https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=John0512&feedformat=atomAoPS Wiki - User contributions [en]2024-03-29T02:20:11ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2024_USAJMO_Problems&diff=2171522024 USAJMO Problems2024-03-20T02:39:30Z<p>John0512: no posting allowed</p>
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<div>__TOC__<br />
stop it. seriously.<br />
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DO NOT PUT ANYTHING HERE UNTIL AFTER DISCUSSION PERMITTED PLEASE<br />
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== See also ==<br />
<br />
{{USAJMO box|year=2024|before=[[2023 USAJMO Problems]]|after=[[2025 USAJMO Problems]]}}<br />
{{MAA Notice}}</div>John0512https://artofproblemsolving.com/wiki/index.php?title=AoPS_Wiki:AoPS_Community_Awards&diff=193006AoPS Wiki:AoPS Community Awards2023-05-11T15:58:36Z<p>John0512: /* Perfect AMC 10 scorers */</p>
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<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 />
* Shyam Narayanan (2015)<br />
* Prafulla Susil Dhariwal(2011,2012)<br />
* Akashnil Dutta(2009,2010,2011)<br />
* Krishanu Roy Sankar (2008)<br />
* Zachary Abel (2006) (AoPS assistant instructor)<br />
* Marco Avila (2006)<br />
* John Berman (2009)<br />
* Zarathustra Brady (2006)<br />
* Robert Cordwell (2005)<br />
* Wenyu Cao (2009, 2011)<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 />
* Bobby Shen (2012, 2013)<br />
* Arnav Tripathy (2006, 2007)<br />
* [[Naoki Sato]] (AoPS instructor)<br />
* Yi Sun (2006)<br />
* [[Valentin Vornicu]] (AoPS/MathLinks webmaster)<br />
* Victor Wang (2013)<br />
* Melanie Wood (1998, 1999) ([[WOOT]] instructor)<br />
* Alex Zhai (2005, 2006, 2007, 2008)<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 />
* Carmela Lao (2009, 2010)<br />
* Calvin Deng (2010, 2012, 2013)<br />
* Allen Yuan (2010)<br />
* In-sung Na (2010)<br />
* Xiaoyu He (2010, 2011)<br />
* Kevin Sun (2013)<br />
<br />
===Perfect Scorers===<br />
*Le Hung Viet Bao (2003)<br />
*Brian Lawrence (2005)<br />
*Alex Zhai (2008)<br />
*Jeck Lim (2012)<br />
*Alex Song (2015)<br />
*Allen Liu (2016)<br />
*Yuan Yao (2016)<br />
*Junghun Ju (2016)<br />
*James Lin (2018)<br />
*Agnijo Banerjee (2018)<br />
*Daniel Zhu (2019)<br />
*Colin Tang (2019)<br />
<br />
=== Gold medalists ===<br />
* Shyam Narayanan (2015)<br />
* Akashnil Dutta(2011)<br />
* Zarathustra Brady (2006)<br />
* Kevin Sun (2014, 2015, 2016)<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 />
* Mark Sellke (2013, 2014)<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 />
* John Berman (2009)<br />
* Xiaoyu He (2010, 2011)<br />
* Wenyu Cao (2011)<br />
* Prafulla Susil Dhariwal (2012)<br />
* Bobby Shen (2012, 2013)<br />
* James Tao (2013, 2014)<br />
* Victor Wang (2013)<br />
* Alex Song (2011, 2012, 2013, 2014, 2015)<br />
* Dongryul Kim (2012, 2013, 2014)<br />
* Allen Liu (2014, 2015, 2016)<br />
* Junghun Ju (2015, 2016)<br />
* Ryan Alweiss (2015)<br />
* Evan Chen (2014)<br />
<br />
=== Silver medalists ===<br />
* Akashnil Dutta (2009,2010)<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 />
* Delong Meng (2009)<br />
* Qinxuan Pan (2009)<br />
* Wenyu Cao (2009)<br />
* Carmela Lao (2010)<br />
* Jafar Jafarov (2006)<br />
* Ali Gurel (1996)<br />
* Micheal Kural (2015)<br />
<br />
=== Bronze medalists ===<br />
* Debdyuti Banerjee (2011)<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 />
* Carmela Lao (2009)<br />
* Jafar Jafarov (2007)<br />
* Kevin Sun (2013)<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 />
* Marianna Mao (2009)<br />
* Anand Natarajan (2009)<br />
* Bowei Liu (2009)<br />
* Daniel Li (2010)<br />
<br />
=== Gold Medalists ===<br />
* Yi Sun (2004)<br />
* Rahul Singh (2007)<br />
* Marianna Mao (2009)<br />
* Anand Natarajan (2009)<br />
* Bowei Liu (2009)<br />
* Daniel Li(2010)<br />
* Eric Schneider<br />
* Kevin Zhou<br />
* Micheal Kural (2016)<br />
<br />
=== Silver Medalists ===<br />
* Sherry Gong (2006)<br />
<br />
== IOI Medalists ==<br />
This list is of AoPSers who have won medals at the [[International Olympiad in Informatics]].<br />
<br />
===Gold Medalists ===<br />
* David Benjamin (2008)<br />
* Brian Hamrick (2009)<br />
* Neal Wu (2008, 2009, 2010)<br />
* Wenyu Cao (2010)<br />
* Scott Wu (2012, 2013, 2014)<br />
* Andrew He (2014, 2015)<br />
* Benjamin Qi (2018, 2019)<br />
<br />
===Silver Medalists ===<br />
* Akashnil Dutta(2011)<br />
* Brian Hamrick (2008)<br />
* Jacob Steinhardt (2008)<br />
* Wenyu Cao (2009)<br />
* Travis Hance (2009)<br />
<br />
===Bronze Medalists ===<br />
* David Benjamin (2007)<br />
<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 />
* Alex Zhu (2012)<br />
* Mitchell Lee (2012)<br />
* Bobby Shen (2012)<br />
* Allen Liu (2015, 2016)<br />
* Luke Robitaille (2019)<br />
<br />
=== Winners ===<br />
* Yakov Berchenko-Kogan (2006)<br />
* Wenyu Cao (2009)<br />
* Calvin Deng (2009, 2012, 2013)<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 />
* Shyam Narayanan (2015)<br />
* Albert Ni (2005)<br />
* [[David Patrick]] (1988) (AoPS instructor)<br />
* David Rolnick (2008) (AoPS assistant instructor)<br />
* [[Richard Rusczyk]] (1989) (AoPS founder)<br />
* Krishanu Sankar (2007,2008)<br />
* Bobby Shen (2013)<br />
* Peng Shi (2006)<br />
* Jacob Steinhardt (2007)<br />
* Yi Sun (2006)<br />
* Arnav Tripathy (2006, 2007)<br />
* Victor Wang (2012, 2013)<br />
* [[Sam Vandervelde]] (1987, 1989) ([[WOOT]] instructor)<br />
* Melanie Wood (1998, 1999) ([[WOOT]] instructor)<br />
* David Yang (2009,2011,2012)<br />
* Alex Zhai (2006, 2007,2008)<br />
* Yufei Zhao (2006)<br />
* David Bay Rush (2009)<br />
* Alex Song (2012,2013,2014,2015)<br />
* Evan Chen (2014)<br />
* James Tao (2014)<br />
* Scott Wu (2014)<br />
* Allen Liu (2014,2015,2016)<br />
* Ankan Bhattacharya (2016)<br />
* Yuan Yao (2016)<br />
* Micheal Kural (2015,2016)<br />
* Vincent Huang (2019)<br />
* Daniel Zhu (2019)<br />
* Colin Tang (2019)<br />
* Edward Wan (2018,2019)<br />
* Vincent Bian (2019)<br />
* Guanpeng Xu (2018,2019)<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 />
* Matthew Ince (2005) (AoPS assistant instructor)<br />
* Daniel Kane (2003, 2004, 2005) (AoPS assistant instructor)<br />
* Kiran Kedlaya (1994, 1995, 1996) ([[AoPS Foundation]] board member)<br />
* Brian Lawrence (2007, 2008, 2010, 2011)<br />
* Mitchell Lee (2012, 2013)<br />
* Evan O'Dorney (2011, 2012, 2013)<br />
* Alexander Schwartz (2000, 2002)<br />
* Mark Sellke (2014)<br />
* Bobby Shen (2013, 2014)<br />
* Jan Siwanowicz (2001) <br />
* Arnav Tripathy (2008)<br />
* Melanie Wood (2002) ([[WOOT]] instructor)<br />
* David Yang (2013, 2014, 2015)<br />
* Dongryul Kim (2016, 2018)<br />
* Shyam Narayanan (2018)<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 />
* Franklyn Wang (2nd Individual, 2017)<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 />
* Allen Liu (2016)<br />
* Jonathan Li (2011)<br />
* Evan O'Dorney (2011)<br />
* Philip Mocz (2008)<br />
* Yihe Dong (2008)<br />
* Qiaochu Yuan (2008)<br />
* Greg Brockman (2007)<br />
* Vincent Huang (2019)<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 />
<br />
* Matthew Babbitt (2013)<br />
* David Benjamin (2006)<br />
* [[Mathew Crawford]] (1992) (AoPS instructor)<br />
* Lewis Chen (2014)<br />
* Calvin Deng (2011)<br />
* Sam Elder (2008)<br />
* [[Sandor Lehoczky]] (1990) (AoPS author)<br />
* Tedrick Leung (2006)<br />
* Andrew Lin (2017)<br />
* Tony Liu (2006)<br />
* Haitao Mao (2008)<br />
* Shyam Narayanan (2012, 2014)<br />
* Vinayak Kumar (2016)<br />
* Bobby Shen (2010, 2011, 2012, 2013)<br />
* [[Richard Rusczyk]] (1989) (AoPS founder)<br />
* [[Sam Vandervelde]] (1988) ([[WOOT]] instructor)<br />
* Alexander Whatley (2012)<br />
* Scott Wu (2012)<br />
* James Tao (2013)<br />
* Guanpeng Xu (2016)<br />
* Joshua Lee (2017)<br />
* Luke Robitaille (2015, 2017, 2019)<br />
* Edward Wan (2018, 2019)<br />
* Adam Ardeishar (2019)<br />
* Brian Liu (2017)<br />
* Kevin Liu (2018)<br />
* Swapnil Garg (2017)<br />
* Benjamin Epstein (2020)<br />
* Jaedon Whyte (2020)<br />
* Sanjana Das (2020)<br />
* Siddharth Srinivasan (2020)<br />
* Richie Hsiung (2021)<br />
* Kevin Zhao (2022)<br />
* MD Nahidul Hasan Sabit (2023)<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 />
* Zachary Abel (2005) (AoPS assistant instructor)<br />
* David Benjamin (2006)<br />
* Wenyu Cao (2009)<br />
* Lewis Chen (2014)<br />
* Sam Elder (2008)<br />
* Ruozhou (Joe) Jia (2003) (AoPS assistant instructor)<br />
* Joel Lewis (2003) <br />
* Daniel Li (2009)<br />
* Jonathan Li (2010)<br />
* Jonathan Lowd (2003) (AoPS assistant instructor)<br />
* Thomas Mildorf (2004) (AoPS assistant instructor)<br />
* Alison Miller (2004) (AoPS assistant instructor)<br />
* Shyam Narayanan (2015)<br />
* Albert Ni (2003) (AoPS instructor)<br />
* Graham O'Donnell (2018)<br />
* Brandon Zhu (2018)<br />
* Ajay Sharma (2004)<br />
* Bobby Shen (2009, 2011)<br />
* Matt Superdock (2009)<br />
* Arnav Tripathy (2006, 2007)<br />
* Qiaochu Yuan (2008)<br />
* Alex Zhai (2007)<br />
* Guanpeng Xu (2016)<br />
* Luke Robitaille (2016, 2017, 2018, 2019)<br />
* Reagan Choi (2019)<br />
* Kevin Liu (2017,2018,2019)<br />
* Joey Heerens (2019)<br />
* Ben Kang (2019)<br />
* Ben Qi (2019)<br />
* Brian Liu (2019)<br />
* Daniel Zhu (2019)<br />
* Andrew Gu (2019)<br />
* Jinha Kim (2020)<br />
* Edward Yu (2021)<br />
* Siddharth Srinivasan (2021)<br />
* Sasidhar Kunapuli (2021, 2022)<br />
* Morgan Lin (2021, 2022)<br />
* Haozhe Yang (2022)<br />
<br />
===Perfect AMC 10 scorers===<br />
The [[AMC 10]] is a challenging examination for students in grades 10 and below administered by the [[American Mathematics Competitions]].<br />
<br />
* Matthew Babbitt (2009)<br />
* Sergei Bernstein (2007)<br />
* Yifan Cao (2005)<br />
* Kevin Chen (2007, 2008, 2009)<br />
* Lewis Chen(2009)<br />
* Yongyi Chen (2009)<br />
* In Young Cho (2007)<br />
* Mario Choi (2007)<br />
* Calvin Deng (2008)<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 />
* Michael Kural (2013)<br />
* Vincent Le (2006)<br />
* Daniel Li (2007)<br />
* Jonathan Li (2009, 2007)<br />
* Kevin Li (2009)<br />
* Patricia Li (2005)<br />
* Carl Lian (2007)<br />
* Sam Lite (2009)<br />
* David Lu (2009)<br />
* Michael Ma (2009, 2010, 2011 in grades 3, 4, 5)<br />
* Thomas Mildorf (2002) (AoPS assistant instructor)<br />
* Anupa Murali (2008)<br />
* Shyam Narayanan (2013)<br />
* Graham O'Donnell (2015, 2016)<br />
* Edwin Peng (2013)<br />
* Max Rosett (2005, 2006) (AoPS assistant instructor)<br />
* Amrit Saxena (2009)<br />
* Eric Schneider (2009)<br />
* Maximilian Schindler (2009)<br />
* Bobby Shen (2009)<br />
* Jeffrey Shen (2008)<br />
* Lilly Shen (2009)<br />
* Richard Spence (2009)<br />
* Swapnil Garg (2013, 2014)<br />
* Kyle Stankowski (2009)<br />
* Michael Tan (2009)<br />
* Michael Tang (2014, 2015)<br />
* Kevin Tian (2009)<br />
* Howard Tong (2005)<br />
* Sam Trabucco (2008)<br />
* Brent Woodhouse (2006, 2007)<br />
* Lawrence Wu (2009)<br />
* David Yang (2009)<br />
* Allen Yuan (2009)<br />
* Peijin Zhang (2009)<br />
* Jonathan Zhou (2007)<br />
* Alex Zhu (2009)<br />
* Alex Wei (2016)<br />
* Vittal Thirumalai (2017)<br />
* Samuel Wang (2017)<br />
* Jack Albright (2017, 2018)<br />
* Andrew Cai (2017, 2018, 2020)<br />
* Reagan Choi (2017, 2018)<br />
* Daniel Mai (2019,2021)<br />
* Daniel Yuan (2017,2018)<br />
* Jessica Wan (2018,2019,2020,2021)<br />
* Ethan Zhou (2019)<br />
* Ben Epstein (2017, 2019)<br />
* Alvin Chen (2019)<br />
* William Chen (2020)<br />
* Samrit Grover (2020)<br />
* Edward Yu (2020)<br />
* Maximus Lu (2020)<br />
* Linus Tang (2020)<br />
* Qiao Zhang (2021)<br />
* Aaron Chen (2021)<br />
* Benjamin Fan (2021) <br />
* Won Jang (2021)<br />
* Dylan Yu (2021)<br />
* Brian Li (2021)<br />
* Andrew Yuan (2021)<br />
* Christopher Qiu (2021) <br />
* George Cao (2021)<br />
* Zani Xu (2021)<br />
* MD Nahidul Hasan Sabit (2022)<br />
* Bryan Guo (2022)<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 />
=='''Perfect USAMO Index'''==<br />
*Lewis Chen (2014)<br />
*Samuel Elder (2009)<br />
*Bobby Shen (2011)<br />
*Gabriel Caroll (1998, 1999, 2000, 2001)<br />
*Allen Liu (2015, 2016)<br />
*Guanpeng Xu (2016)<br />
*Swapnil Garg (2017)<br />
*Luke Robitaille (2019)<br />
*Kevin Liu (2018)<br />
*Adam Ardeishar (2019)<br />
*Pavan Jayaraman (2023)<br />
*MD Nahidul Hasan Sabit (USAJMO 2023)<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 />
* Darryl Wu (2008)<br />
* Bobby Shen (2009)<br />
* Mark Sellke (2010)<br />
* Scott Wu (2011)<br />
* Chad Qian (2012)<br />
* Alec Sun (2013)<br />
* Swapnil Garg (2014)<br />
* Kevin Liu (2015)<br />
* Edward Wan (2016)<br />
* Luke Robitaille (2017, 2018)<br />
* Daniel Mai (2019)<br />
* Marvin Mao (2021)<br />
* Allan Yuan (2022)<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 />
* Andrew Cai (2016, 2017, 2018)<br />
* Christopher Chang (1991, 1992)<br />
* Kevin Chen (2006, 2007)<br />
* Steven Chen (2009, 2010)<br />
* Andrew Chien (2003)<br />
* Peter Chien (2004)<br />
* Mario Choi (2007)<br />
* Joseph Chu (2004)<br />
* Alexander Clifton (2009)<br />
* [[Mathew Crawford]] (1990, 1991) (AoPS instructor)<br />
* Calvin Deng (2009)<br />
* Brian Hamrick (2006)<br />
* Frank Han (2015)<br />
* Adam Hesterberg (2002, 2003)<br />
* Jason Hyun (2008)<br />
* Ruozhou (Joe) Jia (2000) (AoPS assistant instructor)<br />
* Ben Kang (2016)<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 />
* Ray Li (2009)<br />
* Po-Ling Loh (2000)<br />
* Brian Liu (2016)<br />
* David Lu (2008)<br />
* Albert Ni (2002) (AoPS assistant instructor)<br />
* Graham O'Donnell (2014)<br />
* Maximilian Schindler (2009)<br />
* Bobby Shen (2008, 2009)<br />
* Elizabeth Synge (2007)<br />
* Jason Trigg (2002)<br />
* [[Sam Vandervelde]] (1985) ([[WOOT]] instructor)<br />
* Edward Wan (2016)<br />
* Victor Wang (2009)<br />
* Alex Wei (2016)<br />
* Eric Wei (2016)<br />
* Ben Wright (2016)<br />
* Neal Wu (2005, 2006)<br />
* Rolland Wu (2006)<br />
* Xiaoyu He (2008)<br />
* Alex Xu (2016)<br />
* David Yang (2009)<br />
* Daesun Yim (2006)<br />
* Darren Yin (2002)<br />
* Justin Yu (2016)<br />
* Allen Yuan (2007)<br />
* Samuel Zbarsky (2008)<br />
* Alex Zhai (2004)<br />
* Mark Zhang (2005)<br />
* Alan Zhou (2009)<br />
* Mark Sellke (2009, 2010)<br />
* Eugene Chen (2010)<br />
* Lewis Chen (2010)<br />
* Shyam Narayanan (2010, 2011)<br />
* Franklyn Wang (2013)<br />
*Akshaj Kadaveru (2013, 2014)<br />
*Scott Wu (2010, 2011)<br />
*Walker Kroubalkian (2015)<br />
* Reagan Choi (2017, 2018)<br />
* Daniel Yuan (2018)<br />
* Andy Xu (2013, 2015)<br />
* Samuel Wang (2018, 2019)<br />
* Karthik Vedula (2019)<br />
* Alan Kappler (2019)<br />
* Ram Goel (2019)<br />
* Suyash Pandit (2019)<br />
*Ethan Zhou (2019)<br />
* Jessica Wan (2018, 2019)<br />
* William Chen (2018, 2019)<br />
* Eric Shen (2019)<br />
* Allan Yuan (2021, 2022)<br />
* Jiahe Liu (2021, 2022)<br />
* Alexander Wang (2021, 2022)<br />
* Bohan Yao (2021)<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 />
* Maximilian Schindler (2009)<br />
* Alex Song (2010)<br />
<br />
=== National Written Champions ===<br />
* [[Mathew Crawford]] (1990) (AoPS instructor)<br />
* Adam Hesterberg (2003)<br />
* Sergei Bernstein (2005)<br />
* Neal Wu (2006)<br />
* Bobby Shen (2008)<br />
* David Yang (2009)<br />
* Mark Sellke (2010)<br />
* Shyam Narayanan (2011)<br />
* Kevin Liu (2014)<br />
* Andy Xu (2015)<br />
* Edward Wan (2016)<br />
* Luke Robitaille (2017, 2018)<br />
* Jeff Lin (2019)<br />
* Marvin Mao (2021)<br />
* Jiahe Liu (2022)<br />
<br />
=== National Team Champions ===<br />
* Team Virginia: Divya Garg, Brian Hamrick, Daniel Li, Jimmy Clark (2006)<br />
* Team Massachusetts: Alec Sun, James Lin, Michael Ren, Matthew Lipman (2013)<br />
* Team California: Swapnil Garg, Harry Wang, Rajiv Movva, Jeffery Li (2014)<br />
* Team Texas: Benjamin Wright, Luke Robitaille, Andrew Cai, Justin Yu (2016)<br />
* Team Massachusetts: Max Xu, Daniel Mai, Jeff Lin, Kaylee Ji (2019)<br />
* Team New Jersey: Marvin Mao, Alexander Wang, Andrew Lin, Evan Fan (2021)<br />
* Team New Jersey: Alexander Wang, Evan Fan, Maitian Sha, Rithik Gumpu (2022)<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 />
<br />
=== ARML winners ===<br />
* Alex Song (2009, 2011)<br />
* Benjamin Gunby (2010)<br />
* Allen Liu (2012, 2013)<br />
* Darryl Wu (2014)<br />
* Brice Huang (2015)<br />
* Daniel Kim (2016)<br />
* Brian Reinhart (2017)<br />
* Luke Robitaille (2018, 2021, 2022)<br />
* David Chen (2019)<br />
* Chris Qiu (2022)<br />
<br />
=== ARML Perfect Scorers ===<br />
* Alex Song (2011)<br />
* Shyam Narayanan (2012, 2014)<br />
* David Chen (2019)<br />
<br />
=== ARML Top 10 ===<br />
* Zachary Abel (2006) (AoPS assistant instructor)<br />
*Lewis Chen(2011,2012)<br />
* Benjamin Gunby (2010)<br />
* Bobby Shen (2010)<br />
* Michael Tang (2015)<br />
* Seva Tchernov (2007)<br />
* Arnav Tripathy (2007)<br />
* David Yang (2009)<br />
* Daesun Yim (2008)<br />
* Alex Song (2009,2011,2012)<br />
*Bailey Wang (2009)(WOOT Grader)<br />
* Shyam Narayanan (2011, 2012 (Perfect Score), 2014 (Perfect Score), 2015)<br />
* Samuel Wang (2019)<br />
* Andrew Cai (2019)<br />
* Luke Robitaille (2016,2018,2019)<br />
* Daniel Zhu (2015)<br />
* Jason Lee (2015)<br />
* Colin Tang (2015)<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>John0512https://artofproblemsolving.com/wiki/index.php?title=User:John0512&diff=190846User:John05122023-03-12T06:03:04Z<p>John0512: Undo revision 189989 by Ej22 (talk)</p>
<hr />
<div>==2021 AMC 8==<br />
2021 AMC 8 problems and solutions. The test has not been held, and will never be held.<br />
<br />
==Problems==<br />
ERROR: Content not found<br />
==Solutions==<br />
ERROR: Content not found<br />
==Results==<br />
<br />
Highest Score: 0.00<br />
<br />
Distinguished Honor Roll: 0.00<br />
<br />
Honor Roll: 0.00<br />
<br />
Average Score: 0.00<br />
<br />
Standard Deviation: 0.00<br />
<br />
==Unnamed Theorem==<br />
<br />
I have something called the Unnamed Theorem (which I did not name as I have not confirmed that this theorem has not existed before).<br />
<br />
Claim: Given a set <math>S=\{1,2,3\cdots n\}</math> where <math>n</math> is a positive integer, the number of ways to choose a subset of <math>S</math> then permute said subset is <math>\lfloor n!\cdot e\rfloor.</math><br />
<br />
Proof: The number of ways to choose a subset of size <math>i</math> and then permute it is <math>\binom{n}{i}\cdot i!</math>. Therefore, the number of ways to choose any subset of <math>S</math> is <cmath>\sum_{i=0}^n \binom{n}{i}\cdot i! = \sum_{i=0}^n \frac{n!}{(n-i)!}.</cmath> This is also equal to <math>\sum_{i=0}^n \frac{n!}{i!}</math> by symmetry across <math>i=\frac{n}{2}</math>. This is also <math>n! \cdot \sum_{i=0}^n \frac{1}{i!}.</math> Note that <math>e</math> is defined as <math>\sum_{i=0}^\infty \frac{1}{i!}</math>, so our expression becomes <cmath>n!(e-\sum_{i={n+1}}^\infty \frac{1}{i!}).</cmath> We claim that <math>\sum_{i={n+1}}^\infty \frac{1}{i!}<\frac{1}{n!}</math> for all positive integers <math>n</math>.<br />
<br />
Since the reciprocal of a factorial decreases faster than a geometric series, we have that <math>\sum_{i={n+1}}^\infty \frac{1}{i!}<\frac{1}{(n+1)!}+\frac{1}{(n+1)!(n+1)}+\frac{1}{(n+1)!(n+1)^2}\cdots</math>. The right side we can evaluate as <math>\frac{1}{n(n!)}</math>, which is always less than or equal to <math>\frac{1}{n!}</math>. This means that the terms being subtracted are always strictly less than <math>\frac{1}{n!}</math>, so we can simply write it as <cmath>\lfloor n!\cdot e\rfloor.</cmath><br />
<br />
Example: How many ways are there 5 distinct clones of mathicorn to each either accept or reject me, then for me to go through the ones that accepted me in some order?<br />
<br />
Solution to example: This is equivalent to the Unnamed Theorem for <math>n=5</math>, so our answer is <math>\lfloor 120e \rfloor=\boxed{326}</math>.<br />
<br />
Solution 2: Since I am not orz, all 5 clones will reject me, so the answer is <math>\boxed{1}</math>. Note that this contradicts with the answer given by the Unnamed Theorem.</div>John0512https://artofproblemsolving.com/wiki/index.php?title=MathCounts_historical_results&diff=173863MathCounts historical results2022-05-09T21:27:57Z<p>John0512: </p>
<hr />
<div>This is the '''MathCounts historical results''' page. Post [[MathCounts]] statistics and lists of high scorers here so that the MathCounts page doesn't get cluttered.<br />
<br />
<br />
== National Team Champions ==<br />
* 1984: [[Virginia MathCounts]]<br />
* 1985: [[Florida MathCounts]]<br />
* 1986: [[California MathCounts]]<br />
* 1987: [[New York MathCounts]]<br />
* 1988: [[New York MathCounts]]<br />
* 1989: [[North Carolina MathCounts]]<br />
* 1990: [[Ohio MathCounts]]<br />
* 1991: [[Alabama MathCounts]]<br />
* 1992: [[California MathCounts]]<br />
* 1993: [[Kansas MathCounts]]<br />
* 1994: [[Pennsylvania MathCounts]]<br />
* 1995: [[Indiana MathCounts]]<br />
* 1996: [[Pennsylvania MathCounts]]<br />
* 1997: [[Massachusetts MathCounts]]<br />
* 1998: [[Wisconsin MathCounts]]<br />
* 1999: [[Massachusetts MathCounts]]<br />
* 2000: [[California MathCounts]]<br />
* 2001: [[Virginia MathCounts]]<br />
* 2002: [[California MathCounts]]<br />
* 2003: [[California MathCounts]]<br />
* 2004: [[Illinois Mathcounts]]<br />
* 2005: [[Texas MathCounts]]<br />
* 2006: [[Virginia MathCounts]]<br />
* 2007: [[Texas MathCounts]]<br />
* 2008: [[Texas MathCounts]]<br />
* 2009: [[Texas MathCounts]]<br />
* 2012: [[Massachusetts MathCounts]]<br />
* 2013: [[Massachusetts MathCounts]]<br />
* 2014: [[California MathCounts]]<br />
* 2015: [[Indiana MathCounts]]<br />
* 2016: [[Texas MathCounts]]<br />
* 2017: [[Texas MathCounts]]<br />
* 2018: [[Texas Mathcounts]]<br />
* 2019: [[Massachusetts MathCounts]]<br />
* 2020: CANCELLED<br />
* 2021: [[New Jersey MathCounts]]<br />
<br />
== Countdown Round Champions ==<br />
The champions of the Count Down Round are considered the individual national champions of MathCounts.<br />
* 1984: Michael Edwards, [[Texas MathCounts]]<br />
* 1985: Timothy Kokesh, [[Oklahoma MathCounts]]<br />
* 1986: Brian David Ewald, [[Florida MathCounts]]<br />
* 1987: Russell Mann, [[Tennessee MathCounts]]<br />
* 1988: Andrew Schultz, [[Illinois Mathcounts]]<br />
* 1989: Albert Kurz, [[Pennsylvania MathCounts]]<br />
* 1990: Brian Jenkins, [[Arkansas MathCounts]]<br />
* 1991: Jonathan L. Weinstein, [[Massachusetts MathCounts]]<br />
* 1992: Andrei C. Gnepp, [[Ohio MathCounts]]<br />
* 1993: Carleton Bosley, [[Kansas MathCounts]]<br />
* 1994: William O. Engel, [[Illinois Mathcounts]]<br />
* 1995: Richard Reifsnyder, [[Kentucky MathCounts]]<br />
* 1996: Alexander Schwartz, [[Pennsylvania MathCounts]]<br />
* 1997: Zhihao (Howard) Liu, [[Wisconsin MathCounts]]<br />
* 1998: Ricky Liu, [[Massachusetts MathCounts]]<br />
* 1999: Po-Ru Loh, [[Wisconsin MathCounts]]<br />
* 2000: Ruozhou (Joe) Jia, [[Illinois Mathcounts]]<br />
* 2001: Ryan Ko, [[New Jersey MathCounts]]<br />
* 2002: Albert Ni, [[Illinois Mathcounts]]<br />
* 2003: Adam Hesterberg, [[Washington MathCounts]]<br />
* 2004: Gregory Gauthier, [[Illinois Mathcounts]]<br />
* 2005: Neal Wu, [[Louisiana MathCounts]]<br />
* 2006: Daesun Yim, [[New Jersey MathCounts]]<br />
* 2007: Kevin Chen, [[Texas Mathcounts]]<br />
* 2008: Darryl Wu, [[Washington MathCounts]]<br />
* 2009: Bobby Shen, [[Texas MathCounts]]<br />
* 2012: Chad Qian, [[Indiana MathCounts]]<br />
* 2013: Alec Sun, [[Massachusetts MathCounts]]<br />
* 2014: Swapnil Garg [[California MathCounts]]<br />
* 2015: Kevin Liu [[Indiana MathCounts]]<br />
* 2016: Edward Wan [[Washington MathCounts]]<br />
* 2017: Luke Robitaille [[Texas MathCounts]]<br />
* 2018: Luke Robitaille [[Texas MathCounts]]<br />
* 2019: Daniel Mai [[Massachusetts MathCounts]]<br />
* 2020: CANCELLED<br />
* 2022: Allan Yuan [[Alabama MathCounts]]<br />
<br />
== Masters Round Winners ==<br />
