https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=GeronimoStilton&feedformat=atomAoPS Wiki - User contributions [en]2024-03-28T09:43:25ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=USAMO_historical_results&diff=153757USAMO historical results2021-05-15T13:31:22Z<p>GeronimoStilton: </p>
<hr />
<div>This is the '''USAMO historical results''' page. <!-- Post results here so the [[USAMO]] page does not become cluttered. --><br />
<br />
==USAMO Perfect Scorers==<br />
<br />
Since 1996, a perfect score on the [[USAMO]] has been 42 points for 6 problems. Prior to 1996 a perfect score was 100 points across 5 problems.<br />
<br />
*2020:<br />
**Brandon Wang<br />
**Luke Robitaille<br />
*2019:<br />
**Kevin Liu<br />
**Colin Tang<br />
**Brandon Wang<br />
**Daniel Zhu<br />
*2018: None<br />
*2017: None<br />
*2016:<br />
**Allen Liu<br />
*2015:<br />
**Allen Liu<br />
**David Stoner<br />
*2014: <br />
**Joshua Brakensiek<br />
*2013: None<br />
*2012: <br />
**Alex Zhu<br />
**Thomas Swayze<br />
**Xiaoyu He<br />
**Mitchell Lee<br />
**Samuel Zbarsky<br />
*2011:<br />
**David Yang<br />
**Evan O'Dorney<br />
*2010: None<br />
*2008: None<br />
*2007: None<br />
*2006:<br />
**Brian Lawrence<br />
*2005: None<br />
*2004:<br />
**Tiankai Liu<br />
*2003:<br />
**Tiankai Liu<br />
**Po-Ru Loh<br />
*2002: <br />
**Daniel Kane<br />
**Ricky Liu<br />
**Tiankai Liu<br />
**Po-Ru Loh<br />
**Inna Zakharevich<br />
*1996:<br />
**Chris Chang<br />
*1992:<br />
**Kiran Kedlaya<br />
**Lenny Ng<br />
<br />
==USAMO Winners==<br />
<br />
The top 12 scorers on the [[USAMO]] are currently designated as USAMO winners. In the past, this number was smaller.<br />
*2021:<br />
**Daniel Hong<br />
**Daniel Yuan<br />
**Eric Shen<br />
**Gopal Goel<br />
**Jaedon Whyte<br />
**Kevin Cong<br />
**Luke Robitaille<br />
**Maxim Li<br />
**Noah Walsh<br />
**Quanlin Chen<br />
**William Wang<br />
**Xinyang Chen<br />
**Zifan Wang<br />
*2020:<br />
**Ankit Bisain<br />
**Brandon Chen<br />
**Quanlin Chen<br />
**Kevin Cong<br />
**Eric Gan<br />
**Gopal Goel<br />
**Thomas Guo<br />
**Daniel Hong<br />
**Tianze Jiang<br />
**Benjamin Kang<br />
**Maxim Li<br />
**Brian Liu<br />
**Luke Robitaille<br />
**Noah Walsh<br />
**Edward Wan<br />
**Brandon Wang<br />
*2019:<br />
**Vincent Bian<br />
**Milan Haiman<br />
**Vincent Huang<br />
**Kevin Liu<br />
**Luke Robitaille<br />
**Victor Rong<br />
**Carl Schildkraut<br />
**Colin Shanmo Tang<br />
**Edward Wan<br />
**Brandon Wang<br />
**Guanpeng Xu<br />
**Daniel Zhu<br />
*2018:<br />
**Eric Gan<br />
**Thomas Guo<br />
**Vincent Huang<br />
**Joshua Lee<br />
**Michael Ren<br />
**Victor Rong<br />
**Carl Schildkraut<br />
**Mihir Singhal<br />
**Edward Wan<br />
**Brandon Wang<br />
**Guanpeng Xu<br />
**Andrew Yao<br />
*2017:<br />
**Zachary Chroman<br />
**Andrew Gu<br />
**James Lin<br />
**Daniel Liu<br />
**Michael Ren<br />
**Victor Rong<br />
**Ashwin Sah<br />
**Mihir Singhal<br />
**Alec Sun<br />
**Kada Williams<br />
**Yuan Yao<br />
**William Zhao<br />
*2016:<br />
**Ankan Bhattacharya<br />
**Ruidi Cao<br />
**Hongyi Chen<br />
**Jacob Klegar<br />
**James Lin<br />
**Allen Liu<br />
**Junyao Peng<br />
**Kevin Ren<br />
**Mihir Singhal<br />
**Alec Sun<br />
**Kevin Sun<br />
**Yuan Yao<br />
*2015:<br />
**Ryan Alweiss<br />
**Kritkorn Karntikoon<br />
**Michael Kural<br />
**Celine Liang<br />
**Allen Liu<br />
**Yang Liu<br />
**Shyam Narayanan<br />
**Kevin Ren<br />
**Zhuoqun Song<br />
**David Stoner<br />
**Kevin Sun<br />
**Danielle Wang<br />
*2014:<br />
**Joshua Brakensiek<br />
**Evan Chen<br />
**Ravi Jagadeesan<br />
**Allen Liu<br />
**Nipun Pitimanaaree<br />
**Mark Sellke<br />
**Zhuoqun Song<br />
**David Stoner<br />
**Kevin Sun<br />
**James Tao<br />
**Alexander Whatley<br />
**Scott Wu<br />
*2013<br />
**Calvin Deng<br />
**Andrew He<br />
**Ravi Jagadeesan<br />
**Pakawut Jiradilok<br />
**Ray Li<br />
**Kevin Li<br />
**Mark Sellke<br />
**Bobby Shen<br />
**Zhuoqun Song<br />
**David Stoner<br />
**Thomas Swayze<br />
**Victor Wang<br />
*2012:<br />
**Andre Arslan<br />
**Joshua Brakensiek<br />
**Calvin Deng<br />
**Xiaoyu He<br />
**Ravi Jagadeesan<br />
**Mitchell Lee<br />
**Alex Song<br />
**Thomas Swayze<br />
**Victor Wang<br />
**David Yang<br />
**Samuel Zbarsky<br />
**Alex Zhu<br />
*2011:<br />
**Wenyu Cao<br />
**Zijing Gao<br />
**Benjamin Gunby<br />
**Xiaoyu He<br />
**Ravi Jagadeesan<br />
**Spencer Kwon<br />
**Mitchell Lee<br />
**Ray Li<br />
**Evan O'Dorney<br />
**Mark Sellke<br />
**David Yang<br />
**Joy Zheng<br />
*2010:<br />
**Timothy Chu<br />
**Calvin Deng<br />
**Michael Druggan<br />
**Brian Hamrick<br />
**Travis Hance<br />
**Xiaoyu He<br />
**Mitchell Lee<br />
**In Sung Na<br />
**Evan O'Dorney<br />
**Toan Phan<br />
**Hunter Spink<br />
**Allen Yuan<br />
*2009:<br />
**John Berman<br />
**Sergei Bernstein<br />
**Wenyu Cao<br />
**Robin Cheng<br />
**Vlad Firoiu<br />
**Eric Larson<br />
**Delong Meng<br />
**Qinxuan Pan<br />
**Panupong Pasupat<br />
**Toan Phan<br />
**David Rush<br />
**David Yang<br />
*2008:<br />
**David Benjamin<br />
**TaoRan Chen<br />
**Paul Christiano<br />
**Sam Elder<br />
**Shaunak Kishore<br />
**Delong Meng<br />
**Evan O'Dorney<br />
**Qinxuan Pan<br />
**David Rolnick<br />
**Colin Sandon<br />
**Krishanu Sankar<br />
**Alex Zhai<br />
*2007:<br />
**Sergei Bernstein<br />
**Sherry Gong<br />
**Adam Hesterberg<br />
**Eric Larson<br />
**Brian Lawrence<br />
**Tedrick Leung<br />
**Haitao Mao<br />
**Delong Meng<br />
**Krishanu Sankar<br />
**Jacob Steinhardt<br />
**Arnav Tripathy<br />
**Alex Zhai<br />
*2006:<br />
**Brian Lawrence<br />
**Alex Zhai<br />
**Yufei Zhao<br />
**Peng Shi<br />
**Sherry Gong<br />
**Richard McCutchen<br />
**Yi Sun<br />
**Arnav Tripathy<br />
**Taehyeon (Ryan) Ko<br />
**Yi Han<br />
**Yakov Berchenko-Kogan<br />
**Tedrick Leung<br />
*2005:<br />
**Robert Cordwell<br />
**Zhou Fan<br />
**Sherry Gong<br />
**Rishi Gupta<br />
**Hyun Soo Kim<br />
**Brian Lawrence<br />
**Albert Ni<br />
**Natee Pitiwan<br />
**Eric Price<br />
**Peng Shi<br />
**Yi Sun<br />
**Yufei Zhao<br />
*2004:<br />
**Tiankai Liu<br />
**Jae Bae<br />
**Jongmin Baek<br />
**Oleg Golberg<br />
**Matt Ince<br />
**Janos Kramar<br />
**Alison Miller<br />
**Aaron Pixton<br />
**Brian Rice<br />
**Jacob Tsimerman<br />
**Ameya Velingker<br />
**Tony Zhang<br />
*2003:<br />
**Tiankai Liu<br />
**Po Ru Loh<br />
**Boris Alexeev<br />
**Jae Bae<br />
**Daniel Kane<br />
**Anders Kaseorg<br />
**Mark Lipson<br />
**Po Ling Loh<br />
**Aaron Pixton<br />
**Kwokfung Tang<br />
**Tony Zhang<br />
**Yan Zhang<br />
*1998:<br />
**Sasha Schwartz<br />
**Melanie Wood<br />
**Kevin Lacker<br />
**Reid Barton<br />
**Gabriel Carroll<br />
**Paul Valiant<br />
**David Speyer<br />
**David Vickrey<br />
*1997:<br />
**Josh Nichols-Barrer<br />
**Daniel Stronger<br />
**John Clyde<br />
**Carl Bosley<br />
**Li-Chung Chen<br />
**Nathan Curtis<br />
**Davesh Maulik<br />
**Kevin Lacker<br />
*1996:<br />
**Chris Chang<br />
**Alex Saltman<br />
**Josh Nichols-Barrer<br />
**Carl Bosley<br />
**Michael Korn<br />
**Carl Miller<br />
**Nathan Curtis<br />
**Daniel Stronger<br />
*1995:<br />
**Aleksander Khazanov<br />
**Jacob Lurie<br />
**Chris Chang<br />
**Jay Chyung<br />
**Andrei Gnepp<br />
**Josh Nichols-Barrer<br />
**Samit Dasgupta<br />
**Craig Helfgott<br />
*1994:<br />
**Jeremy Bem<br />
**Aleksander Khazanov<br />
**Jonathan Weinstein<br />
**Jacob Lurie<br />
**Noam Shazeer<br />
**Stephen Wang<br />
**Chris Chang<br />
**Jacob Rasmussen<br />
*1989:<br />
**Jordan Ellenberg<br />
**Andrew Kresh<br />
**Jonathan Higa<br />
**[[Richard Rusczyk]]<br />
**Jeff Vanderkam<br />
**[[Sam Vandervelde]]<br />
**David Carlton<br />
<br />
==USAMO Honorable Mentions==<br />
<br />
Other students who finish (or tie to finish) in the top 24 of the [[USAMO]] receive Honorable Mention (often abbreviated HM).<br />
*2019<br />
**Quanlin Chen<br />
**Eric Gan<br />
**Sebastian Jeon<br />
**Tianze Jiang<br />
**Jeffrey Kwan<br />
**Benjamin Qi<br />
**William Zhao<br />
*2018<br />
**Adam Ardeishar<br />
**Hongyi Chen<br />
**Andrew Gu<br />
**Milan Haiman<br />
**Daniel Liu<br />
**Kevin Liu<br />
**Victor Luo<br />
**Benjamin Qi<br />
**Kevin Ren<br />
**Luke Robitaille<br />
**Dilhan Salgado<br />
**Nicholas Sun<br />
**Stan Zhang<br />
**William Zhao<br />
*2017<br />
**Swapnil Garg<br />
**Frank Han<br />
**Vincent Huang<br />
**Daniel Kim<br />
**Joshua Lee<br />
**Michael Ma<br />
**Qi Qi<br />
**Nikhil Reddy<br />
**Dilhan Salgado<br />
**Patrick Wang<br />
**Eric Zhang<br />
**Daniel Zhu<br />
*2016<br />
**Kapil Chandran<br />
**Andrew Gu<br />
**Brian Gu<br />
**Meghal Gupta<br />
**Michael Kural<br />
**Dhruv Medarametla<br />
**Eshaan Nichani<br />
**Brian Reinhart<br />
**Ashwin Sah<br />
**Franklyn Wang<br />
**Alexander Wei<br />
**Allen Yang<br />
**Rachel Zhang<br />
**Stan Zhang<br />
**Ziqi Zhou<br />
*2015<br />
**Demi Guo<br />
**Jacob Gurev<br />
**Kevin Li<br />
**James Lin<br />
**Michael Ma<br />
**Zachary Polansky<br />
**Samuel Reinehr<br />
**Ashwin Sah<br />
**Alexander Whatley<br />
**Jinhao Xu<br />
**Kevin Yang<br />
*2014<br />
**Lewis Chen<br />
**Ernest Chiu<br />
**Samuel Korsky<br />
**Yang Liu<br />
**Sammy Luo<br />
**Alexander Meiburg<br />
**Shyam Narayanan<br />
**Ashwin Sah<br />
**Nat Sothanaphan<br />
**Danielle Wang<br />
**Kevin Yang<br />
*2013:<br />
**Joshua Brakensiek<br />
**Lewis Chen<br />
**Sohail Farhangi<br />
**Kevin Huang<br />
**Allen Liu<br />
**Sammy Luo<br />
**Eric Schneider<br />
**Rachit Singh<br />
**Vikram Sundar<br />
**James Tao<br />
**David Yang<br />
**Kevin Yang<br />
*2011:<br />
**Kevin Chen<br />
**Calvin Deng<br />
**Michael Druggan<br />
**Albert Gu<br />
**Bobby Shen<br />
**Alex Song<br />
**Thomas Swayze<br />
**Tianqi Wu<br />
**Dai Yang<br />
**Kevin Yin<br />
**Allen Yuan<br />
**Samuel Zbarsky<br />
**Brian Zhang<br />
*2010:<br />
**Thomas Swayze<br />
**Kerry Xing<br />
**Tian-Yi Jiang<br />
**Brian Zhang<br />
**Archit Kulkarni<br />
**Patrick Hulin<br />
**Carl Lian<br />
**George Xing<br />
**Potcharapol Suteparuk<br />
**Jason Wu<br />
**Shijie Zheng<br />
**Robin Cheng<br />
**Bowei Liu<br />
**Wenyu Cao<br />
**Kevin Yin<br />
**Albert Gu<br />
*2009:<br />
**Timothy Chu<br />
**Brian Hamrick<br />
**Travis Hance<br />
**Jason Hoch<br />
**Tian-Yi Jiang<br />
**Sam Keller<br />
**Holden Lee<br />
**In-Sung Na<br />
**Ofir Nachum<br />
**Jarno Sun <br />
**Matthew Superdock<br />
**Tong Zhan<br />
**Mark Zhang<br />
*2008:<br />
**John Berman<br />
**Sergei Bernstein<br />
**Gregory Gauthier<br />
**Brian Hamrick<br />
**Eric Larson<br />
**Gaku Liu<br />
**Jeffrey Manning<br />
**Palmer Mebane<br />
**Danny Shi<br />
**Jacob Steinhardt<br />
**Matthew Superdock<br />
**Nicholas Triantafillou<br />
*2007:<br />
**David Benjamin<br />
**Gregory Brockman<br />
**Yingyu Gao<br />
**Yan Li<br />
**Gaku Liu<br />
**Jeffrey Manning<br />
**Palmer Mebane<br />
**Evan O'Dorney<br />
**Alexander Remorov<br />
**Max Rosett<br />
**Qiaochu Yuan<br />
**Bohua Zhan<br />
*2006:<br />
**Joseph Chu<br />
**Zachary Abel<br />
**Zarathustra Brady<br />
**Peter Diao<br />
**Richar Peng<br />
**Adam Hesterberg<br />
**Bohua Zhan<br />
**Kevin Modzelewski<br />
**Daniel Poore<br />
**Lin Fei<br />
**Jason Trigg<br />
**Tony Liu<br />
**Viktoriya Krakovna<br />
*2005:<br />
**Zachary Abel<br />
**Charles Chen<br />
**Evan Dummit<br />
**Adam Hesterberg<br />
**Richard Ho<br />
**Alexander Marcus<br />
**Richard McCutchen<br />
**Thomas Mildorf<br />
**Charles Nathanson<br />
**Keenan Pepper<br />
**Timothy Pollio<br />
*2004:<br />
**Boris Alexeev<br />
**Joshua Batson<br />
**Thomas Belulovich<br />
**Rishi Gupta<br />
**Anders Kaseorg<br />
**Hyun Soo Kim<br />
**John Kim<br />
**Po Ling Loh<br />
**Sira Sriswasdi<br />
**Yi Sun<br />
**Elena Udovina<br />
**Yufei Zhao<br />
*2003:<br />
**Timothy Abbott<br />
**Jeffrey Amos<br />
**Jongmin Baek<br />
**Richard Biggs<br />
**Steven Byrnes<br />
**Sherry Gong<br />
**Matt Ince<br />
**Nathaniel Ince<br />
**Pavel Kamyshev<br />
**Nicholas Ma<br />
**Alison Miller<br />
**Eric Price<br />
**David Stolp<br />
**Sean Ting<br />
**Tongke Xue<br />
*1995<br />
** Mathew Crawford<br />
*1993<br />
** Mathew Crawford<br />
<br />
==See also==<br />
[[Category:Historical results]]</div>GeronimoStiltonhttps://artofproblemsolving.com/wiki/index.php?title=USAMO_historical_results&diff=153756USAMO historical results2021-05-15T13:24:17Z<p>GeronimoStilton: </p>
<hr />
<div>This is the '''USAMO historical results''' page. <!-- Post results here so the [[USAMO]] page does not become cluttered. --><br />
<br />
==USAMO Perfect Scorers==<br />
<br />
Since 1996, a perfect score on the [[USAMO]] has been 42 points for 6 problems. Prior to 1996 a perfect score was 100 points across 5 problems.<br />
<br />
*2020:<br />
**Brandon Wang<br />
**Luke Robitaille<br />
*2019:<br />
**Kevin Liu<br />
**Colin Tang<br />
**Brandon Wang<br />
**Daniel Zhu<br />
*2018: None<br />
*2017: None<br />
*2016:<br />
**Allen Liu<br />
*2015:<br />
**Allen Liu<br />
**David Stoner<br />
*2014: <br />
**Joshua Brakensiek<br />
*2013: None<br />
*2012: <br />
**Alex Zhu<br />
**Thomas Swayze<br />
**Xiaoyu He<br />
**Mitchell Lee<br />
**Samuel Zbarsky<br />
*2011:<br />
**David Yang<br />
**Evan O'Dorney<br />
*2010: None<br />
*2008: None<br />
*2007: None<br />
*2006:<br />
**Brian Lawrence<br />
*2005: None<br />
*2004:<br />
**Tiankai Liu<br />
*2003:<br />
**Tiankai Liu<br />
**Po-Ru Loh<br />
*2002: <br />
**Daniel Kane<br />
**Ricky Liu<br />
**Tiankai Liu<br />
**Po-Ru Loh<br />
**Inna Zakharevich<br />
*1996:<br />
**Chris Chang<br />
*1992:<br />
**Kiran Kedlaya<br />
**Lenny Ng<br />
<br />
==USAMO Winners==<br />
<br />
The top 12 scorers on the [[USAMO]] are currently designated as USAMO winners. In the past, this number was smaller.<br />
*2019:<br />
**Vincent Bian<br />
**Milan Haiman<br />
**Vincent Huang<br />
**Kevin Liu<br />
**Luke Robitaille<br />
**Victor Rong<br />
**Carl Schildkraut<br />
**Colin Shanmo Tang<br />
**Edward Wan<br />
**Brandon Wang<br />
**Guanpeng Xu<br />
**Daniel Zhu<br />
*2018:<br />
**Eric Gan<br />
**Thomas Guo<br />
**Vincent Huang<br />
**Joshua Lee<br />
**Michael Ren<br />
**Victor Rong<br />
**Carl Schildkraut<br />
**Mihir Singhal<br />
**Edward Wan<br />
**Brandon Wang<br />
**Guanpeng Xu<br />
**Andrew Yao<br />
*2017:<br />
**Zachary Chroman<br />
**Andrew Gu<br />
**James Lin<br />
**Daniel Liu<br />
**Michael Ren<br />
**Victor Rong<br />
**Ashwin Sah<br />
**Mihir Singhal<br />
**Alec Sun<br />
**Kada Williams<br />
**Yuan Yao<br />
**William Zhao<br />
*2016:<br />
**Ankan Bhattacharya<br />
**Ruidi Cao<br />
**Hongyi Chen<br />
**Jacob Klegar<br />
**James Lin<br />
**Allen Liu<br />
**Junyao Peng<br />
**Kevin Ren<br />
**Mihir Singhal<br />
**Alec Sun<br />
**Kevin Sun<br />
**Yuan Yao<br />
*2015:<br />
**Ryan Alweiss<br />
**Kritkorn Karntikoon<br />
**Michael Kural<br />
**Celine Liang<br />
**Allen Liu<br />
**Yang Liu<br />
**Shyam Narayanan<br />
**Kevin Ren<br />
**Zhuoqun Song<br />
**David Stoner<br />
**Kevin Sun<br />
**Danielle Wang<br />
*2014:<br />
**Joshua Brakensiek<br />
**Evan Chen<br />
**Ravi Jagadeesan<br />
**Allen Liu<br />
**Nipun Pitimanaaree<br />
**Mark Sellke<br />
**Zhuoqun Song<br />
**David Stoner<br />
**Kevin Sun<br />
**James Tao<br />
**Alexander Whatley<br />
**Scott Wu<br />
*2013<br />
**Calvin Deng<br />
**Andrew He<br />
**Ravi Jagadeesan<br />
**Pakawut Jiradilok<br />
**Ray Li<br />
**Kevin Li<br />
**Mark Sellke<br />
**Bobby Shen<br />
**Zhuoqun Song<br />
**David Stoner<br />
**Thomas Swayze<br />
**Victor Wang<br />
*2012:<br />
**Andre Arslan<br />
**Joshua Brakensiek<br />
**Calvin Deng<br />
**Xiaoyu He<br />
**Ravi Jagadeesan<br />
**Mitchell Lee<br />
**Alex Song<br />
**Thomas Swayze<br />
**Victor Wang<br />
**David Yang<br />
**Samuel Zbarsky<br />
**Alex Zhu<br />
*2011:<br />
**Wenyu Cao<br />
**Zijing Gao<br />
**Benjamin Gunby<br />
**Xiaoyu He<br />
**Ravi Jagadeesan<br />
**Spencer Kwon<br />
**Mitchell Lee<br />
**Ray Li<br />
**Evan O'Dorney<br />
**Mark Sellke<br />
**David Yang<br />
**Joy Zheng<br />
*2010:<br />
**Timothy Chu<br />
**Calvin Deng<br />
**Michael Druggan<br />
**Brian Hamrick<br />
**Travis Hance<br />
**Xiaoyu He<br />
**Mitchell Lee<br />
**In Sung Na<br />
**Evan O'Dorney<br />
**Toan Phan<br />
**Hunter Spink<br />
**Allen Yuan<br />
*2009:<br />
**John Berman<br />
**Sergei Bernstein<br />
**Wenyu Cao<br />
**Robin Cheng<br />
**Vlad Firoiu<br />
**Eric Larson<br />
**Delong Meng<br />
**Qinxuan Pan<br />
**Panupong Pasupat<br />
**Toan Phan<br />
**David Rush<br />
**David Yang<br />
*2008:<br />
**David Benjamin<br />
**TaoRan Chen<br />
**Paul Christiano<br />
**Sam Elder<br />
**Shaunak Kishore<br />
**Delong Meng<br />
**Evan O'Dorney<br />
**Qinxuan Pan<br />
**David Rolnick<br />
**Colin Sandon<br />
**Krishanu Sankar<br />
**Alex Zhai<br />
*2007:<br />
**Sergei Bernstein<br />
**Sherry Gong<br />
**Adam Hesterberg<br />
**Eric Larson<br />
**Brian Lawrence<br />
**Tedrick Leung<br />
**Haitao Mao<br />
**Delong Meng<br />
**Krishanu Sankar<br />
**Jacob Steinhardt<br />
**Arnav Tripathy<br />
**Alex Zhai<br />
*2006:<br />
**Brian Lawrence<br />
**Alex Zhai<br />
**Yufei Zhao<br />
**Peng Shi<br />
**Sherry Gong<br />
**Richard McCutchen<br />
**Yi Sun<br />
**Arnav Tripathy<br />
**Taehyeon (Ryan) Ko<br />
**Yi Han<br />
**Yakov Berchenko-Kogan<br />
**Tedrick Leung<br />
*2005:<br />
**Robert Cordwell<br />
**Zhou Fan<br />
**Sherry Gong<br />
**Rishi Gupta<br />
**Hyun Soo Kim<br />
**Brian Lawrence<br />
**Albert Ni<br />
**Natee Pitiwan<br />
**Eric Price<br />
**Peng Shi<br />
**Yi Sun<br />
**Yufei Zhao<br />
*2004:<br />
**Tiankai Liu<br />
**Jae Bae<br />
**Jongmin Baek<br />
**Oleg Golberg<br />
**Matt Ince<br />
**Janos Kramar<br />
**Alison Miller<br />
**Aaron Pixton<br />
**Brian Rice<br />
**Jacob Tsimerman<br />
**Ameya Velingker<br />
**Tony Zhang<br />
*2003:<br />
**Tiankai Liu<br />
**Po Ru Loh<br />
**Boris Alexeev<br />
**Jae Bae<br />
**Daniel Kane<br />
**Anders Kaseorg<br />
**Mark Lipson<br />
**Po Ling Loh<br />
**Aaron Pixton<br />
**Kwokfung Tang<br />
**Tony Zhang<br />
**Yan Zhang<br />
*1998:<br />
**Sasha Schwartz<br />
**Melanie Wood<br />
**Kevin Lacker<br />
**Reid Barton<br />
**Gabriel Carroll<br />
**Paul Valiant<br />
**David Speyer<br />
**David Vickrey<br />
*1997:<br />
**Josh Nichols-Barrer<br />
**Daniel Stronger<br />
**John Clyde<br />
**Carl Bosley<br />
**Li-Chung Chen<br />
**Nathan Curtis<br />
**Davesh Maulik<br />
**Kevin Lacker<br />
*1996:<br />
**Chris Chang<br />
**Alex Saltman<br />
**Josh Nichols-Barrer<br />
**Carl Bosley<br />
**Michael Korn<br />
**Carl Miller<br />
**Nathan Curtis<br />
**Daniel Stronger<br />
*1995:<br />
**Aleksander Khazanov<br />
**Jacob Lurie<br />
**Chris Chang<br />
**Jay Chyung<br />
**Andrei Gnepp<br />
**Josh Nichols-Barrer<br />
**Samit Dasgupta<br />
**Craig Helfgott<br />
*1994:<br />
**Jeremy Bem<br />
**Aleksander Khazanov<br />
**Jonathan Weinstein<br />
**Jacob Lurie<br />
**Noam Shazeer<br />
**Stephen Wang<br />
**Chris Chang<br />
**Jacob Rasmussen<br />
*1989:<br />
**Jordan Ellenberg<br />
**Andrew Kresh<br />
**Jonathan Higa<br />
**[[Richard Rusczyk]]<br />
**Jeff Vanderkam<br />
**[[Sam Vandervelde]]<br />
**David Carlton<br />
<br />
==USAMO Honorable Mentions==<br />
<br />
Other students who finish (or tie to finish) in the top 24 of the [[USAMO]] receive Honorable Mention (often abbreviated HM).<br />
*2019<br />
**Quanlin Chen<br />
**Eric Gan<br />
**Sebastian Jeon<br />
**Tianze Jiang<br />
**Jeffrey Kwan<br />
**Benjamin Qi<br />
**William Zhao<br />
*2018<br />
**Adam Ardeishar<br />
**Hongyi Chen<br />
**Andrew Gu<br />
**Milan Haiman<br />
**Daniel Liu<br />
**Kevin Liu<br />
**Victor Luo<br />
**Benjamin Qi<br />
**Kevin Ren<br />
**Luke Robitaille<br />
**Dilhan Salgado<br />
**Nicholas Sun<br />
**Stan Zhang<br />
**William Zhao<br />
*2017<br />
**Swapnil Garg<br />
**Frank Han<br />
**Vincent Huang<br />
**Daniel Kim<br />
**Joshua Lee<br />
**Michael Ma<br />
**Qi Qi<br />
**Nikhil Reddy<br />
**Dilhan Salgado<br />
**Patrick Wang<br />
**Eric Zhang<br />
**Daniel Zhu<br />
*2016<br />
**Kapil Chandran<br />
**Andrew Gu<br />
**Brian Gu<br />
**Meghal Gupta<br />
**Michael Kural<br />
**Dhruv Medarametla<br />
**Eshaan Nichani<br />
**Brian Reinhart<br />
**Ashwin Sah<br />
**Franklyn Wang<br />
**Alexander Wei<br />
**Allen Yang<br />
**Rachel Zhang<br />
**Stan Zhang<br />
**Ziqi Zhou<br />
*2015<br />
**Demi Guo<br />
**Jacob Gurev<br />
**Kevin Li<br />
**James Lin<br />
**Michael Ma<br />
**Zachary Polansky<br />
**Samuel Reinehr<br />
**Ashwin Sah<br />
**Alexander Whatley<br />
**Jinhao Xu<br />
**Kevin Yang<br />
*2014<br />
**Lewis Chen<br />
**Ernest Chiu<br />
**Samuel Korsky<br />
**Yang Liu<br />
**Sammy Luo<br />
**Alexander Meiburg<br />
**Shyam Narayanan<br />
**Ashwin Sah<br />
**Nat Sothanaphan<br />
**Danielle Wang<br />
**Kevin Yang<br />
*2013:<br />
**Joshua Brakensiek<br />
**Lewis Chen<br />
**Sohail Farhangi<br />
**Kevin Huang<br />
**Allen Liu<br />
**Sammy Luo<br />
**Eric Schneider<br />
**Rachit Singh<br />
**Vikram Sundar<br />
**James Tao<br />
**David Yang<br />
**Kevin Yang<br />
*2011:<br />
**Kevin Chen<br />
**Calvin Deng<br />
**Michael Druggan<br />
**Albert Gu<br />
**Bobby Shen<br />
**Alex Song<br />
**Thomas Swayze<br />
**Tianqi Wu<br />
**Dai Yang<br />
**Kevin Yin<br />
**Allen Yuan<br />
**Samuel Zbarsky<br />
**Brian Zhang<br />
*2010:<br />
**Thomas Swayze<br />
**Kerry Xing<br />
**Tian-Yi Jiang<br />
**Brian Zhang<br />
**Archit Kulkarni<br />
**Patrick Hulin<br />
**Carl Lian<br />
**George Xing<br />
**Potcharapol Suteparuk<br />
**Jason Wu<br />
**Shijie Zheng<br />
**Robin Cheng<br />
**Bowei Liu<br />
**Wenyu Cao<br />
**Kevin Yin<br />
**Albert Gu<br />
*2009:<br />
**Timothy Chu<br />
**Brian Hamrick<br />
**Travis Hance<br />
**Jason Hoch<br />
**Tian-Yi Jiang<br />
**Sam Keller<br />
**Holden Lee<br />
**In-Sung Na<br />
**Ofir Nachum<br />
**Jarno Sun <br />
**Matthew Superdock<br />
**Tong Zhan<br />
**Mark Zhang<br />
*2008:<br />
**John Berman<br />
**Sergei Bernstein<br />
**Gregory Gauthier<br />
**Brian Hamrick<br />
**Eric Larson<br />
**Gaku Liu<br />
**Jeffrey Manning<br />
**Palmer Mebane<br />
**Danny Shi<br />
**Jacob Steinhardt<br />
**Matthew Superdock<br />
**Nicholas Triantafillou<br />
*2007:<br />
**David Benjamin<br />
**Gregory Brockman<br />
**Yingyu Gao<br />
**Yan Li<br />
**Gaku Liu<br />
**Jeffrey Manning<br />
**Palmer Mebane<br />
**Evan O'Dorney<br />
**Alexander Remorov<br />
**Max Rosett<br />
**Qiaochu Yuan<br />
**Bohua Zhan<br />
*2006:<br />
**Joseph Chu<br />
**Zachary Abel<br />
**Zarathustra Brady<br />
**Peter Diao<br />
**Richar Peng<br />
**Adam Hesterberg<br />
**Bohua Zhan<br />
**Kevin Modzelewski<br />
**Daniel Poore<br />
**Lin Fei<br />
**Jason Trigg<br />
**Tony Liu<br />
**Viktoriya Krakovna<br />
*2005:<br />
**Zachary Abel<br />
**Charles Chen<br />
**Evan Dummit<br />
**Adam Hesterberg<br />
**Richard Ho<br />
**Alexander Marcus<br />
**Richard McCutchen<br />
**Thomas Mildorf<br />
**Charles Nathanson<br />
**Keenan Pepper<br />
**Timothy Pollio<br />
*2004:<br />
