolina State Math Contest], with consideration given to results on the [[AMC]] series of tests and performance in local meets, notably the [[Duke Math Meet]].<br />
<br />
== Historical Results ==<br />
{| class="wikitable" style="text-align:center; width:200px; height:200px;"<br />
! Year<br />
! A/A1<br />
! B/A2<br />
|-<br />
<br />
! 2013 <br />
| 2nd <br />
| 33rd<br />
|-<br />
! 2012 <br />
| 1st <br />
| 25th<br />
|-<br />
! 2011 <br />
| 3rd <br />
| 26th<br />
|-<br />
! 2010 <br />
| 4th <br />
| 40th<br />
|-<br />
! 2009 <br />
| 7th <br />
| 38th<br />
|-<br />
! 2008 <br />
| 3rd <br />
| 39rd<br />
|-<br />
! 2007 <br />
| 4th <br />
| 4th*<br />
|-<br />
! 2006 <br />
| 1st <br />
| 14th*<br />
|-<br />
! 2005 <br />
| 7th <br />
| 25th*<br />
|-<br />
! 2004 <br />
| 14th <br />
| 13th*<br />
|}<br />
''B division performances marked by *''<br />
<br />
Earlier results of the A team may be found [http://www.ncssm.edu/smc/Inform/ncarmlhist.htm here].</div>Not trighttps://artofproblemsolving.com/wiki/index.php?title=2009_AMC_12B_Problems/Problem_19&diff=331852009 AMC 12B Problems/Problem 192010-01-11T05:00:25Z<p>Not trig: /* Solution 3 */</p>
<hr />
<div>== Problem ==<br />
For each positive integer <math>n</math>, let <math>f(n) = n^4 - 360n^2 + 400</math>. What is the sum of all values of <math>f(n)</math> that are prime numbers?<br />
<br />
<math>\textbf{(A)}\ 794\qquad \textbf{(B)}\ 796\qquad \textbf{(C)}\ 798\qquad \textbf{(D)}\ 800\qquad \textbf{(E)}\ 802</math><br />
<br />
== Solution ==<br />
<br />
=== Solution 1 ===<br />
<br />
To find the answer it was enough to play around with <math>f</math>. One can easily find that <math>f(1)=41</math> is a prime, then <math>f</math> becomes negative for <math>n</math> between <math>2</math> and <math>18</math>, and then <math>f(19)=761</math> is again a prime number. And as <math>41 + 761 = 802</math> is already the largest option, the answer must be <math>\boxed{802}</math>.<br />
<br />
=== Solution 2 ===<br />
<br />
We will now show a complete solution, with a proof that no other values are prime. <br />
<br />
Consider the function <math>g(x) = x^2 - 360x + 400</math>, then obviously <math>f(x) = g(x^2)</math>.<br />
<br />
The roots of <math>g</math> are:<br />
<cmath><br />
x_{1,2} <br />
= \frac{ 360 \pm \sqrt{ 360^2 - 4\cdot 400 } }2 <br />
= 180 \pm 80 \sqrt 5<br />
</cmath><br />
<br />
We can then write <math>g(x) = (x - 180 - 80\sqrt 5)(x - 180 - 80\sqrt 5)</math>, and thus <math>f(x) = (x^2 - 180 - 80\sqrt 5)(x^2 - 180 - 80\sqrt 5)</math>.<br />
<br />
We would now like to factor the right hand side further, using the formula <math>(x^2 - y^2) = (x-y)(x+y)</math>. To do this, we need to express both constants as squares of some other constants. Luckily, we have a pretty good idea how they look like.<br />
<br />
We are looking for rational <math>a</math> and <math>b</math> such that <math>(a+b\sqrt 5)^2 = 180 + 80\sqrt 5</math>. Expanding the left hand side and comparing coefficients, we get <math>ab=40</math> and <math>a^2 + 5b^2 = 180</math>. We can easily guess (or compute) the solution <math>a=10</math>, <math>b=4</math>.<br />
<br />
Hence <math>180 + 80\sqrt 5 = (10 + 4\sqrt 5)^2</math>, and we can easily verify that also <math>180 - 80\sqrt 5 = (10 - 4\sqrt 5)^2</math>.<br />
<br />
We now know the complete factorization of <math>f(x)</math>: <br />
<br />
<cmath><br />
f(x) = (x - 10 - 4\sqrt 5)(x + 10 + 4\sqrt 5)(x - 10 + 4\sqrt 5)(x + 10 - 4\sqrt 5)<br />
</cmath><br />
<br />
As the final step, we can now combine the factors in a different way, in order to get rid of the square roots. <br />
<br />
We have <math>(x - 10 - 4\sqrt 5)(x - 10 + 4\sqrt 5) = (x-10)^2 - (4\sqrt 5)^2 = x^2 - 20x + 20</math>,<br />
and <math>(x + 10 - 4\sqrt 5)(x + 10 + 4\sqrt 5) = x^2 + 20x + 20</math>.<br />
<br />
Hence we obtain the factorization <math>f(x) = (x^2 - 20x + 20)(x^2 + 20x + 20)</math>.<br />
<br />
For <math>x\geq 20</math> both terms are positive and larger than one, hence <math>f(x)</math> is not prime. For <math>1<x<19</math> the second factor is positive and the first one is negative, hence <math>f(x)</math> is not a prime. The remaining cases are <math>x=1</math> and <math>x=19</math>. In both cases, <math>f(x)</math> is indeed a prime, and their sum is <math>f(1) + f(19) = 41 + 761 = \boxed{802}</math>.<br />
<br />
=== Solution 3 ===<br />
<br />
Instead of doing the hard work, we can try to guess the factorization. One good approach:<br />
<br />
We can make the observation that <math>f(x)</math> looks similar to <math>(x^2 + 20)^2</math> with the exception of the <math>x^2</math> term. In fact, we have <math>(x^2 + 20)^2 = x^4 + 40x^2 + 400</math>. But then we notice that it differs from the desired expression by a square: <math>f(x) = (x^2 + 20)^2 - 400x^2 = (x^2 + 20)^2 - (20x)^2</math>.<br />
<br />
Now we can use the formula <math>(x^2 - y^2) = (x-y)(x+y)</math> to obtain the same factorization as in the previous solution, without all the work.<br />
<br />
== See Also ==<br />
<br />
{{AMC12 box|year=2009|ab=B|num-b=18|num-a=20}}</div>Not trighttps://artofproblemsolving.com/wiki/index.php?title=MaCh_8&diff=30974MaCh 82009-03-26T01:31:20Z<p>Not trig: </p>
<hr />
<div>The MaCh 8 (Mathematics Challenge 8) is a middle school math contest run by students at the North Carolina School of Science and Math in Durham, NC. It will be held on May 9, 2009.<br />
<br />
<br />
== '''Individual Tests''' ==<br />
<br />
<br />
The MaCh 8 features three individual subject tests in Algebra, Geometry, and Discrete Math, as well as a comprehensive "General" test for students who would rather not specialize. Students take either two of the Subject tests or the General test. Individual tests are short answer: 25 problems for Subject tests, and 50 problems for the General tests. Students are allotted an hour to take each Subject test and 2 hours for the General, which is broken into two sets of 25 with a break in between. Each set of 25 is scored as follows:<br />
<br />
1-5: 2 points each<br />
<br />
6-10: 3 points<br />
<br />
11-15: 4 points<br />
<br />
16-20: 5 points<br />
<br />
21-25: 6 points,<br />
<br />
for a total of 100 points per Subject test and 200 points on the General test.<br />
<br />
Top students in each individual test enter a difficult short-answer tiebreaker. Top-scoring students on each Subject test, the General test, and the tiebreakers receive awards.<br />
<br />
<br />
== '''Team Tests''' ==<br />
<br />
The MaCh 8 offers both Team and Power tests. The Team test is a half-hour, 10-question short answer test; each problem is worth 6 points. The Power test is a 1-hour proof test on a set of related mathematical concepts. Although the Power test does not expect rigorous proofs, a clear and complete explanation is required for full credit.<br />
<br />
Those who do not qualify for the individual tiebreakers participate, as teams, in the 1-hour Mach round. This is a fun round consisting of 100 problems to be solved in the hour allotted; a runner obtains the problems in groups of 10. Later problems are more difficult and are worth more, but once teams turn in a set, they cannot get them back later. The Mach round contributes points to the team sweepstakes total as well.<br />
<br />
The top 5 teams receive awards based on their performance in the individual, Power, Team, and Mach rounds. The rounds are weighted appropriately to promote a balance between team and individual scores.<br />
<br />
<br />
== '''More Information''' ==<br />
<br />
See the contest website, [http://math.ncssm.edu/~rash/mach8/], for more information.</div>Not trighttps://artofproblemsolving.com/wiki/index.php?title=MaCh_8&diff=30654MaCh 82009-03-10T18:58:54Z<p>Not trig: </p>
<hr />
<div>The MaCh 8 (Mathematics Challenge 8) is a middle school math contest run by students at the North Carolina School of Science and Math in Durham, NC. It will be held on April 4, 2009.<br />
<br />
<br />
== '''Individual Tests''' ==<br />
<br />
<br />
The MaCh 8 features three individual subject tests in Algebra, Geometry, and Discrete Math, as well as a comprehensive "General" test for students who would rather not specialize. Students take either two of the Subject tests or the General test. Individual tests are short answer: 25 problems for Subject tests, and 50 problems for the General tests. Students are allotted an hour to take each Subject test and 2 hours for the General, which is broken into two sets of 25 with a break in between. Each set of 25 is scored as follows:<br />
<br />
1-5: 2 points each<br />
<br />
6-10: 3 points<br />
<br />
11-15: 4 points<br />
<br />
16-20: 5 points<br />
<br />
21-25: 6 points,<br />
<br />
for a total of 100 points per Subject test and 200 points on the General test.<br />
<br />
Top students in each individual test enter a difficult short-answer tiebreaker. Top-scoring students on each Subject test, the General test, and the tiebreakers receive awards.<br />
<br />
<br />
== '''Team Tests''' ==<br />
<br />
The MaCh 8 offers both Team and Power tests. The Team test is a half-hour, 10-question short answer test; each problem is worth 6 points. The Power test is a 1-hour proof test on a set of related mathematical concepts. Although the Power test does not expect rigorous proofs, a clear and complete explanation is required for full credit.<br />
<br />
Those who do not qualify for the individual tiebreakers participate, as teams, in the 1-hour Mach round. This is a fun round consisting of 100 problems to be solved in the hour allotted; a runner obtains the problems in groups of 10. Later problems are more difficult and are worth more, but once teams turn in a set, they cannot get them back later. The Mach round contributes points to the team sweepstakes total as well.<br />
<br />
The top 5 teams receive awards based on their performance in the individual, Power, Team, and Mach rounds. The rounds are weighted appropriately to promote a balance between team and individual scores.<br />
<br />
<br />
== '''More Information''' ==<br />
<br />
See the contest website, [http://math.ncssm.edu/~rash/mach8/], for more information.</div>Not trighttps://artofproblemsolving.com/wiki/index.php?title=MaCh_8&diff=30653MaCh 82009-03-10T18:58:38Z<p>Not trig: </p>
<hr />
<div>The MaCh 8 (Mathematics Challenge 8) is a middle school math contest run by students at the North Carolina School of Science and Math in Durham, NC. It will be held on April 4, 2009.<br />
<br />
<br />
== '''Individual Tests''' ==<br />
<br />
<br />
The MaCh 8 features three individual subject tests in Algebra, Geometry, and Discrete Math, as well as a comprehensive "General" test for students who would rather not specialize. Students take either two of the Subject tests or the General test. Individual tests are short answer: 25 problems for Subject tests, and 50 problems for the General tests. Students are allotted an hour to take each Subject test and 2 hours for the General, which is broken into two sets of 25 with a break in between. Each set of 25 is scored as follows:<br />
<br />
1-5: 2 points each<br />
6-10: 3 points<br />
11-15: 4 points<br />
16-20: 5 points<br />
21-25: 6 points,<br />
<br />
for a total of 100 points per Subject test and 200 points on the General test.<br />
<br />
Top students in each individual test enter a difficult short-answer tiebreaker. Top-scoring students on each Subject test, the General test, and the tiebreakers receive awards.<br />
<br />
<br />
== '''Team Tests''' ==<br />
<br />
The MaCh 8 offers both Team and Power tests. The Team test is a half-hour, 10-question short answer test; each problem is worth 6 points. The Power test is a 1-hour proof test on a set of related mathematical concepts. Although the Power test does not expect rigorous proofs, a clear and complete explanation is required for full credit.<br />
<br />
Those who do not qualify for the individual tiebreakers participate, as teams, in the 1-hour Mach round. This is a fun round consisting of 100 problems to be solved in the hour allotted; a runner obtains the problems in groups of 10. Later problems are more difficult and are worth more, but once teams turn in a set, they cannot get them back later. The Mach round contributes points to the team sweepstakes total as well.<br />
<br />
The top 5 teams receive awards based on their performance in the individual, Power, Team, and Mach rounds. The rounds are weighted appropriately to promote a balance between team and individual scores.<br />
<br />
<br />
== '''More Information''' ==<br />
<br />
See the contest website, [http://math.ncssm.edu/~rash/mach8/], for more information.</div>Not trighttps://artofproblemsolving.com/wiki/index.php?title=MaCh_8&diff=30652MaCh 82009-03-10T18:57:54Z<p>Not trig: </p>
<hr />
<div>The MaCh 8 (Mathematics Challenge 8) is a middle school math contest run by students at the North Carolina School of Science and Math in Durham, NC. It will be held on April 4, 2009.<br />