* 1987 Ashley Reiter, [[North Carolina MathCounts]]<br />
* 1990 Brian Jenkins, [[Arkansas MathCounts]]<br />
* 1991 Chris Chang, [[New Jersey MathCounts]]<br />
* 1992 Andrei Gnepp, [[Ohio Mathcounts]]<br />
* 1993 Davesh Maulik, [[Kansas Mathcounts]]<br />
* 1994 Kevin Lacker, [[Ohio Mathcounts]]<br />
* 1999 Po-Ru Loh, [[Wisconsin MathCounts]]<br />
* 2000 Tiankai Liu, [[California MathCounts]]<br />
* 2001 Po-Ling Loh, [[Wisconsin MathCounts]]<br />
* 2002 Jason Trigg, [[Pennsylvania MathCounts]]<br />
* 2003 Brian Lawrence, [[Maryland MathCounts]]<br />
* 2004 Gregory Gauthier, [[Illinois Mathcounts]]<br />
* 2005 Sergei Bernstein, [[Massachusetts MathCounts]]<br />
* 2006 Daniel Li, [[Virginia MathCounts]]<br />
* 2007 Kevin Chen, [[Texas Mathcounts]]<br />
* 2008 Bobby Shen, [[Texas MathCounts]]<br />
* 2009 Maximilian Schindler, [[Missouri Mathcounts]]<br />
<br />
== Written Test Champions ==<br />
* 1989 Lenny Ng, [[North Carolina MathCounts]]<br />
* 1990 Mathew Crawford ([[Alabama MathCounts]]), Daniel Schepler ([[Ohio MathCounts]]), Ravi Shanmugan ([[Kansas MathCounts]])<br />
* 1991 Jonathan L. Weinstein, [[Massachusetts MathCounts]]<br />
* 1992 Jenny Hoffman, [[Connecticut MathCounts]]<br />
* 1993 Davesh Maulik, [[Kansas Mathcounts]]<br />
* 1994 Michael Shulman, [[California MathCounts]]<br />
* 1995 Richard Reifsnyder, [[Kentucky Mathcounts]]<br />
* 1996 Alexander Schwartz, [[Pennsylvania Mathcounts]]<br />
* 1999 Po-Ru Loh, [[Wisconsin MathCounts]]<br />
* 2000 Tiankai Liu, [[California MathCounts]]<br />
* 2001 Ryan Ko, [[New Jersey MathCounts]]<br />
* 2002 Conner Rogers, [[Colorado MathCounts]]<br />
* 2003 Adam Hesterberg, [[Washington MathCounts]]<br />
* 2004 Gregory Gauthier, [[Illinois Mathcounts]]<br />
* 2005 Sergei Bernstein, [[Massachusetts Mathcounts]]<br />
* 2006 Neal Wu, [[Louisiana MathCounts]]<br />
* 2007 Justin Ahmann, [[Indiana MathCounts]]<br />
* 2008 Bobby Shen, [[Texas MathCounts]]<br />
* 2009 David Yang, [[California MathCounts]]<br />
* 2010 Mark Selke [[Indiana MathCounts]], Shyam Narayanan [[Kansas Mathcounts]]<br />
* 2011 Shyam Narayanan [[Kansas Mathcounts]]<br />
* 2013 Alec Sun, [[Massachusetts MathCounts]] <br />
* 2014 Kevin Liu, [[Indiana MathCounts]]<br />
* 2015: Andy Xu [[South Carolina MathCounts]]<br />
* 2016: Edward Wan [[Washington MathCounts]]<br />
* 2017: Luke Robitaille [[Texas MathCounts]]<br />
* 2018: Luke Robitaille [[Texas MathCounts]]<br />
* 2019: Huaye (Jeff) Lin, [[Massachusetts MathCounts]]<br />
* 2020: CANCELLED<br />
* 2021: Marvin Mao [[New Jersey MathCounts]]<br />
* 2022: Jiahe Liu [[Ohio MathCounts]]<br />
<br />
== Most Improved Team ==<br />
* 2005 [[Oklahoma MathCounts]]<br />
* 2006 [[South Carolina MathCounts]]<br />
* 2009 [[New Mexico MathCounts]]<br />
* 2011 [[Kansas Mathcounts]]<br />
* 2014 [[Nevada Mathcounts]]<br />
* 2017 [[Georgia MathCounts]]<br />
* 2018 [[Utah MathCounts]]<br />
* 2019 [[Tennessee MathCounts]]<br />
<br />
== Spirit Stick Winners ==<br />
* 2005 [[New Hampshire MathCounts]]<br />
* 2006 [[Wyoming MathCounts]]<br />
* 2007 [[Pennsylvania MathCounts]]<br />
* 2008 [[New York MathCounts]]<br />
* 2009 [[Virgin Islands MathCounts]]<br />
* 2017 [[Louisiana MathCounts]]<br />
* 2018 [[Texas MathCounts]]<br />
* 2019 [[Kentucky MathCounts]]</div>John0512https://artofproblemsolving.com/wiki/index.php?title=User:John0512&diff=170291User:John05122022-01-26T18:55:51Z<p>John0512: Undo revision 170268 by Jwenslawski (talk)</p>
<hr />
<div>==2021 AMC 8==<br />
2021 AMC 8 problems and solutions. The test has not been held, and will never be held.<br />
<br />
==Problems==<br />
ERROR: Content not found<br />
==Solutions==<br />
ERROR: Content not found<br />
==Results==<br />
<br />
Highest Score: 0.00<br />
<br />
Distinguished Honor Roll: 0.00<br />
<br />
Honor Roll: 0.00<br />
<br />
Average Score: 0.00<br />
<br />
Standard Deviation: 0.00<br />
<br />
==Unnamed Theorem==<br />
<br />
I have something called the Unnamed Theorem (which I did not name as I have not confirmed that this theorem has not existed before).<br />
<br />
Claim: Given a set <math>S=\{1,2,3\cdots n\}</math> where <math>n</math> is a positive integer, the number of ways to choose a subset of <math>S</math> then permute said subset is <math>\lfloor n!\cdot e\rfloor.</math><br />
<br />
Proof: The number of ways to choose a subset of size <math>i</math> and then permute it is <math>\binom{n}{i}\cdot i!</math>. Therefore, the number of ways to choose any subset of <math>S</math> is <cmath>\sum_{i=0}^n \binom{n}{i}\cdot i! = \sum_{i=0}^n \frac{n!}{(n-i)!}.</cmath> This is also equal to <math>\sum_{i=0}^n \frac{n!}{i!}</math> by symmetry across <math>i=\frac{n}{2}</math>. This is also <math>n! \cdot \sum_{i=0}^n \frac{1}{i!}.</math> Note that <math>e</math> is defined as <math>\sum_{i=0}^\infty \frac{1}{i!}</math>, so our expression becomes <cmath>n!(e-\sum_{i={n+1}}^\infty \frac{1}{i!}).</cmath> We claim that <math>\sum_{i={n+1}}^\infty \frac{1}{i!}<\frac{1}{n!}</math> for all positive integers <math>n</math>.<br />
<br />
Since the reciprocal of a factorial decreases faster than a geometric series, we have that <math>\sum_{i={n+1}}^\infty \frac{1}{i!}<\frac{1}{(n+1)!}+\frac{1}{(n+1)!(n+1)}+\frac{1}{(n+1)!(n+1)^2}\cdots</math>. The right side we can evaluate as <math>\frac{1}{n(n!)}</math>, which is always less than or equal to <math>\frac{1}{n!}</math>. This means that the terms being subtracted are always strictly less than <math>\frac{1}{n!}</math>, so we can simply write it as <cmath>\lfloor n!\cdot e\rfloor.</cmath><br />
<br />
Example: How many ways are there 5 distinct clones of mathicorn to each either accept or reject me, then for me to go through the ones that accepted me in some order?<br />
<br />
Solution to example: This is equivalent to the Unnamed Theorem for <math>n=5</math>, so our answer is <math>\lfloor 120e \rfloor=\boxed{326}</math>.<br />
<br />
Solution 2: Since I am not orz, all 5 clones will reject me, so the answer is <math>\boxed{1}</math>. Note that this contradicts with the answer given by the Unnamed Theorem.</div>John0512https://artofproblemsolving.com/wiki/index.php?title=User:John0512&diff=170267User:John05122022-01-26T16:54:15Z<p>John0512: /* Results */</p>
<hr />
<div>==2021 AMC 8==<br />
2021 AMC 8 problems and solutions. The test has not been held, and will never be held.<br />
<br />
==Problems==<br />
ERROR: Content not found<br />
==Solutions==<br />
ERROR: Content not found<br />
==Results==<br />
<br />
Highest Score: 0.00<br />
<br />
Distinguished Honor Roll: 0.00<br />
<br />
Honor Roll: 0.00<br />
<br />
Average Score: 0.00<br />
<br />
Standard Deviation: 0.00<br />
<br />
==Unnamed Theorem==<br />
<br />
I have something called the Unnamed Theorem (which I did not name as I have not confirmed that this theorem has not existed before).<br />
<br />
Claim: Given a set <math>S=\{1,2,3\cdots n\}</math> where <math>n</math> is a positive integer, the number of ways to choose a subset of <math>S</math> then permute said subset is <math>\lfloor n!\cdot e\rfloor.</math><br />
<br />
Proof: The number of ways to choose a subset of size <math>i</math> and then permute it is <math>\binom{n}{i}\cdot i!</math>. Therefore, the number of ways to choose any subset of <math>S</math> is <cmath>\sum_{i=0}^n \binom{n}{i}\cdot i! = \sum_{i=0}^n \frac{n!}{(n-i)!}.</cmath> This is also equal to <math>\sum_{i=0}^n \frac{n!}{i!}</math> by symmetry across <math>i=\frac{n}{2}</math>. This is also <math>n! \cdot \sum_{i=0}^n \frac{1}{i!}.</math> Note that <math>e</math> is defined as <math>\sum_{i=0}^\infty \frac{1}{i!}</math>, so our expression becomes <cmath>n!(e-\sum_{i={n+1}}^\infty \frac{1}{i!}).</cmath> We claim that <math>\sum_{i={n+1}}^\infty \frac{1}{i!}<\frac{1}{n!}</math> for all positive integers <math>n</math>.<br />
<br />
Since the reciprocal of a factorial decreases faster than a geometric series, we have that <math>\sum_{i={n+1}}^\infty \frac{1}{i!}<\frac{1}{(n+1)!}+\frac{1}{(n+1)!(n+1)}+\frac{1}{(n+1)!(n+1)^2}\cdots</math>. The right side we can evaluate as <math>\frac{1}{n(n!)}</math>, which is always less than or equal to <math>\frac{1}{n!}</math>. This means that the terms being subtracted are always strictly less than <math>\frac{1}{n!}</math>, so we can simply write it as <cmath>\lfloor n!\cdot e\rfloor.</cmath><br />
<br />
Example: How many ways are there 5 distinct clones of mathicorn to each either accept or reject me, then for me to go through the ones that accepted me in some order?<br />
<br />
Solution to example: This is equivalent to the Unnamed Theorem for <math>n=5</math>, so our answer is <math>\lfloor 120e \rfloor=\boxed{326}</math>.<br />
<br />
Solution 2: Since I am not orz, all 5 clones will reject me, so the answer is <math>\boxed{1}</math>. Note that this contradicts with the answer given by the Unnamed Theorem.</div>John0512https://artofproblemsolving.com/wiki/index.php?title=User:John0512&diff=170266User:John05122022-01-26T16:53:38Z<p>John0512: /* Unnamed Theorem */</p>
<hr />
<div>==2021 AMC 8==<br />
2021 AMC 8 problems and solutions. The test has not been held, and will never be held.<br />
<br />
==Problems==<br />
ERROR: Content not found<br />
==Solutions==<br />
ERROR: Content not found<br />
==Results==<br />
<br />
Highest Score: 0.00<br />
Distinguished Honor Roll: 0.00<br />
Honor Roll: 0.00<br />
Average Score: 0.00<br />
Standard Deviation: 0.00<br />
<br />
<br />
<br />
==Unnamed Theorem==<br />
<br />
I have something called the Unnamed Theorem (which I did not name as I have not confirmed that this theorem has not existed before).<br />
<br />
Claim: Given a set <math>S=\{1,2,3\cdots n\}</math> where <math>n</math> is a positive integer, the number of ways to choose a subset of <math>S</math> then permute said subset is <math>\lfloor n!\cdot e\rfloor.</math><br />
<br />
Proof: The number of ways to choose a subset of size <math>i</math> and then permute it is <math>\binom{n}{i}\cdot i!</math>. Therefore, the number of ways to choose any subset of <math>S</math> is <cmath>\sum_{i=0}^n \binom{n}{i}\cdot i! = \sum_{i=0}^n \frac{n!}{(n-i)!}.</cmath> This is also equal to <math>\sum_{i=0}^n \frac{n!}{i!}</math> by symmetry across <math>i=\frac{n}{2}</math>. This is also <math>n! \cdot \sum_{i=0}^n \frac{1}{i!}.</math> Note that <math>e</math> is defined as <math>\sum_{i=0}^\infty \frac{1}{i!}</math>, so our expression becomes <cmath>n!(e-\sum_{i={n+1}}^\infty \frac{1}{i!}).</cmath> We claim that <math>\sum_{i={n+1}}^\infty \frac{1}{i!}<\frac{1}{n!}</math> for all positive integers <math>n</math>.<br />
<br />
Since the reciprocal of a factorial decreases faster than a geometric series, we have that <math>\sum_{i={n+1}}^\infty \frac{1}{i!}<\frac{1}{(n+1)!}+\frac{1}{(n+1)!(n+1)}+\frac{1}{(n+1)!(n+1)^2}\cdots</math>. The right side we can evaluate as <math>\frac{1}{n(n!)}</math>, which is always less than or equal to <math>\frac{1}{n!}</math>. This means that the terms being subtracted are always strictly less than <math>\frac{1}{n!}</math>, so we can simply write it as <cmath>\lfloor n!\cdot e\rfloor.</cmath><br />
<br />
Example: How many ways are there 5 distinct clones of mathicorn to each either accept or reject me, then for me to go through the ones that accepted me in some order?<br />
<br />
Solution to example: This is equivalent to the Unnamed Theorem for <math>n=5</math>, so our answer is <math>\lfloor 120e \rfloor=\boxed{326}</math>.<br />
<br />
Solution 2: Since I am not orz, all 5 clones will reject me, so the answer is <math>\boxed{1}</math>. Note that this contradicts with the answer given by the Unnamed Theorem.</div>John0512https://artofproblemsolving.com/wiki/index.php?title=2010_AIME_II_Problems/Problem_12&diff=1664152010 AIME II Problems/Problem 122021-11-26T01:24:51Z<p>John0512: </p>
<hr />
<div>== Problem ==<br />
Two non[[congruent]] integer-sided [[isosceles triangle]]s have the same perimeter and the same area. The ratio of the lengths of the bases of the two triangles is <math>8: 7</math>. Find the minimum possible value of their common [[perimeter]].<br />
<br />
== Solution 1==<br />
Let the first triangle have side lengths <math>a</math>, <math>a</math>, <math>14c</math>, and the second triangle have side lengths <math>b</math>, <math>b</math>, <math>16c</math>, where <math>a, b, 2c \in \mathbb{Z}</math>.<br />
<br />
<br/><br />
Equal perimeter: <br />
<br />
<center><br />
<math>\begin{array}{ccc}<br />
2a+14c&=&2b+16c\\<br />
a+7c&=&b+8c\\<br />
c&=&a-b\\<br />
\end{array}</math><br />
</center><br />
<br/><br />
<br />
Equal Area: <br />
<br />
<center><br />
<math>\begin{array}{cccl}<br />
7c(\sqrt{a^2-(7c)^2})&=&8c(\sqrt{b^2-(8c)^2})&{}\\<br />
7(\sqrt{(a+7c)(a-7c)})&=&8(\sqrt{(b+8c)(b-8c)})&{}\\<br />
7(\sqrt{(a-7c)})&=&8(\sqrt{(b-8c)})&\text{(Note that } a+7c=b+8c)\\<br />
49a-343c&=&64b-512c&{}\\<br />
49a+169c&=&64b&{}\\<br />
49a+169(a-b)&=&64b&\text{(Note that } c=a-b)\\<br />
218a&=&233b&{}\\<br />
\end{array}</math><br />
</center><br />
<br />
Since <math>a</math> and <math>b</math> are integer, the minimum occurs when <math>a=233</math>, <math>b=218</math>, and <math>c=15</math>. Hence, the perimeter is <math>2a+14c=2(233)+14(15)=\boxed{676}</math>.<br />
<br />
== Solution 2==<br />
Let <math>s</math> be the semiperimeter of the two triangles. Also, let the base of the longer triangle be <math>16x</math> and the base of the shorter triangle be <math>14x</math> for some arbitrary factor <math>x</math>. Then, the dimensions of the two triangles must be <math>s-8x,s-8x,16x</math> and <math>s-7x,s-7x,14x</math>. By Heron's Formula, we have <br />
<br />
<center><br />
<cmath>\sqrt{s(8x)(8x)(s-16x)}=\sqrt{s(7x)(7x)(s-14x)}</cmath><br />
<cmath>8\sqrt{s-16x}=7\sqrt{s-14x}</cmath><br />
<cmath>64s-1024x=49s-686x</cmath><br />
<cmath>15s=338x</cmath><br />
</center><br />
Since <math>15</math> and <math>338</math> are coprime, to minimize, we must have <math>s=338</math> and <math>x=15</math>. However, we want the minimum perimeter. This means that we must multiply our minimum semiperimeter by <math>2</math>, which gives us a final answer of <math>\boxed{676}</math>.<br />
<br />
== Solution 3==<br />
<br />
Let the first triangle have sides <math>16n,a,a</math>, so the second has sides <math>14n,a+n,a+n</math>. The height of the first triangle is <math>\frac{7}{8}</math> the height of the second triangle. Therefore, we have <cmath>a^2-64n^2=\frac{49}{64}((a+n)^2-49n^2).</cmath> Multiplying this, we get <cmath>64a^2-4096n^2=49a^2+98an-2352n^2,</cmath> which simplifies to <cmath>15a^2-98an-1744n^2=0.</cmath> Solving this for <math>a</math>, we get <math>a=n\cdot\frac{218}{15}</math>, so <math>n=15</math> and <math>a=218</math> and the perimeter is <math>15\cdot16+218+218=\boxed{676}</math>.<br />
<br />
~john0512<br />
<br />
== See also ==<br />
Video Solution: https://www.youtube.com/watch?v=IUxOyPH8b4o<br />
{{AIME box|year=2010|num-b=11|num-a=13|n=II}}<br />
<br />
[[Category:Intermediate Geometry Problems]]<br />
{{MAA Notice}}</div>John0512https://artofproblemsolving.com/wiki/index.php?title=User:John0512&diff=166123User:John05122021-11-24T18:23:04Z<p>John0512: /* Orzorzorz Number */</p>
<hr />
<div>==Unnamed Theorem==<br />
<br />
I have something called the Unnamed Theorem (which I did not name as I have not confirmed that this theorem has not existed before).<br />
<br />
Claim: Given a set <math>S=\{1,2,3\cdots n\}</math> where <math>n</math> is a positive integer, the number of ways to choose a subset of <math>S</math> then permute said subset is <math>\lfloor n!\cdot e\rfloor.</math><br />
<br />
Proof: The number of ways to choose a subset of size <math>i</math> and then permute it is <math>\binom{n}{i}\cdot i!</math>. Therefore, the number of ways to choose any subset of <math>S</math> is <cmath>\sum_{i=0}^n \binom{n}{i}\cdot i! = \sum_{i=0}^n \frac{n!}{(n-i)!}.</cmath> This is also equal to <math>\sum_{i=0}^n \frac{n!}{i!}</math> by symmetry across <math>i=\frac{n}{2}</math>. This is also <math>n! \cdot \sum_{i=0}^n \frac{1}{i!}.</math> Note that <math>e</math> is defined as <math>\sum_{i=0}^\infty \frac{1}{i!}</math>, so our expression becomes <cmath>n!(e-\sum_{i={n+1}}^\infty \frac{1}{i!}).</cmath> We claim that <math>\sum_{i={n+1}}^\infty \frac{1}{i!}<\frac{1}{n!}</math> for all positive integers <math>n</math>.<br />
<br />
Since the reciprocal of a factorial decreases faster than a geometric series, we have that <math>\sum_{i={n+1}}^\infty \frac{1}{i!}<\frac{1}{(n+1)!}+\frac{1}{(n+1)!(n+1)}+\frac{1}{(n+1)!(n+1)^2}\cdots</math>. The right side we can evaluate as <math>\frac{1}{n(n!)}</math>, which is always less than or equal to <math>\frac{1}{n!}</math>. This means that the terms being subtracted are always strictly less than <math>\frac{1}{n!}</math>, so we can simply write it as <cmath>\lfloor n!\cdot e\rfloor.</cmath><br />
<br />
Example: How many ways are there 5 distinct clones of mathicorn to each either accept or reject me, then for me to go through the ones that accepted me in some order?<br />
<br />
Solution to example: This is equivalent to the Unnamed Theorem for <math>n=5</math>, so our answer is <math>\lfloor 120e \rfloor=\boxed{326}</math>.<br />
<br />
Solution 2: Since I am not orz, all 5 clones will reject me, so the answer is <math>\boxed{1}</math>. Note that this contradicts with the answer given by the Unnamed Theorem.</div>John0512https://artofproblemsolving.com/wiki/index.php?title=User:John0512&diff=163011User:John05122021-10-02T01:57:28Z<p>John0512: stop removing my stuff</p>
<hr />
<div>==Orzorzorz Number==<br />
<br />
This is a number I coined. The exact value of this number is <cmath>\sum_{m=17}^{122}\sum_{k=17}^{122}\sum_{n=17}^{\min(m,k)} (2^{24})^{n-17}\cdot(2^{730})^{\min(m,k)-n}\cdot26^{12(m-17)+12(k-17)},</cmath> and is approximately equal to <math>4.8\cdot10^{26639}</math>. It's also approximately <math>2^{76650}\cdot26^{2520}</math> to 210 significant figures. The formula used to find this number is an approximation of the number of ways mathimathz can happen, however I will not go into further detail about exactly how it was derived. Each one of the numbers in the formula has a good reason, however I am not revealing just yet. :)<br />
<br />
==Unnamed Theorem==<br />
<br />
I have something called the Unnamed Theorem (which I did not name as I have not confirmed that this theorem has not existed before).<br />
<br />
Claim: Given a set <math>S=\{1,2,3\cdots n\}</math> where <math>n</math> is a positive integer, the number of ways to choose a subset of <math>S</math> then permute said subset is <math>\lfloor n!\cdot e\rfloor.</math><br />
<br />
Proof: The number of ways to choose a subset of size <math>i</math> and then permute it is <math>\binom{n}{i}\cdot i!</math>. Therefore, the number of ways to choose any subset of <math>S</math> is <cmath>\sum_{i=0}^n \binom{n}{i}\cdot i! = \sum_{i=0}^n \frac{n!}{(n-i)!}.</cmath> This is also equal to <math>\sum_{i=0}^n \frac{n!}{i!}</math> by symmetry across <math>i=\frac{n}{2}</math>. This is also <math>n! \cdot \sum_{i=0}^n \frac{1}{i!}.</math> Note that <math>e</math> is defined as <math>\sum_{i=0}^\infty \frac{1}{i!}</math>, so our expression becomes <cmath>n!(e-\sum_{i={n+1}}^\infty \frac{1}{i!}).</cmath> We claim that <math>\sum_{i={n+1}}^\infty \frac{1}{i!}<\frac{1}{n!}</math> for all positive integers <math>n</math>.<br />
<br />
Since the reciprocal of a factorial decreases faster than a geometric series, we have that <math>\sum_{i={n+1}}^\infty \frac{1}{i!}<\frac{1}{(n+1)!}+\frac{1}{(n+1)!(n+1)}+\frac{1}{(n+1)!(n+1)^2}\cdots</math>. The right side we can evaluate as <math>\frac{1}{n(n!)}</math>, which is always less than or equal to <math>\frac{1}{n!}</math>. This means that the terms being subtracted are always strictly less than <math>\frac{1}{n!}</math>, so we can simply write it as <cmath>\lfloor n!\cdot e\rfloor.</cmath><br />
<br />
Example: How many ways are there 5 distinct clones of mathicorn to each either accept or reject me, then for me to go through the ones that accepted me in some order?<br />
<br />
Solution to example: This is equivalent to the Unnamed Theorem for <math>n=5</math>, so our answer is <math>\lfloor 120e \rfloor=\boxed{326}</math>.<br />
<br />
Solution 2: Since I am not orz, all 5 clones will reject me, so the answer is <math>\boxed{1}</math>. Note that this contradicts with the answer given by the Unnamed Theorem.</div>John0512https://artofproblemsolving.com/wiki/index.php?title=User:John0512&diff=163000User:John05122021-10-01T22:00:32Z<p>John0512: </p>
<hr />
<div>==Orzorzorz Number==<br />
<br />
This is a number I coined. The exact value of this number is <cmath>\sum_{m=17}^{122}\sum_{k=17}^{122}\sum_{n=17}^{\min(m,k)} (2^{24})^{n-17}\cdot(2^{730})^{\min(m,k)-n}\cdot26^{12(m-17)+12(k-17)},</cmath> and is approximately equal to <math>4.8\cdot10^{26639}</math>. It's also approximately <math>2^{76650}\cdot26^{2520}</math> to 210 significant figures. The formula used to find this number is an approximation of the number of ways mathimathz can happen, however I will not go into further detail about exactly how it was derived. Each one of the numbers in the formula has a good reason, however I am not revealing just yet. :)<br />
<br />
==Unnamed Theorem==<br />
<br />
I have something called the Unnamed Theorem (which I did not name as I have not confirmed that this theorem has not existed before).<br />
<br />
Claim: Given a set <math>S=\{1,2,3\cdots n\}</math> where <math>n</math> is a positive integer, the number of ways to choose a subset of <math>S</math> then permute said subset is <math>\lfloor n!\cdot e\rfloor.</math><br />
<br />
Proof: The number of ways to choose a subset of size <math>i</math> and then permute it is <math>\binom{n}{i}\cdot i!</math>. Therefore, the number of ways to choose any subset of <math>S</math> is <cmath>\sum_{i=0}^n \binom{n}{i}\cdot i! = \sum_{i=0}^n \frac{n!}{(n-i)!}.</cmath> This is also equal to <math>\sum_{i=0}^n \frac{n!}{i!}</math> by symmetry across <math>i=\frac{n}{2}</math>. This is also <math>n! \cdot \sum_{i=0}^n \frac{1}{i!}.</math> Note that <math>e</math> is defined as <math>\sum_{i=0}^\infty \frac{1}{i!}</math>, so our expression becomes <cmath>n!(e-\sum_{i={n+1}}^\infty \frac{1}{i!}).</cmath> We claim that <math>\sum_{i={n+1}}^\infty \frac{1}{i!}<\frac{1}{n!}</math> for all positive integers <math>n</math>.<br />
<br />
Since the reciprocal of a factorial decreases faster than a geometric series, we have that <math>\sum_{i={n+1}}^\infty \frac{1}{i!}<\frac{1}{(n+1)!}+\frac{1}{(n+1)!(n+1)}+\frac{1}{(n+1)!(n+1)^2}\cdots</math>. The right side we can evaluate as <math>\frac{1}{n(n!)}</math>, which is always less than or equal to <math>\frac{1}{n!}</math>. This means that the terms being subtracted are always strictly less than <math>\frac{1}{n!}</math>, so we can simply write it as <cmath>\lfloor n!\cdot e\rfloor.</cmath><br />
<br />
Example: How many ways are there 5 distinct clones of mathicorn to each either accept or reject me, then for me to go through the ones that accepted me in some order?<br />
<br />
Solution to example: This is equivalent to the Unnamed Theorem for <math>n=5</math>, so our answer is <math>\lfloor 120e \rfloor=\boxed{326}</math>.<br />
<br />
Solution 2: Since I am not orz, all 5 clones will reject me, so the answer is <math>\boxed{1}</math>. Note that this contradicts with the answer given by the Unnamed Theorem.</div>John0512https://artofproblemsolving.com/wiki/index.php?title=User:John0512&diff=162991User:John05122021-10-01T15:52:38Z<p>John0512: </p>
<hr />
<div>==Orzorzorz Number==<br />
<br />
This is a number I coined. The exact value of this number is <cmath>\sum_{m=17}^{122}\sum_{k=17}^{122}\sum_{n=17}^{\min(m,k)} (2^{24})^{n-17}\cdot(2^{730})^{\min(m,k)-n}\cdot26^{12(m-17)+12(k-17)},</cmath> and is approximately equal to <math>4.8\cdot10^{26639}</math>. The formula used to find this number is an approximation of the number of ways mathimathz can happen, however I will not go into further detail about exactly how it was derived. Each one of the numbers in the formula has a good reason, however I am not revealing just yet. :)<br />
<br />
==Unnamed Theorem==<br />
