https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Pi37&feedformat=atomAoPS Wiki - User contributions [en]2024-03-28T19:45:34ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2014_USAMO_Problems/Problem_5&diff=667162014 USAMO Problems/Problem 52014-12-30T21:29:14Z<p>Pi37: /* Solution */</p>
<hr />
<div>==Problem==<br />
Let <math>ABC</math> be a triangle with orthocenter <math>H</math> and let <math>P</math> be the second intersection of the circumcircle of triangle <math>AHC</math> with the internal bisector of the angle <math>\angle BAC</math>. Let <math>X</math> be the circumcenter of triangle <math>APB</math> and <math>Y</math> the orthocenter of triangle <math>APC</math>. Prove that the length of segment <math>XY</math> is equal to the circumradius of triangle <math>ABC</math>.<br />
<br />
==Solution==<br />
Let <math>O_1</math> be the center of <math>(AHPC)</math>, <math>O</math> be the center of <math>(ABC)</math>. Note that <math>(O_1)</math> is the reflection of <math>(O)</math> across <math>AC</math>, so <math>AO=AO_1</math>. Additionally<br />
<cmath><br />
\angle AYC=180-\angle APC=180-\angle AHC=\angle B<br />
</cmath><br />
so <math>Y</math> lies on <math>(O)</math>. Now since <math>XO,OO_1,XO_1</math> are perpendicular to <math>AB,AC,</math> and their bisector, <math>XOO_1</math> is isosceles with <math>XO=OO_1</math>, and <math>\angle XOO_1=180-\angle A</math>. Also<br />
<cmath><br />
\angle AOY=2\angle ACY=2(90-\angle PAC)=2(90-\frac{A}{2})=180-\angle A = \angle XOO_1<br />
</cmath><br />
But <math>YO=OA</math> as well, and <math>\angle YOX=\angle AOO_1</math>, so <math>\triangle OYX\cong \triangle OAO_1</math>. Thus <math>XY=AO_1=AO</math>.</div>Pi37https://artofproblemsolving.com/wiki/index.php?title=Geometry/Resources&diff=63306Geometry/Resources2014-09-08T22:34:21Z<p>Pi37: /* Olympiad */</p>
<hr />
<div>Listed below are various geometry resources including books, classes, websites, and computer software.<br />
<br />
== Books ==<br />
===Introductory===<br />
* [http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?page_id=9 The Art of Problem Solving Introduction to Geometry] by Richard Rusczyk<br />
===Intermediate===<br />
* [http://www.amazon.com/exec/obidos/ASIN/0486691543/artofproblems-20 Challenging Problems in Geometry] by Salkind Posamentier<br />
===Olympiad===<br />
* [http://www.amazon.com/Advanced-Euclidean-Geometry-Dover-Mathematics/dp/0486462374 Advanced Euclidean Geometry]by Roger Johnson<br />
* [http://www.amazon.com/Elementary-Vector-Geometry-Seymour-Schuster/dp/0486466728/ref=pd_bbs_sr_1?ie=UTF8&s=books&qid=1217986398&sr=8-1 Elementary Vector Geometry]<br />
* [http://www.keypress.com/catalog/products/supplementals/Prod_AdvancedEuclidean.html Advanced Euclidean Geometry] by Salkind Posamentier<br />
* [http://www.amazon.com/exec/obidos/ASIN/0883856190/artofproblems-20 Geometry Revisited] by Coxeter and Greitzer<br />
* [http://www-math.mit.edu/~kedlaya/geometryunbound/ Geometry Unbound] by Kiran Kedlaya<br />
<br />
== Websites ==<br />
All of these links are outside of [AoPSWiki].<br />
* [http://www.cut-the-knot.org/geometry.shtml Cut-the-Knot's Geometry Section]<br />
* [http://www.ics.uci.edu/~eppstein/junkyard/ The Geometry Junkyard]<br />
* [http://www.artofproblemsolving.com/Forum/index.php?f=4 AoPS-ML Olympiad Geometry Forum]<br />
<br />
== Classes ==<br />
===Introductory===<br />
* The [http://www.artofproblemsolving.com/Classes/classdetails.php?course_id=10 Introduction to Geometry] class is where you can learn introductory geometry.<br />
<br />
=== Olympiad ===<br />
* The [http://www.artofproblemsolving.com/Classes/AoPS_C_WOOT.php Worldwide Online Olympiad Training Program] is designed to help students learn to tackle [[mathematical Olympiad]] problems in topics such as geometry.<br />
<br />
== Software ==<br />
These are all outside links.<br />
* [http://www.geometer.org/geometer/ Geometer]<br />
* [http://www.keypress.com/sketchpad/ The Geometer's Sketchpad]<br />
* [http://www.mit.edu/~ibaran/kseg.html KSEG]<br />
* [http://edu.kde.org/kig/ KIG]<br />
* [http://asymptote.sourceforge.net/ Asymptote]<br />
<br />
== EBooks ==<br />
* Viktor Prasolov - [http://students.imsa.edu/~tliu/Math/planegeo.pdf Plane Geometry]<br />
* Hojoo Lee - [http://myhome.personaldb.net/ideahitme/trig.pdf Triangle Geometry Formulas]<br />
<br />
== See also ==<br />
* [[Geometry]]<br />
* [[Geometry/Introduction | Introductory Geometry]]<br />
* [[Geometry/Intermediate | Intermediate Geometry]]<br />
* [[Geometry/Olympiad | Olympiad Geometry]]<br />
[[Category:Academic resources]]</div>Pi37https://artofproblemsolving.com/wiki/index.php?title=AoPS_Wiki:FAQ&diff=56992AoPS Wiki:FAQ2013-08-27T00:11:46Z<p>Pi37: /* What do some of the acronyms such as "OP" stand for? */</p>
<hr />
<div>{{shortcut|[[A:FAQ]]}}<br />
<br />
This is a community created list of Frequently Asked Questions about Art of Problem Solving. If you have a request to edit or add a question on this page, please make it [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=416&t=414129 here].<br />
<br />
== General==<br />
<br />
<br />
==== Can I change my user name? ====<br />
<br />
:As indicated during the time of your registration, you are unable to change your username. <br />
::[[File:Registration.jpg]]<br />
<br />
====What software does Art of Problem Solving use to run the website?====<br />
<br />
:* Forums: phpBB3<br />
:* Blog: User Blog Mod for phpBB3<br />
:* Search: Sphinx<br />
:* Wiki: MediaWiki<br />
:* Asymptote, Latex, and Geogebra are generated through their respective binary packages<br />
:* Videos: YouTube<br />
<br />
Note that as AoPS does not develop the above software, they are not responsible for the proper functioning of said software. Bug reports and feature requests should be sent to the appropriate developers of the above software.<br />
<br />
== Forums ==<br />
<br />
==== What do the stars under my username next to a forum post represent? ====<br />
<br />
:On the Art of Problem Solving website, under your username, you will find stars, as well as the name of one of the Millenium Problems. The number of stars you have, as well as the name of the Millenium Problem, depends on your post count. Here is the table that determines your "rank."<br />
<br />
:*0 - 19 New Member (Zero Stars)<br />
:*20 - 49 P Versus NP (Half Star)<br />
:*50 - 99 Hodge Conjecture (One Star)<br />
:*100 - 249 Poincare Conjecture (Two Stars)<br />
:*250 - 499 Riemann Hypothesis (Two and Half stars) <br />
:*500 - 999 Yang Mills Theory (Three Stars)<br />
:*1000 - 2499 Navier-Stokes Equation (Four Stars)<br />
:*2500 - <math>\infty</math> Birch & Swinnerton Dyer. (Five Stars)<br />
:*Administrators have six stars.<br />
<br />
==== I got the message "You can not post at this time" when trying to post, why? ====<br />
<br />
:New users are not allowed to post messages with URLs and various other things. Once you have five posts you can post normally.<br />
<br />
====I got the message "Too many messages." when trying to send a private message, why?====<br />
:To prevent PM spam abuse, users with less than five forum posts are limited to four private messages within a forty-eight hour period.<br />
<br />
==== If I make more posts, it means I'm a better user, right? ====<br />
<br />
:Absolutely not. Post quality is far more important than post quantity. Users making a lot of senseless posts are often considered worse users, or spammers.<br />
<br />
==== I have made some posts but my post count did not increase. Why? ====<br />
<br />
:When you post in some of the forums, such as the Test Forum, Mafia Forum, and the Fun Factory, it does not count towards your post count.<br />
<br />
==== When can I rate posts? ====<br />
<br />
:You will be able to rate posts after posting 10 messages.<br />
<br />
==== Who can see my post rating? ====<br />
<br />
:Only you, moderators, and administrators.<br />
<br />
==== I rated a post but the post rating does not appear on the post. Why?====<br />
:One or two ratings is not enough to determine a post's overall rating. Therefore, a post has to receive a preset number of ratings before the overall rating of the post appears.<br />
<br />
==== How does AoPS select moderators? ====<br />
<br />
:When a new moderator is needed in the forums, AoPS administrators first check if any current moderators could serve as a moderator of the forum which needs a moderator. Should none be found, AoPS administrators and/or other moderators scour the forum looking for productive users. They may also ask for suggestions from other moderators or trusted users on the site. Once they have pinpointed a possible candidate based on their long term usage of the site, productive posts in the forum, and having no recent behavioral issues, that user is asked if he or she would like to moderate the forum. <br />