**Boris Alexeev<br />
**Joshua Batson<br />
**Thomas Belulovich<br />
**Rishi Gupta<br />
**Anders Kaseorg<br />
**Hyun Soo Kim<br />
**John Kim<br />
**Po Ling Loh<br />
**Sira Sriswasdi<br />
**Yi Sun<br />
**Elena Udovina<br />
**Yufei Zhao<br />
*2003:<br />
**Timothy Abbott<br />
**Jeffrey Amos<br />
**Jongmin Baek<br />
**Richard Biggs<br />
**Steven Byrnes<br />
**Sherry Gong<br />
**Matt Ince<br />
**Nathaniel Ince<br />
**Pavel Kamyshev<br />
**Nicholas Ma<br />
**Alison Miller<br />
**Eric Price<br />
**David Stolp<br />
**Sean Ting<br />
**Tongke Xue<br />
*1995<br />
** Mathew Crawford<br />
*1993<br />
** Mathew Crawford<br />
<br />
==See also==<br />
[[Category:Historical results]]</div>GeronimoStiltonhttps://artofproblemsolving.com/wiki/index.php?title=Eric_Wu&diff=122546Eric Wu2020-05-17T01:08:02Z<p>GeronimoStilton: </p>
<hr />
<div>Eric Wu is a huge bot from [[North Carolina mathematics competitions|North Carolina]] who has a Texas Pin from Luke Robitaille<br \><br />
His AoPS username is Trex4days and he ships Vivethan<br \><br />
You should see his blog<br \><br />
He goes to [[Triangle Math and Science Academy|Teachers Make Students Autistic]]<br \><br />
He also knows [[brianzjk]] IRL<br \><br />
Oh and also he made history as the first Mathcounts 3-timer from North Carolina<br />
<br />
He is a literal bot on Discord; you can find him at Disboard#0569</div>GeronimoStiltonhttps://artofproblemsolving.com/wiki/index.php?title=Kevin_Zhao&diff=120078Kevin Zhao2020-03-25T13:46:33Z<p>GeronimoStilton: </p>
<hr />
<div>Kevin Zhao is a sophomore who goes to Lexington High School in Lexington, MA. He is very strong at math. He moved to Acton MA in the beginning of third grade, and ever since, studied math very hard! He eventually moved to Lexington MA, where he made many friends and studied deeper and harder in math.<br />
<br />
<br />
Fourth Grade:<br />
<br />
* Math Kangaroo 1st place state, 1st place national (Score 96/96)<br />
<br />
Fifth grade:<br />
<br />
* MOEMS medal, trophy, Gold Pin<br />
* Math Kangaroo 16th place national (Score 103/120)<br />
* RSM Olympiad Silver Medal (Score 11/12)<br />
<br />
Sixth grade:<br />
<br />
* AMC 8 Honor Roll (Score 16)<br />
* MOEMS trophy, Gold Pin<br />
* Math Kangaroo 1st place state, 1st place national (Score 120/120)<br />
* RSM Olympiad (2016 International Math Contest) Gold Medal (Score 12/12)<br />
<br />
Seventh grade:<br />
<br />
* AMC 8 Distinguished Honor Roll (Score 24)<br />
* AMC 10 B Achievement Roll (106.5; He failed AMC 10A with an 81)<br />
* Mathcounts State Qualifer (On Team) (Chapter Score 42: 26+16, did not place) (State score 28: 14+14, 41st place)<br />
* MOEMS trophy, Gold Pin<br />
* Math Kangaroo 1st place state, 1st place national (Score 120/120)<br />
* RSM Olympiad (2016 International Math Contest) Silver Medal (Score 9/12)<br />
* ABMC Accuracy Round tied 1st place (Bumped to fifth in tiebreaker)<br />
* ABMC Overall Tenth place Individual<br />
<br />
Eighth Grade:<br />
<br />
* AMC 8 Distinguished Honor Roll, Perfect Score<br />
* EMCC Overall Eighth place Individual <math>^1</math><br />
* AMC 10 A Distinguished Honor Roll (Score 138)<br />
* AMC 10 B Distinguished Honor Roll (Score 127.5)<br />
* 7 on AIME, USAJMO Index 208/197.5<br />
* Mathcounts State Qualifer (Individual) (Chapter Score 44: 28+16, 5th place written, tied 5th place in countdown) (State Score 41: 27+14, 2nd place written, 3rd place countdown)<br />
* Mathcounts National Qualifier (2nd in state) (State Score 41: 27+14)<math>^2</math><br />
* Mathcounts Nationals 60th place (Score 23: 15+8)<math>^3</math><br />
* LMT Overall Winner (First place Individual)<br />
* ABMC Overall Seventh place Individual<br />
* Math Kangaroo 2nd place state, 3rd place national (Score 116/120)<br />
* RSM Olympiad(2018 International Math Contest) Gold Medal (Score 12/12)<br />
<br />
Ninth grade:<br />
<br />
* MAML (Massachusetts Association of Mathematics Leagues) Olympiad qualifier (Top 100 in state) <br />
* CMIMC tied tenth place in Combinatorics/Computer Science round (Scaled score 28.07, Unscaled score 5/10)<br />
* AMC 10 A Distinguished Honor Roll (Score 132)<br />
* AMC 10 B Honor Roll (Score 115.5 :()<br />
* 8 on AIME, USAJMO index 212/195.5<br />
* USAJMO Qualifier<br />
* MMATHS Yale 7th place individual<br />
* NEAML Perfect Scorer<br />
* ARML tied 23rd (Missed tiebreakers by 1 point :()<br />
* Math Kangaroo 1st place state, 1st place national (Score 120/120)<br />
* Invited to Zakopane Camp run by Math Kangaroo (for 2020)<br />
* M.TIP NT3 survivor, GABRIEL C3 survivor<br />
<br />
Tenth grade:<br />
<br />
* MAML (Massachusetts Association of Mathematics Leagues) Olympiad qualifier (Top 100 in state), MAML Round 1 score 3rd place (Score 144) <br />
* AMC 10 A Distinguished Honor Roll (Score 138)<br />
* AMC 10 B Distinguished Honor Roll (Score 135)<br />
* 10 on AIME, USAJMO index 238/235'<br />
* HMIC Qualifier (Scores at HMMT: 4/4<math>^7</math>/4)<br />
* MML Highest Total Score in State<br />
<br />
<br />
<br />
Kevin Zhao lives in one of the strongest schools at math in New England. His school(s) won a lot of achievements with Kevin.<br />
<br />
2017:<br />
* Mathcounts Chapter 2017 First Place (In their chapter)<br />
* Mathcounts State 2017 Third Place (In MA)<br />
* LMT 2017 Fourth Place <math>^4</math><br />
* ABMC 2017 Second Place <math>^4</math><br />
<br />
2018:<br />
* EMCC 2018 First Place <math>^5</math><br />
* LMT 2018 First Place <math>^4</math><br />
* ABMC 2018 Fifth Place <math>^5</math><br />
<br />
2019:<br />
* MMATHS team Third Place<br />
* NEAML team First Place<br />
* ARML team 13th Place<br />
* PUMAC live 7th Place<br />
<br />
2020:<br />
* USMCA Online Qualifier 13th place<br />
* HMMT February Team 10th Place <math>^5</math><br />
* HMMT February Team Round 9th Place<br />
<br />
' = Not sure yet<br />
<br />
<br />
<br />
Note that Kevin choked the 2018 MATHCOUNTS School round (reason is in his blog), so he did not make his school team in 2018. He made his own team with friends, instead of kids that are good at math, for ABMC in 2018.<math>^6</math><br />
<br />
Kevin moved into Lexington during the summer of 2018, and switched to Lexington High School in the fall for ninth grade.<br />
<br />
"My ultimate goal is to get accepted into Massachusetts Institute of Technology or some other very good college, and from there, study mathematics very deep", said Kevin. He has now finished his third year of PRIMES STEP, a research program held at MIT once a week for kids grade 7-9. He got admitted to his third year without an admission test but got rejected from PRIMES.<br />
<br />
<b>Some more interesting facts</b><br />
<br />
* Kevin's NT3 test score would have increased if it was divided by 2.<br />
* Kevin has <b>15</b> medals and <b>8</b> trophies from math competitions.<br />
* Kevin has scored <math>69 \cdot 2</math> on AMC 10 (most likely) twice in his lifetime, and that is his high score.<br />
* Kevin is now taking AP Calculus BC in school.<br />
<br />
<br />
<br />
<b>Footnotes</b><br />
<br />
*<math>^1</math> Kevin tied for Fourth Place, but Tiebreaker brought him down.<br />
*<math>^2</math> Kevin tied for First Place, but Tiebreaker brought him down.<br />
*<math>^3</math> Kevin inverted two of the answer boxes and got tiebreakered barely out of top 56.<br />
*<math>^4</math> This is both team, and sweepstakes(Overall).<br />
*<math>^5</math> This is Sweepstakes(Overall).<br />
*<math>^6</math> Kevin's math rival registered before him, leaving him out.<br />
*<math>^7</math> Kevin got marked wrong for a question he got correct (probably a handwriting error).</div>GeronimoStiltonhttps://artofproblemsolving.com/wiki/index.php?title=User:Icepenguin&diff=120077User:Icepenguin2020-03-25T13:44:06Z<p>GeronimoStilton: </p>
<hr />
<div>It's fun to have a user page all to yourself.<br />
<br />
Anyway, welcome! If you are a registered user and are reading this (and have not visited this page yet), please add <math>\text{ONE}</math> to the number below.<br />
<br />
<cmath> \text{This page has been visited by } 3 \text{ different users.} \text{Please do not troll.}</cmath><br />
<br />
Also, check me out on the Alcumus Hall of Fame. Current rank: <math>41</math><br />
<br />
If you want to know what number I choose for Greed Control every day, unscramble the following anagram (capital/lowercase not necessarily preserved):<br />
<br />
<cmath>\text{Burn oceanic moth team EFIS}</cmath><br />
<br />
Wall of shame (users who trolled):<br />
<br />
1. Bibear</div>GeronimoStiltonhttps://artofproblemsolving.com/wiki/index.php?title=User:GeronimoStilton&diff=119985User:GeronimoStilton2020-03-23T15:44:35Z<p>GeronimoStilton: Created page with "I am a person."</p>
<hr />
<div>I am a person.</div>GeronimoStiltonhttps://artofproblemsolving.com/wiki/index.php?title=2020_AIME_I_Problems/Problem_14&diff=1192902020 AIME I Problems/Problem 142020-03-13T15:02:23Z<p>GeronimoStilton: </p>
<hr />
<div><br />
== Problem ==<br />
Let <math>P(x)</math> be a quadratic polynomial with complex coefficients whose <math>x^2</math> coefficient is <math>1.</math> Suppose the equation <math>P(P(x))=0</math> has four distinct solutions, <math>x=3,4,a,b.</math> Find the sum of all possible values of <math>(a+b)^2.</math><br />
<br />
== Solution 1 ==<br />
Either <math>P(3) = P(4)</math> or not. We first see that if <math>P(3) = P(4)</math> it's easy to obtain by Vieta's that <math>(a+b)^2 = 49</math>. Now, take <math>P(3) \neq P(4)</math> and WLOG <math>P(3) = P(a), P(4) = P(b)</math>. Now, consider the parabola formed by the graph of <math>P</math>. It has vertex <math>\frac{3+a}{2}</math>. Now, say that <math>P(x) = x^2 - (3+a)x + c</math>. We note <math>P(3)P(4) = c = P(3)(4 - 4a + \frac{8a - 1}{2}) \implies a = \frac{7P(3) + 1}{8}</math>. Now, we note <math>P(4) = \frac{7}{2}</math> by plugging in again. Now, it's easy to find that <math>a = -2.5, b = -3.5</math>, yielding a value of <math>36</math>. Finally, we add <math>49 + 36 = \boxed{085}</math>. ~awang11, charmander3333<br />
<br />
== Solution 2 ==<br />
Let the roots of <math>P(x)</math> be <math>m</math> and <math>n</math>, then we can write <math>P(x)=x^2-(m+n)x+mn</math>. The fact that <math>P(P(x))=0</math> has solutions <math>x=3,4,a,b</math> implies that some combination of <math>2</math> of these are the solution to <math>P(x)=m</math>, and the other <math>2</math> are the solution to <math>P(x)=n</math>. It's fairly easy to see there are only <math>2</math> possible such groupings: <math>P(3)=P(4)=m</math> and <math>P(a)=P(b)=n</math>, or <math>P(3)=P(a)=m</math> and <math>P(4)=P(b)=n</math> (Note that <math>a,b</math> are interchangeable, and so are <math>m</math> and <math>n</math>). We now to casework: <br />
If <math>P(3)=P(4)=m</math>, then <br />
<cmath>9-3(m+n)+mn=16-4(m+n)+mn=m \implies m+n=7</cmath><br />
<cmath>a^2-a(m+n)+mn=b^2-b(m+n)+mn=n \implies a+b=m+n=7</cmath><br />
so this gives <math>(a+b)^2=7^2=49</math>. <br />
Next, if <math>P(3)=P(a)=m</math>, then <br />
<cmath>9-3(m+n)+mn=a^2-a(m+n)+mn=m \implies a+3=m+n</cmath><br />
<cmath>16-4(m+n)+mn=b^2-b(m+n)+mn=n \implies b+4=m+n</cmath><br />
Subtracting the first part of the first equation from the first part of the second equation gives <br />
<cmath>7-(m+n)=n-m \implies 2n=7 \implies n=\frac{7}{2} \implies m=-3</cmath><br />
Hence, <math>a+b=2(m+n)-7=2\cdot \frac{1}{2}-7=-6</math>, and so <math>(a+b)^2=(-6)^2=36</math>. <br />
Therefore, the solution is <math>49+36=\boxed{085}</math> ~ktong<br />
<br />
== Solution 3 ==<br />
Write <math>P(x) = x^2+wx+z</math>. Split the problem into two cases: <math>P(3)\ne P(4)</math> and <math>P(3) = P(4)</math>.<br />
<br />
Case 1: We have <math>P(3) \ne P(4)</math>. We must have <br />
<cmath>w=-P(3)-P(4) = -(9+3w+z)-(16+4w+z) = -25-7w-2z.</cmath><br />
Rearrange and divide through by <math>8</math> to obtain<br />
<cmath>w = \frac{-25-2z}{8}.</cmath><br />
Now, note that<br />
<cmath>z = P(3)P(4) = (9+3w+z)(16+4w+z) = \left(9 + 3\cdot \frac{-25-2z}{8} + z\right)\left(16 + 4 \cdot \frac{-25-2z}{8} + z\right) =</cmath><br />
<cmath>\left(-\frac{3}{8} + \frac{z}{4}\right)\left(\frac{7}{2}\right) = -\frac{21}{16} + \frac{7z}{8}.</cmath><br />
Now, rearrange to get<br />
<cmath>\frac{z}{8} = -\frac{21}{16}</cmath><br />
and thus<br />
<cmath>z = -\frac{21}{2}.</cmath><br />
Substituting this into our equation for <math>w</math> yields <math>w = -\frac{1}{2}</math>. Then, it is clear that <math>P</math> does not have a double root at <math>P(3)</math>, so we must have <math>P(a) = P(3)</math> and <math>P(b) = P(4)</math> or vice versa. This gives <math>3+a = \frac{1}{2}</math> and <math>4+b = \frac{1}{2}</math> or vice versa, implying that <math>a+b = 1-3-4 = -6</math> and <math>(a+b)^2 = 6</math>.<br />
<br />
Case 2: We have <math>P(3) = P(4)</math>. Then, we must have <math>w = -7</math>. It is clear that <math>P(a) = P(b)</math> (we would otherwise get <math>P(a)=P(3)=P(4)</math> implying <math>a \in \{3,4\}</math> or vice versa), so <math>a+b=-w=7</math> and <math>(a+b)^2 = 49</math>.<br />
<br />
Thus, our final answer is <math>49+36=\boxed{085}</math>. ~GeronimoStilton<br />
<br />
==See Also==<br />
<br />
{{AIME box|year=2020|n=I|num-b=13|num-a=15}}<br />
{{MAA Notice}}</div>GeronimoStiltonhttps://artofproblemsolving.com/wiki/index.php?title=2020_AIME_I_Problems/Problem_14&diff=1192892020 AIME I Problems/Problem 142020-03-13T15:01:44Z<p>GeronimoStilton: Fixed latex error</p>
<hr />
<div><br />
== Problem ==<br />
Let <math>P(x)</math> be a quadratic polynomial with complex coefficients whose <math>x^2</math> coefficient is <math>1.</math> Suppose the equation <math>P(P(x))=0</math> has four distinct solutions, <math>x=3,4,a,b.</math> Find the sum of all possible values of <math>(a+b)^2.</math><br />
<br />
== Solution 1 ==<br />
Either <math>P(3) = P(4)</math> or not. We first see that if <math>P(3) = P(4)</math> it's easy to obtain by Vieta's that <math>(a+b)^2 = 49</math>. Now, take <math>P(3) \neq P(4)</math> and WLOG <math>P(3) = P(a), P(4) = P(b)</math>. Now, consider the parabola formed by the graph of <math>P</math>. It has vertex <math>\frac{3+a}{2}</math>. Now, say that <math>P(x) = x^2 - (3+a)x + c</math>. We note <math>P(3)P(4) = c = P(3)(4 - 4a + \frac{8a - 1}{2}) \implies a = \frac{7P(3) + 1}{8}</math>. Now, we note <math>P(4) = \frac{7}{2}</math> by plugging in again. Now, it's easy to find that <math>a = -2.5, b = -3.5</math>, yielding a value of <math>36</math>. Finally, we add <math>49 + 36 = \boxed{085}</math>. ~awang11, charmander3333<br />
<br />
== Solution 2 ==<br />
Let the roots of <math>P(x)</math> be <math>m</math> and <math>n</math>, then we can write <math>P(x)=x^2-(m+n)x+mn</math>. The fact that <math>P(P(x))=0</math> has solutions <math>x=3,4,a,b</math> implies that some combination of <math>2</math> of these are the solution to <math>P(x)=m</math>, and the other <math>2</math> are the solution to <math>P(x)=n</math>. It's fairly easy to see there are only <math>2</math> possible such groupings: <math>P(3)=P(4)=m</math> and <math>P(a)=P(b)=n</math>, or <math>P(3)=P(a)=m</math> and <math>P(4)=P(b)=n</math> (Note that <math>a,b</math> are interchangeable, and so are <math>m</math> and <math>n</math>). We now to casework: <br />
If <math>P(3)=P(4)=m</math>, then <br />
<cmath>9-3(m+n)+mn=16-4(m+n)+mn=m \implies m+n=7</cmath><br />
<cmath>a^2-a(m+n)+mn=b^2-b(m+n)+mn=n \implies a+b=m+n=7</cmath><br />
so this gives <math>(a+b)^2=7^2=49</math>. <br />
Next, if <math>P(3)=P(a)=m</math>, then <br />
<cmath>9-3(m+n)+mn=a^2-a(m+n)+mn=m \implies a+3=m+n</cmath><br />
<cmath>16-4(m+n)+mn=b^2-b(m+n)+mn=n \implies b+4=m+n</cmath><br />
Subtracting the first part of the first equation from the first part of the second equation gives <br />
<cmath>7-(m+n)=n-m \implies 2n=7 \implies n=\frac{7}{2} \implies m=-3</cmath><br />
Hence, <math>a+b=2(m+n)-7=2\cdot \frac{1}{2}-7=-6</math>, and so <math>(a+b)^2=(-6)^2=36</math>. <br />
Therefore, the solution is <math>49+36=\boxed{085}</math> ~ktong<br />
<br />
== Solution 3 ==<br />
Write <math>P(x) = x^2+wx+z</math>. Split the problem into two cases: <math>P(3)\ne P(4)</math> and <math>P(3) = P(4)</math>.<br />
<br />
Case 1: We have <math>P(3) \ne P(4)</math>. We must have <br />
<cmath>w=-P(3)-P(4) = -(9+3w+z)-(16+4w+z) = -25-7w-2z.</cmath><br />
Rearrange and divide through by <math>8</math> to obtain<br />
<cmath>w = \frac{-25-2z}{8}.</cmath><br />
Now, note that<br />
<cmath>z = P(3)P(4) = (9+3w+z)(16+4w+z) = \left(9 + 3\cdot \frac{-25-2z}{8} + z\right)\left(16 + 4 \cdot \frac{-25-2z}{8} + z\right) =</cmath><br />
<cmath>\left(-\frac{3}{8} + \frac{z}{4}\right)\left(\frac{7}{2}\right) = -\frac{21}{16} + \frac{7z}{8}.</cmath><br />
Now, rearrange to get<br />
<cmath>\frac{z}{8} = -\frac{21}{16}</cmath><br />
and thus<br />
<cmath>z = -\frac{21}{2}.</cmath><br />
Substituting this into our equation for <math>w</math> yields <math>w = -\frac{1}{2}</math>. Then, it is clear that <math>P</math> does not have a double root at <math>P(3)</math>, so we must have <math>P(a) = P(3)</math> and <math>P(b) = P(4)</math> or vice versa. This gives <math>3+a = \frac{1}{2}</math> and <math>4+b = \frac{1}{2}</math> or vice versa, implying that <math>a+b = 1-3-4 = -6</math> and <math>(a+b)^2 = 6</math>.<br />
<br />
Case 2: We have <math>P(3) = P(4)</math>. Then, we must have <math>w = -7</math>. It is clear that <math>P(a) = P(b)</math> (we would otherwise get <math>P(a)=P(3)=P(4)</math> implying <math>a \in \{3,4\}</math> or vice versa), so <math>a+b=-w=7</math> and <math>(a+b)^2 = 49</math>.<br />
<br />
Thus, our final answer is <math>49+36=\boxed{085)</math>. ~GeronimoStilton<br />
<br />
==See Also==<br />
<br />
{{AIME box|year=2020|n=I|num-b=13|num-a=15}}<br />
{{MAA Notice}}</div>GeronimoStiltonhttps://artofproblemsolving.com/wiki/index.php?title=2020_AIME_I_Problems/Problem_14&diff=1192882020 AIME I Problems/Problem 142020-03-13T15:01:14Z<p>GeronimoStilton: Added another solution</p>
<hr />
<div><br />
== Problem ==<br />
Let <math>P(x)</math> be a quadratic polynomial with complex coefficients whose <math>x^2</math> coefficient is <math>1.</math> Suppose the equation <math>P(P(x))=0</math> has four distinct solutions, <math>x=3,4,a,b.</math> Find the sum of all possible values of <math>(a+b)^2.</math><br />
<br />
== Solution 1 ==<br />
Either <math>P(3) = P(4)</math> or not. We first see that if <math>P(3) = P(4)</math> it's easy to obtain by Vieta's that <math>(a+b)^2 = 49</math>. Now, take <math>P(3) \neq P(4)</math> and WLOG <math>P(3) = P(a), P(4) = P(b)</math>. Now, consider the parabola formed by the graph of <math>P</math>. It has vertex <math>\frac{3+a}{2}</math>. Now, say that <math>P(x) = x^2 - (3+a)x + c</math>. We note <math>P(3)P(4) = c = P(3)(4 - 4a + \frac{8a - 1}{2}) \implies a = \frac{7P(3) + 1}{8}</math>. Now, we note <math>P(4) = \frac{7}{2}</math> by plugging in again. Now, it's easy to find that <math>a = -2.5, b = -3.5</math>, yielding a value of <math>36</math>. Finally, we add <math>49 + 36 = \boxed{085}</math>. ~awang11, charmander3333<br />
<br />
== Solution 2 ==<br />
Let the roots of <math>P(x)</math> be <math>m</math> and <math>n</math>, then we can write <math>P(x)=x^2-(m+n)x+mn</math>. The fact that <math>P(P(x))=0</math> has solutions <math>x=3,4,a,b</math> implies that some combination of <math>2</math> of these are the solution to <math>P(x)=m</math>, and the other <math>2</math> are the solution to <math>P(x)=n</math>. It's fairly easy to see there are only <math>2</math> possible such groupings: <math>P(3)=P(4)=m</math> and <math>P(a)=P(b)=n</math>, or <math>P(3)=P(a)=m</math> and <math>P(4)=P(b)=n</math> (Note that <math>a,b</math> are interchangeable, and so are <math>m</math> and <math>n</math>). We now to casework: <br />
If <math>P(3)=P(4)=m</math>, then <br />
<cmath>9-3(m+n)+mn=16-4(m+n)+mn=m \implies m+n=7</cmath><br />
<cmath>a^2-a(m+n)+mn=b^2-b(m+n)+mn=n \implies a+b=m+n=7</cmath><br />
so this gives <math>(a+b)^2=7^2=49</math>. <br />
Next, if <math>P(3)=P(a)=m</math>, then <br />
<cmath>9-3(m+n)+mn=a^2-a(m+n)+mn=m \implies a+3=m+n</cmath><br />
<cmath>16-4(m+n)+mn=b^2-b(m+n)+mn=n \implies b+4=m+n</cmath><br />
Subtracting the first part of the first equation from the first part of the second equation gives <br />
<cmath>7-(m+n)=n-m \implies 2n=7 \implies n=\frac{7}{2} \implies m=-3</cmath><br />
Hence, <math>a+b=2(m+n)-7=2\cdot \frac{1}{2}-7=-6</math>, and so <math>(a+b)^2=(-6)^2=36</math>. <br />
Therefore, the solution is <math>49+36=\boxed{085}</math> ~ktong<br />
<br />
== Solution 3 ==<br />
Write <math>P(x) = x^2+wx+z</math>. Split the problem into two cases: <math>P(3)\ne P(4)</math> and <math>P(3) = P(4)</math>.<br />
<br />
Case 1: We have <math>P(3) \ne P(4)</math>. We must have <br />
<cmath>w=-P(3)-P(4) = -(9+3w+z)-(16+4w+z) = -25-7w-2z.</cmath><br />
Rearrange and divide through by <math>8</math> to obtain<br />
<cmath>w = \frac{-25-2z}{8}.</cmath><br />
Now, note that<br />
<cmath>z = P(3)P(4) = (9+3w+z)(16+4w+z) = \left(9 + 3\cdot \frac{-25-2z}{8} + z\right)\left(16 + 4 \cdot \frac{-25-2z}{8} + z\right) =</cmath><br />
<cmath>\left(-\frac{3}{8} + \frac{z}{4}\right)\left(\frac{7}{2}\right) = -\frac{21}{16} + \frac{7z}{8}.</cmath><br />
Now, rearrange to get<br />
<cmath>\frac{z}{8} = -\frac{21}{16}</cmath><br />
and thus<br />
<cmath>z = -\frac{21}{2}.</cmath><br />
Substituting this into our equation for <math>w</math> yields <math>w = -\frac{1}{2}</math>. Then, it is clear that <math>P</math> does not have a double root at <math>P(3)</math>, so we must have <math>P(a) = P(3)</math> and <math>P(b) = P(4)</math> or vice versa. This gives <math>3+a = \frac{1}{2}</math> and <math>4+b = \frac{1}{2}</math> or vice versa, implying that <math>a+b = 1-3-4 = -6</math> and <math>(a+b)^2 = 6</math>.<br />
<br />
Case 2: We have <math>P(3) = P(4)</math>. Then, we must have <math>w = -7</math>. It is clear that <math>P(a) = P(b)</math> (we would otherwise get <math>P(a)=P(3)=P(4)</math> implying <math>a \in \{3,4\}</math> or vice versa), so <math>a+b=-w=7</math> and <math>(a+b)^2 = 49</math>.<br />
<br />
Thus, our final answer is <math>49+36=\fbox{085)</math>. ~GeronimoStilton<br />
<br />
==See Also==<br />
<br />
{{AIME box|year=2020|n=I|num-b=13|num-a=15}}<br />
{{MAA Notice}}</div>GeronimoStiltonhttps://artofproblemsolving.com/wiki/index.php?title=Cauchy-davenport&diff=109898Cauchy-davenport2019-09-20T17:39:06Z<p>GeronimoStilton: </p>
<hr />
<div>The Cauchy-Davenport Theorem states that for all nonempty sets <math>A,B \subseteq \mathbb{Z}/p\mathbb{Z}</math> , we have that <br />
<cmath>|A+B| \geq \min\{|A|+|B|-1,p\},</cmath><br />
where <math>A+B</math> is defined as the set of all <math>c \in \mathbb{Z}/p\mathbb{Z}</math> that can be expressed as <math>a+b</math> for <math>a \in A</math> and <math>b \in B</math>.<br />
<br />
== Proof of the Cauchy-Davenport Theorem by the Combinatorial Nullstellensatz ==<br />
<br />
== Proof by Induction ==<br />
<br />
== Applications of the Cauchy-Davenport Theorem ==</div>GeronimoStiltonhttps://artofproblemsolving.com/wiki/index.php?title=Cauchy-davenport&diff=109897Cauchy-davenport2019-09-20T17:32:13Z<p>GeronimoStilton: </p>
<hr />