<br />
I have something called the Unnamed Theorem (which I did not name as I have not confirmed that this theorem has not existed before).<br />
<br />
Claim: Given a set <math>S=\{1,2,3\cdots n\}</math> where <math>n</math> is a positive integer, the number of ways to choose a subset of <math>S</math> then permute said subset is <math>\lfloor n!\cdot e\rfloor.</math><br />
<br />
Proof: The number of ways to choose a subset of size <math>i</math> and then permute it is <math>\binom{n}{i}\cdot i!</math>. Therefore, the number of ways to choose any subset of <math>S</math> is <cmath>\sum_{i=0}^n \binom{n}{i}\cdot i! = \sum_{i=0}^n \frac{n!}{(n-i)!}.</cmath> This is also equal to <math>\sum_{i=0}^n \frac{n!}{i!}</math> by symmetry across <math>i=\frac{n}{2}</math>. This is also <math>n! \cdot \sum_{i=0}^n \frac{1}{i!}.</math> Note that <math>e</math> is defined as <math>\sum_{i=0}^\infty \frac{1}{i!}</math>, so our expression becomes <cmath>n!(e-\sum_{i={n+1}}^\infty \frac{1}{i!}).</cmath> We claim that <math>\sum_{i={n+1}}^\infty \frac{1}{i!}<\frac{1}{n!}</math> for all positive integers <math>n</math>.<br />
<br />
Since the reciprocal of a factorial decreases faster than a geometric series, we have that <math>\sum_{i={n+1}}^\infty \frac{1}{i!}<\frac{1}{(n+1)!}+\frac{1}{(n+1)!(n+1)}+\frac{1}{(n+1)!(n+1)^2}\cdots</math>. The right side we can evaluate as <math>\frac{1}{n(n!)}</math>, which is always less than or equal to <math>\frac{1}{n!}</math>. This means that the terms being subtracted are always strictly less than <math>\frac{1}{n!}</math>, so we can simply write it as <cmath>\lfloor n!\cdot e\rfloor.</cmath><br />
<br />
Example: How many ways are there 5 distinct clones of mathicorn to each either accept or reject me, then for me to go through the ones that accepted me in some order?<br />
<br />
Solution to example: This is equivalent to the Unnamed Theorem for <math>n=5</math>, so our answer is <math>\lfloor 120e \rfloor=\boxed{326}</math>.<br />
<br />
Solution 2: Since I am not orz, all 5 clones will reject me, so the answer is <math>\boxed{1}</math>. Note that this contradicts with the answer given by the Unnamed Theorem.</div>John0512https://artofproblemsolving.com/wiki/index.php?title=User:John0512&diff=162989User:John05122021-10-01T15:44:52Z<p>John0512: </p>
<hr />
<div>==Orzorzorz Number==<br />
<br />
This is a number I coined. The exact value of this number is <cmath>\sum_{m=17}^{122}\sum_{k=17}^{122}\sum_{n=17}^{\min(m,k)} (2^{24})^{n-17}\cdot(2^{730})^{\min(m,k)-n}\cdot26^{12m+12k},</cmath> and is approximately equal to <math>9.8\cdot10^{27216}</math>. The formula used to find this number is an approximation of the number of ways mathimathz can happen, however I will not go into further detail about exactly how it was derived. Each one of the numbers in the formula has a good reason, however I am not revealing just yet. :)<br />
<br />
==Unnamed Theorem==<br />
<br />
I have something called the Unnamed Theorem (which I did not name as I have not confirmed that this theorem has not existed before).<br />
<br />
Claim: Given a set <math>S=\{1,2,3\cdots n\}</math> where <math>n</math> is a positive integer, the number of ways to choose a subset of <math>S</math> then permute said subset is <math>\lfloor n!\cdot e\rfloor.</math><br />
<br />
Proof: The number of ways to choose a subset of size <math>i</math> and then permute it is <math>\binom{n}{i}\cdot i!</math>. Therefore, the number of ways to choose any subset of <math>S</math> is <cmath>\sum_{i=0}^n \binom{n}{i}\cdot i! = \sum_{i=0}^n \frac{n!}{(n-i)!}.</cmath> This is also equal to <math>\sum_{i=0}^n \frac{n!}{i!}</math> by symmetry across <math>i=\frac{n}{2}</math>. This is also <math>n! \cdot \sum_{i=0}^n \frac{1}{i!}.</math> Note that <math>e</math> is defined as <math>\sum_{i=0}^\infty \frac{1}{i!}</math>, so our expression becomes <cmath>n!(e-\sum_{i={n+1}}^\infty \frac{1}{i!}).</cmath> We claim that <math>\sum_{i={n+1}}^\infty \frac{1}{i!}<\frac{1}{n!}</math> for all positive integers <math>n</math>.<br />
<br />
Since the reciprocal of a factorial decreases faster than a geometric series, we have that <math>\sum_{i={n+1}}^\infty \frac{1}{i!}<\frac{1}{(n+1)!}+\frac{1}{(n+1)!(n+1)}+\frac{1}{(n+1)!(n+1)^2}\cdots</math>. The right side we can evaluate as <math>\frac{1}{n(n!)}</math>, which is always less than or equal to <math>\frac{1}{n!}</math>. This means that the terms being subtracted are always strictly less than <math>\frac{1}{n!}</math>, so we can simply write it as <cmath>\lfloor n!\cdot e\rfloor.</cmath><br />
<br />
Example: How many ways are there 5 distinct clones of mathicorn to each either accept or reject me, then for me to go through the ones that accepted me in some order?<br />
<br />
Solution to example: This is equivalent to the Unnamed Theorem for <math>n=5</math>, so our answer is <math>\lfloor 120e \rfloor=\boxed{326}</math>.<br />
<br />
Solution 2: Since I am not orz, all 5 clones will reject me, so the answer is <math>\boxed{1}</math>. Note that this contradicts with the answer given by the Unnamed Theorem.</div>John0512https://artofproblemsolving.com/wiki/index.php?title=User:John0512&diff=162299User:John05122021-09-15T19:01:47Z<p>John0512: Undo revision 162291 by Asdf334 (talk)</p>
<hr />
<div>I have something called the Unnamed Theorem (which I did not name as I have not confirmed that this theorem has not existed before).<br />
<br />
Claim: Given a set <math>S=\{1,2,3\cdots n\}</math> where <math>n</math> is a positive integer, the number of ways to choose a subset of <math>S</math> then permute said subset is <math>\lfloor n!\cdot e\rfloor.</math><br />
<br />
Proof: The number of ways to choose a subset of size <math>i</math> and then permute it is <math>\binom{n}{i}\cdot i!</math>. Therefore, the number of ways to choose any subset of <math>S</math> is <cmath>\sum_{i=0}^n \binom{n}{i}\cdot i! = \sum_{i=0}^n \frac{n!}{(n-i)!}.</cmath> This is also equal to <math>\sum_{i=0}^n \frac{n!}{i!}</math> by symmetry across <math>i=\frac{n}{2}</math>. This is also <math>n! \cdot \sum_{i=0}^n \frac{1}{i!}.</math> Note that <math>e</math> is defined as <math>\sum_{i=0}^\infty \frac{1}{i!}</math>, so our expression becomes <cmath>n!(e-\sum_{i={n+1}}^\infty \frac{1}{i!}).</cmath> We claim that <math>\sum_{i={n+1}}^\infty \frac{1}{i!}<\frac{1}{n!}</math> for all positive integers <math>n</math>.<br />
<br />
Since the reciprocal of a factorial decreases faster than a geometric series, we have that <math>\sum_{i={n+1}}^\infty \frac{1}{i!}<\frac{1}{(n+1)!}+\frac{1}{(n+1)!(n+1)}+\frac{1}{(n+1)!(n+1)^2}\cdots</math>. The right side we can evaluate as <math>\frac{1}{n(n!)}</math>, which is always less than or equal to <math>\frac{1}{n!}</math>. This means that the terms being subtracted are always strictly less than <math>\frac{1}{n!}</math>, so we can simply write it as <cmath>\lfloor n!\cdot e\rfloor.</cmath><br />
<br />
Example: How many ways are there 5 distinct clones of mathicorn to each either accept or reject me, then for me to go through the ones that accepted me in some order?<br />
<br />
Solution to example: This is equivalent to the Unnamed Theorem for <math>n=5</math>, so our answer is <math>\lfloor 120e \rfloor=\boxed{326}</math>.<br />
<br />
Solution 2: Since I am not orz, all 5 clones will reject me, so the answer is <math>\boxed{1}</math>. Note that this contradicts with the answer given by the Unnamed Theorem.</div>John0512https://artofproblemsolving.com/wiki/index.php?title=User:John0512&diff=162288User:John05122021-09-15T12:00:56Z<p>John0512: </p>
<hr />
<div>I have something called the Unnamed Theorem (which I did not name as I have not confirmed that this theorem has not existed before).<br />
<br />
Claim: Given a set <math>S=\{1,2,3\cdots n\}</math> where <math>n</math> is a positive integer, the number of ways to choose a subset of <math>S</math> then permute said subset is <math>\lfloor n!\cdot e\rfloor.</math><br />
<br />
Proof: The number of ways to choose a subset of size <math>i</math> and then permute it is <math>\binom{n}{i}\cdot i!</math>. Therefore, the number of ways to choose any subset of <math>S</math> is <cmath>\sum_{i=0}^n \binom{n}{i}\cdot i! = \sum_{i=0}^n \frac{n!}{(n-i)!}.</cmath> This is also equal to <math>\sum_{i=0}^n \frac{n!}{i!}</math> by symmetry across <math>i=\frac{n}{2}</math>. This is also <math>n! \cdot \sum_{i=0}^n \frac{1}{i!}.</math> Note that <math>e</math> is defined as <math>\sum_{i=0}^\infty \frac{1}{i!}</math>, so our expression becomes <cmath>n!(e-\sum_{i={n+1}}^\infty \frac{1}{i!}).</cmath> We claim that <math>\sum_{i={n+1}}^\infty \frac{1}{i!}<\frac{1}{n!}</math> for all positive integers <math>n</math>.<br />
<br />
Since the reciprocal of a factorial decreases faster than a geometric series, we have that <math>\sum_{i={n+1}}^\infty \frac{1}{i!}<\frac{1}{(n+1)!}+\frac{1}{(n+1)!(n+1)}+\frac{1}{(n+1)!(n+1)^2}\cdots</math>. The right side we can evaluate as <math>\frac{1}{n(n!)}</math>, which is always less than or equal to <math>\frac{1}{n!}</math>. This means that the terms being subtracted are always strictly less than <math>\frac{1}{n!}</math>, so we can simply write it as <cmath>\lfloor n!\cdot e\rfloor.</cmath><br />
<br />
Example: How many ways are there 5 distinct clones of mathicorn to each either accept or reject me, then for me to go through the ones that accepted me in some order?<br />
<br />
Solution to example: This is equivalent to the Unnamed Theorem for <math>n=5</math>, so our answer is <math>\lfloor 120e \rfloor=\boxed{326}</math>.<br />
<br />
Solution 2: Since I am not orz, all 5 clones will reject me, so the answer is <math>\boxed{1}</math>. Note that this contradicts with the answer given by the Unnamed Theorem.</div>John0512https://artofproblemsolving.com/wiki/index.php?title=User:John0512&diff=162269User:John05122021-09-15T03:40:20Z<p>John0512: </p>
<hr />
<div>I have something called the Unnamed Theorem (which I did not name as I have not confirmed that this theorem has not existed before).<br />
<br />
Claim: Given a set <math>S=\{1,2,3\cdots n\}</math> where <math>n</math> is a positive integer, the number of ways to choose a subset of <math>S</math> then permute said subset is <math>\lfloor n!\cdot e\rfloor.</math><br />
<br />
Proof: The number of ways to choose a subset of size <math>i</math> and then permute it is <math>\binom{n}{i}\cdot i!</math>. Therefore, the number of ways to choose any subset of <math>S</math> is <cmath>\sum_{i=0}^n \binom{n}{i}\cdot i! = \sum_{i=0}^n \frac{n!}{(n-i)!}.</cmath> This is also equal to <math>\sum_{i=0}^n \frac{n!}{i!}</math> by symmetry across <math>i=\frac{n}{2}</math>. This is also <math>n! \cdot \sum_{i=0}^n \frac{1}{i!}.</math> Note that <math>e</math> is defined as <math>\sum_{i=0}^\infty \frac{1}{i!}</math>, so our expression becomes <cmath>n!(e-\sum_{i={n+1}}^\infty \frac{1}{i!}).</cmath> We claim that <math>\sum_{i={n+1}}^\infty \frac{1}{i!}<\frac{1}{n!}</math> for all positive integers <math>n</math>.<br />
<br />
Since the reciprocal of a factorial decreases faster than a geometric series, we have that <math>\sum_{i={n+1}}^\infty \frac{1}{i!}<\frac{1}{(n+1)!}+\frac{1}{(n+1)!(n+1)}+\frac{1}{(n+1)!(n+1)^2}\cdots</math>. The right side we can evaluate as <math>\frac{1}{n(n!)}</math>, which is always less than or equal to <math>\frac{1}{n!}</math>. This means that the terms being subtracted are always strictly less than <math>\frac{1}{n!}</math>, so we can simply write it as <cmath>\lfloor n!\cdot e\rfloor.</cmath><br />
<br />
Example: How many ways are there 5 distinct clones of mathicorn to each either accept or reject me, then for me to go through the ones that accepted me in some order?<br />
<br />
Solution to example: This is equivalent to the Unnamed Theorem for <math>n=5</math>, so our answer is <math>\lfloor 120e \rfloor=\boxed{326}</math>.</div>John0512https://artofproblemsolving.com/wiki/index.php?title=User:John0512&diff=162268User:John05122021-09-15T03:39:18Z<p>John0512: </p>
<hr />
<div>I have something called the Unnamed Theorem (which I did not name as I have not confirmed that this theorem has not existed before).<br />
<br />
Claim: Given a set <math>S=\{1,2,3\cdots n\}</math> where <math>n</math> is a positive integer, the number of ways to choose a subset of <math>S</math> then permute said subset is <math>\lfloor n!\cdot e\rfloor.</math><br />
<br />
Proof: The number of ways to choose a subset of size <math>i</math> and then permute it is <math>\binom{n}{i}\cdot i!</math>. Therefore, the number of ways to choose any subset of <math>S</math> is <cmath>\sum_{i=0}^n \binom{n}{i}\cdot i! = \sum_{i=0}^n \frac{n!}{(n-i)!}.</cmath> This is also equal to <math>\sum_{i=0}^n \frac{n!}{i!}</math> by symmetry across <math>i=\frac{n}{2}</math>. This is also <math>n! \cdot \sum_{i=0}^n \frac{1}{i!}.</math> Note that <math>e</math> is defined as <math>\sum_{i=0}^\infty \frac{1}{i!}</math>, so our expression becomes <cmath>n!(e-\sum_{i={n+1}}^\infty \frac{1}{i!}).</cmath> We claim that <math>\sum_{i={n+1}}^\infty \frac{1}{i!}<\frac{1}{n!}</math> for all positive integers <math>n</math>.<br />
<br />
Since the reciprocal of a factorial decreases faster than a geometric series, we have that <math>\sum_{i={n+1}}^\infty \frac{1}{i!}<\frac{1}{(n+1)!}+\frac{1}{(n+1)!(n+1)}+\frac{1}{(n+1)!(n+1)^2}\cdots</math>. The right side we can evaluate as <math>\frac{1}{n(n!)}</math>, which is always less than or equal to <math>\frac{1}{n!}</math>. This means that the terms being subtracted are always strictly less than <math>\frac{1}{n!}</math>, so we can simply write it as <cmath>\lfloor n!\cdot e\rfloor.</cmath><br />
<br />
Example: How many ways are there 5 clones of mathicorn to each either accept or reject me, then for me to go through the ones that accepted me in some order?<br />
<br />
Solution to example: This is equivalent to the Unnamed Theorem for <math>n=5</math>, so our answer is <math>\lfloor 120e \rfloor=\boxed{326}</math>.</div>John0512https://artofproblemsolving.com/wiki/index.php?title=1983_IMO_Problems/Problem_6&diff=1618351983 IMO Problems/Problem 62021-09-08T02:08:36Z<p>John0512: </p>
<hr />
<div>==Problem 6==<br />
Let <math>a</math>, <math>b</math> and <math>c</math> be the lengths of the sides of a triangle. Prove that<br />
<br />
<math>a^2 b(a-b) + b^2 c(b-c) + c^2 a(c-a) \geq 0</math>.<br />
<br />
Determine when equality occurs.<br />
<br />
==Solution 1==<br />
<br />
By Ravi substitution, let <math>a = y+z</math>, <math>b = z+x</math>, <math>c = x+y</math>. Then, the triangle condition becomes <math>x, y, z > 0</math>. After some manipulation, the inequality becomes:<br />
<br />
<math>xy^3 + yz^3 + zx^3 \geq xyz(x+y+z)</math>.<br />
<br />
By Cauchy, we have: <br />
<br />
<math>(xy^3 + yz^3 + zx^3)(z+x+y) \geq xyz(y+z+x)^2</math> with equality if and only if <math>\frac{xy^3}{z} = \frac{yz^3}{x} =\frac{zx^3}{y}</math>. So the inequality holds with equality if and only if x = y = z. Thus the original inequality has equality if and only if the triangle is equilateral.<br />
<br />
==Solution 2==<br />
<br />
Without loss of generality, let <math>a \geq b \geq c > 0</math>. By Muirhead or by AM-GM, we see that <math>a^3 b + a^3 c + b^3 c + b^3 a + c^3 a + c^3 b \geq 2(a^2 b^2 + a^2 c^2 + b^2 c^2)</math>.<br />
<br />
If we can show that <math>a^3 b + b^3 c+ c^3 a \geq a^3 c + b^3 a + c^3 b</math>, we are done, since then <math>2(a^3 b + b^3 c+ c^3 a ) \geq a^3 b + a^3 c + b^3 c + b^3 a + c^3 a + c^3 b \geq 2(a^2 b^2 + a^2 c^2 + b^2 c^2)</math>, and we can divide by <math>2</math>. <br />
<br />
We first see that, <math>(a^2 + ac + c^2) \geq (b^2 + bc + c^2)</math>, so <math>(a-c)(b-c)(a^2 + ac + c^2) \geq (a-c)(b-c)(b^2 + bc + c^2)</math>.<br />
<br />
Factoring, this becomes <math>(a^3 - c^3)(b-c) \geq (a-c)(b^3 - c^3)</math>. This is the same as:<br />
<br />
<math>(a^3 - c^3)(b-c) + (b^3 - c^3)(c-a) \geq 0</math>.<br />
<br />
Expanding and refactoring, this is equal to <math>a^3 (b-c) + b^3(c-a) + c^3 (a-b) \geq 0</math>. (This step makes more sense going backwards.)<br />
<br />
Expanding this out, we have <br />
<br />
<math>a^3b + b^3 c + c^3 a \geq a^3 c + b^3 a + c^3 b</math>,<br />
<br />
which is the desired result.<br />
<br />
==Solution 3==<br />
Let <math>s</math> be the semiperimeter, <math>\frac{a+b+c}{2}</math>, of the triangle. Then, <math>a=s-\frac{-a+b+c}{2}</math>, <math>b=s-\frac{a-b+c}{2}</math>, and <math>c=s-\frac{a+b-c}{2}</math>. We let <math>x=\frac{-a+b+c}{2},</math> <math>y=\frac{a-b+c}{2}</math>, and <math>z=\frac{a+b-c}{2}.</math> (Note that <math>x,y,z</math> are all positive, since all sides must be shorter than the semiperimeter.) Then, we have <math>a=s-x</math>, <math>b=s-y</math>, and <math>c=s-z</math>. Note that <math>x+y+z=s</math>, so <cmath>a=y+z,b=x+z,c=x+y.</cmath> Plugging this into <cmath>a^2b(a-b)+b^2c(b-c)+c^2a(c-a)\geq0</cmath> and doing some expanding and cancellation, we get <cmath>2x^3z+2xy^3+2yz^3-2x^2yz-2xy^2z-2xyz^2\geq0.</cmath> The fact that each term on the left hand side has at least two variables multiplied motivates us to divide the inequality by <math>2xyz</math>, which we know is positive from earlier so we can maintain the sign of the inequality. This gives <cmath>\frac{x^2}{y}-x+\frac{y^2}{z}-y-z+\frac{z^2}{x}\geq0.</cmath> We move the negative terms to the right, giving <cmath>\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}\geq x+y+z.</cmath> We rewrite this as <cmath>\sum_{cyc}\frac{x^2}{y}\geq\sum_{cyc}rx+(1-r)y.</cmath> where <math>r</math> is any real number. (This works because if we evaulate the cyclic sum, then as long as the coefficients of <math>x</math> and <math>y</math> on the right sum to 1 the right side will be <math>x+y+z</math>.<br />
<br />
Thus, we need to show that there exists a real number <math>r</math> such that <math>\frac{x^2}{y}\geq rx+(1-r)y</math> for all positive <math>x,y</math>. We claim that <math>r=2</math> works. This becomes <math>\frac{x^2}{y}\geq2x-y</math>, and since <math>y</math> is positive we can multiply by <math>y</math> to yield <math>x^2\geq2xy-y^2</math>, or <math>(x-y)^2\geq0</math>, which is obviously true by the Trivial Inequality. Thus, we are done with part (a).<br />
<br />
(To clarify how this works, we have <math>\frac{x^2}{y}\geq 2x-y</math>, <math>\frac{y^2}{z}\geq 2y-z</math>, and <math>\frac{z^2}{x}\geq 2z-x</math>, we add these inequalities to get <math>\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}\geq x+y+z.</math>)<br />
<br />
Equality occurs if and only if <math>\frac{x^2}{y}=2x-y</math>, <math>\frac{y^2}{z}=2y-z</math>, and <math>\frac{z^2}{x}=2z-x</math> at the same time, since if they are not all equal then the left side will be greater than the right side. From there, it is easy to see that the equality case is <math>x=y=z</math>, which is <math>a=b=c</math>.<br />
<br />
~john0512<br />
== See Also == {{IMO box|year=1983|num-b=5|after=Last Problem}}</div>John0512https://artofproblemsolving.com/wiki/index.php?title=User:John0512&diff=160992User:John05122021-08-26T22:43:12Z<p>John0512: </p>
<hr />
<div>i am bad at math</div>John0512https://artofproblemsolving.com/wiki/index.php?title=User:Mathicorn&diff=160991User:Mathicorn2021-08-26T22:42:51Z<p>John0512: Undo revision 160988 by John0512 (talk)</p>
<hr />
<div>Hello<br />
I am mathicorn<br />
Check out my blog (it's the blog in my profile). <br />
I'm not op or pro, although many say so. <br />
My friends are pro <br />
<br />
there's maybe some but none like you</div>John0512https://artofproblemsolving.com/wiki/index.php?title=User:John0512&diff=160989User:John05122021-08-26T21:59:50Z<p>John0512: </p>
<hr />
<div>https://artofproblemsolving.com/wiki/index.php/User:mathicorn</div>John0512https://artofproblemsolving.com/wiki/index.php?title=User:Mathicorn&diff=160988User:Mathicorn2021-08-26T21:59:36Z<p>John0512: </p>
<hr />
<div>https://artofproblemsolving.com/wiki/index.php?title=User:john0512</div>John0512https://artofproblemsolving.com/wiki/index.php?title=2012_USAJMO_Problems/Problem_3&diff=1609682012 USAJMO Problems/Problem 32021-08-26T13:40:40Z<p>John0512: </p>
<hr />
<div>== Problem ==<br />
<br />
Let <math>a</math>, <math>b</math>, <math>c</math> be positive real numbers. Prove that<br />
<cmath>\frac{a^3 + 3b^3}{5a + b} + \frac{b^3 + 3c^3}{5b + c} + \frac{c^3 + 3a^3}{5c + a} \ge \frac{2}{3} (a^2 + b^2 + c^2).</cmath><br />
<br />
==Solution==<br />
<br />
By the [[Cauchy-Schwarz_Inequality|Cauchy-Schwarz]] inequality,<br />
<cmath>[a(5a + b) + b(5b + c) + c(5c + a)] \left( \frac{a^3}{5a + b} + \frac{b^3}{5b + c} + \frac{c^3}{5c + a} \right) \ge (a^2 + b^2 + c^2)^2,</cmath><br />
so<br />
<cmath>\frac{a^3}{5a + b} + \frac{b^3}{5b + c} + \frac{c^3}{5c + a} \ge \frac{(a^2 + b^2 + c^2)^2}{5a^2 + 5b^2 + 5c^2 + ab + ac + bc}.</cmath><br />
Since <math>a^2 + b^2 + c^2 \ge ab + ac + bc</math>,<br />
<cmath>\frac{(a^2 + b^2 + c^2)^2}{5a^2 + 5b^2 + 5c^2 + ab + ac + bc} \ge \frac{(a^2 + b^2 + c^2)^2}{6a^2 + 6b^2 + 6c^2} = \frac{1}{6} (a^2 + b^2 + c^2).</cmath><br />
Hence,<br />
<cmath>\frac{a^3}{5a + b} + \frac{b^3}{5b + c} + \frac{c^3}{5c + a} \ge \frac{1}{6} (a^2 + b^2 + c^2).</cmath><br />
<br />
Again by the Cauchy-Schwarz inequality,<br />
<cmath>[b(5a + b) + c(5b + c) + a(5c + a)] \left( \frac{b^3}{5a + b} + \frac{c^3}{5b + c} + \frac{a^3}{5c + a} \right) \ge (a^2 + b^2 + c^2)^2,</cmath><br />
so<br />
<cmath>\frac{b^3}{5a + b} + \frac{c^3}{5b + c} + \frac{a^3}{5c + a} \ge \frac{(a^2 + b^2 + c^2)^2}{a^2 + b^2 + c^2 + 5ab + 5ac + 5bc}.</cmath><br />
Since <math>a^2 + b^2 + c^2 \ge ab + ac + bc</math>,<br />
<cmath>\frac{(a^2 + b^2 + c^2)^2}{a^2 + b^2 + c^2 + 5ab + 5ac + 5bc} \ge \frac{(a^2 + b^2 + c^2)^2}{6a^2 + 6b^2 + 6c^2} = \frac{1}{6} (a^2 + b^2 + c^2).</cmath><br />
Hence,<br />
<cmath>\frac{b^3}{5a + b} + \frac{c^3}{5b + c} + \frac{a^3}{5c + a} \ge \frac{1}{6} (a^2 + b^2 + c^2).</cmath><br />
<br />
Therefore,<br />
<cmath>\frac{a^3 + 3b^3}{5a + b} + \frac{b^3 + 3c^3}{5b + c} + \frac{c^3 + 3a^3}{5c + a} \ge \frac{1 + 3}{6} (a^2 + b^2 + c^2) = \frac{2}{3} (a^2 + b^2 + c^2).</cmath><br />
==Solution 2==<br />
Titu's Lemma: The sum of multiple fractions in the form <math>\frac{a_n^2}{b_n}</math> where <math>a_n</math> and <math>b_n</math> are sequences of real numbers is greater than of equal to the square of the sum of all <math>a_i</math> divided by the sum of all <math>b_i</math>, where i is a whole number less than n+1. Titu's Lemma can be proved using the Cauchy-Schwarz Inequality after multiplying out the denominator of the RHS.<br />
<br />
Consider the LHS of the proposed inequality. Split up the numerators and multiply both sides of the fraction by either a or 3a to make the LHS<br />
<cmath>\sum_{cyc} \frac {a^4} {5a^2+ab}+\sum_{cyc} \frac {9a^4} {15ac+3a^2}</cmath><br />
(Cyclic notation is a special notation in which all permutations of (a,b,c) is the summation command.)<br />
<br />
Then use Titu's Lemma on all terms: <cmath>\sum_{cyc} \frac {a^4} {5a^2+ab} +\sum_{cyc} \frac {9a^4} {15ac+3a^2} \ge \frac {16(a^2+b^2+c^2)^2} {8(a^2+b^2+c^2)+16(ab+bc+ca)} \ge \frac {16(a^2 + b^2 + c^2)^2}{24(a^2 + b^2 + c^2)} = \frac{2}{3} (a^2 + b^2 + c^2) </cmath> owing to the fact that <math>a^2+b^2+c^2 \ge ab+bc+ca</math>, which is actually equivalent to <math>(a-b)^2 + (b-c)^2 + (c-a)^2 \ge 0</math>!<br />
<br />
==Solution 3==<br />
We proceed to prove that<br />
<cmath>\frac{a^3 + 3b^3}{5a + b} \ge -\frac{1}{36} a^2 + \frac{25}{36} b^2 </cmath><br />
<br />
(then the inequality in question is just the cyclic sum of both sides, since <cmath>\sum_{cyc} (-\frac{1}{36} a^2 + \frac{25}{36} b^2) = \frac{24}{36}\sum_{cyc} a^2 = \frac{2}{3} (a^2+b^2+c^2)</cmath><br />
)<br />
<br />
Indeed, by AP-GP, we have <br />
<br />
<cmath> 41 (a^3 + b^3+b^3) \ge 41 \cdot 3 ab^2 </cmath><br />
<br />
and<br />
<br />
<cmath> b^3 + a^2b \ge 2 ab^2</cmath><br />
<br />
Summing up, we have<br />
<br />
<cmath> 41a^3 + 83b^3 + a^2 b \ge 125 ab^2</cmath><br />
<br />
which is equivalent to:<br />
<br />
<cmath> 36(a^3 + 3b^3) \ge (5a + b)(-a^2 + 25b^2) </cmath><br />
<br />
Dividing <math>36(5a+b)</math> from both sides, the desired inequality is proved.<br />
<br />