<br />
:Less active forums often have no moderator. Inappropriate posts should be reported by users and administrators will take appropriate action.<br />
<br />
:AoPS receives MANY requests to be a moderator. As they receive so many, it is possible that you won't get a response should you request to be one. Also, AoPS very rarely makes someone a mod for asking to be one, so '''please do not ask'''.<br />
<br />
==== I believe a post needs corrective action. What should I do? ====<br />
<br />
:If you believe a post needs moderative action, you may report it by clicking the "!" icon on the bottom-right corner of that post. If it's a minor mistake, you may want to PM the offending user instead and explain how they can make their post better. Usually, you shouldn't publicly post such things on a thread itself, which is called "backseat moderation" and is considered rude.<br />
<br />
==== How long of a non-commented thread is considered reviving? ====<br />
<br />
:If any post is still on-topic and isn't spammy or anything, it isn't considered reviving. The definition of reviving in the Games forum is 1 month. However, everyone has a different period of time that they consider reviving. In general, apply common sense.<br />
<br />
==== Someone is marking all my posts as spam, what should I do? ====<br />
<br />
:It happens to everyone. There's really not much you can do.<br />
<br />
==== Are posts marked spam more often than good? ====<br />
<br />
:No. The most common rating is 6 cubes. We understand that many posts are rated 1 when they shouldn't be. We also know that many posts are rated a 6 when they shouldn't be. It pretty much all averages out in the end. The best way to safeguard yourself is not to complain about it! In fact, most other members cannot see your rating. If you want to make a mark here, let your post quality do the talking.<br />
<br />
==== How do I post images? ====<br />
:There are limited attachment options for posts. Attachments have an overall size limit, and will be deleted as they get old. Attachments also may be deleted during any server move or software upgrade or change. You may instead wish to host images on another site and embed them in to your post using the [img] tags. The general format is [img]{url to image}[/img], excluding the braces. There are a number of image hosting sites, including:<br />
:* [http://imgur.com/ Imgur]<br />
:* [http://photobucket.com Photobucket]<br />
::*In thumbnail view<br />
:::*Hover over image and click the text box labeled IMG code. It will automatically copy to your clipboard<br />
:::*Paste to your message<br />
::*In image view<br />
:::*Look for the '''Links''' box which should appear at the right side of your screen<br />
:::*Click the box labeled IMG Code<br />
:::*Copy the text<br />
:::*Paste to your message<br />
:* [http://imageshack.com ImageShack]<br />
:* [http://minus.com minus.com]<br />
::*Go to image you wish to embed<br />
::*Click the share tab<br />
::*Copy the contents of the Forum Code text<br />
::*Paste to your message<br />
:* [http://bayfiles.com bayfiles.com]<br />
:* [http://picasaweb.google.com Picasa]<br />
:** This will vary by browser and OS, but the process is similar. The provided directions are for Firefox on Windows<br />
:** Go to the image you want to embed<br />
:** Right click on the image<br />
:** Select Copy Image Location<br />
:** Paste into your message, surrounding the pasted text with [img] and [/img] tags<br />
:* [http://www.flickr.com Flickr]<br />
<br />
See also:<br />
[http://www.artofproblemsolving.com/Wiki/index.php/Direct_Image_Link Direct Image Link]<br />
<br />
== Blogs ==<br />
==== How come I can't create a blog? ====<br />
One needs to have at least 5 posts in order to make a blog.<br />
<br />
== Contests ==<br />
==== Where can I find past contest questions and solutions? ====<br />
:In the [http://www.artofproblemsolving.com/Forum/resources.php Contests] section.<br />
<br />
==== How do I get problems onto the contest page? ====<br />
<br />
:Make a topic for each question in the appropriate forum, copy/paste the urls to the National Olympiad. Your problems may eventually be submitted into the Contest page.<br />
<br />
==== Who can I ask to add posts to the contests section? ====<br />
:Any one of the members in the the [http://www.artofproblemsolving.com/Forum/memberlist.php?mode=group&g=417 RManagers] group.<br />
<br />
==== What are the guidelines for posting problems to be added to the contests section? ====<br />
:Refer to the [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=144&t=195579 guidelines in this post].<br />
<br />
==== Why is the wiki missing many contest questions? ====<br />
:Generally, it is because users have not yet posted them onto the wiki (translation difficulties, not having access to the actual problems, lack of interest, etc). If you have a copy, please post the problems in the Community Section! In some cases, however, problems may be missing due to copyright claims from maths organizations. See, for example, [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1391106#p1391106 this post].<br />
<br />
==== What if I find an error on a problem? ====<br />
Please post an accurate description of the problem in [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&t=426693 this thread]<br />
<br />
== LaTeX, Asymptote, GeoGebra ==<br />
==== What is LaTeX, and how do I use it? ====<br />
<br />
:<math>\LaTeX</math> is a typesetting markup language that is useful to produce properly formatted mathematical and scientific expressions.<br />
<br />
==== How can I download LaTeX to use on the forums? ====<br />
<br />
:There are no downloads necessary; the forums and the wiki render LaTeX commands between dollar signs. <br />
<br />
==== How can I download LaTeX for personal use? ====<br />
:You can download TeXstudio [http://texstudio.sourceforge.net here] or TeXnicCenter [http://www.texniccenter.org here]<br />
<br />
==== Where can I find a list of LaTeX commands? ====<br />
:See [[LaTeX:Symbols|here]].<br />
<br />
==== Where can I test LaTeX commands? ====<br />
<br />
:[[A:SAND|Sandbox]] or [http://www.artofproblemsolving.com/Resources/texer.php TeXeR]. <br />
<br />
==== Where can I find examples of Asymptote diagrams and code? ====<br />
<br />
:Search this wiki for the <tt><nowiki><asy></nowiki></tt> tag or the Forums for the <tt><nowiki>[asy]</nowiki></tt> tag. See also [[Asymptote:_Useful_commands_and_their_Output|these examples]] and [[Proofs without words|this article]] (click on the images to obtain the code).<br />
<br />
==== How can I draw 3D diagrams? ====<br />
<br />
:See [[Asymptote: 3D graphics]].<br />
<br />
==== What is the cse5 package? ==== <br />
<br />
:See [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=519&t=149650 here]. The package contains a set of shorthand commands that implement the behavior of usual commands, for example <tt>D()</tt> for <tt>draw()</tt> and <tt>dot()</tt>, and so forth.<br />
<br />
==== What is the olympiad package? ====<br />
<br />
:See [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=519&t=165767 here]. The package contains a set of commands useful for drawing diagrams related to [[:Category:Olympiad Geometry Problems|olympiad geometry problems]].<br />
<br />
==== Can I convert diagrams from GeoGebra to other formats? ====<br />
:It is possible to export GeoGebra to [[Asymptote (Vector Graphics Language)|Asymptote]] (see [[User:Azjps/geogebra|here]]), PsTricks, and PGF/TikZ; and GeoGebra animations into .gif or video files. <br />
<br />
== AoPSWiki ==<br />
==== Is there a guide for wiki syntax? ====<br />
<br />
:See [http://en.wikipedia.org/wiki/Help:Wiki_markup wiki markup], [[AoPSWiki:Tutorial]], and [[Help:Contents]].<br />
<br />
==== What do I do if I see a mistake in the wiki? ====<br />
<br />
:Edit the page and correct the error! You can edit most pages on the wiki. Click the "edit" button on the right sidebar to edit a page.<br />
<br />
==== Why can't I edit the wiki? ====<br />
<br />
You must be a registered user to edit. To be registered, make sure you give a correct email, and activate your account.<br />
<br />
== Miscellaneous ==<br />
==== Is it possible to join the AoPS Staff? ====<br />
<br />
:Yes. Mr. Rusczyk will sometimes hire a small army of college students to work as interns. You must be at least in your second semester of your senior year and be legal to work in the U.S. (at least 16).<br />