<div>The Cauchy-Davenport theorem states that for all nonempty sets <math>A,B \subseteq \mathbb{Z}/p\mathbb{Z}</math> , we have that <br />
<cmath>|A+B| \geq \min\{|A|+|B|-1,p\},</cmath><br />
where <math>A+B</math> is defined as the set of all <math>c \in \mathbb{Z}/p\mathbb{Z}</math> that can be expressed as <math>a+b</math> for <math>a \in A</math> and <math>b \in B</math>.</div>GeronimoStiltonhttps://artofproblemsolving.com/wiki/index.php?title=Cauchy-davenport&diff=109896Cauchy-davenport2019-09-20T17:31:53Z<p>GeronimoStilton: Created page with "The Cauchy-Davenport theorem states that for all nonempty sets <math>A,B \subseteq \mathbb{Z}/p\mathbb{Z}</math> , we have that <cmath>|A+B| \geqslant \min\{|A|+|B|-1,p\},</c..."</p>
<hr />
<div>The Cauchy-Davenport theorem states that for all nonempty sets <math>A,B \subseteq \mathbb{Z}/p\mathbb{Z}</math> , we have that <br />
<cmath>|A+B| \geqslant \min\{|A|+|B|-1,p\},</cmath><br />
where <math>A+B</math> is defined as the set of all <math>c \in \mathbb{Z}/p\mathbb{Z}</math> that can be expressed as <math>a+b</math> for <math>a \in A</math> and <math>b \in B</math>.</div>GeronimoStiltonhttps://artofproblemsolving.com/wiki/index.php?title=USAMO_historical_results&diff=109744USAMO historical results2019-09-09T19:01:38Z<p>GeronimoStilton: </p>
<hr />
<div>This is the '''USAMO historical results''' page. <!-- Post results here so the [[USAMO]] page does not become cluttered. --><br />
<br />
==USAMO Perfect Scorers==<br />
<br />
Since 1996, a perfect score on the [[USAMO]] has been 42 points for 6 problems. Prior to 1996 a perfect score was 100 points across 5 problems.<br />
*2019: Luke Robitaille<br />
*2018: None<br />
*2017: None<br />
*2016:<br />
**Allen Liu<br />
*2015:<br />
**Allen Liu<br />
**David Stoner<br />
*2014: <br />
**Joshua Brakensiek<br />
*2013: None<br />
*2012: <br />
**Alex Zhu<br />
**Thomas Swayze<br />
**Xiaoyu He<br />
**Mitchell Lee<br />
**Samuel Zbarsky<br />
*2011:<br />
**David Yang<br />
**Evan O'Dorney<br />
*2010: None<br />
*2008: None<br />
*2007: None<br />
*2006:<br />
**Brian Lawrence<br />
*2005: None<br />
*2004:<br />
**Tiankai Liu<br />
*2003:<br />
**Tiankai Liu<br />
**Po-Ru Loh<br />
*2002: <br />
**Daniel Kane<br />
**Ricky Liu<br />
**Tiankai Liu<br />
**Po-Ru Loh<br />
**Inna Zakharevich<br />
*1996:<br />
**Chris Chang<br />
*1992:<br />
**Kiran Kedlaya<br />
**Lenny Ng<br />
<br />
==USAMO Winners==<br />
<br />
The top 12 scorers on the [[USAMO]] are currently designated as USAMO winners. In the past, this number was smaller.<br />
*2019:<br />
**Vincent Bian<br />
**Milan Haiman<br />
**Vincent Huang<br />
**Kevin Liu<br />
**Luke Robitaille<br />
**Victor Rong<br />
**Carl Schildkraut<br />
**Colin Shanmo Tang<br />
**Edward Wan<br />
**Brandon Wang<br />
**Guanpeng Xu<br />
**Daniel Zhu<br />
*2018:<br />
**Eric Gan<br />
**Thomas Guo<br />
**Vincent Huang<br />
**Joshua Lee<br />
**Michael Ren<br />
**Victor Rong<br />
**Carl Schildkraut<br />
**Mihir Singhal<br />
**Edward Wan<br />
**Brandon Wang<br />
**Guanpeng Xu<br />
**Andrew Yao<br />
*2017:<br />
**Zachary Chroman<br />
**Andrew Gu<br />
**James Lin<br />
**Daniel Liu<br />
**Michael Ren<br />
**Victor Rong<br />
**Ashwin Sah<br />
**Mihir Singhal<br />
**Alec Sun<br />
**Kada Williams<br />
**Yuan Yao<br />
**William Zhao<br />
*2016:<br />
**Ankan Bhattacharya<br />
**Ruidi Cao<br />
**Hongyi Chen<br />
**Jacob Klegar<br />
**James Lin<br />
**Allen Liu<br />
**Junyao Peng<br />
**Kevin Ren<br />
**Mihir Singhal<br />
**Alec Sun<br />
**Kevin Sun<br />
**Yuan Yao<br />
*2015:<br />
**Ryan Alweiss<br />
**Kritkorn Karntikoon<br />
**Michael Kural<br />
**Celine Liang<br />
**Allen Liu<br />
**Yang Liu<br />
**Shyam Narayanan<br />
**Kevin Ren<br />
**Zhuoqun Song<br />
**David Stoner<br />
**Kevin Sun<br />
**Danielle Wang<br />
*2014:<br />
**Joshua Brakensiek<br />
**Evan Chen<br />
**Ravi Jagadeesan<br />
**Allen Liu<br />
**Nipun Pitimanaaree<br />
**Mark Sellke<br />
**Zhuoqun Song<br />
**David Stoner<br />
**Kevin Sun<br />
**James Tao<br />
**Alexander Whatley<br />
**Scott Wu<br />
*2013<br />
**Calvin Deng<br />
**Andrew He<br />
**Ravi Jagadeesan<br />
**Pakawut Jiradilok<br />
**Ray Li<br />
**Kevin Li<br />
**Mark Sellke<br />
**Bobby Shen<br />
**Zhuoqun Song<br />
**David Stoner<br />
**Thomas Swayze<br />
**Victor Wang<br />
*2012:<br />
**Andre Arslan<br />
**Joshua Brakensiek<br />
**Calvin Deng<br />
**Xiaoyu He<br />
**Ravi Jagadeesan<br />
**Mitchell Lee<br />
**Alex Song<br />
**Thomas Swayze<br />
**Victor Wang<br />
**David Yang<br />
**Samuel Zbarsky<br />
**Alex Zhu<br />
*2011:<br />
**Wenyu Cao<br />
**Zijing Gao<br />
**Benjamin Gunby<br />
**Xiaoyu He<br />
**Ravi Jagadeesan<br />
**Spencer Kwon<br />
**Mitchell Lee<br />
**Ray Li<br />
**Evan O'Dorney<br />
**Mark Sellke<br />
**David Yang<br />
**Joy Zheng<br />
*2010:<br />
**Timothy Chu<br />
**Calvin Deng<br />
**Michael Druggan<br />
**Brian Hamrick<br />
**Travis Hance<br />
**Xiaoyu He<br />
**Mitchell Lee<br />
**In Sung Na<br />
**Evan O'Dorney<br />
**Toan Phan<br />
**Hunter Spink<br />
**Allen Yuan<br />
*2009:<br />
**John Berman<br />
**Sergei Bernstein<br />
**Wenyu Cao<br />
**Robin Cheng<br />
**Vlad Firoiu<br />
**Eric Larson<br />
**Delong Meng<br />
**Qinxuan Pan<br />
**Panupong Pasupat<br />
**Toan Phan<br />
**David Rush<br />
**David Yang<br />
*2008:<br />
**David Benjamin<br />
**TaoRan Chen<br />
**Paul Christiano<br />
**Sam Elder<br />
**Shaunak Kishore<br />
**Delong Meng<br />
**Evan O'Dorney<br />
**Qinxuan Pan<br />
**David Rolnick<br />
**Colin Sandon<br />
**Krishanu Sankar<br />
**Alex Zhai<br />
*2007:<br />
**Sergei Bernstein<br />
**Sherry Gong<br />
**Adam Hesterberg<br />
**Eric Larson<br />
**Brian Lawrence<br />
**Tedrick Leung<br />
**Haitao Mao<br />
**Delong Meng<br />
**Krishanu Sankar<br />
**Jacob Steinhardt<br />
**Arnav Tripathy<br />
**Alex Zhai<br />
*2006:<br />
**Brian Lawrence<br />
**Alex Zhai<br />
**Yufei Zhao<br />
**Peng Shi<br />
**Sherry Gong<br />
**Richard McCutchen<br />
**Yi Sun<br />
**Arnav Tripathy<br />
**Taehyeon (Ryan) Ko<br />
**Yi Han<br />
**Yakov Berchenko-Kogan<br />
**Tedrick Leung<br />
*2005:<br />
**Robert Cordwell<br />
**Zhou Fan<br />
**Sherry Gong<br />
**Rishi Gupta<br />
**Hyun Soo Kim<br />
**Brian Lawrence<br />
**Albert Ni<br />
**Natee Pitiwan<br />
**Eric Price<br />
**Peng Shi<br />
**Yi Sun<br />
**Yufei Zhao<br />
*2004:<br />
**Tiankai Liu<br />
**Jae Bae<br />
**Jongmin Baek<br />
**Oleg Golberg<br />
**Matt Ince<br />
**Janos Kramar<br />
**Alison Miller<br />
**Aaron Pixton<br />
**Brian Rice<br />
**Jacob Tsimerman<br />
**Ameya Velingker<br />
**Tony Zhang<br />
*2003:<br />
**Tiankai Liu<br />
**Po Ru Loh<br />
**Boris Alexeev<br />
**Jae Bae<br />
**Daniel Kane<br />
**Anders Kaseorg<br />
**Mark Lipson<br />
**Po Ling Loh<br />
**Aaron Pixton<br />
**Kwokfung Tang<br />
**Tony Zhang<br />
**Yan Zhang<br />
*1998:<br />
**Sasha Schwartz<br />
**Melanie Wood<br />
**Kevin Lacker<br />
**Reid Barton<br />
**Gabriel Carroll<br />
**Paul Valiant<br />
**David Speyer<br />
**David Vickrey<br />
*1997:<br />
**Josh Nichols-Barrer<br />
**Daniel Stronger<br />
**John Clyde<br />
**Carl Bosley<br />
**Li-Chung Chen<br />
**Nathan Curtis<br />
**Davesh Maulik<br />
**Kevin Lacker<br />
*1996:<br />
**Chris Chang<br />
**Alex Saltman<br />
**Josh Nichols-Barrer<br />
**Carl Bosley<br />
**Michael Korn<br />
**Carl Miller<br />
**Nathan Curtis<br />
**Daniel Stronger<br />
*1995:<br />
**Aleksander Khazanov<br />
**Jacob Lurie<br />
**Chris Chang<br />
**Jay Chyung<br />
**Andrei Gnepp<br />
**Josh Nichols-Barrer<br />
**Samit Dasgupta<br />
**Craig Helfgott<br />
*1994:<br />
**Jeremy Bem<br />
**Aleksander Khazanov<br />
**Jonathan Weinstein<br />
**Jacob Lurie<br />
**Noam Shazeer<br />
**Stephen Wang<br />
**Chris Chang<br />
**Jacob Rasmussen<br />
*1989:<br />
**Jordan Ellenberg<br />
**Andrew Kresh<br />
**Jonathan Higa<br />
**[[Richard Rusczyk]]<br />
**Jeff Vanderkam<br />
**[[Sam Vandervelde]]<br />
**David Carlton<br />
<br />
==USAMO Honorable Mentions==<br />
<br />
Other students who finish (or tie to finish) in the top 24 of the [[USAMO]] receive Honorable Mention (often abbreviated HM).<br />
*2019<br />
**Quanlin Chen<br />
**Eric Gan<br />
**Sebastian Jeon<br />
**Tianze Jiang<br />
**Jeffrey Kwan<br />
**Benjamin Qi<br />
**William Zhao<br />
*2018<br />
**Adam Ardeishar<br />
**Hongyi Chen<br />
**Andrew Gu<br />
**Milan Haiman<br />
**Daniel Liu<br />
**Kevin Liu<br />
**Victor Luo<br />
**Benjamin Qi<br />
**Kevin Ren<br />
**Luke Robitaille<br />
**Dilhan Salgado<br />
**Nicholas Sun<br />
**Stan Zhang<br />
**William Zhao<br />
*2017<br />
**Swapnil Garg<br />
**Frank Han<br />
**Vincent Huang<br />
**Daniel Kim<br />
**Joshua Lee<br />
**Michael Ma<br />
**Qi Qi<br />
**Nikhil Reddy<br />
**Dilhan Salgado<br />
**Patrick Wang<br />
**Eric Zhang<br />
**Daniel Zhu<br />
*2016<br />
**Kapil Chandran<br />
**Andrew Gu<br />
**Brian Gu<br />
**Meghal Gupta<br />
**Michael Kural<br />
**Dhruv Medarametla<br />
**Eshaan Nichani<br />
**Brian Reinhart<br />
**Ashwin Sah<br />
**Franklyn Wang<br />
**Alexander Wei<br />
**Allen Yang<br />
**Stan Zhang<br />
**Ziqi Zhou<br />
*2015<br />
**Demi Guo<br />
**Jacob Gurev<br />
**Kevin Li<br />
**James Lin<br />
**Michael Ma<br />
**Zachary Polansky<br />
**Samuel Reinehr<br />
**Ashwin Sah<br />
**Alexander Whatley<br />
**Jinhao Xu<br />
**Kevin Yang<br />
*2014<br />
**Lewis Chen<br />
**Ernest Chiu<br />
**Samuel Korsky<br />
**Yang Liu<br />
**Sammy Luo<br />
**Alexander Meiburg<br />
**Shyam Narayanan<br />
**Ashwin Sah<br />
**Nat Sothanaphan<br />
**Danielle Wang<br />
**Kevin Yang<br />
*2013:<br />
**Joshua Brakensiek<br />
**Lewis Chen<br />
**Sohail Farhangi<br />
**Kevin Huang<br />
**Allen Liu<br />
**Sammy Luo<br />
**Eric Schneider<br />
**Rachit Singh<br />
**Vikram Sundar<br />
**James Tao<br />
**David Yang<br />
**Kevin Yang<br />
*2011:<br />
**Kevin Chen<br />
**Calvin Deng<br />
**Michael Druggan<br />
**Albert Gu<br />
**Bobby Shen<br />
**Alex Song<br />
**Thomas Swayze<br />
**Tianqi Wu<br />
**Dai Yang<br />
**Kevin Yin<br />
**Allen Yuan<br />
**Samuel Zbarsky<br />
**Brian Zhang<br />
*2010:<br />
**Thomas Swayze<br />
**Kerry Xing<br />
**Tian-Yi Jiang<br />
**Brian Zhang<br />
**Archit Kulkarni<br />
**Patrick Hulin<br />
**Carl Lian<br />
**George Xing<br />
**Potcharapol Suteparuk<br />
**Jason Wu<br />
**Shijie Zheng<br />
**Robin Cheng<br />
**Bowei Liu<br />
**Wenyu Cao<br />
**Kevin Yin<br />
**Albert Gu<br />
*2009:<br />
**Timothy Chu<br />
**Brian Hamrick<br />
**Travis Hance<br />
**Jason Hoch<br />
**Tian-Yi Jiang<br />
**Sam Keller<br />
**Holden Lee<br />
**In-Sung Na<br />
**Ofir Nachum<br />
**Jarno Sun <br />
**Matthew Superdock<br />
**Tong Zhan<br />
**Mark Zhang<br />
*2008:<br />
**John Berman<br />
**Sergei Bernstein<br />
**Gregory Gauthier<br />
**Brian Hamrick<br />
**Eric Larson<br />
**Gaku Liu<br />
**Jeffrey Manning<br />
**Palmer Mebane<br />
**Danny Shi<br />
**Jacob Steinhardt<br />
**Matthew Superdock<br />
**Nicholas Triantafillou<br />
*2007:<br />
**David Benjamin<br />
**Gregory Brockman<br />
**Yingyu Gao<br />
**Yan Li<br />
**Gaku Liu<br />
**Jeffrey Manning<br />
**Palmer Mebane<br />
**Evan O'Dorney<br />
**Alexander Remorov<br />
**Max Rosett<br />
**Qiaochu Yuan<br />
**Bohua Zhan<br />
*2006:<br />
**Joseph Chu<br />
**Zachary Abel<br />
**Zarathustra Brady<br />
**Peter Diao<br />
**Richar Peng<br />
**Adam Hesterberg<br />
**Bohua Zhan<br />
**Kevin Modzelewski<br />
**Daniel Poore<br />
**Lin Fei<br />
**Jason Trigg<br />
**Tony Liu<br />
**Viktoriya Krakovna<br />
*2005:<br />
**Zachary Abel<br />
**Charles Chen<br />
**Evan Dummit<br />
**Adam Hesterberg<br />
**Richard Ho<br />
**Alexander Marcus<br />
**Richard McCutchen<br />
**Thomas Mildorf<br />
**Charles Nathanson<br />
**Keenan Pepper<br />
**Timothy Pollio<br />
*2004:<br />
**Boris Alexeev<br />
**Joshua Batson<br />
**Thomas Belulovich<br />
**Rishi Gupta<br />
**Anders Kaseorg<br />
**Hyun Soo Kim<br />
**John Kim<br />
**Po Ling Loh<br />
**Sira Sriswasdi<br />
**Yi Sun<br />
**Elena Udovina<br />
**Yufei Zhao<br />
*2003:<br />
**Timothy Abbott<br />
**Jeffrey Amos<br />
**Jongmin Baek<br />
**Richard Biggs<br />
**Steven Byrnes<br />
**Sherry Gong<br />
**Matt Ince<br />
**Nathaniel Ince<br />
**Pavel Kamyshev<br />
**Nicholas Ma<br />
**Alison Miller<br />
**Eric Price<br />
**David Stolp<br />
**Sean Ting<br />
**Tongke Xue<br />
*1995<br />
** Mathew Crawford<br />
*1993<br />
** Mathew Crawford<br />
<br />
==See also==<br />
[[Category:Historical results]]</div>GeronimoStiltonhttps://artofproblemsolving.com/wiki/index.php?title=USAMO_historical_results&diff=109743USAMO historical results2019-09-09T18:55:37Z<p>GeronimoStilton: </p>
<hr />
<div>This is the '''USAMO historical results''' page. <!-- Post results here so the [[USAMO]] page does not become cluttered. --><br />
<br />
==USAMO Perfect Scorers==<br />
<br />
Since 1996, a perfect score on the [[USAMO]] has been 42 points for 6 problems. Prior to 1996 a perfect score was 100 points across 5 problems.<br />
*2019: Luke Robitaille<br />
*2018: None<br />
*2017: None<br />
*2016:<br />
**Allen Liu<br />
*2015:<br />
**Allen Liu<br />
**David Stoner<br />
*2014: <br />
**Joshua Brakensiek<br />
*2013: None<br />
*2012: <br />
**Alex Zhu<br />
**Thomas Swayze<br />
**Xiaoyu He<br />
**Mitchell Lee<br />
**Samuel Zbarsky<br />
*2011:<br />
**David Yang<br />
**Evan O'Dorney<br />
*2010: None<br />
*2008: None<br />
*2007: None<br />
*2006:<br />
**Brian Lawrence<br />
*2005: None<br />
*2004:<br />
**Tiankai Liu<br />
*2003:<br />
**Tiankai Liu<br />
**Po-Ru Loh<br />
*2002: <br />
**Daniel Kane<br />
**Ricky Liu<br />
**Tiankai Liu<br />
**Po-Ru Loh<br />
**Inna Zakharevich<br />
*1996:<br />
**Chris Chang<br />
*1992:<br />
**Kiran Kedlaya<br />
**Lenny Ng<br />
<br />
==USAMO Winners==<br />
<br />
The top 12 scorers on the [[USAMO]] are currently designated as USAMO winners. In the past, this number was smaller.<br />
*2019:<br />
**Vincent Bian<br />
**Milan Haiman<br />
**Vincent Huang<br />
**Kevin Liu<br />
**Luke Robitaille<br />
**Victor Rong<br />
**Carl Schildkraut<br />
**Colin Shanmo Tang<br />
**Edward Wan<br />
**Brandon Wang<br />
**Guanpeng Xu<br />
**Daniel Zhu<br />
*2018:<br />
**Eric Gan<br />
**Thomas Guo<br />
**Vincent Huang<br />
**Joshua Lee<br />
**Michael Ren<br />
**Victor Rong<br />
**Carl Schildkraut<br />
**Mihir Singhal<br />
**Edward Wan<br />
**Brandon Wang<br />
**Guanpeng Xu<br />
**Andrew Yao<br />
*2017:<br />
**Zachary Chroman<br />
**Andrew Gu<br />
**James Lin<br />
**Daniel Liu<br />
**Michael Ren<br />
**Victor Rong<br />
**Ashwin Sah<br />
**Mihir Singhal<br />
**Alec Sun<br />
**Kada Williams<br />
**Yuan Yao<br />
**William Zhao<br />
*2016:<br />
**Ankan Bhattacharya<br />
**Ruidi Cao<br />
**Hongyi Chen<br />
**Jacob Klegar<br />
**James Lin<br />
**Allen Liu<br />
**Junyao Peng<br />
**Kevin Ren<br />
**Mihir Singhal<br />
**Alec Sun<br />
**Kevin Sun<br />
**Yuan Yao<br />
*2015:<br />
**Ryan Alweiss<br />
**Kritkorn Karntikoon<br />
**Michael Kural<br />
**Celine Liang<br />
**Allen Liu<br />
**Yang Liu<br />
**Shyam Narayanan<br />
**Kevin Ren<br />
**Zhuoqun Song<br />
**David Stoner<br />
**Kevin Sun<br />
**Danielle Wang<br />
*2014:<br />
**Joshua Brakensiek<br />
**Evan Chen<br />
**Ravi Jagadeesan<br />
**Allen Liu<br />
**Nipun Pitimanaaree<br />
**Mark Sellke<br />
**Zhuoqun Song<br />
**David Stoner<br />
**Kevin Sun<br />
**James Tao<br />
**Alexander Whatley<br />
**Scott Wu<br />
*2013<br />
**Calvin Deng<br />
**Andrew He<br />
**Ravi Jagadeesan<br />
**Pakawut Jiradilok<br />
**Ray Li<br />
**Kevin Li<br />
**Mark Sellke<br />
**Bobby Shen<br />
**Zhuoqun Song<br />
**David Stoner<br />
**Thomas Swayze<br />
**Victor Wang<br />
*2012:<br />
**Andre Arslan<br />
**Joshua Brakensiek<br />
**Calvin Deng<br />
**Xiaoyu He<br />
**Ravi Jagadeesan<br />
**Mitchell Lee<br />
**Alex Song<br />
**Thomas Swayze<br />
**Victor Wang<br />
**David Yang<br />
**Samuel Zbarsky<br />
**Alex Zhu<br />
*2011:<br />
**Wenyu Cao<br />
**Zijing Gao<br />
**Benjamin Gunby<br />
**Xiaoyu He<br />
**Ravi Jagadeesan<br />
**Spencer Kwon<br />
**Mitchell Lee<br />
**Ray Li<br />
**Evan O'Dorney<br />
**Mark Sellke<br />
**David Yang<br />
**Joy Zheng<br />
*2010:<br />
**Timothy Chu<br />
**Calvin Deng<br />
**Michael Druggan<br />
**Brian Hamrick<br />
**Travis Hance<br />
**Xiaoyu He<br />
**Mitchell Lee<br />
**In Sung Na<br />
**Evan O'Dorney<br />
**Toan Phan<br />
**Hunter Spink<br />
**Allen Yuan<br />
*2009:<br />
**John Berman<br />
**Sergei Bernstein<br />
**Wenyu Cao<br />
**Robin Cheng<br />
**Vlad Firoiu<br />
**Eric Larson<br />
**Delong Meng<br />
**Qinxuan Pan<br />
**Panupong Pasupat<br />
**Toan Phan<br />
**David Rush<br />
**David Yang<br />
*2008:<br />
**David Benjamin<br />
**TaoRan Chen<br />
**Paul Christiano<br />
**Sam Elder<br />
**Shaunak Kishore<br />
**Delong Meng<br />
**Evan O'Dorney<br />
**Qinxuan Pan<br />
**David Rolnick<br />
**Colin Sandon<br />
**Krishanu Sankar<br />
**Alex Zhai<br />
*2007:<br />
**Sergei Bernstein<br />
**Sherry Gong<br />
**Adam Hesterberg<br />
**Eric Larson<br />
**Brian Lawrence<br />
**Tedrick Leung<br />
**Haitao Mao<br />
**Delong Meng<br />
**Krishanu Sankar<br />
**Jacob Steinhardt<br />
**Arnav Tripathy<br />
**Alex Zhai<br />
*2006:<br />
**Brian Lawrence<br />
**Alex Zhai<br />
**Yufei Zhao<br />
**Peng Shi<br />
**Sherry Gong<br />
**Richard McCutchen<br />
**Yi Sun<br />
**Arnav Tripathy<br />
**Taehyeon (Ryan) Ko<br />
**Yi Han<br />
**Yakov Berchenko-Kogan<br />
**Tedrick Leung<br />
*2005:<br />
**Robert Cordwell<br />
**Zhou Fan<br />
**Sherry Gong<br />
**Rishi Gupta<br />
**Hyun Soo Kim<br />
**Brian Lawrence<br />
**Albert Ni<br />
**Natee Pitiwan<br />
**Eric Price<br />
**Peng Shi<br />
**Yi Sun<br />
**Yufei Zhao<br />
*2004:<br />
**Tiankai Liu<br />
**Jae Bae<br />
**Jongmin Baek<br />
**Oleg Golberg<br />
**Matt Ince<br />
**Janos Kramar<br />
**Alison Miller<br />
**Aaron Pixton<br />
**Brian Rice<br />
**Jacob Tsimerman<br />
**Ameya Velingker<br />
**Tony Zhang<br />
*1998:<br />
**Sasha Schwartz<br />
**Melanie Wood<br />
**Kevin Lacker<br />
**Reid Barton<br />
**Gabriel Carroll<br />
**Paul Valiant<br />
**David Speyer<br />
**David Vickrey<br />
*1997:<br />
**Josh Nichols-Barrer<br />
**Daniel Stronger<br />
**John Clyde<br />
**Carl Bosley<br />
**Li-Chung Chen<br />
**Nathan Curtis<br />
**Davesh Maulik<br />
**Kevin Lacker<br />
*1996:<br />
**Chris Chang<br />
**Alex Saltman<br />
**Josh Nichols-Barrer<br />
**Carl Bosley<br />
**Michael Korn<br />
**Carl Miller<br />
**Nathan Curtis<br />
**Daniel Stronger<br />
*1995:<br />
**Aleksander Khazanov<br />
**Jacob Lurie<br />
**Chris Chang<br />
**Jay Chyung<br />
**Andrei Gnepp<br />
**Josh Nichols-Barrer<br />
**Samit Dasgupta<br />
**Craig Helfgott<br />
*1994:<br />
**Jeremy Bem<br />
**Aleksander Khazanov<br />
**Jonathan Weinstein<br />
**Jacob Lurie<br />
**Noam Shazeer<br />
**Stephen Wang<br />
**Chris Chang<br />
**Jacob Rasmussen<br />
*1989:<br />
**Jordan Ellenberg<br />
**Andrew Kresh<br />
**Jonathan Higa<br />
**[[Richard Rusczyk]]<br />
**Jeff Vanderkam<br />
**[[Sam Vandervelde]]<br />
**David Carlton<br />
<br />
==USAMO Honorable Mentions==<br />
<br />
Other students who finish (or tie to finish) in the top 24 of the [[USAMO]] receive Honorable Mention (often abbreviated HM).<br />
*2019<br />
**Quanlin Chen<br />
**Eric Gan<br />
**Sebastian Jeon<br />
**Tianze Jiang<br />
**Jeffrey Kwan<br />
**Benjamin Qi<br />
**William Zhao<br />
*2018<br />
**Adam Ardeishar<br />
**Hongyi Chen<br />
**Andrew Gu<br />
**Milan Haiman<br />
**Daniel Liu<br />
**Kevin Liu<br />
**Victor Luo<br />
**Benjamin Qi<br />
**Kevin Ren<br />
**Luke Robitaille<br />
**Dilhan Salgado<br />
**Nicholas Sun<br />
**Stan Zhang<br />
**William Zhao<br />
*2017<br />
**Swapnil Garg<br />
**Frank Han<br />
**Vincent Huang<br />
**Daniel Kim<br />
**Joshua Lee<br />
**Michael Ma<br />
**Qi Qi<br />
**Nikhil Reddy<br />
**Dilhan Salgado<br />
**Patrick Wang<br />
**Eric Zhang<br />
**Daniel Zhu<br />
*2016<br />
**Kapil Chandran<br />
**Andrew Gu<br />
**Brian Gu<br />
**Meghal Gupta<br />
**Michael Kural<br />
**Dhruv Medarametla<br />
**Eshaan Nichani<br />
**Brian Reinhart<br />
**Ashwin Sah<br />
**Franklyn Wang<br />
**Alexander Wei<br />
**Allen Yang<br />
**Stan Zhang<br />
**Ziqi Zhou<br />
*2015<br />
**Demi Guo<br />
**Jacob Gurev<br />
**Kevin Li<br />
**James Lin<br />
**Michael Ma<br />
**Zachary Polansky<br />
**Samuel Reinehr<br />
**Ashwin Sah<br />
**Alexander Whatley<br />
**Jinhao Xu<br />
**Kevin Yang<br />
*2014<br />
**Lewis Chen<br />
**Ernest Chiu<br />
**Samuel Korsky<br />
**Yang Liu<br />
**Sammy Luo<br />
**Alexander Meiburg<br />
**Shyam Narayanan<br />
**Ashwin Sah<br />
**Nat Sothanaphan<br />
**Danielle Wang<br />
**Kevin Yang<br />
*2013:<br />
**Joshua Brakensiek<br />
**Lewis Chen<br />
**Sohail Farhangi<br />
**Kevin Huang<br />
**Allen Liu<br />
**Sammy Luo<br />
**Eric Schneider<br />
**Rachit Singh<br />
**Vikram Sundar<br />
**James Tao<br />
**David Yang<br />
**Kevin Yang<br />
*2011:<br />
**Kevin Chen<br />
**Calvin Deng<br />
**Michael Druggan<br />
**Albert Gu<br />
**Bobby Shen<br />
**Alex Song<br />
**Thomas Swayze<br />
**Tianqi Wu<br />
**Dai Yang<br />
**Kevin Yin<br />
**Allen Yuan<br />
**Samuel Zbarsky<br />
**Brian Zhang<br />
*2010:<br />
**Thomas Swayze<br />
**Kerry Xing<br />
**Tian-Yi Jiang<br />
**Brian Zhang<br />
**Archit Kulkarni<br />
**Patrick Hulin<br />
**Carl Lian<br />
**George Xing<br />
**Potcharapol Suteparuk<br />
**Jason Wu<br />
**Shijie Zheng<br />
**Robin Cheng<br />
**Bowei Liu<br />
**Wenyu Cao<br />
**Kevin Yin<br />
**Albert Gu<br />
*2009:<br />
**Timothy Chu<br />
**Brian Hamrick<br />
**Travis Hance<br />
**Jason Hoch<br />
**Tian-Yi Jiang<br />
**Sam Keller<br />
**Holden Lee<br />
**In-Sung Na<br />
**Ofir Nachum<br />
**Jarno Sun <br />
**Matthew Superdock<br />
**Tong Zhan<br />
**Mark Zhang<br />
*2008:<br />
**John Berman<br />
**Sergei Bernstein<br />
**Gregory Gauthier<br />
**Brian Hamrick<br />
**Eric Larson<br />
**Gaku Liu<br />
**Jeffrey Manning<br />
**Palmer Mebane<br />
**Danny Shi<br />
**Jacob Steinhardt<br />
**Matthew Superdock<br />
**Nicholas Triantafillou<br />
*2007:<br />
**David Benjamin<br />
**Gregory Brockman<br />
**Yingyu Gao<br />
**Yan Li<br />
**Gaku Liu<br />
**Jeffrey Manning<br />
**Palmer Mebane<br />
**Evan O'Dorney<br />
**Alexander Remorov<br />
**Max Rosett<br />
**Qiaochu Yuan<br />
**Bohua Zhan<br />
*2006:<br />
**Joseph Chu<br />
**Zachary Abel<br />
**Zarathustra Brady<br />
**Peter Diao<br />
**Richar Peng<br />
**Adam Hesterberg<br />
**Bohua Zhan<br />
**Kevin Modzelewski<br />
**Daniel Poore<br />
**Lin Fei<br />
**Jason Trigg<br />
**Tony Liu<br />
**Viktoriya Krakovna<br />
*2005:<br />
**Zachary Abel<br />
**Charles Chen<br />
**Evan Dummit<br />
**Adam Hesterberg<br />
**Richard Ho<br />
**Alexander Marcus<br />
**Richard McCutchen<br />
**Thomas Mildorf<br />
**Charles Nathanson<br />
**Keenan Pepper<br />
**Timothy Pollio<br />
*2004:<br />
