https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Puddles+penguin&feedformat=atomAoPS Wiki - User contributions [en]2020-10-21T07:37:22ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=User:Piphi&diff=123092User:Piphi2020-05-27T23:05:09Z<p>Puddles penguin: /* User Count */</p>
<hr />
<div>{{User:Piphi/Template:Header}}<br />
<br><br />
<div style="border:2px solid black; -webkit-border-radius: 10px; background:#dddddd"><br />
==<font color="black" style="font-family: ITC Avant Garde Gothic Std, Verdana"><div style="margin-left:10px">User Count</div></font>==<br />
<div style="margin-left: 10px; margin-bottom:10px"><font color="black">If this is your first time visiting this page, edit it by incrementing the user count below by one.</font></div><br />
<center><font size="100px">69</font></center><br />
</div><br />
<div style="border:2px solid black; background:#cccccc;-webkit-border-radius: 10px; align:center"><br />
<br />
==<font color="black" style="font-family: ITC Avant Garde Gothic Std, Verdana"><div style="margin-left:10px">About Me</div></font>==<br />
<div style="margin-left: 10px; margin-bottom:10px"><font color="black">PM me if you want to find out about some cool things you can do with the AoPS wiki.<br><br><br />
<br />
My main project on the AoPS wiki is [[AoPS_Administrators#Current_Admins | a list of all the AoPS admins]], everyone is welcome to add more admins to the list by clicking [https://artofproblemsolving.com/wiki/index.php?title=AoPS_Administrators&action=edit&section=1 here]. I also added most of the info in the [[Reaper Archives]].</font></div><br />
</div><br />
<div style="border:2px solid black; background:#bbbbbb;-webkit-border-radius: 10px; align:center"><br />
<br />
==<font color="black" style="font-family: ITC Avant Garde Gothic Std, Verdana"><div style="margin-left:10px">[[User:Piphi/Asymptote|Asymptote]]</div></font>==<br />
<div style="margin-left: 10px; margin-bottom:10px">For a complete list of my Asymptote drawings, go [[User:Piphi/Asymptote|here]].</div><br />
</div></div>Puddles penguinhttps://artofproblemsolving.com/wiki/index.php?title=User:Piphi&diff=122209User:Piphi2020-05-09T17:59:05Z<p>Puddles penguin: /* User Count */</p>
<hr />
<div><center>[[File:Piphi-Avatar.png]]</center><br />
<br />
<div style="border:2px solid black; background:#eeeeee;"><br />
::::<font style="font-family: Verdana, sans-serif">[[User:Piphi|Userpage]] | [[User talk:Piphi|Talk]] | [[Special:Contributions/Piphi|Contributions]]</font><br />
</div><br />
<div style="border:2px solid black; background:#dddddd; align:center"><br />
==<font color="black" style="font-family: MV Boli, Verdana">User Count</font>==<br />
<font color="black">If this is your first time visting this page, edit it by incrementing the user count below by one.<br />
<br />
<center><font size="100px">35</font></center><br />
</font> <br />
</div><br />
<div style="border:2px solid black; background:#cccccc; align:center"><br />
<br />
==<font color="black" style="font-family: MV Boli, Verdana">About Me</font>==<br />
<font color="black">PM me if you want to find out about some cool things you can do with the AoPS wiki.<br />
<br />
My main project on the AoPS wiki is [[AoPS_Administrators#Current_Admins | a list of all the AoPS admins]], everyone is welcome to add more admins to the list by clicking [https://artofproblemsolving.com/wiki/index.php?title=AoPS_Administrators&action=edit&section=1 here]. I also added most of the info in the [[Reaper Archives]].</font> <br />
</div><br />
<div style="border:2px solid black; background:#bbbbbb; align:center"><br />
<br />
==<font color="black" style="font-family: MV Boli, Verdana">Asymptote</font>==<br />
<br />
Here is a list of the different drawings I have made using Asymptote.<br />
<br />
* [[User:Piphi/Firefox | Firefox Logo]] (Started April 25th, 2020, Finished April 28th, 2020)<br />
* [[User:Piphi/Eclipse | Eclipse Logo]] (Started April 29th, 2020, Finished April 29th, 2020)<br />
* [[User:Piphi/Screencast | Screencast Logo]] (Started April 29th, 2020, Finished April 29th, 2020)<br />
* [[User:Piphi/Whatsapp | Whatsapp Logo]] (Started May 2nd, 2020, Finished May 2nd, 2020)<br />
* [[User:Piphi/Office 365 | Office 365 Logo]] (Started May 4th, 2020, Finished May 4th, 2020)<br />
<br />
Work In Progress<br />
<br />
* Wolfram Logo (Not Started Yet)</div>Puddles penguinhttps://artofproblemsolving.com/wiki/index.php?title=MATHCOUNTS&diff=120088MATHCOUNTS2020-03-25T17:58:22Z<p>Puddles penguin: /* National Competition Sites */</p>
<hr />
<div>Many AoPS Community members and online school students have been participants at National MATHCOUNTS, including many Nationals Countdown Round participants in the past decade. '''MATHCOUNTS''' is a large national [[mathematics competition]] and [[mathematics coaching]] program that has served millions of middle school students since 1984. Sponsored by the [http://cna.com CNA Foundation], [http://www.nspe.org/ National Society of Professional Engineers], the [http://www.nctm.org/ National Council of Teachers of Mathematics], and others including Art of Problem Solving, the focus of MATHCOUNTS is on mathematical problem solving. Students are eligible for up to three years, but cannot compete beyond their eighth grade year.<br />
<br />
{{Contest Info|name=MATHCOUNTS|region=USA|type=Free Response|difficulty=0.5 - 2.5|breakdown=<u>Countdown</u>: 0.5 (School/Chapter), 1 (State/National)<br><u>Sprint</u>: 1-1.5 (School/Chapter), 2-2.5 (State/National)<br><u>Target:</u> 1.5 (School), 2 (Chapter), 2-2.5 (State/National)}}<br />
<br />
== MATHCOUNTS Resources ==<br />
=== MATHCOUNTS Books ===<br />
Art of Problem Solving's [http://artofproblemsolving.com/store/list/aops-curriculum Introductory subject textbooks] are ideal for students preparing for MATHCOUNTS, as are [http://artofproblemsolving.com/store/item/aops-vol1 AoPS Volume 1] and [http://artofproblemsolving.com/store/item/competition-math Competition Math for Middle School]<br />
<br />
=== MATHCOUNTS Classes ===<br />
Art of Problem Solving hosts a [http://artofproblemsolving.com/school/course/mathcounts-basics Basic] and an [http://artofproblemsolving.com/school/course/mathcounts-advanced Advanced] MATHCOUNTS course. The AoPS Introduction-level subject courses also include a great deal of MATHCOUNTS preparation. Many AoPS instructors are former National MATHCOUNTS Mathletes.<br />
<br />
=== MATHCOUNTS Online ===<br />
* [http://www.mathcounts.org Official MATHCOUNTS Homepage]<br />
* Art of Problem Solving hosts a large [http://artofproblemsolving.com/community/c3_middle_school_math Middle School Math Forum] as well as a private [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=23209 MATHCOUNTS Coaches Forum].<br />
* The AoPS MATHCOUNTS Trainer is available on the [http://artofproblemsolving.com/mathcounts_trainer AoPS website], as well as on the [https://itunes.apple.com/us/app/mathcounts-trainer-math-contest/id1023961880?ls=1&mt=8 iPhone and iPad].<br />
* The free [http://www.artofproblemsolving.com/alcumus AoPS Alcumus learning system] includes thousands of MATHCOUNTS problems.<br />
* [http://artofproblemsolving.com/ftw/ftw.php For the Win!] gives students free Countdown Round-like practice against other AoPS students.<br />
* AoPS founder Richard Rusczyk has created dozens of [http://artofproblemsolving.com/videos/mathcounts MATHCOUNTS Mini video lessons].<br />
* [http://mathcounts.saab.org/ Elias Saab's MATHCOUNTS Preparation Homepage]<br />
* [http://www.unidata.ucar.edu/staff/russ/mathcounts/diaz.html The MATHCOUNTS Bible According to Mr. Diaz]<br />
*[http://www.artofproblemsolving.com/Resources/AoPS_R_A_MATHCOUNTS.php/ Building a Successful MATHCOUNTS Program] by [[Jeff Boyd]], who coached the 2005, 2007, and 2008 National Champion [[Texas MathCounts]] team.<br />
*[http://mathweb.scranton.edu/monks/courses/ProblemSolving/MathCountsPlaybookBW.pdf Coach Monk's MathCounts Playbook]<br />
* MathCounts Minis make hard problems easy<br />
<br />
== MATHCOUNTS Curriculum ==<br />
MATHCOUNTS curriculum includes [[arithmetic]], [[algebra]], [[counting]], [[geometry]], [[number theory]], [[probability]], and [[statistics]]. The focus of MATHCOUNTS curriculum is in developing [[mathematical problem solving]] skills.<br />
<br />
Before 1990, MATHCOUNTS chose particular areas of mathematics to highlight each year before changing the focus of the competition more broadly to problem solving.<br />
<br />
== Past State Team Winners ==<br />
* 1984: Virginia<br />
* 1985: Florida<br />
* 1986: California<br />
* 1987: New York<br />
* 1988: New York<br />
* 1989: North Carolina<br />
* 1990: Ohio<br />
* 1991: Alabama<br />
* 1992: California<br />
* 1993: Kansas<br />
* 1994: Pennsylvania<br />
* 1995: Indiana<br />
* 1996: Wisconsin<br />
* 1997: Massachusetts<br />
* 1998: Wisconsin<br />
* 1999: Massachusetts<br />
* 2000: California<br />