<br />
==== What is the minimum age to be an assistant in an Art of Problem Solving class? ====<br />
<br />
:You must have graduated from high school, or at least be in the second term of your senior year.<br />
<br />
==What do some of the acronyms such as "OP" stand for?==<br />
*'''AFK'''- Away from keyboard<br />
*'''AoPS'''- Art of Problem Solving, the website you're on right now!<br />
*'''AIME'''- American Invitational Mathematics Examination<br />
*'''AMC'''- American Math Competititions<br />
*'''ATM'''- At the Moment<br />
*'''brb'''- Be right Back<br />
*'''BTW'''- By the way<br />
*'''EBWOP'''- Editing by way of post<br />
*'''FTW'''- For the Win, a game on AoPS<br />
*'''gj'''- Good Job<br />
*'''GLHF'''-Good Luck Have Fun<br />
*'''IDK'''-I Don't Know<br />
*'''iff'''-If and only if<br />
*'''IIRC'''- If I recall correctly<br />
*'''IMO'''- In my opinion (or International Math Olympiad, depending on context)<br />
*'''lol'''- Laugh Out Loud<br />
*'''MC'''- Mathcounts, a popular math contest for Middle School students.<br />
*'''MOP'''- Mathematical Olympiad (Summer) Program<br />
*'''OBC'''- Online by computer<br />
*'''OMG'''- Oh My Gosh.<br />
*'''OP'''- Original Poster/Original Post/Original Problem, or Overpowered/Overpowering<br />
*'''QED'''- Quod erat demonstrandum or Which was to be proven<br />
*'''QS&A'''- Questions, Suggestions, and Announcements Forum<br />
*'''sa''' - sa<br />
*'''USAJMO'''- USA Junior Mathematical Olympiad<br />
*'''USAMO'''- USA Mathematical Olympiad<br />
*'''V/LA'''- Vacation or Long Absence<br />
*'''WLOG'''- Without loss of generality<br />
*'''wrt'''- With respect to<br />
*'''tytia'''- Thank you, that is all<br />
*'''xD'''- Bursting Laugh<br />
*'''wtg'''- Way to go<br />
*'''smh'''- Shaking my head<br />
*'''gtg''' - Got to go<br />
*'''rotfl''' - Rolling on the floor laughing<br />
<br />
== FTW! ==<br />
<br />
==== How do you access FTW? ====<br />
You can access FTW by clicking FTW! on the green bar at the top of the page.<br />
<br />
<br />
==== Did FTW miscount my number of games?====<br />
<br />
No! However, the (Overall) rating statistics do not count games with less than 6 problems or less than 15 seconds.<br />
<br />
For example, if you have played 30 games, but not all of them were 6 problems or higher, then you will still be muted.<br />
<br />
== School ==<br />
<br />
==== What if I want to drop out of a class? ====<br />
:For any course with more than 2 classes, students can drop the course any time before the third class begins and receive a full refund. No drops are allowed after the third class has started. To drop the class, go to the My Classes section by clicking the My Classes link at the top-right of the website. Then find the area on the right side of the page that lets you drop the class. A refund will be processed within 10 business days.<br />
<br />
==== What if I miss a class? ====<br />
:There are classroom transcripts available under My Classes, available at the top right of the web site. You can view these transcripts in order to review any missed material. You can also ask questions on the class message board.<br />
<br />
==== Is there audio or video in class? ====<br />
There is no audio or video in the class. The classes are completely text based, in an interactive chat room environment, which allows students to ask questions at any time during the class. In addition to audio and video limiting interactivity, the technology isn't quite there yet for all students to be able to adequately receive streaming audio and video. <br />
<br />
====I feel like joining! What are my class choices? ====<br />
:[http://www.artofproblemsolving.com/School/classlist.php Class List] [http://www.artofproblemsolving.com/School/index.php?page=school.instructors Instructors List]</div>Pi37https://artofproblemsolving.com/wiki/index.php?title=1997_AHSME_Problems/Problem_29&diff=568801997 AHSME Problems/Problem 292013-08-09T17:11:37Z<p>Pi37: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
Call a positive real number special if it has a decimal representation that consists entirely of digits <math>0</math> and <math>7</math>. For example, <math> \frac{700}{99}= 7.\overline{07}= 7.070707\cdots </math> and <math> 77.007 </math> are special numbers. What is the smallest <math>n</math> such that <math>1</math> can be written as a sum of <math>n</math> special numbers?<br />
<br />
<math> \textbf{(A)}\ 7\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\\ \textbf{(E)}\ \text{The number 1 cannot be represented as a sum of finitely many special numbers.} </math><br />
<br />
==Solution==<br />
Define a super-special number to be a number whose decimal expansion only consists of <math>0</math>'s and <math>1</math>'s. The problem is equivalent to finding the number of super-special numbers necessary to add up to <math>\frac{1}{7}=0.142857142857\hdots</math>. This can be done in <math>8</math> numbers if we take<br />
<cmath><br />
0.111111\hdots, 0.011111\hdots, 0.010111\hdots, 0.010111\hdots, 0.000111\hdots, 0.000101\hdots, 0.000101\hdots, 0.000100\hdots<br />
</cmath><br />
Now assume for sake of contradiction that we can do this with strictly less than <math>8</math> super-special numbers (in particular, less than <math>10</math>.) Then the result of the addition won't have any carry over, so each digit is simply the number of super-special numbers which had a <math>1</math> in that place. This means that in order to obtain the <math>8</math> in <math>0.1428\hdots</math>, there must be <math>8</math> super-special numbers, so the answer is <math>\boxed{\textbf{(B)}\ 8}</math>.<br />
<br />
== See also ==<br />
{{AHSME box|year=1997|num-b=28|num-a=30}}<br />
{{MAA Notice}}</div>Pi37https://artofproblemsolving.com/wiki/index.php?title=AoPS_Wiki:AoPS_Community_Awards&diff=52756AoPS Wiki:AoPS Community Awards2013-05-15T05:08:35Z<p>Pi37: /* Perfect AMC 10 scorers */</p>
<hr />
<div>This '''AoPS Community Awards''' page is a celebration of the accomplishments of members of the [[AoPS]] community.<br />
<br />
<br />
== IMO Participants and Medalists ==<br />
This is a list of members of the AoPS community who have competed for their country at the [[International Mathematical Olympiad]].<br />
<br />
=== Participants ===<br />
* Prafulla Susil Dhariwal(2011,2012)<br />
* Akashnil Dutta(2009,2010,2011)<br />
* Alex Zhai (2008) <br />
* Krishanu Roy Sankar (2008)<br />
* Zachary Abel (2006) (AoPS assistant instructor)<br />
* Marco Avila (2006)<br />
* John Berman (2009)<br />
* Zarathustra Brady (2006)<br />
* Robert Cordwell (2005)<br />
* Wenyu Cao (2009, 2011)<br />
* Sherry Gong (2002, 2003, 2004, 2005, 2007)<br />
* Elyot Grant (2005)<br />
* Darij Grinberg (2006)<br />
* Mahbubul Hasan (2005)<br />
* Daniel Kane (AoPS assistant instructor)<br />
* Kiran Kedlaya (1990, 1991, 1992) ([[Art of Problem Solving Foundation]] board member)<br />
* Viktoriya Krakovna (2006)<br />
* Nate Ince (2004) (AoPS assistant instructor)<br />
* Brian Lawrence (2005, 2007) ([[WOOT]] instructor)<br />
* Thomas Mildorf (2005) (AoPS assistant instructor)<br />
* Alison Miller (2004) (AoPS assistant instructor)<br />
* Richard Peng (2005, 2006)<br />
* Eric Price (2005)<br />
* David Rhee (2004, 2005, 2006)<br />
* Peng Shi (2004, 2005, 2006)<br />
* Arne Smeets (2003, 2004)<br />
* Bobby Shen (2012)<br />
* Arnav Tripathy (2006, 2007)<br />
* [[Naoki Sato]] (AoPS instructor)<br />
* Yi Sun (2006)<br />
* [[Valentin Vornicu]] (AoPS/MathLinks webmaster)<br />
* Melanie Wood (1998, 1999) ([[WOOT]] instructor)<br />
* Alex Zhai (2005, 2006, 2007, 2008)<br />
* Yufei Zhao (2004, 2005, 2006)<br />
* Tigran Sloyan(2003,2004,2005,2006,2007)<br />
* Marco Avila (2006)<br />
* Vipul Naik (2003,2004)<br />
* Bhargav Narayanan (2007)<br />
* Tigran Hakobyan (2007)<br />
* Carmela Lao (2009, 2010)<br />
* Calvin Deng (2010, 2012)<br />
* Allen Yuan (2010)<br />
* In-sung Na (2010)<br />
* Xiaoyu He (2010, 2011)<br />
<br />
===Perfect Scorers===<br />
*Brian Lawrence (2005)<br />
*Alex Zhai (2008)<br />
*Jeck Lim (2012)<br />
<br />
=== Gold medalists ===<br />
* Akashnil Dutta(2011)<br />
* Zarathustra Brady (2006)<br />
* Robert Cordwell (2005)<br />
* Darij Grinberg (2006)<br />
* Kiran Kedlaya (1990, 1992) ([[Art of Problem Solving Foundation]] board member)<br />
* Brian Lawrence (2005) ([[WOOT]] instructor)<br />
* Thomas Mildorf (2005) (AoPS assistant instructor)<br />
* Alison Miller (2004) (AoPS assistant instructor)<br />
* Arnav Tripathy (2006)<br />
* Eric Price (2005)<br />