**Boris Alexeev<br />
**Joshua Batson<br />
**Thomas Belulovich<br />
**Rishi Gupta<br />
**Anders Kaseorg<br />
**Hyun Soo Kim<br />
**John Kim<br />
**Po Ling Loh<br />
**Sira Sriswasdi<br />
**Yi Sun<br />
**Elena Udovina<br />
**Yufei Zhao<br />
*1995<br />
** Mathew Crawford<br />
*1993<br />
** Mathew Crawford<br />
<br />
==See also==<br />
[[Category:Historical results]]</div>GeronimoStiltonhttps://artofproblemsolving.com/wiki/index.php?title=USAMO_historical_results&diff=109742USAMO historical results2019-09-09T18:53:16Z<p>GeronimoStilton: </p>
<hr />
<div>This is the '''USAMO historical results''' page. <!-- Post results here so the [[USAMO]] page does not become cluttered. --><br />
<br />
==USAMO Perfect Scorers==<br />
<br />
Since 1996, a perfect score on the [[USAMO]] has been 42 points for 6 problems. Prior to 1996 a perfect score was 100 points across 5 problems.<br />
*2019: Luke Robitaille<br />
*2018: None<br />
*2017: None<br />
*2016:<br />
**Allen Liu<br />
*2015:<br />
**Allen Liu<br />
**David Stoner<br />
*2014: <br />
**Joshua Brakensiek<br />
*2013: None<br />
*2012: <br />
**Alex Zhu<br />
**Thomas Swayze<br />
**Xiaoyu He<br />
**Mitchell Lee<br />
**Samuel Zbarsky<br />
*2011:<br />
**David Yang<br />
**Evan O'Dorney<br />
*2010: None<br />
*2008: None<br />
*2007: None<br />
*2006:<br />
**Brian Lawrence<br />
*2005: None<br />
*2004:<br />
**Tiankai Liu<br />
*2003:<br />
**Tiankai Liu<br />
**Po-Ru Loh<br />
*2002: <br />
**Daniel Kane<br />
**Ricky Liu<br />
**Tiankai Liu<br />
**Po-Ru Loh<br />
**Inna Zakharevich<br />
*1996:<br />
**Chris Chang<br />
*1992:<br />
**Kiran Kedlaya<br />
**Lenny Ng<br />
<br />
==USAMO Winners==<br />
<br />
The top 12 scorers on the [[USAMO]] are currently designated as USAMO winners. In the past, this number was smaller.<br />
*2019:<br />
**Vincent Bian<br />
**Milan Haiman<br />
**Vincent Huang<br />
**Kevin Liu<br />
**Luke Robitaille<br />
**Victor Rong<br />
**Carl Schildkraut<br />
**Colin Shanmo Tang<br />
**Edward Wan<br />
**Brandon Wang<br />
**Guanpeng Xu<br />
**Daniel Zhu<br />
*2018:<br />
**Eric Gan<br />
**Thomas Guo<br />
**Vincent Huang<br />
**Joshua Lee<br />
**Michael Ren<br />
**Victor Rong<br />
**Carl Schildkraut<br />
**Mihir Singhal<br />
**Edward Wan<br />
**Brandon Wang<br />
**Guanpeng Xu<br />
**Andrew Yao<br />
*2017:<br />
**Zachary Chroman<br />
**Andrew Gu<br />
**James Lin<br />
**Daniel Liu<br />
**Michael Ren<br />
**Victor Rong<br />
**Ashwin Sah<br />
**Mihir Singhal<br />
**Alec Sun<br />
**Kada Williams<br />
**Yuan Yao<br />
**William Zhao<br />
*2016:<br />
**Ankan Bhattacharya<br />
**Ruidi Cao<br />
**Hongyi Chen<br />
**Jacob Klegar<br />
**James Lin<br />
**Allen Liu<br />
**Junyao Peng<br />
**Kevin Ren<br />
**Mihir Singhal<br />
**Alec Sun<br />
**Kevin Sun<br />
**Yuan Yao<br />
*2015:<br />
**Ryan Alweiss<br />
**Kritkorn Karntikoon<br />
**Michael Kural<br />
**Celine Liang<br />
**Allen Liu<br />
**Yang Liu<br />
**Shyam Narayanan<br />
**Kevin Ren<br />
**Zhuoqun Song<br />
**David Stoner<br />
**Kevin Sun<br />
**Danielle Wang<br />
*2014:<br />
**Joshua Brakensiek<br />
**Evan Chen<br />
**Ravi Jagadeesan<br />
**Allen Liu<br />
**Nipun Pitimanaaree<br />
**Mark Sellke<br />
**Zhuoqun Song<br />
**David Stoner<br />
**Kevin Sun<br />
**James Tao<br />
**Alexander Whatley<br />
**Scott Wu<br />
*2013<br />
**Calvin Deng<br />
**Andrew He<br />
**Ravi Jagadeesan<br />
**Pakawut Jiradilok<br />
**Ray Li<br />
**Kevin Li<br />
**Mark Sellke<br />
**Bobby Shen<br />
**Zhuoqun Song<br />
**David Stoner<br />
**Thomas Swayze<br />
**Victor Wang<br />
*2012:<br />
**Andre Arslan<br />
**Joshua Brakensiek<br />
**Calvin Deng<br />
**Xiaoyu He<br />
**Ravi Jagadeesan<br />
**Mitchell Lee<br />
**Alex Song<br />
**Thomas Swayze<br />
**Victor Wang<br />
**David Yang<br />
**Samuel Zbarsky<br />
**Alex Zhu<br />
*2011:<br />
**Wenyu Cao<br />
**Zijing Gao<br />
**Benjamin Gunby<br />
**Xiaoyu He<br />
**Ravi Jagadeesan<br />
**Spencer Kwon<br />
**Mitchell Lee<br />
**Ray Li<br />
**Evan O'Dorney<br />
**Mark Sellke<br />
**David Yang<br />
**Joy Zheng<br />
*2010:<br />
**Timothy Chu<br />
**Calvin Deng<br />
**Michael Druggan<br />
**Brian Hamrick<br />
**Travis Hance<br />
**Xiaoyu He<br />
**Mitchell Lee<br />
**In Sung Na<br />
**Evan O'Dorney<br />
**Toan Phan<br />
**Hunter Spink<br />
**Allen Yuan<br />
*2009:<br />
**John Berman<br />
**Sergei Bernstein<br />
**Wenyu Cao<br />
**Robin Cheng<br />
**Vlad Firoiu<br />
**Eric Larson<br />
**Delong Meng<br />
**Qinxuan Pan<br />
**Panupong Pasupat<br />
**Toan Phan<br />
**David Rush<br />
**David Yang<br />
*2008:<br />
**David Benjamin<br />
**TaoRan Chen<br />
**Paul Christiano<br />
**Sam Elder<br />
**Shaunak Kishore<br />
**Delong Meng<br />
**Evan O'Dorney<br />
**Qinxuan Pan<br />
**David Rolnick<br />
**Colin Sandon<br />
**Krishanu Sankar<br />
**Alex Zhai<br />
*2007:<br />
**Sergei Bernstein<br />
**Sherry Gong<br />
**Adam Hesterberg<br />
**Eric Larson<br />
**Brian Lawrence<br />
**Tedrick Leung<br />
**Haitao Mao<br />
**Delong Meng<br />
**Krishanu Sankar<br />
**Jacob Steinhardt<br />
**Arnav Tripathy<br />
**Alex Zhai<br />
*2006:<br />
**Brian Lawrence<br />
**Alex Zhai<br />
**Yufei Zhao<br />
**Peng Shi<br />
**Sherry Gong<br />
**Richard McCutchen<br />
**Yi Sun<br />
**Arnav Tripathy<br />
**Taehyeon (Ryan) Ko<br />
**Yi Han<br />
**Yakov Berchenko-Kogan<br />
**Tedrick Leung<br />
*2005:<br />
**Robert Cordwell<br />
**Zhou Fan<br />
**Sherry Gong<br />
**Rishi Gupta<br />
**Hyun Soo Kim<br />
**Brian Lawrence<br />
**Albert Ni<br />
**Natee Pitiwan<br />
**Eric Price<br />
**Peng Shi<br />
**Yi Sun<br />
**Yufei Zhao<br />
*2004:<br />
**Tiankai Liu<br />
**Jae Bae<br />
**Jongmin Baek<br />
**Oleg Golberg<br />
**Matt Ince<br />
**Janos Kramar<br />
**Alison Miller<br />
**Aaron Pixton<br />
**Brian Rice<br />
**Jacob Tsimerman<br />
**Ameya Velingker<br />
**Tony Zhang<br />
*1998:<br />
**Sasha Schwartz<br />
**Melanie Wood<br />
**Kevin Lacker<br />
**Reid Barton<br />
**Gabriel Carroll<br />
**Paul Valiant<br />
**David Speyer<br />
**David Vickrey<br />
*1997:<br />
**Josh Nichols-Barrer<br />
**Daniel Stronger<br />
**John Clyde<br />
**Carl Bosley<br />
**Li-Chung Chen<br />
**Nathan Curtis<br />
**Davesh Maulik<br />
**Kevin Lacker<br />
*1996:<br />
**Chris Chang<br />
**Alex Saltman<br />
**Josh Nichols-Barrer<br />
**Carl Bosley<br />
**Michael Korn<br />
**Carl Miller<br />
**Nathan Curtis<br />
**Daniel Stronger<br />
*1995:<br />
**Aleksander Khazanov<br />
**Jacob Lurie<br />
**Chris Chang<br />
**Jay Chyung<br />
**Andrei Gnepp<br />
**Josh Nichols-Barrer<br />
**Samit Dasgupta<br />
**Craig Helfgott<br />
*1994:<br />
**Jeremy Bem<br />
**Aleksander Khazanov<br />
**Jonathan Weinstein<br />
**Jacob Lurie<br />
**Noam Shazeer<br />
**Stephen Wang<br />
**Chris Chang<br />
**Jacob Rasmussen<br />
*1989:<br />
**Jordan Ellenberg<br />
**Andrew Kresh<br />
**Jonathan Higa<br />
**[[Richard Rusczyk]]<br />
**Jeff Vanderkam<br />
**[[Sam Vandervelde]]<br />
**David Carlton<br />
<br />
==USAMO Honorable Mentions==<br />
<br />
Other students who finish (or tie to finish) in the top 24 of the [[USAMO]] receive Honorable Mention (often abbreviated HM).<br />
*2018<br />
**Adam Ardeishar<br />
**Hongyi Chen<br />
**Andrew Gu<br />
**Milan Haiman<br />
**Daniel Liu<br />
**Kevin Liu<br />
**Victor Luo<br />
**Benjamin Qi<br />
**Kevin Ren<br />
**Luke Robitaille<br />
**Dilhan Salgado<br />
**Nicholas Sun<br />
**Stan Zhang<br />
**William Zhao<br />
*2017<br />
**Swapnil Garg<br />
**Frank Han<br />
**Vincent Huang<br />
**Daniel Kim<br />
**Joshua Lee<br />
**Michael Ma<br />
**Qi Qi<br />
**Nikhil Reddy<br />
**Dilhan Salgado<br />
**Patrick Wang<br />
**Eric Zhang<br />
**Daniel Zhu<br />
*2016<br />
**Kapil Chandran<br />
**Andrew Gu<br />
**Brian Gu<br />
**Meghal Gupta<br />
**Michael Kural<br />
**Dhruv Medarametla<br />
**Eshaan Nichani<br />
**Brian Reinhart<br />
**Ashwin Sah<br />
**Franklyn Wang<br />
**Alexander Wei<br />
**Allen Yang<br />
**Stan Zhang<br />
**Ziqi Zhou<br />
*2015<br />
**Demi Guo<br />
**Jacob Gurev<br />
**Kevin Li<br />
**James Lin<br />
**Michael Ma<br />
**Zachary Polansky<br />
**Samuel Reinehr<br />
**Ashwin Sah<br />
**Alexander Whatley<br />
**Jinhao Xu<br />
**Kevin Yang<br />
*2014<br />
**Lewis Chen<br />
**Ernest Chiu<br />
**Samuel Korsky<br />
**Yang Liu<br />
**Sammy Luo<br />
**Alexander Meiburg<br />
**Shyam Narayanan<br />
**Ashwin Sah<br />
**Nat Sothanaphan<br />
**Danielle Wang<br />
**Kevin Yang<br />
*2013:<br />
**Joshua Brakensiek<br />
**Lewis Chen<br />
**Sohail Farhangi<br />
**Kevin Huang<br />
**Allen Liu<br />
**Sammy Luo<br />
**Eric Schneider<br />
**Rachit Singh<br />
**Vikram Sundar<br />
**James Tao<br />
**David Yang<br />
**Kevin Yang<br />
*2011:<br />
**Kevin Chen<br />
**Calvin Deng<br />
**Michael Druggan<br />
**Albert Gu<br />
**Bobby Shen<br />
**Alex Song<br />
**Thomas Swayze<br />
**Tianqi Wu<br />
**Dai Yang<br />
**Kevin Yin<br />
**Allen Yuan<br />
**Samuel Zbarsky<br />
**Brian Zhang<br />
*2010:<br />
**Thomas Swayze<br />
**Kerry Xing<br />
**Tian-Yi Jiang<br />
**Brian Zhang<br />
**Archit Kulkarni<br />
**Patrick Hulin<br />
**Carl Lian<br />
**George Xing<br />
**Potcharapol Suteparuk<br />
**Jason Wu<br />
**Shijie Zheng<br />
**Robin Cheng<br />
**Bowei Liu<br />
**Wenyu Cao<br />
**Kevin Yin<br />
**Albert Gu<br />
*2009:<br />
**Timothy Chu<br />
**Brian Hamrick<br />
**Travis Hance<br />
**Jason Hoch<br />
**Tian-Yi Jiang<br />
**Sam Keller<br />
**Holden Lee<br />
**In-Sung Na<br />
**Ofir Nachum<br />
**Jarno Sun <br />
**Matthew Superdock<br />
**Tong Zhan<br />
**Mark Zhang<br />
*2008:<br />
**John Berman<br />
**Sergei Bernstein<br />
**Gregory Gauthier<br />
**Brian Hamrick<br />
**Eric Larson<br />
**Gaku Liu<br />
**Jeffrey Manning<br />
**Palmer Mebane<br />
**Danny Shi<br />
**Jacob Steinhardt<br />
**Matthew Superdock<br />
**Nicholas Triantafillou<br />
*2007:<br />
**David Benjamin<br />
**Gregory Brockman<br />
**Yingyu Gao<br />
**Yan Li<br />
**Gaku Liu<br />
**Jeffrey Manning<br />
**Palmer Mebane<br />
**Evan O'Dorney<br />
**Alexander Remorov<br />
**Max Rosett<br />
**Qiaochu Yuan<br />
**Bohua Zhan<br />
*2006:<br />
**Joseph Chu<br />
**Zachary Abel<br />
**Zarathustra Brady<br />
**Peter Diao<br />
**Richar Peng<br />
**Adam Hesterberg<br />
**Bohua Zhan<br />
**Kevin Modzelewski<br />
**Daniel Poore<br />
**Lin Fei<br />
**Jason Trigg<br />
**Tony Liu<br />
**Viktoriya Krakovna<br />
*2005:<br />
**Zachary Abel<br />
**Charles Chen<br />
**Evan Dummit<br />
**Adam Hesterberg<br />
**Richard Ho<br />
**Alexander Marcus<br />
**Richard McCutchen<br />
**Thomas Mildorf<br />
**Charles Nathanson<br />
**Keenan Pepper<br />
**Timothy Pollio<br />
*2004:<br />
**Boris Alexeev<br />
**Joshua Batson<br />
**Thomas Belulovich<br />
**Rishi Gupta<br />
**Anders Kaseorg<br />
**Hyun Soo Kim<br />
**John Kim<br />
**Po Ling Loh<br />
**Sira Sriswasdi<br />
**Yi Sun<br />
**Elena Udovina<br />
**Yufei Zhao<br />
*1995<br />
** Mathew Crawford<br />
*1993<br />
** Mathew Crawford<br />
<br />
==See also==<br />
[[Category:Historical results]]</div>GeronimoStiltonhttps://artofproblemsolving.com/wiki/index.php?title=AMC_historical_results&diff=95475AMC historical results2018-06-21T20:57:34Z<p>GeronimoStilton: </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 />
==2018==<br />
===AMC 10A===<br />
*Average score: 53.84<br />
*AIME floor: 111<br />
*DHR: 127.5<br />
<br />
===AMC 10B===<br />
*Average score: 57.81<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 />
==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: 60.32<br />
*AIME floor: 96<br />
*DHR: 115.5<br />
<br />
===AMC 12B===<br />
*Average score: 58.35<br />
*AIME floor: 100.5<br />
*DHR: 129<br />
<br />
===AIME I===<br />
*Average score:5.70<br />
*Median score: 6<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: 6<br />
*USAMO cutoff: 221 (AMC 12A), 230.5 (AMC 12B)<br />
*USAJMO cutoff: 219 (AMC 10A), 225 (AMC 10B)<br />
<br />
==2016==<br />
===AMC 10A===<br />
*Average score: 65.3<br />
*AIME floor: 111<br />
*DHR: 120<br />
<br />
===AMC 10B===<br />
*Average score: 65.4<br />
*AIME floor: 111<br />
*DHR: 124.5<br />
<br />
===AMC 12A===<br />
*Average score: 59.06<br />
*AIME floor: 93<br />
*DHR: 111<br />
<br />
===AMC 12B===<br />
*Average score: 67.96<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: 4.33<br />
*Median score: 4<br />
*USAMO cutoff: 205<br />
*USAJMO cutoff: 200<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: 76.10<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: 66.92<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 />
==2014==<br />
===AMC 10A===<br />
*Average score: 63.83<br />
*AIME floor: 120<br />
*DHR: 132<br />
<br />
===AMC 10B===<br />
*Average score: 71.44<br />
*AIME floor: 120<br />
*DHR: 132<br />
<br />
===AMC 12A===<br />
*Average score: 64.01<br />
*AIME floor: 93<br />
*DHR: 109.5<br />
<br />
===AMC 12B===<br />
*Average score: 68.11<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 />
==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 />
==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: <br />
*USAMO cutoff: 204.5<br />
*USAJMO cutoff: 204<br />
<br />
===AIME II===<br />
*Average score: 4.94<br />
*Median score: <br />
*USAMO cutoff: 204.5<br />
*USAJMO cutoff: 204<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: <br />
*USAMO cutoff: 188<br />
*USAJMO cutoff: 179<br />
<br />
===AIME II===<br />
*Average score: 5.47<br />
*Median score: <br />
*USAMO cutoff: 215.5<br />
*USAJMO cutoff: 196.5<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 />
<br />
===AMC 12B===<br />
*Average score: 59.58<br />
*AIME floor: 88.5<br />
<br />
===AIME I===<br />
*Average score: 5.90<br />
*Median score: <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: <br />
*USAMO cutoff: 208.5 (204.5 for non juniors and seniors)<br />
*USAJMO cutoff: 188.5<br />
<br />
==2009==<br />
===AMC 10A===<br />
*Average score: 67.41<br />
*AIME floor: 120<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 />
==2008==<br />
===AMC 10A===<br />
*Average score: <br />
*AIME floor: <br />
<br />
===AMC 10B===<br />
*Average score: <br />
*AIME floor: <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: <br />
*Median score: <br />
*USAMO floor: <br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<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 />
==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 />
==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 />
==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: 120<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 />
==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 />
==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: 64.2<br />
*AIME floor: 120<br />
<br />
===AMC 12===<br />
*Average score: 64.9<br />
*AIME floor: <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: 68.8<br />
*AIME floor:<br />
<br />
===AIME===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:</div>GeronimoStiltonhttps://artofproblemsolving.com/wiki/index.php?title=American_Invitational_Mathematics_Examination&diff=92218American Invitational Mathematics Examination2018-02-22T13:42:14Z<p>GeronimoStilton: </p>
<hr />
<div>The '''American Invitational Mathematics Examination''' ('''AIME''') is the second exam in the series of exams used to challenge bright students on the path toward choosing the team that represents the United States at the [[International Mathematics Olympiad]] (IMO). While most AIME participants are high school students, some bright middle school students also qualify each year.<br />
<br />
High scoring AIME students are invited to take the prestigious [[United States of America Mathematics Olympiad]] (USAMO) for qualification from taking the AMC 12 or [[United States of America Junior Mathematics Olympiad]] (USAJMO) for qualification from taking the AMC 10.<br />
<br />
The AIME is administered by the [[American Mathematics Competitions]] (AMC). [[Art of Problem Solving]] (AoPS) is a proud sponsor of the AMC!<br />
<br />
<br />
== Format ==<br />
<br />
The AIME is a 15 question, 3 hour exam<math>^1</math> taken by high scorers on the [[AMC 10]], [[AMC 12]], and [[USAMTS]] competitions. Qualification through USAMTS only is rare, however. Each answer is an integer from 0 to 999, inclusive, making guessing almost futile. Wrong answers receive no credit, while correct answers receive one point of credit, making the maximum score 15. Problems generally increase in difficulty as the exam progresses - the first few questions are generally AMC 12 level, while the later questions become extremely difficult in comparison. Calculators are not permitted.<br />
<br />
<br />
<math>^1</math> In the first two years (1983 and 1984) there was a 2.5 hour time limit instead of the current 3 hour limit.<br />
<br />
== Curriculum ==<br />
The AIME tests [[mathematical problem solving]] with [[arithmetic]], [[algebra]], [[counting]], [[geometry]], [[number theory]], and [[probability]] and other secondary school math topics. Problems usually require either very creative use of secondary school curriculum, or an understanding as to how different areas of math can be used together to investigate and solve a problem. <br />
<br />
<br />
== Resources ==<br />
=== Links ===<br />
* [http://www.unl.edu/amc/ AMC homepage] and their [http://www.unl.edu/amc/e-exams/e7-aime/aime.shtml AIME page]<br />
* [[AIME Problems and Solutions]] -- A community effort to provide solutions to all AIME problems from which students can learn.<br />
* The [[AoPS]] [http://www.artofproblemsolving.com/Resources/AoPS_R_Contests_AIME.php AIME guide].<br />
* [http://www.artofproblemsolving.com/Forum/index.php?f=133 AMC Forum] for discussion of the AMC and problems from AMC and AIME exams.<br />
* The [http://www.artofproblemsolving.com/Forum/resources.php AoPS Contest Archive] includes problems and solutions from [http://www.artofproblemsolving.com/Forum/resources.php?c=182 most AMC and all AIME exams].<br />
* [[Mock AIME | Mock AIME exams by AoPSers]] -- A wealth of secondary practice materials.<br />
<br />
=== Recommended reading ===<br />
* [http://www.artofproblemsolving.com/Books/AoPS_B_CP_AMC.php Problem and solution books for past AMC exams]. One of these books also includes numerous past AIMEs and solutions.<br />
* Introduction to Counting & Probability by Dr. [[David Patrick]] is recommended for students who qualify for the AIME, but feel they lag behind in their understanding of basic combinatorics and probability relative to their other areas of math. [http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?page_id=3 Information]<br />
* Introduction to Geometry by [[Richard Rusczyk]]. [http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?page_id=9 Information]<br />
* The Art of Problem Solving Volume II by [[Sandor Lehoczky]] and [[Richard Rusczyk]]. [http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?page_id=2 Information].<br />
<br />
<br />
=== AIME Preparation Classes ===<br />
* [[AoPS]] hosts an [http://www.artofproblemsolving.com/Classes/AoPS_C_About.php online school] teaching introductory classes in topics covered by the AIME as well as AIME preparation classes.<br />
* [[AoPS]] holds many free [[Math Jams]], some of which are devoted to discussing problems on the AIME. [http://www.artofproblemsolving.com/Community/AoPS_Y_Math_Jams.php Math Jam Schedule]<br />
<br />
<br />
=== AIME Exams in the AoPSWiki ===<br />
This is a list of all AIME exams in the AoPSWiki. Some of them contain complete questions and solutions, others complete questions, and others are lacking both questions and solutions. Many of these problems and solutions are also available in the [http://www.artofproblemsolving.com/Forum/resources.php?c=182 AoPS Resources] section. If you find problems that are in the Resources section which are not in the AoPSWiki, please consider adding them. Also, if you notice that a problem in the Wiki differs from the original wording, feel free to correct it. Finally, additions to and improvements on the solutions in the AoPSWiki are always welcome.<br />
<br />
* [[1983 AIME]]<br />
* [[1984 AIME]]<br />
* [[1985 AIME]]<br />
* [[1986 AIME]]<br />
* [[1987 AIME]]<br />
* [[1988 AIME]]<br />
* [[1989 AIME]]<br />
* [[1990 AIME]]<br />
* [[1991 AIME]]<br />
* [[1992 AIME]]<br />
* [[1993 AIME]]<br />
* [[1994 AIME]]<br />
* [[1995 AIME]]<br />
* [[1996 AIME]]<br />
* [[1997 AIME]]<br />
* [[1998 AIME]]<br />
* [[1999 AIME]]<br />
* [[2000 AIME I]]<br />
* [[2000 AIME II]]<br />
* [[2001 AIME I]]<br />
* [[2001 AIME II]]<br />
* [[2002 AIME I]]<br />
* [[2002 AIME II]]<br />
* [[2003 AIME I]]<br />
* [[2003 AIME II]]<br />
* [[2004 AIME I]]<br />
* [[2004 AIME II]]<br />
* [[2005 AIME I]]<br />
* [[2005 AIME II]]<br />
* [[2006 AIME I]]<br />
* [[2006 AIME II]]<br />
* [[2007 AIME I]]<br />
* [[2007 AIME II]]<br />
* [[2008 AIME I]]<br />
* [[2008 AIME II]]<br />
* [[2009 AIME I]]<br />
* [[2009 AIME II]]<br />
* [[2010 AIME I]]<br />
* [[2010 AIME II]]<br />
* [[2011 AIME I]]<br />
* [[2011 AIME II]]<br />
* [[2012 AIME I]]<br />
* [[2012 AIME II]]<br />
* [[2013 AIME I]]<br />
* [[2013 AIME II]]<br />
* [[2014 AIME I]]<br />
* [[2014 AIME II]]<br />
* [[2015 AIME I]]<br />
* [[2015 AIME II]]<br />
* [[2016 AIME I]]<br />
* [[2016 AIME II]]<br />
* [[2017 AIME I]]<br />
* [[2017 AIME II]]<br />
<br />
== See also ==<br />
* [[Mathematics competitions]]<br />
* [[ARML]]<br />
* [[Mathematics summer programs]]<br />
* [[Mathematics scholarships]]<br />
<br />
<br />
<br />
[[Category:Mathematics competitions]]</div>GeronimoStiltonhttps://artofproblemsolving.com/wiki/index.php?title=2009_AIME_II_Problems&diff=898902009 AIME II Problems2018-01-20T15:41:13Z<p>GeronimoStilton: </p>
<hr />
<div>{{AIME Problems|year=2009|n=II}}<br />
<br />
== Problem 1 ==<br />
Before starting to paint, Bill had <math>130</math> ounces of blue paint, <math>164</math> ounces of red paint, and <math>188</math> ounces of white paint. Bill painted four equally sized stripes on a wall, making a blue stripe, a red stripe, a white stripe, and a pink stripe. Pink is a mixture of red and white, not necessarily in equal amounts. When Bill finished, he had equal amounts of blue, red, and white paint left. Find the total number of ounces of paint Bill had left.<br />
<br />
[[2009 AIME II Problems/Problem 1|Solution]]<br />
<br />
== Problem 2 ==<br />
Suppose that <math>a</math>, <math>b</math>, and <math>c</math> are positive real numbers such that <math>a^{\log_3 7} = 27</math>, <math>b^{\log_7 11} = 49</math>, and <math>c^{\log_{11}25} = \sqrt{11}</math>. Find<br />
<cmath> a^{(\log_3 7)^2} + b^{(\log_7 11)^2} + c^{(\log_{11} 25)^2}. </cmath><br />
<br />
[[2009 AIME II Problems/Problem 2|Solution]]<br />
<br />
== Problem 3 ==<br />
In rectangle <math>ABCD</math>, <math>AB=100</math>. Let <math>E</math> be the midpoint of <math>\overline{AD}</math>. Given that line <math>AC</math> and line <math>BE</math> are perpendicular, find the greatest integer less than <math>AD</math>.<br />
<br />
[[2009 AIME II Problems/Problem 3|Solution]]<br />
<br />
== Problem 4 ==<br />
A group of children held a grape-eating contest. When the contest was over, the winner had eaten <math>n</math> grapes, and the child in <math>k</math>-th place had eaten <math>n+2-2k</math> grapes. The total number of grapes eaten in the contest was <math>2009</math>. Find the smallest possible value of <math>n</math>.<br />