* 2001: Virginia<br />
* 2002: California<br />
* 2003: California<br />
* 2004: Illinois<br />
* 2005: Texas<br />
* 2006: Virginia<br />
* 2007: Texas<br />
* 2008: Texas<br />
* 2009: Texas<br />
* 2010: California<br />
* 2011: California<br />
* 2012: Massachusetts<br />
* 2013: Massachusetts<br />
* 2014: California<br />
* 2015: Indiana<br />
* 2016: Texas<br />
* 2017: Texas<br />
* 2018: Texas<br />
* 2019: Massachusetts<br />
<br />
== MATHCOUNTS Competition Structure ==<br />
<br />
=== Sprint Round ===<br />
<br />
30 problems are given all at once. Students have 40 minutes to complete the Sprint Round. This round is very fast-paced and requires speed and accuracy as well. The earlier problems are usually the easiest problems in the competition, and the later problems can be as hard as some of the Team Round questions. No calculators are allowed during this round.<br />
<br />
=== Target Round ===<br />
8 problems given 2 at a time. Students have 6 minutes to complete each set of two problems. Students may not go back to previous rounds (or forwards to future rounds) even if they finish before time is called. Calculators are allowed for the Target Round. Usually comprised of one "confidence booster" and one hard problem.<br />
<br />
=== Team Round ===<br />
<br />
10 problems in 20 minutes for a team of 4 students. These problems typically include some of the most difficult problems of the competition. Use of a calculator is allowed (and required for some questions).<br />
<br />
=== Countdown Round ===<br />
High scoring individuals compete head-to-head until a champion is crowned. People compete from off a screen taking 45 seconds or less to finish the problem. The Countdown round is run differently in various different chapter, state, and national competitions. In the national competitions, it is the round that determines the champion. Calculators are not allowed, but scratch paper will be provided.<br />
<br />
<br />
====Chapter and State Competitions====<br />
<br />
In the chapter and state competitions, the countdown round is not mandatory. However, if it is deemed official by the chapter or state, the following format must be used:<br />
<br />
*The 10th place written finisher competes against the 9th place written finisher. A problem is displayed, and both competitors have 45 seconds to answer the question, and the first competitor to correctly answer the question receives one point. The person who gets the most correct out of three questions (not necessarily two out of three) is the winner.<br />
<br />
*The winner of the first round goes up against the 8th place finisher.<br />
<br />
*The winner of the second round goes up against the 7th place finisher.<br />
<br />
This process is continued until the countdown round reaches the top four written competitors. Starting then, the first person to get three questions correct wins (as opposed to the best-out-of-three rule).<br />
<br />
If the countdown round is unofficial, any format may be used. Single-elimination bracket-style tournaments are common.<br />
<br />
====National Competition====<br />
<br />
At the national competition, there are some structural changes to the countdown round. The top 12 (not the top 10) written finishers make it to the countdown round, and the format is changed from a ladder competition to a single elimination tournament where the top four written competitors get a bye. This setup makes it far more likely for a 12th place finisher to become champion, and it makes it less likely for a first place written finisher to become champion, equalizing the field. But even then, a 12th place written competitor will have less of a chance to become champion than the top 4, because the top 4 get a bye. Until the semi-finals, the scoring is best out of five advances.<br />
<br />
At the first round and the second round, the person to correctly answer the most out of 5 questions wins. However, at the semifinals, the rules slightly change&mdash;the first person to correctly answer four questions wins.<br />
<br />
=== Ciphering Round ===<br />
In some states, (most notably Florida) there is an optional ciphering round. Very similar to countdown (in both difficulty and layout), a team sends up a representative to go against all representatives from the other teams. A problem is shown on a screen and students work fast to answer the problem. The students give their answer and after 45 seconds the answer is shown and the answers are checked to see if they are right. The fastest correct answer gets five points, the next fastest gets 4, etc. There are 4 questions per individual and teams send up 4 people. A perfect score is then 80. Often times the questions take clever reading skills. For example, one question was "How much dirt is in a 3 ft by 3 ft by 4 ft hole?" The answer was 0 because there is no dirt in a hole.<br />
<br />
=== Masters Round ===<br />
Top students give in-depth explanations to challenging problems. This round is optional at the state level competition and is mandatory at the national competition (up to 2011). At nationals the top two on the written and countdown participate. <br />
In 2012, it was replaced by the Reel Math Challenge (now called the Math Video Challenge).<br />
<br />
=== Scoring and Ranking ===<br />
An individual's score is their total number of correct sprint round answers plus 2 times their total number of correct target round answers. This total is out of a maximum of <math>30 + 2(8) = 46</math> points.<br />
<br />
A team's score is the individual scores of its members divided by 4 plus 2 points for every correct team round answer, making a team's maximum possible score 66 points. Therefore, it is possible to win with a relatively low team score and a phenomenal individual score, as the team score is only roughly 30% of the total team score. Note that when there are less than four members the score will become less.<br />
<br />
== MATHCOUNTS Competition Levels ==<br />
=== School Competition ===<br />
Students vie for the chance to make their school teams. Problems at this level are generally the easiest and most basic in curriculum.<br />
<br />
=== Chapter Competition ===<br />
Chapter competitions serve as a selection filter for state competitions. A few states don't need to host chapter competitions due to a small population size.<br />
<br />
=== State Competition ===<br />
The top 4 students in each state form the state team for the national competition. The coach of the top school team at the state level is invited to coach the state team at the national competition. Interestingly, the coach of a state team is not necessarily the coach of any of the state's team members.<br />
<br />
=== National Competition ===<br />
==== National Competition Sites ====<br />
For many years, the National MATHCOUNTS competition was held in Washington, D.C. More recently, the competition has changed venues often.<br />
<br />
* The 2020 competition was canceled due to the COVID-19 pandemic<br />
* The 2019 competition was held in Orlando, Florida.<br />
* The 2018 competition was held in Washington, D.C.<br />
* The 2017 competition was held in Orlando, Florida.<br />
* The 2016 competition was held in Washington, D.C.<br />
* The 2015 competition was held in Boston, Massachusetts.<br />
* The 2014 competition was held in Orlando, Florida.<br />
* The 2013 competition was held in Washington, D.C.<br />
* The 2012 competition was held in Orlando, Florida.<br />
* The 2011 competition was held in Washington, D.C.<br />
* The 2009 and 2010 competitions were held in Orlando, Florida.<br />
* The 2008 competition was held in Denver, Colorado.<br />
* The 2007 competition was held in Fort Worth, Texas.<br />
* The 2006 competition was held in Arlington, Virginia.<br />
* The 2005 competition was held in Detroit, Michigan.<br />
* The 2004 competition was held in Washington, D.C.<br />
* The 2002 and 2003 competitions were held in Chicago, Illinois.<br />
<br />
== What comes after MATHCOUNTS? ==<br />
<br />
Give the following competitions a try and take a look at the [[List of United States high school mathematics competitions]].<br />
* [[American Mathematics Competitions]]<br />
* [[American Regions Math League]]<br />
* [[Mandelbrot Competition]]<br />
* [[Mu Alpha Theta]]<br />
<br />
[[Category:Mathematics competitions]]<br />
<br />
== See also... ==<br />
* [[List of national MATHCOUNTS teams]]<br />
* [[Mathematics competition resources]]<br />
* [[Math contest books]]<br />
* [[Math books]]<br />
* [[List of United States middle school mathematics competitions]]<br />
* [[List of United States high school mathematics competitions]]<br />
* [http://www.mathcounts.org/webarticles/anmviewer.asp?a=921&z=71 2006 MATHCOUNTS Countdown Video]<br />
<br />
