https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Thedoge&feedformat=atom AoPS Wiki - User contributions [en] 2021-06-24T02:07:32Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=1959_IMO_Problems/Problem_1&diff=93949 1959 IMO Problems/Problem 1 2018-04-14T22:46:28Z <p>Thedoge: </p> <hr /> <div>== Problem ==<br /> <br /> Prove that the fraction &lt;math&gt;\frac{21n+4}{14n+3}&lt;/math&gt; is irreducible for every natural number &lt;math&gt;n&lt;/math&gt;.<br /> <br /> <br /> == Solutions ==<br /> <br /> === First Solution ===<br /> <br /> <br /> <br /> For this fraction to be reducible there must be a number &lt;math&gt;x&lt;/math&gt; such that &lt;math&gt;x \cdot (14n+3) = (21n+4)&lt;/math&gt;, and a &lt;math&gt;1/x&lt;/math&gt; such that &lt;math&gt;1/x \cdot (21n+4) = (14n+3)&lt;/math&gt;. Since &lt;math&gt;x&lt;/math&gt; can only be one number (&lt;math&gt;x&lt;/math&gt; is a linear term) we only have to evaluate &lt;math&gt;x&lt;/math&gt; for one of these equations. Using the first one, &lt;math&gt;x&lt;/math&gt; would have to equal &lt;math&gt;3/2&lt;/math&gt;. However, &lt;math&gt;3*3/2&lt;/math&gt; results in &lt;math&gt;9/2&lt;/math&gt;, and is not equal to our desired &lt;math&gt;4&lt;/math&gt;. Since there is no &lt;math&gt;x&lt;/math&gt; to make the numerator and denominator equal, we can conclude the fraction is irreducible.<br /> <br /> EDIT: It appears to me that this solution is incorrect because it assumes that if &lt;math&gt;ax + b = cx + d&lt;/math&gt;, then &lt;math&gt;a = c&lt;/math&gt; and &lt;math&gt;b = d&lt;/math&gt; -- the problem says irreducible for ALL &lt;math&gt;n&lt;/math&gt;. If you agree, please remove this solution. 12/17/2017<br /> <br /> EDIT 2: Disregard the first edit, because when comparing two polynomials (linear functions in this case), the coefficients for each term must be equal to eachother. For example, if &lt;math&gt;x&lt;/math&gt; can be any number, then in &lt;math&gt;ax^2+4=6x^2+b&lt;/math&gt;, we must have &lt;math&gt;(a,b)=(6,4)&lt;/math&gt;. Thus, if &lt;math&gt;ax+b=cx+d&lt;/math&gt; for all values &lt;math&gt;x&lt;/math&gt;, then &lt;math&gt;a&lt;/math&gt; must equal &lt;math&gt;c&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; must equal &lt;math&gt;d&lt;/math&gt;.<br /> <br /> Clarification for edit 2: In this problem, we have &lt;math&gt;14nx+3x=21n+4&lt;/math&gt;, with an unknown &lt;math&gt;n&lt;/math&gt;, so we must have &lt;math&gt;14x=21&lt;/math&gt; and &lt;math&gt;3x=4&lt;/math&gt;, which cannot be true for any &lt;math&gt;1&lt;/math&gt; value of &lt;math&gt;x&lt;/math&gt;.<br /> <br /> === Second Solution ===<br /> <br /> Denoting the greatest common divisor of &lt;math&gt;a, b &lt;/math&gt; as &lt;math&gt;(a,b) &lt;/math&gt;, we use the [[Euclidean algorithm]] as follows:<br /> <br /> &lt;center&gt;<br /> &lt;math&gt;( 21n+4, 14n+3 ) = ( 7n+1, 14n+3 ) = ( 7n+1, 1 ) = 1 &lt;/math&gt;<br /> &lt;/center&gt;<br /> <br /> As in the first solution, it follows that &lt;math&gt;\frac{21n+4}{14n+3}&lt;/math&gt; is irreducible. Q.E.D.<br /> <br /> === Third Solution ===<br /> <br /> [[Proof by contradiction]]:<br /> <br /> Assume that &lt;math&gt;\dfrac{14n+3}{21n+4}&lt;/math&gt; is a [[reducible fraction]] where &lt;math&gt;p&lt;/math&gt; is a [[divisor]] of both the [[numerator]] and the [[denominator]]:<br /> <br /> &lt;center&gt;<br /> &lt;math&gt;14n+3\equiv 0\pmod{p} \implies 42n+9\equiv 0\pmod{p}&lt;/math&gt;<br /> <br /> &lt;math&gt;21n+4\equiv 0\pmod{p} \implies 42n+8\equiv 0\pmod{p}&lt;/math&gt;<br /> &lt;/center&gt;<br /> <br /> Subtracting the second [[equation]] from the first [[equation]] we get &lt;math&gt;1\equiv 0\pmod{p}&lt;/math&gt; which is clearly absurd. <br /> <br /> Hence &lt;math&gt;\frac{21n+4}{14n+3}&lt;/math&gt; is irreducible. Q.E.D.<br /> &lt;!--Solution by tonypr--&gt;<br /> <br /> <br /> === Fourth Solution ===<br /> <br /> We notice that:<br /> <br /> &lt;center&gt;<br /> &lt;math&gt;\frac{21n+4}{14n+3} = \frac{(14n+3)+(7n+1)}{14n+3} = 1+\frac{7n+1}{14n+3}&lt;/math&gt;<br /> &lt;/center&gt;<br /> <br /> So it follows that &lt;math&gt;7n+1&lt;/math&gt; and &lt;math&gt;14n+3&lt;/math&gt; must be coprime for every natural number &lt;math&gt;n&lt;/math&gt; for the fraction to be irreducible. Now the problem simplifies to proving &lt;math&gt; \frac{7n+1}{14n+3} &lt;/math&gt; irreducible. We re-write this fraction as:<br /> <br /> &lt;center&gt;<br /> &lt;math&gt; \frac{7n+1}{(7n+1)+(7n+1) + 1} = \frac{7n+1}{2(7n+1)+1}&lt;/math&gt; <br /> &lt;/center&gt;<br /> <br /> Since the denominator &lt;math&gt; 2(7n+1) + 1 &lt;/math&gt; differs from a multiple of the numerator &lt;math&gt;7n+1&lt;/math&gt; by 1, the numerator and the denominator must be relatively prime natural numbers. Hence it follows that &lt;math&gt; \frac{21n+4}{14n+3}&lt;/math&gt; is irreducible.<br /> <br /> Q.E.D<br /> <br /> <br /> ===Fifth Solution===<br /> By Bezout's Theorem, &lt;math&gt;3 \cdot (14n+3) - 2 \cdot (21n + 4) = 1&lt;/math&gt;, so the gcd of the numerator and denominator is 1 and the fraction is irreducible.<br /> <br /> <br /> {{alternate solutions}}<br /> <br /> <br /> == See Also ==<br /> {{IMO box|year=1959|before=First question|num-a=2}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> [[Category:Intermediate Number Theory Problems]]</div> Thedoge https://artofproblemsolving.com/wiki/index.php?title=2015_USAMO_Problems&diff=93893 2015 USAMO Problems 2018-04-12T00:03:24Z <p>Thedoge: /* Problem 6 */</p> <hr /> <div>==Day 1==<br /> ===Problem 1===<br /> Solve in integers the equation<br /> &lt;cmath&gt; x^2+xy+y^2 = \left(\frac{x+y}{3}+1\right)^3. &lt;/cmath&gt;<br /> <br /> [[2015 USAMO Problems/Problem 1|Solution]]<br /> <br /> ===Problem 2===<br /> Quadrilateral &lt;math&gt;APBQ&lt;/math&gt; is inscribed in circle &lt;math&gt;\omega&lt;/math&gt; with &lt;math&gt;\angle P = \angle Q = 90^{\circ}&lt;/math&gt; and &lt;math&gt;AP = AQ &lt; BP&lt;/math&gt;. Let &lt;math&gt;X&lt;/math&gt; be a variable point on segment &lt;math&gt;\overline{PQ}&lt;/math&gt;. Line &lt;math&gt;AX&lt;/math&gt; meets &lt;math&gt;\omega&lt;/math&gt; again at &lt;math&gt;S&lt;/math&gt; (other than &lt;math&gt;A&lt;/math&gt;). Point &lt;math&gt;T&lt;/math&gt; lies on arc &lt;math&gt;AQB&lt;/math&gt; of &lt;math&gt;\omega&lt;/math&gt; such that &lt;math&gt;\overline{XT}&lt;/math&gt; is perpendicular to &lt;math&gt;\overline{AX}&lt;/math&gt;. Let &lt;math&gt;M&lt;/math&gt; denote the midpoint of chord &lt;math&gt;\overline{ST}&lt;/math&gt;. As &lt;math&gt;X&lt;/math&gt; varies on segment &lt;math&gt;\overline{PQ}&lt;/math&gt;, show that &lt;math&gt;M&lt;/math&gt; moves along a circle.<br /> <br /> [[2015 USAMO Problems/Problem 2|Solution]]<br /> <br /> ===Problem 3===<br /> Let &lt;math&gt;S = \{1, 2, ..., n\}&lt;/math&gt;, where &lt;math&gt;n \ge 1&lt;/math&gt;. Each of the &lt;math&gt;2^n&lt;/math&gt; subsets of &lt;math&gt;S&lt;/math&gt; is to be colored red or blue. (The subset itself is assigned a color and not its individual elements.) For any set &lt;math&gt;T \subseteq S&lt;/math&gt;, we then write &lt;math&gt;f(T)&lt;/math&gt; for the number of subsets of T that are blue.<br /> <br /> Determine the number of colorings that satisfy the following condition: for any subsets &lt;math&gt;T_1&lt;/math&gt; and &lt;math&gt;T_2&lt;/math&gt; of &lt;math&gt;S&lt;/math&gt;,<br /> &lt;cmath&gt;f(T_1)f(T_2) = f(T_1 \cup T_2)f(T_1 \cap T_2).&lt;/cmath&gt;<br /> <br /> [[2015 USAMO Problems/Problem 3|Solution]]<br /> <br /> ==Day 2==<br /> <br /> ===Problem 4===<br /> <br /> Steve is piling &lt;math&gt;m\geq 1&lt;/math&gt; indistinguishable stones on the squares of an &lt;math&gt;n\times n&lt;/math&gt; grid. Each square can have an arbitrarily high pile of stones. After he finished piling his stones in some manner, he can then perform stone moves, defined as follows. Consider any four grid squares, which are corners of a rectangle, i.e. in positions &lt;math&gt;(i, k), (i, l), (j, k), (j, l)&lt;/math&gt; for some &lt;math&gt;1\leq i, j, k, l\leq n&lt;/math&gt;, such that &lt;math&gt;i&lt;j&lt;/math&gt; and &lt;math&gt;k&lt;l&lt;/math&gt;. A stone move consists of either removing one stone from each of &lt;math&gt;(i, k)&lt;/math&gt; and &lt;math&gt;(j, l)&lt;/math&gt; and moving them to &lt;math&gt;(i, l)&lt;/math&gt; and &lt;math&gt;(j, k)&lt;/math&gt; respectively, or removing one stone from each of &lt;math&gt;(i, l)&lt;/math&gt; and &lt;math&gt;(j, k)&lt;/math&gt; and moving them to &lt;math&gt;(i, k)&lt;/math&gt; and &lt;math&gt;(j, l)&lt;/math&gt; respectively.<br /> <br /> Two ways of piling the stones are equivalent if they can be obtained from one another by a sequence of stone moves.<br /> <br /> How many different non-equivalent ways can Steve pile the stones on the grid?<br /> <br /> [[2015 USAMO Problems/Problem 4|Solution]]<br /> <br /> ===Problem 5===<br /> Let &lt;math&gt;a, b, c, d, e&lt;/math&gt; be distinct positive integers such that &lt;math&gt;a^4 + b^4 = c^4 + d^4 = e^5&lt;/math&gt;. Show that &lt;math&gt;ac + bd&lt;/math&gt; is a composite number.<br /> <br /> [[2015 USAMO Problems/Problem 5|Solution]]<br /> <br /> ===Problem 6===<br /> Consider &lt;math&gt;0&lt;\lambda&lt;1&lt;/math&gt;, and let &lt;math&gt;A&lt;/math&gt; be a multiset of positive integers. Let &lt;math&gt;A_n=\{a\in A: a\leq n\}&lt;/math&gt;. Assume that for every &lt;math&gt;n\in\mathbb{N}&lt;/math&gt;, the set &lt;math&gt;A_n&lt;/math&gt; contains at most &lt;math&gt;n\lambda&lt;/math&gt; numbers. Show that there are infinitely many &lt;math&gt;n\in\mathbb{N}&lt;/math&gt; for which the sum of the elements in &lt;math&gt;A_n&lt;/math&gt; is at most &lt;math&gt;\frac{n(n+1)}{2}\lambda&lt;/math&gt;. (A multiset is a set-like collection of elements in which order is ignored, but repetition of elements is allowed and multiplicity of elements is significant. For example, multisets &lt;math&gt;\{1, 2, 3\}&lt;/math&gt; and &lt;math&gt;\{2, 1, 3\}&lt;/math&gt; are equivalent, but &lt;math&gt;\{1, 1, 2, 3\}&lt;/math&gt; and &lt;math&gt;\{1, 2, 3\}&lt;/math&gt; differ.)