--[[User:Lightest|Lightest]] 15:31, 7 May 2012 (EDT)<br />
<br />
==Solution 4==<br />
By Cauchy-Schwarz, <br />
<br />
<cmath>\left(\sum_{cyc} \dfrac{a^3}{5a + b} + \dfrac{b^3}{5a + b} + \dfrac{b^3}{5a + b} + \dfrac{b^3}{5a + b} \right) \ge \dfrac{\left( \sum_{cyc} a^2 + b^2 + b^2 + b^2 \right)^2}{ \left( \sum_{cyc} a(5a + b) + b(5a + b) + b(5a + b) + b(5a + b) \right) }</cmath><br />
<br />
<cmath>= \dfrac{\left(4a^2 + 4b^2 + 4c^2\right)^2}{\left(8a^2 + 8b^2 + 8c^2\right) + \left(16ab + 16bc + 16ca\right)}</cmath><br />
<br />
<cmath>\ge \dfrac{16\left(a^2 + b^2 + c^2\right)^2}{\left(8a^2 + 8b^2 + 8c^2\right) + \left(8a^2 + 8b^2\right) + \left(8b^2 + 8c^2\right) + \left(8c^2 + 8a^2\right)}</cmath> (by AM-GM) <cmath>= \dfrac{2}{3}\left(a^2 + b^2 + c^2\right)</cmath> as desired.<br />
<br />
==Solution 5==<br />
<br />
We note that if we can prove that there exists a real number <math>x</math> such that <cmath>\frac{a^3+3b^3}{5a+b}\geq xa^2+(\frac{2}{3}-x)b^2</cmath> for all positive reals <math>a</math> and <math>b</math>, then we are done since both sides are a cyclic sum of <math>a,b</math>, and <math>c</math>. Note that if we multiply both <math>a</math> and <math>b</math> by a constant <math>r</math>, the left and right sides will both multiply by <math>r^2</math> (since the numerator on the left will multiply by <math>r^3</math> and the denominator will multiply by <math>r</math>. This means that if the inequality is satisfied for an ordered pair <math>(a,b)</math>, then it is also satisfied for <math>(ar,br)</math> for any positive real <math>r</math>. Thus, without loss of generality we can let <math>b=1</math>,and our inequality becomes <cmath>\frac{a^3+3}{5a+1}\geq a^2x+\frac{2}{3}-x.</cmath> Expanding this and rearraingingm we get <math>a^3(1-5x)+a^2(-x)+a(5x-\frac{10}{3})+x+\frac{7}{3}\geq0</math>. Since <math>1</math> is a root of the left side, we can factor this as <cmath>(a-1)(a^2(1-5x)+a(1-6x)-x-\frac{7}{3})\geq0.</cmath> Note that for this to always be nonnegative, we must have <math>a^2(1-5x)+a(1-6x)-x-\frac{7}{3}</math> be positive when <math>a>1</math> and negative when <math>a<1</math>. Thus, it must have a root at <math>a=1</math>, so we plug in <math>a=1</math> is a root and find that we must have <math>x=-\frac{1}{36}</math>. When <math>x=-\frac{1}{36}</math>, the expression factors as <math>(a-1)^2(\frac{41}{36}a+\frac{83}{36})\geq0</math> which obviously is nonnegative for all positive reals <math>a</math>, so we are done.<br />
<br />
~john0512, bronzetruck2016<br />
==See Also==<br />
*[[USAJMO Problems and Solutions]]<br />
<br />
{{USAJMO newbox|year=2012|num-b=2|num-a=4}}<br />
{{MAA Notice}}</div>John0512https://artofproblemsolving.com/wiki/index.php?title=2015_AIME_I_Problems/Problem_3&diff=1591662015 AIME I Problems/Problem 32021-07-26T18:23:09Z<p>John0512: </p>
<hr />
<div>==Problem==<br />
<br />
There is a prime number <math>p</math> such that <math>16p+1</math> is the cube of a positive integer. Find <math>p</math>.<br />
<br />
== Video Solution ==<br />
https://youtu.be/3bRjcrkd5mQ?t=1096<br />
<br />
~ pi_is_3.14<br />
<br />
==Solution 1==<br />
<br />
Let the positive integer mentioned be <math>a</math>, so that <math>a^3 = 16p+1</math>. Note that <math>a</math> must be odd, because <math>16p+1</math> is odd.<br />
<br />
Rearrange this expression and factor the left side (this factoring can be done using <math>(a^3-b^3) = (a-b)(a^2+a b+b^2)</math> or synthetic divison once it is realized that <math>a = 1</math> is a root):<br />
<br />
<cmath>\begin{align*}<br />
a^3-1 &= 16p\\<br />
(a-1)(a^2+a+1) &= 16p\\<br />
\end{align*}</cmath><br />
<br />
Because <math>a</math> is odd, <math>a-1</math> is even and <math>a^2+a+1</math> is odd. If <math>a^2+a+1</math> is odd, <math>a-1</math> must be some multiple of <math>16</math>. However, for <math>a-1</math> to be any multiple of <math>16</math> other than <math>16</math> would mean <math>p</math> is not a prime. Therefore, <math>a-1 = 16</math> and <math>a = 17</math>.<br />
<br />
Then our other factor, <math>a^2+a+1</math>, is the prime <math>p</math>:<br />
<br />
<cmath>\begin{align*}<br />
(a-1)(a^2+a+1) &= 16p\\<br />
(17-1)(17^2+17+1) &=16p\\<br />
p = 289+17+1 &= \boxed{307}<br />
\end{align*}</cmath><br />
<br />
==Solution 2 (Similar to 1)==<br />
<br />
Observe that this is the same as <math>16p+1=n^3</math> for some integer <math>n</math>.<br />
So:<br />
<br />
<cmath>\begin{align*}<br />
16p &= n^3-1\\<br />
16p &= n^3-1^3\\<br />
16p &= (n-1)(n^2+n+1)\\<br />
\end{align*}</cmath><br />
<br />
Observe that either <math>p=n-1</math> or <math>p=n^2+n+1</math> because <math>p</math> and <math>16</math> share no factors (<math>p</math> can't be <math>2</math>).<br />
Let <math>p=n-1</math>.<br />
Then:<br />
<br />
<cmath>\begin{align*}<br />
p &= n-1\\<br />
16 &= n^2+n+1\\<br />
n^2+n &= 15\\<br />
n(n+1) &= 15\\<br />
\end{align*}</cmath><br />
<br />
Which is impossible for integer n. So <math>p=n^2+n+1</math> and<br />
<br />
<cmath>\begin{align*}<br />
16 &= n-1\\<br />
n &= 17\\<br />
p &= 17^2+17+1\\<br />
p = 289+17+1 &= \boxed{307}\\<br />
\end{align*}</cmath> - firebolt360<br />
<br />
==Solution 3==<br />
<br />
Since <math>16p+1</math> is odd, let <math>16p+1 = (2a+1)^3</math>. Therefore, <math>16p+1 = (2a+1)^3 = 8a^3+12a^2+6a+1</math>. From this, we get <math>8p=a(4a^2+6a+3)</math>. We know <math>p</math> is a prime number and it is not an even number. Since <math>4a^2+6a+3</math> is an odd number, we know that <math>a=8</math>.<br />
<br />
Therefore, <math>p=4a^2+6a+3=4*8^2+6*8+3=\boxed{307}</math>.<br />
<br />
==Solution 4==<br />
<br />
Let <math>16p+1=a^3</math>. Realize that <math>a</math> congruent to <math>1\mod 4</math>, so let <math>a=4n+1</math>. Expansion, then division by 4, gets <math>16n^3+12n^2+3n=4p</math>. Clearly <math>n=4m</math> for some <math>m</math>. Substitution and another division by 4 gets <math>256m^3+48m^4+3m=p</math>. Since <math>p</math> is prime and there is a factor of <math>m</math> in the LHS, <math>m=1</math>. Therefore, <math>p=\boxed{307}</math>.<br />
<br />
==Solution 5==<br />
<br />
Notice that <math>16p+1</math> must be in the form <math>(a+1)^3 = a^3 + 3a^2 + 3a + 1</math>. Thus <math>16p = a^3 + 3a^2 + 3a</math>, or <math>16p = a\cdot (a^2 + 3a + 3)</math>. Since <math>p</math> must be prime, we either have <math>p = a</math> or <math>a = 16</math>. Upon further inspection and/or using the quadratic formula, we can deduce <math>p \neq a</math>. Thus we have <math>a = 16</math>, and <math>p = 16^2 + 3\cdot 16 + 3 = \boxed{307}</math>.<br />
<br />
==Solution 6==<br />
Notice that the cube 16p+1 is equal to is congruent to 1 mod 16. The only cubic numbers that leave a residue of 1 mod 16 are 1 and 15. <br />
Case one: The cube is of the form 16k+1-->Plugging this in, and taking note that p is prime and has only 1 factor gives p=307<br />
Case two: The cube is of the form 16k+15--> Plugging this in, we quickly realize that this case is invalid, as that implies p is even, and p=2 doesn't work here<br />
<br />
Hence, <math>p=\boxed{307}</math> is our only answer<br />
<br />
<br />
pi_is_3.141<br />
<br />
==Solution 7==<br />
We make a table of <math>x^3</math> (mod 16). We notice that only <math>x\cong1</math> (mod 16) gives <math>x^3\cong1</math> (mod 16). Then, we try <math>16p+1=1</math>, which obviously doesn't work. However, <math>16p+1=4913</math> gives <math>p=\fbox{307}</math>.<br />
<br />
~john0512<br />
<br />
== See also ==<br />
{{AIME box|year=2015|n=I|num-b=2|num-a=4}}<br />
{{MAA Notice}}<br />
[[Category:Introductory Number Theory Problems]]</div>John0512https://artofproblemsolving.com/wiki/index.php?title=2018_AIME_I_Problems/Problem_4&diff=1574642018 AIME I Problems/Problem 42021-07-06T20:51:01Z<p>John0512: /* Solution 12 (Double Angle Identity) */</p>
<hr />
<div>==Problem 4==<br />
In <math>\triangle ABC, AB = AC = 10</math> and <math>BC = 12</math>. Point <math>D</math> lies strictly between <math>A</math> and <math>B</math> on <math>\overline{AB}</math> and point <math>E</math> lies strictly between <math>A</math> and <math>C</math> on <math>\overline{AC}</math> so that <math>AD = DE = EC</math>. Then <math>AD</math> can be expressed in the form <math>\dfrac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. Find <math>p+q</math>.<br />
<br />
==Solution 1==<br />
<math>\cos(A) = \frac{5^2+5^2-6^2}{2*5*5} = \frac{7}{25}</math>. Let <math>M</math> be midpoint of <math>AE</math>, then <math>\frac{7}{25} = \frac{10-x}{2x} \iff x =\frac{250}{39}</math>. So, our answer is <math>250+39=\boxed{289}</math>.<br />
<br />
==Solution 1 (No Trig)==<br />
<center><br />
<asy><br />
import cse5;<br />
unitsize(10mm);<br />
pathpen=black;<br />
dotfactor=3;<br />
<br />
pair B = (0,0), A = (6,8), C = (12,0), D = (2.154,2.872), E = (8.153, 5.128), F=(7.68,5.76), G=(7.077,6.564), H=(5.6,4.3), I=(4.5,6), J=(10,2.66);<br />
pair[] dotted = {A,B,C,D,E,F,G};<br />
<br />
D(A--B);<br />
D(C--B);<br />
D(A--C);<br />
D(D--E);<br />
pathpen=dashed;<br />
D(B--F);<br />
D(D--G);<br />
<br />
dot(dotted);<br />
label("$A$",A,N);<br />
label("$B$",B,SW);<br />
label("$C$",C,SE);<br />
label("$D$",D,NW);<br />
label("$E$",E,NE);<br />
label("$F$",F,NE);<br />
label("$G$",G,NE);<br />
label("$x$",H,NW);<br />
label("$x$",I,NW);<br />
label("$x$",J,NE);<br />
</asy><br />
</center><br />
<br />
We draw the altitude from <math>B</math> to <math>\overline{AC}</math> to get point <math>F</math>. We notice that the triangle's height from <math>A</math> to <math>\overline{BC}</math> is 8 because it is a <math>3-4-5</math> Right Triangle. To find the length of <math>\overline{BF}</math>, we let <math>h</math> represent <math>\overline{BF}</math> and set up an equation by finding two ways to express the area. The equation is <math>(8)(12)=(10)(h)</math>, which leaves us with <math>h=9.6</math>. We then solve for the length <math>\overline{AF}</math>, which is done through pythagorean theorm and get <math>\overline{AF}</math> = <math>2.8</math>. We can now see that <math>\triangle AFB</math> is a <math>7-24-25</math> Right Triangle. Thus, we set <math>\overline{AG}</math> as <math>5-</math><math>\tfrac{x}{2}</math>, and yield that <math>\overline{AD}</math> <math>=</math> <math>(</math> <math>5-</math> <math>\tfrac{x}{2}</math> <math>)</math> <math>(</math> <math>\tfrac{25}{7}</math> <math>)</math>. Now, we can see <math>x</math> = <math>(</math> <math>5-</math> <math>\tfrac{x}{2}</math> <math>)</math> <math>(</math> <math>\tfrac{25}{7}</math> <math>)</math>. Solving this equation, we yield <math>39x=250</math>, or <math>x=</math> <math>\tfrac{250}{39}</math>. Thus, our final answer is <math>250+39=\boxed{289}</math>.<br />
~bluebacon008<br />
<br />
==Solution 2 (Easy Similar Triangles)==<br />
We start by adding a few points to the diagram. Call <math>F</math> the midpoint of <math>AE</math>, and <math>G</math> the midpoint of <math>BC</math>. (Note that <math>DF</math> and <math>AG</math> are altitudes of their respective triangles). We also call <math>\angle BAC = \theta</math>. Since triangle <math>ADE</math> is isosceles, <math>\angle AED = \theta</math>, and <math>\angle ADF = \angle EDF = 90 - \theta</math>. Since <math>\angle DEA = \theta</math>, <math>\angle DEC = 180 - \theta</math> and <math>\angle EDC = \angle ECD = \frac{\theta}{2}</math>. Since <math>FDC</math> is a right triangle, <math>\angle FDC = 180 - 90 - \frac{\theta}{2} = \frac{180-m}{2}</math>. <br />
<br />
Since <math>\angle BAG = \frac{\theta}{2}</math> and <math>\angle ABG = \frac{180-m}{2}</math>, triangles <math>ABG</math> and <math>CDF</math> are similar by Angle-Angle similarity. Using similar triangle ratios, we have <math>\frac{AG}{BG} = \frac{CF}{DF}</math>. <math>AG = 8</math> and <math>BG = 6</math> because there are <math>2</math> <math>6-8-10</math> triangles in the problem. Call <math>AD = x</math>. Then <math>CE = x</math>, <math>AE = 10-x</math>, and <math>EF = \frac{10-x}{2}</math>. Thus <math>CF = x + \frac{10-x}{2}</math>. Our ratio now becomes <math>\frac{8}{6} = \frac{x+ \frac{10-x}{2}}{DF}</math>. Solving for <math>DF</math> gives us <math>DF = \frac{30+3x}{8}</math>. Since <math>DF</math> is a height of the triangle <math>ADE</math>, <math>FE^2 + DF^2 = x^2</math>, or <math>DF = \sqrt{x^2 - (\frac{10-x}{2})^2}</math>. Solving the equation <math>\frac{30+3x}{8} = \sqrt{x^2 - (\frac{10-x}{2})^2}</math> gives us <math>x = \frac{250}{39}</math>, so our answer is <math>250+39 = \boxed{289}</math>.<br />
<br />
==Solution 3 (Algebra w/ Law of Cosines)==<br />
As in the diagram, let <math>DE = x</math>. Consider point <math>G</math> on the diagram shown above. Our goal is to be able to perform Pythagorean Theorem on <math>DG, GC</math>, and <math>DC</math>. Let <math>GE = 10-x</math>. Therefore, it is trivial to see that <math>GC^2 = \Big(x + \frac{10-x}{2}\Big)^2</math> (leave everything squared so that it cancels nicely at the end). By Pythagorean Theorem on Triangle <math>DGE</math>, we know that <math>DG^2 = x^2 - \Big(\frac{10-x}{2}\Big)^2</math>. Finally, we apply Law of Cosines on Triangle <math>DBC</math>. We know that <math>\cos(\angle DBC) = \frac{3}{5}</math>. Therefore, we get that <math>DC^2 = (10-x)^2 + 12^2 - 2(12)(10-x)\frac{3}{5}</math>. We can now do our final calculation:<br />
<cmath><br />
DG^2 + GC^2 = DC^2 \implies x^2 - \Big(\frac{10-x}{2}\Big)^2 + \Big(x + \frac{10-x}{2}\Big)^2 = (10-x)^2 + 12^2 - 2(12)(10-x)\frac{3}{5}<br />
</cmath><br />
After some quick cleaning up, we get<br />
<cmath><br />
30x = \frac{72}{5} + 100 \implies x = \frac{250}{39}<br />
</cmath><br />
Therefore, our answer is <math>250+39=\boxed{289}</math>.<br />
<br />
~awesome1st<br />
<br />
<br />
==Solution 4 (Coordinates)==<br />
Let <math>B = (0, 0)</math>, <math>C = (12, 0)</math>, and <math>A = (6, 8)</math>. Then, let <math>x</math> be in the interval <math>0<x<2</math> and parametrically define <math>D</math> and <math>E</math> as <math>(6-3x, 8-4x)</math> and <math>(12-3x, 4x)</math> respectively. Note that <math>AD = 5x</math>, so <math>DE = 5x</math>. This means that<br />
<cmath>\begin{align*}<br />
\sqrt{36+(8x-8)^2} &= 5x\\<br />
36+(8x-8)^2 &= 25x^2\\<br />
64x^2-128x+100 &= 25x^2\\<br />
39x^2-128x+100 &= 0\\<br />
x &= \dfrac{128\pm\sqrt{16384-15600}}{78}\\<br />
x &= \dfrac{100}{78}, 2\\<br />
\end{align*}</cmath><br />
However, since <math>2</math> is extraneous by definition, <math>x=\dfrac{50}{39}\implies AD = \dfrac{250}{39}\implies\boxed{289}</math> ~ mathwiz0803<br />
<br />
==Solution 5 (Law of Cosines)==<br />
As shown in the diagram, let <math>x</math> denote <math>\overline{AD}</math>. Let us denote the foot of the altitude of <math>A</math> to <math>\overline{BC}</math> as <math>F</math>. Note that <math>\overline{AE}</math> can be expressed as <math>10-x</math> and <math>\triangle{ABF}</math> is a <math>6-8-10</math> triangle . Therefore, <math>\sin(\angle{BAF})=\frac{3}{5}</math> and <math>\cos(\angle{BAF})=\frac{4}{5}</math>. Before we can proceed with the Law of Cosines, we must determine <math>\cos(\angle{BAC})=\cos(2\cdot \angle{BAF})=\cos^2(\angle{BAF})-\sin^2(\angle{BAF})=\frac{7}{25}</math>. Using LOC, we can write the following statement:<br />
<cmath>(\overline{DE})^2=(\overline{AD})^2+\overline{AE}^2-2(\overline{AD})(\overline{AE})\cos(\angle{BAC})\implies</cmath><br />
<cmath>x^2=x^2+(10-x)^2-2(x)(10-x)\left(\frac{7}{25}\right)\implies</cmath><br />
<cmath>0=(10-x)^2-\frac{14x}{25}(10-x)\implies</cmath><br />
<cmath>0=10-x-\frac{14x}{25}\implies</cmath><br />
<cmath>10=\frac{39x}{25}\implies x=\frac{250}{39}</cmath><br />
Thus, the desired answer is <math>\boxed{289}</math> ~ blitzkrieg21<br />
<br />
==Solution 6==<br />
In isosceles triangle, draw the altitude from <math>D</math> onto <math>\overline{AD}</math>. Let the point of intersection be <math>X</math>. Clearly, <math>AE=10-AD</math>, and hence <math>AX=\frac{10-AD}{2}</math>.<br />
<br />
Now, we recognise that the perpendicular from <math>A</math> onto <math>\overline{AD}</math> gives us two <math>6</math>-<math>8</math>-<math>10</math> triangles. So, we calculate <math>\sin \angle ABC=\frac{4}{5}</math> and <math>\cos \angle ABC=\frac{3}{5}</math><br />
<br />
<math>\angle BAC = 180-2\cdot\angle ABC</math>. And hence,<br />
<br />
<math>\cos \angle BAC = \cos \angle (180-2\cdot\angle ABC)<br />
= -\cos (2\cdot\angle ABC)<br />
= \sin^2 \angle ABC - \cos^2 \angle ABC<br />
= 2\cos^2 \angle ABC - 1<br />
= \frac{32}{25}-\frac{25}{25}=\frac{7}{25}</math><br />
<br />
Inspecting <math>\triangle ADX</math> gives us <math>\cos \angle BAC = \frac{\frac{10-x}{2}}{x} = \frac{10-x}{2x}</math><br />
Solving the equation <math>\frac{10-x}{2x}=\frac{7}{25}</math> gives <math>x= \frac{250}{39} \implies\boxed{289}</math><br />
<br />
~novus677<br />
<br />
==Solution 7 (Fastest via Law of Cosines)==<br />
We can have 2 Law of Cosines applied on <math>\angle A</math> (one from <math>\triangle ADE</math> and one from <math>\triangle ABC</math>),<br />
<br />
<math>x^2=x^2+(10-x)^2-2(x)(10-x)\cdot \cos{A}</math> and <math>12^2=10^2+10^2-2(10)(10)\cdot \cos{A}</math><br />
<br />
Solving for <math>\cos{A}</math> in both equations, we get<br />
<br />
<math>\cos{A} = \frac{(10-x)^2}{(2)(10-x)(x)}</math> and <math>cos A = \frac{7}{25} \implies \frac{(10-x)^2}{(2)(10-x)(x)} = \frac{7}{25} \implies x = \frac{250}{39}</math>, so the answer is <math>\boxed {289}</math> <br />
<br />
'''-RootThreeOverTwo'''<br />
<br />
==Solution 8 (Easiest way- Coordinates without bash)==<br />
Let <math>B=(0, 0)</math>, and <math>C=(12, 0)</math>. From there, we know that <math>A=(6, 8)</math>, so line <math>AB</math> is <math>y=\frac{4}{3}x</math>. Hence, <math>D=(a, \frac{4}{3}a)</math> for some <math>a</math>, and <math>BD=\frac{5}{3}a</math> so <math>AD=10-\frac{5}{3}a</math>. Now, notice that by symmetry, <math>E=(6+a, 8-\frac{4}{3}a)</math>, so <math>ED^2=6^2+(8-\frac{8}{3}a)^2</math>. Because <math>AD=ED</math>, we now have <math>(10-\frac{5}{3})^2=6^2+(8-\frac{8}{3}a)^2</math>, which simplifies to <math>\frac{64}{9}a^2-\frac{128}{3}a+100=\frac{25}{9}a^2-\frac{100}{3}a+100</math>, so <math>\frac{39}{9}a=\frac{13}{3}a=\frac{28}{3}</math>, and <math>a=\frac{28}{13}</math>.<br />
It follows that <math>AD=10-\frac{5}{3}\times\frac{28}{13}=10-\frac{140}{39}=\frac{390-140}{39}=\frac{250}{39}</math>, and our answer is <math>250+39=\boxed{289}</math>.<br />
<br />
-Stormersyle<br />
<br />
== Solution 9 Even Faster Law of Cosines(1 variable equation)==<br />
<br />
Doing law of cosines we know that <math>\cos A</math> is <math>\frac{7}{25}.</math>* Dropping the perpendicular from <math>D</math> to <math>AE</math> we get that <cmath>\frac{10-x}{2}=\frac{7x}{25}.</cmath> <br />
Solving for <math>x</math> we get <math>\frac{250}{39}</math> so our answer is <math>289</math>.<br />
<br />
-harsha12345<br />
<br />
* It is good to remember that doubling the smallest angle of a 3-4-5 triangle gives the larger (not right) angle in a 7-24-25 triangle.<br />
<br />
== Solution 10 (Law of Sines)==<br />
<br />
Let's label <math>\angle A = \theta</math> and <math>\angle ECD = \alpha</math>. Using isosceles triangle properties and the triangle angle sum equation, we get <cmath>180-(180-2\theta+\alpha) + \frac{180-\theta}{2} + \left(\frac{180-\theta}{2} - \alpha\right) = 180.</cmath> Solving, we find <math>\theta = 2 \alpha</math>. <br />
<br />
<br />
Relabelling our triangle, we get <math>\angle ABC = 90 - \alpha</math>. Dropping an altitude from <math>A</math> to <math>BC</math> and using the Pythagorean theorem, we find <math>[ABC] = 48</math>. Using the sine area formula, we see <math>\frac12 \cdot 10 \cdot 12 \cdot \sin(90-\alpha) = 48</math>. Plugging in our sine angle cofunction identity, <math>\sin(90-\alpha) = \cos(\alpha)</math>, we get <math>\alpha = \cos{^{-1}}{\frac45}</math>. <br />
<br />
<br />
Now, using the Law of Sines on <math>\triangle ADE</math>, we get <cmath>\frac{\sin{2\alpha}}{\frac{p}{q}} = \frac{\sin{(180-4\alpha)}}{10-\frac{p}{q}}.</cmath> After applying numerous trigonometric and algebraic tricks, identities, and simplifications, such as <math>\sin{(180-4\alpha)}=\sin{4\alpha}</math> and <math>\sin{\left(\cos{^{-1}}{\frac45}\right)} = \frac35</math>, we find <math>\frac{p}{q} = \frac{10\sin{2\alpha}}{\sin{4\alpha}+\sin{2\alpha}} = \frac{250}{39}</math>. <br />
<br />
<br />
<br />
Therefore, our answer is <math>250 + 39 = \boxed{289}</math>.<br />
<br />
<br />
~Tiblis<br />
<br />
== Solution 11 (Trigonometry)==<br />
We start by labelling a few angles (all of them in degrees). Let <math>\angle{BAC}=2\alpha = \angle{AED}, \angle{EDC}=\angle{ECD}=\alpha, \angle{DEC}=180-2\alpha, \angle{BDC}=3\alpha, \angle{DCB}=90-2\alpha, \angle{DBC}=90-\alpha</math>. Also let <math>AD=a</math>. By sine rule in <math>\triangle{ADE},</math> we get <math>\frac{a}{\sin{2\alpha}}=\frac{10-a}{\sin{4\alpha}} \implies \cos{2\alpha}=\frac{5}{a}-\frac{1}{2}</math><br />
Using sine rule in <math>\triangle{ABC}</math>, we get <math>\sin{\alpha}=\frac{3}{5}</math>. Hence we get <math>\cos{2\alpha}=1-2\sin^2{\alpha}=1-\frac{18}{25}=\frac{7}{25}</math>. Hence <math>\frac{5}{a}=\frac{1}{2}+\frac{7}{25}=\frac{39}{50} \implies a=\frac{250}{39}</math>. Therefore, our answer is <math>\boxed{289}</math><br />
<br />
Alternatively, use sine rule in <math>\triangle{BDC}</math>. (It’s easier)<br />
<br />
~Prabh1512<br />
<br />
== Solution 12 (Double Angle Identity)==<br />
<br />
We let <math>AD=x</math>. Then, angle <math>A</math> is <math>2\sin^{-1}(\frac{3}{5})</math> and so is angle <math>AED</math>. We note that <math>AE=10-x</math>. We drop an altitude from <math>D</math> to <math>AE</math>, and we call the foot <math>F</math>. We note that <math>AF=\frac{10-x}{2}</math>. Using the double angle identity, we have <math>\sin(2\sin^{-1}(\frac{3}{5}))=2(\frac{3}{5})(\frac{4}{5})=\frac{24}{25}.</math> Therefore, <math>DG=\frac{24}{25}AD.</math> We now use the Pythagorean Theorem, which gives <math>(\frac{10-x}{2})^2+(\frac{24}{25}x)^2=x^2</math>. Rearranging and simplifying, this becomes <math>429x^2-12500x+62500=0</math>. Using the quadratic formula, this is <math>\frac{12500\pm\sqrt{12500^2-250000\cdot429}}{858}</math>. We take out a <math>10000</math> from the square root and make it a <math>100</math> outside of the square root to make it simpler. We end up with <math>\frac{12500\pm7000}{858}</math>. We note that this must be less than 10 to ensure that <math>10-x</math> is positive. Therefore, we take the minus, and we get <math>\frac{5500}{858}=\frac{250}{39} \implies \fbox{289}.</math><br />
<br />
~john0512<br />
<br />
==Video Solution==<br />
<br />
https://www.youtube.com/watch?v=iE8paW_ICxw<br />
<br />
<br />
https://youtu.be/dI6uZ67Ae2s ~yofro<br />
<br />
==See Also==<br />
{{AIME box|year=2018|n=I|num-b=3|num-a=5}}<br />
{{MAA Notice}}</div>John0512https://artofproblemsolving.com/wiki/index.php?title=2018_AIME_I_Problems/Problem_4&diff=1574632018 AIME I Problems/Problem 42021-07-06T20:49:56Z<p>John0512: /* Solution 12 (Double Angle Identity) */</p>
<hr />
<div>==Problem 4==<br />
In <math>\triangle ABC, AB = AC = 10</math> and <math>BC = 12</math>. Point <math>D</math> lies strictly between <math>A</math> and <math>B</math> on <math>\overline{AB}</math> and point <math>E</math> lies strictly between <math>A</math> and <math>C</math> on <math>\overline{AC}</math> so that <math>AD = DE = EC</math>. Then <math>AD</math> can be expressed in the form <math>\dfrac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. Find <math>p+q</math>.<br />
<br />
==Solution 1==<br />
<math>\cos(A) = \frac{5^2+5^2-6^2}{2*5*5} = \frac{7}{25}</math>. Let <math>M</math> be midpoint of <math>AE</math>, then <math>\frac{7}{25} = \frac{10-x}{2x} \iff x =\frac{250}{39}</math>. So, our answer is <math>250+39=\boxed{289}</math>.<br />
<br />
==Solution 1 (No Trig)==<br />
<center><br />
<asy><br />
import cse5;<br />
unitsize(10mm);<br />
pathpen=black;<br />
dotfactor=3;<br />
<br />
pair B = (0,0), A = (6,8), C = (12,0), D = (2.154,2.872), E = (8.153, 5.128), F=(7.68,5.76), G=(7.077,6.564), H=(5.6,4.3), I=(4.5,6), J=(10,2.66);<br />
pair[] dotted = {A,B,C,D,E,F,G};<br />
<br />
D(A--B);<br />