* Yufei Zhao (2005)<br />
* Alex Zhai (2007, 2008)<br />
*Sherry Gong (2007)<br />
*Krishanu Sankar (2008)<br />
* John Berman (2009)<br />
* Xiaoyu He (2010, 2011)<br />
* Wenyu Cao (2011)<br />
* Prafulla Susil Dhariwal (2012)<br />
* Bobby Shen (2012)<br />
<br />
=== Silver medalists ===<br />
* Akashnil Dutta (2009,2010)<br />
* Zachary Abel (2006) (AoPS assistant instructor)<br />
* Sherry Gong (2004, 2005)<br />
* Nate Ince (2004) (AoPS assistant instructor)<br />
* Kiran Kedlaya (1991) ([[Art of Problem Solving Foundation]] board member)<br />
* Viktoriya Krakovna (2006)<br />
* Hyun Soo Kim (2005) (AoPS assistant instructor)<br />
* Richard Peng (2005)<br />
* David Rhee (2006)<br />
* Naoki Sato (AoPS instructor)<br />
* Peng Shi (2006)<br />
* Arne Smeets (2004)<br />
* Yi Sun (2006)<br />
* [[Sam Vandervelde]] (1989) ([[WOOT]] instructor)<br />
* Melanie Wood (1998, 1999) ([[WOOT]] instructor)<br />
* Alex Zhai (2006)<br />
* Yufei Zhao (2006)<br />
* Tigran Sloyan (2006,2007)<br />
* Vipul Naik (2003,2004)<br />
* Delong Meng (2009)<br />
* Qinxuan Pan (2009)<br />
* Wenyu Cao (2009)<br />
* Carmela Lao (2010)<br />
* Jafar Jafarov (2006)<br />
<br />
=== Bronze medalists ===<br />
* Debdyuti Banerjee (2011)<br />
* Sherry Gong (2003)<br />
* Elyot Grant (2005)<br />
* Richard Peng (2006)<br />
* [[Naoki Sato]] (AoPS instructor)<br />
* [[Valentin Vornicu]] (AoPS/[[MathLinks]] webmaster)<br />
* Yufei Zhao (2004)<br />
* Tigran Sloyan(2004;2005)<br />
* Tigran Hakobyan (2007)<br />
* Carmela Lao (2009)<br />
* Jafar Jafarov (2007)<br />
<br />
== IPhO Participants and Medalists ==<br />
This is a list of members of the AoPS community who have competed for their country at the [[International Physics Olympiad]].<br />
=== Participants ===<br />
* Sherry Gong (2006)<br />
* Yi Sun (2004)<br />
* Arnav Tripathy (2006)<br />
* Marianna Mao (2009)<br />
* Anand Natarajan (2009)<br />
* Bowei Liu (2009)<br />
* Daniel Li (2010)<br />
<br />
=== Gold Medalists ===<br />
* Yi Sun (2004)<br />
* Rahul Singh (2007)<br />
* Marianna Mao (2009)<br />
* Anand Natarajan (2009)<br />
* Bowei Liu (2009)<br />
* Daniel Li(2010)<br />
* Eric Schneider(2012)<br />
* Kevin Zhou (2012)<br />
<br />
=== Silver Medalists ===<br />
* Sherry Gong (2006)<br />
<br />
== IOI Medalists ==<br />
This list is of AoPSers who have won medals at the [[International Olympiad in Informatics]].<br />
<br />
===Gold Medalists ===<br />
* David Benjamin (2008)<br />
* Brian Hamrick (2009)<br />
* Neal Wu (2008, 2009, 2010)<br />
* Wenyu Cao (2010)<br />
<br />
===Silver Medalists ===<br />
* Akashnil Dutta(2011)<br />
* Brian Hamrick (2008)<br />
* Jacob Steinhardt (2008)<br />
* Wenyu Cao (2009)<br />
* Travis Hance (2009)<br />
<br />
===Bronze Medalists ===<br />
* David Benjamin (2007)<br />
<br />
<br />
== USAMO ==<br />
The following AoPSers have won the [[United States of America Mathematical Olympiad]] (USAMO). (Note that the definition of "winner" has changed over the years -- currently it is the top 12 scores on the USAMO, but in the past it has been the top 6 or top 8 scores.)<br />
=== Perfect Scorers ===<br />
* Daniel Kane (AoPS assistant instructor)<br />
* Kiran Kedlaya (1991) ([[Art of Problem Solving Foundation]] board member)<br />
* Brian Lawrence (2006) ([[WOOT]] instructor)<br />
* Alex Zhu (2012)<br />
<br />
=== Winners ===<br />
* Yakov Berchenko-Kogan (2006)<br />
* Wenyu Cao (2009)<br />
* Sherry Gong (2006, 2007)<br />
* Yi Han (2006)<br />
* Adam Hesterberg (2007)<br />
* Daniel Kane (AoPS assistant instructor)<br />
* Kiran Kedlaya (1990, 1991, 1992) ([[Art of Problem Solving Foundation]] board member)<br />
* Brian Lawrence (2005, 2006, 2007) ([[WOOT]] instructor)<br />
* Tedrick Leung (2006, 2007)<br />
* Haitao Mao (2007)<br />
* Richard Mccutchen (2006)<br />
* Albert Ni (2005)<br />
* [[David Patrick]] (1988) (AoPS instructor)<br />
* David Rolnick (2008) (AoPS assistant instructor)<br />
* [[Richard Rusczyk]] (1989) (AoPS founder)<br />
* Krishanu Sankar (2007,2008)<br />
* Peng Shi (2006)<br />
* Jacob Steinhardt (2007)<br />
* Yi Sun (2006)<br />
* Arnav Tripathy (2006, 2007)<br />
* [[Sam Vandervelde]] (1987, 1989) ([[WOOT]] instructor)<br />
* Melanie Wood (1998, 1999) ([[WOOT]] instructor)<br />
* David Yang (2009)<br />
* Alex Zhai (2006, 2007,2008)<br />
* Yufei Zhao (2006)<br />
* David Bay Rush (2009)<br />
<br />
== Putnam Fellows ==<br />
The top 5 students (including ties) on the collegiate [[Putnam Exam|William Lowell Putnam Competition]] are named Putnam Fellows.<br />
* David Ash (1981, 1982, 1983)<br />
* Matthew Ince (2005) (AoPS assistant instructor)<br />
* Daniel Kane (2003, 2004, 2005) (AoPS assistant instructor)<br />
* Kiran Kedlaya (1994, 1995, 1996) ([[AoPS Foundation]] board member)<br />
* Mitchell Lee (2012)<br />
* Evan O'Dorney (2011, 2012)<br />
* Alexander Schwartz (2000, 2002)<br />
* Jan Siwanowicz (2001) <br />
* Melanie Wood (2002) ([[WOOT]] instructor)<br />
* Arnav Tripathy (2008)<br />
* Brian Lawrence (2008)<br />
<br />
== Siemens Competition Winners ==<br />
The annual [[Siemens Competition]] (formerly Siemens-Westinghouse) is a scientific research competition.<br />
* Michael Viscardi (1st Individual, 2005)<br />
* Lucia Mocz (2nd Team, 2006)<br />
<br />
==Intel STS Finalists==<br />
The annual [[Intel Science Talent Search]] is a science competition seeking to find and reward the most scientifically accomplished seniors.<br />
<br />
* Jonathan Li (2011)<br />
*Evan O'Dorney (2011)<br />
* Philip Mocz (2008)<br />
* Yihe Dong (2008)<br />
* Qiaochu Yuan (2008)<br />
* Greg Brockman (2007)<br />
<br />
== Clay Junior Fellows ==<br />
Each year since 2003, the [[Clay Mathematics Institute]] has selected 12 Junior Fellows.<br />
* Thomas Belulovich (2005) (AoPS assistant instructor)<br />
* Atoshi Chowdhury (2003) (AoPS assistant instructor)<br />
* Robert Cordwell (2005)<br />
* Eve Drucker (2003) (AoPS assistant instructor)<br />
* Matthew Ince (2004) (AoPS assistant instructor)<br />
* Nate Ince (2004) (AoPS assistant instructor)<br />
* Hyun Soo Kim (2005) (AoPS assistant instructor)<br />
* Raju Krishnamoorthy (2005)<br />
* Alison Miller (2003) (AoPS assistant instructor)<br />
* Brian Rice (2003) (AoPS assistant instructor)<br />
* Dmitry Taubinski (2005) (AoPS assistant instructor)<br />
* Ameya Velingker (2005)<br />
<br />
<br />
== Perfect AIME Scores ==<br />
Very few students have ever achieved a perfect score on the [[American Invitational Mathematics Examination]] (AIME)<br />
<br />
* David Benjamin (2006)<br />
* [[Mathew Crawford]] (1992) (AoPS instructor)<br />
* Calvin Deng (2011)<br />
* Sam Elder (2008)<br />
* [[Sandor Lehoczky]] (1990) (AoPS author)<br />
* Tedrick Leung (2006)<br />
* Tony Liu (2006)<br />
* Haitao Mao (2008)<br />
* Shyam Narayanan (2012)<br />
* Bobby Shen (2010, 2011, 2012, 2013)<br />
* [[Richard Rusczyk]] (1989) (AoPS founder)<br />
* [[Sam Vandervelde]] (1988) ([[WOOT]] instructor)<br />
* Alexander Whatley (2012)<br />
* Scott Wu (2012)<br />
<br />
== Perfect AMC Scores ==<br />
=== Perfect AMC 12 Scores ===<br />
The [[AMC 12]] is a challenging examination for students in grades 12 and below administered by the [[American Mathematics Competitions]].<br />
* Zachary Abel (2005) (AoPS assistant instructor)<br />
* David Benjamin (2006)<br />
* Wenyu Cao (2009)<br />
* Sam Elder (2008)<br />
* Ruozhou (Joe) Jia (2003) (AoPS assistant instructor)<br />
* Joel Lewis (2003) <br />
* Jonathan Li (2010)<br />
*Daniel Li (2009)<br />
* Jonathan Lowd (2003) (AoPS assistant instructor)<br />
* Thomas Mildorf (2004) (AoPS assistant instructor)<br />
* Alison Miller (2004) (AoPS assistant instructor)<br />
* Albert Ni (2003) (AoPS instructor)<br />
* Ajay Sharma (2004)<br />
* Bobby Shen (2009, 2011)<br />
* Matt Superdock (2009)<br />
* Arnav Tripathy (2006, 2007)<br />
* Qiaochu Yuan (2008)<br />
* Alex Zhai (2007)<br />
<br />
===Perfect AMC 10 scorers===<br />
The [[AMC 10]] is a challenging examination for students in grades 10 and below administered by the [[American Mathematics Competitions]].<br />
* Yongyi Chen (2009)<br />
* Matthew Babbitt (2009)<br />
* Sergei Bernstein (2007)<br />
* Yifan Cao (2005)<br />
* Kevin Chen (2007, 2008, 2009)<br />
*Lewis Chen(2009)<br />
* In Young Cho (2007)<br />
* Mario Choi (2007)<br />
* Calvin Deng (2008)<br />
* Billy Dorminy (2007)<br />
* Zhou Fan (2005)<br />
* Albert Gu (2007)<br />
* Robin He (2007)<br />
* Keone Hon (2005)<br />
* Susan Hu (2005)<br />
* Lyndon Ji (2008)<br />
* Sam Keller (2007)<br />
* Vincent Le (2006)<br />
* Daniel Li (2007)<br />