[[Category:Introductory mathematics competitions]]</div>Puddles penguinhttps://artofproblemsolving.com/wiki/index.php?title=MATHCOUNTS&diff=105791MATHCOUNTS2019-05-13T19:14:23Z<p>Puddles penguin: /* National Competition Sites */</p>
<hr />
<div>Many AoPS Community members and online school students have been participants at National MATHCOUNTS, including many Nationals Countdown Round participants in the past decade. '''MATHCOUNTS''' is a large national [[mathematics competition]] and [[mathematics coaching]] program that has served millions of middle school students since 1984. Sponsored by the [http://cna.com CNA Foundation], [http://www.nspe.org/ National Society of Professional Engineers], the [http://www.nctm.org/ National Council of Teachers of Mathematics], and others including Art of Problem Solving, the focus of MATHCOUNTS is on mathematical problem solving. Students are eligible for up to three years, but cannot compete beyond their eighth grade year.<br />
<br />
{{Contest Info|name=MATHCOUNTS|region=USA|type=Free Response|difficulty=0.5 - 2.5|breakdown=<u>Countdown</u>: 0.5 (School/Chapter), 1 (State/National)<br><u>Sprint</u>: 1-1.5 (School/Chapter), 2-2.5 (State/National)<br><u>Target:</u> 1.5 (School), 2 (Chapter), 2-2.5 (State/National)}}<br />
<br />
== MATHCOUNTS Resources ==<br />
=== MATHCOUNTS Books ===<br />
Art of Problem Solving's [http://artofproblemsolving.com/store/list/aops-curriculum Introductory subject textbooks] are ideal for students preparing for MATHCOUNTS, as are [http://artofproblemsolving.com/store/item/aops-vol1 AoPS Volume 1] and [http://artofproblemsolving.com/store/item/competition-math Competition Math for Middle School]<br />
<br />
=== MATHCOUNTS Classes ===<br />
Art of Problem Solving hosts a [http://artofproblemsolving.com/school/course/mathcounts-basics Basic] and an [http://artofproblemsolving.com/school/course/mathcounts-advanced Advanced] MATHCOUNTS course. The AoPS Introduction-level subject courses also include a great deal of MATHCOUNTS preparation. Many AoPS instructors are former National MATHCOUNTS Mathletes.<br />
<br />
=== MATHCOUNTS Online ===<br />
* [http://www.mathcounts.org Official MATHCOUNTS Homepage]<br />
* Art of Problem Solving hosts a large [http://artofproblemsolving.com/community/c3_middle_school_math Middle School Math Forum] as well as a private [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=23209 MATHCOUNTS Coaches Forum].<br />
* The AoPS MATHCOUNTS Trainer is available on the [http://artofproblemsolving.com/mathcounts_trainer AoPS website], as well as on the [https://itunes.apple.com/us/app/mathcounts-trainer-math-contest/id1023961880?ls=1&mt=8 iPhone and iPad].<br />
* The free [http://www.artofproblemsolving.com/alcumus AoPS Alcumus learning system] includes thousands of MATHCOUNTS problems.<br />
* [http://artofproblemsolving.com/ftw/ftw.php For the Win!] gives students free Countdown Round-like practice against other AoPS students.<br />
* AoPS founder Richard Rusczyk has created dozens of [http://artofproblemsolving.com/videos/mathcounts MATHCOUNTS Mini video lessons].<br />
* [http://mathcounts.saab.org/ Elias Saab's MATHCOUNTS Preparation Homepage]<br />
* [http://www.unidata.ucar.edu/staff/russ/mathcounts/diaz.html The MATHCOUNTS Bible According to Mr. Diaz]<br />
*[http://www.artofproblemsolving.com/Resources/AoPS_R_A_MATHCOUNTS.php/ Building a Successful MATHCOUNTS Program] by [[Jeff Boyd]], who coached the 2005, 2007, and 2008 National Champion [[Texas MathCounts]] team.<br />
*[http://mathweb.scranton.edu/monks/courses/ProblemSolving/MathCountsPlaybookBW.pdf Coach Monk's MathCounts Playbook]<br />
* MathCounts Minis make hard problems easy<br />
<br />
== MATHCOUNTS Curriculum ==<br />
MATHCOUNTS curriculum includes [[arithmetic]], [[algebra]], [[counting]], [[geometry]], [[number theory]], [[probability]], and [[statistics]]. The focus of MATHCOUNTS curriculum is in developing [[mathematical problem solving]] skills.<br />
<br />
Before 1990, MATHCOUNTS chose particular areas of mathematics to highlight each year before changing the focus of the competition more broadly to problem solving.<br />
<br />
== Past State Team Winners ==<br />
* 1984: Virginia<br />
* 1985: Florida<br />
* 1986: California<br />
* 1987: New York<br />
* 1988: New York<br />
* 1989: North Carolina<br />
* 1990: Ohio<br />
* 1991: Alabama<br />
* 1992: California<br />
* 1993: Kansas<br />
* 1994: Pennsylvania<br />
* 1995: Indiana<br />
* 1996: Wisconsin<br />
* 1997: Massachusetts<br />
* 1998: Wisconsin<br />
* 1999: Massachusetts<br />
* 2000: California<br />
* 2001: Virginia<br />
* 2002: California<br />
* 2003: California<br />
* 2004: Illinois<br />
* 2005: Texas<br />
* 2006: Virginia<br />
* 2007: Texas<br />
* 2008: Texas<br />
* 2009: Texas<br />
* 2010: California<br />
* 2011: California<br />
* 2012: Massachusetts<br />
* 2013: Massachusetts<br />
* 2014: California<br />
* 2015: Indiana<br />
* 2016: Texas<br />
* 2017: Texas<br />
* 2018: Texas<br />
* 2019: Massachusetts<br />
<br />
== MATHCOUNTS Competition Structure ==<br />
<br />
=== Sprint Round ===<br />
<br />
30 problems are given all at once. Students have 40 minutes to complete the Sprint Round. This round is very fast-paced and requires speed and accuracy as well. The earlier problems are usually the easiest problems in the competition, and the later problems can be as hard as some of the Team Round questions. No calculators are allowed during this round.<br />
<br />
=== Target Round ===<br />
8 problems given 2 at a time. Students have 6 minutes to complete each set of two problems. Students may not go back to previous rounds (or forwards to future rounds) even if they finish before time is called. Calculators are allowed for the Target Round. Usually comprised of one "confidence booster" and one hard problem.<br />
<br />
=== Team Round ===<br />
<br />
10 problems in 20 minutes for a team of 4 students. These problems typically include some of the most difficult problems of the competition. Use of a calculator is allowed (and required for some questions).<br />
<br />
=== Countdown Round ===<br />
High scoring individuals compete head-to-head until a champion is crowned. People compete from off a screen taking 45 seconds or less to finish the problem. The Countdown round is run differently in various different chapter, state, and national competitions. In the national competitions, it is the round that determines the champion. Calculators are not allowed, but scratch paper will be provided.<br />
<br />
<br />
====Chapter and State Competitions====<br />
<br />
In the chapter and state competitions, the countdown round is not mandatory. However, if it is deemed official by the chapter or state, the following format must be used:<br />
<br />
*The 10th place written finisher competes against the 9th place written finisher. A problem is displayed, and both competitors have 45 seconds to answer the question, and the first competitor to correctly answer the question receives one point. The person who gets the most correct out of three questions (not necessarily two out of three) is the winner.<br />
<br />
*The winner of the first round goes up against the 8th place finisher.<br />
<br />
*The winner of the second round goes up against the 7th place finisher.<br />
<br />
This process is continued until the countdown round reaches the top four written competitors. Starting then, the first person to get three questions correct wins (as opposed to the best-out-of-three rule).<br />
<br />
If the countdown round is unofficial, any format may be used. Single-elimination bracket-style tournaments are common.<br />
<br />
====National Competition====<br />
<br />
At the national competition, there are some structural changes to the countdown round. The top 12 (not the top 10) written finishers make it to the countdown round, and the format is changed from a ladder competition to a single elimination tournament where the top four written competitors get a bye. This setup makes it far more likely for a 12th place finisher to become champion, and it makes it less likely for a first place written finisher to become champion, equalizing the field. But even then, a 12th place written competitor will have less of a chance to become champion than the top 4, because the top 4 get a bye. Until the semi-finals, the scoring is best out of five advances.<br />
<br />
At the first round and the second round, the person to correctly answer the most out of 5 questions wins. However, at the semifinals, the rules slightly change&mdash;the first person to correctly answer four questions wins.<br />
<br />
=== Ciphering Round ===<br />