<br /> <br /> <br /> {{USAMO newbox|year= 2015 |before=[[2014 USAMO]]|after=[[2016 USAMO]]}}<br /> <br /> <br /> {{MAA Notice}}</div> Thedoge https://artofproblemsolving.com/wiki/index.php?title=AoPS_Wiki:FAQ&diff=87762 AoPS Wiki:FAQ 2017-10-10T01:50:03Z <p>Thedoge: /* AoPS Acronyms */</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&amp;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 /> <br /> ==== Something looks weird (e.g., blurry, missing line, etc.) ====<br /> <br /> : This is likely due to your browser zoom level. Please make sure your zoom level is at 100% for correct rendering of the web page.<br /> <br /> ==== Can I make more than one account?==== <br /> <br /> :Short answer: No.<br /> <br /> :Long answer: Multiple accounts (multis) are banned on AoPS. Having more than one account leads to issues of not remembering on what account you did what. Using multiple accounts to &quot;game&quot; the system, (e.g. increase rating for posts or in online games) will lead to bans on all accounts associated to you. If you have already made additional accounts, please choose one account and stop using the others.<br /> <br /> ====What software does Art of Problem Solving use to run the website?====<br /> <br /> :* Search: Solr<br /> :* Wiki: MediaWiki<br /> :* Asymptote and Latex are generated through their respective binary packages<br /> :* Videos: YouTube<br /> <br /> :All other parts of the website are custom built.<br /> <br /> ====Can you make an AoPS App?====<br /> <br /> :There are currently no plans to build an app. Any app built would work almost identically to the website, and not be any faster, so there is little value in building an app.<br /> <br /> == Forums ==<br /> <br /> ====How do I create a forum?====<br /> <br /> :To create an AoPS forum, an user must be on the AoPS community for at least 2 weeks. To create a forum, hover over the community tab, then click &quot;My AoPS.&quot; You should now see your avatar, and a list of your friends. Now, click on the &quot;My Forums&quot; tab. There, you would be able to see which forum you moderate or administrate, as well as the private forums you can access. Click on the &quot;+&quot; button at the top right. This should lead you to a forum creating page.<br /> <br /> ==== How do I format my post, e.g. bold text, add URLs, etc.? ====<br /> <br /> :AoPS is based on a markup language called BBCode. A tutorial of its functions on AoPS and how to use them can be found [[BBCode:Tutorial|here]].<br /> <br /> ==== How do I hide content in the forums? ====<br /> :Wrap the content you want to hide in [hide] tags.<br /> [hide]Content[/hide]<br /> :If you want to customize the label, instead of saying &quot;Click here to reveal hidden text&quot;, you can do something like:<br /> [hide=Label to display]Content[/hide]<br /> <br /> ==== I got the message &quot;You can not post at this time&quot; 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 &quot;Too many messages.&quot; 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, Fun Factory, and most user created forums, the post does not count towards your overall post count.<br /> <br /> ==== The time the post was posted seems wrong ====<br /> <br /> :Posts may say something weird for when posted, &quot;such as 5 minutes ago&quot; or some time in the future when you just posted. This is due to your system clock being incorrect. Please update your system clock to the correct time. Many operating systems have an option to keep the clock accurate automatically. Mac users may wish to check [http://www.macinstruct.com/node/92 Synchronize your Mac's Clock With A Time Server]. You can also check [http://www.time.gov the US offical time.]<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 scout 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 &quot;!&quot; icon on the upper-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 &quot;backseat moderation&quot; and is considered rude and unproductive.<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, it isn't considered reviving. However, everyone has a different period of time that they consider reviving. In general, apply common sense.<br /> <br /> ==== How do I post images? ====<br /> <br /> :While AoPS forums have the ability to attach images, we do not generally recommend doing so, as we can not guarantee the images will be available through upgrades, restorations, etc. We also have limited disk space which causes us to remove attachments from time to time. Therefore, we recommend using a third party image hosting solution. There are many options, but we recommend imgur.com.<br /> <br /> :* Go to imgur.com<br /> :* Click upload images<br /> :* Follow the on screen instructions to upload the image to imgur.<br /> :* After uploading, you'll be presented the uploaded image along with links on the left.<br /> :* Find the link labeled BBCode (message boards &amp; forums)<br /> :* Copy the code, and paste into your post. It will look something like:<br /> <br /> [img]http://i.imgur.com/aBcDeFgH.jpg[/img]<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 /> ==== How do I make my blog look nice? ====<br /> :Many AoPSers make their blogs look awesome by applying [[CSS|CSS]], which is a high-level stylesheet language. This can be done by typing CSS code into the CSS box in the Blog Control Panel.<br /> <br /> <br /> == Alcumus ==<br /> <br /> ==== How is rating computed? ====<br /> :The rating is more of a prediction of what percentage of problems in the topic the Alcumus engine believes you will get correct. As you get more and more correct, the rating will go up slower and slower. However, if you are predicted to get most correct, and you miss one or two problems, the rating, or prediction of percentage correct, will go down.<br /> <br /> ==== I am stuck on a problem, changing the topic does not change the problem. ====<br /> :Alcumus provides review problems to make sure you still recall information learned from the past. You are not able to skip these problems. You will need to answer the problem before moving on.<br /> <br /> ==== Why can't I change topics? ====<br /> :Alcumus provides review problems to make sure you still recall information learned from the past. You are not able to skip these problems. You will need to answer the problem before the topic changes to the currently selected topic.<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 /> ==== 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&amp;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&amp;t=426693 this thread]. If the problem is on the wiki, you can edit it yourself.<br /> <br /> == LaTeX and Asymptote ==<br /> ==== What is LaTeX, and how do I use it? ====<br /> <br /> :&lt;math&gt;\LaTeX&lt;/math&gt; 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]. You can also use our [http://artofproblemsolving.com/community/c67_test_forum Test Forum].<br /> <br /> ==== Where can I find examples of Asymptote diagrams and code? ====<br /> <br /> :Search this wiki for the &lt;tt&gt;&lt;nowiki&gt;&lt;asy&gt;&lt;/nowiki&gt;&lt;/tt&gt; tag or the Forums for the &lt;tt&gt;&lt;nowiki&gt;[asy]&lt;/nowiki&gt;&lt;/tt&gt; 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&amp;t=149650 here]. The package contains a set of shorthand commands that implement the behavior of usual commands, for example &lt;tt&gt;D()&lt;/tt&gt; for &lt;tt&gt;draw()&lt;/tt&gt; and &lt;tt&gt;dot()&lt;/tt&gt;, and so forth.<br /> <br /> ==== What is the olympiad package? ====<br /> <br /> :See [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=519&amp;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 /> == 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 &quot;edit&quot; 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 on AoPS 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, graders, and teaching assistants. 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 /> ==AoPS Acronyms==<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 Competitions<br /> *'''ATM'''- At the Moment<br /> *'''brb'''- Be right Back<br /> *'''BTW'''- By the way<br /> *'''CEMC''' - Centre for Mathematics and Computing<br /> *'''C&amp;P or C+P or CP''' - Counting and Probability or Contests and Programs<br /> *'''EBWOP'''- Editing by way of post<br /> *'''FTW'''- For the Win, a game on AoPS<br /> *'''gg'''- Good Game<br /> *'''gj'''- Good Job<br /> *'''glhf'''-Good Luck Have Fun<br /> *'''gtg''' - Got to go<br /> *'''HSM''' - High School Math Forum<br /> *'''ID(R)K'''-I Don't (Really) Know<br /> *'''iff'''-If and only if<br /> *'''IIRC'''- If I recall correctly<br /> *'''IKR'''- I know right?<br /> *'''IMO'''- In my opinion (or International Math Olympiad, depending on context)<br /> *'''JMO'''- United States of America Junior Mathematical Olympiad<br /> *'''lol'''- Laugh Out Loud<br /> *'''MC'''- Mathcounts, a popular math contest for Middle School students.