D(C--B);<br />
D(A--C);<br />
D(D--E);<br />
pathpen=dashed;<br />
D(B--F);<br />
D(D--G);<br />
<br />
dot(dotted);<br />
label("$A$",A,N);<br />
label("$B$",B,SW);<br />
label("$C$",C,SE);<br />
label("$D$",D,NW);<br />
label("$E$",E,NE);<br />
label("$F$",F,NE);<br />
label("$G$",G,NE);<br />
label("$x$",H,NW);<br />
label("$x$",I,NW);<br />
label("$x$",J,NE);<br />
</asy><br />
</center><br />
<br />
We draw the altitude from <math>B</math> to <math>\overline{AC}</math> to get point <math>F</math>. We notice that the triangle's height from <math>A</math> to <math>\overline{BC}</math> is 8 because it is a <math>3-4-5</math> Right Triangle. To find the length of <math>\overline{BF}</math>, we let <math>h</math> represent <math>\overline{BF}</math> and set up an equation by finding two ways to express the area. The equation is <math>(8)(12)=(10)(h)</math>, which leaves us with <math>h=9.6</math>. We then solve for the length <math>\overline{AF}</math>, which is done through pythagorean theorm and get <math>\overline{AF}</math> = <math>2.8</math>. We can now see that <math>\triangle AFB</math> is a <math>7-24-25</math> Right Triangle. Thus, we set <math>\overline{AG}</math> as <math>5-</math><math>\tfrac{x}{2}</math>, and yield that <math>\overline{AD}</math> <math>=</math> <math>(</math> <math>5-</math> <math>\tfrac{x}{2}</math> <math>)</math> <math>(</math> <math>\tfrac{25}{7}</math> <math>)</math>. Now, we can see <math>x</math> = <math>(</math> <math>5-</math> <math>\tfrac{x}{2}</math> <math>)</math> <math>(</math> <math>\tfrac{25}{7}</math> <math>)</math>. Solving this equation, we yield <math>39x=250</math>, or <math>x=</math> <math>\tfrac{250}{39}</math>. Thus, our final answer is <math>250+39=\boxed{289}</math>.<br />
~bluebacon008<br />
<br />
==Solution 2 (Easy Similar Triangles)==<br />
We start by adding a few points to the diagram. Call <math>F</math> the midpoint of <math>AE</math>, and <math>G</math> the midpoint of <math>BC</math>. (Note that <math>DF</math> and <math>AG</math> are altitudes of their respective triangles). We also call <math>\angle BAC = \theta</math>. Since triangle <math>ADE</math> is isosceles, <math>\angle AED = \theta</math>, and <math>\angle ADF = \angle EDF = 90 - \theta</math>. Since <math>\angle DEA = \theta</math>, <math>\angle DEC = 180 - \theta</math> and <math>\angle EDC = \angle ECD = \frac{\theta}{2}</math>. Since <math>FDC</math> is a right triangle, <math>\angle FDC = 180 - 90 - \frac{\theta}{2} = \frac{180-m}{2}</math>. <br />
<br />
Since <math>\angle BAG = \frac{\theta}{2}</math> and <math>\angle ABG = \frac{180-m}{2}</math>, triangles <math>ABG</math> and <math>CDF</math> are similar by Angle-Angle similarity. Using similar triangle ratios, we have <math>\frac{AG}{BG} = \frac{CF}{DF}</math>. <math>AG = 8</math> and <math>BG = 6</math> because there are <math>2</math> <math>6-8-10</math> triangles in the problem. Call <math>AD = x</math>. Then <math>CE = x</math>, <math>AE = 10-x</math>, and <math>EF = \frac{10-x}{2}</math>. Thus <math>CF = x + \frac{10-x}{2}</math>. Our ratio now becomes <math>\frac{8}{6} = \frac{x+ \frac{10-x}{2}}{DF}</math>. Solving for <math>DF</math> gives us <math>DF = \frac{30+3x}{8}</math>. Since <math>DF</math> is a height of the triangle <math>ADE</math>, <math>FE^2 + DF^2 = x^2</math>, or <math>DF = \sqrt{x^2 - (\frac{10-x}{2})^2}</math>. Solving the equation <math>\frac{30+3x}{8} = \sqrt{x^2 - (\frac{10-x}{2})^2}</math> gives us <math>x = \frac{250}{39}</math>, so our answer is <math>250+39 = \boxed{289}</math>.<br />
<br />
==Solution 3 (Algebra w/ Law of Cosines)==<br />
As in the diagram, let <math>DE = x</math>. Consider point <math>G</math> on the diagram shown above. Our goal is to be able to perform Pythagorean Theorem on <math>DG, GC</math>, and <math>DC</math>. Let <math>GE = 10-x</math>. Therefore, it is trivial to see that <math>GC^2 = \Big(x + \frac{10-x}{2}\Big)^2</math> (leave everything squared so that it cancels nicely at the end). By Pythagorean Theorem on Triangle <math>DGE</math>, we know that <math>DG^2 = x^2 - \Big(\frac{10-x}{2}\Big)^2</math>. Finally, we apply Law of Cosines on Triangle <math>DBC</math>. We know that <math>\cos(\angle DBC) = \frac{3}{5}</math>. Therefore, we get that <math>DC^2 = (10-x)^2 + 12^2 - 2(12)(10-x)\frac{3}{5}</math>. We can now do our final calculation:<br />
<cmath><br />
DG^2 + GC^2 = DC^2 \implies x^2 - \Big(\frac{10-x}{2}\Big)^2 + \Big(x + \frac{10-x}{2}\Big)^2 = (10-x)^2 + 12^2 - 2(12)(10-x)\frac{3}{5}<br />
</cmath><br />
After some quick cleaning up, we get<br />
<cmath><br />
30x = \frac{72}{5} + 100 \implies x = \frac{250}{39}<br />
</cmath><br />
Therefore, our answer is <math>250+39=\boxed{289}</math>.<br />
<br />
~awesome1st<br />
<br />
<br />
==Solution 4 (Coordinates)==<br />
Let <math>B = (0, 0)</math>, <math>C = (12, 0)</math>, and <math>A = (6, 8)</math>. Then, let <math>x</math> be in the interval <math>0<x<2</math> and parametrically define <math>D</math> and <math>E</math> as <math>(6-3x, 8-4x)</math> and <math>(12-3x, 4x)</math> respectively. Note that <math>AD = 5x</math>, so <math>DE = 5x</math>. This means that<br />
<cmath>\begin{align*}<br />
\sqrt{36+(8x-8)^2} &= 5x\\<br />
36+(8x-8)^2 &= 25x^2\\<br />
64x^2-128x+100 &= 25x^2\\<br />
39x^2-128x+100 &= 0\\<br />
x &= \dfrac{128\pm\sqrt{16384-15600}}{78}\\<br />
x &= \dfrac{100}{78}, 2\\<br />
\end{align*}</cmath><br />
However, since <math>2</math> is extraneous by definition, <math>x=\dfrac{50}{39}\implies AD = \dfrac{250}{39}\implies\boxed{289}</math> ~ mathwiz0803<br />
<br />
==Solution 5 (Law of Cosines)==<br />
As shown in the diagram, let <math>x</math> denote <math>\overline{AD}</math>. Let us denote the foot of the altitude of <math>A</math> to <math>\overline{BC}</math> as <math>F</math>. Note that <math>\overline{AE}</math> can be expressed as <math>10-x</math> and <math>\triangle{ABF}</math> is a <math>6-8-10</math> triangle . Therefore, <math>\sin(\angle{BAF})=\frac{3}{5}</math> and <math>\cos(\angle{BAF})=\frac{4}{5}</math>. Before we can proceed with the Law of Cosines, we must determine <math>\cos(\angle{BAC})=\cos(2\cdot \angle{BAF})=\cos^2(\angle{BAF})-\sin^2(\angle{BAF})=\frac{7}{25}</math>. Using LOC, we can write the following statement:<br />
<cmath>(\overline{DE})^2=(\overline{AD})^2+\overline{AE}^2-2(\overline{AD})(\overline{AE})\cos(\angle{BAC})\implies</cmath><br />
<cmath>x^2=x^2+(10-x)^2-2(x)(10-x)\left(\frac{7}{25}\right)\implies</cmath><br />
<cmath>0=(10-x)^2-\frac{14x}{25}(10-x)\implies</cmath><br />
<cmath>0=10-x-\frac{14x}{25}\implies</cmath><br />
<cmath>10=\frac{39x}{25}\implies x=\frac{250}{39}</cmath><br />
Thus, the desired answer is <math>\boxed{289}</math> ~ blitzkrieg21<br />
<br />
==Solution 6==<br />
In isosceles triangle, draw the altitude from <math>D</math> onto <math>\overline{AD}</math>. Let the point of intersection be <math>X</math>. Clearly, <math>AE=10-AD</math>, and hence <math>AX=\frac{10-AD}{2}</math>.<br />
<br />
Now, we recognise that the perpendicular from <math>A</math> onto <math>\overline{AD}</math> gives us two <math>6</math>-<math>8</math>-<math>10</math> triangles. So, we calculate <math>\sin \angle ABC=\frac{4}{5}</math> and <math>\cos \angle ABC=\frac{3}{5}</math><br />
<br />
<math>\angle BAC = 180-2\cdot\angle ABC</math>. And hence,<br />
<br />
<math>\cos \angle BAC = \cos \angle (180-2\cdot\angle ABC)<br />
= -\cos (2\cdot\angle ABC)<br />
= \sin^2 \angle ABC - \cos^2 \angle ABC<br />
= 2\cos^2 \angle ABC - 1<br />
= \frac{32}{25}-\frac{25}{25}=\frac{7}{25}</math><br />
<br />
Inspecting <math>\triangle ADX</math> gives us <math>\cos \angle BAC = \frac{\frac{10-x}{2}}{x} = \frac{10-x}{2x}</math><br />
Solving the equation <math>\frac{10-x}{2x}=\frac{7}{25}</math> gives <math>x= \frac{250}{39} \implies\boxed{289}</math><br />
<br />
~novus677<br />
<br />
==Solution 7 (Fastest via Law of Cosines)==<br />
We can have 2 Law of Cosines applied on <math>\angle A</math> (one from <math>\triangle ADE</math> and one from <math>\triangle ABC</math>),<br />
<br />
<math>x^2=x^2+(10-x)^2-2(x)(10-x)\cdot \cos{A}</math> and <math>12^2=10^2+10^2-2(10)(10)\cdot \cos{A}</math><br />
<br />
Solving for <math>\cos{A}</math> in both equations, we get<br />
<br />
<math>\cos{A} = \frac{(10-x)^2}{(2)(10-x)(x)}</math> and <math>cos A = \frac{7}{25} \implies \frac{(10-x)^2}{(2)(10-x)(x)} = \frac{7}{25} \implies x = \frac{250}{39}</math>, so the answer is <math>\boxed {289}</math> <br />
<br />
'''-RootThreeOverTwo'''<br />
<br />
==Solution 8 (Easiest way- Coordinates without bash)==<br />
Let <math>B=(0, 0)</math>, and <math>C=(12, 0)</math>. From there, we know that <math>A=(6, 8)</math>, so line <math>AB</math> is <math>y=\frac{4}{3}x</math>. Hence, <math>D=(a, \frac{4}{3}a)</math> for some <math>a</math>, and <math>BD=\frac{5}{3}a</math> so <math>AD=10-\frac{5}{3}a</math>. Now, notice that by symmetry, <math>E=(6+a, 8-\frac{4}{3}a)</math>, so <math>ED^2=6^2+(8-\frac{8}{3}a)^2</math>. Because <math>AD=ED</math>, we now have <math>(10-\frac{5}{3})^2=6^2+(8-\frac{8}{3}a)^2</math>, which simplifies to <math>\frac{64}{9}a^2-\frac{128}{3}a+100=\frac{25}{9}a^2-\frac{100}{3}a+100</math>, so <math>\frac{39}{9}a=\frac{13}{3}a=\frac{28}{3}</math>, and <math>a=\frac{28}{13}</math>.<br />
It follows that <math>AD=10-\frac{5}{3}\times\frac{28}{13}=10-\frac{140}{39}=\frac{390-140}{39}=\frac{250}{39}</math>, and our answer is <math>250+39=\boxed{289}</math>.<br />
<br />
-Stormersyle<br />
<br />
== Solution 9 Even Faster Law of Cosines(1 variable equation)==<br />
<br />
Doing law of cosines we know that <math>\cos A</math> is <math>\frac{7}{25}.</math>* Dropping the perpendicular from <math>D</math> to <math>AE</math> we get that <cmath>\frac{10-x}{2}=\frac{7x}{25}.</cmath> <br />
Solving for <math>x</math> we get <math>\frac{250}{39}</math> so our answer is <math>289</math>.<br />
<br />
-harsha12345<br />
<br />
* It is good to remember that doubling the smallest angle of a 3-4-5 triangle gives the larger (not right) angle in a 7-24-25 triangle.<br />
<br />
== Solution 10 (Law of Sines)==<br />
<br />
Let's label <math>\angle A = \theta</math> and <math>\angle ECD = \alpha</math>. Using isosceles triangle properties and the triangle angle sum equation, we get <cmath>180-(180-2\theta+\alpha) + \frac{180-\theta}{2} + \left(\frac{180-\theta}{2} - \alpha\right) = 180.</cmath> Solving, we find <math>\theta = 2 \alpha</math>. <br />
<br />
<br />
Relabelling our triangle, we get <math>\angle ABC = 90 - \alpha</math>. Dropping an altitude from <math>A</math> to <math>BC</math> and using the Pythagorean theorem, we find <math>[ABC] = 48</math>. Using the sine area formula, we see <math>\frac12 \cdot 10 \cdot 12 \cdot \sin(90-\alpha) = 48</math>. Plugging in our sine angle cofunction identity, <math>\sin(90-\alpha) = \cos(\alpha)</math>, we get <math>\alpha = \cos{^{-1}}{\frac45}</math>. <br />
<br />
<br />
Now, using the Law of Sines on <math>\triangle ADE</math>, we get <cmath>\frac{\sin{2\alpha}}{\frac{p}{q}} = \frac{\sin{(180-4\alpha)}}{10-\frac{p}{q}}.</cmath> After applying numerous trigonometric and algebraic tricks, identities, and simplifications, such as <math>\sin{(180-4\alpha)}=\sin{4\alpha}</math> and <math>\sin{\left(\cos{^{-1}}{\frac45}\right)} = \frac35</math>, we find <math>\frac{p}{q} = \frac{10\sin{2\alpha}}{\sin{4\alpha}+\sin{2\alpha}} = \frac{250}{39}</math>. <br />
<br />
<br />
<br />
Therefore, our answer is <math>250 + 39 = \boxed{289}</math>.<br />
<br />
<br />
~Tiblis<br />
<br />
== Solution 11 (Trigonometry)==<br />
We start by labelling a few angles (all of them in degrees). Let <math>\angle{BAC}=2\alpha = \angle{AED}, \angle{EDC}=\angle{ECD}=\alpha, \angle{DEC}=180-2\alpha, \angle{BDC}=3\alpha, \angle{DCB}=90-2\alpha, \angle{DBC}=90-\alpha</math>. Also let <math>AD=a</math>. By sine rule in <math>\triangle{ADE},</math> we get <math>\frac{a}{\sin{2\alpha}}=\frac{10-a}{\sin{4\alpha}} \implies \cos{2\alpha}=\frac{5}{a}-\frac{1}{2}</math><br />
Using sine rule in <math>\triangle{ABC}</math>, we get <math>\sin{\alpha}=\frac{3}{5}</math>. Hence we get <math>\cos{2\alpha}=1-2\sin^2{\alpha}=1-\frac{18}{25}=\frac{7}{25}</math>. Hence <math>\frac{5}{a}=\frac{1}{2}+\frac{7}{25}=\frac{39}{50} \implies a=\frac{250}{39}</math>. Therefore, our answer is <math>\boxed{289}</math><br />
<br />
Alternatively, use sine rule in <math>\triangle{BDC}</math>. (It’s easier)<br />
<br />
~Prabh1512<br />
<br />
== Solution 12 (Double Angle Identity)==<br />
<br />
We let <math>AD=x</math>. Then, angle <math>A</math> is <math>2\sin^{-1}(\frac{3}{5})</math> and so is angle <math>AED</math>. We note that <math>AE=10-x</math>. We drop an altitude from <math>D</math> to <math>AE</math>, and we call the foot <math>F</math>. We note that <math>AF=\frac{10-x}{2}</math>. Using the double angle identity, we have <math>\sin(2\sin^{-1}(\frac{3}{5}))=2(\frac{3}{5})(\frac{4}{5})=\frac{24}{25}.</math> Therefore, <math>DG=\frac{24}{25}AD.</math> We now use the Pythagorean Theorem, which gives <math>(\frac{10-x}{2})^2+(\frac{24}{25}x)^2=x^2</math>. Rearranging and simplifying, this becomes <math>429x^2-12500x+62500=0</math>. Using the quadratic formula, this is <math>\frac{12500\pm\sqrt{12500^2-250000\cdot429}}{858}</math>. We take out a <math>10000</math> from the square root and make it a <math>100</math> outside of the square root to make it simpler. We end up with <math>\frac{12500\pm7000}{858}</math>. We note that this must be less than 10 to ensure that <math>10-x</math> is positive. Therefore, we take the minus, and we get <math>\frac{5500}{858}=\frac{250}{39} -> \fbox{289}.</math><br />
<br />
~john0512<br />
<br />
==Video Solution==<br />
<br />
https://www.youtube.com/watch?v=iE8paW_ICxw<br />
<br />
<br />
https://youtu.be/dI6uZ67Ae2s ~yofro<br />
<br />
==See Also==<br />
{{AIME box|year=2018|n=I|num-b=3|num-a=5}}<br />
{{MAA Notice}}</div>John0512https://artofproblemsolving.com/wiki/index.php?title=2018_AIME_I_Problems/Problem_4&diff=1574622018 AIME I Problems/Problem 42021-07-06T20:49:31Z<p>John0512: /* Solution 12 (Double Angle Identity) */</p>
<hr />
<div>==Problem 4==<br />
In <math>\triangle ABC, AB = AC = 10</math> and <math>BC = 12</math>. Point <math>D</math> lies strictly between <math>A</math> and <math>B</math> on <math>\overline{AB}</math> and point <math>E</math> lies strictly between <math>A</math> and <math>C</math> on <math>\overline{AC}</math> so that <math>AD = DE = EC</math>. Then <math>AD</math> can be expressed in the form <math>\dfrac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. Find <math>p+q</math>.<br />
<br />
==Solution 1==<br />
<math>\cos(A) = \frac{5^2+5^2-6^2}{2*5*5} = \frac{7}{25}</math>. Let <math>M</math> be midpoint of <math>AE</math>, then <math>\frac{7}{25} = \frac{10-x}{2x} \iff x =\frac{250}{39}</math>. So, our answer is <math>250+39=\boxed{289}</math>.<br />
<br />
==Solution 1 (No Trig)==<br />
<center><br />
<asy><br />
import cse5;<br />
unitsize(10mm);<br />
pathpen=black;<br />
dotfactor=3;<br />
<br />
pair B = (0,0), A = (6,8), C = (12,0), D = (2.154,2.872), E = (8.153, 5.128), F=(7.68,5.76), G=(7.077,6.564), H=(5.6,4.3), I=(4.5,6), J=(10,2.66);<br />
pair[] dotted = {A,B,C,D,E,F,G};<br />
<br />
D(A--B);<br />
D(C--B);<br />
D(A--C);<br />
D(D--E);<br />
pathpen=dashed;<br />
D(B--F);<br />
D(D--G);<br />
<br />
dot(dotted);<br />
label("$A$",A,N);<br />
label("$B$",B,SW);<br />
label("$C$",C,SE);<br />
label("$D$",D,NW);<br />
label("$E$",E,NE);<br />
label("$F$",F,NE);<br />
label("$G$",G,NE);<br />
label("$x$",H,NW);<br />
label("$x$",I,NW);<br />
label("$x$",J,NE);<br />
</asy><br />
</center><br />
<br />
We draw the altitude from <math>B</math> to <math>\overline{AC}</math> to get point <math>F</math>. We notice that the triangle's height from <math>A</math> to <math>\overline{BC}</math> is 8 because it is a <math>3-4-5</math> Right Triangle. To find the length of <math>\overline{BF}</math>, we let <math>h</math> represent <math>\overline{BF}</math> and set up an equation by finding two ways to express the area. The equation is <math>(8)(12)=(10)(h)</math>, which leaves us with <math>h=9.6</math>. We then solve for the length <math>\overline{AF}</math>, which is done through pythagorean theorm and get <math>\overline{AF}</math> = <math>2.8</math>. We can now see that <math>\triangle AFB</math> is a <math>7-24-25</math> Right Triangle. Thus, we set <math>\overline{AG}</math> as <math>5-</math><math>\tfrac{x}{2}</math>, and yield that <math>\overline{AD}</math> <math>=</math> <math>(</math> <math>5-</math> <math>\tfrac{x}{2}</math> <math>)</math> <math>(</math> <math>\tfrac{25}{7}</math> <math>)</math>. Now, we can see <math>x</math> = <math>(</math> <math>5-</math> <math>\tfrac{x}{2}</math> <math>)</math> <math>(</math> <math>\tfrac{25}{7}</math> <math>)</math>. Solving this equation, we yield <math>39x=250</math>, or <math>x=</math> <math>\tfrac{250}{39}</math>. Thus, our final answer is <math>250+39=\boxed{289}</math>.<br />
~bluebacon008<br />
<br />
==Solution 2 (Easy Similar Triangles)==<br />
We start by adding a few points to the diagram. Call <math>F</math> the midpoint of <math>AE</math>, and <math>G</math> the midpoint of <math>BC</math>. (Note that <math>DF</math> and <math>AG</math> are altitudes of their respective triangles). We also call <math>\angle BAC = \theta</math>. Since triangle <math>ADE</math> is isosceles, <math>\angle AED = \theta</math>, and <math>\angle ADF = \angle EDF = 90 - \theta</math>. Since <math>\angle DEA = \theta</math>, <math>\angle DEC = 180 - \theta</math> and <math>\angle EDC = \angle ECD = \frac{\theta}{2}</math>. Since <math>FDC</math> is a right triangle, <math>\angle FDC = 180 - 90 - \frac{\theta}{2} = \frac{180-m}{2}</math>. <br />
<br />
Since <math>\angle BAG = \frac{\theta}{2}</math> and <math>\angle ABG = \frac{180-m}{2}</math>, triangles <math>ABG</math> and <math>CDF</math> are similar by Angle-Angle similarity. Using similar triangle ratios, we have <math>\frac{AG}{BG} = \frac{CF}{DF}</math>. <math>AG = 8</math> and <math>BG = 6</math> because there are <math>2</math> <math>6-8-10</math> triangles in the problem. Call <math>AD = x</math>. Then <math>CE = x</math>, <math>AE = 10-x</math>, and <math>EF = \frac{10-x}{2}</math>. Thus <math>CF = x + \frac{10-x}{2}</math>. Our ratio now becomes <math>\frac{8}{6} = \frac{x+ \frac{10-x}{2}}{DF}</math>. Solving for <math>DF</math> gives us <math>DF = \frac{30+3x}{8}</math>. Since <math>DF</math> is a height of the triangle <math>ADE</math>, <math>FE^2 + DF^2 = x^2</math>, or <math>DF = \sqrt{x^2 - (\frac{10-x}{2})^2}</math>. Solving the equation <math>\frac{30+3x}{8} = \sqrt{x^2 - (\frac{10-x}{2})^2}</math> gives us <math>x = \frac{250}{39}</math>, so our answer is <math>250+39 = \boxed{289}</math>.<br />
<br />
==Solution 3 (Algebra w/ Law of Cosines)==<br />
As in the diagram, let <math>DE = x</math>. Consider point <math>G</math> on the diagram shown above. Our goal is to be able to perform Pythagorean Theorem on <math>DG, GC</math>, and <math>DC</math>. Let <math>GE = 10-x</math>. Therefore, it is trivial to see that <math>GC^2 = \Big(x + \frac{10-x}{2}\Big)^2</math> (leave everything squared so that it cancels nicely at the end). By Pythagorean Theorem on Triangle <math>DGE</math>, we know that <math>DG^2 = x^2 - \Big(\frac{10-x}{2}\Big)^2</math>. Finally, we apply Law of Cosines on Triangle <math>DBC</math>. We know that <math>\cos(\angle DBC) = \frac{3}{5}</math>. Therefore, we get that <math>DC^2 = (10-x)^2 + 12^2 - 2(12)(10-x)\frac{3}{5}</math>. We can now do our final calculation:<br />
<cmath><br />
DG^2 + GC^2 = DC^2 \implies x^2 - \Big(\frac{10-x}{2}\Big)^2 + \Big(x + \frac{10-x}{2}\Big)^2 = (10-x)^2 + 12^2 - 2(12)(10-x)\frac{3}{5}<br />
</cmath><br />
After some quick cleaning up, we get<br />
<cmath><br />
30x = \frac{72}{5} + 100 \implies x = \frac{250}{39}<br />
</cmath><br />
Therefore, our answer is <math>250+39=\boxed{289}</math>.<br />
<br />
~awesome1st<br />
<br />
<br />
==Solution 4 (Coordinates)==<br />
Let <math>B = (0, 0)</math>, <math>C = (12, 0)</math>, and <math>A = (6, 8)</math>. Then, let <math>x</math> be in the interval <math>0<x<2</math> and parametrically define <math>D</math> and <math>E</math> as <math>(6-3x, 8-4x)</math> and <math>(12-3x, 4x)</math> respectively. Note that <math>AD = 5x</math>, so <math>DE = 5x</math>. This means that<br />
<cmath>\begin{align*}<br />
\sqrt{36+(8x-8)^2} &= 5x\\<br />
36+(8x-8)^2 &= 25x^2\\<br />
64x^2-128x+100 &= 25x^2\\<br />
39x^2-128x+100 &= 0\\<br />
x &= \dfrac{128\pm\sqrt{16384-15600}}{78}\\<br />
x &= \dfrac{100}{78}, 2\\<br />
\end{align*}</cmath><br />
However, since <math>2</math> is extraneous by definition, <math>x=\dfrac{50}{39}\implies AD = \dfrac{250}{39}\implies\boxed{289}</math> ~ mathwiz0803<br />
<br />
==Solution 5 (Law of Cosines)==<br />
As shown in the diagram, let <math>x</math> denote <math>\overline{AD}</math>. Let us denote the foot of the altitude of <math>A</math> to <math>\overline{BC}</math> as <math>F</math>. Note that <math>\overline{AE}</math> can be expressed as <math>10-x</math> and <math>\triangle{ABF}</math> is a <math>6-8-10</math> triangle . Therefore, <math>\sin(\angle{BAF})=\frac{3}{5}</math> and <math>\cos(\angle{BAF})=\frac{4}{5}</math>. Before we can proceed with the Law of Cosines, we must determine <math>\cos(\angle{BAC})=\cos(2\cdot \angle{BAF})=\cos^2(\angle{BAF})-\sin^2(\angle{BAF})=\frac{7}{25}</math>. Using LOC, we can write the following statement:<br />
<cmath>(\overline{DE})^2=(\overline{AD})^2+\overline{AE}^2-2(\overline{AD})(\overline{AE})\cos(\angle{BAC})\implies</cmath><br />
<cmath>x^2=x^2+(10-x)^2-2(x)(10-x)\left(\frac{7}{25}\right)\implies</cmath><br />
<cmath>0=(10-x)^2-\frac{14x}{25}(10-x)\implies</cmath><br />
<cmath>0=10-x-\frac{14x}{25}\implies</cmath><br />
<cmath>10=\frac{39x}{25}\implies x=\frac{250}{39}</cmath><br />
Thus, the desired answer is <math>\boxed{289}</math> ~ blitzkrieg21<br />
<br />
==Solution 6==<br />
In isosceles triangle, draw the altitude from <math>D</math> onto <math>\overline{AD}</math>. Let the point of intersection be <math>X</math>. Clearly, <math>AE=10-AD</math>, and hence <math>AX=\frac{10-AD}{2}</math>.<br />
<br />
Now, we recognise that the perpendicular from <math>A</math> onto <math>\overline{AD}</math> gives us two <math>6</math>-<math>8</math>-<math>10</math> triangles. So, we calculate <math>\sin \angle ABC=\frac{4}{5}</math> and <math>\cos \angle ABC=\frac{3}{5}</math><br />
<br />
<math>\angle BAC = 180-2\cdot\angle ABC</math>. And hence,<br />
<br />
<math>\cos \angle BAC = \cos \angle (180-2\cdot\angle ABC)<br />
= -\cos (2\cdot\angle ABC)<br />
= \sin^2 \angle ABC - \cos^2 \angle ABC<br />
= 2\cos^2 \angle ABC - 1<br />
= \frac{32}{25}-\frac{25}{25}=\frac{7}{25}</math><br />
<br />
Inspecting <math>\triangle ADX</math> gives us <math>\cos \angle BAC = \frac{\frac{10-x}{2}}{x} = \frac{10-x}{2x}</math><br />
Solving the equation <math>\frac{10-x}{2x}=\frac{7}{25}</math> gives <math>x= \frac{250}{39} \implies\boxed{289}</math><br />
<br />
~novus677<br />
<br />
==Solution 7 (Fastest via Law of Cosines)==<br />
We can have 2 Law of Cosines applied on <math>\angle A</math> (one from <math>\triangle ADE</math> and one from <math>\triangle ABC</math>),<br />
<br />
<math>x^2=x^2+(10-x)^2-2(x)(10-x)\cdot \cos{A}</math> and <math>12^2=10^2+10^2-2(10)(10)\cdot \cos{A}</math><br />
<br />
Solving for <math>\cos{A}</math> in both equations, we get<br />
<br />
<math>\cos{A} = \frac{(10-x)^2}{(2)(10-x)(x)}</math> and <math>cos A = \frac{7}{25} \implies \frac{(10-x)^2}{(2)(10-x)(x)} = \frac{7}{25} \implies x = \frac{250}{39}</math>, so the answer is <math>\boxed {289}</math> <br />
<br />
'''-RootThreeOverTwo'''<br />
<br />
==Solution 8 (Easiest way- Coordinates without bash)==<br />
Let <math>B=(0, 0)</math>, and <math>C=(12, 0)</math>. From there, we know that <math>A=(6, 8)</math>, so line <math>AB</math> is <math>y=\frac{4}{3}x</math>. Hence, <math>D=(a, \frac{4}{3}a)</math> for some <math>a</math>, and <math>BD=\frac{5}{3}a</math> so <math>AD=10-\frac{5}{3}a</math>. Now, notice that by symmetry, <math>E=(6+a, 8-\frac{4}{3}a)</math>, so <math>ED^2=6^2+(8-\frac{8}{3}a)^2</math>. Because <math>AD=ED</math>, we now have <math>(10-\frac{5}{3})^2=6^2+(8-\frac{8}{3}a)^2</math>, which simplifies to <math>\frac{64}{9}a^2-\frac{128}{3}a+100=\frac{25}{9}a^2-\frac{100}{3}a+100</math>, so <math>\frac{39}{9}a=\frac{13}{3}a=\frac{28}{3}</math>, and <math>a=\frac{28}{13}</math>.<br />