In some states, (most notably Florida) there is an optional ciphering round. Very similar to countdown (in both difficulty and layout), a team sends up a representative to go against all representatives from the other teams. A problem is shown on a screen and students work fast to answer the problem. The students give their answer and after 45 seconds the answer is shown and the answers are checked to see if they are right. The fastest correct answer gets five points, the next fastest gets 4, etc. There are 4 questions per individual and teams send up 4 people. A perfect score is then 80. Often times the questions take clever reading skills. For example, one question was "How much dirt is in a 3 ft by 3 ft by 4 ft hole?" The answer was 0 because there is no dirt in a hole.<br />
<br />
=== Masters Round ===<br />
Top students give in-depth explanations to challenging problems. This round is optional at the state level competition and is mandatory at the national competition (up to 2011). At nationals the top two on the written and countdown participate. <br />
In 2012, it was replaced by the Reel Math Challenge (now called the Math Video Challenge).<br />
<br />
=== Scoring and Ranking ===<br />
An individual's score is their total number of correct sprint round answers plus 2 times their total number of correct target round answers. This total is out of a maximum of <math>30 + 2(8) = 46</math> points.<br />
<br />
A team's score is the individual scores of its members divided by 4 plus 2 points for every correct team round answer, making a team's maximum possible score 66 points. Therefore, it is possible to win with a relatively low team score and a phenomenal individual score, as the team score is only roughly 30% of the total team score. Note that when there are less than four members the score will become less.<br />
<br />
== MATHCOUNTS Competition Levels ==<br />
=== School Competition ===<br />
Students vie for the chance to make their school teams. Problems at this level are generally the easiest and most basic in curriculum.<br />
<br />
=== Chapter Competition ===<br />
Chapter competitions serve as a selection filter for state competitions. A few states don't need to host chapter competitions due to a small population size.<br />
<br />
=== State Competition ===<br />
The top 4 students in each state form the state team for the national competition. The coach of the top school team at the state level is invited to coach the state team at the national competition. Interestingly, the coach of a state team is not necessarily the coach of any of the state's team members.<br />
<br />
=== National Competition ===<br />
==== National Competition Sites ====<br />
For many years, the National MATHCOUNTS competition was held in Washington, D.C. More recently, the competition has changed venues often.<br />
<br />
* The 2019 competition was held in Orlando, Florida.<br />
* The 2018 competition was held in Washington, D.C.<br />
* The 2017 competition was held in Orlando, Florida.<br />
* The 2016 competition was held in Washington, D.C.<br />
* The 2015 competition was held in Boston, Massachusetts.<br />
* The 2014 competition was held in Orlando, Florida.<br />
* The 2013 competition was held in Washington, D.C.<br />
* The 2012 competition was held in Orlando, Florida.<br />
* The 2011 competition was held in Washington, D.C.<br />
* The 2009 and 2010 competitions were held in Orlando, Florida.<br />
* The 2008 competition was held in Denver, Colorado.<br />
* The 2007 competition was held in Fort Worth, Texas.<br />
* The 2006 competition was held in Arlington, Virginia.<br />
* The 2005 competition was held in Detroit, Michigan.<br />
* The 2004 competition was held in Washington, D.C.<br />
* The 2002 and 2003 competitions were held in Chicago, Illinois.<br />
<br />
== What comes after MATHCOUNTS? ==<br />
<br />
Give the following competitions a try and take a look at the [[List of United States high school mathematics competitions]].<br />
* [[American Mathematics Competitions]]<br />
* [[American Regions Math League]]<br />
* [[Mandelbrot Competition]]<br />
* [[Mu Alpha Theta]]<br />
<br />
[[Category:Mathematics competitions]]<br />
<br />
== See also... ==<br />
* [[List of national MATHCOUNTS teams]]<br />
* [[Mathematics competition resources]]<br />
* [[Math contest books]]<br />
* [[Math books]]<br />
* [[List of United States middle school mathematics competitions]]<br />
* [[List of United States high school mathematics competitions]]<br />
* [http://www.mathcounts.org/webarticles/anmviewer.asp?a=921&z=71 2006 MATHCOUNTS Countdown Video]<br />
<br />
[[Category:Introductory mathematics competitions]]</div>Puddles penguinhttps://artofproblemsolving.com/wiki/index.php?title=MATHCOUNTS&diff=105790MATHCOUNTS2019-05-13T19:14:01Z<p>Puddles penguin: /* Past State Team Winners */</p>
<hr />
<div>Many AoPS Community members and online school students have been participants at National MATHCOUNTS, including many Nationals Countdown Round participants in the past decade. '''MATHCOUNTS''' is a large national [[mathematics competition]] and [[mathematics coaching]] program that has served millions of middle school students since 1984. Sponsored by the [http://cna.com CNA Foundation], [http://www.nspe.org/ National Society of Professional Engineers], the [http://www.nctm.org/ National Council of Teachers of Mathematics], and others including Art of Problem Solving, the focus of MATHCOUNTS is on mathematical problem solving. Students are eligible for up to three years, but cannot compete beyond their eighth grade year.<br />
<br />
{{Contest Info|name=MATHCOUNTS|region=USA|type=Free Response|difficulty=0.5 - 2.5|breakdown=<u>Countdown</u>: 0.5 (School/Chapter), 1 (State/National)<br><u>Sprint</u>: 1-1.5 (School/Chapter), 2-2.5 (State/National)<br><u>Target:</u> 1.5 (School), 2 (Chapter), 2-2.5 (State/National)}}<br />
<br />
== MATHCOUNTS Resources ==<br />
=== MATHCOUNTS Books ===<br />
Art of Problem Solving's [http://artofproblemsolving.com/store/list/aops-curriculum Introductory subject textbooks] are ideal for students preparing for MATHCOUNTS, as are [http://artofproblemsolving.com/store/item/aops-vol1 AoPS Volume 1] and [http://artofproblemsolving.com/store/item/competition-math Competition Math for Middle School]<br />
<br />
=== MATHCOUNTS Classes ===<br />
Art of Problem Solving hosts a [http://artofproblemsolving.com/school/course/mathcounts-basics Basic] and an [http://artofproblemsolving.com/school/course/mathcounts-advanced Advanced] MATHCOUNTS course. The AoPS Introduction-level subject courses also include a great deal of MATHCOUNTS preparation. Many AoPS instructors are former National MATHCOUNTS Mathletes.<br />
<br />
=== MATHCOUNTS Online ===<br />
* [http://www.mathcounts.org Official MATHCOUNTS Homepage]<br />
* Art of Problem Solving hosts a large [http://artofproblemsolving.com/community/c3_middle_school_math Middle School Math Forum] as well as a private [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=23209 MATHCOUNTS Coaches Forum].<br />
* The AoPS MATHCOUNTS Trainer is available on the [http://artofproblemsolving.com/mathcounts_trainer AoPS website], as well as on the [https://itunes.apple.com/us/app/mathcounts-trainer-math-contest/id1023961880?ls=1&mt=8 iPhone and iPad].<br />
* The free [http://www.artofproblemsolving.com/alcumus AoPS Alcumus learning system] includes thousands of MATHCOUNTS problems.<br />
* [http://artofproblemsolving.com/ftw/ftw.php For the Win!] gives students free Countdown Round-like practice against other AoPS students.<br />
* AoPS founder Richard Rusczyk has created dozens of [http://artofproblemsolving.com/videos/mathcounts MATHCOUNTS Mini video lessons].<br />
* [http://mathcounts.saab.org/ Elias Saab's MATHCOUNTS Preparation Homepage]<br />
* [http://www.unidata.ucar.edu/staff/russ/mathcounts/diaz.html The MATHCOUNTS Bible According to Mr. Diaz]<br />
*[http://www.artofproblemsolving.com/Resources/AoPS_R_A_MATHCOUNTS.php/ Building a Successful MATHCOUNTS Program] by [[Jeff Boyd]], who coached the 2005, 2007, and 2008 National Champion [[Texas MathCounts]] team.<br />
*[http://mathweb.scranton.edu/monks/courses/ProblemSolving/MathCountsPlaybookBW.pdf Coach Monk's MathCounts Playbook]<br />
* MathCounts Minis make hard problems easy<br />
<br />
== MATHCOUNTS Curriculum ==<br />
MATHCOUNTS curriculum includes [[arithmetic]], [[algebra]], [[counting]], [[geometry]], [[number theory]], [[probability]], and [[statistics]]. The focus of MATHCOUNTS curriculum is in developing [[mathematical problem solving]] skills.<br />
<br />
Before 1990, MATHCOUNTS chose particular areas of mathematics to highlight each year before changing the focus of the competition more broadly to problem solving.<br />
<br />
== Past State Team Winners ==<br />
* 1984: Virginia<br />
* 1985: Florida<br />
* 1986: California<br />
* 1987: New York<br />
* 1988: New York<br />
* 1989: North Carolina<br />
* 1990: Ohio<br />
* 1991: Alabama<br />
* 1992: California<br />
* 1993: Kansas<br />
* 1994: Pennsylvania<br />
* 1995: Indiana<br />
* 1996: Wisconsin<br />
* 1997: Massachusetts<br />
* 1998: Wisconsin<br />
* 1999: Massachusetts<br />
* 2000: California<br />