* Jonathan Li (2009, 2007)<br />
* Kevin Li (2009)<br />
* Patricia Li (2005)<br />
* Carl Lian (2007)<br />
* Sam Lite (2009)<br />
* David Lu (2009)<br />
* Michael Ma (2009, 2010, 2011 in grades 3, 4, 5)<br />
* Thomas Mildorf (2002) (AoPS assistant instructor)<br />
* Anupa Murali (2008)<br />
* Shyam Narayanan (2013)<br />
* Max Rosett (2005, 2006) (AoPS assistant instructor)<br />
* Amrit Saxena (2009)<br />
* Maximilian Schindler (2009)<br />
* Eric Schneider (2009)<br />
* Bobby Shen (2009)<br />
* Lilly Shen (2009)<br />
* Jeffrey Shen (2008)<br />
* Kyle Stankowski (2009)<br />
* Michael Tan (2009)<br />
* Kevin Tian (2009)<br />
* Howard Tong (2005)<br />
* Sam Trabucco (2008)<br />
* Brent Woodhouse (2006, 2007)<br />
* Lawrence Wu (2009)<br />
* Allen Yuan (2009)<br />
* David Yang (2009)<br />
* Peijin Zhang (2009)<br />
* Jonathan Zhou (2007)<br />
* Alex Zhu (2009)<br />
* Michael Kural (2013)<br />
<br />
=== Perfect AHSME Scores ===<br />
The [[American High School Mathematics Examination]] (AHSME) was the predecessor of the AMC 12.<br />
* Christopher Chang (1994, 1995, 1996)<br />
* [[Mathew Crawford]] (1994, 1995) (AoPS instructor)<br />
* [[David Patrick]] (1988) (AoPS instructor)<br />
<br />
=='''Perfect USAMO Index'''==<br />
*Samuel Elder (2009)<br />
*Bobby Shen (2011)<br />
*Gabriel Caroll (1998, 1999, 2000, 2001)<br />
<br />
== MATHCOUNTS ==<br />
[[MathCounts]] is the premier middle school [[mathematics competition]] in the U.S.<br />
=== National Champions ===<br />
* Ruozhou (Joe) Jia (2000) (AoPS assistant instructor)<br />
* Albert Ni (2002) (AoPS instructor)<br />
* Adam Hesterberg (2003)<br />
* Neal Wu (2005)<br />
* Daesun Yim (2006)<br />
* Kevin Chen (2007)<br />
* Darryl Wu (2008)<br />
* Bobby Shen (2009)<br />
* Mark Sellke (2010)<br />
* Scott Wu (2011)<br />
* Chad Qian (2012)<br />
<br />
=== National Top 12 ===<br />
* Ashley Reiter Ahlin (1987) ([[WOOT]] instructor)<br />
* Andrew Ardito (2005, 2006)<br />
* David Benjamin (2004, 2005)<br />
* Nathan Benjamin (2005, 2006)<br />
* Wenyu Cao (2007)<br />
* Christopher Chang (1991, 1992)<br />
* Kevin Chen (2006, 2007)<br />
* Steven Chen (2009, 2010)<br />
* Andrew Chien (2003)<br />
* Peter Chien (2004)<br />
* Mario Choi (2007)<br />
* Joseph Chu (2004)<br />
* Alexander Clifton (2009)<br />
* [[Mathew Crawford]] (1990, 1991) (AoPS instructor)<br />
* Calvin Deng (2009)<br />
* Brian Hamrick (2006)<br />
* Adam Hesterberg (2002, 2003)<br />
* Jason Hyun (2008)<br />
* Ruozhou (Joe) Jia (2000) (AoPS assistant instructor)<br />
* Sam Keller (2006)<br />
* Shaunak Kishore (2003, 2004)<br />
* Kiran Kota (2005)<br />
* Brian Lawrence (2003) ([[WOOT]] instructor)<br />
* Karlanna Lewis (2005)<br />
* Daniel Li (2006)<br />
* Patricia Li (2005)<br />
* Ray Li (2009)<br />
* Poh-Ling Loh (2000)<br />
* David Lu (2008)<br />
* Albert Ni (2002) (AoPS assistant instructor)<br />
* Maximilian Schindler (2009)<br />
* Bobby Shen (2008, 2009)<br />
* Elizabeth Synge (2007)<br />
* Jason Trigg (2002)<br />
* [[Sam Vandervelde]] (1985) ([[WOOT]] instructor)<br />
* Victor Wang (2009)<br />
* Neal Wu (2005, 2006)<br />
* Rolland Wu (2006)<br />
* Xiaoyu He (2008)<br />
* David Yang (2009)<br />
* Daesun Yim (2006)<br />
* Darren Yin (2002)<br />
* Allen Yuan (2007)<br />
* Samuel Zbarsky (2008)<br />
* Alex Zhai (2004)<br />
* Mark Zhang (2005)<br />
* Alan Zhou (2009)<br />
* Mark Sellke (2009, 2010)<br />
* Eugene Chen (2010)<br />
* Lewis Chen (2010)<br />
* Shyam Narayanan (2010,2011)<br />
<br />
=== Masters Round Champions ===<br />
* Christopher Chang (1991)<br />
* Brian Lawrence (2003) ([[WOOT]] instructor)<br />
* Sergei Bernstein (2005)<br />
* Daniel Li (2006)<br />
* Kevin Chen (2007)<br />
* Bobby Shen (2008)<br />
* Maximilian Schindler (2009)<br />
* Alex Song (2010)<br />
<br />
=== National Test Champions ===<br />
* [[Mathew Crawford]] (1990) (AoPS instructor)<br />
* Adam Hesterberg (2003)<br />
* Sergei Bernstein (2005)<br />
* Neal Wu (2006)<br />
* Bobby Shen (2008)<br />
* David Yang (2009)<br />
* Mark Sellke (2010)<br />
*Shyam Narayanan (2011)<br />
<br />
== Harvard-MIT Math Tournament ==<br />
<br />
The [[HMMT]] 2007 winning team, the "WOOTlings", consisted entirely of [[WOOT]]ers:<br />
<br />
* Wenyu Cao<br />
* Eric Chang<br />
* Jeremy Hahn<br />
* Alex Kandell<br />
* Adeel Khan<br />
* Sathish Nagappan<br />
* Krishanu Roy Sankar<br />
* Patrick Tenorio<br />
<br />
== ARML ==<br />
<br />
<br />
=== ARML winners ===<br />
* Alex Song (2009, 2011)<br />
* Benjamin Gunby (2010)<br />
* Allen Liu (2012)<br />
<br />
=== ARML Perfect Scorers ===<br />
* Alex Song (2011)<br />
* Shyam Narayanan (2012)<br />
<br />
=== ARML Top 10 ===<br />
* Zachary Abel (2006) (AoPS assistant instructor)<br />
*Lewis Chen(2011,2012)<br />
* Benjamin Gunby (2010)<br />
* Bobby Shen (2010)<br />
* Seva Tchernov (2007)<br />
* Arnav Tripathy (2007)<br />
* David Yang (2009)<br />
* Daesun Yim (2008)<br />
* Alex Song (2009,2011,2012)<br />
*Bailey Wang (2009)(WOOT Grader)<br />
* Shyam Narayanan (2011,2012 (Perfect Score))<br />
<br />
== See also ==<br />
* [[Academic competitions]]<br />
* [[Mathematics competitions]]<br />
* [[Mathematics competition resources]]<br />
* [[Academic scholarships]]<br />
<br />
<br />
<br />
[[Category:Art of Problem Solving]]</div>Pi37https://artofproblemsolving.com/wiki/index.php?title=Connecticut_mathematics_competitions&diff=52551Connecticut mathematics competitions2013-04-15T21:53:29Z<p>Pi37: /* Connecticut high school mathematics competitions */</p>
<hr />
<div>==Connecticut middle school mathematics competitions==<br />
<br />
[[AoPS]] hosts [http://www.artofproblemsolving.com/Forum/index.php?f=298 middle school math forums] where students can discuss contest problems and mathematics.<br />
<br />
* [[Connecticut MathCounts]] is part of the national [[MathCounts]] competition.<br />
<br />
<br />
==Connecticut high school mathematics competitions==<br />
<br />
[[AoPS]] hosts [http://www.artofproblemsolving.com/Forum/index.php?f=214 high school math forums] where students can discuss contest problems and mathematics.<br />
<br />
*[[Connecticut State Association of Mathematics Leagues (CSAML)]] [http://studentweb.choate.edu/mathteam/CSAML.HTM website]<br />
*[[Greater New Haven Mathematics League (GNHML)]]<br />
*[[Fairfield County Math League]] [http://fcml.wikispaces.com/ website]<br />
<br />
== National math contests ==<br />
* [[List of United States elementary school mathematics competitions]] for national contests.<br />
* [[List of United States middle school mathematics competitions]] for national contests.<br />
* [[List of United States high school mathematics competitions]] for national contests.<br />
<br />
<br />
== See also ==<br />
* [[List of mathematics competitions]]<br />
* [[Mathematics competitions resources]]</div>Pi37https://artofproblemsolving.com/wiki/index.php?title=User:Pi37&diff=50449User:Pi372013-01-03T21:36:23Z<p>Pi37: Created page with "My username is pi37, not Pi37 I like geometry, G>N>A>C Also sa"</p>
<hr />
<div>My username is pi37, not Pi37<br />
<br />
I like geometry, G>N>A>C<br />
<br />
Also sa</div>Pi37https://artofproblemsolving.com/wiki/index.php?title=User:Binomial-Theorem&diff=49005User:Binomial-Theorem2012-11-01T01:01:20Z<p>Pi37: /* Feel free to edit this wiki page to insert whatever you want to, love poems, hate stories, whatever you want to, just put it in bold :). */</p>
<hr />
<div>=Welcome to '''nsun's''' wiki page=<br />
<br />
I '''loathe''' '''eating''' as '''little''' as I '''may'''. I'll '''put''' the rest of this page with '''randomness''' that '''everybody''' will get unless you are certain people. <br />
<br />
== Feel free to edit this wiki page to insert whatever you want to, love poems, hate stories, whatever you want to, just put it in bold :). ==<br />
<br />
<br />
<br />
'''I am one of the few moderators with a well-known multi''' (See [[User:DSquared]] for more information.)<br />
^congratulations, we're all proud of you<br />
'''Note my awesomeness -- Anonymous'''<br />
<br />
'''laimonib'''<br />
<br />
'''meroeht'''<br />
<br />
'''uaiwehguwahgioawhrbioawneiofnilenlbuaneifdkjsgniaunberiusgfjkc'''<br />
'''<br />
"Cool Story Bro"''' '''No tell it again?'''<br />
<br />
'''qwertyuiopasdfghjklzxcvbnm hi dsquared this is spamspamspamspamspamspamspamspamspam :P'''<br />
<br />
'''I love you.'''<br />
<br />
=Guess who said it game=<br />
Canada is cool. <br />
<br />
Don't ******* curse! '''This actually sounds like it could be my theater teacher. -- apple'''<br />
<br />
Wait, we get cheerleaders AND asian girls AT THE SAME CAMP!!!!???? :) <br />