<br />
'''Individual Tests'''<br />
The MaCh 8 features three individual subject tests in Algebra, Geometry, and Discrete Math, as well as a comprehensive "General" test for students who would rather not specialize. Students take either two of the Subject tests or the General test. Individual tests are short answer: 25 problems for Subject tests, and 50 problems for the General tests. Students are allotted an hour to take each Subject test and 2 hours for the General, which is broken into two sets of 25 with a break in between. Each set of 25 is scored as follows:<br />
<br />
#1-5: 2 points each<br />
#6-10: 3 points<br />
#11-15: 4 points<br />
#16-20: 5 points<br />
#21-25: 6 points,<br />
<br />
for a total of 100 points per Subject test and 200 points on the General test.<br />
<br />
Top students in each individual test enter a difficult short-answer tiebreaker. Top-scoring students on each Subject test, the General test, and the tiebreakers receive awards.<br />
<br />
'''Team Tests'''<br />
The MaCh 8 offers both Team and Power tests. The Team test is a half-hour, 10-question short answer test; each problem is worth 6 points. The Power test is a 1-hour proof test on a set of related mathematical concepts. Although the Power test does not expect rigorous proofs, a clear and complete explanation is required for full credit.<br />
<br />
Those who do not qualify for the individual tiebreakers participate, as teams, in the 1-hour Mach round. This is a fun round consisting of 100 problems to be solved in the hour allotted; a runner obtains the problems in groups of 10. Later problems are more difficult and are worth more, but once teams turn in a set, they cannot get them back later. The Mach round contributes points to the team sweepstakes total as well.<br />
<br />
The top 5 teams receive awards based on their performance in the individual, Power, Team, and Mach rounds. The rounds are weighted appropriately to promote a balance between team and individual scores.<br />
<br />
'''More Information'''<br />
See the contest website, [http://math.ncssm.edu/~rash/mach8/], for more information.</div>Not trighttps://artofproblemsolving.com/wiki/index.php?title=MaCh_8&diff=30651MaCh 82009-03-10T18:57:11Z<p>Not trig: New page: The MaCh 8 (Mathematics Challenge 8) is a middle school math contest run by students at the North Carolina School of Science and Math in Durham, NC. It will be held on April 4, 2009. [[In...</p>
<hr />
<div>The MaCh 8 (Mathematics Challenge 8) is a middle school math contest run by students at the North Carolina School of Science and Math in Durham, NC. It will be held on April 4, 2009.<br />
<br />
[[Individual Tests]]<br />
The MaCh 8 features three individual subject tests in Algebra, Geometry, and Discrete Math, as well as a comprehensive "General" test for students who would rather not specialize. Students take either two of the Subject tests or the General test. Individual tests are short answer: 25 problems for Subject tests, and 50 problems for the General tests. Students are allotted an hour to take each Subject test and 2 hours for the General, which is broken into two sets of 25 with a break in between. Each set of 25 is scored as follows:<br />
<br />
#1-5: 2 points each<br />
#6-10: 3 points<br />
#11-15: 4 points<br />
#16-20: 5 points<br />
#21-25: 6 points,<br />
<br />
for a total of 100 points per Subject test and 200 points on the General test.<br />
<br />
Top students in each individual test enter a difficult short-answer tiebreaker. Top-scoring students on each Subject test, the General test, and the tiebreakers receive awards.<br />
<br />
[[Team Tests]]<br />
The MaCh 8 offers both Team and Power tests. The Team test is a half-hour, 10-question short answer test; each problem is worth 6 points. The Power test is a 1-hour proof test on a set of related mathematical concepts. Although the Power test does not expect rigorous proofs, a clear and complete explanation is required for full credit.<br />
<br />
Those who do not qualify for the individual tiebreakers participate, as teams, in the 1-hour Mach round. This is a fun round consisting of 100 problems to be solved in the hour allotted; a runner obtains the problems in groups of 10. Later problems are more difficult and are worth more, but once teams turn in a set, they cannot get them back later. The Mach round contributes points to the team sweepstakes total as well.<br />
<br />
The top 5 teams receive awards based on their performance in the individual, Power, Team, and Mach rounds. The rounds are weighted appropriately to promote a balance between team and individual scores.<br />
<br />
[[More Information]]<br />
See the contest website, [http://math.ncssm.edu/~rash/mach8/], for more information.</div>Not trighttps://artofproblemsolving.com/wiki/index.php?title=AoPS_Wiki:AoPS_Community_Awards&diff=28111AoPS Wiki:AoPS Community Awards2008-09-30T15:25:57Z<p>Not trig: /* Perfect AIME Scores */</p>
<hr />
<div>This '''AoPS Community Awards''' page is a celebration of the accomplishments of members of the [[AoPS]] community.<br />
<br />
<br />
== IMO Participants and Medalists ==<br />
This is a list of members of the AoPS community who have competed for their country at the [[International Mathematical Olympiad]].<br />
<br />
=== Participants ===<br />
* Zachary Abel (2006) (AoPS assistant instructor)<br />
* Marco Avila (2006)<br />
* Zarathustra Brady (2006)<br />
* Robert Cordwell (2005)<br />
* Sherry Gong (2002, 2003, 2004, 2005, 2007)<br />
* Elyot Grant (2005)<br />
* Darij Grinberg (2006)<br />
* Mahbubul Hasan (2005)<br />
* Daniel Kane (AoPS assistant instructor)<br />
* Kiran Kedlaya (1990, 1991, 1992) ([[Art of Problem Solving Foundation]] board member)<br />
* Viktoriya Krakovna (2006)<br />
* Nate Ince (2004) (AoPS assistant instructor)<br />
* Brian Lawrence (2005, 2007) ([[WOOT]] instructor)<br />
* Thomas Mildorf (2005) (AoPS assistant instructor)<br />
* Alison Miller (2004) (AoPS assistant instructor)<br />
* Richard Peng (2005, 2006)<br />
* Eric Price (2005)<br />
* David Rhee (2004, 2005, 2006)<br />
* Peng Shi (2004, 2005, 2006)<br />
* Arne Smeets (2003, 2004)<br />
* Arnav Tripathy (2006, 2007)<br />
* [[Naoki Sato]] (AoPS instructor)<br />
* Yi Sun (2006)<br />
* [[Valentin Vornicu]] (AoPS/MathLinks webmaster)<br />
* Melanie Wood (1998, 1999) ([[WOOT]] instructor)<br />
* Alex Zhai (2005, 2006, 2007)<br />
* Yufei Zhao (2004, 2005, 2006)<br />
* Tigran Sloyan(2003,2004,2005,2006,2007)<br />
* Marco Avila (2006)<br />
* Vipul Naik (2003,2004)<br />
* Bhargav Narayanan (2007)<br />
* Tigran Hakobyan (2007)<br />
<br />
===Perfect Scorers===<br />
*Brian Lawrence (2005)<br />
*Alex Zhai (2008)<br />
<br />
=== Gold medalists ===<br />
* Zarathustra Brady (2006)<br />
* Robert Cordwell (2005)<br />
* Darij Grinberg (2006)<br />
* Kiran Kedlaya (1990, 1992) ([[Art of Problem Solving Foundation]] board member)<br />
* Brian Lawrence (2005) ([[WOOT]] instructor)<br />
* Thomas Mildorf (2005) (AoPS assistant instructor)<br />
* Alison Miller (2004) (AoPS assistant instructor)<br />
* Arnav Tripathy (2006)<br />
* Eric Price (2005)<br />
* Yufei Zhao (2005)<br />
* Alex Zhai (2007, 2008)<br />
*Sherry Gong (2007)<br />
*Krishanu Sankar (2008)<br />
<br />
=== Silver medalists ===<br />
* Zachary Abel (2006) (AoPS assistant instructor)<br />
* Sherry Gong (2004, 2005)<br />
* Nate Ince (2004) (AoPS assistant instructor)<br />
* Kiran Kedlaya (1991) ([[Art of Problem Solving Foundation]] board member)<br />
* Viktoriya Krakovna (2006)<br />
* Hyun Soo Kim (2005) (AoPS assistant instructor)<br />
* Richard Peng (2005)<br />
* David Rhee (2006)<br />
* Naoki Sato (AoPS instructor)<br />
* Peng Shi (2006)<br />
* Arne Smeets (2004)<br />
* Yi Sun (2006)<br />
* [[Sam Vandervelde]] (1989) ([[WOOT]] instructor)<br />
* Melanie Wood (1998, 1999) ([[WOOT]] instructor)<br />
* Alex Zhai (2006)<br />
* Yufei Zhao (2006)<br />
* Tigran Sloyan(2006,2007)<br />
* Vipul Naik (2003,2004)<br />
<br />
=== Bronze medalists ===<br />
* Sherry Gong (2003)<br />
* Elyot Grant (2005)<br />
* Richard Peng (2006)<br />
* [[Naoki Sato]] (AoPS instructor)<br />
* [[Valentin Vornicu]] (AoPS/[[MathLinks]] webmaster)<br />
* Yufei Zhao (2004)<br />
* Tigran Sloyan(2004;2005)<br />
* Tigran Hakobyan (2007)<br />
<br />
== IPhO Participants and Medalists ==<br />
This is a list of members of the AoPS community who have competed for their country at the [[International Physics Olympiad]].<br />
=== Participants ===<br />
* Sherry Gong (2006)<br />
* Yi Sun (2004)<br />
* Arnav Tripathy (2006)<br />
<br />
=== Gold Medalists ===<br />
* Yi Sun (2004)<br />
* Rahul Singh (2007)<br />
<br />
=== Silver Medalists ===<br />
* Sherry Gong (2006)<br />
<br />
== USAMO ==<br />
The following AoPSers have won the [[United States of America Mathematical Olympiad]] (USAMO). (Note that the definition of "winner" has changed over the years -- currently it is the top 12 scores on the USAMO, but in the past it has been the top 6 or top 8 scores.)<br />
=== Perfect Scorers ===<br />
* Daniel Kane (AoPS assistant instructor)<br />
* Kiran Kedlaya (1991) ([[Art of Problem Solving Foundation]] board member)<br />
* Brian Lawrence (2006) ([[WOOT]] instructor)<br />
<br />
=== Winners ===<br />
* Yakov Berchenko-Kogan (2006)<br />
* Sherry Gong (2006, 2007)<br />
* Yi Han (2006)<br />
* Adam Hesterberg (2007)<br />
* Daniel Kane (AoPS assistant instructor)<br />
* Kiran Kedlaya (1990, 1991, 1992) ([[Art of Problem Solving Foundation]] board member)<br />
* Brian Lawrence (2005, 2006, 2007) ([[WOOT]] instructor)<br />
* Tedrick Leung (2006, 2007)<br />
* Haitao Mao (2007)<br />
* Richard Mccutchen (2006)<br />
* Albert Ni (2005)<br />
* [[David Patrick]] (1988) (AoPS instructor)<br />
* [[Richard Rusczyk]] (1989) (AoPS founder)<br />
* Krishanu Sankar (2007,2008)<br />
* Peng Shi (2006)<br />
* Jacob Steinhardt (2007)<br />
* Yi Sun (2006)<br />
* Arnav Tripathy (2006, 2007)<br />
* [[Sam Vandervelde]] (1987, 1989) ([[WOOT]] instructor)<br />
* Melanie Wood (1998, 1999) ([[WOOT]] instructor)<br />
* Alex Zhai (2006, 2007,2008)<br />
* Yufei Zhao (2006)<br />
<br />
== Putnam Fellows ==<br />
The top 5 students (including ties) on the collegiate [[Putnam Exam|William Lowell Putnam Competition]] are named Putnam Fellows.<br />
* David Ash (1981, 1982, 1983)<br />
* Daniel Kane (2003, 2004, 2005) (AoPS assistant instructor)<br />
* Kiran Kedlaya (1994, 1995, 1996) ([[AoPS Foundation]] board member)<br />
* Matthew Ince (2005) (AoPS assistant instructor)<br />
* Alexander Schwartz (2000, 2002)<br />
* Jan Siwanowicz (2001) <br />
* Melanie Wood (2002) ([[WOOT]] instructor)<br />
*Arnav Tripathy (2008)<br />
*Brian Lawrence (2008)<br />
<br />
== Siemens Competition Winners ==<br />
The annual [[Siemens Competition]] (formerly Siemens-Westinghouse) is a scientific research competition.<br />
* Michael Viscardi (1st Individual,2005)<br />
* Lucia Mocz (2nd Team, 2006)<br />
<br />
==Intel STS Finalists==<br />
The annual [[Intel Science Talent Search]] is a science competition seeking to find and reward the most scientifically accomplished seniors.<br />
<br />
* Philip Mocz (2008)<br />
* Yihe Dong (2008)<br />
* Qiaochu Yuan (2008)<br />
* Greg Brockman (2007)<br />
<br />
== Clay Junior Fellows ==<br />
Each year since 2003, the [[Clay Mathematics Institute]] has selected 12 Junior Fellows.<br />
* Thomas Belulovich (2005) (AoPS assistant instructor)<br />
* Atoshi Chowdhury (2003) (AoPS assistant instructor)<br />
* Robert Cordwell (2005)<br />
* Eve Drucker (2003) (AoPS assistant instructor)<br />
* Matthew Ince (2004) (AoPS assistant instructor)<br />
* Nate Ince (2004) (AoPS assistant instructor)<br />
* Hyun Soo Kim (2005) (AoPS assistant instructor)<br />
* Raju Krishnamoorthy (2005)<br />
* Alison Miller (2003) (AoPS assistant instructor)<br />
* Brian Rice (2003) (AoPS assistant instructor)<br />
* Dmitry Taubinski (2005) (AoPS assistant instructor)<br />
* Ameya Velingker (2005)<br />
<br />
<br />
== Perfect AIME Scores ==<br />
Very few students have ever achieved a perfect score on the [[American Invitational Mathematics Examination]] (AIME)<br />
* David Benjamin (2006)<br />
* [[Mathew Crawford]] (1992) (AoPS instructor)<br />
* [[Sandor Lehoczky]] (1990) (AoPS author)<br />
* Tedrick Leung (2006)<br />
* Tony Liu (2006)<br />
* [[Richard Rusczyk]] (1989) (AoPS founder)<br />
* [[Sam Vandervelde]] (1988) ([[WOOT]] instructor)<br />
* Sam Elder (2008)<br />
* Haitao Mao (2008)<br />
<br />
== Perfect AMC Scores ==<br />
=== Perfect AMC 12 Scores ===<br />
The [[AMC 12]] is a challenging examination for students in grades 12 and below administered by the [[American Mathematics Competitions]].<br />
* David Benjamin (2006)<br />
* Zachary Abel (2005) (AoPS assistant instructor)<br />
* Ruozhou (Joe) Jia (2003) (AoPS assistant instructor)<br />
* Joel Lewis (2003) <br />