<br /> *'''mod(s)'''- Moderator(s)<br /> *'''MOEMS'''- Math Olympiads for Elementary and Middle Schools<br /> *'''MO(S)P'''- Mathematical Olympiad (Summer) Program<br /> *'''MSM'''- Middle School Math Forum<br /> *'''NIMO&quot;-National Internet Math Olympiad<br /> *'''NT'''- Number Theory<br /> *'''OBC'''- Online by computer<br /> *'''OMG'''- Oh My Gosh<br /> *'''OMO'''-Online Math Olympiad<br /> *'''OP'''- Original Poster/Original Post/Original Problem, or Overpowered/Overpowering<br /> *'''QED'''- Quod erat demonstrandum, Latin for &quot;Which was to be proven&quot;; some English mathematicians use it as an acronym for Quite Elegantly Done<br /> *'''QS&amp;A'''- Questions, Suggestions, and Announcements Forum<br /> *'''ro(t)fl''' - Rolling on the floor laughing<br /> *'''smh''' - Shaking my head<br /> *'''sqrt''' - Square root<br /> *'''Sticky'''- A post pinned to the top of a forum - a thing no one reads but really should read<br /> *'''ToS'''- Terms of Service - a thing no one reads but really should read<br /> *'''USA(J)MO'''- USA (Junior) Mathematical Olympiad<br /> *'''V/LA'''- Vacation or Long Absence/Limited Access<br /> *'''WLOG'''- Without loss of generality<br /> *'''wrt'''- With respect to<br /> *'''wtg''' - Way to go<br /> *'''tytia'''- Thank you, that is all<br /> *'''xD'''- Bursting Laugh<br /> <br /> == FTW! ==<br /> <br /> Please see the [//artofproblemsolving.com/ftw/faq For the Win! FAQ] for many common questions.<br /> <br /> == School ==<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. Don't worry, though, classroom participation usually isn't weighted heavily.<br /> <br /> ==== Is there audio or video in class? ====<br /> :There is generally no audio or video in the class. The classes are generally text and image 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 and being less pedagogically effective, the technology isn't quite there yet for all students to be able to adequately receive streaming audio and video.<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 /> ==== For my homework, there is suppose to be a green bar but it's orange, why? ====<br /> <br /> :For the bar to turn green, the writing problem must be attempted. Once it is attempted the bar will turn green.<br /> <br /> ==== I need more time for my homework, what should I do? ====<br /> <br /> :There is a &quot;Request Extension&quot; button in the homework tab of your class. This will automatically extend the due date to 2 days after the normal deadline. If you want more time you need to ask for it in the little comment box, stating the reason why you want the extension, and how much time you want. This request will be looked at by the teachers and they will decide if you get the extension or not. Note that you can only use this button 3 times.<br /> :Otherwise, you can send an email to extensions@aops.com with your username, class name and ID, if known, and reason for extension. Someone should get back to you within a couple days.</div> Thedoge https://artofproblemsolving.com/wiki/index.php?title=2017_USAJMO_Problems/Problem_6&diff=85344 2017 USAJMO Problems/Problem 6 2017-04-21T00:39:04Z <p>Thedoge: /* See also */</p> <hr /> <div>==Problem==<br /> Let &lt;math&gt;P_1, \ldots, P_{2n}&lt;/math&gt; be &lt;math&gt;2n&lt;/math&gt; distinct points on the unit circle &lt;math&gt;x^2 + y^2 = 1&lt;/math&gt; other than &lt;math&gt;(1,0)&lt;/math&gt;. Each point is colored either red or blue, with exactly &lt;math&gt;n&lt;/math&gt; of them red and exactly &lt;math&gt;n&lt;/math&gt; of them blue. Let &lt;math&gt;R_1, \ldots, R_n&lt;/math&gt; be any ordering of the red points. Let &lt;math&gt;B_1&lt;/math&gt; be the nearest blue point to &lt;math&gt;R_1&lt;/math&gt; traveling counterclockwise around the circle starting from &lt;math&gt;R_1&lt;/math&gt;. Then let &lt;math&gt;B_2&lt;/math&gt; be the nearest of the remaining blue points to &lt;math&gt;R_2&lt;/math&gt; traveling counterclockwise around the circle from &lt;math&gt;R_2&lt;/math&gt;, and so on, until we have labeled all the blue points &lt;math&gt;B_1, \ldots, B_n&lt;/math&gt;. Show that the number of counterclockwise arcs of the form &lt;math&gt;R_i \rightarrow B_i&lt;/math&gt; that contain the point &lt;math&gt;(1,0)&lt;/math&gt; is independent of the way we chose the ordering &lt;math&gt;R_1, \ldots, R_n&lt;/math&gt; of the red points. <br /> <br /> ==Solution==<br /> <br /> {{MAA Notice}}<br /> <br /> ==See also==<br /> {{USAJMO newbox|year=2017|num-b=2|aftertext=|after=Last Problem}}</div> Thedoge https://artofproblemsolving.com/wiki/index.php?title=2017_USAJMO_Problems/Problem_6&diff=85343 2017 USAJMO Problems/Problem 6 2017-04-21T00:38:40Z <p>Thedoge: </p> <hr /> <div>==Problem==<br /> Let &lt;math&gt;P_1, \ldots, P_{2n}&lt;/math&gt; be &lt;math&gt;2n&lt;/math&gt; distinct points on the unit circle &lt;math&gt;x^2 + y^2 = 1&lt;/math&gt; other than &lt;math&gt;(1,0)&lt;/math&gt;. Each point is colored either red or blue, with exactly &lt;math&gt;n&lt;/math&gt; of them red and exactly &lt;math&gt;n&lt;/math&gt; of them blue. Let &lt;math&gt;R_1, \ldots, R_n&lt;/math&gt; be any ordering of the red points. Let &lt;math&gt;B_1&lt;/math&gt; be the nearest blue point to &lt;math&gt;R_1&lt;/math&gt; traveling counterclockwise around the circle starting from &lt;math&gt;R_1&lt;/math&gt;. Then let &lt;math&gt;B_2&lt;/math&gt; be the nearest of the remaining blue points to &lt;math&gt;R_2&lt;/math&gt; traveling counterclockwise around the circle from &lt;math&gt;R_2&lt;/math&gt;, and so on, until we have labeled all the blue points &lt;math&gt;B_1, \ldots, B_n&lt;/math&gt;. Show that the number of counterclockwise arcs of the form &lt;math&gt;R_i \rightarrow B_i&lt;/math&gt; that contain the point &lt;math&gt;(1,0)&lt;/math&gt; is independent of the way we chose the ordering &lt;math&gt;R_1, \ldots, R_n&lt;/math&gt; of the red points. <br /> <br /> ==Solution==<br /> <br /> {{MAA Notice}}<br /> <br /> ==See also==<br /> {{USAJMO newbox|year=2017|num-b=2|aftertext=|after=LastProblem}}</div> Thedoge https://artofproblemsolving.com/wiki/index.php?title=2017_USAJMO_Problems/Problem_6&diff=85342 2017 USAJMO Problems/Problem 6 2017-04-21T00:36:49Z <p>Thedoge: Created page with &quot;==Problem== Let &lt;math&gt;P_1, \ldots, P_{2n}&lt;/math&gt; be &lt;math&gt;2n&lt;/math&gt; distinct points on the unit circle &lt;math&gt;x^2 + y^2 = 1&lt;/math&gt; other than &lt;math&gt;(1,0)&lt;/math&gt;. Each point is...&quot;</p> <hr /> <div>==Problem==<br /> Let &lt;math&gt;P_1, \ldots, P_{2n}&lt;/math&gt; be &lt;math&gt;2n&lt;/math&gt; distinct points on the unit circle &lt;math&gt;x^2 + y^2 = 1&lt;/math&gt; other than &lt;math&gt;(1,0)&lt;/math&gt;. Each point is colored either red or blue, with exactly &lt;math&gt;n&lt;/math&gt; of them red and exactly &lt;math&gt;n&lt;/math&gt; of them blue. Let &lt;math&gt;R_1, \ldots, R_n&lt;/math&gt; be any ordering of the red points. Let &lt;math&gt;B_1&lt;/math&gt; be the nearest blue point to &lt;math&gt;R_1&lt;/math&gt; traveling counterclockwise around the circle starting from &lt;math&gt;R_1&lt;/math&gt;. Then let &lt;math&gt;B_2&lt;/math&gt; be the nearest of the remaining blue points to &lt;math&gt;R_2&lt;/math&gt; traveling counterclockwise around the circle from &lt;math&gt;R_2&lt;/math&gt;, and so on, until we have labeled all the blue points &lt;math&gt;B_1, \ldots, B_n&lt;/math&gt;. Show that the number of counterclockwise arcs of the form &lt;math&gt;R_i \rightarrow B_i&lt;/math&gt; that contain the point &lt;math&gt;(1,0)&lt;/math&gt; is independent of the way we chose the ordering &lt;math&gt;R_1, \ldots, R_n&lt;/math&gt; of the red points. <br /> <br /> ==Solution==</div> Thedoge https://artofproblemsolving.com/wiki/index.php?title=2017_USAJMO_Problems/Problem_4&diff=85341 2017 USAJMO Problems/Problem 4 2017-04-21T00:36:14Z <p>Thedoge: /* See also */</p> <hr /> <div>==Problem==<br /> Are there any triples &lt;math&gt;(a,b,c)&lt;/math&gt; of positive integers such that &lt;math&gt;(a-2)(b-2)(c-2) + 12&lt;/math&gt; is prime that properly divides the number &lt;math&gt;a^2 + b^2 + c^2 + abc - 2017&lt;/math&gt;?<br /> <br /> ==Solution==<br /> <br /> {{MAA Notice}}<br /> <br /> ==See also==<br /> {{USAJMO newbox|year=2017|num-b=3|num-a=5}}</div> Thedoge https://artofproblemsolving.com/wiki/index.php?title=2017_USAJMO_Problems/Problem_4&diff=85340 2017 USAJMO Problems/Problem 4 2017-04-21T00:35:58Z <p>Thedoge: Created page with &quot;==Problem== Are there any triples &lt;math&gt;(a,b,c)&lt;/math&gt; of positive integers such that &lt;math&gt;(a-2)(b-2)(c-2) + 12&lt;/math&gt; is prime that properly divides the number &lt;math&gt;a^2 + b...&quot;</p> <hr /> <div>==Problem==<br /> Are there any triples &lt;math&gt;(a,b,c)&lt;/math&gt; of positive integers such that &lt;math&gt;(a-2)(b-2)(c-2) + 12&lt;/math&gt; is prime that properly divides the number &lt;math&gt;a^2 + b^2 + c^2 + abc - 2017&lt;/math&gt;?<br /> <br /> ==Solution==<br /> <br /> {{MAA Notice}}<br /> <br /> ==See also==<br /> {{USAJMO newbox|year=2017|num-b=1|num-a=3}}</div> Thedoge https://artofproblemsolving.com/wiki/index.php?title=2017_USAJMO_Problems/Problem_5&diff=85339 2017 USAJMO Problems/Problem 5 2017-04-21T00:35:29Z <p>Thedoge: /* See also */</p> <hr /> <div>==Problem==<br /> <br /> Let &lt;math&gt;O&lt;/math&gt; and &lt;math&gt;H&lt;/math&gt; be the circumcenter and the orthocenter of an acute triangle &lt;math&gt;ABC&lt;/math&gt;. Points &lt;math&gt;M&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt; lie on side &lt;math&gt;BC&lt;/math&gt; such that &lt;math&gt;BM = CM&lt;/math&gt; and &lt;math&gt;\angle BAD = \angle CAD&lt;/math&gt;. Ray &lt;math&gt;MO&lt;/math&gt; intersects the circumcircle of triangle &lt;math&gt;BHC&lt;/math&gt; in point &lt;math&gt;N&lt;/math&gt;. Prove that &lt;math&gt;\angle ADO = \angle HAN&lt;/math&gt;. <br /> <br /> ==Solution==<br /> <br /> {{MAA Notice}}<br /> <br /> ==See also==<br /> {{USAJMO newbox|year=2017|num-b=4|num-a=6}}</div> Thedoge https://artofproblemsolving.com/wiki/index.php?title=2017_USAJMO_Problems/Problem_5&diff=85338 2017 USAJMO Problems/Problem 5 2017-04-21T00:35:16Z <p>Thedoge: </p> <hr /> <div>==Problem==<br /> <br /> Let &lt;math&gt;O&lt;/math&gt; and &lt;math&gt;H&lt;/math&gt; be the circumcenter and the orthocenter of an acute triangle &lt;math&gt;ABC&lt;/math&gt;. Points &lt;math&gt;M&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt; lie on side &lt;math&gt;BC&lt;/math&gt; such that &lt;math&gt;BM = CM&lt;/math&gt; and &lt;math&gt;\angle BAD = \angle CAD&lt;/math&gt;. Ray &lt;math&gt;MO&lt;/math&gt; intersects the circumcircle of triangle &lt;math&gt;BHC&lt;/math&gt; in point &lt;math&gt;N&lt;/math&gt;. Prove that &lt;math&gt;\angle ADO = \angle HAN&lt;/math&gt;. <br /> <br /> ==Solution==<br /> <br /> {{MAA Notice}}<br /> <br /> ==See also==<br /> {{USAJMO newbox|year=2017|num-b=1|num-a=3}}</div> Thedoge https://artofproblemsolving.com/wiki/index.php?title=2017_USAJMO_Problems/Problem_5&diff=85337 2017 USAJMO Problems/Problem 5 2017-04-21T00:34:08Z <p>Thedoge: Created page with &quot;==Problem== Let &lt;math&gt;O&lt;/math&gt; and &lt;math&gt;H&lt;/math&gt; be the circumcenter and the orthocenter of an acute triangle &lt;math&gt;ABC&lt;/math&gt;. Points &lt;math&gt;M&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt; lie...