<br />
I think I may be dead before the time you are ready for this class. <br />
<br />
Wait, what? <br />
<br />
Justin is irreducible.<br />
<br />
Anybody that doesn't understand this material please go get me markers.<br />
<br />
Don't worry about there being violence in this movie, it's only rated R.<br />
<br />
PLEASE HELP ME BEFORE NEW YORK COMES I"M IN MY PAJAMAS AND MY ROOMMATE LOCKED ME OUT OF MY ROOM!!!!!!!!!!!<br />
<br />
'''TROLLOLOLOLOLOLOLOLOLOLOLOLOLOLOLOLOLOLOLOLOLOLOLOLOLOLOLOLOLOLOLOLOLOLOLOLOLOLOLOLOLOLOLOLOLOLOLOLOLOLOLOLOLOLOLOLOLOLOLOLOLOLOLOLOLOLOLOLOLOLOL- ksun'''<br />
<br />
'''Your wiki page is going to spam up real fast. Have fun. '''<br />
<br />
'''RAWR Justin Catch!'''<br />
<br />
'''FEEL MY FAIRIES!!!!!'''<br />
justin's a nb troll -.- also veryhormonal</div>Pi37https://artofproblemsolving.com/wiki/index.php?title=2012_USAMO_Problems/Problem_4&diff=466282012 USAMO Problems/Problem 42012-04-26T18:12:34Z<p>Pi37: /* Solution */</p>
<hr />
<div>== Problem ==<br />
<br />
Find all functions <math>f : \mathbb{Z}^+ \to \mathbb{Z}^+</math> (where <math>\mathbb{Z}^+</math> is the set of positive integers) such that <math>f(n!) = f(n)!</math> for all positive integers <math>n</math> and such that <math>m - n</math> divides <math>f(m) - f(n)</math> for all distinct positive integers <math>m</math>, <math>n</math>.<br />
<br />
==Solution==<br />
Note that <math>f(1)=f(1!)=f(1)!</math> and <math>f(2)=f(2!)=f(2)!</math>, so <math>f(1)=1</math> or <math>2</math> and similarly for <math>f(2)</math>. We always have<br />
<cmath><br />
n\cdot n!=(n+1)!-n!|f(n+1)!-f(n)!<br />
</cmath><br />
(1)<br />
<br />
Now if <math>f(n)=1</math> for any <math>n\ge 2</math>, then <math>f(k)=1</math> for all <math>k\ge n</math>. This follows because then <math>f(n+1)!\equiv 1 \mod n\cdot n!</math>, which is only possible if <math>f(n+1)=1</math>, and the rest follows by induction.<br />
We now divide into cases:<br />
<br />
'''Case 1:''' <math>f(1)=f(2)=1</math> <br />
<br />
This gives <math>f(n)=1</math> always from the previous claim, which is a solution. <br />
<br />
'''Case 2:''' <math>f(1)=2, f(2)=1</math> <br />
<br />
This implies <math>f(n)=1</math> for all <math>n\ge 2</math>, but this does not satisfy the initial conditions. Indeed, we would have<br />
<cmath><br />
3-1|f(3)-f(1)<br />
</cmath><br />
so <math>2|-1</math>, a contradiction.<br />
<br />
'''Case 3:''' <math>f(1)=1</math>, <math>f(2)=2</math><br />
<br />
We claim <math>f(n)=n</math> always by induction. The bases cases are already shown. If <math>f(k)=k</math>, then by (1) we have<br />
<cmath><br />
f(k+1)\equiv k! \mod k\cdot k!<br />
</cmath><br />
Which gives <math>f(k+1)<2k</math> (otherwise <math>f(k+1)!\equiv 0 \mod k\cdot k!</math>). Also we have<br />
<cmath><br />
k+1-1|f(k+1)-f(1)<br />
</cmath><br />
so <math>f(k+1)\equiv 1 \mod k</math>. This gives the solutions <math>f(k+1)=1</math> (obviously impossible) and <math>f(k+1)=k+1</math>. Then by induction, this always holds. Note that this also satisfies the requirements.<br />
<br />
'''Case 4:''' <math>f(1)=f(2)=2</math><br />
<br />
We claim <math>f(n)=2</math> by a similar induction. Again if <math>f(k)=2</math>, then by (1) we have<br />
<cmath><br />
f(k+1)\equiv 2 \mod k\cdot k!<br />
</cmath><br />
Which gives <math>f(k+1)<2k</math> similarly. Also note that<br />
<cmath><br />
k+1-1|f(k+1)-2<br />
</cmath><br />
<cmath><br />
k+1-2|f(k+1)-2<br />
</cmath><br />
so <math>f(k+1)\equiv 2 \mod k(k-1)</math>. Then the only possible solution is <math>f(k+1)=2</math>. By induction this always holds, and note that this satisfies the requirements.<br />
<br />
The solutions are <math>\boxed{f(n)=1, f(n)=2, f(n)=n}</math>.<br />
<br />
==See also==<br />
*[[USAMO Problems and Solutions]]<br />
<br />
{{USAMO newbox|year=2012|num-b=3|num-a=5}}</div>Pi37https://artofproblemsolving.com/wiki/index.php?title=2009_AIME_II_Problems/Problem_15&diff=377892009 AIME II Problems/Problem 152011-03-27T02:29:47Z<p>Pi37: /* Solution */</p>
<hr />
<div>== Problem ==<br />
Let <math>\overline{MN}</math> be a diameter of a circle with diameter 1. 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=\frac{3}5</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 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, 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 />
== Solution ==<br />
(For some reason, I can't submit LaTeX for this page.)<br />
<br />
Let <math>O</math> be the center of the circle. Define <math>\angle{MOC}=t</math>, <math>\angle{BOA}=2a</math>, and let <math>BC</math> and <math>AC</math> intersect <math>MN</math> at points <math>X</math> and <math>Y</math>, respectively. We will express the length of <math>XY</math> as a function of <math>t</math> and maximize that function in the interval <math>[0, \pi]</math>.<br />
<br />
Let <math>C'</math> be the foot of the perpendicular from <math>C</math> to <math>MN</math>. We compute <math>XY</math> as follows.<br />
<br />
(a) By the Extended Law of Sines in triangle <math>ABC</math>, we have<br />
<br />
<cmath>CA</cmath><br />
<br />
<cmath>= \sin\angle{ABC}</cmath><br />
<br />
<cmath>= \sin\left(\frac{\widehat{AN} + \widehat{NC}}{2}\right)</cmath><br />
<br />
<cmath>= \sin\left(\frac{\frac{\pi}{2} + (\pi-t)}{2}\right)</cmath><br />
<br />
<cmath>= \sin\left(\frac{3\pi}{4} - \frac{t}{2}\right)</cmath><br />
<br />
<cmath>= \sin\left(\frac{\pi}{4} + \frac{t}{2}\right)</cmath><br />
<br />
(b) Note that <math>CC' = CO\sin(t) = \left(\frac{1}{2}\right)\sin(t)</math> and <math>AO = \frac{1}{2}</math>. Since <math>CC'Y</math> and <math>AOY</math> are similar right triangles, we have <math>CY/AY = CC'/AO = \sin(t)</math>, and hence,<br />
<br />
<cmath>CY/CA</cmath><br />
<br />
<cmath>= \frac{CY}{CY + AY}</cmath><br />
<br />
<cmath>= \frac{\sin(t)}{1 + \sin(t)}</cmath><br />
<br />
<cmath>= \frac{\sin(t)}{\sin\left(\frac{\pi}{2}\right) + \sin(t)}</cmath><br />
<br />
<cmath>= \frac{\sin(t)}{2\sin\left(\frac{\pi}{4} + \frac{t}{2}\right)\cos\left(\frac{\pi}{4} - \frac{t}{2}\right)}</cmath><br />
<br />
(c) We have <math>\angle{XCY} = \frac{\widehat{AB}}{2}=a</math> and <math>\angle{CXY} = \frac{\widehat{MB}+\widehat{CN}}{2} = \frac{\left(\frac{\pi}{2} - 2a\right) + (\pi - t)}{2} = \frac{3\pi}{4} - a - \frac{t}{2}</math>, and hence by the Law of Sines,<br />
<br />
<cmath>XY/CY</cmath><br />
<br />
<cmath>= \frac{\sin\angle{XCY}}{\sin\angle{CXY}}</cmath><br />
<br />
<cmath>= \frac{\sin(a)}{\sin\left(\frac{3\pi}{4} - a - \frac{t}{2}\right)}</cmath><br />
<br />
<cmath>= \frac{\sin(a)}{\sin\left(\frac{\pi}{4} + a + \frac{t}{2}\right)}</cmath><br />
<br />
(d) Multiplying (a), (b), and (c), we have<br />
<br />
<cmath>XY</cmath><br />
<br />
<cmath>= CA * (CY/CA) * (XY/CY)</cmath><br />
<br />
<cmath>= \frac{\sin(t)\sin(a)}{2\cos\left(\frac{\pi}{4} - \frac{t}{2}\right)\sin\left(\frac{\pi}{4} + a + \frac{t}{2}\right)}</cmath><br />
<br />
<cmath>= \frac{\sin(t)\sin(a)}{\sin\left(\frac{\pi}{2} + a\right) + \sin(a + t)}</cmath><br />
<br />
<cmath>= \sin(a)\times\frac{\sin(t)}{\sin(t + a) + \cos(a)}</cmath>,<br />
<br />
which is a function of <math>t</math> (and the constant <math>a</math>). Differentiating this with respect to <math>t</math> yields<br />
<br />
<cmath>\sin(a)\times\frac{\cos(t)(\sin(t + a) + \cos(a)) - \sin(t)\cos(t + a)}{(\sin(t + a) + \cos(a))^2}</cmath>,<br />
<br />
and the numerator of this is<br />
<br />
<cmath>\sin(a) \times(\sin(t + a)\cos(t) - \cos(t + a)\sin(t) + \cos(a)\cos(t))</cmath> <br />
<cmath>= \sin(a) \times (\sin(a) + \cos(a)\cos(t))</cmath>,<br />
<br />
which vanishes when <math>\sin(a) + \cos(a)\cos(t) = 0</math>. Therefore, the length of <math>XY</math> is maximized when <math>t=t'</math>, where <math>t'</math> is the value in <math>[0, \pi]</math> that satisfies <math>\cos(t') = -\tan(a)</math>.<br />
<br />
Note that<br />
<br />
<cmath>\frac{1 - \tan(a)}{1 + \tan(a)} = \tan\left(\frac{\pi}{4} - a\right) = \tan((\widehat{MB})/2) = \tan\angle{MNB} = \frac{3}{4}</cmath>,<br />
<br />
so <math>\tan(a) = \frac{1}{7}</math>. We compute<br />
<br />
<cmath>\sin(a) = \frac{\sqrt{2}}{10}</cmath><br />
<br />
<cmath>\cos(a) = \frac{7\sqrt{2}}{10}</cmath><br />
<br />
<cmath>\cos(t') = -\tan(a) = -\frac{1}{7}</cmath><br />
<br />
<cmath>\sin(t') = \frac{4\sqrt{3}}{7}</cmath><br />
<br />
<cmath>\sin(t' + a)=\sin(t')\cos(a) + \cos(t')\sin(a) = \frac{28\sqrt{6} - \sqrt{2}}{70}</cmath>,<br />
<br />