* Jonathan Lowd (2003) (AoPS assistant instructor)<br />
* Thomas Mildorf (2004) (AoPS assistant instructor)<br />
* Alison Miller (2004) (AoPS assistant instructor)<br />
* Albert Ni (2003) (AoPS instructor)<br />
* Ajay Sharma (2004)<br />
* Arnav Tripathy (2006, 2007)<br />
* Alex Zhai (2007)<br />
<br />
=== Perfect AMC 10 Scores ===<br />
The [[AMC 10]] is a challenging examination for students in grades 10 and below administered by the [[American Mathematics Competitions]].<br />
* Sergei Bernstein (2007)<br />
* Yifan Cao (2005)<br />
* Kevin Chen (2007)<br />
* In Young Cho (2007)<br />
* Mario Choi (2007)<br />
* Billy Dorminy (2007)<br />
* Zhou Fan (2005)<br />
* Albert Gu (2007)<br />
* Robin He (2007)<br />
* Keone Hon (2005)<br />
* Susan Hu (2005)<br />
* Lyndon Ji (2008)<br />
* Sam Keller (2007)<br />
* Vincent Le (2006)<br />
* Daniel Li (2007)<br />
* Johnny Li (2007)<br />
* Patricia Li (2005)<br />
* Carl Lian (2007)<br />
* Thomas Mildorf (2002) (AoPS assistant instructor)<br />
* Anupa Murali (2008)<br />
* Howard Tong (2005)<br />
* Sam Trabucco (2008)<br />
* Brent Woodhouse (2006, 2007)<br />
* Jonathan Zhou (2007)<br />
<br />
=== Perfect AHSME Scores ===<br />
The [[American High School Mathematics Examination]] (AHSME) was the predecessor of the AMC 12.<br />
* Christopher Chang (1994, 1995, 1996)<br />
* [[Mathew Crawford]] (1994, 1995) (AoPS instructor)<br />
* [[David Patrick]] (1988) (AoPS instructor)<br />
<br />
<br />
== MATHCOUNTS ==<br />
[[MathCounts]] is the premier middle school [[mathematics competition]] in the U.S.<br />
=== National Champions ===<br />
* Ruozhou (Joe) Jia (2000) (AoPS assistant instructor)<br />
* Albert Ni (2002) (AoPS instructor)<br />
* Adam Hesterberg (2003)<br />
* Neal Wu (2005)<br />
* Daesun Yim (2006)<br />
* Kevin Chen (2007)<br />
<br />
=== National Top 12 ===<br />
* Ashley Reiter Ahlin (1987) ([[WOOT]] instructor)<br />
* Andrew Ardito (2005, 2006)<br />
* David Benjamin (2004, 2005)<br />
* Nathan Benjamin (2005, 2006)<br />
* Wenyu Cao (2007)<br />
* Christopher Chang (1991, 1992)<br />
* Kevin Chen (2006, 2007)<br />
* Andrew Chien (2003)<br />
* Peter Chien (2004)<br />
* Mario Choi (2007)<br />
* Joseph Chu (2004)<br />
* [[Mathew Crawford]] (1990, 1991) (AoPS instructor)<br />
* Brian Hamrick (2006)<br />
* Adam Hesterberg (2002, 2003)<br />
* Ruozhou (Joe) Jia (2000) (AoPS assistant instructor)<br />
* Sam Keller (2006)<br />
* Shaunak Kishore (2003, 2004)<br />
* Kiran Kota (2005)<br />
* Brian Lawrence (2003) ([[WOOT]] instructor)<br />
* Karlanna Lewis (2005)<br />
* Daniel Li (2006)<br />
* Patricia Li (2005)<br />
* Poh-Ling Loh (2000)<br />
* Albert Ni (2002) (AoPS assistant instructor)<br />
* Elizabeth Synge (2007)<br />
* Jason Trigg (2002)<br />
* [[Sam Vandervelde]] (1985) ([[WOOT]] instructor)<br />
* Neal Wu (2005, 2006)<br />
* Rolland Wu (2006)<br />
* Daesun Yim (2006)<br />
* Darren Yin (2002)<br />
* Allen Yuan (2007)<br />
* Alex Zhai (2004)<br />
* Mark Zhang (2005)<br />
<br />
=== Masters Round Champions ===<br />
* Christopher Chang (1991)<br />
* Brian Lawrence (2003) ([[WOOT]] instructor)<br />
* Sergei Bernstein (2005)<br />
* Daniel Li (2006)<br />
* Kevin Chen (2007)<br />
* Bobby Shen (2008)<br />
<br />
=== National Test Champions ===<br />
* [[Mathew Crawford]] (1990) (AoPS instructor)<br />
* Adam Hesterberg (2003)<br />
* Sergei Bernstein (2005)<br />
* Neal Wu (2006)<br />
* Bobby Shen (2008)<br />
<br />
== Harvard-MIT Math Tournament ==<br />
<br />
The [[HMMT]] 2007 winning team, the "WOOTlings", consisted entirely of [[WOOT]]ers:<br />
<br />
* Wenyu Cao<br />
* Eric Chang<br />
* Jeremy Hahn<br />
* Alex Kandell<br />
* Adeel Khan<br />
* Sathish Nagappan<br />
* Krishanu Roy Sankar<br />
* Patrick Tenorio<br />
<br />
== ARML ==<br />
<br />
=== ARML winners ===<br />
<br />
=== ARML Top 10 ===<br />
* Zachary Abel (2006) (AoPS assistant instructor)<br />
* Arnav Tripathy (2007)<br />
<br />
== See also ==<br />
* [[Academic competitions]]<br />
* [[Mathematics competitions]]<br />
* [[Mathematics competition resources]]<br />
* [[Academic scholarships]]<br />
<br />
<br />
<br />
[[Category:Art of Problem Solving]]</div>Not trighttps://artofproblemsolving.com/wiki/index.php?title=De_Moivre%27s_Theorem&diff=27894De Moivre's Theorem2008-09-10T17:15:54Z<p>Not trig: </p>
<hr />
<div>{{WotWAnnounce|week=September 5- September 11}}<br />
<br />
'''DeMoivre's Theorem''' is a very useful theorem in the mathematical fields of [[complex numbers]]. It allows complex numbers in [[polar form]] to be easily raised to certain powers. It states that for <math>x\in\mathbb{R}</math> and <math>n\in\mathbb{Z}</math>, <math>\left(\cos x+i\sin x\right)^n=\cos(nx)+i\sin(nx)</math>.<br />
<br />
== Proof ==<br />
This is one proof of De Moivre's theorem by [[induction]].<br />
<br />
*If <math>n>0</math>, for <math>n=1</math>, the case is obviously true.<br />
<br />
:Assume true for the case <math>n=k</math>. Now, the case of <math>n=k+1</math>:<br />
<br />
:[[Image:DeMoivreInductionP1.gif]]<br />
<br />
:Therefore, the result is true for all positive integers <math>n</math>.<br />
<br />
*If <math>n=0</math>, the formula holds true because <math>\cos(0x)+i\sin (0x)=1+i0=1</math>. Since <math>z^0=1</math>, the equation holds true.<br />
<br />
*If <math>n<0</math>, one must consider <math>n=-m</math> when <math>m</math> is a positive integer.<br />
<br />
:[[Image:DeMoivreInductionP2.gif]]<br />
<br />
And thus, the formula proves true for all integral values of <math>n</math>. <math>\Box</math><br />
<br />
Note that from the functional equation <math>f(x)^n = f(nx)</math> where <math>f(x) = \cos x + i\sin x</math>, we see that <math>f(x)</math> behaves like an exponential function. Indeed, [[Euler's formula]] states that <math>e^{ix} = \cos x+i\sin x\right</math>. This extends De Moivre's theorem to all <math>n\in \mathbb{R}</math>.<br />
<br />
[[Category:Theorems]]<br />
[[Category:Complex numbers]]</div>Not trighttps://artofproblemsolving.com/wiki/index.php?title=De_Moivre%27s_Theorem&diff=27893De Moivre's Theorem2008-09-10T17:14:54Z<p>Not trig: </p>
<hr />
<div>{{WotWAnnounce|week=September 5- September 11}}<br />
<br />
'''DeMoivre's Theorem''' is a very useful theorem in the mathematical fields of [[complex numbers]]. It allows complex numbers in [[polar form]] to be easily raised to certain powers. It states that for <math>x\in\mathbb{R}</math> and <math>n\in\mathbb{N}</math>, <math>\left(\cos x+i\sin x\right)^n=\cos(nx)+i\sin(nx)</math>.<br />
<br />
== Proof ==<br />
This is one proof of De Moivre's theorem by [[induction]].<br />
<br />
*If <math>n>0</math>, for <math>n=1</math>, the case is obviously true.<br />
<br />
:Assume true for the case <math>n=k</math>. Now, the case of <math>n=k+1</math>:<br />
<br />
:[[Image:DeMoivreInductionP1.gif]]<br />
<br />
:Therefore, the result is true for all positive integers <math>n</math>.<br />
<br />
*If <math>n=0</math>, the formula holds true because <math>\cos(0x)+i\sin (0x)=1+i0=1</math>. Since <math>z^0=1</math>, the equation holds true.<br />
<br />
*If <math>n<0</math>, one must consider <math>n=-m</math> when <math>m</math> is a positive integer.<br />
<br />
:[[Image:DeMoivreInductionP2.gif]]<br />
<br />
And thus, the formula proves true for all integral values of <math>n</math>. <math>\Box</math><br />
<br />
Note that from the functional equation <math>f(x)^n = f(nx)</math> where <math>f(x) = \cos x + i\sin x</math>, we see that <math>f(x)</math> behaves like an exponential function. Indeed, [[Euler's formula]] states that <math>e^{ix} = \cos x+i\sin x\right</math>. This extends De Moivre's theorem to all <math>n\in \mathbb{R}</math>.<br />
<br />
[[Category:Theorems]]<br />
[[Category:Complex numbers]]</div>Not trighttps://artofproblemsolving.com/wiki/index.php?title=De_Moivre%27s_Theorem&diff=27892De Moivre's Theorem2008-09-10T17:14:02Z<p>Not trig: </p>
<hr />
<div>{{WotWAnnounce|week=September 5- September 11}}<br />
<br />
'''DeMoivre's Theorem''' is a very useful theorem in the mathematical fields of [[complex numbers]]. It allows complex numbers in [[polar form]] to be easily raised to certain powers. It states that for <math>x\in\mathbb{R}</math> and <math>n\in\mathbb{N}</math>, <math>\left(\cos x+i\sin x\right)^n=\cos(nx)+i\sin(nx)</math>.<br />
<br />
== Proof ==<br />
This is one proof of De Moivre's theorem by [[induction]].<br />
<br />
*If <math>n>0</math>, for <math>n=1</math>, the case is obviously true.<br />
<br />
:Assume true for the case <math>n=k</math>. Now, the case of <math>n=k+1</math>:<br />
<br />
:[[Image:DeMoivreInductionP1.gif]]<br />
<br />
:Therefore, the result is true for all positive integers <math>n</math>.<br />
<br />
*If <math>n=0</math>, the formula holds true because <math>\cos(0x)+i\sin (0x)=1+i0=1</math>. Since <math>z^0=1</math>, the equation holds true.<br />
<br />
*If <math>n<0</math>, one must consider <math>n=-m</math> when <math>m</math> is a positive integer.<br />
<br />
:[[Image:DeMoivreInductionP2.gif]]<br />
<br />
And thus, the formula proves true for all integral values of <math>n</math>. <math>\Box</math><br />
<br />
Note that from the functional equation <math>f(x)^n = f(nx)</math> where <math>f(x) = \cos x + i\sin x</math>, we see that <math>f(x)</math> behaves like an exponential function. Indeed, by [[Euler's formula]] (<math>e^{ix} = \cos x+i\sin x\right</math>). This extends De Moivre's theorem to all <math>n\in \mathbb{R}</math>.<br />
<br />
[[Category:Theorems]]<br />
[[Category:Complex numbers]]</div>Not trighttps://artofproblemsolving.com/wiki/index.php?title=De_Moivre%27s_Theorem&diff=27891De Moivre's Theorem2008-09-10T17:13:12Z<p>Not trig: </p>
<hr />
<div>{{WotWAnnounce|week=September 5- September 11}}<br />
<br />
'''DeMoivre's Theorem''' is a very useful theorem in the mathematical fields of [[complex numbers]]. It allows complex numbers in [[polar form]] to be easily raised to certain powers. It states that for <math>x\in\mathbb{R}</math> and <math>n\in\mathbb{N}</math>, <math>\left(\cos x+i\sin x\right)^n=\cos(nx)+i\sin(nx)</math>.<br />
<br />
== Proof ==<br />
This is one proof of De Moivre's theorem by [[induction]].<br />
<br />
*If <math>n>0</math>, for <math>n=1</math>, the case is obviously true.<br />
<br />
:Assume true for the case <math>n=k</math>. Now, the case of <math>n=k+1</math>:<br />
<br />
:[[Image:DeMoivreInductionP1.gif]]<br />
<br />
:Therefore, the result is true for all positive integers <math>n</math>.<br />
<br />
*If <math>n=0</math>, the formula holds true because <math>\cos(0x)+i\sin (0x)=1+i0=1</math>. Since <math>z^0=1</math>, the equation holds true.<br />
<br />
*If <math>n<0</math>, one must consider <math>n=-m</math> when <math>m</math> is a positive integer.<br />
<br />
:[[Image:DeMoivreInductionP2.gif]]<br />
<br />
And thus, the formula proves true for all integral values of <math>n</math>. <math>\Box</math><br />
<br />
Note that from the functional equation <math>f(x) = f(nx)</math> where <math>f(x) = \cos x + i\sin x</math>, we see that <math>f(x)</math> behaves like an exponential function. Indeed, by [[Euler's formula]] (<math>e^{ix} = \cos x+i\sin x\right</math>). This extends De Moivre's theorem to all <math>n\in \mathbb{R}</math>.<br />
<br />
[[Category:Theorems]]<br />
[[Category:Complex numbers]]</div>Not trighttps://artofproblemsolving.com/wiki/index.php?title=North_Carolina_MathCounts&diff=27612North Carolina MathCounts2008-08-27T20:05:11Z<p>Not trig: </p>
<hr />
<div>North Carolina often fields a top 10 team at national [[MathCounts]] competitions. In recent years, [[Harold Reiter]] has organized practice sessions for the national teams from NC and SC at the [[University of North Carolina at Charlotte]]. <br />
<br />
The top four students from the written round at the state competition are selected for the national team. The coach of the winning school team is offered the opportunity to coach the state team at the national competition.<br />
<br />
<br />
== National Team Awards and Placement ==<br />
<br />
* 2007 8th<br />
* 2006 9th<br />
* 2005 10th<br />
* 2003 5th<br />
* 1998 5th<br />
* 1989 1st<br />
<br />
== NC State Team Champions ==<br />
<br />
* 2007 - Arendell Parrott Academy, Kinston<br />
* 2006 - Martin Middle School, Raleigh<br />
* 2005 - [http://web2k.wsfcs.k12.nc.us/hanesms/academics/mcounts.htm Hanes Middle School], Winston-Salem<br />
* 2004 - Ligon Middle School, Raleigh<br />
* 2003 - Guy B. Phillips Middle School, Chapel Hill<br />
* 2002 - Martin Middle School, Raleigh<br />
* 2001 - Ligon Middle School, Raleigh<br />
* 2000 - Durham School of Arts, Durham<br />
* 1999 - Martin Middle School, Raleigh<br />
* 1998 - Charlotte Latin, Charlotte<br />
<br />
* 1987 - Charlotte Latin, Charlotte<br />
<br />
== NC State Team Members (top 57 placement) ==<br />
<br />
* 2007 - Jenny Chen (35), Gray Symon (37), Brendan Fletcher (44), David Lucia, Coach: Tonya McLawhorn<br />