&quot;</p> <hr /> <div>==Problem==<br /> <br /> Let &lt;math&gt;O&lt;/math&gt; and &lt;math&gt;H&lt;/math&gt; be the circumcenter and the orthocenter of an acute triangle &lt;math&gt;ABC&lt;/math&gt;. Points &lt;math&gt;M&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt; lie on side &lt;math&gt;BC&lt;/math&gt; such that &lt;math&gt;BM = CM&lt;/math&gt; and &lt;math&gt;\angle BAD = \angle CAD&lt;/math&gt;. Ray &lt;math&gt;MO&lt;/math&gt; intersects the circumcircle of triangle &lt;math&gt;BHC&lt;/math&gt; in point &lt;math&gt;N&lt;/math&gt;. Prove that &lt;math&gt;\angle ADO = \angle HAN&lt;/math&gt;. <br /> <br /> ==Solution==</div> Thedoge https://artofproblemsolving.com/wiki/index.php?title=2017_USAJMO_Problems&diff=85336 2017 USAJMO Problems 2017-04-21T00:32:44Z <p>Thedoge: </p> <hr /> <div>==Day 1==<br /> <br /> Note: For any geometry problem whose statement begins with an asterisk (&lt;math&gt;*&lt;/math&gt;), the first page of the solution must be a large, in-scale, clearly labeled diagram. Failure to meet this requirement will result in an automatic 1-point deduction. <br /> <br /> ===Problem 1===<br /> Prove that there are infinitely many distinct pairs &lt;math&gt;(a,b)&lt;/math&gt; of relatively prime positive integers &lt;math&gt;a &gt; 1&lt;/math&gt; and &lt;math&gt;b &gt; 1&lt;/math&gt; such that &lt;math&gt;a^b + b^a&lt;/math&gt; is divisible by &lt;math&gt;a + b.&lt;/math&gt; <br /> <br /> [[2017 USAJMO Problems/Problem 1|Solution]]<br /> <br /> ===Problem 2===<br /> Consider the equation <br /> &lt;cmath&gt;\left(3x^3 + xy^2 \right) \left(x^2y + 3y^3 \right) = (x-y)^7.&lt;/cmath&gt;<br /> <br /> (a) Prove that there are infinitely many pairs &lt;math&gt;(x,y)&lt;/math&gt; of positive integers satisfying the equation. <br /> <br /> (b) Describe all pairs &lt;math&gt;(x,y)&lt;/math&gt; of positive integers satisfying the equation. <br /> <br /> [[2017 USAJMO Problems/Problem 2|Solution]]<br /> <br /> ===Problem 3===<br /> (&lt;math&gt;*&lt;/math&gt;) Let &lt;math&gt;ABC&lt;/math&gt; be an equilateral triangle and let &lt;math&gt;P&lt;/math&gt; be a point on its circumcircle. Let lines &lt;math&gt;PA&lt;/math&gt; and &lt;math&gt;BC&lt;/math&gt; intersect at &lt;math&gt;D&lt;/math&gt;; let lines &lt;math&gt;PB&lt;/math&gt; and &lt;math&gt;CA&lt;/math&gt; intersect at &lt;math&gt;E&lt;/math&gt;; and let lines &lt;math&gt;PC&lt;/math&gt; and &lt;math&gt;AB&lt;/math&gt; intersect at &lt;math&gt;F&lt;/math&gt;. Prove that the area of triangle &lt;math&gt;DEF&lt;/math&gt; is twice the area of triangle &lt;math&gt;ABC&lt;/math&gt;.<br /> <br /> [[2017 USAJMO Problems/Problem 3|Solution]]<br /> <br /> ==Day 2==<br /> <br /> Note: For any geometry problem whose statement begins with an asterisk (&lt;math&gt;*&lt;/math&gt;), the first page of the solution must be a large, in-scale, clearly labeled diagram. Failure to meet this requirement will result in an automatic 1-point deduction. <br /> <br /> ===Problem 4===<br /> Are there any triples &lt;math&gt;(a,b,c)&lt;/math&gt; of positive integers such that &lt;math&gt;(a-2)(b-2)(c-2) + 12&lt;/math&gt; is prime that properly divides the number &lt;math&gt;a^2 + b^2 + c^2 + abc - 2017&lt;/math&gt;?<br /> <br /> [[2017 USAJMO Problems/Problem 4|Solution]]<br /> <br /> ===Problem 5===<br /> (&lt;math&gt;*&lt;/math&gt;) Let &lt;math&gt;O&lt;/math&gt; and &lt;math&gt;H&lt;/math&gt; be the circumcenter and the orthocenter of an acute triangle &lt;math&gt;ABC&lt;/math&gt;. Points &lt;math&gt;M&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt; lie on side &lt;math&gt;BC&lt;/math&gt; such that &lt;math&gt;BM = CM&lt;/math&gt; and &lt;math&gt;\angle BAD = \angle CAD&lt;/math&gt;. Ray &lt;math&gt;MO&lt;/math&gt; intersects the circumcircle of triangle &lt;math&gt;BHC&lt;/math&gt; in point &lt;math&gt;N&lt;/math&gt;. Prove that &lt;math&gt;\angle ADO = \angle HAN&lt;/math&gt;. <br /> <br /> [[2017 USAJMO Problems/Problem 5|Solution]]<br /> <br /> ===Problem 6===<br /> Let &lt;math&gt;P_1, \ldots, P_{2n}&lt;/math&gt; be &lt;math&gt;2n&lt;/math&gt; distinct points on the unit circle &lt;math&gt;x^2 + y^2 = 1&lt;/math&gt; other than &lt;math&gt;(1,0)&lt;/math&gt;. Each point is colored either red or blue, with exactly &lt;math&gt;n&lt;/math&gt; of them red and exactly &lt;math&gt;n&lt;/math&gt; of them blue. Let &lt;math&gt;R_1, \ldots, R_n&lt;/math&gt; be any ordering of the red points. Let &lt;math&gt;B_1&lt;/math&gt; be the nearest blue point to &lt;math&gt;R_1&lt;/math&gt; traveling counterclockwise around the circle starting from &lt;math&gt;R_1&lt;/math&gt;. Then let &lt;math&gt;B_2&lt;/math&gt; be the nearest of the remaining blue points to &lt;math&gt;R_2&lt;/math&gt; traveling counterclockwise around the circle from &lt;math&gt;R_2&lt;/math&gt;, and so on, until we have labeled all the blue points &lt;math&gt;B_1, \ldots, B_n&lt;/math&gt;. Show that the number of counterclockwise arcs of the form &lt;math&gt;R_i \rightarrow B_i&lt;/math&gt; that contain the point &lt;math&gt;(1,0)&lt;/math&gt; is independent of the way we chose the ordering &lt;math&gt;R_1, \ldots, R_n&lt;/math&gt; of the red points. <br /> <br /> [[2017 USAJMO Problems/Problem 6|Solution]]<br /> <br /> {{MAA Notice}}<br /> <br /> {{USAJMO newbox|year= 2017 |before=[[2016 USAJMO]]|after=[[2018 USAJMO]]}}</div> Thedoge https://artofproblemsolving.com/wiki/index.php?title=2017_USAJMO_Problems&diff=85335 2017 USAJMO Problems 2017-04-21T00:32:19Z <p>Thedoge: </p> <hr /> <div>==Day 1==<br /> <br /> Note: For any geometry problem whose statement begins with an asterisk (&lt;math&gt;*&lt;/math&gt;), the first page of the solution must be a large, in-scale, clearly labeled diagram. Failure to meet this requirement will result in an automatic 1-point deduction. <br /> <br /> ===Problem 1===<br /> Prove that there are infinitely many distinct pairs &lt;math&gt;(a,b)&lt;/math&gt; of relatively prime positive integers &lt;math&gt;a &gt; 1&lt;/math&gt; and &lt;math&gt;b &gt; 1&lt;/math&gt; such that &lt;math&gt;a^b + b^a&lt;/math&gt; is divisible by &lt;math&gt;a + b.&lt;/math&gt; <br /> <br /> [[2017 USAJMO Problems/Problem 1|Solution]]<br /> <br /> ===Problem 2===<br /> Consider the equation <br /> &lt;cmath&gt;\left(3x^3 + xy^2 \right) \left(x^2y + 3y^3 \right) = (x-y)^7.&lt;/cmath&gt;<br /> <br /> (a) Prove that there are infinitely many pairs &lt;math&gt;(x,y)&lt;/math&gt; of positive integers satisfying the equation. <br /> <br /> (b) Describe all pairs &lt;math&gt;(x,y)&lt;/math&gt; of positive integers satisfying the equation. <br /> <br /> [[2017 USAJMO Problems/Problem 2|Solution]]<br /> ===Problem 3===<br /> (&lt;math&gt;*&lt;/math&gt;) Let &lt;math&gt;ABC&lt;/math&gt; be an equilateral triangle and let &lt;math&gt;P&lt;/math&gt; be a point on its circumcircle. Let lines &lt;math&gt;PA&lt;/math&gt; and &lt;math&gt;BC&lt;/math&gt; intersect at &lt;math&gt;D&lt;/math&gt;; let lines &lt;math&gt;PB&lt;/math&gt; and &lt;math&gt;CA&lt;/math&gt; intersect at &lt;math&gt;E&lt;/math&gt;; and let lines &lt;math&gt;PC&lt;/math&gt; and &lt;math&gt;AB&lt;/math&gt; intersect at &lt;math&gt;F&lt;/math&gt;. Prove that the area of triangle &lt;math&gt;DEF&lt;/math&gt; is twice the area of triangle &lt;math&gt;ABC&lt;/math&gt;.<br /> <br /> [[2017 USAJMO Problems/Problem 3|Solution]]<br /> <br /> ==Day 2==<br /> <br /> Note: For any geometry problem whose statement begins with an asterisk (&lt;math&gt;*&lt;/math&gt;), the first page of the solution must be a large, in-scale, clearly labeled diagram. Failure to meet this requirement will result in an automatic 1-point deduction. <br /> <br /> ===Problem 4===<br /> Are there any triples &lt;math&gt;(a,b,c)&lt;/math&gt; of positive integers such that &lt;math&gt;(a-2)(b-2)(c-2) + 12&lt;/math&gt; is prime that properly divides the number &lt;math&gt;a^2 + b^2 + c^2 + abc - 2017&lt;/math&gt;?