* 2001: Virginia<br />
* 2002: California<br />
* 2003: California<br />
* 2004: Illinois<br />
* 2005: Texas<br />
* 2006: Virginia<br />
* 2007: Texas<br />
* 2008: Texas<br />
* 2009: Texas<br />
* 2010: California<br />
* 2011: California<br />
* 2012: Massachusetts<br />
* 2013: Massachusetts<br />
* 2014: California<br />
* 2015: Indiana<br />
* 2016: Texas<br />
* 2017: Texas<br />
* 2018: Texas<br />
* 2019: Massachusetts<br />
<br />
== MATHCOUNTS Competition Structure ==<br />
<br />
=== Sprint Round ===<br />
<br />
30 problems are given all at once. Students have 40 minutes to complete the Sprint Round. This round is very fast-paced and requires speed and accuracy as well. The earlier problems are usually the easiest problems in the competition, and the later problems can be as hard as some of the Team Round questions. No calculators are allowed during this round.<br />
<br />
=== Target Round ===<br />
8 problems given 2 at a time. Students have 6 minutes to complete each set of two problems. Students may not go back to previous rounds (or forwards to future rounds) even if they finish before time is called. Calculators are allowed for the Target Round. Usually comprised of one "confidence booster" and one hard problem.<br />
<br />
=== Team Round ===<br />
<br />
10 problems in 20 minutes for a team of 4 students. These problems typically include some of the most difficult problems of the competition. Use of a calculator is allowed (and required for some questions).<br />
<br />
=== Countdown Round ===<br />
High scoring individuals compete head-to-head until a champion is crowned. People compete from off a screen taking 45 seconds or less to finish the problem. The Countdown round is run differently in various different chapter, state, and national competitions. In the national competitions, it is the round that determines the champion. Calculators are not allowed, but scratch paper will be provided.<br />
<br />
<br />
====Chapter and State Competitions====<br />
<br />
In the chapter and state competitions, the countdown round is not mandatory. However, if it is deemed official by the chapter or state, the following format must be used:<br />
<br />
*The 10th place written finisher competes against the 9th place written finisher. A problem is displayed, and both competitors have 45 seconds to answer the question, and the first competitor to correctly answer the question receives one point. The person who gets the most correct out of three questions (not necessarily two out of three) is the winner.<br />
<br />
*The winner of the first round goes up against the 8th place finisher.<br />
<br />
*The winner of the second round goes up against the 7th place finisher.<br />
<br />
This process is continued until the countdown round reaches the top four written competitors. Starting then, the first person to get three questions correct wins (as opposed to the best-out-of-three rule).<br />
<br />
If the countdown round is unofficial, any format may be used. Single-elimination bracket-style tournaments are common.<br />
<br />
====National Competition====<br />
<br />
At the national competition, there are some structural changes to the countdown round. The top 12 (not the top 10) written finishers make it to the countdown round, and the format is changed from a ladder competition to a single elimination tournament where the top four written competitors get a bye. This setup makes it far more likely for a 12th place finisher to become champion, and it makes it less likely for a first place written finisher to become champion, equalizing the field. But even then, a 12th place written competitor will have less of a chance to become champion than the top 4, because the top 4 get a bye. Until the semi-finals, the scoring is best out of five advances.<br />
<br />
At the first round and the second round, the person to correctly answer the most out of 5 questions wins. However, at the semifinals, the rules slightly change&mdash;the first person to correctly answer four questions wins.<br />
<br />
=== Ciphering Round ===<br />
In some states, (most notably Florida) there is an optional ciphering round. Very similar to countdown (in both difficulty and layout), a team sends up a representative to go against all representatives from the other teams. A problem is shown on a screen and students work fast to answer the problem. The students give their answer and after 45 seconds the answer is shown and the answers are checked to see if they are right. The fastest correct answer gets five points, the next fastest gets 4, etc. There are 4 questions per individual and teams send up 4 people. A perfect score is then 80. Often times the questions take clever reading skills. For example, one question was "How much dirt is in a 3 ft by 3 ft by 4 ft hole?" The answer was 0 because there is no dirt in a hole.<br />
<br />
=== Masters Round ===<br />
Top students give in-depth explanations to challenging problems. This round is optional at the state level competition and is mandatory at the national competition (up to 2011). At nationals the top two on the written and countdown participate. <br />
In 2012, it was replaced by the Reel Math Challenge (now called the Math Video Challenge).<br />
<br />
=== Scoring and Ranking ===<br />
An individual's score is their total number of correct sprint round answers plus 2 times their total number of correct target round answers. This total is out of a maximum of <math>30 + 2(8) = 46</math> points.<br />
<br />
A team's score is the individual scores of its members divided by 4 plus 2 points for every correct team round answer, making a team's maximum possible score 66 points. Therefore, it is possible to win with a relatively low team score and a phenomenal individual score, as the team score is only roughly 30% of the total team score. Note that when there are less than four members the score will become less.<br />
<br />
== MATHCOUNTS Competition Levels ==<br />
=== School Competition ===<br />
Students vie for the chance to make their school teams. Problems at this level are generally the easiest and most basic in curriculum.<br />
<br />
=== Chapter Competition ===<br />
Chapter competitions serve as a selection filter for state competitions. A few states don't need to host chapter competitions due to a small population size.<br />
<br />
=== State Competition ===<br />
The top 4 students in each state form the state team for the national competition. The coach of the top school team at the state level is invited to coach the state team at the national competition. Interestingly, the coach of a state team is not necessarily the coach of any of the state's team members.<br />
<br />
=== National Competition ===<br />
==== National Competition Sites ====<br />
For many years, the National MATHCOUNTS competition was held in Washington, D.C. More recently, the competition has changed venues often.<br />
<br />
* The 2019 competition will be held in Orlando, Florida.<br />
* The 2018 competition was held in Washington, D.C.<br />
* The 2017 competition was held in Orlando, Florida.<br />
* The 2016 competition was held in Washington, D.C.<br />
* The 2015 competition was held in Boston, Massachusetts.<br />
* The 2014 competition was held in Orlando, Florida.<br />
* The 2013 competition was held in Washington, D.C.<br />
* The 2012 competition was held in Orlando, Florida.<br />
* The 2011 competition was held in Washington, D.C.<br />
* The 2009 and 2010 competitions were held in Orlando, Florida.<br />
* The 2008 competition was held in Denver, Colorado.<br />
* The 2007 competition was held in Fort Worth, Texas.<br />
* The 2006 competition was held in Arlington, Virginia.<br />
* The 2005 competition was held in Detroit, Michigan.<br />
* The 2004 competition was held in Washington, D.C.<br />
* The 2002 and 2003 competitions were held in Chicago, Illinois.<br />
<br />
== What comes after MATHCOUNTS? ==<br />
<br />
Give the following competitions a try and take a look at the [[List of United States high school mathematics competitions]].<br />
* [[American Mathematics Competitions]]<br />
* [[American Regions Math League]]<br />
* [[Mandelbrot Competition]]<br />
* [[Mu Alpha Theta]]<br />
<br />
[[Category:Mathematics competitions]]<br />
<br />
== See also... ==<br />
* [[List of national MATHCOUNTS teams]]<br />
* [[Mathematics competition resources]]<br />
* [[Math contest books]]<br />
* [[Math books]]<br />
* [[List of United States middle school mathematics competitions]]<br />
* [[List of United States high school mathematics competitions]]<br />
* [http://www.mathcounts.org/webarticles/anmviewer.asp?a=921&z=71 2006 MATHCOUNTS Countdown Video]<br />
<br />
[[Category:Introductory mathematics competitions]]</div>Puddles penguinhttps://artofproblemsolving.com/wiki/index.php?title=2019_AIME_I_Problems/Problem_8&diff=1045812019 AIME I Problems/Problem 82019-03-16T17:52:39Z<p>Puddles penguin: /* Solution 1 */</p>