* 2006 - Eric "Pom" McCabe (22), Brendan Fletcher, Gray Symon (48), Akash Ganapathi, Coach: Linda Harvel (Southern Middle)<br />
* 2005 - Eric "Pom" McCabe (13), John Berman (45), Bryce Taylor, Kevin Lang, Coach: Greg Taylor<br />
* 2004 - Ray Wang (44), Vivek Bhattacharya, Matthew Adams, Kevin Lang, Coach: Cathy Whelan<br />
* 2003 - Arnav Tripathy (6), Mikhail Lavrov (16), Michael Lin, Vivek Bhattacharya, Coach: Marla McCrea<br />
* 2002 - Mikhail Lavrov, Chris Hoersting, Sunny Xi, Amy Wen, Coach: Jeanette Rojeski<br />
* 2001 - Jeremy Diepenbrock, Jeff Tang, Paul Huang, Sunny Xi, Coach: Pat Heald<br />
* 2000 - John Shoun, Larry Wise, Charles Yu, Justin Lo, Coach: Robin Barbour<br />
* 1999 - Morgan Brown, Edward Su, Jonathan Campbell, Don Swartz, Coach: Lucy Kay<br />
* 1998 - Anders Kaseorg (17), Paul Wrayno, Eric Sundstrom (38), Chris Goulette, Coach: Caroline Wolfe<br />
* 1997 - Anders Kaseorg,?,?,?<br />
<br />
* 1989 - Lenny Ng (1),?,?,?<br />
<br />
* 1987 - Ashley Reiter (3),?,?,?, Coach: Caroline Wolfe<br />
* 1986 - Jeff VanderKam (2),?,?,?</div>Not trighttps://artofproblemsolving.com/wiki/index.php?title=2008_USAMO_Problems/Problem_2&diff=259782008 USAMO Problems/Problem 22008-05-12T01:09:42Z<p>Not trig: </p>
<hr />
<div>== Problem ==<br />
(''Zuming Feng'') Let <math>ABC</math> be an acute, [[scalene]] triangle, and let <math>M</math>, <math>N</math>, and <math>P</math> be the midpoints of <math>\overline{BC}</math>, <math>\overline{CA}</math>, and <math>\overline{AB}</math>, respectively. Let the [[perpendicular]] [[bisect]]ors of <math>\overline{AB}</math> and <math>\overline{AC}</math> intersect ray <math>AM</math> in points <math>D</math> and <math>E</math> respectively, and let lines <math>BD</math> and <math>CE</math> intersect in point <math>F</math>, inside of triangle <math>ABC</math>. Prove that points <math>A</math>, <math>N</math>, <math>F</math>, and <math>P</math> all lie on one circle.<br />
<br />
__TOC__<br />
== Solution ==<br />
=== Solution 1 (synthetic) ===<br />
<center><asy><br />
/* setup and variables */<br />
size(280);<br />
pathpen = black + linewidth(0.7); pointpen = black; pen s = fontsize(8);<br />
pair B=(0,0),C=(5,0),A=(1,4); /* A.x > C.x/2 */<br />
/* construction and drawing */<br />
pair P=(A+B)/2,M=(B+C)/2,N=(A+C)/2,D=IP(A--M,P--P+5*(P-bisectorpoint(A,B))),E=IP(A--M,N--N+5*(bisectorpoint(A,C)-N)),F=IP(B--B+5*(D-B),C--C+5*(E-C)),O=circumcenter(A,B,C);<br />
D(MP("A",A,(0,1),s)--MP("B",B,SW,s)--MP("C",C,SE,s)--A--MP("M",M,s));<br />
D(B--D(MP("D",D,NE,s))--MP("P",P,(-1,0),s)--D(MP("O",O,(0,1),s)));<br />
D(D(MP("E",E,SW,s))--MP("N",N,(1,0),s));<br />
D(C--D(MP("F",F,NW,s))); <br />
D(B--O--C,linetype("4 4")+linewidth(0.7));<br />
D(M--N,linetype("4 4")+linewidth(0.7));<br />
D(rightanglemark(A,P,D,3.5));D(rightanglemark(A,N,E,3.5));<br />
D(anglemark(B,A,C)); MP("y",A,(0,-6));MP("z",A,(4,-6));<br />
D(anglemark(B,F,C,4),linewidth(0.6));D(anglemark(B,O,C,4),linewidth(0.6));<br />
picture p = new picture;<br />
draw(p,circumcircle(B,O,C),linetype("1 4")+linewidth(0.7));<br />
clip(p,B+(-5,0)--B+(-5,A.y+2)--C+(5,A.y+2)--C+(5,0)--cycle); add(p);<br />
<br />
/* D(circumcircle(A,P,N),linetype("4 4")+linewidth(0.7)); */<br />
</asy></center><br />
<br />
[[Without loss of generality]] <math>AB < AC</math>. The intersection of <math>NE</math> and <math>PD</math> is <math>O</math>, the circumcenter of <math>\triangle ABC</math>. <br />
<br />
Let <math>\angle BAM = y</math> and <math>\angle CAM = z</math>. Note <math>D</math> lies on the perpendicular bisector of <math>AB</math>, so <math>AD = BD</math>. So <math>\angle FBC = \angle B - \angle ABD = B - y</math>. Similarly, <math>\angle FCE = C - z</math>, so <math>\angle BFC = 180 - (B + C) + (y + z) = 2A</math>. Notice that <math>\angle BOC</math> intercepts the minor arc <math>BC</math> in the [[circumcircle]] of <math>\triangle ABC</math>, which is double <math>\angle A</math>. Hence <math>\angle BFC = \angle BOC</math>, so <math>FOBC</math> is cyclic. <br />
<br />
<br />
''Lemma 1'': <math>\triangle FEO</math> is directly similar to <math>\triangle NEM</math><br />
<cmath><br />
\angle OFE = \angle OFC = \angle OBC = \frac {1}{2}\cdot (180 - 2A) = 90 - A<br />
</cmath><br />
since <math>F</math>, <math>E</math>, <math>C</math> are collinear, <math>FOBC</math> is cyclic, and <math>OB = OC</math>. Also<br />
<cmath><br />
\angle ENM = 90 - \angle MNC = 90 - A<br />
</cmath><br />
because <math>NE\perp AC</math>, and <math>MNP</math> is the medial triangle of <math>\triangle ABC</math> so <math>AB \parallel MN</math>. Hence <math>\angle OFE = \angle ENM</math>.<br />
<br />
Notice that <math>\angle AEN = 90 - z = \angle CEN</math> since <math>NE\perp BC</math>. <math>\angle FED = \angle MEC = 2z</math>. Then<br />
<cmath><br />
\angle FEO = \angle FED + \angle AEN = \angle CEM + \angle CEN = \angle NEM<br />
</cmath><br />
Hence <math>\angle FEO = \angle NEM</math>. <br />
<br />
Hence <math>\triangle FEO</math> is similar to <math>\triangle NEM</math> by AA similarity. It is easy to see that they are oriented such that they are directly similar. ''End Lemma 1.''<br />
<br />
<center><asy><br />
/* setup and variables */<br />
size(280);<br />
pathpen = black + linewidth(0.7); pointpen = black; pen s = fontsize(8);<br />
pair B=(0,0),C=(5,0),A=(1,4); /* A.x > C.x/2 */<br />
/* construction and drawing */<br />
pair P=(A+B)/2,M=(B+C)/2,N=(A+C)/2,D=IP(A--M,P--P+5*(P-bisectorpoint(A,B))),E=IP(A--M,N--N+5*(bisectorpoint(A,C)-N)),F=IP(B--B+5*(D-B),C--C+5*(E-C)),O=circumcenter(A,B,C);<br />
D(MP("A",A,(0,1),s)--MP("B",B,SW,s)--MP("C",C,SE,s)--A--MP("M",M,s));<br />
D(B--D(MP("D",D,NE,s))--MP("P",P,(-1,0),s)--D(MP("O",O,(1,0),s)));<br />
D(D(MP("E",E,SW,s))--MP("N",N,(1,0),s));<br />
D(C--D(MP("F",F,NW,s))); <br />
D(B--O--C,linetype("4 4")+linewidth(0.7));<br />
D(F--N); D(O--M);<br />
D(rightanglemark(A,P,D,3.5));D(rightanglemark(A,N,E,3.5));<br />
<br />
/* commented from above asy<br />
D(circumcircle(A,P,N),linetype("4 4")+linewidth(0.7)); <br />
D(anglemark(B,A,C)); MP("y",A,(0,-6));MP("z",A,(4,-6));<br />
D(anglemark(B,F,C,4),linewidth(0.6));D(anglemark(B,O,C,4),linewidth(0.6));<br />
picture p = new picture;<br />
draw(p,circumcircle(B,O,C),linetype("1 4")+linewidth(0.7));<br />
clip(p,B+(-5,0)--B+(-5,A.y+2)--C+(5,A.y+2)--C+(5,0)--cycle); add(p);<br />
*/<br />
</asy></center><br />
<br />
By the similarity in Lemma 1, <math>FE: EO = NE: EM\implies FE: EN = OE: NM</math>. <math>\angle FEN = \angle OEM</math> so <math>\triangle FEN\sim\triangle OEM</math> by SAS similarity. Hence<br />
<cmath><br />
\angle EMO = \angle ENF = \angle ONF<br />
</cmath><br />
Using essentially the same angle chasing, we can show that <math>\triangle PDM</math> is directly similar to <math>\triangle FMO</math>. It follows that <math>\triangle PDF</math> is directly similar to <math>MDO</math>. So<br />
<cmath><br />
\angle EMO = \angle DMO = \angle DPF = \angle OPF<br />
</cmath><br />
Hence <math>\angle OPF = \angle ONF</math>, so <math>FONP</math> is cyclic. In other words, <math>F</math> lies on the circumcircle of <math>\triangle PON</math>. Note that <math>\angle ONA = \angle OPA = 90</math>, so <math>APON</math> is cyclic. In other words, <math>A</math> lies on the circumcircle of <math>\triangle PON</math>. <math>A</math>, <math>P</math>, <math>N</math>, <math>O</math>, and <math>F</math> all lie on the circumcircle of <math>\triangle PON</math>. Hence <math>A</math>, <math>P</math>, <math>F</math>, and <math>N</math> lie on a circle, as desired.<br />
<br />
=== Solution 2 (synthetic) ===<br />
Hint: consider <math>CF</math> intersection with <math>PM</math>; show that the resulting intersection lies on the desired circle. {{incomplete|solution}}<br />
<br />
=== Solution 3 (synthetic) ===<br />
This solution utilizes the ''phantom point method.'' Clearly, APON are cyclic because <math>\angle OPA = \angle ONA = 90</math>. Let the circumcircles of triangles <math>APN</math> and <math>BOC</math> intersect at <math>F'</math> and <math>O</math>. <br />
<br />
Lemma. If <math>A,B,C</math> are points on circle <math>\omega</math> with center <math>O</math>, and the tangents to <math>\omega</math> at <math>B,C</math> intersect at <math>Q</math>, then <math>AP</math> is the symmedian from <math>A</math> to <math>BC</math>.<br />
<br />
This is fairly easy to prove (as H, O are isogonal conjugates, plus using SAS similarity), but the author lacks time to write it up fully, and will do so soon.<br />
<br />
It is easy to see <math>Q</math> (the intersection of ray <math>OM</math> and the circumcircle of <math>\triangle BOC</math>) is colinear with <math>A</math> and <math>F'</math>, and because line <math>OM</math> is the diameter of that circle, <math>\angle QBO = \angle QCO = 90</math>, so <math>Q</math> is the point <math>Q</math> in the lemma; hence, we may apply the lemma. From here, it is simple angle-chasing to show that <math>F'</math> satisfies the original construction for <math>F</math>, showing <math>F=F'</math>; we are done. {{incomplete|solution}}<br />
<br />
=== Solution 4 (trigonometric) ===<br />
By the [[Law of Sines]], <math>\frac {\sin\angle BAM}{\sin\angle CAM} = \frac {\sin B}{\sin C} = \frac bc = \frac {b/AF}{c/AF} = \frac {\sin\angle AFC\cdot\sin\angle ABF}{\sin\angle ACF\cdot\sin\angle AFB}</math>. Since <math>\angle ABF = \angle ABD = \angle BAD = \angle BAM</math> and similarly <math>\angle ACF = \angle CAM</math>, we cancel to get <math>\sin\angle AFC = \sin\angle AFB</math>. Obviously, <math>\angle AFB + \angle AFC > 180^\circ</math> so <math>\angle AFC = \angle AFB</math>.<br />
<br />
Then <math>\angle FAB + \angle ABF = 180^\circ - \angle AFB = 180^\circ - \angle AFC = \angle FAC + \angle ACF</math> and <math>\angle ABF + \angle ACF = \angle A = \angle FAB + \angle FAC</math>. Subtracting these two equations, <math>\angle FAB - \angle FCA = \angle FCA - \angle FAB</math> so <math>\angle BAF = \angle ACF</math>. Therefore, <math>\triangle ABF\sim\triangle CAF</math> (by AA similarity), so a spiral similarity centered at <math>F</math> takes <math>B</math> to <math>A</math> and <math>A</math> to <math>C</math>. Therefore, it takes the midpoint of <math>\overline{BA}</math> to the midpoint of <math>\overline{AC}</math>, or <math>P</math> to <math>N</math>. So <math>\angle APF = \angle CNF = 180^\circ - \angle ANF</math> and <math>APFN</math> is cyclic.<br />
<br />
=== Solution 5 (isogonal conjugates) ===<br />
<center><asy><br />
/* setup and variables */<br />
size(280);<br />
pathpen = black + linewidth(0.7); pointpen = black; pen s = fontsize(8);<br />
pair B=(0,0),C=(5,0),A=(4,4); /* A.x > C.x/2 */<br />
/* construction and drawing */<br />
pair P=(A+B)/2,M=(B+C)/2,N=(A+C)/2,D=IP(A--M,P--P+5*(P-bisectorpoint(A,B))),E=IP(A--M,N--N+5*(bisectorpoint(A,C)-N)),F=IP(B--B+5*(D-B),C--C+5*(E-C)),O=circumcenter(A,B,C);<br />
D(MP("A",A,(0,1),s)--MP("B",B,SW,s)--MP("C",C,SE,s)--A--MP("M",M,s));<br />
D(C--D(MP("E",E,NW,s))--MP("N",N,(1,0),s)--D(MP("O",O,SW,s)));<br />
D(D(MP("D",D,SE,s))--MP("P",P,W,s));<br />
D(B--D(MP("F",F,s))); D(O--A--F,linetype("4 4")+linewidth(0.7));<br />
D(MP("O'",circumcenter(A,P,N),NW,s));<br />
D(circumcircle(A,P,N),linetype("4 4")+linewidth(0.7));<br />
D(rightanglemark(A,P,D,3.5));D(rightanglemark(A,N,E,3.5));<br />
picture p = new picture;<br />
draw(p,circumcircle(B,O,C),linetype("1 4")+linewidth(0.7));<br />
draw(p,circumcircle(A,B,C),linetype("1 4")+linewidth(0.7));<br />
clip(p,B+(-5,0)--B+(-5,A.y+2)--C+(5,A.y+2)--C+(5,0)--cycle); add(p);<br />
</asy></center><br />
<br />
Construct <math>T</math> on <math>AM</math> such that <math>\angle BCT = \angle ACF</math>. Then <math>\angle BCT = \angle CAM</math>. Then <math>\triangle AMC\sim\triangle CMT</math>, so <math>\frac {AM}{CM} = \frac {CM}{TM}</math>, or <math>\frac {AM}{BM} = \frac {BM}{TM}</math>. Then <math>\triangle AMB\sim\triangle BMT</math>, so <math>\angle CBT = \angle BAM = \angle FBA</math>. Then we have<br />
<br />
<math>\angle CBT = \angle ABF</math> and <math>\angle BCT = \angle ACF</math>. So <math>T</math> and <math>F</math> are [[isogonal conjugate|isogonally conjugate]]. Thus <math>\angle BAF = \angle CAM</math>. Then<br />
<br />
<math>\angle AFB = 180 - \angle ABF - \angle BAF = 180 - \angle BAM - \angle CAM = 180 - \angle BAC</math>.<br />
<br />
If <math>O</math> is the [[circumcenter]] of <math>\triangle ABC</math> then <math>\angle BFC = 2\angle BAC = \angle BOC</math> so <math>BFOC</math> is cyclic. Then <math>\angle BFO = 180 - \angle BOC = 180 - (90 - \angle BAC) = 90 + \angle BAC</math>.<br />
<br />
Then <math>\angle AFO = 360 - \angle AFB - \angle BFO = 360 - (180 - \angle BAC) - (90 + \angle BAC) = 90</math>. Then <math>\triangle AFO</math> is a right triangle.<br />
<br />