<br /> <br /> [[2017 USAJMO Problems/Problem 4|Solution]]<br /> ===Problem 5===<br /> (&lt;math&gt;*&lt;/math&gt;) Let &lt;math&gt;O&lt;/math&gt; and &lt;math&gt;H&lt;/math&gt; be the circumcenter and the orthocenter of an acute triangle &lt;math&gt;ABC&lt;/math&gt;. Points &lt;math&gt;M&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt; lie on side &lt;math&gt;BC&lt;/math&gt; such that &lt;math&gt;BM = CM&lt;/math&gt; and &lt;math&gt;\angle BAD = \angle CAD&lt;/math&gt;. Ray &lt;math&gt;MO&lt;/math&gt; intersects the circumcircle of triangle &lt;math&gt;BHC&lt;/math&gt; in point &lt;math&gt;N&lt;/math&gt;. Prove that &lt;math&gt;\angle ADO = \angle HAN&lt;/math&gt;. <br /> <br /> [[2017 USAJMO Problems/Problem 5|Solution]]<br /> ===Problem 6===<br /> Let &lt;math&gt;P_1, \ldots, P_{2n}&lt;/math&gt; be &lt;math&gt;2n&lt;/math&gt; distinct points on the unit circle &lt;math&gt;x^2 + y^2 = 1&lt;/math&gt; other than &lt;math&gt;(1,0)&lt;/math&gt;. Each point is colored either red or blue, with exactly &lt;math&gt;n&lt;/math&gt; of them red and exactly &lt;math&gt;n&lt;/math&gt; of them blue. Let &lt;math&gt;R_1, \ldots, R_n&lt;/math&gt; be any ordering of the red points. Let &lt;math&gt;B_1&lt;/math&gt; be the nearest blue point to &lt;math&gt;R_1&lt;/math&gt; traveling counterclockwise around the circle starting from &lt;math&gt;R_1&lt;/math&gt;. Then let &lt;math&gt;B_2&lt;/math&gt; be the nearest of the remaining blue points to &lt;math&gt;R_2&lt;/math&gt; traveling counterclockwise around the circle from &lt;math&gt;R_2&lt;/math&gt;, and so on, until we have labeled all the blue points &lt;math&gt;B_1, \ldots, B_n&lt;/math&gt;. Show that the number of counterclockwise arcs of the form &lt;math&gt;R_i \rightarrow B_i&lt;/math&gt; that contain the point &lt;math&gt;(1,0)&lt;/math&gt; is independent of the way we chose the ordering &lt;math&gt;R_1, \ldots, R_n&lt;/math&gt; of the red points. <br /> <br /> [[2017 USAJMO Problems/Problem 6|Solution]]<br /> {{MAA Notice}}<br /> <br /> {{USAJMO newbox|year= 2017 |before=[[2016 USAJMO]]|after=[[2018 USAJMO]]}}</div> Thedoge https://artofproblemsolving.com/wiki/index.php?title=2017_USAJMO&diff=85334 2017 USAJMO 2017-04-21T00:30:08Z <p>Thedoge: </p> <hr /> <div>The test was held on April 19th and 20th, 2017. The first link will contain the full set of test problems. The rest will contain each individual problem and its solution.<br /> <br /> [[2017 USAJMO Problems]]<br /> * [[2017 USAJMO Problems/Problem 1]]<br /> * [[2017 USAJMO Problems/Problem 2]]<br /> * [[2017 USAJMO Problems/Problem 3]]<br /> * [[2017 USAJMO Problems/Problem 4]]<br /> * [[2017 USAJMO Problems/Problem 5]]<br /> * [[2017 USAJMO Problems/Problem 6]]<br /> <br /> == See Also ==<br /> * [[Mathematics competitions]]<br /> * [[Mathematics competition resources]]<br /> * [[Math books]]<br /> * [[USAJMO]]<br /> <br /> {{USAJMO newbox|year= 2017 |before=[[2016 USAJMO]]|after=[[2018 USAJMO]]}}</div> Thedoge https://artofproblemsolving.com/wiki/index.php?title=2017_USAJMO_Problems/Problem_3&diff=85311 2017 USAJMO Problems/Problem 3 2017-04-19T23:51:45Z <p>Thedoge: /* Solution */</p> <hr /> <div>==Problem==<br /> (&lt;math&gt;*&lt;/math&gt;) Let &lt;math&gt;ABC&lt;/math&gt; be an equilateral triangle and let &lt;math&gt;P&lt;/math&gt; be a point on its circumcircle. Let lines &lt;math&gt;PA&lt;/math&gt; and &lt;math&gt;PB&lt;/math&gt; intersect at &lt;math&gt;D&lt;/math&gt;; let lines &lt;math&gt;PB&lt;/math&gt; and &lt;math&gt;CA&lt;/math&gt; intersect at &lt;math&gt;E&lt;/math&gt;; and let lines &lt;math&gt;PC&lt;/math&gt; and &lt;math&gt;AB&lt;/math&gt; intersect at &lt;math&gt;F&lt;/math&gt;. Prove that the area of triangle &lt;math&gt;DEF&lt;/math&gt; is twice that of triangle &lt;math&gt;ABC&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> size(3inch);<br /> pair A = (0, 3sqrt(3)), B = (-3,0), C = (3,0), P = (0, -sqrt(3)), D = (0, 0), E1 = (6, -3sqrt(3)), F = (-6, -3sqrt(3)), O = (0, sqrt(3));<br /> draw(Circle(O, 2sqrt(3)), black);<br /> draw(A--B--C--cycle);<br /> draw(B--E1--C);<br /> draw(C--F--B);<br /> draw(A--P);<br /> draw(D--E1--F--cycle, dashed);<br /> label(&quot;A&quot;, A, N);<br /> label(&quot;B&quot;, B, W);<br /> label(&quot;C&quot;, C, E);<br /> label(&quot;P&quot;, P, S);<br /> label(&quot;D&quot;, D, NW);<br /> label(&quot;E&quot;, E1, SE);<br /> label(&quot;F&quot;, F, SW);<br /> &lt;/asy&gt;<br /> <br /> ==Solution==<br /> WLOG, let &lt;math&gt;AB = 1&lt;/math&gt;. Let &lt;math&gt;[ABD] = X, [ACD] = Y&lt;/math&gt;, and &lt;math&gt;\angle BAD = \theta&lt;/math&gt;. After some angle chasing, we find that &lt;math&gt;\angle BCF \cong \angle BCE \cong \theta&lt;/math&gt; and &lt;math&gt;\angle FBC \cong \angle BCE \cong 120^{\circ}&lt;/math&gt;. Therefore, &lt;math&gt;\triangle FBC&lt;/math&gt; ~ &lt;math&gt;\triangle BCE&lt;/math&gt;. <br /> <br /> ------------------------------------------------------------------------------------------------------<br /> <br /> Lemma 1: If &lt;math&gt;BF = k&lt;/math&gt;, then &lt;math&gt;CE = \frac 1k&lt;/math&gt;.<br /> This lemma results directly from the fact that &lt;math&gt;\triangle FBC&lt;/math&gt; ~ &lt;math&gt;\triangle BCE&lt;/math&gt;; &lt;math&gt;\frac{BF}{BC} = \frac{BF}{1} = \frac{BC}{CE} = \frac{1}{CE}&lt;/math&gt;, or &lt;math&gt;CE = \frac{1}{BF}&lt;/math&gt;. <br /> <br /> ------------------------------------------------------------------------------------------------------<br /> <br /> Lemma 2: &lt;math&gt;[AEF] = (k+\frac 1k + 2)(X+Y)&lt;/math&gt;.<br /> We see that &lt;math&gt;[AEF] = (X+Y) \frac{[AEF]}{[ABC]} = (k+1)(1+\frac 1k)(X+Y) = (k + \frac 1k + 2)(X+Y)&lt;/math&gt;, as desired. <br /> <br /> ------------------------------------------------------------------------------------------------------<br /> <br /> Lemma 3: &lt;math&gt;\frac{X}{Y} = k&lt;/math&gt;.<br /> We see that <br /> &lt;cmath&gt;\frac XY = \frac{\frac 12 (AB)(AD) \sin(\theta)}{\frac 12 (AC)(AD) \sin (60^{\circ} - \theta)} = \frac{\sin(\theta)}{\sin (60^{\circ} - \theta)}.&lt;/cmath&gt;<br /> However, after some angle chasing and by the Law of Sines in &lt;math&gt;\triangle BCF&lt;/math&gt;, we have &lt;math&gt;\frac{k}{\sin(\theta)} = \frac{1}{\sin(60^{\circ} - \theta)}&lt;/math&gt;, or &lt;math&gt;k = \frac{\sin(\theta)}{\sin (60^{\circ} - \theta)}&lt;/math&gt;, which implies the result. <br /> <br /> ------------------------------------------------------------------------------------------------------<br /> <br /> By the area lemma, we have &lt;math&gt;[BDF] = kX&lt;/math&gt; and &lt;math&gt;[CDF] = \frac{Y}{k}&lt;/math&gt;. <br /> <br /> We see that &lt;math&gt;[DEF] = [AEF] - [ABC] - [BDF] - [DCE] = Xk + Yk + \frac Xk + \frac Yk + 2X + 2Y - X - Y - Xk - \frac Yk = X + Y + \frac Xk + yk&lt;/math&gt;. Thus, it suffices to show that &lt;math&gt;X + Y + \frac Xk + Yk = 2X + 2Y&lt;/math&gt;, or &lt;math&gt;\frac Xk + Yk = X + Y&lt;/math&gt;. Rearranging, we find this to be equivalent to &lt;math&gt;\frac XY = k&lt;/math&gt;, which is Lemma 3, so the result has been proven.<br /> <br /> {{MAA Notice}}<br /> <br /> ==See also==<br /> {{USAJMO newbox|year=2017|num-b=2|num-a=4}}</div> Thedoge https://artofproblemsolving.com/wiki/index.php?title=2017_USAJMO_Problems/Problem_1&diff=85307 2017 USAJMO Problems/Problem 1 2017-04-19T23:32:38Z <p>Thedoge: /* Solution 1 */</p> <hr /> <div>== Problem ==<br /> <br /> Prove that there are infinitely many distinct pairs &lt;math&gt;(a,b)&lt;/math&gt; of relatively prime integers &lt;math&gt;a&gt;1&lt;/math&gt; and &lt;math&gt;b&gt;1&lt;/math&gt; such that &lt;math&gt;a^b+b^a&lt;/math&gt; is divisible by &lt;math&gt;a+b&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> Let &lt;math&gt;a = 2n-1&lt;/math&gt; and &lt;math&gt;b = 2n+1&lt;/math&gt;. We see that &lt;math&gt;(2n \pm 1)^2 = 4n^2-4n+1 \equiv 1 \pmod{4n}&lt;/math&gt;. Therefore, we have &lt;math&gt;(2n+1)^{2n-1} + (2n-1)^{2n+1} \equiv 2n + 1 + 2n - 1 = 4n \equiv 0 \pmod{4n}&lt;/math&gt;, as desired. <br /> <br /> (Credits to laegolas)<br /> <br /> ==Solution 2==<br /> Let &lt;math&gt;x&lt;/math&gt; be any odd number above 1<br /> <br /> <br /> {{MAA Notice}}<br /> <br /> ==See also==<br /> {{USAJMO newbox|year=2017|beforetext=|before=First Problem|num-a=2}}</div> Thedoge https://artofproblemsolving.com/wiki/index.php?title=2017_USAJMO_Problems/Problem_2&diff=85301 2017 USAJMO Problems/Problem 2 2017-04-19T23:24:21Z <p>Thedoge: /* Solution */</p> <hr /> <div>==Problem:==<br /> Consider the equation <br /> &lt;cmath&gt;\left(3x^3 + xy^2 \right) \left(x^2y + 3y^3 \right) = (x-y)^7.&lt;/cmath&gt;<br /> <br /> (a) Prove that there are infinitely many pairs &lt;math&gt;(x,y)&lt;/math&gt; of positive integers satisfying the equation. <br /> <br /> (b) Describe all pairs &lt;math&gt;(x,y)&lt;/math&gt; of positive integers satisfying the equation.<br /> <br /> ==Solution==<br /> <br /> {{MAA Notice}}<br /> <br /> ==See also==<br /> {{USAJMO newbox|year=2017|num-b=1|num-a=3}}</div> Thedoge https://artofproblemsolving.com/wiki/index.php?title=2017_USAJMO_Problems/Problem_3&diff=85300 2017 USAJMO Problems/Problem 3 2017-04-19T23:17:26Z <p>Thedoge: </p> <hr /> <div>==Problem==<br /> (&lt;math&gt;*&lt;/math&gt;) Let &lt;math&gt;ABC&lt;/math&gt; be an equilateral triangle and let &lt;math&gt;P&lt;/math&gt; be a point on its circumcircle. Let lines &lt;math&gt;PA&lt;/math&gt; and &lt;math&gt;PB&lt;/math&gt; intersect at &lt;math&gt;D&lt;/math&gt;; let lines &lt;math&gt;PB&lt;/math&gt; and &lt;math&gt;CA&lt;/math&gt; intersect at &lt;math&gt;E&lt;/math&gt;; and let lines &lt;math&gt;PC&lt;/math&gt; and &lt;math&gt;AB&lt;/math&gt; intersect at &lt;math&gt;F&lt;/math&gt;. Prove that the area of triangle &lt;math&gt;DEF&lt;/math&gt; is twice that of triangle &lt;math&gt;ABC&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> size(3inch);<br /> pair A = (0, 3sqrt(3)), B = (-3,0), C = (3,0), P = (0, -sqrt(3)), D = (0, 0), E1 = (6, -3sqrt(3)), F = (-6, -3sqrt(3)), O = (0, sqrt(3));<br /> draw(Circle(O, 2sqrt(3)), black);<br /> draw(A--B--C--cycle);<br /> draw(B--E1--C);<br /> draw(C--F--B);<br /> draw(A--P);<br /> draw(D--E1--F--cycle, dashed);<br /> label(&quot;A&quot;, A, N);<br /> label(&quot;B&quot;, B, W);<br /> label(&quot;C&quot;, C, E);<br /> label(&quot;P&quot;, P, S);<br /> label(&quot;D&quot;, D, NW);<br /> label(&quot;E&quot;, E1, SE);<br /> label(&quot;F&quot;, F, SW);<br /> &lt;/asy&gt;<br /> <br /> ==Solution==<br /> <br /> {{MAA Notice}}<br /> <br /> ==See also==<br /> {{USAJMO newbox|year=2017|num-b=2|num-a=4}}</div> Thedoge https://artofproblemsolving.com/wiki/index.php?title=2017_USAJMO_Problems/Problem_1&diff=85299 2017 USAJMO Problems/Problem 1 2017-04-19T23:16:06Z <p>Thedoge: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> Prove that there are infinitely many distinct pairs &lt;math&gt;(a,b)&lt;/math&gt; of relatively prime integers &lt;math&gt;a&gt;1&lt;/math&gt; and &lt;math&gt;b&gt;1&lt;/math&gt; such that &lt;math&gt;a^b+b^a&lt;/math&gt; is divisible by &lt;math&gt;a+b&lt;/math&gt;.