<hr />
<div>==Problem 8==<br />
Let <math>x</math> be a real number such that <math>\sin^{10}x+\cos^{10} x = \tfrac{11}{36}</math>. Then <math>\sin^{12}x+\cos^{12} x = \tfrac{m}{n}</math> where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.<br />
<br />
==Solution 1==<br />
<br />
We can substitute <math>y = \sin^2{x}</math>. Since we know that <math>\cos^2{x}=1-\sin^2{x}</math>, we can do some simplification. <br />
<br />
This yields <math>y^5+(1-y)^5=\frac{11}{36}</math>. From this, we can substitute again to get some cancellation through binomials. If we let <math>z=1/2-y</math>, we can simplify the equation to <math>(1/2+z)^5+(1/2-z)^5=\frac{11}{36}</math>. After using binomial theorem, this simplifies to <math>\frac{1}{16}(80z^4+40z^2+1)=11/36</math>. If we use the quadratic theorem, we obtain that <math>z^2=\pm\frac{1}{12}</math>, so <math>z=\pm\frac{1}{2\sqrt{3}}</math>. By plugging z into <math>(1/2-z)^6+(1/2+z)^6</math> (which is equal to <math>\sin^{12}{x}+\cos^{12}{x}</math>, we can either use binomial theorem or sum of cubes to simplify, and we end up with <math>\frac{13}{54}</math>. Therefore, the answer is <math>\boxed{067}</math>.<br />
<br />
eric2020, inspired by Tommy2002<br />
<br />
==Solution 2==<br />
<br />
First, for simplicity, let <math>a=\sin{x}</math> and <math>b=\cos{x}</math>. Note that <math>a^2+b^2=1</math>. We then bash the rest of the problem out. Take the tenth power of this expression and get <math>a^{10}+b^{10}+5a^2b^2(a^6+b^6)+10a^4b^4(a^2+b^2)=\frac{11}{36}+5a^2b^2(a^6+b^6)+10a^4b^4=1</math>. Note that we also have <math>\frac{11}{36}=a^{10}+b^{10}=(a^{10}+b^{10})(a^2+b^2)=a^{12}+b^{12}+a^2b^2(a^8+b^8)</math>. So, it suffices to compute <math>a^2b^2(a^8+b^8)</math>. Let <math>y=a^2b^2</math>. We have from cubing <math>a^2+b^2=1</math> that <math>a^6+b^6+3a^2b^2(a^2+b^2)=1</math> or <math>a^6+b^6=1-3y</math>. Next, using <math>\frac{11}{36}+5a^2b^2(a^6+b^6)+10a^4b^4=1</math>, we get <math>a^2b^2(a^6+b^6)+2a^4b^4=\frac{5}{36}</math> or <math>y(1-3y)+2y^2=y-y^2=\frac{5}{36}</math>. Solving gives <math>y=\frac{5}{6}</math> or <math>y=\frac{1}{6}</math>. Clearly <math>y=\frac{5}{6}</math> is extraneous, so <math>y=\frac{1}{6}</math>. Now note that <math>a^4+b^4=(a^2+b^2)-2a^2b^2=\frac{2}{3}</math>, and <math>a^8+b^8=(a^4+b^4)^2-2a^4b^4=\frac{4}{9}-\frac{1}{18}=\frac{7}{18}</math>. Thus we finally get <math>a^{12}+b^{12}=\frac{11}{36}-\frac{7}{18}*\frac{1}{6}=\frac{13}{54}</math>, giving <math>\boxed{067}</math>.<br />
<br />
-Emathmaster<br />
<br />
==Solution 3 (Newton Sums)==<br />
Newton sums is basically constructing the powers of the roots of the polynomials instead of deconstructing them which was done in solution 2. Let <math>\sin^2x</math> and <math>\cos^2x</math> be the roots of some polynomial <math>F(a)</math>. Then, <math>F(a)=a^2-a+b</math> for some <math>b=\sin^2x\cdot\cos^2x</math>.<br />
<br />
Let <math>S_k=\left(\sin^2x\right)^k+\left(\cos^2x\right)^k</math>. We want to find <math>S_6</math>. Clearly <math>S_1=1</math> and <math>S_2=1-2b</math>. Newton sums tells us that <math>S_k-S_{k-1}+bS_{k-2}=0\Rightarrow S_k=S_{k-1}-bS_{k-2}</math> where <math>k\ge 3</math> for our polynomial <math>F(a)</math>. <br />
<br />
Bashing, we have<br />
<cmath>S_3=S_2-bS_1\Rightarrow S_3=(1-2b)-b(1)=1-3b</cmath><br />
<cmath>S_4=S_3-bS_2\Rightarrow S_4=(1-3b)-b(1-2b)=2b^2-4b+1</cmath><br />
<cmath>S_5=S_4-bS_3\Rightarrow S_5=(2b^2-4b+1)-b(1-3b)=5b^2-5b+1=\frac{11}{36}</cmath><br />
<br />
Thus <cmath>5b^2-5b+1=\frac{11}{36}\Rightarrow 5b^2-5b+\frac{25}{36}=0, 36b^2-36b+5=0, (6b-1)(6b-5)=0</cmath><br />
<math>b=\frac{1}{6} \text{ or } \frac{5}{6}</math>. Clearly, <math>\sin^2x\cdot\cos^2x\not=\frac{5}{6}</math> so <math>\sin^2x\cdot\cos^2x=b=\frac{1}{6}</math>.<br />
<br />
Note <math>S_4=\frac{7}{18}</math>. Solving for <math>S_6</math>, we get <math>S_6=S_5-\frac{1}{6}S_4=\frac{13}{54}</math>. Finally, <math>13+54=\boxed{067}</math>.<br />
<br />
==Solution 4 (Fun)==<br />
We let <math>a = \sin^2(x)</math> and <math>b = \cos^2(x)</math>, so we have <math>a+b=1</math> and <math>a^5 + b^5 = \frac{11}{36}</math>. Noticing that <math>ab</math> might be a useful value to find, we let <math>c = ab</math>. Then we can work our way up to find <math>c</math>.<br />
<cmath>a + b = 1</cmath><br />
<cmath>(a+b)(a+b) = 1(a+b)</cmath><br />
<cmath>a^2 + 2ab + b^2 = 1</cmath><br />
<cmath>a^2 + b^2 = -2c + 1</cmath><br />
<cmath>(a+b)(a^2 + b^2) = -2c + 1</cmath><br />
<cmath>a^3 + ab^2 + a^2b + b^3 = -2c + 1</cmath><br />
<cmath>a^3 + b^3 + ab(a + b) = -2c + 1</cmath><br />
<cmath>a^3 + b^3 + c = -2c + 1</cmath><br />
<cmath>a^3 + b^3 = -3c + 1</cmath><br />
<cmath>(a + b)(a^3 + b^3) = -3c + 1</cmath><br />
<cmath>a^4 + ab^3 + a^3b + b^4 = -3c + 1</cmath><br />
<cmath>a^4 + b^4 + ab(a^2 + b^2) = -3c + 1</cmath><br />
<cmath>a^4 + b^4 + c(-2c + 1) = -3c + 1</cmath><br />
<cmath>a^4 + b^4 - 2c^2 + c = -3c + 1</cmath><br />
<cmath>a^4 + b^4 = 2c^2 - 4c + 1</cmath><br />
<cmath>a^5 + ab^4 + a^4b + b^5 = 2c^2 - 4c + 1</cmath><br />
<cmath>a^5 + b^5 + ab(a^3 + b^3) = 2c^2 - 4c + 1</cmath><br />
<cmath>\frac{11}{36} + c(-3c + 1) = 2c^2 - 4c + 1</cmath><br />
<cmath>\frac{11}{36} = 5c^2 - 5c + 1</cmath><br />
<cmath>5c^2 - 5c + \frac{25}{36} = 0</cmath><br />
using quadform you get <math>c = \frac{1}{6}</math> or <math>c = \frac{5}{6}</math>. Since <math>c = \sin^2(x)\cos^2(x) = (\sin(x)\cos(x))^2 = (\frac{\sin(2x)}{2})^2</math>, and since <math>\sin(2x)</math> can't exceed 1, <math>c</math> can't exceed <math>(\frac{1}{2})^2 = \frac{1}{4}</math>. Clearly, <math>c = \frac{1}{6}</math>. And finally,<br />
<cmath>a^5 + b^5 = (a + b)(a^5 + b^5)</cmath><br />
<cmath>a^5 + b^5 = a^6 + ab^5 + a^5b + b^6</cmath><br />
<cmath>\frac{11}{36} = a^6 + b^6 + c(a^4 + b^4)</cmath><br />
looking back to previous results, we see that <math>a^4 + b^4 = 2c^2 - 4c + 1 = \frac{14}{36}</math> (it's easier not to simplify the fraction).<br />
<cmath>\frac{11}{36} = a^6 + b^6 + c(\frac{14}{36}) = a^6 + b^6 + \frac{14}{216}</cmath><br />
<cmath>a^6 + b^6 = \frac{11}{36} - \frac{14}{216} = \frac{13}{54}</cmath><br />
which yields the answer <math>\boxed{067}</math>.<br />
<br />
~PCampbell<br />
<br />
==See Also==<br />
{{AIME box|year=2019|n=I|num-b=7|num-a=9}}<br />
{{MAA Notice}}</div>Puddles penguinhttps://artofproblemsolving.com/wiki/index.php?title=2011_AIME_I_Problems/Problem_4&diff=1037822011 AIME I Problems/Problem 42019-02-24T01:36:55Z<p>Puddles penguin: /* Solution 2 */</p>
<hr />
<div>== Problem 4 ==<br />
In triangle <math>ABC</math>, <math>AB=125</math>, <math>AC=117</math> and <math>BC=120</math>. The angle bisector of angle <math>A</math> intersects <math> \overline{BC} </math> at point <math>L</math>, and the angle bisector of angle <math>B</math> intersects <math> \overline{AC} </math> at point <math>K</math>. Let <math>M</math> and <math>N</math> be the feet of the perpendiculars from <math>C</math> to <math> \overline{BK}</math> and <math> \overline{AL}</math>, respectively. Find <math>MN</math>.<br />
<br />
<br />
== Solution 1 == <br />
Extend <math>{CM}</math> and <math>{CN}</math> such that they intersect line <math>{AB}</math> at points <math>P</math> and <math>Q</math>, respectively. <br />
Since <math>{BM}</math> is the angle bisector of angle B, and <math>{CM}</math> is perpendicular to <math>{BM}</math>, so <math>BP=BC=120</math>, M is the midpoint of <math>{CP}</math>. For the same reason, <math>AQ=AC=117</math>,N is the midpoint of <math>{CQ}</math>.<br />
Hence <math>MN=\frac{PQ}{2}</math>. But <math>PQ=BP+AQ-AB=120+117-125=112</math>, so <math>MN=\boxed{056}</math>.<br />
<br />
== Solution 2 ==<br />
[There seem to be some mislabeled points going on here but the idea is sound.]<br />
Let <math>I</math> be the incenter of <math>ABC</math>. Now, since <math>AM \perp LC</math> and <math>AN \perp KB</math>, we have <math>AMIN</math> is a cyclic quadrilateral. Consequently, <math>\frac{MN}{\sin \angle MIN} = 2R = AI</math>. Since <math>\angle MIN = 90 - \frac{\angle BAC}{2} = \cos \angle IAK</math>, we have that <math>MN = AI \cdot \cos \angle IAK</math>. Letting <math>X</math> be the point of contact of the incircle of <math>ABC</math> with side <math>AC</math>, we have <math>AX=MN</math> thus <math>MN=\frac{117+120-125}{2}=\boxed{056}</math><br />
<br />
== Solution 3 (Bash) == <br />