Now by the [[homothety]] centered at <math>A</math> with ratio <math>\frac {1}{2}</math>, <math>B</math> is taken to <math>P</math> and <math>C</math> is taken to <math>N</math>. Thus <math>O</math> is taken to the circumcenter of <math>\triangle APN</math> and is the midpoint of <math>AO</math>, which is also the circumcenter of <math>\triangle AFO</math>, so <math>A,P,N,F,O</math> all lie on a circle.<br />
<br />
=== Solution 6 (symmedians) ===<br />
Median <math>AM</math> of a triangle <math>ABC</math> implies <math>\frac {\sin{BAM}}{\sin{CAM}} = \frac {\sin{B}}{\sin{C}}</math>.<br />
Trig ceva for <math>F</math> shows that <math>AF</math> is a symmedian.<br />
Then <math>FP</math> is a median, use the lemma again to show that <math>AFP = C</math>, and similarly <math>AFN = B</math>, so you're done. {{incomplete|solution}}<br />
<br />
=== Solution 7 (inversion) ===<br />
{{image}}<br />
<center><asy><br />
size(280);<br />
pathpen = black + linewidth(0.7); pointpen = black; pen s = fontsize(8);<br />
pair B=(0,0),C=(5,0),A=(4,4); /* A.x > C.x/2 */<br />
real r = 1.2; /* inversion radius */<br />
<br />
/* construction and drawing */<br />
pair P=(A+B)/2,M=(B+C)/2,N=(A+C)/2,D=IP(A--M,P--P+5*(P-bisectorpoint(A,B))),E=IP(A--M,N--N+5*(bisectorpoint(A,C)-N)),F=IP(B--B+5*(D-B),C--C+5*(E-C)),O=circumcenter(A,B,C);<br />
D(MP("A",A,(0,1),s)--MP("B",B,SW,s)--MP("C",C,SE,s)--A--MP("M",M,s));<br />
D(C--D(MP("E",E,NW,s))--MP("N",N,(1,0),s)--D(MP("O",O,SW,s)));<br />
D(D(MP("D",D,SE,s))--MP("P",P,W,s));<br />
D(B--D(MP("F",F,s))); <br />
D(rightanglemark(A,P,D,3.5));D(rightanglemark(A,N,E,3.5));<br />
<br />
D(CR(A,r));<br />
pair Pa = A + (P-A)/(r*r); D(MP("P'",Pa,NW,s));<br />
<br />
</asy></center><br />
We consider an [[inversion]] by an arbitrary [[radius]] about <math>A</math>. We want to show that <math>P', F',</math> and <math>N'</math> are [[collinear]]. Notice that <math>D', A,</math> and <math>P'</math> lie on a circle with center <math>B'</math>, and similarly for the other side. We also have that <math>B', D', F', A</math> form a cyclic quadrilateral, and similarly for the other side. By angle chasing, we can prove that <math>A B' F' C'</math> is a [[parallelogram]], indicating that <math>F'</math> is the midpoint of <math>P'N'</math>. {{incomplete|solution}}<br />
<br />
=== Solution 8 (analytical) ===<br />
We let <math>A</math> be at the [[origin]], <math>B</math> be at the point <math>(a,0)</math>, and <math>C</math> be at the point <math>(b,c):\ b<a</math>. Then the equation of the perpendicular bisector of <math>\overline{AB}</math> is <math>x = a/2</math>, and {{incomplete|solution}}<br />
<br />
{{alternate solutions}}<br />
<br />
== Resources ==<br />
{{USAMO newbox|year=2008|num-b=1|num-a=3}}<br />
<br />
* <url>viewtopic.php?t=202907 Discussion on AoPS/MathLinks</url><br />
<br />
[[Category:Olympiad Geometry Problems]]</div>Not trighttps://artofproblemsolving.com/wiki/index.php?title=2008_USAMO_Problems/Problem_2&diff=259772008 USAMO Problems/Problem 22008-05-12T01:08:44Z<p>Not trig: </p>
<hr />
<div>== Problem ==<br />
(''Zuming Feng'') Let <math>ABC</math> be an acute, [[scalene]] triangle, and let <math>M</math>, <math>N</math>, and <math>P</math> be the midpoints of <math>\overline{BC}</math>, <math>\overline{CA}</math>, and <math>\overline{AB}</math>, respectively. Let the [[perpendicular]] [[bisect]]ors of <math>\overline{AB}</math> and <math>\overline{AC}</math> intersect ray <math>AM</math> in points <math>D</math> and <math>E</math> respectively, and let lines <math>BD</math> and <math>CE</math> intersect in point <math>F</math>, inside of triangle <math>ABC</math>. Prove that points <math>A</math>, <math>N</math>, <math>F</math>, and <math>P</math> all lie on one circle.<br />
<br />
__TOC__<br />
== Solution ==<br />
=== Solution 1 (synthetic) ===<br />
<center><asy><br />
/* setup and variables */<br />
size(280);<br />
pathpen = black + linewidth(0.7); pointpen = black; pen s = fontsize(8);<br />
pair B=(0,0),C=(5,0),A=(1,4); /* A.x > C.x/2 */<br />
/* construction and drawing */<br />
pair P=(A+B)/2,M=(B+C)/2,N=(A+C)/2,D=IP(A--M,P--P+5*(P-bisectorpoint(A,B))),E=IP(A--M,N--N+5*(bisectorpoint(A,C)-N)),F=IP(B--B+5*(D-B),C--C+5*(E-C)),O=circumcenter(A,B,C);<br />
D(MP("A",A,(0,1),s)--MP("B",B,SW,s)--MP("C",C,SE,s)--A--MP("M",M,s));<br />
D(B--D(MP("D",D,NE,s))--MP("P",P,(-1,0),s)--D(MP("O",O,(0,1),s)));<br />
D(D(MP("E",E,SW,s))--MP("N",N,(1,0),s));<br />
D(C--D(MP("F",F,NW,s))); <br />
D(B--O--C,linetype("4 4")+linewidth(0.7));<br />
D(M--N,linetype("4 4")+linewidth(0.7));<br />
D(rightanglemark(A,P,D,3.5));D(rightanglemark(A,N,E,3.5));<br />
D(anglemark(B,A,C)); MP("y",A,(0,-6));MP("z",A,(4,-6));<br />
D(anglemark(B,F,C,4),linewidth(0.6));D(anglemark(B,O,C,4),linewidth(0.6));<br />
picture p = new picture;<br />
draw(p,circumcircle(B,O,C),linetype("1 4")+linewidth(0.7));<br />
clip(p,B+(-5,0)--B+(-5,A.y+2)--C+(5,A.y+2)--C+(5,0)--cycle); add(p);<br />
<br />
/* D(circumcircle(A,P,N),linetype("4 4")+linewidth(0.7)); */<br />
</asy></center><br />
<br />
[[Without loss of generality]] <math>AB < AC</math>. The intersection of <math>NE</math> and <math>PD</math> is <math>O</math>, the circumcenter of <math>\triangle ABC</math>. <br />
<br />
Let <math>\angle BAM = y</math> and <math>\angle CAM = z</math>. Note <math>D</math> lies on the perpendicular bisector of <math>AB</math>, so <math>AD = BD</math>. So <math>\angle FBC = \angle B - \angle ABD = B - y</math>. Similarly, <math>\angle FCE = C - z</math>, so <math>\angle BFC = 180 - (B + C) + (y + z) = 2A</math>. Notice that <math>\angle BOC</math> intercepts the minor arc <math>BC</math> in the [[circumcircle]] of <math>\triangle ABC</math>, which is double <math>\angle A</math>. Hence <math>\angle BFC = \angle BOC</math>, so <math>FOBC</math> is cyclic. <br />
<br />
<br />
''Lemma 1'': <math>\triangle FEO</math> is directly similar to <math>\triangle NEM</math><br />
<cmath><br />
\angle OFE = \angle OFC = \angle OBC = \frac {1}{2}\cdot (180 - 2A) = 90 - A<br />
</cmath><br />
since <math>F</math>, <math>E</math>, <math>C</math> are collinear, <math>FOBC</math> is cyclic, and <math>OB = OC</math>. Also<br />
<cmath><br />
\angle ENM = 90 - \angle MNC = 90 - A<br />
</cmath><br />
because <math>NE\perp AC</math>, and <math>MNP</math> is the medial triangle of <math>\triangle ABC</math> so <math>AB \parallel MN</math>. Hence <math>\angle OFE = \angle ENM</math>.<br />
<br />
Notice that <math>\angle AEN = 90 - z = \angle CEN</math> since <math>NE\perp BC</math>. <math>\angle FED = \angle MEC = 2z</math>. Then<br />
<cmath><br />
\angle FEO = \angle FED + \angle AEN = \angle CEM + \angle CEN = \angle NEM<br />
</cmath><br />
Hence <math>\angle FEO = \angle NEM</math>. <br />
<br />
Hence <math>\triangle FEO</math> is similar to <math>\triangle NEM</math> by AA similarity. It is easy to see that they are oriented such that they are directly similar. ''End Lemma 1.''<br />
<br />
<center><asy><br />
/* setup and variables */<br />
size(280);<br />
pathpen = black + linewidth(0.7); pointpen = black; pen s = fontsize(8);<br />
pair B=(0,0),C=(5,0),A=(1,4); /* A.x > C.x/2 */<br />
/* construction and drawing */<br />
pair P=(A+B)/2,M=(B+C)/2,N=(A+C)/2,D=IP(A--M,P--P+5*(P-bisectorpoint(A,B))),E=IP(A--M,N--N+5*(bisectorpoint(A,C)-N)),F=IP(B--B+5*(D-B),C--C+5*(E-C)),O=circumcenter(A,B,C);<br />
D(MP("A",A,(0,1),s)--MP("B",B,SW,s)--MP("C",C,SE,s)--A--MP("M",M,s));<br />
D(B--D(MP("D",D,NE,s))--MP("P",P,(-1,0),s)--D(MP("O",O,(1,0),s)));<br />
D(D(MP("E",E,SW,s))--MP("N",N,(1,0),s));<br />
D(C--D(MP("F",F,NW,s))); <br />
D(B--O--C,linetype("4 4")+linewidth(0.7));<br />
D(F--N); D(O--M);<br />
D(rightanglemark(A,P,D,3.5));D(rightanglemark(A,N,E,3.5));<br />
<br />
/* commented from above asy<br />
D(circumcircle(A,P,N),linetype("4 4")+linewidth(0.7)); <br />
D(anglemark(B,A,C)); MP("y",A,(0,-6));MP("z",A,(4,-6));<br />
D(anglemark(B,F,C,4),linewidth(0.6));D(anglemark(B,O,C,4),linewidth(0.6));<br />
picture p = new picture;<br />
draw(p,circumcircle(B,O,C),linetype("1 4")+linewidth(0.7));<br />
clip(p,B+(-5,0)--B+(-5,A.y+2)--C+(5,A.y+2)--C+(5,0)--cycle); add(p);<br />
*/<br />
</asy></center><br />
<br />
By the similarity in Lemma 1, <math>FE: EO = NE: EM\implies FE: EN = OE: NM</math>. <math>\angle FEN = \angle OEM</math> so <math>\triangle FEN\sim\triangle OEM</math> by SAS similarity. Hence<br />
<cmath><br />
\angle EMO = \angle ENF = \angle ONF<br />
</cmath><br />
Using essentially the same angle chasing, we can show that <math>\triangle PDM</math> is directly similar to <math>\triangle FMO</math>. It follows that <math>\triangle PDF</math> is directly similar to <math>MDO</math>. So<br />
<cmath><br />
\angle EMO = \angle DMO = \angle DPF = \angle OPF<br />
</cmath><br />
Hence <math>\angle OPF = \angle ONF</math>, so <math>FONP</math> is cyclic. In other words, <math>F</math> lies on the circumcircle of <math>\triangle PON</math>. Note that <math>\angle ONA = \angle OPA = 90</math>, so <math>APON</math> is cyclic. In other words, <math>A</math> lies on the circumcircle of <math>\triangle PON</math>. <math>A</math>, <math>P</math>, <math>N</math>, <math>O</math>, and <math>F</math> all lie on the circumcircle of <math>\triangle PON</math>. Hence <math>A</math>, <math>P</math>, <math>F</math>, and <math>N</math> lie on a circle, as desired.<br />
<br />
=== Solution 2 (synthetic) ===<br />
Hint: consider <math>CF</math> intersection with <math>PM</math>; show that the resulting intersection lies on the desired circle. {{incomplete|solution}}<br />
<br />
=== Solution 3 (synthetic) ===<br />
This solution utilizes the ''phantom point method.'' Clearly, APON are cyclic because <math>\angle OPA = \angle ONA = 90</math>. Let the circumcircles of triangles <math>APN</math> and <math>BOC</math> intersect at <math>F'</math> and <math>O</math>. <br />
<br />
Lemma. If <math>A,B,C</math> are points on circle <math>\omega</math> with center <math>O</math>, and the tangents to <math>\omega</math> at <math>B,C</math> intersect at <math>Q</math>, then <math>AP</math> is the symmedian from <math>A</math> to <math>BC</math>.<br />
<br />
This is fairly easy to prove (as H, O are isogonal conjugates, plus using SAS similarity), but the author lacks time to write it up fully, and will do so soon.<br />
<br />
It is easy to see <math>Q</math> (the intersection of ray <math>OM</math> and the circumcircle of <math>\triangle BOC</math>) is colinear with <math>A</math> and <math>F'</math>, and because line <math>OM</math> is the diameter of that circle, <math>\angle QBO = \angle QCO = 90</math>, so <math>Q</math> is the point <math>Q</math> in the lemma; hence, we may apply the lemma. From here, it is simple angle-chasing to show that <math>F'</math> satisfies the original construction for <math>F</math>, showing <math>F=F'</math>; we are done. {{incomplete|solution}}<br />
<br />
=== Solution 4 (trigonometric) ===<br />
By the [[Law of Sines]], <math>\frac {\sin\angle BAM}{\sin\angle CAM} = \frac {\sin B}{\sin C} = \frac bc = \frac {b/AF}{c/AF} = \frac {\sin\angle AFC\cdot\sin\angle ABF}{\sin\angle ACF\cdot\sin\angle AFB}</math>. Since <math>\angle ABF = \angle ABD = \angle BAD = \angle BAM</math> and similarly <math>\angle ACF = \angle CAM</math>, we cancel to get <math>\sin\angle AFC = \sin\angle AFB</math>. Obviously, <math>\angle AFB + \angle AFC > 180^\circ</math> so <math>\angle AFC = \angle AFB</math>.<br />
<br />
Then <math>\angle FAB + \angle ABF = 180^\circ - \angle AFB = 180^\circ - \angle AFC = \angle FAC + \angle ACF</math> and <math>\angle ABF + \angle ACF = \angle A = \angle FAB + \angle FAC</math>. Subtracting these two equations, <math>\angle FAB - \angle FCA = \angle FCA - \angle FAB</math> so <math>\angle BAF = \angle ACF</math>. Therefore, <math>\triangle ABF\sim\triangle CAF</math> (by AA similarity), so a spiral similarity centered at <math>F</math> takes <math>B</math> to <math>A</math> and <math>A</math> to <math>C</math>. Therefore, it takes the midpoint of <math>\overline{BA}</math> to the midpoint of <math>\overline{AC}</math>, or <math>P</math> to <math>N</math>. So <math>\angle APF = \angle CNF = 180^\circ - \angle ANF</math> and <math>APFN</math> is cyclic.<br />
<br />
=== Solution 5 (isogonal conjugates) ===<br />
<center><asy><br />
/* setup and variables */<br />
size(280);<br />
pathpen = black + linewidth(0.7); pointpen = black; pen s = fontsize(8);<br />
pair B=(0,0),C=(5,0),A=(4,4); /* A.x > C.x/2 */<br />
/* construction and drawing */<br />
pair P=(A+B)/2,M=(B+C)/2,N=(A+C)/2,D=IP(A--M,P--P+5*(P-bisectorpoint(A,B))),E=IP(A--M,N--N+5*(bisectorpoint(A,C)-N)),F=IP(B--B+5*(D-B),C--C+5*(E-C)),O=circumcenter(A,B,C);<br />
D(MP("A",A,(0,1),s)--MP("B",B,SW,s)--MP("C",C,SE,s)--A--MP("M",M,s));<br />
D(C--D(MP("E",E,NW,s))--MP("N",N,(1,0),s)--D(MP("O",O,SW,s)));<br />
D(D(MP("D",D,SE,s))--MP("P",P,W,s));<br />
D(B--D(MP("F",F,s))); D(O--A--F,linetype("4 4")+linewidth(0.7));<br />
D(MP("O'",circumcenter(A,P,N),NW,s));<br />
D(circumcircle(A,P,N),linetype("4 4")+linewidth(0.7));<br />
D(rightanglemark(A,P,D,3.5));D(rightanglemark(A,N,E,3.5));<br />
picture p = new picture;<br />