<br /> <br /> ==Solution==<br /> Let &lt;math&gt;a = 2^n - 1&lt;/math&gt; and &lt;math&gt;b = 2^n + 1&lt;/math&gt;. We see that &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are relatively prime (they are consecutive positive odd integers). <br /> <br /> ---------------------------<br /> Lemma: &lt;math&gt;(2^k + 1)^{-1} \equiv 2^k + 1 \pmod{2^{k+1}}&lt;/math&gt;. <br /> <br /> Since every number has a unique modular inverse, the lemma is equivalent to proving that &lt;math&gt;(2^k+1)^2 \equiv 1 \pmod{2^{k+1}}&lt;/math&gt;. Expanding, we have the result. <br /> <br /> ----------------------------<br /> <br /> Substituting for &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;, we have <br /> &lt;cmath&gt;(2^k+1)^{2^k-1} + (2^k-1)^{2^k+1} \equiv 2^k - 1 + 2^k + 1 \equiv 0 \pmod{2^{k+1}},&lt;/cmath&gt;<br /> where we use our lemma and the Euler totient theorem: &lt;math&gt;a^{\phi{n}} \equiv 1 \pmod{n}&lt;/math&gt; when &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime. <br /> <br /> {{MAA Notice}}<br /> <br /> ==See also==<br /> {{USAJMO newbox|year=2017|beforetext=|before=First Problem|num-a=2}}</div> Thedoge https://artofproblemsolving.com/wiki/index.php?title=2017_USAJMO_Problems/Problem_2&diff=85298 2017 USAJMO Problems/Problem 2 2017-04-19T23:15:41Z <p>Thedoge: /* Solution */</p> <hr /> <div>==Problem:==<br /> Consider the equation <br /> &lt;cmath&gt;\left(3x^3 + xy^2 \right) \left(x^2y + 3y^3 \right) = (x-y)^7.&lt;/cmath&gt;<br /> <br /> (a) Prove that there are infinitely many pairs &lt;math&gt;(x,y)&lt;/math&gt; of positive integers satisfying the equation. <br /> <br /> (b) Describe all pairs &lt;math&gt;(x,y)&lt;/math&gt; of positive integers satisfying the equation.<br /> <br /> ==Solution==<br /> Part a: Let &lt;math&gt;y = an&lt;/math&gt; and &lt;math&gt;x = a(n + 1)&lt;/math&gt;. Substituting, we have <br /> &lt;cmath&gt;a^7 = a^6 \left(3(n+1)^3 + (n+1)n^2 \right) \left(3n^3 + n(n+1)^2 \right).&lt;/cmath&gt;<br /> Therefore, we have <br /> &lt;cmath&gt;a = \left(3(n+1)^3 + (n+1)n^2 \right) \left(3n^3 + n(n+1)^2 \right),&lt;/cmath&gt;<br /> which implies that there is a solution for every positive integer &lt;math&gt;n&lt;/math&gt;. <br /> <br /> {{MAA Notice}}<br /> <br /> ==See also==<br /> {{USAJMO newbox|year=2017|num-b=1|num-a=3}}</div> Thedoge https://artofproblemsolving.com/wiki/index.php?title=2017_USAJMO_Problems/Problem_2&diff=85297 2017 USAJMO Problems/Problem 2 2017-04-19T23:15:31Z <p>Thedoge: /* Solution */</p> <hr /> <div>==Problem:==<br /> Consider the equation <br /> &lt;cmath&gt;\left(3x^3 + xy^2 \right) \left(x^2y + 3y^3 \right) = (x-y)^7.&lt;/cmath&gt;<br /> <br /> (a) Prove that there are infinitely many pairs &lt;math&gt;(x,y)&lt;/math&gt; of positive integers satisfying the equation. <br /> <br /> (b) Describe all pairs &lt;math&gt;(x,y)&lt;/math&gt; of positive integers satisfying the equation.<br /> <br /> ==Solution==<br /> Part a: Let &lt;math&gt;y = an&lt;/math&gt; and &lt;math&gt;x = a(n + 1)&lt;/math&gt;. Substituting, we have <br /> &lt;cmath&gt;a^7 = a^6 \left(3(n+1)^3 + (n+1)n^2 \right) \left(3n^3 + n(n+1)^2 \right).&lt;/cmath&gt;<br /> Therefore, we have <br /> &lt;cmath&gt;a = \left(3(n+1)^3 + (n+1)n^2 \right) \left(3n^3 + n(n+1)^2 \right),&lt;/cmath&gt;<br /> which implies that there is a solution for every positive integer &lt;math&gt;n&lt;/math&gt;<br /> <br /> {{MAA Notice}}<br /> <br /> ==See also==<br /> {{USAJMO newbox|year=2017|num-b=1|num-a=3}}</div> Thedoge https://artofproblemsolving.com/wiki/index.php?title=2017_USAJMO_Problems/Problem_3&diff=85296 2017 USAJMO Problems/Problem 3 2017-04-19T23:14:14Z <p>Thedoge: /* Solution */</p> <hr /> <div>==Problem==<br /> (&lt;math&gt;*&lt;/math&gt;) Let &lt;math&gt;ABC&lt;/math&gt; be an equilateral triangle and let &lt;math&gt;P&lt;/math&gt; be a point on its circumcircle. Let lines &lt;math&gt;PA&lt;/math&gt; and &lt;math&gt;PB&lt;/math&gt; intersect at &lt;math&gt;D&lt;/math&gt;; let lines &lt;math&gt;PB&lt;/math&gt; and &lt;math&gt;CA&lt;/math&gt; intersect at &lt;math&gt;E&lt;/math&gt;; and let lines &lt;math&gt;PC&lt;/math&gt; and &lt;math&gt;AB&lt;/math&gt; intersect at &lt;math&gt;F&lt;/math&gt;. Prove that the area of triangle &lt;math&gt;DEF&lt;/math&gt; is twice that of triangle &lt;math&gt;ABC&lt;/math&gt;.<br /> <br /> ==Solution==<br /> <br /> &lt;asy&gt;<br /> size(5inch);<br /> pair A = (0, 3sqrt(3)), B = (-3,0), C = (3,0), P = (0, -sqrt(3)), D = (0, 0), E1 = (6, -3sqrt(3)), F = (-6, -3sqrt(3)), O = (0, sqrt(3));<br /> draw(Circle(O, 2sqrt(3)), black);<br /> draw(A--B--C--cycle);<br /> draw(B--E1--C);<br /> draw(C--F--B);<br /> draw(A--P);<br /> draw(D--E1--F--cycle, dashed);<br /> label(&quot;A&quot;, A, N);<br /> label(&quot;B&quot;, B, W);<br /> label(&quot;C&quot;, C, E);<br /> label(&quot;P&quot;, P, S);<br /> label(&quot;D&quot;, D, NW);<br /> label(&quot;E&quot;, E1, SE);<br /> label(&quot;F&quot;, F, SW);<br /> &lt;/asy&gt;<br /> {{MAA Notice}}<br /> <br /> ==See also==<br /> {{USAJMO newbox|year=2017|num-b=2|num-a=4}}</div> Thedoge https://artofproblemsolving.com/wiki/index.php?title=2017_USAJMO_Problems/Problem_3&diff=85295 2017 USAJMO Problems/Problem 3 2017-04-19T23:13:26Z <p>Thedoge: </p> <hr /> <div>==Problem==<br /> (&lt;math&gt;*&lt;/math&gt;) Let &lt;math&gt;ABC&lt;/math&gt; be an equilateral triangle and let &lt;math&gt;P&lt;/math&gt; be a point on its circumcircle. Let lines &lt;math&gt;PA&lt;/math&gt; and &lt;math&gt;PB&lt;/math&gt; intersect at &lt;math&gt;D&lt;/math&gt;; let lines &lt;math&gt;PB&lt;/math&gt; and &lt;math&gt;CA&lt;/math&gt; intersect at &lt;math&gt;E&lt;/math&gt;; and let lines &lt;math&gt;PC&lt;/math&gt; and &lt;math&gt;AB&lt;/math&gt; intersect at &lt;math&gt;F&lt;/math&gt;. Prove that the area of triangle &lt;math&gt;DEF&lt;/math&gt; is twice that of triangle &lt;math&gt;ABC&lt;/math&gt;.<br /> <br /> ==Solution==<br /> <br /> {{MAA Notice}}<br /> <br /> ==See also==<br /> {{USAJMO newbox|year=2017|num-b=2|num-a=4}}</div> Thedoge https://artofproblemsolving.com/wiki/index.php?title=2017_USAJMO_Problems/Problem_3&diff=85294 2017 USAJMO Problems/Problem 3 2017-04-19T23:13:01Z <p>Thedoge: Created page with &quot;==Problem== (&lt;math&gt;*&lt;/math&gt;) Let &lt;math&gt;ABC&lt;/math&gt; be an equilateral triangle and let &lt;math&gt;P&lt;/math&gt; be a point on its circumcircle. Let lines &lt;math&gt;PA&lt;/math&gt; and &lt;math&gt;PB&lt;/mat...&quot;</p> <hr /> <div>==Problem==<br /> (&lt;math&gt;*&lt;/math&gt;) Let &lt;math&gt;ABC&lt;/math&gt; be an equilateral triangle and let &lt;math&gt;P&lt;/math&gt; be a point on its circumcircle. Let lines &lt;math&gt;PA&lt;/math&gt; and &lt;math&gt;PB&lt;/math&gt; intersect at &lt;math&gt;D&lt;/math&gt;; let lines &lt;math&gt;PB&lt;/math&gt; and &lt;math&gt;CA&lt;/math&gt; intersect at &lt;math&gt;E&lt;/math&gt;; and let lines &lt;math&gt;PC&lt;/math&gt; and &lt;math&gt;AB&lt;/math&gt; intersect at &lt;math&gt;F&lt;/math&gt;. Prove that the area of triangle &lt;math&gt;DEF&lt;/math&gt; is twice that of triangle &lt;math&gt;ABC&lt;/math&gt;.</div> Thedoge https://artofproblemsolving.com/wiki/index.php?title=2017_USAJMO_Problems/Problem_2&diff=85293 2017 USAJMO Problems/Problem 2 2017-04-19T23:12:31Z <p>Thedoge: /* Problem: */</p> <hr /> <div>==Problem:==<br /> Consider the equation <br /> &lt;cmath&gt;\left(3x^3 + xy^2 \right) \left(x^2y + 3y^3 \right) = (x-y)^7.&lt;/cmath&gt;<br /> <br /> (a) Prove that there are infinitely many pairs &lt;math&gt;(x,y)&lt;/math&gt; of positive integers satisfying the equation. <br /> <br /> (b) Describe all pairs &lt;math&gt;(x,y)&lt;/math&gt; of positive integers satisfying the equation.<br /> <br /> ==Solution==<br /> <br /> <br /> {{MAA Notice}}<br /> <br /> ==See also==<br /> {{USAJMO newbox|year=2017|num-b=1|num-a=3}}</div> Thedoge https://artofproblemsolving.com/wiki/index.php?title=2017_USAJMO_Problems/Problem_1&diff=85292 2017 USAJMO Problems/Problem 1 2017-04-19T23:12:05Z <p>Thedoge: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> Prove that there are infinitely many distinct pairs &lt;math&gt;(a,b)&lt;/math&gt; of relatively prime integers &lt;math&gt;a&gt;1&lt;/math&gt; and &lt;math&gt;b&gt;1&lt;/math&gt; such that &lt;math&gt;a^b+b^a&lt;/math&gt; is divisible by &lt;math&gt;a+b&lt;/math&gt;.