Project <math>I</math> onto <math>AC</math> and <math>BC</math> as <math>D</math> and <math>E</math>. <math>ID</math> and <math>IE</math> are both in-radii of <math>\triangle ABC</math> so we get right triangles with legs <math>r</math> (the in-radius length) and <math>s - c = 56</math>. Since <math>IC</math> is the hypotenuse for the 4 triangles (<math>\triangle INC, \triangle IMC, \triangle IDC,</math> and <math>\triangle IEC</math>), <math>C, D, M, I, N, E</math> are con-cyclic on a circle we shall denote as <math>\omega</math> which is also the circumcircle of <math>\triangle CMN</math> and <math>\triangle CDE</math>. To find <math>MN</math>, we can use the Law of Cosines on <math>\angle MON \implies MN^2 = 2R^2(1 - \cos{2\angle MCN})</math> where <math>O</math> is the center of <math>\omega</math>. Now, the circumradius <math>R</math> can be found with Pythagorean Theorem with <math>\triangle CDI</math> or <math>\triangle CEI</math>: <math>r^2 + 56^2 = (2R)^2</math>. To find <math>r</math>, we can use the formula <math>rs = [ABC]</math> and by Heron's, <math>[ABC] = \sqrt{181 \cdot 61 \cdot 56 \cdot 64} \implies r = \sqrt{\frac{61 \cdot 56 \cdot 64}{181}} \implies 2R^2 = \frac{393120}{181}</math>. To find <math>\angle MCN</math>, we can find <math>\angle MIN</math> since <math>\angle MCN = 180 - \angle MIN</math>. <math>\angle MIN = \angle MIC + \angle NIC = 180 - \angle BIC + 180 - \angle AIC = 180 - (180 - \frac{\angle A + \angle C}{2}) + 180 - (180 - \frac{\angle B + \angle C}{2}) = \frac{\angle A + \angle B + \angle C}{2} + \frac{\angle C}{2}</math>. Thus, <math>\angle MCN = 180 - \frac{\angle A + \angle B + \angle C}{2} - \frac{\angle C}{2}</math> and since <math>\angle A + \angle B + \angle C = 180</math>, we have <math>\angle A + \angle B + \angle C - \frac{\angle A + \angle B + \angle C}{2} - \frac{\angle C}{2} = \frac{\angle A + \angle B}{2}</math>. Plugging this into our Law of Cosines formula gives <math>MN^2 = 2R^2(1 - \cos{\angle A + \angle B}) = 2R^2(1 + \cos{\angle C})</math>. To find <math>\cos{\angle C}</math>, we use LoC on <math>\triangle ABC \implies cos{\angle C} = \frac{120^2 + 117^2 - 125^2}{2 \cdot 117 \cdot 120} = \frac{41 \cdot 19}{117 \cdot 15}</math>. Our formula now becomes <math>MN^2 = \frac{393120}{181} + \frac{2534}{15 \cdot 117}</math>. After simplifying, we get <math>MN^2 = 3136 \implies MN = \boxed{056}</math>.<br />
<br />
--lucasxia01<br />
<br />
== Solution 4==<br />
<br />
Because <math>\angle CMI = \angle CNI = 90</math>, <math>CMIN</math> is cyclic. <br />
<br />
Ptolemy on CMIN:<br />
<br />
<math>CN*MI+CM*IN=CI*MN</math><br />
<br />
<math>CI^2(\cos \angle ICN \sin \angle ICM + \cos \angle ICM \sin \angle ICN) = CI * MN</math><br />
<br />
<math>MN = CI \sin \angle MCN</math> by angle addition formula.<br />
<br />
<math>\angle MCN = 180 - \angle MIN = 90 - \angle BCI</math>. <br />
<br />
Let <math>H</math> be where the incircle touches <math>BC</math>, then <math>CI \cos \angle BCI = CH = \frac{a+b-c}{2}</math>.<br />
<math>a=120, b=117, c=125</math>, for a final answer of <math>\boxed{056}</math>.<br />
<br />
== See also ==<br />
{{AIME box|year=2011|n=I|num-b=3|num-a=5}}<br />
{{MAA Notice}}</div>Puddles penguinhttps://artofproblemsolving.com/wiki/index.php?title=2011_AIME_I_Problems/Problem_4&diff=1037812011 AIME I Problems/Problem 42019-02-24T01:36:28Z<p>Puddles penguin: /* Solution 1 */</p>
<hr />
<div>== Problem 4 ==<br />
In triangle <math>ABC</math>, <math>AB=125</math>, <math>AC=117</math> and <math>BC=120</math>. The angle bisector of angle <math>A</math> intersects <math> \overline{BC} </math> at point <math>L</math>, and the angle bisector of angle <math>B</math> intersects <math> \overline{AC} </math> at point <math>K</math>. Let <math>M</math> and <math>N</math> be the feet of the perpendiculars from <math>C</math> to <math> \overline{BK}</math> and <math> \overline{AL}</math>, respectively. Find <math>MN</math>.<br />
<br />
<br />
== Solution 1 == <br />
Extend <math>{CM}</math> and <math>{CN}</math> such that they intersect line <math>{AB}</math> at points <math>P</math> and <math>Q</math>, respectively. <br />
Since <math>{BM}</math> is the angle bisector of angle B, and <math>{CM}</math> is perpendicular to <math>{BM}</math>, so <math>BP=BC=120</math>, M is the midpoint of <math>{CP}</math>. For the same reason, <math>AQ=AC=117</math>,N is the midpoint of <math>{CQ}</math>.<br />
Hence <math>MN=\frac{PQ}{2}</math>. But <math>PQ=BP+AQ-AB=120+117-125=112</math>, so <math>MN=\boxed{056}</math>.<br />
<br />
== Solution 2 ==<br />
[There seem to be some mislabeled points going on here but the idea is sound.]<br />
Let <math>I</math> be the incenter of <math>ABC</math>. Now, since <math>AM \perp LC</math> and <math>AN \perp KB</math>, we have <math>AMIN</math> is a cyclic quadrilateral. Consequently, <math>\frac{MN}{\sin \angle MIN} = 2R = AI</math>. Since <math>\angle MIN = 90 - \frac{\angle BAC}{2} = \cos \angle IAK</math>, we have that <math>MN = AI \cdot \cos \angle IAK</math>. Letting <math>X</math> be the point of contact of the incircle of <math>ABC</math> with side <math>AC</math>, we have <math>AX=MN</math> thus <math>MN=\frac{117+120-125}{2}=\boxed{56}</math><br />
<br />
== Solution 3 (Bash) == <br />
Project <math>I</math> onto <math>AC</math> and <math>BC</math> as <math>D</math> and <math>E</math>. <math>ID</math> and <math>IE</math> are both in-radii of <math>\triangle ABC</math> so we get right triangles with legs <math>r</math> (the in-radius length) and <math>s - c = 56</math>. Since <math>IC</math> is the hypotenuse for the 4 triangles (<math>\triangle INC, \triangle IMC, \triangle IDC,</math> and <math>\triangle IEC</math>), <math>C, D, M, I, N, E</math> are con-cyclic on a circle we shall denote as <math>\omega</math> which is also the circumcircle of <math>\triangle CMN</math> and <math>\triangle CDE</math>. To find <math>MN</math>, we can use the Law of Cosines on <math>\angle MON \implies MN^2 = 2R^2(1 - \cos{2\angle MCN})</math> where <math>O</math> is the center of <math>\omega</math>. Now, the circumradius <math>R</math> can be found with Pythagorean Theorem with <math>\triangle CDI</math> or <math>\triangle CEI</math>: <math>r^2 + 56^2 = (2R)^2</math>. To find <math>r</math>, we can use the formula <math>rs = [ABC]</math> and by Heron's, <math>[ABC] = \sqrt{181 \cdot 61 \cdot 56 \cdot 64} \implies r = \sqrt{\frac{61 \cdot 56 \cdot 64}{181}} \implies 2R^2 = \frac{393120}{181}</math>. To find <math>\angle MCN</math>, we can find <math>\angle MIN</math> since <math>\angle MCN = 180 - \angle MIN</math>. <math>\angle MIN = \angle MIC + \angle NIC = 180 - \angle BIC + 180 - \angle AIC = 180 - (180 - \frac{\angle A + \angle C}{2}) + 180 - (180 - \frac{\angle B + \angle C}{2}) = \frac{\angle A + \angle B + \angle C}{2} + \frac{\angle C}{2}</math>. Thus, <math>\angle MCN = 180 - \frac{\angle A + \angle B + \angle C}{2} - \frac{\angle C}{2}</math> and since <math>\angle A + \angle B + \angle C = 180</math>, we have <math>\angle A + \angle B + \angle C - \frac{\angle A + \angle B + \angle C}{2} - \frac{\angle C}{2} = \frac{\angle A + \angle B}{2}</math>. Plugging this into our Law of Cosines formula gives <math>MN^2 = 2R^2(1 - \cos{\angle A + \angle B}) = 2R^2(1 + \cos{\angle C})</math>. To find <math>\cos{\angle C}</math>, we use LoC on <math>\triangle ABC \implies cos{\angle C} = \frac{120^2 + 117^2 - 125^2}{2 \cdot 117 \cdot 120} = \frac{41 \cdot 19}{117 \cdot 15}</math>. Our formula now becomes <math>MN^2 = \frac{393120}{181} + \frac{2534}{15 \cdot 117}</math>. After simplifying, we get <math>MN^2 = 3136 \implies MN = \boxed{056}</math>.<br />
<br />
--lucasxia01<br />
<br />
== Solution 4==<br />
<br />
Because <math>\angle CMI = \angle CNI = 90</math>, <math>CMIN</math> is cyclic. <br />
<br />
Ptolemy on CMIN:<br />
<br />
<math>CN*MI+CM*IN=CI*MN</math><br />
<br />
<math>CI^2(\cos \angle ICN \sin \angle ICM + \cos \angle ICM \sin \angle ICN) = CI * MN</math><br />
<br />
<math>MN = CI \sin \angle MCN</math> by angle addition formula.<br />
<br />
<math>\angle MCN = 180 - \angle MIN = 90 - \angle BCI</math>. <br />
<br />
Let <math>H</math> be where the incircle touches <math>BC</math>, then <math>CI \cos \angle BCI = CH = \frac{a+b-c}{2}</math>.<br />
<math>a=120, b=117, c=125</math>, for a final answer of <math>\boxed{056}</math>.<br />
<br />
== See also ==<br />
{{AIME box|year=2011|n=I|num-b=3|num-a=5}}<br />
{{MAA Notice}}</div>Puddles penguinhttps://artofproblemsolving.com/wiki/index.php?title=2011_AIME_I_Problems/Problem_4&diff=1037802011 AIME I Problems/Problem 42019-02-24T01:35:58Z<p>Puddles penguin: /* Solution 3 (Bash) */</p>
<hr />
<div>== Problem 4 ==<br />
In triangle <math>ABC</math>, <math>AB=125</math>, <math>AC=117</math> and <math>BC=120</math>. The angle bisector of angle <math>A</math> intersects <math> \overline{BC} </math> at point <math>L</math>, and the angle bisector of angle <math>B</math> intersects <math> \overline{AC} </math> at point <math>K</math>. Let <math>M</math> and <math>N</math> be the feet of the perpendiculars from <math>C</math> to <math> \overline{BK}</math> and <math> \overline{AL}</math>, respectively. Find <math>MN</math>.<br />
<br />
<br />
== Solution 1 == <br />
Extend <math>{CM}</math> and <math>{CN}</math> such that they intersect line <math>{AB}</math> at points <math>P</math> and <math>Q</math>, respectively. <br />