so the maximum length of <math>XY</math> is <math>\sin(a)\times\frac{\sin(t')}{\sin(t' + a) + \cos(a)} = 7 - 4sqrt(3)</math>, and the answer is <math>7 + 4 + 3 = \boxed{014}</math>.</div>Pi37https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12B_Problems/Problem_7&diff=372822011 AMC 12B Problems/Problem 72011-03-06T22:35:46Z<p>Pi37: /* See also */</p>
<hr />
<div>==Problem==<br />
Let <math>x</math> and <math>y</math> be two-digit positive integers with mean <math>60</math>. What is the maximum value of the ratio <math>\frac{x}{y}</math>?<br />
<br />
<math>\textbf{(A)}\ 3 \qquad \textbf{(B)}\ \frac{33}{7} \qquad \textbf{(C)}\ \frac{39}{7} \qquad \textbf{(D)}\ 9 \qquad \textbf{(E)}\ \frac{99}{10}</math><br />
==Solution==<br />
If <math>x</math> and <math>y</math> have a mean of <math>60</math>, then <math>\frac{x+y}{2}=60</math> and <math>x+y=120</math>. To maximize <math>\frac{x}{y}</math>, we need to maximize <math>x</math> and minimize <math>y</math>. Since they are both two-digit positive integers, the maximum of <math>x</math> is <math>99</math> which gives <math>y=21</math>. <math>y</math> cannot be decreased because doing so would increase <math>x</math>, so this gives the maximum value of <math>\frac{x}{y}</math>, which is <math>\frac{99}{21}=\boxed{\frac{33}{7}\ \textbf{(B)}}</math><br />
==See also==<br />
{{AMC12 box|year=2011|num-b=6|num-a=8|ab=B}}</div>Pi37https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12B_Problems/Problem_5&diff=372802011 AMC 12B Problems/Problem 52011-03-06T22:33:07Z<p>Pi37: /* See also */</p>
<hr />
<div>== Problem ==<br />
Let <math>N</math> be the second smallest positive integer that is divisible by every positive integer less than <math>7</math>. What is the sum of the digits of <math>N</math>?<br />
<br />
<math>\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 9</math><br />
==Solution==<br />
<math>N</math> must be divisible by every positive integer less than <math>7</math>, or <math>1, 2, 3, 4, 5,</math> and <math>6</math>. Each number that is divisible by each of these is is a multiple of their least common multiple. <math>LCM(1,2,3,4,5,6)=60</math>, so each number divisible by these is a multiple of <math>60</math>. The smallest multiple of <math>60</math> is clearly <math>60</math>, so the second smallest multiple of <math>60</math> is <math>2\times60=120</math>. Therefore, the sum of the digits of <math>N</math> is <math>1+2+0=\boxed{3\ \textbf{(A)}}</math><br />
<br />
==See also==<br />
{{AMC12 box|year=2011|num-b=4|num-a=6|ab=B}}</div>Pi37https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12B_Problems/Problem_7&diff=372782011 AMC 12B Problems/Problem 72011-03-06T18:49:37Z<p>Pi37: Created page with '==Problem== Let <math>x</math> and <math>y</math> be two-digit positive integers with mean <math>60</math>. What is the maximum value of the ratio <math>\frac{x}{y}</math>? <mat…'</p>
<hr />
<div>==Problem==<br />
Let <math>x</math> and <math>y</math> be two-digit positive integers with mean <math>60</math>. What is the maximum value of the ratio <math>\frac{x}{y}</math>?<br />
<br />
<math>\textbf{(A)}\ 3 \qquad \textbf{(B)}\ \frac{33}{7} \qquad \textbf{(C)}\ \frac{39}{7} \qquad \textbf{(D)}\ 9 \qquad \textbf{(E)}\ \frac{99}{10}</math><br />
==Solution==<br />
If <math>x</math> and <math>y</math> have a mean of <math>60</math>, then <math>\frac{x+y}{2}=60</math> and <math>x+y=120</math>. To maximize <math>\frac{x}{y}</math>, we need to maximize <math>x</math> and minimize <math>y</math>. Since they are both two-digit positive integers, the maximum of <math>x</math> is <math>99</math> which gives <math>y=21</math>. <math>y</math> cannot be decreased because doing so would increase <math>x</math>, so this gives the maximum value of <math>\frac{x}{y}</math>, which is <math>\frac{99}{21}=\boxed{\frac{33}{7}\ \textbf{(B)}}</math><br />
==See also==</div>Pi37https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12B_Problems/Problem_5&diff=372772011 AMC 12B Problems/Problem 52011-03-06T18:36:40Z<p>Pi37: /* Solution */</p>
<hr />
<div>== Problem ==<br />
Let <math>N</math> be the second smallest positive integer that is divisible by every positive integer less than <math>7</math>. What is the sum of the digits of <math>N</math>?<br />
<br />
<math>\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 9</math><br />
==Solution==<br />
<math>N</math> must be divisible by every positive integer less than <math>7</math>, or <math>1, 2, 3, 4, 5,</math> and <math>6</math>. Each number that is divisible by each of these is is a multiple of their least common multiple. <math>LCM(1,2,3,4,5,6)=60</math>, so each number divisible by these is a multiple of <math>60</math>. The smallest multiple of <math>60</math> is clearly <math>60</math>, so the second smallest multiple of <math>60</math> is <math>2\times60=120</math>. Therefore, the sum of the digits of <math>N</math> is <math>1+2+0=\boxed{3\ \textbf{(A)}}</math><br />
<br />
==See also==</div>Pi37https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12B_Problems/Problem_5&diff=372762011 AMC 12B Problems/Problem 52011-03-06T18:36:25Z<p>Pi37: /* Solution */</p>
<hr />
<div>== Problem ==<br />
Let <math>N</math> be the second smallest positive integer that is divisible by every positive integer less than <math>7</math>. What is the sum of the digits of <math>N</math>?<br />
<br />
<math>\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 9</math><br />
==Solution==<br />
<math>N</math> must be divisible by every positive integer less than <math>7</math>, or <math>1, 2, 3, 4, 5,</math> and <math>6</math>. Each number that is divisible by each of these is is a multiple of their least common multiple. <math>LCM(1,2,3,4,5,6)=60</math>, so each number divisible by these is a multiple of <math>60</math>. The smallest multiple of <math>60</math> is clearly <math>60</math>, so the second smallest multiple of <math>60</math> is <math>2\times60=120</math>. Therefore, the sum of the digits of <math>N</math> is <math>1+2+0=\boxed{3\ \textbf{(A)}}</math><br />
==See also==</div>Pi37https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12B_Problems/Problem_5&diff=372752011 AMC 12B Problems/Problem 52011-03-06T18:35:28Z<p>Pi37: /* Problem */</p>
<hr />
<div>== Problem ==<br />
Let <math>N</math> be the second smallest positive integer that is divisible by every positive integer less than <math>7</math>. What is the sum of the digits of <math>N</math>?<br />
<br />
<math>\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 9</math><br />
==Solution==<br />
<math>N</math> must be divisible by every positive integer less than <math>7</math>, or <math>1, 2, 3, 4, 5,</math> and <math>6</math>. Each number that is divisible by each of these is is a multiple of their least common multiple. <math>LCM(1,2,3,4,5,6)=60</math>, so each number divisible by these is a multiple of <math>60</math>. The smallest multiple of <math>60</math> is clearly <math>60</math>, so the second smallest multiple of <math>60</math> is <math>2\times60=120</math>. Therefore, the sum of the digits of <math>N</math> is <math>1+2+0=\boxed{3\ \textbf{(A)}}</math></div>Pi37https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12B_Problems/Problem_5&diff=372742011 AMC 12B Problems/Problem 52011-03-06T18:23:21Z<p>Pi37: /* Problem */</p>
<hr />
<div>== Problem ==<br />
Let <math>N</math> be the second smallest positive integer that is divisible by every positive integer less than <math>7</math>. What is the sum of the digits of <math>N</math>?<br />
<br />
<math>\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 9</math></div>Pi37https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12B_Problems/Problem_4&diff=372732011 AMC 12B Problems/Problem 42011-03-06T18:19:16Z<p>Pi37: /* Solution */</p>
<hr />
<div>== Problem ==<br />
In multiplying two positive integers <math>a</math> and <math>b</math>, Ron reversed the digits of the two-digit number <math>a</math>. His erroneous product was <math>161.</math> What is the correct value of the product of <math>a</math> and <math>b</math>? <br />
<br />
<math><br />
\textbf{(A)}\ 116 \qquad<br />
\textbf{(B)}\ 161 \qquad<br />
\textbf{(C)}\ 204 \qquad<br />
\textbf{(D)}\ 214 \qquad<br />
\textbf{(E)}\ 224 </math><br />
<br />
<br />
== Solution ==<br />
Taking the prime factorization of <math>161</math> reveals that it is equal to <math>23*7.</math> Therefore, the only ways to represent <math>161</math> as a product of two positive integers is <math>161*1</math> and <math>23*7.</math> Because neither <math>161</math> nor <math>1</math> is a two-digit number, we know that <math>a</math> and <math>b</math> are <math>23</math> and <math>7.</math> Because <math>23</math> is a two-digit number, we know that a, with its two digits reversed, gives <math>23.</math> Therefore, <math>a = 32</math> and <math>b = 7.</math> Multiplying our two correct values of <math>a</math> and <math>b</math> yields<br />