<br /> <br /> ==Solution==<br /> Let &lt;math&gt;a = 2^n - 1&lt;/math&gt; and &lt;math&gt;b = 2^n + 1&lt;/math&gt;. We see that &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are relatively prime (they are consecutive positive odd integers). <br /> <br /> ---------------------------<br /> Lemma: &lt;math&gt;(2^k + 1)^{-1} \equiv 2^k + 1 \pmod{2^{k+1}}&lt;/math&gt;. <br /> <br /> Since every number has a unique modular inverse, the lemma is equivalent to proving that &lt;math&gt;(2^k+1)^2 \equiv 1 \pmod{2^{k+1}}&lt;/math&gt;. Expanding, we have the result. <br /> <br /> ----------------------------<br /> <br /> Substituting for &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;, we have <br /> &lt;cmath&gt;(2^k+1)^{2^k-1} + (2^k-1)^{2^k+1} \equiv 2^k - 1 + 2^k + 1 \equiv 0 \pmod{2^{k+1}},&lt;/cmath&gt;<br /> where we use our lemma and the Euler totient theorem: &lt;math&gt;a^\phi{n} \equiv 1 \pmod{n}&lt;/math&gt; when &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime. <br /> <br /> {{MAA Notice}}<br /> <br /> ==See also==<br /> {{USAJMO newbox|year=2017|beforetext=|before=First Problem|num-a=2}}</div> Thedoge https://artofproblemsolving.com/wiki/index.php?title=2017_USAJMO_Problems/Problem_1&diff=85291 2017 USAJMO Problems/Problem 1 2017-04-19T23:11:46Z <p>Thedoge: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> Prove that there are infinitely many distinct pairs &lt;math&gt;(a,b)&lt;/math&gt; of relatively prime integers &lt;math&gt;a&gt;1&lt;/math&gt; and &lt;math&gt;b&gt;1&lt;/math&gt; such that &lt;math&gt;a^b+b^a&lt;/math&gt; is divisible by &lt;math&gt;a+b&lt;/math&gt;.<br /> <br /> ==Solution==<br /> Let &lt;math&gt;a = 2^n - 1&lt;/math&gt; and &lt;math&gt;b = 2^n + 1&lt;/math&gt;. We see that &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are relatively prime (they are consecutive positive odd integers). <br /> <br /> Lemma: &lt;math&gt;(2^k + 1)^{-1} \equiv 2^k + 1 \pmod{2^{k+1}}&lt;/math&gt;. <br /> <br /> Since every number has a unique modular inverse, the lemma is equivalent to proving that &lt;math&gt;(2^k+1)^2 \equiv 1 \pmod{2^{k+1}}&lt;/math&gt;. Expanding, we have the result. <br /> <br /> Substituting for &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;, we have <br /> &lt;cmath&gt;(2^k+1)^{2^k-1} + (2^k-1)^{2^k+1} \equiv 2^k - 1 + 2^k + 1 \equiv 0 \pmod{2^{k+1}},&lt;/cmath&gt;<br /> where we use our lemma and the Euler totient theorem: &lt;math&gt;a^\phi{n} \equiv 1 \pmod{n}&lt;/math&gt; when &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime. <br /> <br /> {{MAA Notice}}<br /> <br /> ==See also==<br /> {{USAJMO newbox|year=2017|beforetext=|before=First Problem|num-a=2}}</div> Thedoge https://artofproblemsolving.com/wiki/index.php?title=2017_USAJMO_Problems/Problem_2&diff=85289 2017 USAJMO Problems/Problem 2 2017-04-19T23:11:14Z <p>Thedoge: /* Solution */</p> <hr /> <div>==Problem:==<br /> Prove that there are infinitely many distinct pairs &lt;math&gt;(a,b)&lt;/math&gt; of relatively prime positive integers &lt;math&gt;a &gt; 1&lt;/math&gt; and &lt;math&gt;b &gt; 1&lt;/math&gt; such that &lt;math&gt;a^b + b^a&lt;/math&gt; is divisible by &lt;math&gt;a + b&lt;/math&gt;. <br /> <br /> ==Solution==<br /> <br /> <br /> {{MAA Notice}}<br /> <br /> ==See also==<br /> {{USAJMO newbox|year=2017|num-b=1|num-a=3}}</div> Thedoge https://artofproblemsolving.com/wiki/index.php?title=2017_USAJMO_Problems/Problem_2&diff=85288 2017 USAJMO Problems/Problem 2 2017-04-19T23:10:50Z <p>Thedoge: </p> <hr /> <div>==Problem:==<br /> Prove that there are infinitely many distinct pairs &lt;math&gt;(a,b)&lt;/math&gt; of relatively prime positive integers &lt;math&gt;a &gt; 1&lt;/math&gt; and &lt;math&gt;b &gt; 1&lt;/math&gt; such that &lt;math&gt;a^b + b^a&lt;/math&gt; is divisible by &lt;math&gt;a + b&lt;/math&gt;. <br /> <br /> ==Solution==<br /> Let &lt;math&gt;a = 2^n - 1&lt;/math&gt; and &lt;math&gt;b = 2^n + 1&lt;/math&gt;. We see that &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are relatively prime (they are consecutive positive odd integers). <br /> <br /> Lemma: &lt;math&gt;(2^k + 1)^{-1} \equiv 2^k + 1 \pmod{2^{k+1}}&lt;/math&gt;. <br /> <br /> Since every number has a unique modular inverse, the lemma is equivalent to proving that &lt;math&gt;(2^k+1)^2 \equiv 1 \pmod{2^{k+1}}&lt;/math&gt;. Expanding, we have the result. <br /> <br /> Substituting for &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;, we have <br /> &lt;cmath&gt;(2^k+1)^{2^k-1} + (2^k-1)^{2^k+1} \equiv 2^k - 1 + 2^k + 1 \equiv 0 \pmod{2^{k+1}},&lt;/cmath&gt;<br /> where we use our lemma and the Euler totient theorem: &lt;math&gt;a^\phi{n} \equiv 1 \pmod{n}&lt;/math&gt; when &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime. <br /> <br /> {{MAA Notice}}<br /> <br /> ==See also==<br /> {{USAJMO newbox|year=2017|num-b=1|num-a=3}}</div> Thedoge https://artofproblemsolving.com/wiki/index.php?title=2017_USAJMO_Problems/Problem_2&diff=85287 2017 USAJMO Problems/Problem 2 2017-04-19T23:08:46Z <p>Thedoge: Created page with &quot;{{MAA Notice}} ==See also== {{USAJMO newbox|year=2017|num-b=1|num-a=3}}&quot;</p> <hr /> <div>{{MAA Notice}}<br /> <br /> ==See also==<br /> {{USAJMO newbox|year=2017|num-b=1|num-a=3}}</div> Thedoge https://artofproblemsolving.com/wiki/index.php?title=2017_USAJMO_Problems&diff=85286 2017 USAJMO Problems 2017-04-19T23:07:34Z <p>Thedoge: </p> <hr /> <div>==Day 1==<br /> <br /> Note: For any geometry problem whose statement begins with an asterisk (&lt;math&gt;*&lt;/math&gt;), the first page of the solution must be a large, in-scale, clearly labeled diagram. Failure to meet this requirement will result in an automatic 1-point deduction. <br /> <br /> ===Problem 1===<br /> Prove that there are infinitely many distinct pairs &lt;math&gt;(a,b)&lt;/math&gt; of relatively prime positive integers &lt;math&gt;a &gt; 1&lt;/math&gt; and &lt;math&gt;b &gt; 1&lt;/math&gt; such that &lt;math&gt;a^b + b^a&lt;/math&gt; is divisible by &lt;math&gt;a + b&lt;/math&gt;. <br /> <br /> [[2017 USAJMO Problems/Problem 1|Solution]]<br /> ===Problem 2===<br /> Consider the equation <br /> &lt;cmath&gt;\left(3x^3 + xy^2 \right) \left(x^2y + 3y^3 \right) = (x-y)^7.&lt;/cmath&gt;<br /> <br /> (a) Prove that there are infinitely many pairs &lt;math&gt;(x,y)&lt;/math&gt; of positive integers satisfying the equation. <br /> <br /> (b) Describe all pairs &lt;math&gt;(x,y)&lt;/math&gt; of positive integers satisfying the equation. <br /> <br /> [[2017 USAJMO Problems/Problem 2|Solution]]<br /> ===Problem 3===<br /> (&lt;math&gt;*&lt;/math&gt;) Let &lt;math&gt;ABC&lt;/math&gt; be an equilateral triangle and let &lt;math&gt;P&lt;/math&gt; be a point on its circumcircle. Let lines &lt;math&gt;PA&lt;/math&gt; and &lt;math&gt;PB&lt;/math&gt; intersect at &lt;math&gt;D&lt;/math&gt;; let lines &lt;math&gt;PB&lt;/math&gt; and &lt;math&gt;CA&lt;/math&gt; intersect at &lt;math&gt;E&lt;/math&gt;; and let lines &lt;math&gt;PC&lt;/math&gt; and &lt;math&gt;AB&lt;/math&gt; intersect at &lt;math&gt;F&lt;/math&gt;. Prove that the area of triangle &lt;math&gt;DEF&lt;/math&gt; is twice the area of triangle &lt;math&gt;ABC&lt;/math&gt;.<br /> <br /> [[2017 USAJMO Problems/Problem 3|Solution]]<br /> <br /> ==Day 2==<br /> ===Problem 4===<br /> <br /> ===Problem 5===<br /> <br /> ===Problem 6===<br /> <br /> {{MAA Notice}}<br /> <br /> {{USAJMO newbox|year= 2017 |before=[[2016 USAJMO]]|after=[[2018 USAJMO]]}}</div> Thedoge https://artofproblemsolving.com/wiki/index.php?title=2017_USAJMO_Problems&diff=85284 2017 USAJMO Problems 2017-04-19T23:02:55Z <p>Thedoge: /* Problem 3 */</p> <hr /> <div>==Day 1==<br /> <br /> Note: For any geometry problem whose statement begins with an asterisk (&lt;math&gt;*&lt;/math&gt;), the first page of the solution must be a large, in-scale, clearly labeled diagram. Failure to meet this requirement will result in an automatic 1-point deduction. <br /> <br /> ===Problem 1===<br /> Prove that there are infinitely many distinct pairs &lt;math&gt;(a,b)&lt;/math&gt; of relatively prime positive integers &lt;math&gt;a &gt; 1&lt;/math&gt; and &lt;math&gt;b &gt; 1&lt;/math&gt; such that &lt;math&gt;a^b + b^a&lt;/math&gt; is divisible by &lt;math&gt;a + b&lt;/math&gt;. <br /> <br /> ===Problem 2===<br /> Consider the equation <br /> &lt;cmath&gt;\left(3x^3 + xy^2 \right) \left(x^2y + 3y^3 \right) = (x-y)^7.&lt;/cmath&gt;<br /> <br /> (a) Prove that there are infinitely many pairs &lt;math&gt;(x,y)&lt;/math&gt; of positive integers satisfying the equation. <br /> <br /> (b) Describe all pairs &lt;math&gt;(x,y)&lt;/math&gt; of positive integers satisfying the equation. <br /> <br /> ===Problem 3===<br /> (&lt;math&gt;*&lt;/math&gt;) Let &lt;math&gt;ABC&lt;/math&gt; be an equilateral triangle and let &lt;math&gt;P&lt;/math&gt; be a point on its circumcircle. Let lines &lt;math&gt;PA&lt;/math&gt; and &lt;math&gt;PB&lt;/math&gt; intersect at &lt;math&gt;D&lt;/math&gt;; let lines &lt;math&gt;PB&lt;/math&gt; and &lt;math&gt;CA&lt;/math&gt; intersect at &lt;math&gt;E&lt;/math&gt;; and let lines &lt;math&gt;PC&lt;/math&gt; and &lt;math&gt;AB&lt;/math&gt; intersect at &lt;math&gt;F&lt;/math&gt;. Prove that the area of triangle &lt;math&gt;DEF&lt;/math&gt; is twice the area of triangle &lt;math&gt;ABC&lt;/math&gt;.<br /> <br /> ==Day 2==<br /> ===Problem 4===<br /> <br /> ===Problem 5===<br /> <br /> ===Problem 6===</div> Thedoge https://artofproblemsolving.com/wiki/index.php?title=2017_USAJMO_Problems&diff=85283 2017 USAJMO Problems 2017-04-19T23:02:06Z <p>Thedoge: Created page with &quot;==Day 1== Note: For any geometry problem whose statement begins with an asterisk (&lt;math&gt;*&lt;/math&gt;), the first page of the solution must be a large, in-scale, clearly labeled d...&quot;</p> <hr /> <div>==Day 1==<br /> <br /> Note: For any geometry problem whose statement begins with an asterisk (&lt;math&gt;*&lt;/math&gt;), the first page of the solution must be a large, in-scale, clearly labeled diagram. Failure to meet this requirement will result in an automatic 1-point deduction. <br /> <br /> ===Problem 1===<br /> Prove that there are infinitely many distinct pairs &lt;math&gt;(a,b)&lt;/math&gt; of relatively prime positive integers &lt;math&gt;a &gt; 1&lt;/math&gt; and &lt;math&gt;b &gt; 1&lt;/math&gt; such that &lt;math&gt;a^b + b^a&lt;/math&gt; is divisible by &lt;math&gt;a + b&lt;/math&gt;. <br /> <br /> ===Problem 2===<br /> Consider the equation <br /> &lt;cmath&gt;\left(3x^3 + xy^2 \right) \left(x^2y + 3y^3 \right) = (x-y)^7.