<br />
[[2009 AIME II Problems/Problem 4|Solution]]<br />
<br />
== Problem 5 ==<br />
Equilateral triangle <math>T</math> is inscribed in circle <math>A</math>, which has radius <math>10</math>. Circle <math>B</math> with radius <math>3</math> is internally tangent to circle <math>A</math> at one vertex of <math>T</math>. Circles <math>C</math> and <math>D</math>, both with radius <math>2</math>, are internally tangent to circle <math>A</math> at the other two vertices of <math>T</math>. Circles <math>B</math>, <math>C</math>, and <math>D</math> are all externally tangent to circle <math>E</math>, which has radius <math>\dfrac mn</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.<br />
<br />
<asy><br />
unitsize(3mm);<br />
defaultpen(linewidth(.8pt));<br />
dotfactor=4;<br />
<br />
pair A=(0,0), D=8*dir(330), C=8*dir(210), B=7*dir(90);<br />
pair Ep=(0,4-27/5);<br />
pair[] dotted={A,B,C,D,Ep};<br />
<br />
draw(Circle(A,10));<br />
draw(Circle(B,3));<br />
draw(Circle(C,2));<br />
draw(Circle(D,2));<br />
draw(Circle(Ep,27/5));<br />
<br />
dot(dotted);<br />
label("$E$",Ep,E);<br />
label("$A$",A,W);<br />
label("$B$",B,W);<br />
label("$C$",C,W);<br />
label("$D$",D,E);<br />
</asy><br />
<br />
[[2009 AIME II Problems/Problem 5|Solution]]<br />
<br />
== Problem 6 ==<br />
Let <math>m</math> be the number of five-element subsets that can be chosen from the set of the first <math>14</math> natural numbers so that at least two of the five numbers are consecutive. Find the remainder when <math>m</math> is divided by <math>1000</math>.<br />
<br />
[[2009 AIME II Problems/Problem 6|Solution]]<br />
<br />
== Problem 7 ==<br />
Define <math>n!!</math> to be <math>n(n-2)(n-4)\cdots 3\cdot 1</math> for <math>n</math> odd and <math>n(n-2)(n-4)\cdots 4\cdot 2</math> for <math>n</math> even. When <math>\sum_{i=1}^{2009} \frac{(2i-1)!!}{(2i)!!}</math> is expressed as a fraction in lowest terms, its denominator is <math>2^ab</math> with <math>b</math> odd. Find <math>\dfrac{ab}{10}</math>.<br />
<br />
[[2009 AIME II Problems/Problem 7|Solution]]<br />
<br />
== Problem 8 ==<br />
Dave rolls a fair six-sided die until a six appears for the first time. Independently, Linda rolls a fair six-sided die until a six appears for the first time. Let <math>m</math> and <math>n</math> be relatively prime positive integers such that <math>\dfrac mn</math> is the probability that the number of times Dave rolls his die is equal to or within one of the number of times Linda rolls her die. Find <math>m+n</math>.<br />
<br />
[[2009 AIME II Problems/Problem 8|Solution]]<br />
<br />
== Problem 9 ==<br />
Let <math>m</math> be the number of solutions in positive integers to the equation <math>4x+3y+2z=2009</math>, and let <math>n</math> be the number of solutions in positive integers to the equation <math>4x+3y+2z=2000</math>. Find the remainder when <math>m-n</math> is divided by <math>1000</math>.<br />
<br />
[[2009 AIME II Problems/Problem 9|Solution]]<br />
<br />
== Problem 10 ==<br />
Four lighthouses are located at points <math>A</math>, <math>B</math>, <math>C</math>, and <math>D</math>. The lighthouse at <math>A</math> is <math>5</math> kilometers from the lighthouse at <math>B</math>, the lighthouse at <math>B</math> is <math>12</math> kilometers from the lighthouse at <math>C</math>, and the lighthouse at <math>A</math> is <math>13</math> kilometers from the lighthouse at <math>C</math>. To an observer at <math>A</math>, the angle determined by the lights at <math>B</math> and <math>D</math> and the angle determined by the lights at <math>C</math> and <math>D</math> are equal. To an observer at <math>C</math>, the angle determined by the lights at <math>A</math> and <math>B</math> and the angle determined by the lights at <math>D</math> and <math>B</math> are equal. The number of kilometers from <math>A</math> to <math>D</math> is given by <math>\frac{p\sqrt{r}}{q}</math>, where <math>p</math>, <math>q</math>, and <math>r</math> are relatively prime positive integers, and <math>r</math> is not divisible by the square of any prime. Find <math>p+q+r</math>.<br />
<br />
[[2009 AIME II Problems/Problem 10|Solution]]<br />
<br />
== Problem 11 ==<br />
For certain pairs <math>(m,n)</math> of positive integers with <math>m\geq n</math> there are exactly <math>50</math> distinct positive integers <math>k</math> such that <math>|\log m - \log k| < \log n</math>. Find the sum of all possible values of the product <math>mn</math>.<br />
<br />
[[2009 AIME II Problems/Problem 11|Solution]]<br />
<br />
== Problem 12 ==<br />
From the set of integers <math>\{1,2,3,\dots,2009\}</math>, choose <math>k</math> pairs <math>\{a_i,b_i\}</math> with <math>a_i<b_i</math> so that no two pairs have a common element. Suppose that all the sums <math>a_i+b_i</math> are distinct and less than or equal to <math>2009</math>. Find the maximum possible value of <math>k</math>.<br />
<br />
[[2009 AIME II Problems/Problem 12|Solution]]<br />
<br />
== Problem 13 ==<br />
Let <math>A</math> and <math>B</math> be the endpoints of a semicircular arc of radius <math>2</math>. The arc is divided into seven congruent arcs by six equally spaced points <math>C_1,C_2,\dots,C_6</math>. All chords of the form <math>\overline{AC_i}</math> or <math>\overline{BC_i}</math> are drawn. Let <math>n</math> be the product of the lengths of these twelve chords. Find the remainder when <math>n</math> is divided by <math>1000</math>.<br />
<br />
[[2009 AIME II Problems/Problem 13|Solution]]<br />
<br />
== Problem 14 ==<br />
The sequence <math>(a_n)</math> satisfies <math>a_0=0</math> and <math>a_{n + 1} = \frac{8}{5}a_n + \frac{6}{5}\sqrt{4^n - a_n^2}</math> for <math>n \geq 0</math>. Find the greatest integer less than or equal to <math>a_{10}</math>.<br />
<br />
[[2009 AIME II Problems/Problem 14|Solution]]<br />
<br />
== Problem 15 ==<br />
Let <math>\overline{MN}</math> be a diameter of a circle with diameter <math>1</math>. Let <math>A</math> and <math>B</math> be points on one of the semicircular arcs determined by <math>\overline{MN}</math> such that <math>A</math> is the midpoint of the semicircle and <math>MB=\dfrac 35</math>. Point <math>C</math> lies on the other semicircular arc. Let <math>d</math> be the length of the line segment whose endpoints are the intersections of diameter <math>\overline{MN}</math> with the chords <math>\overline{AC}</math> and <math>\overline{BC}</math>. The largest possible value of <math>d</math> can be written in the form <math>r-s\sqrt t</math>, where <math>r</math>, <math>s</math>, and <math>t</math> are positive integers and <math>t</math> is not divisible by the square of any prime. Find <math>r+s+t</math>.<br />
<br />
[[2009 AIME II Problems/Problem 15|Solution]]<br />
<br />
== See also ==<br />
* [[American Invitational Mathematics Examination]]<br />
* [[AIME Problems and Solutions]]<br />
* [[Mathematics competition resources]]<br />
{{MAA Notice}}</div>GeronimoStiltonhttps://artofproblemsolving.com/wiki/index.php?title=2004_AMC_12B_Problems/Problem_23&diff=898852004 AMC 12B Problems/Problem 232018-01-20T01:10:43Z<p>GeronimoStilton: </p>
<hr />
<div>== Problem ==<br />
The [[polynomial]] <math>x^3 - 2004 x^2 + mx + n</math> has [[integer]] coefficients and three distinct positive zeros. Exactly one of these is an integer, and it is the sum of the other two. How many values of <math>n</math> are possible?<br />
<br />
<math>\mathrm{(A)}\ 250,\!000<br />
\qquad\mathrm{(B)}\ 250,\!250<br />
\qquad\mathrm{(C)}\ 250,\!500<br />
\qquad\mathrm{(D)}\ 250,\!750<br />
\qquad\mathrm{(E)}\ 251,\!000</math><br />
<br />
== Solution 1 ==<br />
Let the roots be <math>r,s,r + s</math>, and let <math>t = rs</math>. Then<br />
<br />
<center><math>(x - r)(x - s)(x - (r + s))</math> <math>= x^3 - (r + s + r + s) x^2 + (rs + r(r + s) + s(r + s))x - rs(r + s) = 0</math></center><br />
<br />
and by matching coefficients, <math>2(r + s) = 2004 \Longrightarrow r + s = 1002</math>. Then our polynomial looks like <br />
<cmath>x^3 - 2004x^2 + (t + 1002^2)x - 1002t = 0</cmath><br />
and we need the number of possible products <math>t = rs = r(1002 - r)</math>. <br />
<br />
Since <math>r > 0</math> and <math>t > 0</math>, it follows that <math>0 < t = r(1002-r) < 501^2 = 251001</math>, with the endpoints not achievable because the roots must be distinct. Because <math>r</math> cannot be an integer, there are <math>251000 - 500 = 250,\!500\ \mathrm{(C)}</math> possible values of <math>n = -1002t</math>.<br />
<br />
== Solution 2 ==<br />
<br />
Letting the roots be <math>r</math>, <math>s</math>, and <math>t</math>, where <math>t = r+s</math>, we see that by Vieta's Formula's, <math>2004 = r+s+t = t + t = 2t</math>, and so <math>t = 1002</math>. Therefore, <math>x-1002</math> is a factor of <math>x^3 - 2004x^2 + mx + n</math>. Letting <math>x = 0</math> gives that <math>1002 \mid n</math> because <math>x - 1002 \mid x^3 - 2004x^2 + mx + n</math>. Letting <math>n = 1002a</math> and noting that <math>x^3 - 2004x^2 + mx + n = (x-1002)(x^2 - bx + a)</math> for some <math>b</math>, we see that <math>b</math> is the sum of the roots of <math>x^2 - bx + a</math>, <math>r</math> and <math>s</math>, and so <math>b = 1002</math>. Now, we have that <math>x^2 - 1002x + a</math> has roots <math>r</math> and <math>s</math>, and we wish to find the number of possible values of <math>a</math>. By the quadratic formula, we see that <cmath>\frac{1002 \pm \sqrt{1002^2 - 4a}}{2} = 501 \pm \sqrt{501^2 - a}</cmath> are the two values of noninteger positive real numbers <math>r</math> and <math>s</math>, neither of which is equal to <math>1002</math>. This information gives us that <math>0 < 501^2 - a < 501^2</math>, and so since <math>501^2 - a</math> is evidently not a square, we have <math>501^2 - 1 - 500 = 251,001 - 500 - 1 = 250,\!500\ \mathrm{(C)}</math> possible values of <math>n = 1002a</math>. <br />
<br />
== See also ==<br />
{{AMC12 box|year=2004|ab=B|num-b=22|num-a=24}}<br />
<br />
[[Category:Intermediate Algebra Problems]]<br />
{{MAA Notice}}</div>GeronimoStiltonhttps://artofproblemsolving.com/wiki/index.php?title=2004_AMC_12B_Problems/Problem_23&diff=898842004 AMC 12B Problems/Problem 232018-01-20T01:10:16Z<p>GeronimoStilton: </p>
<hr />
<div>== Problem ==<br />
The [[polynomial]] <math>x^3 - 2004 x^2 + mx + n</math> has [[integer]] coefficients and three distinct positive zeros. Exactly one of these is an integer, and it is the sum of the other two. How many values of <math>n</math> are possible?<br />
<br />
<math>\mathrm{(A)}\ 250,\!000<br />
\qquad\mathrm{(B)}\ 250,\!250<br />
\qquad\mathrm{(C)}\ 250,\!500<br />
\qquad\mathrm{(D)}\ 250,\!750<br />
\qquad\mathrm{(E)}\ 251,\!000</math><br />
<br />
== Solution 1 ==<br />
Let the roots be <math>r,s,r + s</math>, and let <math>t = rs</math>. Then<br />
<br />
<center><math>(x - r)(x - s)(x - (r + s))</math> <math>= x^3 - (r + s + r + s) x^2 + (rs + r(r + s) + s(r + s))x - rs(r + s) = 0</math></center><br />
<br />
and by matching coefficients, <math>2(r + s) = 2004 \Longrightarrow r + s = 1002</math>. Then our polynomial looks like <br />
<cmath>x^3 - 2004x^2 + (t + 1002^2)x - 1002t = 0</cmath><br />
and we need the number of possible products <math>t = rs = r(1002 - r)</math>. <br />
<br />
Since <math>r > 0</math> and <math>t > 0</math>, it follows that <math>0 < t = r(1002-r) < 501^2 = 251001</math>, with the endpoints not achievable because the roots must be distinct. Because <math>r</math> cannot be an integer, there are <math>251000 - 500 = 250,\!500\ \mathrm{(C)}</math> possible values of <math>n = -1002t</math>.<br />
<br />
== Solution 2 ==<br />
<br />
Letting the roots be <math>r</math>, <math>s</math>, and <math>t</math>, where <math>t = r+s</math>, we see that by Vieta's Formula's, <math>2004 = r+s+t = t + t = 2t</math>, and so <math>t = 1002</math>. Therefore, <math>x-1002</math> is a factor of <math>x^3 - 2004x^2 + mx + n</math>. Letting <math>x = 0</math> gives that <math>1002 \mid n</math> because <math>x - 1002 \mid x^3 - 2004x^2 + mx + n</math>. Letting <math>n = 1002a</math> and noting that <math>x^3 - 2004x^2 + mx + n = (x-1002)(x^2 - bx + a)</math> for some <math>b</math>, we see that <math>b</math> is the sum of the roots of <math>x^2 - bx + a</math>, <math>r</math> and <math>s</math>, and so <math>b = 1002</math>. Now, we have that <math>x^2 - 1002x + a</math> has roots <math>r</math> and <math>s</math>, and we wish to find the number of possible values of <math>a</math>. By the quadratic formula, we see that <cmath>\frac{1002 \pm \sqrt{1002^2 - 4a}}{2} = 501 \pm \sqrt{501^2 - a}</cmath> are the two values of noninteger positive real numbers <math>r</math> and <math>s</math>, neither of which is equal to <math>1002</math>. This information gives us that <math>0 < 501^2 - a < 501^2</math>, and so since <math>501^2 - a</math> is evidently not a square, we have <math>501^2 - 1 - 500 = 251001 - 500 - 1 = 250\!500\ \mathrm{(C)}</math> possible values of <math>n = 1002a</math>. <br />
<br />
== See also ==<br />
{{AMC12 box|year=2004|ab=B|num-b=22|num-a=24}}<br />
<br />
[[Category:Intermediate Algebra Problems]]<br />
{{MAA Notice}}</div>GeronimoStiltonhttps://artofproblemsolving.com/wiki/index.php?title=AoPS_Wiki:FAQ&diff=89029AoPS Wiki:FAQ2017-12-21T18:54:35Z<p>GeronimoStilton: /* For my homework, there is suppose to be a green bar but it's orange, why? */</p>
<hr />
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:Otherwise, you can send an email to extensions@aops.com with your username, class name and ID (the number in the class page URL after https://artofproblemsolving.com/class/) and reason for extension. Someone should get back to you within a couple days.</div>GeronimoStiltonhttps://artofproblemsolving.com/wiki/index.php?title=2017_USAMO_Problems/Problem_1&diff=878122017 USAMO Problems/Problem 12017-10-15T01:08:25Z<p>GeronimoStilton: /* Solution 2 */</p>
<hr />
<div>== Problem ==<br />
<br />
Prove that there are infinitely many distinct pairs <math>(a,b)</math> of relatively prime positive integers <math>a>1</math> and <math>b>1</math> such that <math>a^b+b^a</math> is divisible by <math>a+b</math>.<br />
<br />
== Solution 1 ==<br />
Let <math>n=a+b</math>. Since <math>gcd(a,b)=1</math>, we know <math>gcd(a,n)=1</math>. We can rewrite the condition as <br />
<br />
<cmath>a^{n-a}+(n-a)^a \equiv 0 \mod{n}</cmath><br />
<cmath>a^{n-a}\equiv-(-a)^a \mod{n}</cmath><br />
Assume <math>a</math> is odd. Since we need to prove an infinite number of pairs exist, it suffices to show that infinitely many pairs with <math>a</math> odd exist.<br />
<br />
Then we have <br />
<cmath>a^{n-a}\equiv a^a \mod{n}</cmath><br />
<cmath>1 \equiv a^{2a-n} \mod{n}</cmath><br />
<br />
We know by Euler's theorem that <math>a^{\varphi(n)} \equiv 1 \mod{n}</math>, so if <math>2a-n=\varphi(n)</math> we will have the required condition.<br />
<br />
This means <math>a=\frac{n+\varphi(n)}{2}</math>. Let <math>n=2p</math> where <math>p</math> is a prime, <math>p\equiv 1\mod{4}</math>. Then <math>\varphi(n) = 2p*\left(1-\frac{1}{2}\right)\left(1-\frac{1}{p}\right) = p-1</math>, so <cmath>a = \frac{2p+p-1}{2} = \frac{3p-1}{2}</cmath><br />
Note the condition that <math>p\equiv 1\mod{4}</math> guarantees that <math>a</math> is odd, since <math>3p-1 \equiv 2\mod{4}</math><br />
<br />
This makes <math>b = \frac{p+1}{2}</math>. Now we need to show that <math>a</math> and <math>b</math> are relatively prime. We see that<br />
<cmath>gcd\left(\frac{3p-1}{2},\frac{p+1}2\right)=\frac{gcd(3p-1,p+1)}{2}</cmath><br />
<cmath>=\frac{gcd(p+1,4)}{2}=\frac22=1</cmath><br />
By the Euclidean Algorithm.<br />
<br />
Therefore, for all primes <math>p \equiv 1\mod{4}</math>, the pair <math>\left(\frac{3p-1}{2},\frac{p+1}{2}\right)</math> satisfies the criteria, so infinitely many such pairs exist.<br />
<br />
== Solution 2 ==<br />
<br />
Take <math>a=2n-1, b=2n+1, n\geq 2</math>. It is obvious (use the Euclidean Algorithm, if you like), that <math>\gcd(a,b)=1</math>, and that <math>a,b>1</math>.<br />
<br />
Note that<br />
<br />
<cmath>a^2 = 4n^2-4n+1 \equiv 1 (\bmod 4n)</cmath><br />
<br />
<cmath>b^2 = 4n^2+4n+1 \equiv 1 (\bmod 4n)</cmath><br />
<br />
So<br />
<br />
<cmath>a^b+b^a = a(a^2)^n+b(b^2)^{n-1} \equiv </cmath> <cmath>a\cdot 1^n + b\cdot 1^{n-1} \equiv a+b = 4n \equiv 0 (\bmod 4n)</cmath><br />
<br />
Since <math>a+b=4n</math>, all such pairs work, and we are done.</div>GeronimoStiltonhttps://artofproblemsolving.com/wiki/index.php?title=1986_AHSME_Problems/Problem_19&diff=877541986 AHSME Problems/Problem 192017-10-10T00:45:41Z<p>GeronimoStilton: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
A park is in the shape of a regular hexagon <math>2</math> km on a side. Starting at a corner, <br />
Alice walks along the perimeter of the park for a distance of <math>5</math> km. <br />
How many kilometers is she from her starting point?<br />
<br />
<math>\textbf{(A)}\ \sqrt{13}\qquad<br />
\textbf{(B)}\ \sqrt{14}\qquad<br />
\textbf{(C)}\ \sqrt{15}\qquad<br />
\textbf{(D)}\ \sqrt{16}\qquad<br />
\textbf{(E)}\ \sqrt{17} </math><br />
<br />
==Solution==<br />
<br />
We imagine this problem on a coordinate plane and let Alice's starting position be the origin. We see that she will travel along two edges and then go halfway along a third. Therefore, her new <math>x</math>-coordinate will be <math>1 + 2 + \frac{1}{2} = \frac{7}{2}</math> because she travels along a distance of <math>2 \cdot \frac{1}{2} = 1</math> km because of the side relationships of an equilateral triangle, then <math>2</math> km because the line is parallel to the <math>x</math>-axis, and the remaining distance is <math>\frac{1}{2}</math> km because she went halfway along and because of the logic for the first part of her route. For her <math>y</math>-coordinate, we can use similar logic to find that the coordinate is <math>\sqrt{3} + 0 - \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2}</math>. Therefore, her distance is <cmath>\sqrt{\left(\frac{7}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} = \sqrt{\frac{49}{4} + \frac{3}{4}} = \sqrt{\frac{52}{4}} = \sqrt{13},</cmath> giving an answer of <math>\boxed{A}</math>.<br />
<br />
== See also ==<br />
{{AHSME box|year=1986|num-b=18|num-a=20}} <br />
<br />
[[Category: Intermediate Geometry Problems]]<br />
{{MAA Notice}}</div>GeronimoStiltonhttps://artofproblemsolving.com/wiki/index.php?title=1986_AHSME_Problems/Problem_18&diff=877521986 AHSME Problems/Problem 182017-10-10T00:01:34Z<p>GeronimoStilton: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
A plane intersects a right circular cylinder of radius <math>1</math> forming an ellipse. <br />
If the major axis of the ellipse of <math>50\%</math> longer than the minor axis, the length of the major axis is<br />
<br />
<math>\textbf{(A)}\ 1\qquad<br />
\textbf{(B)}\ \frac{3}{2}\qquad<br />
\textbf{(C)}\ 2\qquad<br />
\textbf{(D)}\ \frac{9}{4}\qquad<br />
\textbf{(E)}\ 3 </math> <br />
<br />
==Solution==<br />
<br />
We note that we can draw the minor axis to see that because the minor axis is the minimum distance between two opposite points on the ellipse, we can draw a line through two opposite points of the cylinder, and so the minor axis is <math>2(1) = 2</math>. Therefore, our answer is <math>2(1.5) = 3</math>, and so our answer is <math>\boxed{E}</math>.<br />
<br />
== See also ==<br />
{{AHSME box|year=1986|num-b=17|num-a=19}} <br />
<br />
[[Category: Intermediate Geometry Problems]]<br />
{{MAA Notice}}</div>GeronimoStiltonhttps://artofproblemsolving.com/wiki/index.php?title=1986_AHSME_Problems/Problem_20&diff=877491986 AHSME Problems/Problem 202017-10-09T23:50:40Z<p>GeronimoStilton: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
<br />
Suppose <math>x</math> and <math>y</math> are inversely proportional and positive. If <math>x</math> increases by <math>p\%</math>, then <math>y</math> decreases by <br />
<br />
<math>\textbf{(A)}\ p\%\qquad<br />
\textbf{(B)}\ \frac{p}{1+p}\%\qquad<br />
\textbf{(C)}\ \frac{100}{p}\%\qquad<br />
\textbf{(D)}\ \frac{p}{100+p}\%\qquad<br />
\textbf{(E)}\ \frac{100p}{100+p}\% </math> <br />
<br />
==Solution==<br />
<br />
We see that <math>x</math> is multiplied by <math>\frac{100+p}{100}</math> when it is increased by <math>p</math>%. Therefore, <math>y</math> is multiplied by <math>\frac{100}{100+p}</math>, and so it is decreased by <math>\frac{p}{100+p}</math> times itself. Therefore, it is decreased by <math>\frac{100p}{100+p}</math>%, and so the answer is <math>\boxed{E}</math>.<br />
<br />
== See also ==<br />
{{AHSME box|year=1986|num-b=19|num-a=21}} <br />
<br />
[[Category: Intermediate Algebra Problems]]<br />
{{MAA Notice}}</div>GeronimoStiltonhttps://artofproblemsolving.com/wiki/index.php?title=1986_AHSME_Problems/Problem_20&diff=877481986 AHSME Problems/Problem 202017-10-09T23:50:14Z<p>GeronimoStilton: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
<br />
Suppose <math>x</math> and <math>y</math> are inversely proportional and positive. If <math>x</math> increases by <math>p\%</math>, then <math>y</math> decreases by <br />
<br />
<math>\textbf{(A)}\ p\%\qquad<br />
\textbf{(B)}\ \frac{p}{1+p}\%\qquad<br />
\textbf{(C)}\ \frac{100}{p}\%\qquad<br />
\textbf{(D)}\ \frac{p}{100+p}\%\qquad<br />
\textbf{(E)}\ \frac{100p}{100+p}\% </math> <br />
<br />
==Solution==<br />
<br />
We see that <math>x</math> is multiplied by <math>\frac{100+p}{100}</math> when it is increased by <math>p%</math>. Therefore, <math>y</math> is multiplied by <math>\frac{100}{100+p}</math>, and so it is decreased by <math>\frac{p}{100+p}</math> times itself. Therefore, it is decreased by <math>\frac{100p}{100+p}%</math>, and so the answer is <math>\boxed{E}</math>.<br />
<br />
== See also ==<br />
{{AHSME box|year=1986|num-b=19|num-a=21}} <br />
<br />
[[Category: Intermediate Algebra Problems]]<br />
{{MAA Notice}}</div>GeronimoStiltonhttps://artofproblemsolving.com/wiki/index.php?title=2007_AMC_12A_Problems/Problem_25&diff=876882007 AMC 12A Problems/Problem 252017-10-05T15:05:38Z<p>GeronimoStilton: /* Solution 1 */</p>
<hr />
<div>== Problem ==<br />
Call a set of integers ''spacy'' if it contains no more than one out of any three consecutive integers. How many [[subset]]s of <math>\{1,2,3,\ldots,12\},</math> including the [[empty set]], are spacy?<br />
<br />
<math>\mathrm{(A)}\ 121 \qquad \mathrm{(B)}\ 123 \qquad \mathrm{(C)}\ 125 \qquad \mathrm{(D)}\ 127 \qquad \mathrm{(E)}\ 129</math><br />
<br />
__TOC__<br />
<br />
== Solution ==<br />
=== Solution 1 ===<br />
Let <math>S_{n}</math> denote the number of spacy subsets of <math>\{ 1, 2, ... n \}</math>. We have <math>S_{0} = 1, S_{1} = 2, S_{2} = 3</math>. <br />
<br />
The spacy subsets of <math>S_{n + 1}</math> can be divided into two groups:<br />
*<math>A = </math> those not containing <math>n + 1</math>. Clearly <math>|A|=S_{n}</math>. <br />
*<math>B = </math> those containing <math>n + 1</math>. We have <math>|B|=S_{n - 2}</math>, since removing <math>n + 1</math> from any set in <math>B</math> produces a spacy set with all elements at most equal to <math>n - 2,</math> and each such spacy set can be constructed from exactly one spacy set in <math>B</math>.<br />
Hence,<br />
<br />