draw(p,circumcircle(B,O,C),linetype("1 4")+linewidth(0.7));<br />
draw(p,circumcircle(A,B,C),linetype("1 4")+linewidth(0.7));<br />
clip(p,B+(-5,0)--B+(-5,A.y+2)--C+(5,A.y+2)--C+(5,0)--cycle); add(p);<br />
</asy></center><br />
<br />
Construct <math>T</math> on <math>AM</math> such that <math>\angle BCT = \angle ACF</math>. Then <math>\angle BCT = \angle CAM</math>. Then <math>\triangle AMC\sim\triangle CMT</math>, so <math>\frac {AM}{CM} = \frac {CM}{TM}</math>, or <math>\frac {AM}{BM} = \frac {BM}{TM}</math>. Then <math>\triangle AMB\sim\triangle BMT</math>, so <math>\angle CBT = \angle BAM = \angle FBA</math>. Then we have<br />
<br />
<math>\angle CBT = \angle ABF</math> and <math>\angle BCT = \angle ACF</math>. So <math>T</math> and <math>F</math> are [[isogonal conjugate|isogonally conjugate]]. Thus <math>\angle BAF = \angle CAM</math>. Then<br />
<br />
<math>\angle AFB = 180 - \angle ABF - \angle BAF = 180 - \angle BAM - \angle CAM = 180 - \angle BAC</math>.<br />
<br />
If <math>O</math> is the [[circumcenter]] of <math>\triangle ABC</math> then <math>\angle BFC = 2\angle BAC = \angle BOC</math> so <math>BFOC</math> is cyclic. Then <math>\angle BFO = 180 - \angle BOC = 180 - (90 - \angle BAC) = 90 + \angle BAC</math>.<br />
<br />
Then <math>\angle AFO = 360 - \angle AFB - \angle BFO = 360 - (180 - \angle BAC) - (90 + \angle BAC) = 90</math>. Then <math>\triangle AFO</math> is a right triangle.<br />
<br />
Now by the [[homothety]] centered at <math>A</math> with ratio <math>\frac {1}{2}</math>, <math>B</math> is taken to <math>P</math> and <math>C</math> is taken to <math>N</math>. Thus <math>O</math> is taken to the circumcenter of <math>\triangle APN</math> and is the midpoint of <math>AO</math>, which is also the circumcenter of <math>\triangle AFO</math>, so <math>A,P,N,F,O</math> all lie on a circle.<br />
<br />
=== Solution 6 (symmedians) ===<br />
Median <math>AM</math> of a triangle <math>ABC</math> implies <math>\frac {\sin{BAM}}{\sin{CAM}} = \frac {\sin{B}}{\sin{C}}</math>.<br />
Trig ceva for <math>F</math> shows that <math>AF</math> is a symmedian.<br />
Then <math>FP</math> is a median, use the lemma again to show that <math>AFP = C</math>, and similarly <math>AFN = B</math>, so you're done. {{incomplete|solution}}<br />
<br />
=== Solution 7 (inversion) ===<br />
{{image}}<br />
<center><asy><br />
size(280);<br />
pathpen = black + linewidth(0.7); pointpen = black; pen s = fontsize(8);<br />
pair B=(0,0),C=(5,0),A=(4,4); /* A.x > C.x/2 */<br />
real r = 1.2; /* inversion radius */<br />
<br />
/* construction and drawing */<br />
pair P=(A+B)/2,M=(B+C)/2,N=(A+C)/2,D=IP(A--M,P--P+5*(P-bisectorpoint(A,B))),E=IP(A--M,N--N+5*(bisectorpoint(A,C)-N)),F=IP(B--B+5*(D-B),C--C+5*(E-C)),O=circumcenter(A,B,C);<br />
D(MP("A",A,(0,1),s)--MP("B",B,SW,s)--MP("C",C,SE,s)--A--MP("M",M,s));<br />
D(C--D(MP("E",E,NW,s))--MP("N",N,(1,0),s)--D(MP("O",O,SW,s)));<br />
D(D(MP("D",D,SE,s))--MP("P",P,W,s));<br />
D(B--D(MP("F",F,s))); <br />
D(rightanglemark(A,P,D,3.5));D(rightanglemark(A,N,E,3.5));<br />
<br />
D(CR(A,r));<br />
pair Pa = A + (P-A)/(r*r); D(MP("P'",Pa,NW,s));<br />
<br />
</asy></center><br />
We consider an [[inversion]] by an arbitrary [[radius]] about <math>A</math>. We want to show that <math>P', F',</math> and <math>N'</math> are [[collinear]]. Notice that <math>D', A,</math> and <math>P'</math> lie on a circle with center <math>B'</math>, and similarly for the other side. We also have that <math>B', D', F', A</math> form a cyclic quadrilateral, and similarly for the other side. By angle chasing, we can prove that <math>A B' F' C'</math> is a [[parallelogram]], indicating that <math>F'</math> is the midpoint of <math>P'N'</math>. {{incomplete|solution}}<br />
<br />
=== Solution 7 (analytical) ===<br />
We let <math>A</math> be at the [[origin]], <math>B</math> be at the point <math>(a,0)</math>, and <math>C</math> be at the point <math>(b,c):\ b<a</math>. Then the equation of the perpendicular bisector of <math>\overline{AB}</math> is <math>x = a/2</math>, and {{incomplete|solution}}<br />
<br />
{{alternate solutions}}<br />
<br />
== Resources ==<br />
{{USAMO newbox|year=2008|num-b=1|num-a=3}}<br />
<br />
* <url>viewtopic.php?t=202907 Discussion on AoPS/MathLinks</url><br />
<br />
[[Category:Olympiad Geometry Problems]]</div>Not trighttps://artofproblemsolving.com/wiki/index.php?title=IMO_Shortlist&diff=24597IMO Shortlist2008-04-07T22:50:10Z<p>Not trig: </p>
<hr />
<div>{{WotWAnnounce|week=March 28-April 5}}<br />
Before the [[IMO]] is held, the IMO Problem Selection Committee takes the '''IMO LongList''' (the list of all proposed problems from all participating countries) and chooses between twenty and thirty problems. This list of twenty to thirty problems is known as the '''IMO Shortlist''', or the '''ISL'''. The final six [[IMO]] problems are chosen from the '''IMO Shortlist'''. A problem that is chosen by the IMO Problem Selection Committee for the '''IMO Shortlist''' is often referred to as having been '''shortlisted'''. The shortlist problems are currently split into their respective subjects (Algebra, Combinatorics, Geometry, and Number Theory, abbreviated A, C, G, N respectively.)<br />
{{stub}}</div>Not trighttps://artofproblemsolving.com/wiki/index.php?title=AoPS_Wiki:AoPS_Community_Awards&diff=20527AoPS Wiki:AoPS Community Awards2007-12-02T22:16:02Z<p>Not trig: /* Winners */</p>
<hr />
<div>This '''AoPS Community Awards''' page is a celebration of the accomplishments of members of the [[AoPS]] community.<br />
<br />
<br />
== IMO Participants and Medalists ==<br />
This is a list of members of the AoPS community who have competed for their country at the [[International Mathematical Olympiad]].<br />
<br />
=== Participants ===<br />
* Zachary Abel (2006) (AoPS assistant instructor)<br />
* Marco Avila (2006)<br />
* Zarathustra Brady (2006)<br />
* Robert Cordwell (2005)<br />
* Sherry Gong (2002, 2003, 2004, 2005, 2007)<br />
* Elyot Grant (2005)<br />
* Darij Grinberg (2006)<br />
* Mahbubul Hasan (2005)<br />
* Daniel Kane (AoPS assistant instructor)<br />
* Kiran Kedlaya (1990, 1991, 1992) ([[Art of Problem Solving Foundation]] board member)<br />
* Viktoriya Krakovna (2006)<br />
* Nate Ince (2004) (AoPS assistant instructor)<br />
* Brian Lawrence (2005, 2007) ([[WOOT]] instructor)<br />
* Thomas Mildorf (2005) (AoPS assistant instructor)<br />
* Alison Miller (2004) (AoPS assistant instructor)<br />
* Richard Peng (2005, 2006)<br />
* Eric Price (2005)<br />
* David Rhee (2004, 2005, 2006)<br />
* Peng Shi (2004, 2005, 2006)<br />
* Arne Smeets (2003, 2004)<br />
* Arnav Tripathy (2006, 2007)<br />
* [[Naoki Sato]] (AoPS instructor)<br />
* Yi Sun (2006)<br />
* [[Valentin Vornicu]] (AoPS/MathLinks webmaster)<br />
* Melanie Wood (1998, 1999) ([[WOOT]] instructor)<br />
* Alex Zhai (2006, 2007)<br />
* Yufei Zhao (2004, 2005, 2006)<br />
* Tigran Sloyan(2003,2004,2005,2006,2007)<br />
* Marco Avila (2006)<br />
* Vipul Naik (2003,2004)<br />
* Bhargav Narayanan (2007)<br />
* Tigran Hakobyan (2007)<br />
<br />
=== Gold medalists ===<br />
* Zarathustra Brady (2006)<br />
* Robert Cordwell (2005)<br />
* Darij Grinberg (2006)<br />
* Kiran Kedlaya (1990, 1992) ([[Art of Problem Solving Foundation]] board member)<br />
* Brian Lawrence (2005) ([[WOOT]] instructor)<br />
* Thomas Mildorf (2005) (AoPS assistant instructor)<br />
* Alison Miller (2004) (AoPS assistant instructor)<br />
* Arnav Tripathy (2006)<br />
* Eric Price (2005)<br />
* Yufei Zhao (2005)<br />
* Alex Zhai (2007)<br />
<br />
=== Silver medalists ===<br />
* Zachary Abel (2006) (AoPS assistant instructor)<br />
* Sherry Gong (2004, 2005)<br />
* Nate Ince (2004) (AoPS assistant instructor)<br />
* Kiran Kedlaya (1991) ([[Art of Problem Solving Foundation]] board member)<br />
* Viktoriya Krakovna (2006)<br />
* Hyun Soo Kim (2005) (AoPS assistant instructor)<br />
* Richard Peng (2005)<br />
* David Rhee (2006)<br />
* Naoki Sato (AoPS instructor)<br />
* Peng Shi (2006)<br />
* Arne Smeets (2004)<br />
* Yi Sun (2006)<br />
* [[Sam Vandervelde]] (1989) ([[WOOT]] instructor)<br />
* Melanie Wood (1998, 1999) ([[WOOT]] instructor)<br />
* Alex Zhai (2006)<br />
* Yufei Zhao (2006)<br />
* Tigran Sloyan(2006,2007)<br />
* Vipul Naik (2003,2004)<br />
<br />
=== Bronze medalists ===<br />
* Sherry Gong (2003)<br />
* Elyot Grant (2005)<br />
* Richard Peng (2006)<br />
* [[Naoki Sato]] (AoPS instructor)<br />
* [[Valentin Vornicu]] (AoPS/[[MathLinks]] webmaster)<br />
* Yufei Zhao (2004)<br />
* Tigran Sloyan(2004;2005)<br />
* Tigran Hakobyan (2007)<br />
<br />
== IPhO Participants and Medalists ==<br />
This is a list of members of the AoPS community who have competed for their country at the [[International Physics Olympiad]].<br />
=== Participants ===<br />
* Sherry Gong (2006)<br />
* Yi Sun (2004)<br />
* Arnav Tripathy (2006)<br />
<br />
=== Gold Medalists ===<br />
* Yi Sun (2004)<br />
* Rahul Singh (2007)<br />
<br />
=== Silver Medalists ===<br />
* Sherry Gong (2006)<br />
<br />
== USAMO ==<br />
The following AoPSers have won the [[United States of America Mathematical Olympiad]] (USAMO). (Note that the definition of "winner" has changed over the years -- currently it is the top 12 scores on the USAMO, but in the past it has been the top 6 or top 8 scores.)<br />
=== Perfect Scorers ===<br />
* Daniel Kane (AoPS assistant instructor)<br />
* Kiran Kedlaya (1991) ([[Art of Problem Solving Foundation]] board member)<br />
* Brian Lawrence (2006) ([[WOOT]] instructor)<br />
<br />
=== Winners ===<br />
* Yakov Berchenko-Kogan (2006)<br />
* Sherry Gong (2006, 2007)<br />
* Yi Han (2006)<br />
* Adam Hesterberg (2007)<br />
* Daniel Kane (AoPS assistant instructor)<br />
* Kiran Kedlaya (1990, 1991, 1992) ([[Art of Problem Solving Foundation]] board member)<br />
* Brian Lawrence (2005, 2006, 2007) ([[WOOT]] instructor)<br />
* Tedrick Leung (2006, 2007)<br />
* Haitao Mao (2007)<br />
* Richard Mccutchen (2006)<br />
* Albert Ni (2005)<br />
* [[David Patrick]] (1988) (AoPS instructor)<br />
* [[Richard Rusczyk]] (1989) (AoPS founder)<br />
* Krishanu Sankar (2007)<br />
* Peng Shi (2006)<br />
* Jacob Steinhardt (2007)<br />
* Yi Sun (2006)<br />
* Arnav Tripathy (2006, 2007)<br />
* [[Sam Vandervelde]] (1987, 1989) ([[WOOT]] instructor)<br />
* Melanie Wood (1998, 1999) ([[WOOT]] instructor)<br />
* Alex Zhai (2006, 2007)<br />
* Yufei Zhao (2006)<br />
<br />
== Putnam Fellows ==<br />
The top 5 students (including ties) on the collegiate [[Putnam Exam|William Lowell Putnam Competition]] are named Putnam Fellows.<br />
* David Ash (1981, 1982, 1983)<br />
* Daniel Kane (2003, 2004, 2005) (AoPS assistant instructor)<br />
* Kiran Kedlaya (1994, 1995, 1996) ([[AoPS Foundation]] board member)<br />
* Matthew Ince (2005) (AoPS assistant instructor)<br />
* Alexander Schwartz (2000, 2002)<br />
* Jan Siwanowicz (2001) <br />
* Melanie Wood (2002) ([[WOOT]] instructor)<br />
<br />
== Siemens Competition Winners ==<br />
The annual [[Siemens Competition]] (formerly Siemens-Westinghouse) is a scientific research competition.<br />
* Michael Viscardi (1st Individual,2005)<br />
* Lucia Mocz (2nd Team, 2006)<br />
<br />
== Clay Junior Fellows ==<br />
Each year since 2003, the [[Clay Mathematics Institute]] has selected 12 Junior Fellows.<br />
* Thomas Belulovich (2005) (AoPS assistant instructor)<br />
* Atoshi Chowdhury (2003) (AoPS assistant instructor)<br />
* Robert Cordwell (2005)<br />
* Eve Drucker (2003) (AoPS assistant instructor)<br />
* Matthew Ince (2004) (AoPS assistant instructor)<br />
* Nate Ince (2004) (AoPS assistant instructor)<br />
* Hyun Soo Kim (2005) (AoPS assistant instructor)<br />
* Raju Krishnamoorthy (2005)<br />
* Alison Miller (2003) (AoPS assistant instructor)<br />
* Brian Rice (2003) (AoPS assistant instructor)<br />
* Dmitry Taubinski (2005) (AoPS assistant instructor)<br />
* Ameya Velingker (2005)<br />
<br />
<br />
== Perfect AIME Scores ==<br />
Very few students have ever achieved a perfect score on the [[American Invitational Mathematics Examination]] (AIME)<br />
* David Benjamin (2006)<br />
* [[Mathew Crawford]] (1992) (AoPS instructor)<br />
* [[Sandor Lehoczky]] (1990) (AoPS author)<br />
* Tedrick Leung (2006)<br />
* Tony Liu (2006)<br />
* [[Richard Rusczyk]] (1989) (AoPS founder)<br />
* [[Sam Vandervelde]] (1988) ([[WOOT]] instructor)<br />
<br />
== Perfect AMC Scores ==<br />
=== Perfect AMC 12 Scores ===<br />
The [[AMC 12]] is a challenging examination for students in grades 12 and below administered by the [[American Mathematics Competitions]].<br />
* David Benjamin (2006)<br />
* Zachary Abel (2005) (AoPS assistant instructor)<br />
* Ruozhou (Joe) Jia (2003) (AoPS assistant instructor)<br />
* Joel Lewis (2003) <br />
* Jonathan Lowd (2003) (AoPS assistant instructor)<br />
* Thomas Mildorf (2004) (AoPS assistant instructor)<br />
* Alison Miller (2004) (AoPS assistant instructor)<br />