It follows that <math>AD=10-\frac{5}{3}\times\frac{28}{13}=10-\frac{140}{39}=\frac{390-140}{39}=\frac{250}{39}</math>, and our answer is <math>250+39=\boxed{289}</math>.<br />
<br />
-Stormersyle<br />
<br />
== Solution 9 Even Faster Law of Cosines(1 variable equation)==<br />
<br />
Doing law of cosines we know that <math>\cos A</math> is <math>\frac{7}{25}.</math>* Dropping the perpendicular from <math>D</math> to <math>AE</math> we get that <cmath>\frac{10-x}{2}=\frac{7x}{25}.</cmath> <br />
Solving for <math>x</math> we get <math>\frac{250}{39}</math> so our answer is <math>289</math>.<br />
<br />
-harsha12345<br />
<br />
* It is good to remember that doubling the smallest angle of a 3-4-5 triangle gives the larger (not right) angle in a 7-24-25 triangle.<br />
<br />
== Solution 10 (Law of Sines)==<br />
<br />
Let's label <math>\angle A = \theta</math> and <math>\angle ECD = \alpha</math>. Using isosceles triangle properties and the triangle angle sum equation, we get <cmath>180-(180-2\theta+\alpha) + \frac{180-\theta}{2} + \left(\frac{180-\theta}{2} - \alpha\right) = 180.</cmath> Solving, we find <math>\theta = 2 \alpha</math>. <br />
<br />
<br />
Relabelling our triangle, we get <math>\angle ABC = 90 - \alpha</math>. Dropping an altitude from <math>A</math> to <math>BC</math> and using the Pythagorean theorem, we find <math>[ABC] = 48</math>. Using the sine area formula, we see <math>\frac12 \cdot 10 \cdot 12 \cdot \sin(90-\alpha) = 48</math>. Plugging in our sine angle cofunction identity, <math>\sin(90-\alpha) = \cos(\alpha)</math>, we get <math>\alpha = \cos{^{-1}}{\frac45}</math>. <br />
<br />
<br />
Now, using the Law of Sines on <math>\triangle ADE</math>, we get <cmath>\frac{\sin{2\alpha}}{\frac{p}{q}} = \frac{\sin{(180-4\alpha)}}{10-\frac{p}{q}}.</cmath> After applying numerous trigonometric and algebraic tricks, identities, and simplifications, such as <math>\sin{(180-4\alpha)}=\sin{4\alpha}</math> and <math>\sin{\left(\cos{^{-1}}{\frac45}\right)} = \frac35</math>, we find <math>\frac{p}{q} = \frac{10\sin{2\alpha}}{\sin{4\alpha}+\sin{2\alpha}} = \frac{250}{39}</math>. <br />
<br />
<br />
<br />
Therefore, our answer is <math>250 + 39 = \boxed{289}</math>.<br />
<br />
<br />
~Tiblis<br />
<br />
== Solution 11 (Trigonometry)==<br />
We start by labelling a few angles (all of them in degrees). Let <math>\angle{BAC}=2\alpha = \angle{AED}, \angle{EDC}=\angle{ECD}=\alpha, \angle{DEC}=180-2\alpha, \angle{BDC}=3\alpha, \angle{DCB}=90-2\alpha, \angle{DBC}=90-\alpha</math>. Also let <math>AD=a</math>. By sine rule in <math>\triangle{ADE},</math> we get <math>\frac{a}{\sin{2\alpha}}=\frac{10-a}{\sin{4\alpha}} \implies \cos{2\alpha}=\frac{5}{a}-\frac{1}{2}</math><br />
Using sine rule in <math>\triangle{ABC}</math>, we get <math>\sin{\alpha}=\frac{3}{5}</math>. Hence we get <math>\cos{2\alpha}=1-2\sin^2{\alpha}=1-\frac{18}{25}=\frac{7}{25}</math>. Hence <math>\frac{5}{a}=\frac{1}{2}+\frac{7}{25}=\frac{39}{50} \implies a=\frac{250}{39}</math>. Therefore, our answer is <math>\boxed{289}</math><br />
<br />
Alternatively, use sine rule in <math>\triangle{BDC}</math>. (It’s easier)<br />
<br />
~Prabh1512<br />
<br />
== Solution 12 (Double Angle Identity)==<br />
<br />
We let <math>AD=x</math>. Then, angle <math>A</math> is <math>2\sin^{-1}(\frac{3}{5})</math> and so is angle <math>AED</math>. We note that <math>AE=10-x</math>. We drop an altitude from <math>D</math> to <math>AE</math>, and we call the foot <math>F</math>. We note that <math>AF=\frac{10-x}{2}</math>. Using the double angle identity, we have <math>\sin(2sin^{-1}(\frac{3}{5}))=2(\frac{3}{5})(\frac{4}{5})=\frac{24}{25}.</math> Therefore, <math>DG=\frac{24}{25}AD.</math> We now use the Pythagorean Theorem, which gives <math>(\frac{10-x}{2})^2+(\frac{24}{25}x)^2=x^2</math>. Rearranging and simplifying, this becomes <math>429x^2-12500x+62500=0</math>. Using the quadratic formula, this is <math>\frac{12500\pm\sqrt{12500^2-250000\cdot429}}{858}</math>. We take out a <math>10000</math> from the square root and make it a <math>100</math> outside of the square root to make it simpler. We end up with <math>\frac{12500\pm7000}{858}</math>. We note that this must be less than 10 to ensure that <math>10-x</math> is positive. Therefore, we take the minus, and we get <math>\frac{5500}{858}=\frac{250}{39} -> \fbox{289}.</math><br />
<br />
~john0512<br />
<br />
==Video Solution==<br />
<br />
https://www.youtube.com/watch?v=iE8paW_ICxw<br />
<br />
<br />
https://youtu.be/dI6uZ67Ae2s ~yofro<br />
<br />
==See Also==<br />
{{AIME box|year=2018|n=I|num-b=3|num-a=5}}<br />
{{MAA Notice}}</div>John0512https://artofproblemsolving.com/wiki/index.php?title=2018_AIME_I_Problems/Problem_4&diff=1574612018 AIME I Problems/Problem 42021-07-06T20:46:43Z<p>John0512: /* Solution 12 (Double Angle Identity) */</p>
<hr />
<div>==Problem 4==<br />
In <math>\triangle ABC, AB = AC = 10</math> and <math>BC = 12</math>. Point <math>D</math> lies strictly between <math>A</math> and <math>B</math> on <math>\overline{AB}</math> and point <math>E</math> lies strictly between <math>A</math> and <math>C</math> on <math>\overline{AC}</math> so that <math>AD = DE = EC</math>. Then <math>AD</math> can be expressed in the form <math>\dfrac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. Find <math>p+q</math>.<br />
<br />
==Solution 1==<br />
<math>\cos(A) = \frac{5^2+5^2-6^2}{2*5*5} = \frac{7}{25}</math>. Let <math>M</math> be midpoint of <math>AE</math>, then <math>\frac{7}{25} = \frac{10-x}{2x} \iff x =\frac{250}{39}</math>. So, our answer is <math>250+39=\boxed{289}</math>.<br />
<br />
==Solution 1 (No Trig)==<br />
<center><br />
<asy><br />
import cse5;<br />
unitsize(10mm);<br />
pathpen=black;<br />
dotfactor=3;<br />
<br />
pair B = (0,0), A = (6,8), C = (12,0), D = (2.154,2.872), E = (8.153, 5.128), F=(7.68,5.76), G=(7.077,6.564), H=(5.6,4.3), I=(4.5,6), J=(10,2.66);<br />
pair[] dotted = {A,B,C,D,E,F,G};<br />
<br />
D(A--B);<br />
D(C--B);<br />
D(A--C);<br />
D(D--E);<br />
pathpen=dashed;<br />
D(B--F);<br />
D(D--G);<br />
<br />
dot(dotted);<br />
label("$A$",A,N);<br />
label("$B$",B,SW);<br />
label("$C$",C,SE);<br />
label("$D$",D,NW);<br />
label("$E$",E,NE);<br />
label("$F$",F,NE);<br />
label("$G$",G,NE);<br />
label("$x$",H,NW);<br />
label("$x$",I,NW);<br />
label("$x$",J,NE);<br />
</asy><br />
</center><br />
<br />
We draw the altitude from <math>B</math> to <math>\overline{AC}</math> to get point <math>F</math>. We notice that the triangle's height from <math>A</math> to <math>\overline{BC}</math> is 8 because it is a <math>3-4-5</math> Right Triangle. To find the length of <math>\overline{BF}</math>, we let <math>h</math> represent <math>\overline{BF}</math> and set up an equation by finding two ways to express the area. The equation is <math>(8)(12)=(10)(h)</math>, which leaves us with <math>h=9.6</math>. We then solve for the length <math>\overline{AF}</math>, which is done through pythagorean theorm and get <math>\overline{AF}</math> = <math>2.8</math>. We can now see that <math>\triangle AFB</math> is a <math>7-24-25</math> Right Triangle. Thus, we set <math>\overline{AG}</math> as <math>5-</math><math>\tfrac{x}{2}</math>, and yield that <math>\overline{AD}</math> <math>=</math> <math>(</math> <math>5-</math> <math>\tfrac{x}{2}</math> <math>)</math> <math>(</math> <math>\tfrac{25}{7}</math> <math>)</math>. Now, we can see <math>x</math> = <math>(</math> <math>5-</math> <math>\tfrac{x}{2}</math> <math>)</math> <math>(</math> <math>\tfrac{25}{7}</math> <math>)</math>. Solving this equation, we yield <math>39x=250</math>, or <math>x=</math> <math>\tfrac{250}{39}</math>. Thus, our final answer is <math>250+39=\boxed{289}</math>.<br />
~bluebacon008<br />
<br />
==Solution 2 (Easy Similar Triangles)==<br />
We start by adding a few points to the diagram. Call <math>F</math> the midpoint of <math>AE</math>, and <math>G</math> the midpoint of <math>BC</math>. (Note that <math>DF</math> and <math>AG</math> are altitudes of their respective triangles). We also call <math>\angle BAC = \theta</math>. Since triangle <math>ADE</math> is isosceles, <math>\angle AED = \theta</math>, and <math>\angle ADF = \angle EDF = 90 - \theta</math>. Since <math>\angle DEA = \theta</math>, <math>\angle DEC = 180 - \theta</math> and <math>\angle EDC = \angle ECD = \frac{\theta}{2}</math>. Since <math>FDC</math> is a right triangle, <math>\angle FDC = 180 - 90 - \frac{\theta}{2} = \frac{180-m}{2}</math>. <br />
<br />
Since <math>\angle BAG = \frac{\theta}{2}</math> and <math>\angle ABG = \frac{180-m}{2}</math>, triangles <math>ABG</math> and <math>CDF</math> are similar by Angle-Angle similarity. Using similar triangle ratios, we have <math>\frac{AG}{BG} = \frac{CF}{DF}</math>. <math>AG = 8</math> and <math>BG = 6</math> because there are <math>2</math> <math>6-8-10</math> triangles in the problem. Call <math>AD = x</math>. Then <math>CE = x</math>, <math>AE = 10-x</math>, and <math>EF = \frac{10-x}{2}</math>. Thus <math>CF = x + \frac{10-x}{2}</math>. Our ratio now becomes <math>\frac{8}{6} = \frac{x+ \frac{10-x}{2}}{DF}</math>. Solving for <math>DF</math> gives us <math>DF = \frac{30+3x}{8}</math>. Since <math>DF</math> is a height of the triangle <math>ADE</math>, <math>FE^2 + DF^2 = x^2</math>, or <math>DF = \sqrt{x^2 - (\frac{10-x}{2})^2}</math>. Solving the equation <math>\frac{30+3x}{8} = \sqrt{x^2 - (\frac{10-x}{2})^2}</math> gives us <math>x = \frac{250}{39}</math>, so our answer is <math>250+39 = \boxed{289}</math>.<br />
<br />
==Solution 3 (Algebra w/ Law of Cosines)==<br />
As in the diagram, let <math>DE = x</math>. Consider point <math>G</math> on the diagram shown above. Our goal is to be able to perform Pythagorean Theorem on <math>DG, GC</math>, and <math>DC</math>. Let <math>GE = 10-x</math>. Therefore, it is trivial to see that <math>GC^2 = \Big(x + \frac{10-x}{2}\Big)^2</math> (leave everything squared so that it cancels nicely at the end). By Pythagorean Theorem on Triangle <math>DGE</math>, we know that <math>DG^2 = x^2 - \Big(\frac{10-x}{2}\Big)^2</math>. Finally, we apply Law of Cosines on Triangle <math>DBC</math>. We know that <math>\cos(\angle DBC) = \frac{3}{5}</math>. Therefore, we get that <math>DC^2 = (10-x)^2 + 12^2 - 2(12)(10-x)\frac{3}{5}</math>. We can now do our final calculation:<br />
<cmath><br />
DG^2 + GC^2 = DC^2 \implies x^2 - \Big(\frac{10-x}{2}\Big)^2 + \Big(x + \frac{10-x}{2}\Big)^2 = (10-x)^2 + 12^2 - 2(12)(10-x)\frac{3}{5}<br />
</cmath><br />
After some quick cleaning up, we get<br />
<cmath><br />
30x = \frac{72}{5} + 100 \implies x = \frac{250}{39}<br />
</cmath><br />
Therefore, our answer is <math>250+39=\boxed{289}</math>.<br />
<br />
~awesome1st<br />
<br />
<br />
==Solution 4 (Coordinates)==<br />
Let <math>B = (0, 0)</math>, <math>C = (12, 0)</math>, and <math>A = (6, 8)</math>. Then, let <math>x</math> be in the interval <math>0<x<2</math> and parametrically define <math>D</math> and <math>E</math> as <math>(6-3x, 8-4x)</math> and <math>(12-3x, 4x)</math> respectively. Note that <math>AD = 5x</math>, so <math>DE = 5x</math>. This means that<br />
<cmath>\begin{align*}<br />
\sqrt{36+(8x-8)^2} &= 5x\\<br />
36+(8x-8)^2 &= 25x^2\\<br />
64x^2-128x+100 &= 25x^2\\<br />
39x^2-128x+100 &= 0\\<br />
x &= \dfrac{128\pm\sqrt{16384-15600}}{78}\\<br />
x &= \dfrac{100}{78}, 2\\<br />
\end{align*}</cmath><br />
However, since <math>2</math> is extraneous by definition, <math>x=\dfrac{50}{39}\implies AD = \dfrac{250}{39}\implies\boxed{289}</math> ~ mathwiz0803<br />
<br />
==Solution 5 (Law of Cosines)==<br />
As shown in the diagram, let <math>x</math> denote <math>\overline{AD}</math>. Let us denote the foot of the altitude of <math>A</math> to <math>\overline{BC}</math> as <math>F</math>. Note that <math>\overline{AE}</math> can be expressed as <math>10-x</math> and <math>\triangle{ABF}</math> is a <math>6-8-10</math> triangle . Therefore, <math>\sin(\angle{BAF})=\frac{3}{5}</math> and <math>\cos(\angle{BAF})=\frac{4}{5}</math>. Before we can proceed with the Law of Cosines, we must determine <math>\cos(\angle{BAC})=\cos(2\cdot \angle{BAF})=\cos^2(\angle{BAF})-\sin^2(\angle{BAF})=\frac{7}{25}</math>. Using LOC, we can write the following statement:<br />
<cmath>(\overline{DE})^2=(\overline{AD})^2+\overline{AE}^2-2(\overline{AD})(\overline{AE})\cos(\angle{BAC})\implies</cmath><br />
<cmath>x^2=x^2+(10-x)^2-2(x)(10-x)\left(\frac{7}{25}\right)\implies</cmath><br />
<cmath>0=(10-x)^2-\frac{14x}{25}(10-x)\implies</cmath><br />
<cmath>0=10-x-\frac{14x}{25}\implies</cmath><br />
<cmath>10=\frac{39x}{25}\implies x=\frac{250}{39}</cmath><br />
Thus, the desired answer is <math>\boxed{289}</math> ~ blitzkrieg21<br />
<br />
==Solution 6==<br />
In isosceles triangle, draw the altitude from <math>D</math> onto <math>\overline{AD}</math>. Let the point of intersection be <math>X</math>. Clearly, <math>AE=10-AD</math>, and hence <math>AX=\frac{10-AD}{2}</math>.<br />
<br />
Now, we recognise that the perpendicular from <math>A</math> onto <math>\overline{AD}</math> gives us two <math>6</math>-<math>8</math>-<math>10</math> triangles. So, we calculate <math>\sin \angle ABC=\frac{4}{5}</math> and <math>\cos \angle ABC=\frac{3}{5}</math><br />
<br />
<math>\angle BAC = 180-2\cdot\angle ABC</math>. And hence,<br />
<br />
<math>\cos \angle BAC = \cos \angle (180-2\cdot\angle ABC)<br />
= -\cos (2\cdot\angle ABC)<br />
= \sin^2 \angle ABC - \cos^2 \angle ABC<br />
= 2\cos^2 \angle ABC - 1<br />
= \frac{32}{25}-\frac{25}{25}=\frac{7}{25}</math><br />
<br />
Inspecting <math>\triangle ADX</math> gives us <math>\cos \angle BAC = \frac{\frac{10-x}{2}}{x} = \frac{10-x}{2x}</math><br />
Solving the equation <math>\frac{10-x}{2x}=\frac{7}{25}</math> gives <math>x= \frac{250}{39} \implies\boxed{289}</math><br />
<br />
~novus677<br />
<br />
==Solution 7 (Fastest via Law of Cosines)==<br />
We can have 2 Law of Cosines applied on <math>\angle A</math> (one from <math>\triangle ADE</math> and one from <math>\triangle ABC</math>),<br />
<br />
<math>x^2=x^2+(10-x)^2-2(x)(10-x)\cdot \cos{A}</math> and <math>12^2=10^2+10^2-2(10)(10)\cdot \cos{A}</math><br />
<br />
Solving for <math>\cos{A}</math> in both equations, we get<br />
<br />
<math>\cos{A} = \frac{(10-x)^2}{(2)(10-x)(x)}</math> and <math>cos A = \frac{7}{25} \implies \frac{(10-x)^2}{(2)(10-x)(x)} = \frac{7}{25} \implies x = \frac{250}{39}</math>, so the answer is <math>\boxed {289}</math> <br />
<br />
'''-RootThreeOverTwo'''<br />
<br />
==Solution 8 (Easiest way- Coordinates without bash)==<br />
Let <math>B=(0, 0)</math>, and <math>C=(12, 0)</math>. From there, we know that <math>A=(6, 8)</math>, so line <math>AB</math> is <math>y=\frac{4}{3}x</math>. Hence, <math>D=(a, \frac{4}{3}a)</math> for some <math>a</math>, and <math>BD=\frac{5}{3}a</math> so <math>AD=10-\frac{5}{3}a</math>. Now, notice that by symmetry, <math>E=(6+a, 8-\frac{4}{3}a)</math>, so <math>ED^2=6^2+(8-\frac{8}{3}a)^2</math>. Because <math>AD=ED</math>, we now have <math>(10-\frac{5}{3})^2=6^2+(8-\frac{8}{3}a)^2</math>, which simplifies to <math>\frac{64}{9}a^2-\frac{128}{3}a+100=\frac{25}{9}a^2-\frac{100}{3}a+100</math>, so <math>\frac{39}{9}a=\frac{13}{3}a=\frac{28}{3}</math>, and <math>a=\frac{28}{13}</math>.<br />
It follows that <math>AD=10-\frac{5}{3}\times\frac{28}{13}=10-\frac{140}{39}=\frac{390-140}{39}=\frac{250}{39}</math>, and our answer is <math>250+39=\boxed{289}</math>.<br />
<br />
-Stormersyle<br />
<br />
== Solution 9 Even Faster Law of Cosines(1 variable equation)==<br />
<br />
Doing law of cosines we know that <math>\cos A</math> is <math>\frac{7}{25}.</math>* Dropping the perpendicular from <math>D</math> to <math>AE</math> we get that <cmath>\frac{10-x}{2}=\frac{7x}{25}.</cmath> <br />
Solving for <math>x</math> we get <math>\frac{250}{39}</math> so our answer is <math>289</math>.<br />
<br />
-harsha12345<br />
<br />
* It is good to remember that doubling the smallest angle of a 3-4-5 triangle gives the larger (not right) angle in a 7-24-25 triangle.<br />
<br />
== Solution 10 (Law of Sines)==<br />
<br />
Let's label <math>\angle A = \theta</math> and <math>\angle ECD = \alpha</math>. Using isosceles triangle properties and the triangle angle sum equation, we get <cmath>180-(180-2\theta+\alpha) + \frac{180-\theta}{2} + \left(\frac{180-\theta}{2} - \alpha\right) = 180.</cmath> Solving, we find <math>\theta = 2 \alpha</math>. <br />
<br />
<br />
Relabelling our triangle, we get <math>\angle ABC = 90 - \alpha</math>. Dropping an altitude from <math>A</math> to <math>BC</math> and using the Pythagorean theorem, we find <math>[ABC] = 48</math>. Using the sine area formula, we see <math>\frac12 \cdot 10 \cdot 12 \cdot \sin(90-\alpha) = 48</math>. Plugging in our sine angle cofunction identity, <math>\sin(90-\alpha) = \cos(\alpha)</math>, we get <math>\alpha = \cos{^{-1}}{\frac45}</math>. <br />
<br />
<br />
Now, using the Law of Sines on <math>\triangle ADE</math>, we get <cmath>\frac{\sin{2\alpha}}{\frac{p}{q}} = \frac{\sin{(180-4\alpha)}}{10-\frac{p}{q}}.</cmath> After applying numerous trigonometric and algebraic tricks, identities, and simplifications, such as <math>\sin{(180-4\alpha)}=\sin{4\alpha}</math> and <math>\sin{\left(\cos{^{-1}}{\frac45}\right)} = \frac35</math>, we find <math>\frac{p}{q} = \frac{10\sin{2\alpha}}{\sin{4\alpha}+\sin{2\alpha}} = \frac{250}{39}</math>. <br />
<br />
<br />
<br />
Therefore, our answer is <math>250 + 39 = \boxed{289}</math>.<br />
<br />
<br />
~Tiblis<br />
<br />
== Solution 11 (Trigonometry)==<br />
We start by labelling a few angles (all of them in degrees). Let <math>\angle{BAC}=2\alpha = \angle{AED}, \angle{EDC}=\angle{ECD}=\alpha, \angle{DEC}=180-2\alpha, \angle{BDC}=3\alpha, \angle{DCB}=90-2\alpha, \angle{DBC}=90-\alpha</math>. Also let <math>AD=a</math>. By sine rule in <math>\triangle{ADE},</math> we get <math>\frac{a}{\sin{2\alpha}}=\frac{10-a}{\sin{4\alpha}} \implies \cos{2\alpha}=\frac{5}{a}-\frac{1}{2}</math><br />
Using sine rule in <math>\triangle{ABC}</math>, we get <math>\sin{\alpha}=\frac{3}{5}</math>. Hence we get <math>\cos{2\alpha}=1-2\sin^2{\alpha}=1-\frac{18}{25}=\frac{7}{25}</math>. Hence <math>\frac{5}{a}=\frac{1}{2}+\frac{7}{25}=\frac{39}{50} \implies a=\frac{250}{39}</math>. Therefore, our answer is <math>\boxed{289}</math><br />
<br />
Alternatively, use sine rule in <math>\triangle{BDC}</math>. (It’s easier)<br />
<br />
~Prabh1512<br />
<br />
== Solution 12 (Double Angle Identity)==<br />
<br />
We let AD=x. Then, angle A is <math>2\sin^{-1}(\frac{3}{5})</math> and so is angle <math>AED</math>. We note that <math>AE=10-x</math>. We drop an altitude from <math>D</math> to <math>AE</math>, and we call the foot <math>F</math>. We note that <math>AF=\frac{10-x}{2}</math>. Using the double angle identity, we have <math>\sin(2sin^{-1}(\frac{3}{5}))=2(\frac{3}{5})(\frac{4}{5})=\frac{24}{25}.</math> Therefore, <math>DF=\frac{24}{25}AD.</math> We now use the Pythagorean Theorem, which gives <math>(\frac{10-x}{2})^2+(\frac{24}{25}x)^2=x^2</math>. Rearranging and simplifying, this becomes <math>429x^2-12500x+62500=0</math>. Using the quadratic formula, this is <math>\frac{12500\pm\sqrt{12500^2-250000\cdot429}}{858}</math>. We take out a <math>10000</math> from the square root and make it a <math>100</math> outside of the square root to make it simpler. We end up with <math>\frac{12500\pm7000}{858}</math>. We note that this must be less than 10 to ensure that <math>10-x</math> is positive. Therefore, we take the minus, and we get <math>\frac{5500}{858}=\frac{250}{39} -> \fbox{289}.</math><br />
<br />
~john0512<br />
<br />
==Video Solution==<br />
<br />
https://www.youtube.com/watch?v=iE8paW_ICxw<br />
<br />
<br />
https://youtu.be/dI6uZ67Ae2s ~yofro<br />
<br />
==See Also==<br />
{{AIME box|year=2018|n=I|num-b=3|num-a=5}}<br />
{{MAA Notice}}</div>John0512https://artofproblemsolving.com/wiki/index.php?title=User:John0512&diff=153596User:John05122021-05-12T14:55:38Z<p>John0512: </p>
<hr />
<div>is bad at math<br />
<br />
only got 7th on mathcounts nationals 2021<br />
<br />
also missed jmo by 1 problem, im just that bad<br />
<br />
also who is this person<br />
<br />
https://artofproblemsolving.com/community/user/784626</div>John0512https://artofproblemsolving.com/wiki/index.php?title=User:John0512&diff=153595User:John05122021-05-12T14:49:39Z<p>John0512: </p>
<hr />
<div>is bad at math<br />
<br />
only got 7th on mathcounts nationals 2021<br />
<br />
also missed jmo by 1 problem, im just that bad<br />
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yas v4913 orz orz orz</div>John0512https://artofproblemsolving.com/wiki/index.php?title=2018_AIME_I_Problems/Problem_4&diff=1487222018 AIME I Problems/Problem 42021-03-06T21:04:53Z<p>John0512: /* Solution 12 (Double Angle Identity) */</p>
<hr />
<div>==Problem 4==<br />
In <math>\triangle ABC, AB = AC = 10</math> and <math>BC = 12</math>. Point <math>D</math> lies strictly between <math>A</math> and <math>B</math> on <math>\overline{AB}</math> and point <math>E</math> lies strictly between <math>A</math> and <math>C</math> on <math>\overline{AC}</math> so that <math>AD = DE = EC</math>. Then <math>AD</math> can be expressed in the form <math>\dfrac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. Find <math>p+q</math>.<br />
<br />
==Solution 1==<br />
<math>\cos(A) = \frac{5^2+5^2-6^2}{2*5*5} = \frac{7}{25}</math>. Let <math>M</math> be midpoint of <math>AE</math>, then <math>\frac{7}{25} = \frac{10-x}{2x} \iff x =\frac{250}{39}</math>. So, our answer is <math>250+39=\boxed{289}</math>.<br />
<br />
==Solution 1 (No Trig)==<br />
<center><br />
<asy><br />
import cse5;<br />
unitsize(10mm);<br />
pathpen=black;<br />
dotfactor=3;<br />
<br />
pair B = (0,0), A = (6,8), C = (12,0), D = (2.154,2.872), E = (8.153, 5.128), F=(7.68,5.76), G=(7.077,6.564), H=(5.6,4.3), I=(4.5,6), J=(10,2.66);<br />
pair[] dotted = {A,B,C,D,E,F,G};<br />
<br />
D(A--B);<br />
D(C--B);<br />
D(A--C);<br />
D(D--E);<br />
pathpen=dashed;<br />
D(B--F);<br />
D(D--G);<br />
<br />
dot(dotted);<br />
label("$A$",A,N);<br />
label("$B$",B,SW);<br />
label("$C$",C,SE);<br />
label("$D$",D,NW);<br />
label("$E$",E,NE);<br />
label("$F$",F,NE);<br />
label("$G$",G,NE);<br />
label("$x$",H,NW);<br />
label("$x$",I,NW);<br />
label("$x$",J,NE);<br />
</asy><br />
</center><br />
<br />
We draw the altitude from <math>B</math> to <math>\overline{AC}</math> to get point <math>F</math>. We notice that the triangle's height from <math>A</math> to <math>\overline{BC}</math> is 8 because it is a <math>3-4-5</math> Right Triangle. To find the length of <math>\overline{BF}</math>, we let <math>h</math> represent <math>\overline{BF}</math> and set up an equation by finding two ways to express the area. The equation is <math>(8)(12)=(10)(h)</math>, which leaves us with <math>h=9.6</math>. We then solve for the length <math>\overline{AF}</math>, which is done through pythagorean theorm and get <math>\overline{AF}</math> = <math>2.8</math>. We can now see that <math>\triangle AFB</math> is a <math>7-24-25</math> Right Triangle. Thus, we set <math>\overline{AG}</math> as <math>5-</math><math>\tfrac{x}{2}</math>, and yield that <math>\overline{AD}</math> <math>=</math> <math>(</math> <math>5-</math> <math>\tfrac{x}{2}</math> <math>)</math> <math>(</math> <math>\tfrac{25}{7}</math> <math>)</math>. Now, we can see <math>x</math> = <math>(</math> <math>5-</math> <math>\tfrac{x}{2}</math> <math>)</math> <math>(</math> <math>\tfrac{25}{7}</math> <math>)</math>. Solving this equation, we yield <math>39x=250</math>, or <math>x=</math> <math>\tfrac{250}{39}</math>. Thus, our final answer is <math>250+39=\boxed{289}</math>.<br />
~bluebacon008<br />
<br />
==Solution 2 (Easy Similar Triangles)==<br />