<div style="text-align:center;"><math>S_{n + 1} = S_{n} + S_{n - 2}</math></div><br />
<br />
From this [[recursion]], we find that<br />
<br />
{| class="wikitable" border="1px solid"<br />
|-<br />
| <math>S(0)</math> || <math>S(1)</math> || <math>S(2)</math> || <math>S(3)</math> || <math>S(4)</math> || <math>S(5)</math> || <math>S(6)</math> || <math>S(7)</math> || <math>S(8)</math> || <math>S(9)</math> || <math>S(10)</math> || <math>S(11)</math> || <math>S(12)</math> || <br />
|-<br />
| 1 || 2 || 3 || 4 || 6 || 9 || 13 || 19 || 28 || 41 || 60 || 88 || 129<br />
|}<br />
And so the answer is <math>E</math>, <math>129</math>.<br />
<br />
=== Solution 2 ===<br />
Since each of the elements of the subsets must be spaced at least two apart, a divider counting argument can be used.<br />
<br />
From the set <math>\{1,2,3,4,5,6,7,8,9,10,11,12\}</math> we choose at most four numbers. Let those numbers be represented by balls. Between each of the balls there are at least two dividers. So for example, o | | o | | o | | o | | represents <math>{1,4,7,10}</math>.<br />
<br />
For subsets of size <math>k</math> there must be <math>2(k - 1)</math> dividers between the balls, leaving <math>12 - k - 2(k - 1) = 12 - 3k + 2</math> dividers to be be placed in <math>k + 1</math> spots between the balls. The number of way this can be done is <math>\binom{(12 - 3k + 2) + (k + 1) - 1}k = \binom{12 - 2k + 2}k</math>.<br />
<br />
Therefore, the number of spacy subsets is <math>\binom 64 + \binom 83 + \binom{10}2 + \binom{12}1 + \binom{14}0 = 129</math>.<br />
=== Solution 3 ===<br />
A shifting argument is also possible, and is similar in spirit to Solution 2. Clearly we can have at most <math>4</math> elements. Given any arrangment, we subract <math>2i-2</math> from the <math>i-th</math> element in our subset, when the elements are arranged in increasing order. This creates a [[bijection]] with the number of size <math>k</math> subsets of the set of the first <math>14-2k</math> positive integers. For instance, the arrangment o | | o | | o | | | o | corresponds to the arrangment o o o | o |. Notice that there is no longer any restriction on consectutive numbers. Therefore, we can easily plug in the possible integers 0, 1, 2, 3, 4, 5 for <math>k</math>: <math>{14 \choose 0} + {12 \choose 1} + {10 \choose 2} + {8 \choose 3} + {6 \choose 4} = \boxed{129}</math><br />
<br />
In general, the number of subsets of a set with <math>n</math> element and with no <math>k</math> consecutive numbers is <math>\sum^{\lfloor{\frac{n}{k}}\rfloor}_{i=0}{{n-(k-1)(i-1) \choose i}}</math>.<br />
<br />
== See also ==<br />
{{AMC12 box|year=2007|ab=A|num-b=24|after=Last question}}<br />
<br />
[[Category:Intermediate Combinatorics Problems]]<br />
{{MAA Notice}}</div>GeronimoStiltonhttps://artofproblemsolving.com/wiki/index.php?title=2007_AMC_12A_Problems/Problem_25&diff=876872007 AMC 12A Problems/Problem 252017-10-05T15:03:16Z<p>GeronimoStilton: /* Solution 3 */</p>
<hr />
<div>== Problem ==<br />
Call a set of integers ''spacy'' if it contains no more than one out of any three consecutive integers. How many [[subset]]s of <math>\{1,2,3,\ldots,12\},</math> including the [[empty set]], are spacy?<br />
<br />
<math>\mathrm{(A)}\ 121 \qquad \mathrm{(B)}\ 123 \qquad \mathrm{(C)}\ 125 \qquad \mathrm{(D)}\ 127 \qquad \mathrm{(E)}\ 129</math><br />
<br />
__TOC__<br />
<br />
== Solution ==<br />
=== Solution 1 ===<br />
Let <math>S_{n}</math> denote the number of spacy subsets of <math>\{ 1, 2, ... n \}</math>. We have <math>S_{0} = 1, S_{1} = 2, S_{2} = 3</math>. <br />
<br />
The spacy subsets of <math>S_{n + 1}</math> can be divided into two groups:<br />
*<math>A = </math> those not containing <math>n + 1</math>. Clearly <math>|A|=S_{n}</math>. <br />
*<math>B = </math> those containing <math>n + 1</math>. We have <math>|B|=S_{n - 2}</math>, since removing <math>n + 1</math> from any set in <math>B</math> produces a spacy set with all elements at most equal to <math>n - 2,</math> and each such spacy set can be constructed from exactly one spacy set in <math>B</math>.<br />
Hence,<br />
<br />
<div style="text-align:center;"><math>S_{n + 1} = S_{n} + S_{n - 2}</math></div><br />
<br />
From this [[recursion]], we find that<br />
<br />
{| class="wikitable" border="1px solid"<br />
|-<br />
| <math>S(0)</math> || <math>S(1)</math> || <math>S(2)</math> || <math>S(3)</math> || <math>S(4)</math> || <math>S(5)</math> || <math>S(6)</math> || <math>S(7)</math> || <math>S(8)</math> || <math>S(9)</math> || <math>S(10)</math> || <math>S(11)</math> || <math>S(12)</math> || <br />
|-<br />
| 1 || 2 || 3 || 4 || 6 || 9 || 13 || 19 || 28 || 41 || 60 || 88 || 129<br />
|}<br />
<br />
=== Solution 2 ===<br />
Since each of the elements of the subsets must be spaced at least two apart, a divider counting argument can be used.<br />
<br />
From the set <math>\{1,2,3,4,5,6,7,8,9,10,11,12\}</math> we choose at most four numbers. Let those numbers be represented by balls. Between each of the balls there are at least two dividers. So for example, o | | o | | o | | o | | represents <math>{1,4,7,10}</math>.<br />
<br />
For subsets of size <math>k</math> there must be <math>2(k - 1)</math> dividers between the balls, leaving <math>12 - k - 2(k - 1) = 12 - 3k + 2</math> dividers to be be placed in <math>k + 1</math> spots between the balls. The number of way this can be done is <math>\binom{(12 - 3k + 2) + (k + 1) - 1}k = \binom{12 - 2k + 2}k</math>.<br />
<br />
Therefore, the number of spacy subsets is <math>\binom 64 + \binom 83 + \binom{10}2 + \binom{12}1 + \binom{14}0 = 129</math>.<br />
=== Solution 3 ===<br />
A shifting argument is also possible, and is similar in spirit to Solution 2. Clearly we can have at most <math>4</math> elements. Given any arrangment, we subract <math>2i-2</math> from the <math>i-th</math> element in our subset, when the elements are arranged in increasing order. This creates a [[bijection]] with the number of size <math>k</math> subsets of the set of the first <math>14-2k</math> positive integers. For instance, the arrangment o | | o | | o | | | o | corresponds to the arrangment o o o | o |. Notice that there is no longer any restriction on consectutive numbers. Therefore, we can easily plug in the possible integers 0, 1, 2, 3, 4, 5 for <math>k</math>: <math>{14 \choose 0} + {12 \choose 1} + {10 \choose 2} + {8 \choose 3} + {6 \choose 4} = \boxed{129}</math><br />
<br />
In general, the number of subsets of a set with <math>n</math> element and with no <math>k</math> consecutive numbers is <math>\sum^{\lfloor{\frac{n}{k}}\rfloor}_{i=0}{{n-(k-1)(i-1) \choose i}}</math>.<br />
<br />
== See also ==<br />
{{AMC12 box|year=2007|ab=A|num-b=24|after=Last question}}<br />
<br />
[[Category:Intermediate Combinatorics Problems]]<br />
{{MAA Notice}}</div>GeronimoStiltonhttps://artofproblemsolving.com/wiki/index.php?title=1985_AHSME_Problems/Problem_27&diff=876811985 AHSME Problems/Problem 272017-10-03T16:44:56Z<p>GeronimoStilton: /* Problem */</p>
<hr />
<div>==Problem==<br />
Consider a sequence <math> x_1, x_2, x_3, \cdots </math> defined by<br />
<br />
<math> x_1=\sqrt[3]{3} </math><br />
<br />
<math> x_2=\sqrt[3]{3}^{\sqrt[3]{3}} </math><br />
<br />
and in general<br />
<br />
<math> x_n=(x_{n-1})^{\sqrt[3]{3}} </math> for <math> n>1 </math>.<br />
<br />
What is the smallest value of <math> n </math> for which <math> x_n </math> is an [[integer]]?<br />
<br />
<math> \mathrm{(A)\ } 2 \qquad \mathrm{(B) \ }3 \qquad \mathrm{(C) \ } 4 \qquad \mathrm{(D) \ } 9 \qquad \mathrm{(E) \ }27 </math><br />
<br />
==Solution==<br />
First, we will use induction to prove that <math> x_n=\sqrt[3]{3}^{\left(\sqrt[3]{3}^{n-1}\right)} </math><br />
<br />
We see that <math> x_1=\sqrt[3]{3}=\sqrt[3]{3}^{\left(\sqrt[3]{3}^0\right)} </math>. This is our base case.<br />
<br />
Now, we have <math> x_n=(x_{n-1})^{\sqrt[3]{3}}=\left(\sqrt[3]{3}^{\left(\sqrt[3]{3}^{n-2}\right)}\right)^{\sqrt[3]{3}}=\sqrt[3]{3}^{\left(\sqrt[3]{3}\cdot\sqrt[3]{3}^{n-2}\right)}=\sqrt[3]{3}^{\left(\sqrt[3]{3}^{n-1}\right)} </math>. Thus the induction is complete.<br />
<br />
We now get rid of the cubed roots by introducing fractions into the exponents. <br />
<br />
<math> x_n=\sqrt[3]{3}^{\left(\sqrt[3]{3}^{n-1}\right)}=\sqrt[3]{3}^{\left(3^{\left(\frac{n-1}{3}\right)}\right)}=3^{\left(\frac{1}{3}\cdot3^{\left(\frac{n-1}{3}\right)}\right)}=3^{\left(3^{\left(\frac{n-4}{3}\right)}\right)} </math>.<br />
<br />
Notice that since <math> 3 </math> isn't a perfect power, <math> x_n </math> is integral if and only if the exponent, <math> 3^{\left(\frac{n-4}{3}\right)} </math>, is integral. By the same logic, this is integeral if and only if <math> \frac{n-4}{3} </math> is integral. We can now clearly see that the smallest positive value of <math> n </math> for which this is integral is <math> 4, \boxed{\text{C}} </math>.<br />
<br />
==See Also==<br />
{{AHSME box|year=1985|num-b=26|num-a=28}}<br />
{{MAA Notice}}</div>GeronimoStiltonhttps://artofproblemsolving.com/wiki/index.php?title=1985_AHSME_Problems/Problem_25&diff=876801985 AHSME Problems/Problem 252017-10-03T15:24:05Z<p>GeronimoStilton: /* Solution */</p>
<hr />
<div>==Problem==<br />
The [[volume]] of a certain rectangular solid is <math> 8 \text{cm}^3 </math>, its total [[surface area]] is <math> 32 \text{cm}^2 </math>, and its three dimensions are in [[geometric progression]]. The sums of the lengths in cm of all the edges of this solid is<br />
<br />
<math> \mathrm{(A)\ } 28 \qquad \mathrm{(B) \ }32 \qquad \mathrm{(C) \ } 36 \qquad \mathrm{(D) \ } 40 \qquad \mathrm{(E) \ }44 </math><br />
<br />
==Solution 1==<br />
Let the side lengths be <math> \frac{b}{r}, b, br </math>. Thus, the volume is <math> \left(\frac{b}{r}\right)(b)(br)=b^3=8 </math>, so <math> b=2 </math> and the side lengths can be written as <math> \frac{2}{r}, 2, 2r </math>.<br />
<br />
The surface area is <math> 2\left(\frac{2}{r}\right)(2)+2\left(\frac{2}{r}\right)(2r)+2(2)(2r)=32 </math><br />
<br />
<math> \frac{8}{r}+8+8r=32 </math><br />
<br />
<math> r+\frac{1}{r}=3 </math><br />
<br />
<math> r^2-3r+1=0 </math><br />
<br />
<math> r=\frac{3\pm\sqrt{5}}{2} </math><br />
<br />
Both values of <math> r </math> give the same side length, the only difference is that one makes them count up and one makes them count down. We pick <math> r=\frac{3+\sqrt{5}}{2} </math>. (The solution proceeds the same had we picked <math> r=\frac{3-\sqrt{5}}{2} </math>). Thus, the side lengths are <br />
<br />
<math> \frac{2}{\frac{3+\sqrt{5}}{2}}, 2, 2\left(\frac{3+\sqrt{5}}{2}\right) </math><br />
<br />
<math> \frac{4}{3+\sqrt{5}}, 2, 3+\sqrt{5} </math><br />
<br />
<math> 3-\sqrt{5}, 2, 3+\sqrt{5} </math><br />
<br />
We have the sum of the distinct side lengths is <math> 3-\sqrt{5}+2+3+\sqrt{5}=8 </math>, and since each side length repeats <math> 4 </math> times, the total sum is <math> 4(8)=32, \boxed{\text{B}} </math>.<br />
<br />
===Solution 2===<br />
<br />
We see let the side lengths be <math>b</math>, <math>rb</math>, and <math>r^2b</math> because they form an arithmetic progression. Therefore, we have that <math>8 = r^3b^3 = (rb)^3</math>. Therefore, <math>rb = 2</math>. The next piece of information tells us that <cmath>2rb^2 + 2r^2b^2 + 2r^3b^2 = 32.</cmath> Dividing both sides by <math>2rb = 4</math>, which is clearly nonzero, as it is a side length, we see that <math>b + rb + r^2b = 8</math>. For finding the sum of the side lengths, we will need to obtain <math>4b + 4rb + 4r^2b</math> though, and so the answer is <math>4 \times 8 = 32</math>, <math>\boxed{B}</math>.<br />
<br />
==See Also==<br />
{{AHSME box|year=1985|num-b=24|num-a=26}}<br />
{{MAA Notice}}</div>GeronimoStiltonhttps://artofproblemsolving.com/wiki/index.php?title=1985_AHSME_Problems/Problem_16&diff=876791985 AHSME Problems/Problem 162017-10-03T14:52:16Z<p>GeronimoStilton: /* Solution 2 */</p>
<hr />
<div>==Problem==<br />
If <math> A=20^\circ </math> and <math> B=25^\circ </math>, then the value of <math> (1+\tan A)(1+\tan B) </math> is<br />
<br />
<math> \mathrm{(A)\ } \sqrt{3} \qquad \mathrm{(B) \ }2 \qquad \mathrm{(C) \ } 1+\sqrt{2} \qquad \mathrm{(D) \ } 2(\tan A+\tan B) \qquad \mathrm{(E) \ }\text{none of these} </math><br />
<br />
==Solution==<br />
===Solution 1===<br />
First, let's leave everything in variables and see if we can simplify <math> (1+\tan A)(1+\tan B) </math>.<br />
<br />
<br />
We can write everything in terms of sine and cosine to get <math> \left(\frac{\cos A}{\cos A}+\frac{\sin A}{\cos A}\right)\left(\frac{\cos B}{\cos B}+\frac{\sin B}{\cos B}\right)=\frac{(\sin A+\cos A)(\sin B+\cos B)}{\cos A\cos B} </math>.<br />
<br />
<br />
<br />
We can multiply out the numerator to get <math> \frac{\sin A\sin B+\cos A\cos B+\sin A\cos B+\sin B\cos A}{\cos A\cos B} </math>.<br />
<br />
<br />
It may seem at first that we've made everything more complicated, however, we can recognize the numerator from the angle sum formulas:<br />
<br />
<br />
<math> \cos(A-B)=\sin A\sin B+\cos A\cos B </math><br />
<br />
<math> \sin(A+B)=\sin A\cos B+\sin B\cos A </math><br />
<br />
<br />
Therefore, our fraction is equal to <math> \frac{\cos(A-B)+\sin(A+B)}{\cos A\cos B} </math>.<br />
<br />
<br />
We can also use the product-to-sum formula<br />
<br />
<math> \cos A\cos B=\frac{1}{2}(\cos(A-B)+\cos(A+B)) </math> to simplify the denominator:<br />
<br />
<br />
<math> \frac{\cos(A-B)+\sin(A+B)}{\frac{1}{2}(\cos(A-B)+\cos(A+B))} </math>.<br />
<br />
<br />
But now we seem stuck. However, we can note that since <math> A+B=45^\circ </math>, we have <math> \sin(A+B)=\cos(A+B) </math>, so we get<br />
<br />
<br />
<math> \frac{\cos(A-B)+\sin(A+B)}{\frac{1}{2}(\cos(A-B)+\sin(A+B))} </math><br />
<br />
<br />
<math> \frac{1}{\frac{1}{2}} </math><br />
<br />
<math> 2, \boxed{\text{B}} </math><br />
<br />
Note that we only used the fact that <math> \sin(A+B)=\cos(A+B) </math>, so we have in fact not just shown that <math> (1+\tan A)(1+\tan B)=2 </math> for <math> A=20^\circ </math> and <math> B=25^\circ </math>, but for all <math> A, B </math> such that <math> A+B=45^\circ+n180^\circ </math>, for integer <math> n </math>.<br />
<br />
<br />
===Solution 2===<br />
<br />
We can see that <math>25^o+20^o=45^o</math>. We also know that <math>\tan 45=1</math>. First, let us expand <math>(1+\tan A)(1+\tan B)</math>.<br />
<br />
We get <math>1+\tan A+\tan B+\tan A\tan B</math>. <br />
<br />
Now, let us look at <math>\tan45=\tan(20+25)</math>.<br />
<br />
By the <math>\tan</math> sum formula, we know that <math>\tan45=\dfrac{\text{tan A}+\text{tan B}}{1- \text{tan A} \text{tan B}}</math><br />
<br />
Then, since <math>\tan 45=1</math>, we can see that <math>\tan A+\tan B=1-\tan A\tan B</math><br />
<br />
Then <math>1=\tan A+\tan B+\tan A\tan B</math><br />
<br />
Thus, the sum become <math>1+1=2</math> and the answer is <math>\fbox{\text{(B)}}</math><br />
<br />
===Solution 3===<br />
<br />
Let's write out the expression in terms of sine and cosine, so that we may see that it is equal to <cmath>\left(1+\frac{\sin 20^\circ}{\cos 20^\circ}\right)\left(1+\frac{\sin 25^\circ}{\cos 25^\circ}\right) = \left(\frac{\cos 20^\circ}{\cos 20^\circ}+\frac{\sin 20^\circ}{\cos 20^\circ}\right)\left(\frac{\cos 25^\circ}{\cos 25^\circ}+\frac{\sin 25^\circ}{\cos 25^\circ}\right) = </cmath> <cmath>\frac{\cos 20^\circ \cos 25^\circ + \cos 20^\circ \sin 25^\circ + \sin 20^\circ \cos 25^\circ + \sin 20^\circ \sin 25^\circ}{\cos 20^\circ \cos 25^\circ}.</cmath> Clearly, that is equal to <cmath>1 + \frac{\cos 20^\circ \sin 25^\circ + \sin 20^\circ \cos 25^\circ + \sin 20^\circ \sin 25^\circ}{\cos 20^\circ \cos 25^\circ}.</cmath> Now, we note that <cmath>\cos 20^\circ \sin 25^\circ + \sin 20^\circ \cos 25^\circ</cmath> is equal to <math>\sin 45^\circ</math>. Now, we would like to get <math>\sin 20^\circ \sin 25^\circ</math> in the denominator. What springs to mind is the fact that <cmath>\cos 20^\circ \cos 25^\circ - \sin 20^\circ \sin 25^\circ = \cos 45^\circ.</cmath> Therefore, we can express the desired value as <cmath>1 + \frac{\sin 45^\circ + \sin 20^\circ \sin 25^\circ}{\cos 45^\circ + \sin 20^\circ \sin 25^\circ}.</cmath> Because <math>\cos 45^\circ = \sin 45^\circ</math>, we see that the fractional part is <math>1</math>, and so the sum is <math>1 + 1 = 2</math>, which brings us to the answer <math>\boxed{B}</math>.<br />
<br />
==See Also==<br />
{{AHSME box|year=1985|num-b=15|num-a=17}}<br />
{{MAA Notice}}</div>GeronimoStiltonhttps://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12B_Problems/Problem_15&diff=838622017 AMC 12B Problems/Problem 152017-02-16T23:07:27Z<p>GeronimoStilton: /* Solution */</p>
<hr />
<div>==Problem 15==<br />
Let <math>ABC</math> be an equilateral triangle. Extend side <math>\overline{AB}</math> beyond <math>B</math> to a point <math>B'</math> so that <math>BB'=3AB</math>. Similarly, extend side <math>\overline{BC}</math> beyond <math>C</math> to a point <math>C'</math> so that <math>CC'=3BC</math>, and extend side <math>\overline{CA}</math> beyond <math>A</math> to a point <math>A'</math> so that <math>AA'=3CA</math>. What is the ratio of the area of <math>\triangle A'B'C'</math> to the area of <math>\triangle ABC</math>?<br />
<br />
<math>\textbf{(A)}\ 9:1\qquad\textbf{(B)}\ 16:1\qquad\textbf{(C)}\ 25:1\qquad\textbf{(D)}\ 36:1\qquad\textbf{(E)}\ 37:1</math><br />
<br />
<br />
==Solution==<br />
Solution by HydroQuantum<br />
<br />
Let <math>AB=BC=CA=x</math>.<br />
<br />
<br />
Recall The Law of Cosines. Letting <math>A'B'=B'C'=C'A'=y</math>, <cmath>y^2=(3x)^2+(x+3x)^2-2(3x)(x+3x)(cos120) = </cmath> <cmath>(3x)^2+(4x)^2-2(3x)(4x)(cos120)=9x^2+16x^2-24cos120=25x^2+12x^2=37x^2.</cmath> Since both <math>\triangle ABC</math> and <math>\triangle A'B'C'</math> are both equilateral triangles, they must be similar due to <math>AA</math> similarity. This means that <math>\frac{A'B'}{AB}</math> <math>=</math> <math>\frac{B'C'}{BC}</math> <math>=</math> <math>\frac{C'A'}{CA}</math> <math>=</math> <math>\frac{[\triangle A'B'C']}{[\triangle ABC]}</math> <math>=</math> <math>\frac{37}{1}</math>.<br />
<br />
<br />
Therefore, our answer is <math>\boxed{\textbf{(E) }37:1}</math>.<br />
<br />
<br />
[[2017 AMC 10B Problems/Problem 19|2017 AMC 10B Problem 19 Solution]]</div>GeronimoStiltonhttps://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12B_Problems/Problem_5&diff=838592017 AMC 12B Problems/Problem 52017-02-16T23:04:13Z<p>GeronimoStilton: /* Problem 5 */</p>
<hr />
<div>==Problem 5==<br />
<br />
The data set <math>[6, 19, 33, 33, 39, 41, 41, 43, 51, 57]</math> has median <math>Q_2 = 40</math>, first quartile <math>Q_1 = 33</math>, and third quartile <math>Q_3 = 43</math>. An outlier in a data set is a value that is more than <math>1.5</math> times the interquartile range below the first quartle (<math>Q_1</math>) or more than <math>1.5</math> times the interquartile range above the third quartile (<math>Q_3</math>), where the interquartile range is defined as <math>Q_3 - Q_1</math>. How many outliers does this data set have?<br />
<br />
<math>\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4</math><br />
<br />
==Solution==<br />
<br />
==See Also==<br />
{{AMC12 box|year=2017|ab=B|num-b=4|num-a=6}}<br />
{{MAA Notice}}</div>GeronimoStiltonhttps://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12B_Problems&diff=838572017 AMC 12B Problems2017-02-16T22:52:42Z<p>GeronimoStilton: /* Problem 7 */</p>
<hr />
<div>WORK IN PROGRESS<br />
<br />
{{AMC12 Problems|year=2017|ab=B}}<br />
<br />
==Problem 1==<br />
<br />
Kymbrea's comic book collection currently has <math>30</math> comic books in it, and she is adding to her collection at the rate of <math>2</math> comic books per month. LaShawn's collection currently has <math>10</math> comic books in it, and he is adding to his collection at the rate of <math>6</math> comic books per month. After how many months will LaShawn's collection have twice as many comic books as Kymbrea's?<br />
<br />
<math>\textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 20\qquad\textbf{(E)}\ 25</math><br />
<br />
==Problem 2==<br />
<br />
Real numbers <math>x</math>, <math>y</math>, and <math>z</math> satify the inequalities<br />
<math>0<x<1</math>, <math>-1<y<0</math>, and <math>1<z<2</math>.<br />
Which of the following numbers is necessarily positive?<br />
<br />
<math>\textbf{(A)}\ y+x^2\qquad\textbf{(B)}\ y+xz\qquad\textbf{(C)}\ y+y^2\qquad\textbf{(D)}\ y+2y^2\qquad\textbf{(E)}\ y+z</math><br />
<br />
[[2017 AMC 10B Problems/Problem 3|Solution]]<br />
<br />
==Problem 3==<br />
<br />
Supposed that <math>x</math> and <math>y</math> are nonzero real numbers such that <math>\frac{3x+y}{x-3y}=-2</math>. What is the value of <math>\frac{x+3y}{3x-y}</math>?<br />
<br />
<math>\textbf{(A)}\ -3\qquad\textbf{(B)}\ -1\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ 3</math><br />
<br />
[[2017 AMC 10B Problems/Problem 4|Solution]]<br />
<br />
==Problem 4==<br />
Samia set off on her bicycle to visit her friend, traveling at an average speed of <math>17</math> kilometers per hour. When she had gone half the distance to her friend's house, a tire went flat, and she walked the rest of the way at <math>5</math> kilometers per hour. In all it took her <math>44</math> minutes to reach her friend's house. In kilometers rounded to the nearest tenth, how far did Samia walk?<br />
<br />
<math>\textbf{(A)}\ 2.0\qquad\textbf{(B)}\ 2.2\qquad\textbf{(C)}\ 2.8\qquad\textbf{(D)}\ 3.4\qquad\textbf{(E)}\ 4.4</math><br />
<br />
[[2017 AMC 12B Problems/Problem 4|Solution]]<br />
<br />
==Problem 5==<br />
<br />
==Problem 6==<br />
<br />
==Problem 7==<br />
The functions <math>\sin(x)</math> and <math>\cos(x)</math> are periodic with least period <math>2\pi</math>. What is the least period of the function <math>\cos(\sin(x))</math>?<br />
<br />
<math>\textbf{(A)}\ \frac{\sqrt{3}-1}{2}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ \frac{\sqrt{5}-1}{2} \qquad\textbf{(D)}\ \frac{\sqrt{2}}{2} \qquad\textbf{(E)}\ \frac{\sqrt{6}-1}{2}</math><br />
<br />
[[2017 AMC 12B Problems/Problem 8|Solution]]<br />
<br />
We simply note that <math>\cos(\sin(x)) = \cos(-\sin(x)) = \cos(\sin(x + \pi))</math>, and so the period is <math>\pi</math>.<br />
<br />
==Problem 8==<br />
<br />
==Problem 9==<br />
<br />
==Problem 10==<br />
At Typico High School, <math>60\%</math> of the students like dancing, and the rest dislike it. Of those who like dancing, <math>80\%</math> say that they like it, and the rest say that they dislike it. Of those who dislike dancing, <math>90\%</math> say that they dislike it, and the rest say that they like it. What fraction of students who say they dislike dancing actually like it?<br />
<br />
<math>\textbf{(A)}\ 10\%\qquad\textbf{(B)}\ 12\%\qquad\textbf{(C)}\ 20\%\qquad\textbf{(D)}\ 25\%\qquad\textbf{(E)}\ 33\frac{1}{3}\%</math><br />
<br />
[[2017 AMC 12B Problems/Problem 10|Solution]]<br />
<br />
==Problem 11==<br />
Call a positive integer <math>monotonous</math> if it is a one-digit number or its digits, when read from left to right, form either a strictly increasing or a strictly decreasing sequence. For example, <math>3</math>, <math>23578</math>, and <math>987620</math> are monotonous, but <math>88</math>, <math>7434</math>, and <math>23557</math> are not. How many monotonous positive integers are there?<br />
<br />