* Albert Ni (2003) (AoPS instructor)<br />
* Ajay Sharma (2004)<br />
* Arnav Tripathy (2006, 2007)<br />
* Alex Zhai (2007)<br />
<br />
=== Perfect AMC 10 Scores ===<br />
The [[AMC 10]] is a challenging examination for students in grades 10 and below administered by the [[American Mathematics Competitions]].<br />
* Sergei Bernstein (2007)<br />
* Yifan Cao (2005)<br />
* Kevin Chen (2007)<br />
* In Young Cho (2007)<br />
* Mario Choi (2007)<br />
* Billy Dorminy (2007)<br />
* Zhou Fan (2005)<br />
* Albert Gu (2007)<br />
* Robin He (2007)<br />
* Keone Hon (2005)<br />
* Susan Hu (2005)<br />
* Sam Keller (2007)<br />
* Vincent Le (2006)<br />
* Daniel Li (2007)<br />
* Johnny Li (2007)<br />
* Patricia Li (2005)<br />
* Carl Lian (2007)<br />
* Thomas Mildorf (2002) (AoPS assistant instructor)<br />
* Howard Tong (2005)<br />
* Brent Woodhouse (2006, 2007)<br />
* Jonathan Zhou (2007)<br />
<br />
=== Perfect AHSME Scores ===<br />
The [[American High School Mathematics Examination]] (AHSME) was the predecessor of the AMC 12.<br />
* Christopher Chang (1994, 1995, 1996)<br />
* [[Mathew Crawford]] (1994, 1995) (AoPS instructor)<br />
* [[David Patrick]] (1988) (AoPS instructor)<br />
<br />
<br />
== MATHCOUNTS ==<br />
[[MathCounts]] is the premier middle school [[mathematics competition]] in the U.S.<br />
=== National Champions ===<br />
* Ruozhou (Joe) Jia (2000) (AoPS assistant instructor)<br />
* Albert Ni (2002) (AoPS instructor)<br />
* Adam Hesterberg (2003)<br />
* Neal Wu (2005)<br />
* Daesun Yim (2006)<br />
* Kevin Chen (2007)<br />
<br />
=== National Top 12 ===<br />
* Ashley Reiter Ahlin (1987) ([[WOOT]] instructor)<br />
* Andrew Ardito (2005, 2006)<br />
* David Benjamin (2004, 2005)<br />
* Nathan Benjamin (2005, 2006)<br />
* Wenyu Cao (2007)<br />
* Christopher Chang (1991, 1992)<br />
* Kevin Chen (2006, 2007)<br />
* Andrew Chien (2003)<br />
* Peter Chien (2004)<br />
* Mario Choi (2007)<br />
* Joseph Chu (2004)<br />
* [[Mathew Crawford]] (1990, 1991) (AoPS instructor)<br />
* Brian Hamrick (2006)<br />
* Adam Hesterberg (2002, 2003)<br />
* Ruozhou (Joe) Jia (2000) (AoPS assistant instructor)<br />
* Sam Keller (2006)<br />
* Shaunak Kishore (2003, 2004)<br />
* Kiran Kota (2005)<br />
* Brian Lawrence (2003) ([[WOOT]] instructor)<br />
* Karlanna Lewis (2005)<br />
* Daniel Li (2006)<br />
* Patricia Li (2005)<br />
* Poh-Ling Loh (2000)<br />
* Albert Ni (2002) (AoPS assistant instructor)<br />
* Elizabeth Synge (2007)<br />
* Jason Trigg (2002)<br />
* [[Sam Vandervelde]] (1985) ([[WOOT]] instructor)<br />
* Neal Wu (2005, 2006)<br />
* Rolland Wu (2006)<br />
* Daesun Yim (2006)<br />
* Darren Yin (2002)<br />
* Allen Yuan (2007)<br />
* Alex Zhai (2004)<br />
* Mark Zhang (2005)<br />
<br />
=== Masters Round Champions ===<br />
* Christopher Chang (1991)<br />
* Brian Lawrence (2003) ([[WOOT]] instructor)<br />
* Sergei Bernstein (2005)<br />
* Daniel Li (2006)<br />
* Kevin Chen (2007)<br />
<br />
=== National Test Champions ===<br />
* [[Mathew Crawford]] (1990) (AoPS instructor)<br />
* Adam Hesterberg (2003)<br />
* Sergei Bernstein (2005)<br />
* Neal Wu (2006)<br />
<br />
== Harvard-MIT Math Tournament ==<br />
<br />
The [[HMMT]] 2007 winning team, the "WOOTlings", consisted entirely of [[WOOT]]ers:<br />
<br />
* Wenyu Cao<br />
* Eric Chang<br />
* Jeremy Hahn<br />
* Alex Kandell<br />
* Adeel Khan<br />
* Sathish Nagappan<br />
* Krishanu Roy Sankar<br />
* Patrick Tenorio<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
== See also ==<br />
* [[Academic competitions]]<br />
* [[Mathematics competitions]]<br />
* [[Mathematics competition resources]]<br />
* [[Academic scholarships]]<br />
<br />
<br />
<br />
[[Category:Art of Problem Solving]]</div>Not trighttps://artofproblemsolving.com/wiki/index.php?title=AoPS_Wiki:AoPS_Community_Awards&diff=20526AoPS Wiki:AoPS Community Awards2007-12-02T22:14:47Z<p>Not trig: /* Gold medalists */</p>
<hr />
<div>This '''AoPS Community Awards''' page is a celebration of the accomplishments of members of the [[AoPS]] community.<br />
<br />
<br />
== IMO Participants and Medalists ==<br />
This is a list of members of the AoPS community who have competed for their country at the [[International Mathematical Olympiad]].<br />
<br />
=== Participants ===<br />
* Zachary Abel (2006) (AoPS assistant instructor)<br />
* Marco Avila (2006)<br />
* Zarathustra Brady (2006)<br />
* Robert Cordwell (2005)<br />
* Sherry Gong (2002, 2003, 2004, 2005, 2007)<br />
* Elyot Grant (2005)<br />
* Darij Grinberg (2006)<br />
* Mahbubul Hasan (2005)<br />
* Daniel Kane (AoPS assistant instructor)<br />
* Kiran Kedlaya (1990, 1991, 1992) ([[Art of Problem Solving Foundation]] board member)<br />
* Viktoriya Krakovna (2006)<br />
* Nate Ince (2004) (AoPS assistant instructor)<br />
* Brian Lawrence (2005, 2007) ([[WOOT]] instructor)<br />
* Thomas Mildorf (2005) (AoPS assistant instructor)<br />
* Alison Miller (2004) (AoPS assistant instructor)<br />
* Richard Peng (2005, 2006)<br />
* Eric Price (2005)<br />
* David Rhee (2004, 2005, 2006)<br />
* Peng Shi (2004, 2005, 2006)<br />
* Arne Smeets (2003, 2004)<br />
* Arnav Tripathy (2006, 2007)<br />
* [[Naoki Sato]] (AoPS instructor)<br />
* Yi Sun (2006)<br />
* [[Valentin Vornicu]] (AoPS/MathLinks webmaster)<br />
* Melanie Wood (1998, 1999) ([[WOOT]] instructor)<br />
* Alex Zhai (2006, 2007)<br />
* Yufei Zhao (2004, 2005, 2006)<br />
* Tigran Sloyan(2003,2004,2005,2006,2007)<br />
* Marco Avila (2006)<br />
* Vipul Naik (2003,2004)<br />
* Bhargav Narayanan (2007)<br />
* Tigran Hakobyan (2007)<br />
<br />
=== Gold medalists ===<br />
* Zarathustra Brady (2006)<br />
* Robert Cordwell (2005)<br />
* Darij Grinberg (2006)<br />
* Kiran Kedlaya (1990, 1992) ([[Art of Problem Solving Foundation]] board member)<br />
* Brian Lawrence (2005) ([[WOOT]] instructor)<br />
* Thomas Mildorf (2005) (AoPS assistant instructor)<br />
* Alison Miller (2004) (AoPS assistant instructor)<br />
* Arnav Tripathy (2006)<br />
* Eric Price (2005)<br />
* Yufei Zhao (2005)<br />
* Alex Zhai (2007)<br />
<br />
=== Silver medalists ===<br />
* Zachary Abel (2006) (AoPS assistant instructor)<br />
* Sherry Gong (2004, 2005)<br />
* Nate Ince (2004) (AoPS assistant instructor)<br />
* Kiran Kedlaya (1991) ([[Art of Problem Solving Foundation]] board member)<br />
* Viktoriya Krakovna (2006)<br />
* Hyun Soo Kim (2005) (AoPS assistant instructor)<br />
* Richard Peng (2005)<br />
* David Rhee (2006)<br />
* Naoki Sato (AoPS instructor)<br />
* Peng Shi (2006)<br />
* Arne Smeets (2004)<br />
* Yi Sun (2006)<br />
* [[Sam Vandervelde]] (1989) ([[WOOT]] instructor)<br />
* Melanie Wood (1998, 1999) ([[WOOT]] instructor)<br />
* Alex Zhai (2006)<br />
* Yufei Zhao (2006)<br />
* Tigran Sloyan(2006,2007)<br />
* Vipul Naik (2003,2004)<br />
<br />
=== Bronze medalists ===<br />
* Sherry Gong (2003)<br />
* Elyot Grant (2005)<br />
* Richard Peng (2006)<br />
* [[Naoki Sato]] (AoPS instructor)<br />
* [[Valentin Vornicu]] (AoPS/[[MathLinks]] webmaster)<br />
* Yufei Zhao (2004)<br />
* Tigran Sloyan(2004;2005)<br />
* Tigran Hakobyan (2007)<br />
<br />
== IPhO Participants and Medalists ==<br />
This is a list of members of the AoPS community who have competed for their country at the [[International Physics Olympiad]].<br />
=== Participants ===<br />
* Sherry Gong (2006)<br />
* Yi Sun (2004)<br />
* Arnav Tripathy (2006)<br />
<br />
=== Gold Medalists ===<br />
* Yi Sun (2004)<br />
* Rahul Singh (2007)<br />
<br />
=== Silver Medalists ===<br />
* Sherry Gong (2006)<br />
<br />
== USAMO ==<br />
The following AoPSers have won the [[United States of America Mathematical Olympiad]] (USAMO). (Note that the definition of "winner" has changed over the years -- currently it is the top 12 scores on the USAMO, but in the past it has been the top 6 or top 8 scores.)<br />
=== Perfect Scorers ===<br />
* Daniel Kane (AoPS assistant instructor)<br />
* Kiran Kedlaya (1991) ([[Art of Problem Solving Foundation]] board member)<br />
* Brian Lawrence (2006) ([[WOOT]] instructor)<br />
<br />
=== Winners ===<br />
* Yakov Berchenko-Kogan (2006)<br />
* Sherry Gong (2006, 2007)<br />
* Yi Han (2006)<br />
* Adam Hesterberg (2007)<br />
* Daniel Kane (AoPS assistant instructor)<br />
* Kiran Kedlaya (1990, 1991, 1992) ([[Art of Problem Solving Foundation]] board member)<br />
* Brian Lawrence (2005, 2006, 2007) ([[WOOT]] instructor)<br />
* Tedrick Leung (2006, 2007)<br />
* Richard Mccutchen (2006)<br />
* Albert Ni (2005)<br />
* [[David Patrick]] (1988) (AoPS instructor)<br />
* [[Richard Rusczyk]] (1989) (AoPS founder)<br />
* Krishanu Sankar (2007)<br />
* Peng Shi (2006)<br />
* Jacob Steinhardt<br />
* Yi Sun (2006)<br />
* Arnav Tripathy (2006, 2007)<br />
* [[Sam Vandervelde]] (1987, 1989) ([[WOOT]] instructor)<br />
* Melanie Wood (1998, 1999) ([[WOOT]] instructor)<br />
* Alex Zhai (2006, 2007)<br />
* Yufei Zhao (2006)<br />
<br />
== Putnam Fellows ==<br />
The top 5 students (including ties) on the collegiate [[Putnam Exam|William Lowell Putnam Competition]] are named Putnam Fellows.<br />
* David Ash (1981, 1982, 1983)<br />
* Daniel Kane (2003, 2004, 2005) (AoPS assistant instructor)<br />
* Kiran Kedlaya (1994, 1995, 1996) ([[AoPS Foundation]] board member)<br />
* Matthew Ince (2005) (AoPS assistant instructor)<br />
* Alexander Schwartz (2000, 2002)<br />
* Jan Siwanowicz (2001) <br />
* Melanie Wood (2002) ([[WOOT]] instructor)<br />
<br />
== Siemens Competition Winners ==<br />
The annual [[Siemens Competition]] (formerly Siemens-Westinghouse) is a scientific research competition.<br />
* Michael Viscardi (1st Individual,2005)<br />
* Lucia Mocz (2nd Team, 2006)<br />
<br />
== Clay Junior Fellows ==<br />
Each year since 2003, the [[Clay Mathematics Institute]] has selected 12 Junior Fellows.<br />
* Thomas Belulovich (2005) (AoPS assistant instructor)<br />
* Atoshi Chowdhury (2003) (AoPS assistant instructor)<br />
* Robert Cordwell (2005)<br />
* Eve Drucker (2003) (AoPS assistant instructor)<br />
* Matthew Ince (2004) (AoPS assistant instructor)<br />
* Nate Ince (2004) (AoPS assistant instructor)<br />
* Hyun Soo Kim (2005) (AoPS assistant instructor)<br />
* Raju Krishnamoorthy (2005)<br />
* Alison Miller (2003) (AoPS assistant instructor)<br />
* Brian Rice (2003) (AoPS assistant instructor)<br />
* Dmitry Taubinski (2005) (AoPS assistant instructor)<br />
* Ameya Velingker (2005)<br />
<br />
<br />
== Perfect AIME Scores ==<br />
Very few students have ever achieved a perfect score on the [[American Invitational Mathematics Examination]] (AIME)<br />
* David Benjamin (2006)<br />
* [[Mathew Crawford]] (1992) (AoPS instructor)<br />
* [[Sandor Lehoczky]] (1990) (AoPS author)<br />
* Tedrick Leung (2006)<br />
* Tony Liu (2006)<br />
* [[Richard Rusczyk]] (1989) (AoPS founder)<br />
* [[Sam Vandervelde]] (1988) ([[WOOT]] instructor)<br />
<br />
== Perfect AMC Scores ==<br />
=== Perfect AMC 12 Scores ===<br />
The [[AMC 12]] is a challenging examination for students in grades 12 and below administered by the [[American Mathematics Competitions]].<br />
* David Benjamin (2006)<br />
* Zachary Abel (2005) (AoPS assistant instructor)<br />
* Ruozhou (Joe) Jia (2003) (AoPS assistant instructor)<br />
* Joel Lewis (2003) <br />
* Jonathan Lowd (2003) (AoPS assistant instructor)<br />
* Thomas Mildorf (2004) (AoPS assistant instructor)<br />
* Alison Miller (2004) (AoPS assistant instructor)<br />
* Albert Ni (2003) (AoPS instructor)<br />
* Ajay Sharma (2004)<br />
* Arnav Tripathy (2006, 2007)<br />
* Alex Zhai (2007)<br />
<br />
=== Perfect AMC 10 Scores ===<br />
The [[AMC 10]] is a challenging examination for students in grades 10 and below administered by the [[American Mathematics Competitions]].<br />
* Sergei Bernstein (2007)<br />
* Yifan Cao (2005)<br />
* Kevin Chen (2007)<br />
* In Young Cho (2007)<br />
* Mario Choi (2007)<br />
* Billy Dorminy (2007)<br />
* Zhou Fan (2005)<br />
* Albert Gu (2007)<br />
* Robin He (2007)<br />
* Keone Hon (2005)<br />
* Susan Hu (2005)<br />
* Sam Keller (2007)<br />
* Vincent Le (2006)<br />
* Daniel Li (2007)<br />
* Johnny Li (2007)<br />
* Patricia Li (2005)<br />
* Carl Lian (2007)<br />
* Thomas Mildorf (2002) (AoPS assistant instructor)<br />
* Howard Tong (2005)<br />
* Brent Woodhouse (2006, 2007)<br />
* Jonathan Zhou (2007)<br />
<br />
=== Perfect AHSME Scores ===<br />
The [[American High School Mathematics Examination]] (AHSME) was the predecessor of the AMC 12.<br />
* Christopher Chang (1994, 1995, 1996)<br />
* [[Mathew Crawford]] (1994, 1995) (AoPS instructor)<br />
* [[David Patrick]] (1988) (AoPS instructor)<br />
<br />
<br />
== MATHCOUNTS ==<br />
[[MathCounts]] is the premier middle school [[mathematics competition]] in the U.S.<br />
=== National Champions ===<br />
* Ruozhou (Joe) Jia (2000) (AoPS assistant instructor)<br />
* Albert Ni (2002) (AoPS instructor)<br />
* Adam Hesterberg (2003)<br />