We start by adding a few points to the diagram. Call <math>F</math> the midpoint of <math>AE</math>, and <math>G</math> the midpoint of <math>BC</math>. (Note that <math>DF</math> and <math>AG</math> are altitudes of their respective triangles). We also call <math>\angle BAC = \theta</math>. Since triangle <math>ADE</math> is isosceles, <math>\angle AED = \theta</math>, and <math>\angle ADF = \angle EDF = 90 - \theta</math>. Since <math>\angle DEA = \theta</math>, <math>\angle DEC = 180 - \theta</math> and <math>\angle EDC = \angle ECD = \frac{\theta}{2}</math>. Since <math>FDC</math> is a right triangle, <math>\angle FDC = 180 - 90 - \frac{\theta}{2} = \frac{180-m}{2}</math>. <br />
<br />
Since <math>\angle BAG = \frac{\theta}{2}</math> and <math>\angle ABG = \frac{180-m}{2}</math>, triangles <math>ABG</math> and <math>CDF</math> are similar by Angle-Angle similarity. Using similar triangle ratios, we have <math>\frac{AG}{BG} = \frac{CF}{DF}</math>. <math>AG = 8</math> and <math>BG = 6</math> because there are <math>2</math> <math>6-8-10</math> triangles in the problem. Call <math>AD = x</math>. Then <math>CE = x</math>, <math>AE = 10-x</math>, and <math>EF = \frac{10-x}{2}</math>. Thus <math>CF = x + \frac{10-x}{2}</math>. Our ratio now becomes <math>\frac{8}{6} = \frac{x+ \frac{10-x}{2}}{DF}</math>. Solving for <math>DF</math> gives us <math>DF = \frac{30+3x}{8}</math>. Since <math>DF</math> is a height of the triangle <math>ADE</math>, <math>FE^2 + DF^2 = x^2</math>, or <math>DF = \sqrt{x^2 - (\frac{10-x}{2})^2}</math>. Solving the equation <math>\frac{30+3x}{8} = \sqrt{x^2 - (\frac{10-x}{2})^2}</math> gives us <math>x = \frac{250}{39}</math>, so our answer is <math>250+39 = \boxed{289}</math>.<br />
<br />
==Solution 3 (Algebra w/ Law of Cosines)==<br />
As in the diagram, let <math>DE = x</math>. Consider point <math>G</math> on the diagram shown above. Our goal is to be able to perform Pythagorean Theorem on <math>DG, GC</math>, and <math>DC</math>. Let <math>GE = 10-x</math>. Therefore, it is trivial to see that <math>GC^2 = \Big(x + \frac{10-x}{2}\Big)^2</math> (leave everything squared so that it cancels nicely at the end). By Pythagorean Theorem on Triangle <math>DGE</math>, we know that <math>DG^2 = x^2 - \Big(\frac{10-x}{2}\Big)^2</math>. Finally, we apply Law of Cosines on Triangle <math>DBC</math>. We know that <math>\cos(\angle DBC) = \frac{3}{5}</math>. Therefore, we get that <math>DC^2 = (10-x)^2 + 12^2 - 2(12)(10-x)\frac{3}{5}</math>. We can now do our final calculation:<br />
<cmath><br />
DG^2 + GC^2 = DC^2 \implies x^2 - \Big(\frac{10-x}{2}\Big)^2 + \Big(x + \frac{10-x}{2}\Big)^2 = (10-x)^2 + 12^2 - 2(12)(10-x)\frac{3}{5}<br />
</cmath><br />
After some quick cleaning up, we get<br />
<cmath><br />
30x = \frac{72}{5} + 100 \implies x = \frac{250}{39}<br />
</cmath><br />
Therefore, our answer is <math>250+39=\boxed{289}</math>.<br />
<br />
~awesome1st<br />
<br />
<br />
==Solution 4 (Coordinates)==<br />
Let <math>B = (0, 0)</math>, <math>C = (12, 0)</math>, and <math>A = (6, 8)</math>. Then, let <math>x</math> be in the interval <math>0<x<2</math> and parametrically define <math>D</math> and <math>E</math> as <math>(6-3x, 8-4x)</math> and <math>(12-3x, 4x)</math> respectively. Note that <math>AD = 5x</math>, so <math>DE = 5x</math>. This means that<br />
<cmath>\begin{align*}<br />
\sqrt{36+(8x-8)^2} &= 5x\\<br />
36+(8x-8)^2 &= 25x^2\\<br />
64x^2-128x+100 &= 25x^2\\<br />
39x^2-128x+100 &= 0\\<br />
x &= \dfrac{128\pm\sqrt{16384-15600}}{78}\\<br />
x &= \dfrac{100}{78}, 2\\<br />
\end{align*}</cmath><br />
However, since <math>2</math> is extraneous by definition, <math>x=\dfrac{50}{39}\implies AD = \dfrac{250}{39}\implies\boxed{289}</math> ~ mathwiz0803<br />
<br />
==Solution 5 (Law of Cosines)==<br />
As shown in the diagram, let <math>x</math> denote <math>\overline{AD}</math>. Let us denote the foot of the altitude of <math>A</math> to <math>\overline{BC}</math> as <math>F</math>. Note that <math>\overline{AE}</math> can be expressed as <math>10-x</math> and <math>\triangle{ABF}</math> is a <math>6-8-10</math> triangle . Therefore, <math>\sin(\angle{BAF})=\frac{3}{5}</math> and <math>\cos(\angle{BAF})=\frac{4}{5}</math>. Before we can proceed with the Law of Cosines, we must determine <math>\cos(\angle{BAC})=\cos(2\cdot \angle{BAF})=\cos^2(\angle{BAF})-\sin^2(\angle{BAF})=\frac{7}{25}</math>. Using LOC, we can write the following statement:<br />
<cmath>(\overline{DE})^2=(\overline{AD})^2+\overline{AE}^2-2(\overline{AD})(\overline{AE})\cos(\angle{BAC})\implies</cmath><br />
<cmath>x^2=x^2+(10-x)^2-2(x)(10-x)\left(\frac{7}{25}\right)\implies</cmath><br />
<cmath>0=(10-x)^2-\frac{14x}{25}(10-x)\implies</cmath><br />
<cmath>0=10-x-\frac{14x}{25}\implies</cmath><br />
<cmath>10=\frac{39x}{25}\implies x=\frac{250}{39}</cmath><br />
Thus, the desired answer is <math>\boxed{289}</math> ~ blitzkrieg21<br />
<br />
==Solution 6==<br />
In isosceles triangle, draw the altitude from <math>D</math> onto <math>\overline{AD}</math>. Let the point of intersection be <math>X</math>. Clearly, <math>AE=10-AD</math>, and hence <math>AX=\frac{10-AD}{2}</math>.<br />
<br />
Now, we recognise that the perpendicular from <math>A</math> onto <math>\overline{AD}</math> gives us two <math>6</math>-<math>8</math>-<math>10</math> triangles. So, we calculate <math>\sin \angle ABC=\frac{4}{5}</math> and <math>\cos \angle ABC=\frac{3}{5}</math><br />
<br />
<math>\angle BAC = 180-2\cdot\angle ABC</math>. And hence,<br />
<br />
<math>\cos \angle BAC = \cos \angle (180-2\cdot\angle ABC)<br />
= -\cos (2\cdot\angle ABC)<br />
= \sin^2 \angle ABC - \cos^2 \angle ABC<br />
= 2\cos^2 \angle ABC - 1<br />
= \frac{32}{25}-\frac{25}{25}=\frac{7}{25}</math><br />
<br />
Inspecting <math>\triangle ADX</math> gives us <math>\cos \angle BAC = \frac{\frac{10-x}{2}}{x} = \frac{10-x}{2x}</math><br />
Solving the equation <math>\frac{10-x}{2x}=\frac{7}{25}</math> gives <math>x= \frac{250}{39} \implies\boxed{289}</math><br />
<br />
~novus677<br />
<br />
==Solution 7 (Fastest via Law of Cosines)==<br />
We can have 2 Law of Cosines applied on <math>\angle A</math> (one from <math>\triangle ADE</math> and one from <math>\triangle ABC</math>),<br />
<br />
<math>x^2=x^2+(10-x)^2-2(x)(10-x)\cdot \cos{A}</math> and <math>12^2=10^2+10^2-2(10)(10)\cdot \cos{A}</math><br />
<br />
Solving for <math>\cos{A}</math> in both equations, we get<br />
<br />
<math>\cos{A} = \frac{(10-x)^2}{(2)(10-x)(x)}</math> and <math>cos A = \frac{7}{25} \implies \frac{(10-x)^2}{(2)(10-x)(x)} = \frac{7}{25} \implies x = \frac{250}{39}</math>, so the answer is <math>\boxed {289}</math> <br />
<br />
'''-RootThreeOverTwo'''<br />
<br />
==Solution 8 (Easiest way- Coordinates without bash)==<br />
Let <math>B=(0, 0)</math>, and <math>C=(12, 0)</math>. From there, we know that <math>A=(6, 8)</math>, so line <math>AB</math> is <math>y=\frac{4}{3}x</math>. Hence, <math>D=(a, \frac{4}{3}a)</math> for some <math>a</math>, and <math>BD=\frac{5}{3}a</math> so <math>AD=10-\frac{5}{3}a</math>. Now, notice that by symmetry, <math>E=(6+a, 8-\frac{4}{3}a)</math>, so <math>ED^2=6^2+(8-\frac{8}{3}a)^2</math>. Because <math>AD=ED</math>, we now have <math>(10-\frac{5}{3})^2=6^2+(8-\frac{8}{3}a)^2</math>, which simplifies to <math>\frac{64}{9}a^2-\frac{128}{3}a+100=\frac{25}{9}a^2-\frac{100}{3}a+100</math>, so <math>\frac{39}{9}a=\frac{13}{3}a=\frac{28}{3}</math>, and <math>a=\frac{28}{13}</math>.<br />
It follows that <math>AD=10-\frac{5}{3}\times\frac{28}{13}=10-\frac{140}{39}=\frac{390-140}{39}=\frac{250}{39}</math>, and our answer is <math>250+39=\boxed{289}</math>.<br />
<br />
-Stormersyle<br />
<br />
== Solution 9 Even Faster Law of Cosines(1 variable equation)==<br />
<br />
Doing law of cosines we know that <math>\cos A</math> is <math>\frac{7}{25}.</math>* Dropping the perpendicular from <math>D</math> to <math>AE</math> we get that <cmath>\frac{10-x}{2}=\frac{7x}{25}.</cmath> <br />
Solving for <math>x</math> we get <math>\frac{250}{39}</math> so our answer is <math>289</math>.<br />
<br />
-harsha12345<br />
<br />
* It is good to remember that doubling the smallest angle of a 3-4-5 triangle gives the larger (not right) angle in a 7-24-25 triangle.<br />
<br />
== Solution 10 (Law of Sines)==<br />
<br />
Let's label <math>\angle A = \theta</math> and <math>\angle ECD = \alpha</math>. Using isosceles triangle properties and the triangle angle sum equation, we get <cmath>180-(180-2\theta+\alpha) + \frac{180-\theta}{2} + \left(\frac{180-\theta}{2} - \alpha\right) = 180.</cmath> Solving, we find <math>\theta = 2 \alpha</math>. <br />
<br />
<br />
Relabelling our triangle, we get <math>\angle ABC = 90 - \alpha</math>. Dropping an altitude from <math>A</math> to <math>BC</math> and using the Pythagorean theorem, we find <math>[ABC] = 48</math>. Using the sine area formula, we see <math>\frac12 \cdot 10 \cdot 12 \cdot \sin(90-\alpha) = 48</math>. Plugging in our sine angle cofunction identity, <math>\sin(90-\alpha) = \cos(\alpha)</math>, we get <math>\alpha = \cos{^{-1}}{\frac45}</math>. <br />
<br />
<br />
Now, using the Law of Sines on <math>\triangle ADE</math>, we get <cmath>\frac{\sin{2\alpha}}{\frac{p}{q}} = \frac{\sin{(180-4\alpha)}}{10-\frac{p}{q}}.</cmath> After applying numerous trigonometric and algebraic tricks, identities, and simplifications, such as <math>\sin{(180-4\alpha)}=\sin{4\alpha}</math> and <math>\sin{\left(\cos{^{-1}}{\frac45}\right)} = \frac35</math>, we find <math>\frac{p}{q} = \frac{10\sin{2\alpha}}{\sin{4\alpha}+\sin{2\alpha}} = \frac{250}{39}</math>. <br />
<br />
<br />
<br />
Therefore, our answer is <math>250 + 39 = \boxed{289}</math>.<br />
<br />
<br />
~Tiblis<br />
<br />
== Solution 11 (Trigonometry)==<br />
We start by labelling a few angles (all of them in degrees). Let <math>\angle{BAC}=2\alpha = \angle{AED}, \angle{EDC}=\angle{ECD}=\alpha, \angle{DEC}=180-2\alpha, \angle{BDC}=3\alpha, \angle{DCB}=90-2\alpha, \angle{DBC}=90-\alpha</math>. Also let <math>AD=a</math>. By sine rule in <math>\triangle{ADE},</math> we get <math>\frac{a}{\sin{2\alpha}}=\frac{10-a}{\sin{4\alpha}} \implies \cos{2\alpha}=\frac{5}{a}-\frac{1}{2}</math><br />
Using sine rule in <math>\triangle{ABC}</math>, we get <math>\sin{\alpha}=\frac{3}{5}</math>. Hence we get <math>\cos{2\alpha}=1-2\sin^2{\alpha}=1-\frac{18}{25}=\frac{7}{25}</math>. Hence <math>\frac{5}{a}=\frac{1}{2}+\frac{7}{25}=\frac{39}{50} \implies a=\frac{250}{39}</math>. Therefore, our answer is <math>\boxed{289}</math><br />
<br />
Alternatively, use sine rule in <math>\triangle{BDC}</math>. (It’s easier)<br />
<br />
~Prabh1512<br />
<br />
== Solution 12 (Double Angle Identity)==<br />
<br />
We let AD=x. Then, angle A is <math>2\sin^{-1}(\frac{3}{5})</math> and so is angle <math>AED</math>. We note that <math>AE=10-x</math>. We drop an altitude from <math>D</math> to <math>AE</math>, and we call the foot <math>F</math>. We note that <math>AF=\frac{10-x}{2}</math>. Using the double angle identity, we have <math>\sin(2sin^{-1}(\frac{3}{5}))=2(\frac{3}{5})(\frac{4}{5})=\frac{24}{25}.</math> Therefore, <math>DF=\frac{24}{25}AD.</math> We now use the Pythagorean Theorem, which gives <math>(\frac{10-x}{2})^2+(\frac{24}{25}x)^2=x^2</math>. Rearranging and simplifying, this becomes <math>429x^2-12500x+62500=0</math>. Using the quadratic formula, this is <math>\frac{12500\pm\sqrt{12500^2-250000\cdot429}}{858}</math>. We take out a <math>10000</math> from the square root and make it a <math>100</math> outside of the square root to make it simpler. We end up with <math>\frac{12500\pm7500}{858}</math>. We note that this must be less than 10 to ensure that <math>10-x</math> is positive. Therefore, we take the minus, and we get <math>\frac{5500}{858}=\frac{250}{39} -> \fbox{289}.</math><br />
<br />
~john0512<br />
<br />
==Video Solution==<br />
<br />
https://www.youtube.com/watch?v=iE8paW_ICxw<br />
<br />
<br />
https://youtu.be/dI6uZ67Ae2s ~yofro<br />
<br />
==See Also==<br />
{{AIME box|year=2018|n=I|num-b=3|num-a=5}}<br />
{{MAA Notice}}</div>John0512https://artofproblemsolving.com/wiki/index.php?title=2018_AIME_I_Problems/Problem_4&diff=1487212018 AIME I Problems/Problem 42021-03-06T21:04:31Z<p>John0512: </p>
<hr />
<div>==Problem 4==<br />
In <math>\triangle ABC, AB = AC = 10</math> and <math>BC = 12</math>. Point <math>D</math> lies strictly between <math>A</math> and <math>B</math> on <math>\overline{AB}</math> and point <math>E</math> lies strictly between <math>A</math> and <math>C</math> on <math>\overline{AC}</math> so that <math>AD = DE = EC</math>. Then <math>AD</math> can be expressed in the form <math>\dfrac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. Find <math>p+q</math>.<br />
<br />
==Solution 1==<br />
<math>\cos(A) = \frac{5^2+5^2-6^2}{2*5*5} = \frac{7}{25}</math>. Let <math>M</math> be midpoint of <math>AE</math>, then <math>\frac{7}{25} = \frac{10-x}{2x} \iff x =\frac{250}{39}</math>. So, our answer is <math>250+39=\boxed{289}</math>.<br />
<br />
==Solution 1 (No Trig)==<br />
<center><br />
<asy><br />
import cse5;<br />
unitsize(10mm);<br />
pathpen=black;<br />
dotfactor=3;<br />
<br />
pair B = (0,0), A = (6,8), C = (12,0), D = (2.154,2.872), E = (8.153, 5.128), F=(7.68,5.76), G=(7.077,6.564), H=(5.6,4.3), I=(4.5,6), J=(10,2.66);<br />
pair[] dotted = {A,B,C,D,E,F,G};<br />
<br />
D(A--B);<br />
D(C--B);<br />
D(A--C);<br />
D(D--E);<br />
pathpen=dashed;<br />
D(B--F);<br />
D(D--G);<br />
<br />
dot(dotted);<br />
label("$A$",A,N);<br />
label("$B$",B,SW);<br />
label("$C$",C,SE);<br />
label("$D$",D,NW);<br />
label("$E$",E,NE);<br />
label("$F$",F,NE);<br />
label("$G$",G,NE);<br />
label("$x$",H,NW);<br />
label("$x$",I,NW);<br />
label("$x$",J,NE);<br />
</asy><br />
</center><br />
<br />
We draw the altitude from <math>B</math> to <math>\overline{AC}</math> to get point <math>F</math>. We notice that the triangle's height from <math>A</math> to <math>\overline{BC}</math> is 8 because it is a <math>3-4-5</math> Right Triangle. To find the length of <math>\overline{BF}</math>, we let <math>h</math> represent <math>\overline{BF}</math> and set up an equation by finding two ways to express the area. The equation is <math>(8)(12)=(10)(h)</math>, which leaves us with <math>h=9.6</math>. We then solve for the length <math>\overline{AF}</math>, which is done through pythagorean theorm and get <math>\overline{AF}</math> = <math>2.8</math>. We can now see that <math>\triangle AFB</math> is a <math>7-24-25</math> Right Triangle. Thus, we set <math>\overline{AG}</math> as <math>5-</math><math>\tfrac{x}{2}</math>, and yield that <math>\overline{AD}</math> <math>=</math> <math>(</math> <math>5-</math> <math>\tfrac{x}{2}</math> <math>)</math> <math>(</math> <math>\tfrac{25}{7}</math> <math>)</math>. Now, we can see <math>x</math> = <math>(</math> <math>5-</math> <math>\tfrac{x}{2}</math> <math>)</math> <math>(</math> <math>\tfrac{25}{7}</math> <math>)</math>. Solving this equation, we yield <math>39x=250</math>, or <math>x=</math> <math>\tfrac{250}{39}</math>. Thus, our final answer is <math>250+39=\boxed{289}</math>.<br />
~bluebacon008<br />
<br />
==Solution 2 (Easy Similar Triangles)==<br />
We start by adding a few points to the diagram. Call <math>F</math> the midpoint of <math>AE</math>, and <math>G</math> the midpoint of <math>BC</math>. (Note that <math>DF</math> and <math>AG</math> are altitudes of their respective triangles). We also call <math>\angle BAC = \theta</math>. Since triangle <math>ADE</math> is isosceles, <math>\angle AED = \theta</math>, and <math>\angle ADF = \angle EDF = 90 - \theta</math>. Since <math>\angle DEA = \theta</math>, <math>\angle DEC = 180 - \theta</math> and <math>\angle EDC = \angle ECD = \frac{\theta}{2}</math>. Since <math>FDC</math> is a right triangle, <math>\angle FDC = 180 - 90 - \frac{\theta}{2} = \frac{180-m}{2}</math>. <br />
<br />
Since <math>\angle BAG = \frac{\theta}{2}</math> and <math>\angle ABG = \frac{180-m}{2}</math>, triangles <math>ABG</math> and <math>CDF</math> are similar by Angle-Angle similarity. Using similar triangle ratios, we have <math>\frac{AG}{BG} = \frac{CF}{DF}</math>. <math>AG = 8</math> and <math>BG = 6</math> because there are <math>2</math> <math>6-8-10</math> triangles in the problem. Call <math>AD = x</math>. Then <math>CE = x</math>, <math>AE = 10-x</math>, and <math>EF = \frac{10-x}{2}</math>. Thus <math>CF = x + \frac{10-x}{2}</math>. Our ratio now becomes <math>\frac{8}{6} = \frac{x+ \frac{10-x}{2}}{DF}</math>. Solving for <math>DF</math> gives us <math>DF = \frac{30+3x}{8}</math>. Since <math>DF</math> is a height of the triangle <math>ADE</math>, <math>FE^2 + DF^2 = x^2</math>, or <math>DF = \sqrt{x^2 - (\frac{10-x}{2})^2}</math>. Solving the equation <math>\frac{30+3x}{8} = \sqrt{x^2 - (\frac{10-x}{2})^2}</math> gives us <math>x = \frac{250}{39}</math>, so our answer is <math>250+39 = \boxed{289}</math>.<br />
<br />
==Solution 3 (Algebra w/ Law of Cosines)==<br />
As in the diagram, let <math>DE = x</math>. Consider point <math>G</math> on the diagram shown above. Our goal is to be able to perform Pythagorean Theorem on <math>DG, GC</math>, and <math>DC</math>. Let <math>GE = 10-x</math>. Therefore, it is trivial to see that <math>GC^2 = \Big(x + \frac{10-x}{2}\Big)^2</math> (leave everything squared so that it cancels nicely at the end). By Pythagorean Theorem on Triangle <math>DGE</math>, we know that <math>DG^2 = x^2 - \Big(\frac{10-x}{2}\Big)^2</math>. Finally, we apply Law of Cosines on Triangle <math>DBC</math>. We know that <math>\cos(\angle DBC) = \frac{3}{5}</math>. Therefore, we get that <math>DC^2 = (10-x)^2 + 12^2 - 2(12)(10-x)\frac{3}{5}</math>. We can now do our final calculation:<br />
<cmath><br />
DG^2 + GC^2 = DC^2 \implies x^2 - \Big(\frac{10-x}{2}\Big)^2 + \Big(x + \frac{10-x}{2}\Big)^2 = (10-x)^2 + 12^2 - 2(12)(10-x)\frac{3}{5}<br />
</cmath><br />
After some quick cleaning up, we get<br />
<cmath><br />
30x = \frac{72}{5} + 100 \implies x = \frac{250}{39}<br />
</cmath><br />
Therefore, our answer is <math>250+39=\boxed{289}</math>.<br />
<br />
~awesome1st<br />
<br />
<br />
==Solution 4 (Coordinates)==<br />
Let <math>B = (0, 0)</math>, <math>C = (12, 0)</math>, and <math>A = (6, 8)</math>. Then, let <math>x</math> be in the interval <math>0<x<2</math> and parametrically define <math>D</math> and <math>E</math> as <math>(6-3x, 8-4x)</math> and <math>(12-3x, 4x)</math> respectively. Note that <math>AD = 5x</math>, so <math>DE = 5x</math>. This means that<br />
<cmath>\begin{align*}<br />
\sqrt{36+(8x-8)^2} &= 5x\\<br />
36+(8x-8)^2 &= 25x^2\\<br />
64x^2-128x+100 &= 25x^2\\<br />
39x^2-128x+100 &= 0\\<br />
x &= \dfrac{128\pm\sqrt{16384-15600}}{78}\\<br />
x &= \dfrac{100}{78}, 2\\<br />
\end{align*}</cmath><br />
However, since <math>2</math> is extraneous by definition, <math>x=\dfrac{50}{39}\implies AD = \dfrac{250}{39}\implies\boxed{289}</math> ~ mathwiz0803<br />
<br />
==Solution 5 (Law of Cosines)==<br />
As shown in the diagram, let <math>x</math> denote <math>\overline{AD}</math>. Let us denote the foot of the altitude of <math>A</math> to <math>\overline{BC}</math> as <math>F</math>. Note that <math>\overline{AE}</math> can be expressed as <math>10-x</math> and <math>\triangle{ABF}</math> is a <math>6-8-10</math> triangle . Therefore, <math>\sin(\angle{BAF})=\frac{3}{5}</math> and <math>\cos(\angle{BAF})=\frac{4}{5}</math>. Before we can proceed with the Law of Cosines, we must determine <math>\cos(\angle{BAC})=\cos(2\cdot \angle{BAF})=\cos^2(\angle{BAF})-\sin^2(\angle{BAF})=\frac{7}{25}</math>. Using LOC, we can write the following statement:<br />
<cmath>(\overline{DE})^2=(\overline{AD})^2+\overline{AE}^2-2(\overline{AD})(\overline{AE})\cos(\angle{BAC})\implies</cmath><br />
<cmath>x^2=x^2+(10-x)^2-2(x)(10-x)\left(\frac{7}{25}\right)\implies</cmath><br />
<cmath>0=(10-x)^2-\frac{14x}{25}(10-x)\implies</cmath><br />
<cmath>0=10-x-\frac{14x}{25}\implies</cmath><br />
<cmath>10=\frac{39x}{25}\implies x=\frac{250}{39}</cmath><br />
Thus, the desired answer is <math>\boxed{289}</math> ~ blitzkrieg21<br />
<br />
==Solution 6==<br />
In isosceles triangle, draw the altitude from <math>D</math> onto <math>\overline{AD}</math>. Let the point of intersection be <math>X</math>. Clearly, <math>AE=10-AD</math>, and hence <math>AX=\frac{10-AD}{2}</math>.<br />
<br />
Now, we recognise that the perpendicular from <math>A</math> onto <math>\overline{AD}</math> gives us two <math>6</math>-<math>8</math>-<math>10</math> triangles. So, we calculate <math>\sin \angle ABC=\frac{4}{5}</math> and <math>\cos \angle ABC=\frac{3}{5}</math><br />
<br />
<math>\angle BAC = 180-2\cdot\angle ABC</math>. And hence,<br />
<br />
<math>\cos \angle BAC = \cos \angle (180-2\cdot\angle ABC)<br />
= -\cos (2\cdot\angle ABC)<br />
= \sin^2 \angle ABC - \cos^2 \angle ABC<br />
= 2\cos^2 \angle ABC - 1<br />
= \frac{32}{25}-\frac{25}{25}=\frac{7}{25}</math><br />
<br />
Inspecting <math>\triangle ADX</math> gives us <math>\cos \angle BAC = \frac{\frac{10-x}{2}}{x} = \frac{10-x}{2x}</math><br />
Solving the equation <math>\frac{10-x}{2x}=\frac{7}{25}</math> gives <math>x= \frac{250}{39} \implies\boxed{289}</math><br />
<br />
~novus677<br />
<br />
==Solution 7 (Fastest via Law of Cosines)==<br />
We can have 2 Law of Cosines applied on <math>\angle A</math> (one from <math>\triangle ADE</math> and one from <math>\triangle ABC</math>),<br />
<br />
<math>x^2=x^2+(10-x)^2-2(x)(10-x)\cdot \cos{A}</math> and <math>12^2=10^2+10^2-2(10)(10)\cdot \cos{A}</math><br />
<br />
Solving for <math>\cos{A}</math> in both equations, we get<br />
<br />
<math>\cos{A} = \frac{(10-x)^2}{(2)(10-x)(x)}</math> and <math>cos A = \frac{7}{25} \implies \frac{(10-x)^2}{(2)(10-x)(x)} = \frac{7}{25} \implies x = \frac{250}{39}</math>, so the answer is <math>\boxed {289}</math> <br />
<br />
'''-RootThreeOverTwo'''<br />
<br />
==Solution 8 (Easiest way- Coordinates without bash)==<br />
Let <math>B=(0, 0)</math>, and <math>C=(12, 0)</math>. From there, we know that <math>A=(6, 8)</math>, so line <math>AB</math> is <math>y=\frac{4}{3}x</math>. Hence, <math>D=(a, \frac{4}{3}a)</math> for some <math>a</math>, and <math>BD=\frac{5}{3}a</math> so <math>AD=10-\frac{5}{3}a</math>. Now, notice that by symmetry, <math>E=(6+a, 8-\frac{4}{3}a)</math>, so <math>ED^2=6^2+(8-\frac{8}{3}a)^2</math>. Because <math>AD=ED</math>, we now have <math>(10-\frac{5}{3})^2=6^2+(8-\frac{8}{3}a)^2</math>, which simplifies to <math>\frac{64}{9}a^2-\frac{128}{3}a+100=\frac{25}{9}a^2-\frac{100}{3}a+100</math>, so <math>\frac{39}{9}a=\frac{13}{3}a=\frac{28}{3}</math>, and <math>a=\frac{28}{13}</math>.<br />
It follows that <math>AD=10-\frac{5}{3}\times\frac{28}{13}=10-\frac{140}{39}=\frac{390-140}{39}=\frac{250}{39}</math>, and our answer is <math>250+39=\boxed{289}</math>.<br />
<br />
-Stormersyle<br />
<br />
== Solution 9 Even Faster Law of Cosines(1 variable equation)==<br />
<br />
Doing law of cosines we know that <math>\cos A</math> is <math>\frac{7}{25}.</math>* Dropping the perpendicular from <math>D</math> to <math>AE</math> we get that <cmath>\frac{10-x}{2}=\frac{7x}{25}.</cmath> <br />
Solving for <math>x</math> we get <math>\frac{250}{39}</math> so our answer is <math>289</math>.<br />
<br />
-harsha12345<br />
<br />
* It is good to remember that doubling the smallest angle of a 3-4-5 triangle gives the larger (not right) angle in a 7-24-25 triangle.<br />
<br />
== Solution 10 (Law of Sines)==<br />
<br />
Let's label <math>\angle A = \theta</math> and <math>\angle ECD = \alpha</math>. Using isosceles triangle properties and the triangle angle sum equation, we get <cmath>180-(180-2\theta+\alpha) + \frac{180-\theta}{2} + \left(\frac{180-\theta}{2} - \alpha\right) = 180.</cmath> Solving, we find <math>\theta = 2 \alpha</math>. <br />
<br />
<br />
Relabelling our triangle, we get <math>\angle ABC = 90 - \alpha</math>. Dropping an altitude from <math>A</math> to <math>BC</math> and using the Pythagorean theorem, we find <math>[ABC] = 48</math>. Using the sine area formula, we see <math>\frac12 \cdot 10 \cdot 12 \cdot \sin(90-\alpha) = 48</math>. Plugging in our sine angle cofunction identity, <math>\sin(90-\alpha) = \cos(\alpha)</math>, we get <math>\alpha = \cos{^{-1}}{\frac45}</math>. <br />
<br />
<br />