<br />
<cmath> a*b = 32*7 = </cmath><br />
<br />
<cmath> = \boxed{224\ \(\textbf{(E)}} </cmath><br />
<br />
== See also ==<br />
{{AMC12 box|year=2011|before=Problem 3|num-a=5|ab=B}}</div>Pi37https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12B_Problems/Problem_3&diff=372722011 AMC 12B Problems/Problem 32011-03-06T18:18:39Z<p>Pi37: /* Solution */</p>
<hr />
<div>== Problem ==<br />
LeRoy and Bernardo went on a week-long trip together and agreed to share the costs equally. Over the week, each of them paid for various joint expenses such as gasoline and car rental. At the end of the trip it turned out that LeRoy had paid A dollars and Bernardo had paid B dollars, where <math>A < B.</math> How many dollars must LeRoy give to Bernardo so that they share the costs equally?<br />
<br />
<math><br />
\textbf{(A)}\ \frac{A+B}{2} \qquad<br />
\textbf{(B)}\ \frac{A-B}{2} \qquad<br />
\textbf{(C)}\ \frac{B-A}{2} \qquad<br />
\textbf{(D)}\ B-A \qquad<br />
\textbf{(E)}\ A+B </math><br />
<br />
<br />
== Solution ==<br />
The total amount of money that was spent during the trip was<br />
<cmath> A + B </cmath><br />
So each person should pay<br />
<cmath> \frac{A+B}{2} </cmath><br />
if they were to share the costs equally. Because LeRoy has already paid <math>A</math> dollars of his part, he still has to pay<br />
<cmath> \frac{A+B}{2} - A = </cmath><br />
<cmath> = \boxed{\frac{B-A}{2}\ \(\textbf{(C)}} </cmath><br />
<br />
== See also ==<br />
{{AMC12 box|year=2011|before=Problem 2|num-a=4|ab=B}}</div>Pi37https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12B_Problems/Problem_2&diff=372712011 AMC 12B Problems/Problem 22011-03-06T18:18:02Z<p>Pi37: /* Solution */</p>
<hr />
<div>== Problem ==<br />
Josanna's test scores to date are <math>90, 80, 70, 60,</math> and <math>85.</math> Her goal is to raise her test average at least <math>3</math> points with her next test. What is the minimum test score she would need to accomplish this goal?<br />
<br />
<math><br />
\textbf{(A)}\ 80 \qquad<br />
\textbf{(B)}\ 82 \qquad<br />
\textbf{(C)}\ 85 \qquad<br />
\textbf{(D)}\ 90 \qquad<br />
\textbf{(E)}\ 95 </math><br />
<br />
<br />
== Solution ==<br />
Take the average of her current test scores, which is<br />
<cmath> \frac{90+80+70+60+85}{5} = \frac{385}{5} = 77 </cmath><br />
<br />
This means that she wants her test average after the sixth test to be <math>80.</math> Let <math>x</math> be the score that Josanna receives on her sixth test. Thus, our equation is<br />
<br />
<cmath> \frac{90+80+70+60+85+x}{6} = 80 </cmath><br />
<br />
<cmath> 385+x = 480 </cmath><br />
<br />
<cmath> x = \boxed{95\ \(\textbf{(E)}} </cmath><br />
<br />
== See also ==<br />
{{AMC12 box|year=2011|before=Problem 1|num-a=3|ab=B}}</div>Pi37https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_12A_Problems/Problem_24&diff=370352006 AMC 12A Problems/Problem 242011-02-21T01:39:10Z<p>Pi37: </p>
<hr />
<div>== Problem ==<br />
<br />
The expression <br />
<br />
<math>(x+y+z)^{2006}+(x-y-z)^{2006}</math><br />
<br />
is simplified by expanding it and combining like terms. How many terms are in the simplified expression?<br />
<br />
<math> \mathrm{(A) \ } 6018\qquad \mathrm{(B) \ } 671,676\qquad \mathrm{(C) \ } 1,007,514\qquad \mathrm{(D) \ } 1,008,016\qquad\mathrm{(E) \ } 2,015,028</math><br />
<br />
== Solution 1==<br />
By the [[Multinomial Theorem]], the summands can be written as<br />
<br />
<cmath>\sum_{a+b+c=2006}{\frac{2006!}{a!b!c!}x^ay^bz^c}</cmath><br />
<br />
and<br />
<br />
<cmath>\sum_{a+b+c=2006}{\frac{2006!}{a!b!c!}x^a(-y)^b(-z)^c},</cmath><br />
<br />
respectively. Since the coefficients of like terms are the same in each expression, each like term either cancel one another out or the coefficient doubles. In each expansion there are:<br />
<br />
<cmath>{2006+2\choose 2} = 2015028</cmath><br />
<br />
terms without cancellation. For any term in the second expansion to be negative, the parity of the exponents of <math>y</math> and <math>z</math> must be opposite. Now we find a pattern:<br />
<br />
if the exponent of <math>y</math> is 1, the exponent of <math>z</math> can be all even integers up to 2004, so 1003 terms.<br />
<br />
if the exponent of <math>y</math> is 3, the exponent of <math>z</math> can go up to 2002, so 1002 terms.<br />
<br />
<math>\vdots</math><br />
<br />
if the exponent of <math>y</math> is 2005, then <math>z</math> can only be 0. So 1 term.<br />
<br />
add them up we get <math>\frac{1003\cdot1004}{2}</math> terms. However, we can switch the exponents of <math>y</math> and <math>z</math> and these terms will still have a negative sign. So there are a total of <math>1003\cdot1004</math> negative terms.<br />
<br />
Subtract this number from 2015028 we obtain <math>D. 1008016</math> as our answer.<br />
<br />
<br />
Alternatively, we can use a generating function to solve this problem. <br />
The goal is to find the generating function for the number of unique terms in the simplified expression (in terms of <math>k</math>). In other words, we want to find <math>f(x)</math> where the coefficient of <math>x^k</math> equals the number of unique terms in <math>(x+y+z)^k + (x-y-z)^k</math>.<br />
<br />
<br />
First, we note that all unique terms in the expression have the form, <math>Cx^ay^bz^c</math>, where <math>a+b+c=k</math> and <math>C</math> is some constant. Therefore, the generating function for the MAXIMUM number of unique terms possible in the simplified expression of <math>(x+y+z)^k + (x-y-z)^k</math> is<br />
<cmath>(1+x+x^2+x^3\cdots)^3 = \frac{1}{(1-x)^3}</cmath><br />
<br />
<br />
Secondly, we note that a certain number of terms of the form, <math>Cx^ay^bz^c</math>, do not appear in the simplified version of our expression because those terms cancel. Specifically, we observe that terms cancel when <math>1 \equiv b+c\text{ (mod }2\text{)}</math> because every unique term is of the form:<br />
<cmath>\binom{k}{a,b,c}x^ay^bz^c+(-1)^{b+c}\binom{k}{a,b,c}x^ay^bz^c</cmath><br />
for all possible <math>a,b,c</math>.<br />
<br />
<br />
Since the generating function for the maximum number of unique terms is already known, it is logical that we want to find the generating function for the number of terms that cancel, also in terms of <math>k</math>. With some thought, we see that this desired generating function is the following:<br />
<cmath>2(x+x^3+x^5\cdots)(1+x^2+x^4\cdots)(1+x+x^2+x^3\cdots) = \frac{2x}{(1-x)^3(1+x)^2}</cmath><br />
<br />
<br />
Now, we want to subtract the latter from the former in order to get the generating function for the number of unique terms in <math>(x+y+z)^k + (x-y-z)^k</math>, our initial goal:<br />
<cmath>\frac{1}{(1-x)^3}-\frac{2x}{(1-x)^3(1+x)^2} = \frac{x^2+1}{(1-x)^3(1+x)^2}</cmath><br />
which equals<br />
<cmath>(x^2+1)(1+x+x^2\cdots)^3(1-x+x^2-x^3\cdots)^2</cmath><br />
<br />
<br />
The coefficient of <math>x^{2006}</math> of the above expression equals<br />
<cmath>\sum_{a=0}^{2006}\binom{2+a}{2}\binom{1+2006-a}{1}(-1)^a + \sum_{a=0}^{2004}\binom{2+a}{2}\binom{1+2004-a}{1}(-1)^a</cmath><br />
<br />
<br />
Evaluating the expression, we get <math>1008016</math>, as expected.<br />
<br />
== Solution 2 ==<br />
<br />
Define <math>P</math> such that <math>P=y+z</math>. Then the expression in the problem becomes:<br />
<math>(x+P)^{2006}+(x-P)^{2006}</math>. <br />
<br />
Expanding this using binomial theorem gives <math>x^n+P*x^{n-1}+...+P^{n-1}*x+P^n+x^n-P*x^{n-1}+...-P^{n-1}*x+P^n</math> (we may omit the coefficients, as we are seeking for the number of terms, not the terms themselves). <br />
<br />
Simplifying gives: <math>2(x^n+x^{n-2}*P^2+...+x^2*P^{n-2}+x^n)</math>. We can also take out all the 2 and all the x terms, as they will not affect the answer.<br />
<br />
Thus, we must find the number of terms in this expression:<br />
<br />
<math>1+P^2+P^4+...+P^{2004}+P^{2006}</math>.<br />
<br />
Because <math>P=y+z</math>, <math>P^n</math> will have n+1 terms, by the binomial theorem. Thus, the expression will have <math>1+3+5+...+2005+2007</math> terms. <br />
<br />
We can easily find this sum by noting that this is equal to the sum of the first 1004 consecutive odd integers, or <math>1004^2</math>, which gives us <math>1,008,016</math>, or C.<br />
<br />
== See also ==<br />
* [[2006 AMC 12A Problems]]<br />
<br />
{{AMC12 box|year=2006|ab=A|num-b=23|num-a=25}}<br />
<br />
[[Category:Introductory Algebra Problems]]<br />
[[Category:Introductory Combinatorics Problems]]</div>Pi37