* Neal Wu (2005)<br />
* Daesun Yim (2006)<br />
* Kevin Chen (2007)<br />
<br />
=== National Top 12 ===<br />
* Ashley Reiter Ahlin (1987) ([[WOOT]] instructor)<br />
* Andrew Ardito (2005, 2006)<br />
* David Benjamin (2004, 2005)<br />
* Nathan Benjamin (2005, 2006)<br />
* Wenyu Cao (2007)<br />
* Christopher Chang (1991, 1992)<br />
* Kevin Chen (2006, 2007)<br />
* Andrew Chien (2003)<br />
* Peter Chien (2004)<br />
* Mario Choi (2007)<br />
* Joseph Chu (2004)<br />
* [[Mathew Crawford]] (1990, 1991) (AoPS instructor)<br />
* Brian Hamrick (2006)<br />
* Adam Hesterberg (2002, 2003)<br />
* Ruozhou (Joe) Jia (2000) (AoPS assistant instructor)<br />
* Sam Keller (2006)<br />
* Shaunak Kishore (2003, 2004)<br />
* Kiran Kota (2005)<br />
* Brian Lawrence (2003) ([[WOOT]] instructor)<br />
* Karlanna Lewis (2005)<br />
* Daniel Li (2006)<br />
* Patricia Li (2005)<br />
* Poh-Ling Loh (2000)<br />
* Albert Ni (2002) (AoPS assistant instructor)<br />
* Elizabeth Synge (2007)<br />
* Jason Trigg (2002)<br />
* [[Sam Vandervelde]] (1985) ([[WOOT]] instructor)<br />
* Neal Wu (2005, 2006)<br />
* Rolland Wu (2006)<br />
* Daesun Yim (2006)<br />
* Darren Yin (2002)<br />
* Allen Yuan (2007)<br />
* Alex Zhai (2004)<br />
* Mark Zhang (2005)<br />
<br />
=== Masters Round Champions ===<br />
* Christopher Chang (1991)<br />
* Brian Lawrence (2003) ([[WOOT]] instructor)<br />
* Sergei Bernstein (2005)<br />
* Daniel Li (2006)<br />
* Kevin Chen (2007)<br />
<br />
=== National Test Champions ===<br />
* [[Mathew Crawford]] (1990) (AoPS instructor)<br />
* Adam Hesterberg (2003)<br />
* Sergei Bernstein (2005)<br />
* Neal Wu (2006)<br />
<br />
== Harvard-MIT Math Tournament ==<br />
<br />
The [[HMMT]] 2007 winning team, the "WOOTlings", consisted entirely of [[WOOT]]ers:<br />
<br />
* Wenyu Cao<br />
* Eric Chang<br />
* Jeremy Hahn<br />
* Alex Kandell<br />
* Adeel Khan<br />
* Sathish Nagappan<br />
* Krishanu Roy Sankar<br />
* Patrick Tenorio<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
== See also ==<br />
* [[Academic competitions]]<br />
* [[Mathematics competitions]]<br />
* [[Mathematics competition resources]]<br />
* [[Academic scholarships]]<br />
<br />
<br />
<br />
[[Category:Art of Problem Solving]]</div>Not trighttps://artofproblemsolving.com/wiki/index.php?title=ITest&diff=16585ITest2007-09-23T16:20:37Z<p>Not trig: </p>
<hr />
<div>The '''iTest''' (formerly the [[American High School Internet Mathematics Competition]]) is a team-based math competition testing high school students on [[algebra]], [[geometry]], pre-[[calculus]], [[probability]], [[logic]], and other mathematical topics typically encountered in a high school mathematics curriculum (excluding calculus).<br />
<br />
It consists of 25 multiple-choice questions (with 1 answer choice on the first problem, 2 on the second, etc.), 25 short-answer questions, and 10 Ultimate questions, which are much like relay questions in that each Ultimate question depends on the answer to the previous ones. There are 4 Tiebreaker proof questions as well.<br />
<br />
==External links==<br />
*[http://www.theitest.com iTest home page]<br />
<br />
{{stub}}<br />
[[Category:Mathematics competitions]]</div>Not trighttps://artofproblemsolving.com/wiki/index.php?title=Titu_Andreescu&diff=16013Titu Andreescu2007-09-09T21:49:00Z<p>Not_trig: </p>
<hr />
<div>Born in Romania, '''Titu Andreescu''' is a former [[U.S. IMO]] team coach. He coached the "perfect team" in 1994. He is an author of many [[math books]] and is one of the directors of the [[AwesomeMath]] summer program.<br />
<br />
{{stub}}</div>Not_trighttps://artofproblemsolving.com/wiki/index.php?title=Implicit_differentiation&diff=15957Implicit differentiation2007-09-09T13:17:47Z<p>Not_trig: </p>
<hr />
<div>Implicit differentiation is differentiating both sides of an implicit equation with respect to one of the variables. The dependent variable is treated as a function of the independent variable and is differentiated with the chain rule.</div>Not_trighttps://artofproblemsolving.com/wiki/index.php?title=International_Mathematical_Olympiad&diff=15928International Mathematical Olympiad2007-09-09T00:38:38Z<p>Not_trig: /* External Links */</p>
<hr />
<div>The '''International Mathematical Olympiad''' is the pinnacle of all high school [[mathematics competition]]s. Each year, countries from around the world send a team of 6 students to compete in a grueling competition.<br />
<br />
== Format of the Competition ==<br />
The competition takes place over 2 consecutive days. Each day 3 problems are given to the students to work on for 4.5 hours.<br />
<br />
=== Scoring ===<br />
<br />
Scoring on each problem is done on a 0-7 scale (inclusive and integers only). Full credit is only given for complete, correct solutions. Each solution is intended to be in the form of a [[proof writing|mathematical proof]]. Since there are 6 problems, a perfect score is 42 points.<br />
<br />
=== Awards ===<br />
Medals and honorable mentions are given out.<br />
<br />
* Gold - the top 1/12 of individual scores.<br />
* Silver - the next 2/12 of individual scores.<br />
* Bronze - the next 3/12 of individual scores.<br />
* Honorable mention - any student who receives a score of 7 on any one problem but did not receive a medal.<br />
<br />
=== Team Competition ===<br />
There is no official team competition. Unofficially, however, the scores of each team are compared each year where a team's score is the sum of their individual scores.<br />
<br />
== History ==<br />
<br />
The IMO started in 1959 as a competition among Eastern European countries. Since then, it has evolved into the premier international competition in mathematics.<br />
<br />
== See also ==<br />
* [[IMO Problems and Solutions, with authors]]<br />
* [[Mathematics competition resources]]<br />
* [[Math books]]<br />
* [[Mathematics scholarships]]<br />
* [[Worldwide Online Olympiad Training]]<br />
<br />
== External Links ==<br />
* [http://www.imo2007.edu.vn/ IMO 2007 Vietnam]<br />
* [http://imo2006.dmfa.si/ IMO 2006 Slovenia]<br />
* [http://www.artofproblemsolving.com/Forum/index.php?f=87 AoPS-Mathlinks IMO Forum]<br />
* [http://www.artofproblemsolving.com/Forum/resources.php AoPS-Mathlinks Olympiad Resources]<br />
<br />
{{stub}}<br />
<br />
<br />
<br />
[[Category:Mathematics competitions]]</div>Not_trighttps://artofproblemsolving.com/wiki/index.php?title=2007_AIME_II_Problems/Problem_3&diff=143212007 AIME II Problems/Problem 32007-03-31T23:29:29Z<p>Not_trig: </p>
<hr />
<div>== Problem ==<br />
[[Square]] <math>ABCD</math> has side length <math>13</math>, and [[point]]s <math>E</math> and <math>F</math> are exterior to the square such that <math>BE=DF=5</math> and <math>AE=CF=12</math>. Find <math>\displaystyle EF^{2}</math>.<br />
<br />
<div style="text-align:center;">[[Image:2007 AIME II-3.png]]</div><br />
<br />
== Solution ==<br />
<br />
<br />
== '''Solution 1.''' ==<br />
<br />
Extend <math>\overline{AE}, \overline{DF}</math> and <math>\overline{BE}, \overline{CF}</math> to their points of intersection. Since <math>\triangle ABE \cong \triangle CDF</math> and are both <math>5-12-13</math> [[right triangle]]s, we can come to the conclusion that the two new triangles are also congruent to these two (use [[ASA]], as we know all the sides are <math>13</math> and the angles are mostly complementary). Thus, we create a [[square]] with sides <math>5 + 12 = 17</math>.<br />
<br />
<div style='text-align:center;'>[[Image:2007 AIME II-3b.PNG]]</div><br />
<br />
<math>\overline{EF}</math> is the diagonal of the square, with length <math>17\sqrt{2}</math>; the answer is <math>EF^2 = (17\sqrt{2})^2 = 578</math>.<br />
<br />
<br />
== '''Solution 2.''' ==<br />
<br />
<br />
A slightly more analytic/brute-force approach:<br />
<br />
[[Image:AIME II prob10 bruteforce.PNG ]]<br />
<br />
Drop perpendiculars from <math>E</math> and <math>F</math> to <math>I</math> and <math>J</math>, respectively; construct right triangle <math>EKF</math> with right angle at K and <math>EK || BC</math>. Since <math>2[CDF]=DF*CF=CD*JF</math>, we have <math>JF=5\times12/13 = \frac{60}{13}</math>. Similarly, <math>EI=\frac{60}{13}</math>. Since <math>\triangle DJF \sim \triangle DFC</math>, we have <math>DJ=\frac{5JF}{12}=\frac{25}{13}</math>.<br />
<br />
Now, we see that <math>FK=DC-(DJ+IB)=DC-2DJ=13-\frac{50}{13}=\frac{119}{13}</math>. Also, <math>EK=BC+(JF+IE)=BC+2JF=13+\frac{120}{13}=\frac{289}{13}</math>. By the Pythagorean Theorem, we have <math>EF=\sqrt{\left(\frac{289}{13}\right)^2+\left(\frac{119}{13} \right)^2}=\frac{\sqrt{(17^2)(17^2+7^2)}}{13}=\frac{17\sqrt{338}}{13}=\frac{17(13\sqrt{2})}{13}=17\sqrt{2}</math>. Therefore, <math>EF^2=(17\sqrt{2})^2=578</math>.<br />
<br />
== See also ==<br />
{{AIME box|year=2007|n=II|num-b=2|num-a=4}}<br />
<br />
[[Category:Intermediate Geometry Problems]]</div>Not_trighttps://artofproblemsolving.com/wiki/index.php?title=File:AIME_II_prob10_bruteforce.PNG&diff=14320File:AIME II prob10 bruteforce.PNG2007-03-31T23:27:53Z<p>Not_trig: </p>
<hr />
<div></div>Not_trighttps://artofproblemsolving.com/wiki/index.php?title=2007_AIME_II_Problems/Problem_3&diff=143192007 AIME II Problems/Problem 32007-03-31T23:27:20Z<p>Not_trig: </p>
<hr />
<div>== Problem ==<br />
[[Square]] <math>ABCD</math> has side length <math>13</math>, and [[point]]s <math>E</math> and <math>F</math> are exterior to the square such that <math>BE=DF=5</math> and <math>AE=CF=12</math>. Find <math>\displaystyle EF^{2}</math>.<br />
<br />
<div style="text-align:center;">[[Image:2007 AIME II-3.png]]</div><br />
<br />
== Solution ==<br />
<br />
<br />
== '''Solution 1.''' ==<br />
<br />
Extend <math>\overline{AE}, \overline{DF}</math> and <math>\overline{BE}, \overline{CF}</math> to their points of intersection. Since <math>\triangle ABE \cong \triangle CDF</math> and are both <math>5-12-13</math> [[right triangle]]s, we can come to the conclusion that the two new triangles are also congruent to these two (use [[ASA]], as we know all the sides are <math>13</math> and the angles are mostly complementary). Thus, we create a [[square]] with sides <math>5 + 12 = 17</math>.<br />
<br />
<div style='text-align:center;'>[[Image:2007 AIME II-3b.PNG]]</div><br />
<br />
<math>\overline{EF}</math> is the diagonal of the square, with length <math>17\sqrt{2}</math>; the answer is <math>EF^2 = (17\sqrt{2})^2 = 578</math>.<br />
<br />
<br />
== '''Solution 2.''' ==<br />
<br />
<br />
A slightly more analytic/brute-force approach:<br />
<br />
Drop perpendiculars from <math>E</math> and <math>F</math> to <math>I</math> and <math>J</math>, respectively; construct right triangle <math>EKF</math> with right angle at K and <math>EK || BC</math>. Since <math>2[CDF]=DF*CF=CD*JF</math>, we have <math>JF=5\times12/13 = \frac{60}{13}</math>. Similarly, <math>EI=\frac{60}{13}</math>. Since <math>\triangle DJF \sim \triangle DFC</math>, we have <math>DJ=\frac{5JF}{12}=\frac{25}{13}</math>.<br />
<br />
Now, we see that <math>FK=DC-(DJ+IB)=DC-2DJ=13-\frac{50}{13}=\frac{119}{13}</math>. Also, <math>EK=BC+(JF+IE)=BC+2JF=13+\frac{120}{13}=\frac{289}{13}</math>. By the Pythagorean Theorem, we have <math>EF=\sqrt{\left(\frac{289}{13}\right)^2+\left(\frac{119}{13} \right)^2}=\frac{\sqrt{(17^2)(17^2+7^2)}}{13}=\frac{17\sqrt{338}}{13}=\frac{17(13\sqrt{2})}{13}=17\sqrt{2}</math>. Therefore, <math>EF^2=(17\sqrt{2})^2=578</math>.<br />
<br />
== See also ==<br />
{{AIME box|year=2007|n=II|num-b=2|num-a=4}}<br />
<br />
[[Category:Intermediate Geometry Problems]]</div>Not_trighttps://artofproblemsolving.com/wiki/index.php?title=Talk:IMO_Problems_and_Solutions,_with_authors&diff=10466Talk:IMO Problems and Solutions, with authors2006-10-29T03:03:18Z<p>Not_trig: </p>
<hr />
<div>I created the page [[IMO Problems and Solutions]] today, but I have now realized that the coexistance of that page and this one is probably redundant. Therefore I propose to merge the pages. Any second opinions? &mdash;[[User:Boy Soprano II|Boy Soprano II]] 20:26, 25 July 2006 (EDT)<br />
<br />
Okay; I'm merging them. &mdash;[[User:Boy Soprano II|Boy Soprano II]] 22:42, 26 July 2006 (EDT)<br />
<br />
Could we add links to the problems themselves? <br />
[[User:not_trig|not_trig]] 23:03, 28 Octover 2006 (EDT)</div>Not_trighttps://artofproblemsolving.com/wiki/index.php?title=Talk:IMO_Problems_and_Solutions,_with_authors&diff=10465Talk:IMO Problems and Solutions, with authors2006-10-29T03:02:25Z<p>Not_trig: </p>
<hr />
<div>I created the page [[IMO Problems and Solutions]] today, but I have now realized that the coexistance of that page and this one is probably redundant. Therefore I propose to merge the pages. Any second opinions? &mdash;[[User:Boy Soprano II|Boy Soprano II]] 20:26, 25 July 2006 (EDT)<br />
<br />
Okay; I'm merging them. &mdash;[[User:Boy Soprano II|Boy Soprano II]] 22:42, 26 July 2006 (EDT)<br />
<br />
Could we add links to the problems themselves?</div>Not_trig