Now, using the Law of Sines on <math>\triangle ADE</math>, we get <cmath>\frac{\sin{2\alpha}}{\frac{p}{q}} = \frac{\sin{(180-4\alpha)}}{10-\frac{p}{q}}.</cmath> After applying numerous trigonometric and algebraic tricks, identities, and simplifications, such as <math>\sin{(180-4\alpha)}=\sin{4\alpha}</math> and <math>\sin{\left(\cos{^{-1}}{\frac45}\right)} = \frac35</math>, we find <math>\frac{p}{q} = \frac{10\sin{2\alpha}}{\sin{4\alpha}+\sin{2\alpha}} = \frac{250}{39}</math>. <br />
<br />
<br />
<br />
Therefore, our answer is <math>250 + 39 = \boxed{289}</math>.<br />
<br />
<br />
~Tiblis<br />
<br />
== Solution 11 (Trigonometry)==<br />
We start by labelling a few angles (all of them in degrees). Let <math>\angle{BAC}=2\alpha = \angle{AED}, \angle{EDC}=\angle{ECD}=\alpha, \angle{DEC}=180-2\alpha, \angle{BDC}=3\alpha, \angle{DCB}=90-2\alpha, \angle{DBC}=90-\alpha</math>. Also let <math>AD=a</math>. By sine rule in <math>\triangle{ADE},</math> we get <math>\frac{a}{\sin{2\alpha}}=\frac{10-a}{\sin{4\alpha}} \implies \cos{2\alpha}=\frac{5}{a}-\frac{1}{2}</math><br />
Using sine rule in <math>\triangle{ABC}</math>, we get <math>\sin{\alpha}=\frac{3}{5}</math>. Hence we get <math>\cos{2\alpha}=1-2\sin^2{\alpha}=1-\frac{18}{25}=\frac{7}{25}</math>. Hence <math>\frac{5}{a}=\frac{1}{2}+\frac{7}{25}=\frac{39}{50} \implies a=\frac{250}{39}</math>. Therefore, our answer is <math>\boxed{289}</math><br />
<br />
Alternatively, use sine rule in <math>\triangle{BDC}</math>. (It’s easier)<br />
<br />
~Prabh1512<br />
<br />
== Solution 12 (Double Angle Identity)==<br />
<br />
We let AD=x. Then, angle A is <math>2\sin^{-1}(\frac{3}{5})</math> and so is angle <math>AED</math>. We note that <math>AE=10-x</math>. We drop an altitude from <math>D</math> to <math>AE</math>, and we call the foot <math>F</math>. We note that <math>AF=\frac{10-x}{2}</math>. Using the double angle identity, we have <math>\sin(2sin^{-1}(\frac{3}{5}))=2(\frac{3}{5})(\frac{4}{5})=\frac{24}{25}.</math> Therefore, <math>DF=\frac{24}{25}AD.</math> We now use the Pythagorean Theorem, which gives <math>(\frac{10-x}{2})^2+(\frac{24}{25}x)^2=x^2</math>. Rearranging and simplifying, this becomes <math>429x^2-12500x+62500=0</math>. Using the quadratic formula, this is <math>\frac{12500\pm\sqrt{12500^2-250000\cdot429}}{858}</math>. We take out a <math>10000</math> from the square root and make it a <math>100</math> outside of the square root to make it simpler. We end up with <math>\frac{12500\pm7500}{858}</math>. We note that this must be less than 10 to ensure that <math>10-x</math> is positive. Therefore, we take the minus, and we get <math>\frac{5500}{858}=\frac{250}{39} -> \fbox{289}.</math><br />
==Video Solution==<br />
<br />
https://www.youtube.com/watch?v=iE8paW_ICxw<br />
<br />
<br />
https://youtu.be/dI6uZ67Ae2s ~yofro<br />
<br />
==See Also==<br />
{{AIME box|year=2018|n=I|num-b=3|num-a=5}}<br />
{{MAA Notice}}</div>John0512https://artofproblemsolving.com/wiki/index.php?title=AMC_historical_results&diff=148641AMC historical results2021-03-05T23:48:27Z<p>John0512: /* AMC 10B */</p>
<hr />
<div><!-- Post AMC statistics and lists of high scorers here so that the AMC page doesn't get cluttered. --><br />
This is the '''AMC historical results''' page. This page should include results for the [[AIME]] as well. For [[USAMO]] results, see [[USAMO historical results]].<br />
==Explanations of Awards==<br />
*Average score: Average score of all participants, regardless of age, grade level, gender, and region.<br />
*AIME floor: Before 2020, only the top 2.5% in AMC 10 and 5% in AMC 12 were invited to participate in AIME; in 2020 and after, the MAA invites more students to take AIME, so the AIME cutoffs are lower than the AMC 10 top 2.5% and AMC 12 top 5% cutoffs. In other words, in 2019 and before, the AIME floor is equivalent to the top 2.5% in AMC 10 and 5% in AMC 12 cutoffs; in 2020 and after, the AIME floor is more like the top 3-5% in AMC 10 and 5-10% in AMC 12 cutoffs.<br />
*Distinction: Only exists in 2020 and after. The top 2.5% in AMC 10 and 5% in AMC 12. In previous years, known as the honor roll and equivalent to the AIME floor.<br />
*Distinguished Honor Roll: The top 1% in either competition.<br />
<br />
==2021==<br />
<br />
===AMC 10A===<br />
*Average score: 72.5<br />
*AIME floor: 103.5<br />
<br />
===AMC 10B===<br />
*Average score: 72.2<br />
*AIME floor: 102<br />
*Distinction: 108 <br />
*Distinguished Honor Roll: 126<br />
<br />
===AMC 12A===<br />
*Average score: <br />
*AIME floor: 93<br />
*Distinction: 114<br />
*Distinguished Honor Roll: 133.5<br />
<br />
===AMC 12B===<br />
<br />
*Average score: Not Announced<br />
*AIME floor: 91.5<br />
*Distinction: 109.5<br />
*Distinguished Honor Roll: 132<br />
<br />
===AIME I===<br />
*Average score: <br />
*Median score: <br />
*USAMO cutoff: <br />
*USAJMO cutoff:<br />
===AIME II===<br />
*Average score: <br />
*Median score: <br />
*USAMO cutoff: <br />
*USAJMO cutoff:<br />
===AMC 8===<br />
*Average score:<br />
*Honor Roll:<br />
*DHR:<br />
<br />
==2020==<br />
===AMC 10A===<br />
*Average score: 64.29<br />
*AIME floor: 103.5<br />
*Distinction: 105<br />
*Distinguished Honor Roll: 124.5<br />
<br />
===AMC 10B===<br />
*Average score: 61.22<br />
*AIME floor: 102<br />
*Distinction: 103.5 <br />
*Distinguished Honor Roll: 120<br />
<br />
===AMC 12A===<br />
*Average score: 61.42<br />
*AIME floor: 87<br />
*Distinction: 100.5<br />
*Distinguished Honor Roll: 123<br />
<br />
===AMC 12B===<br />
*Average score: 60.47<br />
*AIME floor: 87<br />
*Distinction: 97.5<br />
*Distinguished Honor Roll: 120<br />
<br />
===AIME I===<br />
*Average score: 5.69<br />
*Median score: 6<br />
*USAMO cutoff: 233.5 (AMC 12A), 235 (AMC 12B)<br />
*USAJMO cutoff: 229.5 (AMC 10A), 230 (AMC 10B)<br />
<br />
===AIME II===<br />
Due to COVID-19, the 2020 AIME II was administered online and referred to as the AOIME.<br />
*Average score: 6.13<br />
*Median score: 6<br />
*USAMO cutoff: 234 (AMC 12A), 234.5 (AMC 12B)<br />
*USAJMO cutoff: 233.5 (AMC 10A), 229.5 (AMC 10B)<br />
<br />
===AMC 8===<br />
*Average score: 10.00<br />
*Honor Roll: 18<br />
*DHR: 21<br />
<br />
==2019==<br />
===AMC 10A===<br />
*Average score: 51.66<br />
*Honor roll: 96<br />
*AIME floor: 103.5<br />
*DHR: 123<br />
<br />
===AMC 10B===<br />
*Average score: 58.42<br />
*Honor roll: 102<br />
*AIME floor: 108<br />
*Distinguished Honor Roll: 121.5<br />
<br />
===AMC 12A===<br />
*Average score: 49.22<br />
*AIME floor: 84<br />
*DHR: 121.5<br />
<br />
===AMC 12B===<br />
*Average score: 56.73<br />
*AIME floor: 94.5 <br />
*DHR: 123<br />
<br />
===AIME I===<br />
*Average score: 5.88<br />
*Median score: 6<br />
*USAMO cutoff: 220 (AMC 12A), 230.5 (AMC 12B)<br />
*USAJMO cutoff: 209.5 (AMC 10A), 216 (AMC 10B)<br />
<br />
===AIME II===<br />
*Average score: 6.47<br />
*Median score: 6<br />
*USAMO cutoff: 230.5 (AMC 12A), 236 (AMC 12B)<br />
*USAJMO cutoff: 216.5 (AMC 10A), 220.5 (AMC 10B)<br />
<br />
===AMC 8===<br />
*Average score: 9.43<br />
*Honor roll: 19<br />
*DHR: 23<br />
<br />
==2018==<br />
===AMC 10A===<br />
*Average score: 53.84<br />
*Honor roll: 100.5<br />
*AIME floor: 111<br />
*DHR: 127.5<br />
<br />
===AMC 10B===<br />
*Average score: 57.81<br />
*Honor roll: 97.5<br />
*AIME floor: 108<br />
*DHR: 123<br />
<br />
===AMC 12A===<br />
*Average score: 56.36<br />
*AIME floor: 93<br />
*DHR: 120<br />
<br />
===AMC 12B===<br />
*Average score: 57.85<br />
*AIME floor: 99<br />
*DHR: 126<br />
<br />
===AIME I===<br />
*Average score: 5.09<br />
*Median score: 5<br />
*USAMO cutoff: 215 (AMC 12A), 235 (AMC 12B)<br />
*USAJMO cutoff: 222 (AMC 10A), 212 (AMC 10B)<br />
<br />
===AIME II===<br />
*Average score: 5.48<br />
*Median score: 5<br />
*USAMO cutoff: 216 (AMC 12A), 230.5 (AMC 12B)<br />
*USAJMO cutoff: 222 (AMC 10A), 212 (AMC 10B)<br />
<br />
===AMC 8===<br />
*Average score: 8.51<br />
*Honor roll: 15<br />
*DHR: 19<br />
<br />
==2017==<br />
===AMC 10A===<br />
*Average score: 59.33<br />
*AIME floor: 112.5<br />
*DHR: 127.5<br />
<br />
===AMC 10B===<br />
*Average score: 66.56<br />
*AIME floor: 120<br />
*DHR: 136.5<br />
<br />
===AMC 12A===<br />
*Average score: 59.66<br />
*AIME floor: 96<br />
*DHR: 115.5<br />
<br />
===AMC 12B===<br />
*Average score: 58.35<br />
*AIME floor: 100<br />
*DHR: 129<br />
<br />
===AIME I===<br />
*Average score: 5.69<br />
*Median score: 5<br />
*USAMO cutoff: 225 (AMC 12A), 235 (AMC 12B)<br />
*USAJMO cutoff: 224.5 (AMC 10A), 233 (AMC 10B)<br />
<br />
===AIME II===<br />
*Average score: 5.64<br />
*Median score: 5<br />
*USAMO cutoff: 221 (AMC 12A), 230.5 (AMC 12B)<br />
*USAJMO cutoff: 219 (AMC 10A), 225 (AMC 10B)<br />
<br />
===AMC 8===<br />
*Average score: 8.96<br />
*Honor roll: 17<br />
*DHR: 20<br />
<br />
==2016==<br />
===AMC 10A===<br />
*Average score: 65.31<br />
*AIME floor: 110<br />
*DHR: 120<br />
<br />
===AMC 10B===<br />
*Average score: 65.40<br />
*AIME floor: 110<br />
*DHR: 124.5<br />
<br />
===AMC 12A===<br />
*Average score: 60.32<br />
*AIME floor: 93<br />
*DHR: 111<br />
<br />
===AMC 12B===<br />
*Average score: 68.65<br />
*AIME floor: 100.5<br />
*DHR: 127.5<br />
<br />
===AIME I===<br />
*Average score: 5.83<br />
*Median score: 6<br />
*USAMO cutoff: 220<br />
*USAJMO cutoff: 210.5<br />
<br />
===AIME II===<br />
*Average score: 4.43<br />
*Median score: 4<br />
*USAMO cutoff: 205<br />
*USAJMO cutoff: 200<br />
<br />
===AMC 8===<br />
*Average score: 9.36<br />
*Honor roll: 18<br />
*DHR: 22<br />
<br />
==2015==<br />
===AMC 10A===<br />
*Average score: 73.39<br />
*AIME floor: 106.5<br />
*DHR: 115.5<br />
<br />
===AMC 10B===<br />
*Average score: 76.09<br />
*AIME floor: 120<br />
*DHR: 132<br />
<br />
===AMC 12A===<br />
*Average score: 69.90<br />
*AIME floor: 99<br />
*DHR: 117<br />
<br />
===AMC 12B===<br />
*Average score: 66.88<br />
*AIME floor: 100.5<br />
*DHR: 126<br />
<br />
===AIME I===<br />
*Average score: 5.29<br />
*Median score: 5<br />
*USAMO cutoff: 219.0<br />
*USAJMO cutoff: 213.0<br />
<br />
===AIME II===<br />
*Average score: 6.63<br />
*Median score: 6<br />
*USAMO cutoff: 229.0<br />
*USAJMO cutoff: 223.5<br />
<br />
===AMC 8===<br />
*Average score: 8.55<br />
*Honor roll: 16<br />
*DHR: 21<br />
<br />
==2014==<br />
===AMC 10A===<br />
*Average score: 63.34<br />
*AIME floor: 120<br />
*DHR: 132<br />
<br />
===AMC 10B===<br />
*Average score: 71.42<br />
*AIME floor: 120<br />
*DHR: 132<br />
<br />
===AMC 12A===<br />
*Average score: 63.60<br />
*AIME floor: 93<br />
*DHR: 109.5<br />
<br />
===AMC 12B===<br />
*Average score: 68.12<br />
*AIME floor: 100.5<br />
*DHR: 121.5<br />
<br />
===AIME I===<br />
*Average score: 4.88<br />
*Median score: 5<br />
*USAMO cutoff: 211.5<br />
*USAJMO cutoff: 211<br />
<br />
===AIME II===<br />
*Average score: 5.49<br />
*Median score: 5<br />
*USAMO cutoff: 211.5<br />
*USAJMO cutoff: 211<br />
<br />
===AMC 8===<br />
*Average score: 11.43<br />
*Honor roll: 19<br />
*DHR: 23<br />
<br />
==2013==<br />
===AMC 10A===<br />
*Average score: 72.50<br />
*AIME floor: 108<br />
*DHR: 117<br />
<br />
===AMC 10B===<br />
*Average score: 72.62<br />
*AIME floor: 120<br />
*DHR: 129<br />
<br />
===AMC 12A===<br />
*Average score: 65.06<br />
*AIME floor: 88.5<br />
*DHR: 106.5<br />
<br />
===AMC 12B===<br />
*Average score: 64.21<br />
*AIME floor: 93<br />
*DHR: 108<br />
<br />
===AIME I===<br />
*Average score: 4.69<br />
*Median score: 4<br />
*USAMO cutoff: 209<br />
*USAJMO cutoff: 210.5<br />
<br />
===AIME II===<br />
*Average score: 6.56<br />
*Median score: 6<br />
*USAMO cutoff: 209<br />
*USAJMO cutoff: 210.5<br />
<br />
===AMC 8===<br />
*Average score: 10.69<br />
*Honor roll: 18<br />
*DHR: 22<br />
<br />
==2012==<br />
===AMC 10A===<br />
*Average score: 72.51<br />
*AIME floor: 115.5<br />
*DHR: 121.5<br />
<br />
===AMC 10B===<br />
*Average score: 76.59<br />
*AIME floor: 120<br />
*DHR: 133.5<br />
<br />
===AMC 12A===<br />
*Average score: 64.62<br />
*AIME floor: 94.5<br />
*DHR: 109.5<br />
<br />
===AMC 12B===<br />
*Average score: 70.08<br />
*AIME floor: 99<br />
*DHR: 114<br />
<br />
===AIME I===<br />
*Average score: 5.13<br />
*Median score: 5<br />
*USAMO cutoff: 204.5<br />
*USAJMO cutoff: 204<br />
<br />
===AIME II===<br />
*Average score: 4.94<br />
*Median score: 5<br />
*USAMO cutoff: 204.5<br />
*USAJMO cutoff: 204<br />
<br />
===AMC 8===<br />
*Average score: 10.67<br />
*Honor roll: 18<br />
*DHR: 22<br />
<br />
==2011==<br />
===AMC 10A===<br />
*Average score: 64.24<br />
*AIME floor: 117<br />
*DHR: 129<br />
<br />
===AMC 10B===<br />
*Average score: 71.78<br />
*AIME floor: 117<br />
*DHR: 133.5<br />
<br />
===AMC 12A===<br />
*Average score: 65.38<br />
*AIME floor: 93<br />
*DHR: 112.5<br />
<br />
===AMC 12B===<br />
*Average score: 64.71<br />
*AIME floor: 97.5<br />
*DHR: 121.5<br />
<br />
===AIME I===<br />
*Average score: 2.23<br />
*Median score: 2<br />
*USAMO cutoff: 188<br />
*USAJMO cutoff: 179<br />
<br />
===AIME II===<br />
*Average score: 5.47<br />
*Median score: 5<br />
*USAMO cutoff: 215.5<br />
*USAJMO cutoff: 196.5<br />
<br />
===AMC 8===<br />
*Average score: 10.75<br />
*Honor roll: 17<br />
*DHR: 22<br />
<br />
==2010==<br />
===AMC 10A===<br />
*Average score: 68.11<br />
*AIME floor: 115.5<br />
<br />
===AMC 10B===<br />
*Average score: 68.57<br />
*AIME floor: 118.5<br />
<br />
===AMC 12A===<br />
*Average score: 61.02<br />
*AIME floor: 88.5<br />
*DHR cutoff: 108<br />
<br />
===AMC 12B===<br />
*Average score: 59.98<br />
*AIME floor: 88.5<br />
*DHR cutoff: 109.5<br />
<br />
===AIME I===<br />
*Average score: 5.90<br />
*Median score: 6<br />
*USAMO cutoff: 208.5 (204.5 for non juniors and seniors)<br />
*USAJMO cutoff: 188.5<br />
<br />
===AIME II===<br />
*Average score: 3.39<br />
*Median score: 3<br />
*USAMO cutoff: 208.5 (204.5 for non juniors and seniors)<br />
*USAJMO cutoff: 188.5<br />
<br />
===AMC 8===<br />
*Average score: 9.59<br />
*Honor roll: 17<br />
*DHR: 22<br />
<br />
==2009==<br />
===AMC 10A===<br />
*Average score: 67.41<br />
*AIME floor: 117<br />
<br />
===AMC 10B===<br />
*Average score: 74.73<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 66.37<br />
*AIME floor: 97.5<br />
<br />
===AMC 12B===<br />
*Average score: 71.88<br />
*AIME floor: 100 (Top 5% (1.00))<br />
<br />
===AIME I===<br />
*Average score: 4.17<br />
*Median score: 4<br />
*USAMO floor: <br />
<br />
===AIME II===<br />
*Average score: 3.27<br />
*Median score: 3<br />
*USAMO floor:<br />
<br />
===AMC 8===<br />
*Average score: 10.28<br />
*Honor roll: 17<br />
*DHR: 20<br />
<br />
==2008==<br />
===AMC 10A===<br />
*Average score: 60.25<br />
*AIME floor: 117<br />
<br />
===AMC 10B===<br />
*Average score: <br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 65.6<br />
*AIME floor: 97.5<br />
<br />
===AMC 12B===<br />
*Average score: 68.9<br />
*AIME floor: 97.5<br />
<br />
===AIME I===<br />
*Average score: 4.77<br />
*Median score: 4<br />
*USAMO floor: <br />
<br />
===AIME II===<br />
*Average score: 5.27<br />
*Median score: 5<br />
*USAMO floor:<br />
<br />
===AMC 8===<br />
*Average score: 11.45<br />
*Honor roll: 19<br />
*DHR: 22<br />
<br />
==2007==<br />
<br />
===AMC 10A===<br />
*Average score: 67.9<br />
*AIME floor: 117<br />
<br />
===AMC 10B=== <br />
*Average score: 61.5<br />
*AIME floor: 115.5<br />
<br />
===AMC 12A===<br />
*Average score: 66.8<br />
*AIME floor: 97.5<br />
<br />
===AMC 12B===<br />
*Average score: 73.1<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score: 5<br />
*Median score: 3<br />
*USAMO floor: 6<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AMC 8===<br />
*Average score: 9.87<br />
*Honor roll: 17<br />
*DHR: 21<br />
<br />
==2006==<br />
===AMC 10A===<br />
*Average score: 79.0<br />
*AIME floor: 120<br />
<br />
===AMC 10B===<br />
*Average score: 68.5<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 85.7<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 85.5<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score: <br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AMC 8===<br />
*Honor roll: 17<br />
*DHR: 21<br />
<br />
==2005==<br />
===AMC 10A===<br />
*Average score: 74.0<br />
*AIME floor: 120<br />
<br />
===AMC 10B===<br />
*Average score: 79.0<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 78.7<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 83.4<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AMC 8===<br />
*Honor roll: 16<br />
*DHR: 20<br />
<br />
==2004==<br />
===AMC 10A===<br />
*Average score: 69.1<br />
*AIME floor: 110<br />
<br />
===AMC 10B===<br />
*Average score: 80.4<br />
*AIME floor: 115<br />
<br />
===AMC 12A===<br />
*Average score: 73.9<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 84.5<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AMC 8===<br />
*Honor roll: 17<br />
*DHR: 21<br />
<br />
==2003==<br />
===AMC 10A===<br />
*Average score: 74.4<br />
*AIME floor: 119<br />
<br />
===AMC 10B===<br />
*Average score: 79.6<br />
*AIME floor: 121<br />
<br />
===AMC 12A===<br />
*Average score: 77.8<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 76.6<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AMC 8===<br />
*Honor roll: 18<br />
*DHR: 22<br />
<br />
==2002==<br />
===AMC 10A===<br />
*Average score: 68.5<br />
*AIME floor: 115<br />
<br />
===AMC 10B===<br />
*Average score: 74.9<br />
*AIME floor: 118<br />
<br />
===AMC 12A===<br />
*Average score: 72.7<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 80.8<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2001==<br />
===AMC 10===<br />
*Average score: 67.8<br />
*AIME floor: 116<br />
<br />
===AMC 12===<br />
*Average score: 56.6<br />
*AIME floor: 84<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2000==<br />
===AMC 10===<br />
*Average score: <math>64.2</math><br />
*AIME floor: <math>110</math><br />
<br />
===AMC 12===<br />
*Average score: <math>64.9</math><br />
*AIME floor: <math>92</math><br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==1999==<br />
===AHSME===<br />
*Average score: <math>68.8</math><br />
*AIME floor:<br />
<br />
===AIME===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:</div>John0512https://artofproblemsolving.com/wiki/index.php?title=User:Sugar_rush&diff=148241User:Sugar rush2021-03-02T18:22:29Z<p>John0512: /* User Count */</p>
<hr />
<div>==<font color="black" style="font-family: ITC Avant Garde Gothic Std, Verdana"><div style="margin-left:10px">User Count</div></font>==<br />
<div style="margin-left: 10px; margin-bottom:10px"><font color="black">If this is your first time visiting this page, edit it by incrementing the user count below by one.</font></div><br />
<center><font size="120px”>22</font></center><br />
</div><br />
<div style="border:2px solid black; background:#ccccff;-webkit-border-radius: 10px; align:center"><br />
<br />
==<font color="black" style="font-family: ITC Avant Garde Gothic Std, Verdana"><div style="margin-left:10px">About Me</div></font>==<br />
<div style="margin-left: 10px; margin-bottom:10px"><font color="black">sugar_rush is 14 years old.<br><br />
sugar_rush was born on January 12, 2007.<br />
<br />
sugar_rush is in 8th grade.<br />
<br />
sugar_rush is a competitive runner and pro at math.<br />
<br />
Got 96 on AMC 12A (although there are debates on whether it's 90 or 96) and 109.5 on AMC 10B<br />
<br />
</font></div><br />
</div><br />
<div style="border:2px solid black; background:#bbbbbb;-webkit-border-radius: 10px; align:center"><br />
<br />
==<font color="black" style="font-family: ITC Avant Garde Gothic Std, Verdana"><div style="margin-left:10px">Goals</div></font>==<br />
<div style="margin-left: 10px; margin-right: 10px; margin-bottom:10px">Qualify for AIME and get at least 6 on 2021 AIME I (never qualified for AIME before)<br />
<br />
</div><br />
</div><br />
Also: here is this guy's youtube channel: https://www.youtube.com/c/PunxsutawnyPhil</div>John0512https://artofproblemsolving.com/wiki/index.php?title=User:Piphi&diff=143554User:Piphi2021-01-28T04:05:34Z<p>John0512: /* User Count */</p>
<hr />
<div>{{User:Piphi/Template:Header}}<br />
<br><br />
__NOTOC__<div style="border:2px solid black; -webkit-border-radius: 10px; background:#F0F2F3"><br />
==<font color="black" style="font-family: ITC Avant Garde Gothic Std, Verdana"><div style="margin-left:10px">User Count</div></font>==<br />
<div style="margin-left: 10px; margin-bottom:10px"><font color="black">If this is your first time visiting this page, edit it by incrementing the user count below by one.</font></div><br />
<center><font size="100px">447</font></center><br />
</div><br />
<div style="border:2px solid black; background:#919293;-webkit-border-radius: 10px; align:center"><br />
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==<font color="black" style="font-family: ITC Avant Garde Gothic Std, Verdana"><div style="margin-left:10px">About Me</div></font>==<br />
<div style="margin-left: 10px; margin-bottom:10px"><font color="black"><font color="black"><font style="font-size: 26px;line-height: 45px;font-weight: 700;">I am currently postbanned, to talk to me, go [[User_talk:Piphi|here]].</font><br />
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Piphi is the creator of the [[User:Piphi/Games|AoPS Wiki Games by Piphi]], the future of games on AoPS.<br><br />
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Piphi started the signature trend at around May 2020.<br><br />
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Piphi has been very close to winning multiple [[Greed Control]] games, piphi placed 5th in game #18 and 2nd in game #19. Thanks to piphi, Greed Control games have started to be kept track of. Piphi made a spreadsheet that has all of Greed Control history [https://artofproblemsolving.com/community/c19451h2126208p15569802 here].<br><br />
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Piphi also found out who won [[Reaper]] games #1 and #2 as seen [https://artofproblemsolving.com/community/c19451h1826745p15526330 here].<br><br />
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Piphi created the [[AoPS Administrators]] page, added most of the AoPS Admins to it, and created the scrollable table.<br><br />
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Piphi has also added a lot of the info that is in the [[Reaper Archives]].<br><br />
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Piphi has a side-project that is making the Wiki's [[Main Page]] look better, you can check that out [[User:Piphi/AoPS Wiki|here]].<br><br />
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Piphi published Greed Control Game 19 statistics [https://artofproblemsolving.com/community/c19451h2126212 here].<br />
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Piphi has a post that was made an announcement in a official AoPS Forum [https://artofproblemsolving.com/community/c68h2175116 here].<br />
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Piphi is a proud member of [https://artofproblemsolving.com/community/c562043 The Interuniversal GMAAS Society].<br />
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Piphi has won 2 gold trophies at the [https://artofproblemsolving.com/community/c1124279 Asymptote Competition] and is now part of the staff.<br />
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==<font color="#f0f2f3" style="font-family: ITC Avant Garde Gothic Std, Verdana"><div style="margin-left:10px">Goals</div></font>==<br />
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You can check out more goals/statistics [[User:Piphi/Statistics|here]].<br />
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A User Count of 500<br />
{{User:Piphi/Template:Progress_Bar|88.8|width=100%}}<br />
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200 subpages of [[User:Piphi]]<br />
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200 signups for [[User:Piphi/Games|AoPS Wiki Games by Piphi]]<br />
{{User:Piphi/Template:Progress_Bar|49.5|width=100%}}<br />
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Make 10,000 edits<br />
{{User:Piphi/Template:Progress_Bar|21.33|width=100%}}</font></div><br />
</div></div>John0512