<math>\textbf{(A)}\ 1024\qquad\textbf{(B)}\ 1524\qquad\textbf{(C)}\ 1533\qquad\textbf{(D)}\ 1536\qquad\textbf{(E)}\ 2048</math><br />
<br />
[[2017 AMC 12B Problems/Problem 11|Solution]]<br />
<br />
==Problem 12==<br />
<br />
==Problem 13==<br />
<br />
==Problem 14==<br />
<br />
==Problem 15==<br />
Let <math>ABC</math> be an equilateral triangle. Extend side <math>\overline{AB}</math> beyond <math>B</math> to a point <math>B'</math> so that <math>BB'=3AB</math>. Similarly, extend side <math>\overline{BC}</math> beyond <math>C</math> to a point <math>C'</math> so that <math>CC'=3BC</math>, and extend side <math>\overline{CA}</math> beyond <math>A</math> to a point <math>A'</math> so that <math>AA'=3CA</math>. What is the ratio of the area of <math>\triangle A'B'C'</math> to the area of <math>\triangle ABC</math>?<br />
<br />
<math>\textbf{(A)}\ 9:1\qquad\textbf{(B)}\ 16:1\qquad\textbf{(C)}\ 25:1\qquad\textbf{(D)}\ 36:1\qquad\textbf{(E)}\ 37:1</math><br />
<br />
[[2017 AMC 12B Problems/Problem 15|Solution]]<br />
<br />
==Problem 16==<br />
The number <math>21!=51,090,942,171,709,440,000</math> has over <math>60,000</math> positive integer divisors. One of them is chosen at random. What is the probability that it is odd?<br />
<br />
<math>\textbf{(A)}\ \frac{1}{21} \qquad \textbf{(B)}\ \frac{1}{19} \qquad \textbf{(C)}\ \frac{1}{18} \qquad \textbf{(D)}\ \frac{1}{2} \qquad \textbf{(E)}\ \frac{11}{21}</math><br />
<br />
[[2017 AMC 12B Problems/Problem 16|Solution]]<br />
<br />
==Problem 17==<br />
<br />
==Problem 18==<br />
The diameter <math>AB</math> of a circle of radius <math>2</math> is extended to a point <math>D</math> outside the circle so that <math>BD=3</math>. Point <math>E</math> is chosen so that <math>ED=5</math> and line <math>ED</math> is perpendicular to line <math>AD</math>. Segment <math>AE</math> intersects the circle at a point <math>C</math> between <math>A</math> and <math>E</math>. What is the area of <math>\triangle <br />
ABC</math>?<br />
<br />
<math>\textbf{(A)}\ \frac{120}{37}\qquad\textbf{(B)}\ \frac{140}{39}\qquad\textbf{(C)}\ \frac{145}{39}\qquad\textbf{(D)}\ \frac{140}{37}\qquad\textbf{(E)}\ \frac{120}{31}</math><br />
<br />
[[2017 AMC 12B Problems/Problem 18|Solution]]<br />
<br />
==Problem 19==<br />
Let <math>N=123456789101112\dots4344</math> be the <math>79</math>-digit number that is formed by writing the integers from <math>1</math> to <math>44</math> in order, one after the other. What is the remainder when <math>N</math> is divided by <math>45</math>?<br />
<br />
<math>\textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 44</math><br />
<br />
[[2017 AMC 12B Problems/Problem 19|Solution]]<br />
<br />
==Problem 20==<br />
<br />
==Problem 21==<br />
<br />
==Problem 22==<br />
<br />
==Problem 23==<br />
<br />
==Problem 24==<br />
<br />
==Problem 25==<br />
<br />
==See also==<br />
{{AMC12 box|year=2017|ab=B|before=[[2017 AMC 12A Problems]]|after=[[2018 AMC 12A Problems]]}}<br />
{{MAA Notice}}</div>GeronimoStiltonhttps://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12B_Problems/Problem_1&diff=838532017 AMC 12B Problems/Problem 12017-02-16T22:46:02Z<p>GeronimoStilton: /* Solution */</p>
<hr />
<div>==Problem 1==<br />
<br />
Kymbrea's comic book collection currently has <math>30</math> comic books in it, and she is adding to her collection at the rate of <math>2</math> comic books per month. LaShawn's collection currently has <math>10</math> comic books in it, and he is adding to his collection at the rate of <math>6</math> comic books per month. After how many months will LaShawn's collection have twice as many comic books as Kymbrea's?<br />
<br />
<math>\textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 20\qquad\textbf{(E)}\ 25</math><br />
<br />
==Solution==<br />
<br />
Kymbrea has <math>30</math> comic books initially and every month, she adds two. This can be represented as <math>30 + 2x</math> where x is the number of months elapsed. LaShawn's collection, similarly, is <math>10 + 6x</math>. To find when LaShawn will have twice the number of comic books as Kymbrea, we solve for x with the equation <math>2(2x + 30) = 6x + 10</math> and get <math>x = 25</math> <math>\boxed{\textbf{E}}</math>.<br />
<br />
==See Also==<br />
{{AMC12 box|year=2017|ab=B|before=First Problem|num-a=2}}<br />
{{MAA Notice}}</div>GeronimoStiltonhttps://artofproblemsolving.com/wiki/index.php?title=Graph_(graph_theory)&diff=82496Graph (graph theory)2017-01-22T15:02:15Z<p>GeronimoStilton: /* Complementrary Graphs */</p>
<hr />
<div>In [[graph theory]], a '''graph''' is a (usually [[finite]]) [[empty set | nonempty]] [[set]] of [[vertex|vertices]] that are joined by a number (possibly zero) of [[edge]]s. Graphs are frequently represented graphically, with the vertices as points and the edges as smooth curves joining pairs of vertices. <br />
<br />
{{image}}<br />
<br />
Formally, a graph <math>G</math> is a pair, <math>G = (V, E)</math>, of a set <math>V</math> of vertices together with a [[class]] of subsets <math>E</math> made up of pairs of elements from <math>V</math>. Note that this definition describes ''simple, loopless'' graphs: there is at most one edge joining two vertices, no edge may join a vertex to itself, and the edges are not directed. For graphs with multiple edges, see [[multigraph]]. If the edges are directed, then <math>E</math> may be defined using ordered pairs from the [[product set]] <math>V \times V</math>.<br />
<br />
==Important Related Definitions==<br />
* If <math>v \in V</math>, <math>e \in E</math> and <math>v \in e</math> then we say <math>e</math> and <math>v</math> are ''incident.'' If <math>e, f \in E</math> and <math>v \in e, f</math> we say the edges <math>e</math> and <math>f</math> are ''coincident'' at <math>v</math>.<br />
* The number of edges in <math>E</math> containing <math>v</math> is the ''degree'' of <math>v</math> and is often denoted <math>d(v)</math>.<br />
* A vertex <math>v</math> is ''isolated'' if <math>d(v) = 0</math>, i.e. if there are no edges incident to <math>v</math>.<br />
* If <math>G_1 = (V_1, E_1)</math> and <math>G_2 = (V_2, E_2)</math> are graphs such that <math>V_2 \subseteq V_1</math> and <math>E_2 \subseteq E_1</math> then we say <math>G_2</math> is a ''subgraph'' of <math>G_1</math>. If <math>E_2 = \{\{v_1, v_2\} \mid \{v_1, v_2\} \in E \textrm{ and } v_1, v_2 \in V_2</math> (informally, if <math>E_2</math> contains all those edges of <math>E_1</math> whose vertices are in <math>V_2</math>) then we say that <math>G_2</math> is an ''induced subgraph'' of <math>G_1</math>.<br />
<br />
==Types of Graphs and Subgraphs==<br />
===Complete Graph or Clique===<br />
A ''complete graph'' is a graph in which there is an edge joining every pair of vertices is connected. The complete graph on <math>n</math> vertices is denoted <math>K_n</math>. If <math>H</math> is a complete subgraph of <math>G</math> then the vertices of <math>H</math> are said to form a ''clique'' in <math>G</math>.<br />
<br />
===Complementary Graphs===<br />
If <math>G_1 = (V, E_1)</math> and <math>G_2 = (V, E_2)</math> are two graphs on the same vertex set such that <math>G = (V, E_1 \cup E_2)</math> is a complete graph and <math>E_1 \cap E_2 = \emptyset</math> then <math>G_2</math> is said to be the ''complement'' of <math>G_1</math> and vice-versa.<br />
<br />
===Null Graph or Independent Set===<br />
A ''null graph'' (or ''independent set'') is the complement of a complete graph. Equivalently, a null graph is a graph in which every vertex is isolated. When drawn in the usual fashion, a null graph is simply a collection of scattered points (the vertices) with no edges connecting them. The terminology "independent set" is used most frequently to refer to a subgraph. In other words, one says <math>V_1 \subseteq V_1</math> is an independent set in <math>G= (V, E)</math> if and only if <math>V_1</math> is a clique in the complement of <math>G</math>.<br />
<br />
===Paths and Cycles===<br />
A ''path'' in a graph <math>G = (V, E)</math> is a sequence <math>v_0, e_1, v_1, \ldots, e_n, v_n</math> such that <math>v_i \in V</math>, <math>e_i \in E</math> and <math>e_i = \{v_{i - 1}, v_i\}</math> for all <math>i</math>. A ''cycle'' is a path in which the initial and final vertices are the same.<br />
<br />
===Connected Graph===<br />
A graph is ''connected'' if any two vertices can be connected by a path. That is, there are no isolated vertices with no paths coming from them, nor can the vertex set be partitioned into two parts with no edge between them.<br />
<br />
===Planar Graphs===<br />
A graph is said to be planar if it can be drawn in a [[plane]] with no intersecting edges. For example, <math>K_1,K_2,K_3,</math> and <math>K_4</math> are planar.<br />
<br />
{{image}}<br />
<br />
In a planar graph, we can define faces of the graph, or the smallest regions bounded by edges. (An alternate definition is the regions bounded by edges which do not have any edges going through them.) Note that the area outside the planar graph is also a face, called the unbounded face. The degree of the face is the number of edges that bound the face. (Note that the same term is used for vertices, which can become confusing)<br />
<br />
All planar graphs have dual graphs, which involve turning the planes of one graph into vertices, and the vertics into planes, with edges connecting if two planes are adjacent. The dual of the dual of a graph is returns the original graph.<br />
<br />
An interesting result is [[Euler's Polyhedral Formula]], which states that in a planar graph with <math>V</math> vertices, <math>E</math> edges, and <math>F</math> faces, then<br />
<cmath>V-E+F=2</cmath><br />
The proof of this is simple using induction, but the derivation of the formula is much trickier.<br />
<br />
Other interesting results for planar graphs are that:<br />
*<math>E\le 3V-6</math><br />
*if the the sum of the degrees of the faces of the graph is <math>F</math>, then <math>F\le \frac{2}{3}E</math>.<br />
<br />
===Bipartite graph===<br />
A graph is called bipartite if its vertex set can be split into two disjoint subsets such that no edge exists within each subset. A graph is bipartite if and only if it has no odd cycles, which is related to the Two Color Theorem. Bipartite graphs have many applications including matching problems.<br />
===Euler Trail===<br />
A Euler trail is a graph where it is possible to form a trail which uses all the edges. A Euler trail has at most two vertices with odd degrees. The sum of all the degrees of the vertices equals twice the number of edges in the graph. <br />
===Trees and Forests===<br />
A ''forest'' is a graph which does not have any cycles. A ''[[tree (graph theory) | tree]]'' is a connected forest.<br />
<br />
===Weighted Graphs===<br />
The edges of a graph can have weights assigned to them that represent some abstract relative value.<br />
<br />
===Hypergraph===<br />
A hypergraph is an extension of the concept of a graph where the edges can encompass more than two vertices, and essentially become sets themselves. Hypergraph theory is often difficult to visualize, and thus is often studied based on the sets that make it up.<br />
<br />
<br />
==See Also==<br />
*[[Graph theory]]<br />
*[[Euler's Polyhedral Formula]]<br />
<br />
<br />
[[Category:Definition]]<br />
[[Category:Graph theory]]</div>GeronimoStiltonhttps://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_25&diff=814812016 AMC 8 Problems/Problem 252016-11-23T17:33:25Z<p>GeronimoStilton: </p>
<hr />
<div>A semicircle is inscribed in an isosceles triangle with base <math>16</math> and height <math>15</math> so that the diameter of the semicircle is contained in the base of the triangle as shown. What is the radius of the semicircle?<br />
<br />
<asy>draw((0,0)--(8,15)--(16,0)--(0,0));<br />
draw(arc((8,0),7.0588,0,180));</asy><br />
<br />
<math>\textbf{(A) }4 \sqrt{3}\qquad\textbf{(B) } \dfrac{120}{17}\qquad\textbf{(C) }10\qquad\textbf{(D) }\dfrac{17\sqrt{2}}{2}\qquad \textbf{(E)} \dfrac{17\sqrt{3}}{2}</math><br />
<br />
<br />
==Solution==<br />
Draw the altitude from the top of the triangle to its base, dividing the isosceles triangle into two right triangles with height <math>15</math> and base <math>\frac{16}{2} = 8</math>. The Pythagorean triple <math>8</math>-<math>15</math>-<math>17</math> tells us that these triangles have hypotenuses of <math>17</math>. <br />
<br />
Now draw an altitude of one of the smaller right triangles, starting from the foot of the first altitude we drew (which is also the center of the circle that contains the semicircle) and going to the hypotenuse of the right triangle. This segment is both an altitude of the right triangle as well as the radius of the semicircle (this is because tangent lines to circles, such as the hypotenuse touching the semicircle, are always perpendicular to the radii of the circles drawn to the point of tangency). Let this segment's length be <math>r</math>.<br />
<br />
The area of the entire isosceles triangle is <math>\frac{(16)(15)}{2} = 120</math>, so the area of each of the two congruent right triangles it gets split into is <math>\frac{120}{2} = 60</math>. We can also find the area of one of the two congruent right triangles by using its hypotenuse as its base and the radius of the semicircle, the altitude we drew, as its height. Then the area of the triangle is <math>\frac{17r}{2}</math>. Thus we can write the equation <math>\frac{17r}{2} = 60</math>, so <math>17r = 120</math>, so <math>r = \boxed{\textbf{(B) }\frac{120}{17}}</math>.<br />
<br />
{{AMC8 box|year=2016|num-b=24|after=Last Problem}}<br />
{{MAA Notice}}</div>GeronimoStiltonhttps://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_6&diff=814792016 AMC 8 Problems/Problem 62016-11-23T17:15:37Z<p>GeronimoStilton: </p>
<hr />
<div>The following bar graph represents the length (in letters) of the names of 19 people. What is the median length of these names?<br />
<br />
<math>\textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad \textbf{(E) }7</math><br />
==Solution==<br />
We first notice that the median name will be the <math>10^{\mbox{th}}</math> name. We take all the <math>3</math> letter names away from the list to see that the <math>3^{\mbox{rd}}</math> name in the new table is the desired length. Since there are <math>3</math> names that are <math>4</math> letters long, the median name length is <math>\textbf{(B)} 4</math>. <br />
<br />
{{AMC8 box|year=2016|num-b=5|num-a=7}}</div>GeronimoStiltonhttps://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_6&diff=814782016 AMC 8 Problems/Problem 62016-11-23T17:15:07Z<p>GeronimoStilton: /* Solution */</p>
<hr />
<div>The following bar graph represents the length (in letters) of the names of 19 people. What is the median length of these names?<br />
<br />
<math>\textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad \textbf{(E) }7</math><br />
==Solution==<br />
We first notice that the median name will be the <math>10^{\mbox{th}}</math> name. We take all the <math>3</math> letter names away from the list to see that the <math>3^{\mbox{rd}}</math> name in the new table is the desired length. Since there are <math>3</math> names that are <math>4</math> letters long, the median name length is <math>(B) 4</math>. <br />
<br />
{{AMC8 box|year=2016|num-b=5|num-a=7}}</div>GeronimoStiltonhttps://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_6&diff=814772016 AMC 8 Problems/Problem 62016-11-23T17:14:33Z<p>GeronimoStilton: </p>
<hr />
<div>The following bar graph represents the length (in letters) of the names of 19 people. What is the median length of these names?<br />
<br />
<math>\textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad \textbf{(E) }7</math><br />
==Solution==<br />
We first notice that the median name will be the <math>10^{\mbox{th}}</math> name. We subtract all the <math>3</math> letter names from the list to see that the <math>3^{\mbox{rd}}</math> name in the new table is the desired length. Since there are <math>3</math> names that are <math>4</math> letters long, the median name length is <math>(B) 4</math>. <br />
<br />
{{AMC8 box|year=2016|num-b=5|num-a=7}}</div>GeronimoStiltonhttps://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_6&diff=814762016 AMC 8 Problems/Problem 62016-11-23T17:13:26Z<p>GeronimoStilton: </p>
<hr />
<div>The following bar graph represents the length (in letters) of the names of 19 people. What is the median length of these names?<br />
<br />
<math>\textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad \textbf{(E) }7</math><br />
==Solution==<br />
We first notice that the median name will be the <math>10-</math>th name. We subtract all the <math>3</math> letter names from the list to see that the <math>3</math>rd name in the new table is the desired length. Since there are <math>3</math> names that are <math>4</math> letters long, the median name length is <math>(B) 4</math>. <br />
<br />
{{AMC8 box|year=2016|num-b=5|num-a=7}}</div>GeronimoStiltonhttps://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_6&diff=814752016 AMC 8 Problems/Problem 62016-11-23T17:12:00Z<p>GeronimoStilton: </p>
<hr />
<div>The following bar graph represents the length (in letters) of the names of 19 people. What is the median length of these names?<br />
<br />
<br />
==Solution==<br />
We first notice that the median name will be the <math>10-</math>th name. We subtract all the <math>3</math> letter names from the list to see that the <math>3</math>rd name in the new table is the desired length. Since there are <math>3</math> names that are <math>4</math> letters long, the median name length is <math>(B) 4</math>. <br />
<br />
{{AMC8 box|year=2016|num-b=5|num-a=7}}</div>GeronimoStiltonhttps://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_6&diff=814742016 AMC 8 Problems/Problem 62016-11-23T17:11:32Z<p>GeronimoStilton: </p>
<hr />
<div>The following bar graph represents the length (in letters) of the names of 19 people. What is the median length of these names?<br />
<br />
<br />
==Solution==<br />
We first notice that the median name will be the <math>10-</math>th name. We subtract all the <math>3-</math>letter names from the list to see that the <math>3</math>rd name in the new table is the desired length. Since there are <math>3</math> names that are <math>4-</math>letters long, the median name length is <math>(B) 4</math>. <br />
<br />
{{AMC8 box|year=2016|num-b=5|num-a=7}}</div>GeronimoStiltonhttps://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_19&diff=814692016 AMC 8 Problems/Problem 192016-11-23T17:02:04Z<p>GeronimoStilton: </p>
<hr />
<div>The sum of <math>25</math> consecutive even integers is <math>10,000</math>. What is the largest of these <math>25</math> consecutive integers?<br />
<br />
<math>\textbf{(A)}\mbox{ }360\qquad\textbf{(B)}\mbox{ }388\qquad\textbf{(C)}\mbox{ }412\qquad\textbf{(D)}\mbox{ }416\qquad\textbf{(E)}\mbox{ }424</math><br />
<br />
==Solution==<br />
Let <math>n</math> be the 13th consecutive even integer that's being added up. By doing this, we can see that the sum of all 25 even numbers will simply by <math>25n</math> since <math>(n-2k)+\dots+(n-4)+(n-2)+(n)+(n+2)+(n+4)+ \dots +(n+2k)=25n</math>. Now, <math>25n=10000 \rightarrow n=400</math> Remembering that this is the 13th integer, we wish to find the 25th, which is <math>400+2(25-13)=\textbf{(E)}424</math>.<br />
{{AMC8 box|year=2016|num-b=18|num-a=20}}<br />
{{MAA Notice}}</div>GeronimoStiltonhttps://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems&diff=814682016 AMC 8 Problems2016-11-23T16:50:37Z<p>GeronimoStilton: </p>
<hr />
<div>==Problem 1==<br />
<br />
<br />
The longest professional tennis match ever played lasted a total of <math>11</math> hours and <math>5</math> minutes. How many minutes was this?<br />
<br />
<math>\textbf{(A) }605\qquad\textbf{(B) }655\qquad\textbf{(C) }665\qquad\textbf{(D) }1005\qquad \textbf{(E) }1105</math><br />
<br />
[[2016 AMC 8 Problems/Problem 1|Solution<br />
]]<br />
<br />
==Problem 2==<br />
<br />
In rectangle <math>ABCD</math>, <math>AB=6</math> and <math>AD=8</math>. Point <math>M</math> is the midpoint of <math>\overline{AD}</math>. What is the area of <math>\triangle AMC</math>?<br />
<br />
<math>\textbf{(A) }12\qquad\textbf{(B) }15\qquad\textbf{(C) }18\qquad\textbf{(D) }20\qquad \textbf{(E) }24</math><br />
<br />
[[2016 AMC 8 Problems/Problem 2|Solution<br />
]]<br />
<br />
==Problem 3==<br />
<br />
Four students take an exam. Three of their scores are <math>70, 80,</math> and <math>90</math>. If the average of their four scores is <math>70</math>, then what is the remaining score?<br />
<br />
<math>\textbf{(A) }40\qquad\textbf{(B) }50\qquad\textbf{(C) }55\qquad\textbf{(D) }60\qquad \textbf{(E) }70</math><br />
<br />
[[2016 AMC 8 Problems/Problem 3|Solution<br />
]]<br />
<br />
==Problem 4==<br />
<br />
When Cheenu was a boy he could run <math>15</math> miles in <math>3</math> hours and <math>30</math> minutes. As an old man he can now walk <math>10</math> miles in <math>4</math> hours. How many minutes longer does it take for him to walk a mile now compared to when he was a boy?<br />
<br />
<math>\textbf{(A) }6\qquad\textbf{(B) }10\qquad\textbf{(C) }15\qquad\textbf{(D) }18\qquad \textbf{(E) }30</math><br />
<br />
[[2016 AMC 8 Problems/Problem 4|Solution<br />
]]<br />
<br />
==Problem 5==<br />
<br />
The number <math>N</math> is a two-digit number.<br />
<br />
• When <math>N</math> is divided by <math>9</math>, the remainder is <math>1</math>.<br />
<br />
• When <math>N</math> is divided by <math>10</math>, the remainder is <math>3</math>.<br />
<br />
What is the remainder when <math>N</math> is divided by <math>11</math>?<br />
<br />
<br />
<math>\textbf{(A) }0\qquad\textbf{(B) }2\qquad\textbf{(C) }4\qquad\textbf{(D) }5\qquad \textbf{(E) }7</math><br />
<br />
[[2016 AMC 8 Problems/Problem 5|Solution<br />
]]<br />
<br />
==Problem 7==<br />
<br />
Which of the following numbers is not a perfect square?<br />
<br />
<math>\textbf{(A) }1^{2016}\qquad\textbf{(B) }2^{2017}\qquad\textbf{(C) }3^{2018}\qquad\textbf{(D) }4^{2019}\qquad \textbf{(E) }5^{2020}</math><br />
<br />
[[2016 AMC 8 Problems/Problem 7|Solution<br />
]]<br />
<br />
==Problem 8==<br />
<br />
Find the value of the expression<br />
<cmath>100-98+96-94+92-90+\cdots+8-6+4-2.</cmath><math>\textbf{(A) }20\qquad\textbf{(B) }40\qquad\textbf{(C) }50\qquad\textbf{(D) }80\qquad \textbf{(E) }100</math><br />
<br />
[[2016 AMC 8 Problems/Problem 8|Solution<br />
]]<br />
<br />
==Problem 9==<br />
<br />
What is the sum of the distinct prime integer divisors of <math>2016</math>?<br />
<br />
<math>\textbf{(A) }9\qquad\textbf{(B) }12\qquad\textbf{(C) }16\qquad\textbf{(D) }49\qquad \textbf{(E) }63</math><br />
<br />
[[2016 AMC 8 Problems/Problem 9|Solution<br />
]]<br />
<br />
==Problem 10==<br />
<br />
Suppose that <math>a * b</math> means <math>3a-b.</math> What is the value of <math>x</math> if<br />
<cmath>2 * (5 * x)=1</cmath><br />
<math>\textbf{(A) }\frac{1}{10} \qquad\textbf{(B) }2\qquad\textbf{(C) }\frac{10}{3} \qquad\textbf{(D) }10\qquad \textbf{(E) }14</math><br />
<br />
[[2016 AMC 8 Problems/Problem 10|Solution<br />
]]<br />
<br />
==Problem 11==<br />
<br />
Determine how many two-digit numbers satisfy the following property: when the number is added to the number obtained by reversing its digits, the sum is <math>132.</math><br />