Since <math>{BM}</math> is the angle bisector of angle B, and <math>{CM}</math> is perpendicular to <math>{BM}</math>, so <math>BP=BC=120</math>, M is the midpoint of <math>{CP}</math>. For the same reason, <math>AQ=AC=117</math>,N is the midpoint of <math>{CQ}</math>.<br />
Hence <math>MN=\frac{PQ}{2}</math>. But <math>PQ=BP+AQ-AB=120+117-125=112</math>, so <math>MN=\boxed{56}</math>.<br />
<br />
== Solution 2 ==<br />
[There seem to be some mislabeled points going on here but the idea is sound.]<br />
Let <math>I</math> be the incenter of <math>ABC</math>. Now, since <math>AM \perp LC</math> and <math>AN \perp KB</math>, we have <math>AMIN</math> is a cyclic quadrilateral. Consequently, <math>\frac{MN}{\sin \angle MIN} = 2R = AI</math>. Since <math>\angle MIN = 90 - \frac{\angle BAC}{2} = \cos \angle IAK</math>, we have that <math>MN = AI \cdot \cos \angle IAK</math>. Letting <math>X</math> be the point of contact of the incircle of <math>ABC</math> with side <math>AC</math>, we have <math>AX=MN</math> thus <math>MN=\frac{117+120-125}{2}=\boxed{56}</math><br />
<br />
== Solution 3 (Bash) == <br />
Project <math>I</math> onto <math>AC</math> and <math>BC</math> as <math>D</math> and <math>E</math>. <math>ID</math> and <math>IE</math> are both in-radii of <math>\triangle ABC</math> so we get right triangles with legs <math>r</math> (the in-radius length) and <math>s - c = 56</math>. Since <math>IC</math> is the hypotenuse for the 4 triangles (<math>\triangle INC, \triangle IMC, \triangle IDC,</math> and <math>\triangle IEC</math>), <math>C, D, M, I, N, E</math> are con-cyclic on a circle we shall denote as <math>\omega</math> which is also the circumcircle of <math>\triangle CMN</math> and <math>\triangle CDE</math>. To find <math>MN</math>, we can use the Law of Cosines on <math>\angle MON \implies MN^2 = 2R^2(1 - \cos{2\angle MCN})</math> where <math>O</math> is the center of <math>\omega</math>. Now, the circumradius <math>R</math> can be found with Pythagorean Theorem with <math>\triangle CDI</math> or <math>\triangle CEI</math>: <math>r^2 + 56^2 = (2R)^2</math>. To find <math>r</math>, we can use the formula <math>rs = [ABC]</math> and by Heron's, <math>[ABC] = \sqrt{181 \cdot 61 \cdot 56 \cdot 64} \implies r = \sqrt{\frac{61 \cdot 56 \cdot 64}{181}} \implies 2R^2 = \frac{393120}{181}</math>. To find <math>\angle MCN</math>, we can find <math>\angle MIN</math> since <math>\angle MCN = 180 - \angle MIN</math>. <math>\angle MIN = \angle MIC + \angle NIC = 180 - \angle BIC + 180 - \angle AIC = 180 - (180 - \frac{\angle A + \angle C}{2}) + 180 - (180 - \frac{\angle B + \angle C}{2}) = \frac{\angle A + \angle B + \angle C}{2} + \frac{\angle C}{2}</math>. Thus, <math>\angle MCN = 180 - \frac{\angle A + \angle B + \angle C}{2} - \frac{\angle C}{2}</math> and since <math>\angle A + \angle B + \angle C = 180</math>, we have <math>\angle A + \angle B + \angle C - \frac{\angle A + \angle B + \angle C}{2} - \frac{\angle C}{2} = \frac{\angle A + \angle B}{2}</math>. Plugging this into our Law of Cosines formula gives <math>MN^2 = 2R^2(1 - \cos{\angle A + \angle B}) = 2R^2(1 + \cos{\angle C})</math>. To find <math>\cos{\angle C}</math>, we use LoC on <math>\triangle ABC \implies cos{\angle C} = \frac{120^2 + 117^2 - 125^2}{2 \cdot 117 \cdot 120} = \frac{41 \cdot 19}{117 \cdot 15}</math>. Our formula now becomes <math>MN^2 = \frac{393120}{181} + \frac{2534}{15 \cdot 117}</math>. After simplifying, we get <math>MN^2 = 3136 \implies MN = \boxed{056}</math>.<br />
<br />
--lucasxia01<br />
<br />
== Solution 4==<br />
<br />
Because <math>\angle CMI = \angle CNI = 90</math>, <math>CMIN</math> is cyclic. <br />
<br />
Ptolemy on CMIN:<br />
<br />
<math>CN*MI+CM*IN=CI*MN</math><br />
<br />
<math>CI^2(\cos \angle ICN \sin \angle ICM + \cos \angle ICM \sin \angle ICN) = CI * MN</math><br />
<br />
<math>MN = CI \sin \angle MCN</math> by angle addition formula.<br />
<br />
<math>\angle MCN = 180 - \angle MIN = 90 - \angle BCI</math>. <br />
<br />
Let <math>H</math> be where the incircle touches <math>BC</math>, then <math>CI \cos \angle BCI = CH = \frac{a+b-c}{2}</math>.<br />
<math>a=120, b=117, c=125</math>, for a final answer of <math>\boxed{056}</math>.<br />
<br />
== See also ==<br />
{{AIME box|year=2011|n=I|num-b=3|num-a=5}}<br />
{{MAA Notice}}</div>Puddles penguinhttps://artofproblemsolving.com/wiki/index.php?title=User:Puddles_penguin&diff=100182User:Puddles penguin2019-01-08T01:22:57Z<p>Puddles penguin: Created page with "An aops user"</p>
<hr />
<div>An aops user</div>Puddles penguinhttps://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10B_Problems/Problem_2&diff=993312009 AMC 10B Problems/Problem 22018-12-08T20:27:45Z<p>Puddles penguin: /* Solution */</p>
<hr />
<div>== Problem ==<br />
<br />
Which of the following is equal to <math>\dfrac{\frac{1}{3}-\frac{1}{4}}{\frac{1}{2}-\frac{1}{3}}</math>?<br />
<br />
<math><br />
\text{(A) } \frac 14<br />
\qquad<br />
\text{(B) } \frac 13<br />
\qquad<br />
\text{(C) } \frac 12<br />
\qquad<br />
\text{(D) } \frac 23<br />
\qquad<br />
\text{(E) } \frac 34<br />
</math><br />
<br />
== Solution ==<br />
<br />
Multiplying the numerator and the denominator by the same value does not change the value of the fraction.<br />
We can multiply both by <math>12</math>, getting <math>\dfrac{4-3}{6-4} = \boxed{\dfrac 12}</math>.<br />
<br />
Alternately, we can directly compute that the numerator is <math>\dfrac 1{12}</math>, the denominator is <math>\dfrac 16</math>, and hence their ratio is <math>\boxed{(C)\frac {1} {2}}</math>.<br />
<br />
== See Also ==<br />
<br />
{{AMC10 box|year=2009|ab=B|num-b=1|num-a=3}}<br />
{{MAA Notice}}</div>Puddles penguinhttps://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10A_Problems/Problem_22&diff=989872010 AMC 10A Problems/Problem 222018-11-24T00:17:46Z<p>Puddles penguin: /* Solution 2 (Using the Answer Choices) */</p>
<hr />
<div>==Problem==<br />
Eight points are chosen on a circle, and chords are drawn connecting every pair of points. No three chords intersect in a single point inside the circle. How many triangles with all three vertices in the interior of the circle are created?<br />
<br />
<math>\textbf{(A)}\ 28 \qquad \textbf{(B)}\ 56 \qquad \textbf{(C)}\ 70 \qquad \textbf{(D)}\ 84 \qquad \textbf{(E)}\ 140</math><br />
<br />
==Solution 1==<br />
<br />
To choose a chord, we know that two points must be chosen. This implies that for three chords to create a triangle and not intersect at a single point, six points need to be chosen. We also know that for any six points we pick, there is only 1 way to connect the chords such that a triangle is formed in the circle's interior. Therefore, the answer is <math>{{8}\choose{6}}</math>, which is equivalent to <math>\boxed{\textbf{(A) }28}</math>.<br />
<br />
==Solution 2 (Using the Answer Choices)==<br />
There are a total of <math>\dbinom{8}{3}=56</math> total triangles that can be made out of these chords. We know that the amount of triangles which have all their vertices inside the circle has to be less than this, to the answer can only be <math>\boxed{\textbf{(A) }28}</math>.<br />
In general, the number of triangles that are in the interior will always be <math>\binom{n}{6}</math> (can you prove it?)<br />
<br />
==See Also==<br />
{{AMC10 box|year=2010|ab=A|num-b=21|num-a=23}}<br />
{{MAA Notice}}</div>Puddles penguinhttps://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10A_Problems/Problem_22&diff=989852010 AMC 10A Problems/Problem 222018-11-24T00:16:28Z<p>Puddles penguin: /* Solution 2 (Using the Answer Choices) */</p>
<hr />
<div>==Problem==<br />
Eight points are chosen on a circle, and chords are drawn connecting every pair of points. No three chords intersect in a single point inside the circle. How many triangles with all three vertices in the interior of the circle are created?<br />
<br />
<math>\textbf{(A)}\ 28 \qquad \textbf{(B)}\ 56 \qquad \textbf{(C)}\ 70 \qquad \textbf{(D)}\ 84 \qquad \textbf{(E)}\ 140</math><br />
<br />
==Solution 1==<br />
<br />
To choose a chord, we know that two points must be chosen. This implies that for three chords to create a triangle and not intersect at a single point, six points need to be chosen. We also know that for any six points we pick, there is only 1 way to connect the chords such that a triangle is formed in the circle's interior. Therefore, the answer is <math>{{8}\choose{6}}</math>, which is equivalent to <math>\boxed{\textbf{(A) }28}</math>.<br />
<br />
==Solution 2 (Using the Answer Choices)==<br />
There are a total of <math>\dbinom{8}{3}=56</math> total triangles that can be made out of these chords. We know that the amount of triangles which have all their vertices inside the circle has to be less than this, to the answer can only be <math>\boxed{\textbf{(A) }28}</math>.<br />
<br />
==See Also==<br />
{{AMC10 box|year=2010|ab=A|num-b=21|num-a=23}}<br />
{{MAA Notice}}</div>Puddles penguin