<br />
<math>\textbf{(A) }5\qquad\textbf{(B) }7\qquad\textbf{(C) }9\qquad\textbf{(D) }11\qquad \textbf{(E) }12</math><br />
<br />
[[2016 AMC 8 Problems/Problem 11|Solution<br />
]]<br />
<br />
==Problem 12==<br />
<br />
Jefferson Middle School has the same number of boys and girls. Three-fourths of the girls and two-thirds of the boys went on a field trip. What fraction of the students were girls?<br />
<br />
<math>\textbf{(A) }\frac{1}{2}\qquad\textbf{(B) }\frac{9}{17}\qquad\textbf{(C) }\frac{7}{13}\qquad\textbf{(D) }\frac{2}{3}\qquad \textbf{(E) }\frac{14}{15}</math><br />
<br />
[[2016 AMC 8 Problems/Problem 12|Solution<br />
]]<br />
<br />
==Problem 13==<br />
<br />
Two different numbers are randomly selected from the set <math>{ - 2, -1, 0, 3, 4, 5}</math> and multiplied together. What is the probability that the product is <math>0</math>?<br />
<br />
<math>\textbf{(A) }\dfrac{1}{6}\qquad\textbf{(B) }\dfrac{1}{5}\qquad\textbf{(C) }\dfrac{1}{4}\qquad\textbf{(D) }\dfrac{1}{3}\qquad \textbf{(E) }\dfrac{1}{2}</math><br />
<br />
[[2016 AMC 8 Problems/Problem 13|Solution<br />
]]<br />
<br />
==Problem 14==<br />
<br />
Karl's car uses a gallon of gas every <math>35</math> miles, and his gas tank holds <math>14</math> gallons when it is full. One day, Karl started with a full tank of gas, drove <math>350</math> miles, bought <math>8</math> gallons of gas, and continued driving to his destination. When he arrived, his gas tank was half full. How many miles did Karl drive that day? <br />
<br />
<math>\textbf{(A)}\mbox{ }525\qquad\textbf{(B)}\mbox{ }560\qquad\textbf{(C)}\mbox{ }595\qquad\textbf{(D)}\mbox{ }665\qquad\textbf{(E)}\mbox{ }735</math><br />
<br />
[[2016 AMC 8 Problems/Problem 14|Solution<br />
]]<br />
<br />
==Problem 15==<br />
<br />
What is the largest power of <math>2</math> that is a divisor of <math>13^4 - 11^4</math>?<br />
<br />
<math>\textbf{(A)}\mbox{ }8\qquad \textbf{(B)}\mbox{ }16\qquad \textbf{(C)}\mbox{ }32\qquad \textbf{(D)}\mbox{ }64\qquad \textbf{(E)}\mbox{ }128</math><br />
<br />
[[2016 AMC 8 Problems/Problem 15|Solution<br />
]]<br />
<br />
==Problem 16==<br />
<br />
Annie and Bonnie are running laps around a <math>400</math>-meter oval track. They started together, but Annie has pulled ahead, because she runs <math>25\%</math> faster than Bonnie. How many laps will Annie have run when she first passes Bonnie?<br />
<br />
<math>\textbf{(A) }1\dfrac{1}{4}\qquad\textbf{(B) }3\dfrac{1}{3}\qquad\textbf{(C) }4\qquad\textbf{(D) }5\qquad \textbf{(E) }25</math><br />
<br />
[[2016 AMC 8 Problems/Problem 16|Solution<br />
]]<br />
<br />
==Problem 17==<br />
<br />
An ATM password at Fred's Bank is composed of four digits from <math>0</math> to <math>9</math>, with repeated digits allowable. If no password may begin with the sequence <math>9,1,1,</math> then how many passwords are possible?<br />
<br />
<math>\textbf{(A)}\mbox{ }30\qquad\textbf{(B)}\mbox{ }7290\qquad\textbf{(C)}\mbox{ }9000\qquad\textbf{(D)}\mbox{ }9990\qquad\textbf{(E)}\mbox{ }9999</math><br />
<br />
[[2016 AMC 8 Problems/Problem 17|Solution<br />
]]<br />
<br />
==Problem 18==<br />
<br />
In an All-Area track meet, <math>216</math> sprinters enter a <math>100-</math>meter dash competition. The track has <math>6</math> lanes, so only <math>6</math> sprinters can compete at a time. At the end of each race, the five non-winners are eliminated, and the winner will compete again in a later race. How many races are needed to determine the champion sprinter?<br />
<br />
<math>\textbf{(A)}\mbox{ }36\qquad\textbf{(B)}\mbox{ }42\qquad\textbf{(C)}\mbox{ }43\qquad\textbf{(D)}\mbox{ }60\qquad\textbf{(E)}\mbox{ }72</math><br />
<br />
[[2016 AMC 8 Problems/Problem 18|Solution<br />
]]<br />
<br />
==Problem 19==<br />
<br />
The sum of <math>25</math> consecutive even integers is <math>10,000</math>. What is the largest of these <math>25</math> consecutive integers?<br />
<br />
<math>\textbf{(A)}\mbox{ }360\qquad\textbf{(B)}\mbox{ }388\qquad\textbf{(C)}\mbox{ }412\qquad\textbf{(D)}\mbox{ }416\qquad\textbf{(E)}\mbox{ }424</math><br />
<br />
[[2016 AMC 8 Problems/Problem 19|Solution<br />
]]<br />
<br />
==Problem 20==<br />
<br />
The least common multiple of <math>a</math> and <math>b</math> is <math>12</math>, and the least common multiple of <math>b</math> and <math>c</math> is <math>15</math>. What is the least possible value of the least common multiple of <math>a</math> and <math>c</math>?<br />
<br />
<math>\textbf{(A) }20\qquad\textbf{(B) }30\qquad\textbf{(C) }60\qquad\textbf{(D) }120\qquad \textbf{(E) }180</math><br />
<br />
[[2016 AMC 8 Problems/Problem 20|Solution<br />
]]<br />
<br />
==Problem 21==<br />
<br />
A box contains 3 red chips and 2 green chips. Chips are drawn randomly, one at a time without replacement, until all 3 of the reds are drawn or until both green chips are drawn. What is the probability that the 3 reds are drawn?<br />
<br />
<math>\textbf{(A) }\dfrac{3}{10}\qquad\textbf{(B) }\dfrac{2}{5}\qquad\textbf{(C) }\dfrac{1}{2}\qquad\textbf{(D) }\dfrac{3}{5}\qquad \textbf{(E) }\dfrac{7}{10}</math><br />
<br />
[[2016 AMC 8 Problems/Problem 21|Solution<br />
]]<br />
<br />
==Problem 22==<br />
<br />
Rectangle <math>DEFA</math> below is a <math>3 \times 4</math> rectangle with <math>DC=CB=BA</math>. What is the area of the "bat wings" (shaded area)?<br />
<asy><br />
draw((0,0)--(3,0)--(3,4)--(0,4)--(0,0)--(2,4)--(3,0));<br />
draw((3,0)--(1,4)--(0,0));<br />
fill((0,0)--(1,4)--(1.5,3)--cycle, black);<br />
fill((3,0)--(2,4)--(1.5,3)--cycle, black);<br />
</asy><br />
<br />
<math>\textbf{(A) }2\qquad\textbf{(B) }2 \frac{1}{2}\qquad\textbf{(C) }3\qquad\textbf{(D) }3 \frac{1}{2}\qquad \textbf{(E) }5</math><br />
<br />
[[2016 AMC 8 Problems/Problem 22|Solution<br />
]]<br />
<br />
==Problem 23==<br />
<br />
Two congruent circles centered at points <math>A</math> and <math>B</math> each pass through the other circle's center. The line containing both <math>A</math> and <math>B</math> is extended to intersect the circles at points <math>C</math> and <math>D</math>. The circles intersect at two points, one of which is <math>E</math>. What is the degree measure of <math>\angle CED</math>?<br />
<br />
<math>\textbf{(A) }90\qquad\textbf{(B) }105\qquad\textbf{(C) }120\qquad\textbf{(D) }135\qquad \textbf{(E) }150</math><br />
<br />
[[2016 AMC 8 Problems/Problem 23|Solution<br />
]]<br />
<br />
==Problem 24==<br />
<br />
The digits <math>1</math>, <math>2</math>, <math>3</math>, <math>4</math>, and <math>5</math> are each used once to write a five-digit number <math>PQRST</math>. The three-digit number <math>PQR</math> is divisible by <math>4</math>, the three-digit number <math>QRS</math> is divisible by <math>5</math>, and the three-digit number <math>RST</math> is divisible by <math>3</math>. What is <math>P</math>?<br />
<br />
<math>\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad \textbf{(E) }5</math><br />
<br />
[[2016 AMC 8 Problems/Problem 24|Solution<br />
]]<br />
<br />
==Problem 25== <br />
<br />
A semicircle is inscribed in an isosceles triangle with base <math>16</math> and height <math>15</math> so that the diameter of the semicircle is contained in the base of the triangle as shown. What is the radius of the semicircle?<br />
<br />
<asy>draw((0,0)--(8,15)--(16,0)--(0,0));<br />
draw(arc((8,0),7.0588,0,180));</asy><br />
<br />
<math>\textbf{(A) }4 \sqrt{3}\qquad\textbf{(B) } \dfrac{120}{17}\qquad\textbf{(C) }10\qquad\textbf{(D) }\dfrac{17\sqrt{2}}{2}\qquad \textbf{(E)} \dfrac{17\sqrt{3}}{2}</math><br />
<br />
[[2016 AMC 8 Problems/Problem 25|Solution<br />
]]<br />
<br />
{{MAA Notice}}</div>GeronimoStiltonhttps://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_19&diff=814672016 AMC 8 Problems/Problem 192016-11-23T16:49:32Z<p>GeronimoStilton: </p>
<hr />
<div>The sum of <math>25</math> consecutive even integers is <math>10,000</math>. What is the largest of these <math>25</math> consecutive integers?<br />
<br />
<math>\textbf{(A)}\mbox{ }360\qquad\textbf{(B)}\mbox{ }388\qquad\textbf{(C)}\mbox{ }412\qquad\textbf{(D)}\mbox{ }416\qquad\textbf{(E)}\mbox{ }424</math><br />
<br />
==Solution==<br />
Let <math>n</math> be the 13th consecutive even integer that's being added up. By doing this, we can see that the sum of all 25 even numbers will simply by <math>25n</math> since <math>(n-2k)+\dots+(n-4)+(n-2)+(n)+(n+2)+(n+4)+ \dots +(n+2k)=25n</math>. Now, <math>25n=10000 \rightarrow n=400</math> Remembering that this is the 13th integer, we wish to find the 25th, which is <math>400+25-13=\textbf{(C)}412</math>.<br />
{{AMC8 box|year=2016|num-b=18|num-a=20}}<br />
{{MAA Notice}}</div>GeronimoStiltonhttps://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_19&diff=814662016 AMC 8 Problems/Problem 192016-11-23T16:48:07Z<p>GeronimoStilton: </p>
<hr />
<div>The sum of <math>25</math> consecutive even integers is <math>10,000</math>. What is the largest of these <math>25</math> consecutive integers?<br />
<br />
<math>\textbf{(A)}\mbox{ }360\qquad\textbf{(B)}\mbox{ }388\qquad\textbf{(C)}\mbox{ }412\qquad\textbf{(D)}\mbox{ }416\qquad\textbf{(E)}\mbox{ }424</math><br />
<br />
==Solution==<br />
Let <math>n</math> be the 13th consecutive even integer that's being added up. By doing this, we can see that the sum of all 25 even numbers will simply by <math>25n</math> since <math>(n-2k)+\dots+(n-4)+(n-2)+(n)+(n+2)+(n+4)+ \dots +(n+2k)=25n</math>. Now, <math>25n=10000 \rightarrow n=400</math> Remembering that this is the 13th integer, we wish to find the 25th, which is <math>400+2 \cdot (15-13)=424</math>.<br />
{{AMC8 box|year=2016|num-b=18|num-a=20}}<br />
{{MAA Notice}}</div>GeronimoStiltonhttps://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems&diff=814652016 AMC 8 Problems2016-11-23T16:46:33Z<p>GeronimoStilton: </p>
<hr />
<div>==Problem 1==<br />
<br />
<br />
The longest professional tennis match ever played lasted a total of <math>11</math> hours and <math>5</math> minutes. How many minutes was this?<br />
<br />
<math>\textbf{(A) }605\qquad\textbf{(B) }655\qquad\textbf{(C) }665\qquad\textbf{(D) }1005\qquad \textbf{(E) }1105</math><br />
<br />
[[2016 AMC 8 Problems/Problem 1|Solution<br />
]]<br />
<br />
==Problem 2==<br />
<br />
In rectangle <math>ABCD</math>, <math>AB=6</math> and <math>AD=8</math>. Point <math>M</math> is the midpoint of <math>\overline{AD}</math>. What is the area of <math>\triangle AMC</math>?<br />
<br />
<math>\textbf{(A) }12\qquad\textbf{(B) }15\qquad\textbf{(C) }18\qquad\textbf{(D) }20\qquad \textbf{(E) }24</math><br />
<br />
[[2016 AMC 8 Problems/Problem 2|Solution<br />
]]<br />
<br />
==Problem 3==<br />
<br />
Four students take an exam. Three of their scores are <math>70, 80,</math> and <math>90</math>. If the average of their four scores is <math>70</math>, then what is the remaining score?<br />
<br />
<math>\textbf{(A) }40\qquad\textbf{(B) }50\qquad\textbf{(C) }55\qquad\textbf{(D) }60\qquad \textbf{(E) }70</math><br />
<br />
[[2016 AMC 8 Problems/Problem 3|Solution<br />
]]<br />
<br />
==Problem 4==<br />
<br />
When Cheenu was a boy he could run <math>15</math> miles in <math>3</math> hours and <math>30</math> minutes. As an old man he can now walk <math>10</math> miles in <math>4</math> hours. How many minutes longer does it take for him to walk a mile now compared to when he was a boy?<br />
<br />
<math>\textbf{(A) }6\qquad\textbf{(B) }10\qquad\textbf{(C) }15\qquad\textbf{(D) }18\qquad \textbf{(E) }30</math><br />
<br />
[[2016 AMC 8 Problems/Problem 4|Solution<br />
]]<br />
<br />
==Problem 5==<br />
<br />
The number <math>N</math> is a two-digit number.<br />
<br />
• When <math>N</math> is divided by <math>9</math>, the remainder is <math>1</math>.<br />
<br />
• When <math>N</math> is divided by <math>10</math>, the remainder is <math>3</math>.<br />
<br />
What is the remainder when <math>N</math> is divided by <math>11</math>?<br />
<br />
<br />
<math>\textbf{(A) }0\qquad\textbf{(B) }2\qquad\textbf{(C) }4\qquad\textbf{(D) }5\qquad \textbf{(E) }7</math><br />
<br />
[[2016 AMC 8 Problems/Problem 5|Solution<br />
]]<br />
<br />
==Problem 7==<br />
<br />
Which of the following numbers is not a perfect square?<br />
<br />
<math>\textbf{(A) }1^{2016}\qquad\textbf{(B) }2^{2017}\qquad\textbf{(C) }3^{2018}\qquad\textbf{(D) }4^{2019}\qquad \textbf{(E) }5^{2020}</math><br />
<br />
[[2016 AMC 8 Problems/Problem 7|Solution<br />
]]<br />
<br />
==Problem 8==<br />
<br />
Find the value of the expression<br />
<cmath>100-98+96-94+92-90+\cdots+8-6+4-2.</cmath><math>\textbf{(A) }20\qquad\textbf{(B) }40\qquad\textbf{(C) }50\qquad\textbf{(D) }80\qquad \textbf{(E) }100</math><br />
<br />
[[2016 AMC 8 Problems/Problem 8|Solution<br />
]]<br />
<br />
==Problem 9==<br />
<br />
What is the sum of the distinct prime integer divisors of <math>2016</math>?<br />
<br />
<math>\textbf{(A) }9\qquad\textbf{(B) }12\qquad\textbf{(C) }16\qquad\textbf{(D) }49\qquad \textbf{(E) }63</math><br />
<br />
[[2016 AMC 8 Problems/Problem 9|Solution<br />
]]<br />
<br />
==Problem 10==<br />
<br />
Suppose that <math>a * b</math> means <math>3a-b.</math> What is the value of <math>x</math> if<br />
<cmath>2 * (5 * x)=1</cmath><br />
<math>\textbf{(A) }\frac{1}{10} \qquad\textbf{(B) }2\qquad\textbf{(C) }\frac{10}{3} \qquad\textbf{(D) }10\qquad \textbf{(E) }14</math><br />
<br />
[[2016 AMC 8 Problems/Problem 10|Solution<br />
]]<br />
<br />
==Problem 11==<br />
<br />
Determine how many two-digit numbers satisfy the following property: when the number is added to the number obtained by reversing its digits, the sum is <math>132.</math><br />
<br />
<math>\textbf{(A) }5\qquad\textbf{(B) }7\qquad\textbf{(C) }9\qquad\textbf{(D) }11\qquad \textbf{(E) }12</math><br />
<br />
[[2016 AMC 8 Problems/Problem 11|Solution<br />
]]<br />
<br />
==Problem 12==<br />
<br />
Jefferson Middle School has the same number of boys and girls. Three-fourths of the girls and two-thirds of the boys went on a field trip. What fraction of the students were girls?<br />
<br />
<math>\textbf{(A) }\frac{1}{2}\qquad\textbf{(B) }\frac{9}{17}\qquad\textbf{(C) }\frac{7}{13}\qquad\textbf{(D) }\frac{2}{3}\qquad \textbf{(E) }\frac{14}{15}</math><br />
<br />
[[2016 AMC 8 Problems/Problem 12|Solution<br />
]]<br />
<br />
==Problem 13==<br />
<br />
Two different numbers are randomly selected from the set <math>{ - 2, -1, 0, 3, 4, 5}</math> and multiplied together. What is the probability that the product is <math>0</math>?<br />
<br />
<math>\textbf{(A) }\dfrac{1}{6}\qquad\textbf{(B) }\dfrac{1}{5}\qquad\textbf{(C) }\dfrac{1}{4}\qquad\textbf{(D) }\dfrac{1}{3}\qquad \textbf{(E) }\dfrac{1}{2}</math><br />
<br />
[[2016 AMC 8 Problems/Problem 13|Solution<br />
]]<br />
<br />
==Problem 14==<br />
<br />
Karl's car uses a gallon of gas every <math>35</math> miles, and his gas tank holds <math>14</math> gallons when it is full. One day, Karl started with a full tank of gas, drove <math>350</math> miles, bought <math>8</math> gallons of gas, and continued driving to his destination. When he arrived, his gas tank was half full. How many miles did Karl drive that day? <br />
<br />
<math>\textbf{(A)}\mbox{ }525\qquad\textbf{(B)}\mbox{ }560\qquad\textbf{(C)}\mbox{ }595\qquad\textbf{(D)}\mbox{ }665\qquad\textbf{(E)}\mbox{ }735</math><br />
<br />
[[2016 AMC 8 Problems/Problem 14|Solution<br />
]]<br />
<br />
==Problem 15==<br />
<br />
What is the largest power of <math>2</math> that is a divisor of <math>13^4 - 11^4</math>?<br />
<br />
<math>\textbf{(A)}\mbox{ }8\qquad \textbf{(B)}\mbox{ }16\qquad \textbf{(C)}\mbox{ }32\qquad \textbf{(D)}\mbox{ }64\qquad \textbf{(E)}\mbox{ }128</math><br />
<br />
[[2016 AMC 8 Problems/Problem 15|Solution<br />
]]<br />
<br />
==Problem 16==<br />
<br />
Annie and Bonnie are running laps around a <math>400</math>-meter oval track. They started together, but Annie has pulled ahead, because she runs <math>25\%</math> faster than Bonnie. How many laps will Annie have run when she first passes Bonnie?<br />
<br />
<math>\textbf{(A) }1\dfrac{1}{4}\qquad\textbf{(B) }3\dfrac{1}{3}\qquad\textbf{(C) }4\qquad\textbf{(D) }5\qquad \textbf{(E) }25</math><br />
<br />
[[2016 AMC 8 Problems/Problem 16|Solution<br />
]]<br />
<br />
==Problem 17==<br />
<br />
An ATM password at Fred's Bank is composed of four digits from <math>0</math> to <math>9</math>, with repeated digits allowable. If no password may begin with the sequence <math>9,1,1,</math> then how many passwords are possible?<br />
<br />
<math>\textbf{(A)}\mbox{ }30\qquad\textbf{(B)}\mbox{ }7290\qquad\textbf{(C)}\mbox{ }9000\qquad\textbf{(D)}\mbox{ }9990\qquad\textbf{(E)}\mbox{ }9999</math><br />
<br />
[[2016 AMC 8 Problems/Problem 17|Solution<br />
]]<br />
<br />
==Problem 18==<br />
<br />
In an All-Area track meet, <math>216</math> sprinters enter a <math>100-</math>meter dash competition. The track has <math>6</math> lanes, so only <math>6</math> sprinters can compete at a time. At the end of each race, the five non-winners are eliminated, and the winner will compete again in a later race. How many races are needed to determine the champion sprinter?<br />
<br />
<math>\textbf{(A)}\mbox{ }36\qquad\textbf{(B)}\mbox{ }42\qquad\textbf{(C)}\mbox{ }43\qquad\textbf{(D)}\mbox{ }60\qquad\textbf{(E)}\mbox{ }72</math><br />
<br />
[[2016 AMC 8 Problems/Problem 18|Solution<br />
]]<br />
<br />
==Problem 20==<br />
<br />
The least common multiple of <math>a</math> and <math>b</math> is <math>12</math>, and the least common multiple of <math>b</math> and <math>c</math> is <math>15</math>. What is the least possible value of the least common multiple of <math>a</math> and <math>c</math>?<br />
<br />
<math>\textbf{(A) }20\qquad\textbf{(B) }30\qquad\textbf{(C) }60\qquad\textbf{(D) }120\qquad \textbf{(E) }180</math><br />
<br />
[[2016 AMC 8 Problems/Problem 20|Solution<br />
]]<br />
<br />
==Problem 21==<br />
<br />
A box contains 3 red chips and 2 green chips. Chips are drawn randomly, one at a time without replacement, until all 3 of the reds are drawn or until both green chips are drawn. What is the probability that the 3 reds are drawn?<br />
<br />
<math>\textbf{(A) }\dfrac{3}{10}\qquad\textbf{(B) }\dfrac{2}{5}\qquad\textbf{(C) }\dfrac{1}{2}\qquad\textbf{(D) }\dfrac{3}{5}\qquad \textbf{(E) }\dfrac{7}{10}</math><br />
<br />
[[2016 AMC 8 Problems/Problem 21|Solution<br />
]]<br />
<br />
==Problem 22==<br />
<br />
Rectangle <math>DEFA</math> below is a <math>3 \times 4</math> rectangle with <math>DC=CB=BA</math>. What is the area of the "bat wings" (shaded area)?<br />
<asy><br />
draw((0,0)--(3,0)--(3,4)--(0,4)--(0,0)--(2,4)--(3,0));<br />
draw((3,0)--(1,4)--(0,0));<br />
fill((0,0)--(1,4)--(1.5,3)--cycle, black);<br />
fill((3,0)--(2,4)--(1.5,3)--cycle, black);<br />
</asy><br />
<br />
<math>\textbf{(A) }2\qquad\textbf{(B) }2 \frac{1}{2}\qquad\textbf{(C) }3\qquad\textbf{(D) }3 \frac{1}{2}\qquad \textbf{(E) }5</math><br />
<br />
[[2016 AMC 8 Problems/Problem 22|Solution<br />
]]<br />
<br />
==Problem 23==<br />
<br />
Two congruent circles centered at points <math>A</math> and <math>B</math> each pass through the other circle's center. The line containing both <math>A</math> and <math>B</math> is extended to intersect the circles at points <math>C</math> and <math>D</math>. The circles intersect at two points, one of which is <math>E</math>. What is the degree measure of <math>\angle CED</math>?<br />
<br />
<math>\textbf{(A) }90\qquad\textbf{(B) }105\qquad\textbf{(C) }120\qquad\textbf{(D) }135\qquad \textbf{(E) }150</math><br />
<br />
[[2016 AMC 8 Problems/Problem 23|Solution<br />
]]<br />
<br />
==Problem 24==<br />
<br />
The digits <math>1</math>, <math>2</math>, <math>3</math>, <math>4</math>, and <math>5</math> are each used once to write a five-digit number <math>PQRST</math>. The three-digit number <math>PQR</math> is divisible by <math>4</math>, the three-digit number <math>QRS</math> is divisible by <math>5</math>, and the three-digit number <math>RST</math> is divisible by <math>3</math>. What is <math>P</math>?<br />
<br />
<math>\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad \textbf{(E) }5</math><br />
<br />
[[2016 AMC 8 Problems/Problem 24|Solution<br />
]]<br />
<br />
==Problem 25== <br />
<br />
A semicircle is inscribed in an isosceles triangle with base <math>16</math> and height <math>15</math> so that the diameter of the semicircle is contained in the base of the triangle as shown. What is the radius of the semicircle?<br />
<br />
<asy>draw((0,0)--(8,15)--(16,0)--(0,0));<br />
draw(arc((8,0),7.0588,0,180));</asy><br />
<br />
<math>\textbf{(A) }4 \sqrt{3}\qquad\textbf{(B) } \dfrac{120}{17}\qquad\textbf{(C) }10\qquad\textbf{(D) }\dfrac{17\sqrt{2}}{2}\qquad \textbf{(E)} \dfrac{17\sqrt{3}}{2}</math><br />
<br />
[[2016 AMC 8 Problems/Problem 25|Solution<br />
]]<br />
<br />
{{MAA Notice}}</div>GeronimoStiltonhttps://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_18&diff=814642016 AMC 8 Problems/Problem 182016-11-23T16:43:07Z<p>GeronimoStilton: </p>
<hr />
<div>In an All-Area track meet, <math>216</math> sprinters enter a <math>100-</math>meter dash competition. The track has <math>6</math> lanes, so only <math>6</math> sprinters can compete at a time. At the end of each race, the five non-winners are eliminated, and the winner will compete again in a later race. How many races are needed to determine the champion sprinter?<br />
<br />
<math>\textbf{(A)}\mbox{ }36\qquad\textbf{(B)}\mbox{ }42\qquad\textbf{(C)}\mbox{ }43\qquad\textbf{(D)}\mbox{ }60\qquad\textbf{(E)}\mbox{ }72</math><br />
==Solution==<br />
From any <math>n-</math>th race, only <math>\frac{1}{6}</math> will continue on. Since we wish to find the total number of races, a column representing the races over time is ideal.<br />
Starting with the first race:<br />
<cmath>\frac{216}{6}=36</cmath><br />
<cmath>\frac{36}{6}=6</cmath><br />
<cmath>\frac{6}{6}=1</cmath><br />
Adding all of the numbers in the second column yields <math>43 \rightarrow \boxed{C}</math><br />
{{AMC8 box|year=2016|num-b=17|num-a=19}}<br />
{{MAA Notice}}\end{align}</div>GeronimoStiltonhttps://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_18&diff=814632016 AMC 8 Problems/Problem 182016-11-23T16:42:45Z<p>GeronimoStilton: </p>
<hr />
<div>In an All-Area track meet, <math>216</math> sprinters enter a <math>100-</math>meter dash competition. The track has <math>6</math> lanes, so only <math>6</math> sprinters can compete at a time. At the end of each race, the five non-winners are eliminated, and the winner will compete again in a later race. How many races are needed to determine the champion sprinter?<br />
<br />
<br />
<math>\textbf{(A)}\mbox{ }36\qquad\textbf{(B)}\mbox{ }42\qquad\textbf{(C)}\mbox{ }43\qquad\textbf{(D)}\mbox{ }60\qquad\textbf{(E)}\mbox{ }72</math><br />
==Solution==<br />
From any <math>n-</math>th race, only <math>\frac{1}{6}</math> will continue on. Since we wish to find the total number of races, a column representing the races over time is ideal.<br />
Starting with the first race:<br />
<cmath>\frac{216}{6}=36</cmath><br />
<cmath>\frac{36}{6}=6</cmath><br />
<cmath>\frac{6}{6}=1</cmath><br />
Adding all of the numbers in the second column yields <math>43 \rightarrow \boxed{C}</math><br />
{{AMC8 box|year=2016|num-b=17|num-a=19}}<br />
{{MAA Notice}}\end{align}</div>GeronimoStilton