&lt;/cmath&gt;<br /> <br /> (a) Prove that there are infinitely many pairs &lt;math&gt;(x,y)&lt;/math&gt; of positive integers satisfying the equation. <br /> <br /> (b) Describe all pairs &lt;math&gt;(x,y)&lt;/math&gt; of positive integers satisfying the equation. <br /> <br /> ===Problem 3===<br /> (&lt;math&gt;*&lt;/math&gt;) Let &lt;math&gt;ABC&lt;/math&gt; be an equilateral triangle and let &lt;math&gt;P&lt;/math&gt; be a point on its circumcircle. Let lines &lt;math&gt;PA&lt;/math&gt; and &lt;math&gt;PB&lt;/math&gt; intersect at &lt;math&gt;D&lt;/math&gt;; let lines &lt;math&gt;PB&lt;/math&gt; and &lt;math&gt;CA&lt;/math&gt; intersect at &lt;math&gt;E&lt;/math&gt;; and let lines &lt;math&gt;PC&lt;/math&gt; and &lt;math&gt;AB&lt;/math&gt; intersect at &lt;math&gt;F&lt;/math&gt;. Prove that the area of triangle &lt;math&gt;DEF&lt;/math&gt; is twice that of triangle &lt;math&gt;ABC&lt;/math&gt;. <br /> <br /> ==Day 2==<br /> ===Problem 4===<br /> <br /> ===Problem 5===<br /> <br /> ===Problem 6===</div> Thedoge https://artofproblemsolving.com/wiki/index.php?title=2017_AIME_II_Problems/Problem_8&diff=85060 2017 AIME II Problems/Problem 8 2017-03-28T21:36:16Z <p>Thedoge: /* Solution 1 */</p> <hr /> <div>==Problem==<br /> Find the number of positive integers &lt;math&gt;n&lt;/math&gt; less than &lt;math&gt;2017&lt;/math&gt; such that &lt;cmath&gt;1+n+\frac{n^2}{2!}+\frac{n^3}{3!}+\frac{n^4}{4!}+\frac{n^5}{5!}+\frac{n^6}{6!}&lt;/cmath&gt; is an integer.<br /> <br /> ==Solution 1==<br /> Writing the last two terms with a common denominator, we have &lt;math&gt;\frac{6n^5+n^6}{720} \implies \frac{n^5(6+n)}{720}&lt;/math&gt; By inspection. this yields that &lt;math&gt;n \equiv 0, 24 \pmod{30}&lt;/math&gt;. Therefore, we get the final answer of &lt;math&gt;67 + 67 = \boxed{134}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> Taking out the &lt;math&gt;1+n&lt;/math&gt; part of the expression and writing the remaining terms under a common denominator, we get &lt;math&gt;\frac{1}{720}(n^6+6n^5+30n^4+120n^3+360n^2)&lt;/math&gt;. Therefore the expression &lt;math&gt;n^6+6n^5+30n^4+120n^3+360n^2&lt;/math&gt; must equal &lt;math&gt;720m&lt;/math&gt; for some positive integer &lt;math&gt;m&lt;/math&gt;.<br /> Taking both sides mod &lt;math&gt;2&lt;/math&gt;, the result is &lt;math&gt;n^6 \equiv 0 \pmod{2}&lt;/math&gt;. Therefore &lt;math&gt;n&lt;/math&gt; must be even. If &lt;math&gt;n&lt;/math&gt; is even, that means &lt;math&gt;n&lt;/math&gt; can be written in the form &lt;math&gt;2a&lt;/math&gt; where &lt;math&gt;a&lt;/math&gt; is a positive integer. Replacing &lt;math&gt;n&lt;/math&gt; with &lt;math&gt;2a&lt;/math&gt; in the expression, &lt;math&gt;64a^6+192a^5+480a^4+960a^3+1440a^2&lt;/math&gt; is divisible by &lt;math&gt;16&lt;/math&gt; because each coefficient is divisible by &lt;math&gt;16&lt;/math&gt;. Therefore, if &lt;math&gt;n&lt;/math&gt; is even, &lt;math&gt;n^6+6n^5+30n^4+120n^3+360n^2&lt;/math&gt; is divisible by &lt;math&gt;16&lt;/math&gt;.<br /> <br /> Taking the equation &lt;math&gt;n^6+6n^5+30n^4+120n^3+360n^2=720m&lt;/math&gt; mod &lt;math&gt;3&lt;/math&gt;, the result is &lt;math&gt;n^6 \equiv 0 \pmod{3}&lt;/math&gt;. Therefore &lt;math&gt;n&lt;/math&gt; must be a multiple of &lt;math&gt;3&lt;/math&gt;. If &lt;math&gt;n&lt;/math&gt; is a multiple of three, that means &lt;math&gt;n&lt;/math&gt; can be written in the form &lt;math&gt;3b&lt;/math&gt; where &lt;math&gt;b&lt;/math&gt; is a positive integer. Replacing &lt;math&gt;n&lt;/math&gt; with &lt;math&gt;3b&lt;/math&gt; in the expression, &lt;math&gt;729b^6+1458b^5+2430b^4+3240b^3+3240b^2&lt;/math&gt; is divisible by &lt;math&gt;9&lt;/math&gt; because each coefficient is divisible by &lt;math&gt;9&lt;/math&gt;. Therefore, if &lt;math&gt;n&lt;/math&gt; is a multiple of &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;n^6+6n^5+30n^4+120n^3+360n^2&lt;/math&gt; is divisibly by &lt;math&gt;9&lt;/math&gt;.<br /> <br /> Taking the equation &lt;math&gt;n^6+6n^5+30n^4+120n^3+360n^2=720m&lt;/math&gt; mod &lt;math&gt;5&lt;/math&gt;, the result is &lt;math&gt;n^6+n^5 \equiv 0 \pmod{5}&lt;/math&gt;. The only values of &lt;math&gt;n (\text{mod }5)&lt;/math&gt; that satisfy the equation are &lt;math&gt;n\equiv0(\text{mod }5)&lt;/math&gt; and &lt;math&gt;n\equiv4(\text{mod }5)&lt;/math&gt;. Therefore if &lt;math&gt;n&lt;/math&gt; is &lt;math&gt;0&lt;/math&gt; or &lt;math&gt;4&lt;/math&gt; mod &lt;math&gt;5&lt;/math&gt;, &lt;math&gt;n^6+6n^5+30n^4+120n^3+360n^2&lt;/math&gt; will be a multiple of &lt;math&gt;5&lt;/math&gt;.<br /> <br /> The only way to get the expression &lt;math&gt;n^6+6n^5+30n^4+120n^3+360n^2&lt;/math&gt; to be divisible by &lt;math&gt;720=16 \cdot 9 \cdot 5&lt;/math&gt; is to have &lt;math&gt;n \equiv 0 \pmod{2}&lt;/math&gt;, &lt;math&gt;n \equiv 0 \pmod{3}&lt;/math&gt;, and &lt;math&gt;n \equiv 0 \text{ or } 4 \pmod{5}&lt;/math&gt;. By the Chinese Remainder Theorem or simple guessing and checking, we see &lt;math&gt;n\equiv0,24 \pmod{30}&lt;/math&gt;. Because no numbers between &lt;math&gt;2011&lt;/math&gt; and &lt;math&gt;2017&lt;/math&gt; are equivalent to &lt;math&gt;0&lt;/math&gt; or &lt;math&gt;24&lt;/math&gt; mod &lt;math&gt;30&lt;/math&gt;, the answer is &lt;math&gt;\frac{2010}{30}\times2=\boxed{134}&lt;/math&gt;.<br /> <br /> =See Also=<br /> {{AIME box|year=2017|n=II|num-b=7|num-a=9}}<br /> {{MAA Notice}}</div> Thedoge https://artofproblemsolving.com/wiki/index.php?title=2003_AIME_I_Problems/Problem_15&diff=85038 2003 AIME I Problems/Problem 15 2017-03-27T00:44:23Z <p>Thedoge: /* Solution 3 */</p> <hr /> <div>== Problem ==<br /> In &lt;math&gt; \triangle ABC, AB = 360, BC = 507, &lt;/math&gt; and &lt;math&gt; CA = 780. &lt;/math&gt; Let &lt;math&gt; M &lt;/math&gt; be the [[midpoint]] of &lt;math&gt; \overline{CA}, &lt;/math&gt; and let &lt;math&gt; D &lt;/math&gt; be the point on &lt;math&gt; \overline{CA} &lt;/math&gt; such that &lt;math&gt; \overline{BD} &lt;/math&gt; bisects angle &lt;math&gt; ABC. &lt;/math&gt; Let &lt;math&gt; F &lt;/math&gt; be the point on &lt;math&gt; \overline{BC} &lt;/math&gt; such that &lt;math&gt; \overline{DF} \perp \overline{BD}. &lt;/math&gt; Suppose that &lt;math&gt; \overline{DF} &lt;/math&gt; meets &lt;math&gt; \overline{BM} &lt;/math&gt; at &lt;math&gt; E. &lt;/math&gt; The ratio &lt;math&gt; DE: EF &lt;/math&gt; can be written in the form &lt;math&gt; m/n, &lt;/math&gt; where &lt;math&gt; m &lt;/math&gt; and &lt;math&gt; n &lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt; m + n. &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> In the following, let the name of a point represent the mass located there. Since we are looking for a ratio, we assume that &lt;math&gt;AB=120&lt;/math&gt;, &lt;math&gt;BC=169&lt;/math&gt;, and &lt;math&gt;CA=260&lt;/math&gt; in order to simplify our computations.<br /> <br /> First, reflect point &lt;math&gt;F&lt;/math&gt; over angle bisector &lt;math&gt;BD&lt;/math&gt; to a point &lt;math&gt;F'&lt;/math&gt;.<br /> &lt;center&gt;&lt;asy&gt;<br /> size(400); pointpen = black; pathpen = black+linewidth(0.7); <br /> pair A=(0,0),C=(7.8,0),B=IP(CR(A,3.6),CR(C,5.07)), M=(A+C)/2, Da = bisectorpoint(A,B,C), D=IP(B--B+(Da-B)*10,A--C), F=IP(D--D+10*(B-D)*dir(270),B--C), E=IP(B--M,D--F);pair Fprime=2*D-F; /* scale down by 100x */<br /> D(MP(&quot;A&quot;,A,NW)--MP(&quot;B&quot;,B,N)--MP(&quot;C&quot;,C)--cycle); D(B--D(MP(&quot;D&quot;,D))--D(MP(&quot;F&quot;,F,NE))); D(B--D(MP(&quot;M&quot;,M)));D(A--MP(&quot;F'&quot;,Fprime,SW)--D); MP(&quot;E&quot;,E,NE); D(rightanglemark(F,D,B,4)); MP(&quot;390&quot;,(M+C)/2); MP(&quot;390&quot;,(M+C)/2); MP(&quot;360&quot;,(A+B)/2,NW); MP(&quot;507&quot;,(B+C)/2,NE);<br /> &lt;/asy&gt;&lt;/center&gt;<br /> As &lt;math&gt;BD&lt;/math&gt; is an angle bisector of both triangles &lt;math&gt;BAC&lt;/math&gt; and &lt;math&gt;BF'F&lt;/math&gt;, we know that &lt;math&gt;F'&lt;/math&gt; lies on &lt;math&gt;AB&lt;/math&gt;. We can now balance triangle &lt;math&gt;BF'C&lt;/math&gt; at point &lt;math&gt;D&lt;/math&gt; using mass points.<br /> <br /> By the [[Angle Bisector Theorem]], we can place [[mass points]] on &lt;math&gt;C,D,A&lt;/math&gt; of &lt;math&gt;120,\,289,\,169&lt;/math&gt; respectively. Thus, a mass of &lt;math&gt;\frac {289}{2}&lt;/math&gt; belongs at both &lt;math&gt;F&lt;/math&gt; and &lt;math&gt;F'&lt;/math&gt; because BD is a median of triangle &lt;math&gt;BF'F&lt;/math&gt; . Therefore, &lt;math&gt;CB/FB=\frac{289}{240}&lt;/math&gt;.<br /> <br /> Now, we reassign mass points to determine &lt;math&gt;FE/FD&lt;/math&gt;. This setup involves &lt;math&gt;\triangle CFD&lt;/math&gt; and [[transversal]] &lt;math&gt;MEB&lt;/math&gt;. For simplicity, put masses of &lt;math&gt;240&lt;/math&gt; and &lt;math&gt;289&lt;/math&gt; at &lt;math&gt;C&lt;/math&gt; and &lt;math&gt;F&lt;/math&gt; respectively. To find the mass we should put at &lt;math&gt;D&lt;/math&gt;, we compute &lt;math&gt;CM/MD&lt;/math&gt;. Applying the Angle Bisector Theorem again and using the fact &lt;math&gt;M&lt;/math&gt; is a midpoint of &lt;math&gt;AC&lt;/math&gt;, we find<br /> &lt;cmath&gt;<br /> \frac {MD}{CM} = \frac {\frac{169}{289}\cdot 260 - 130}{130} = \frac {49}{289}<br /> &lt;/cmath&gt;<br /> At this point we could find the mass at &lt;math&gt;D&lt;/math&gt; but it's unnecessary.<br /> &lt;cmath&gt;<br /> \frac {DE}{EF} = \frac {F}{D} = \frac {F}{C}\cdot\frac {C}{D} = \frac {289}{240}\cdot\frac {49}{289} = \boxed{\frac {49}{240}}<br /> &lt;/cmath&gt;<br /> and the answer is &lt;math&gt;49 + 240 = \boxed{289}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> By the Angle Bisector Theorem, we know that &lt;math&gt;[CBD]=\frac{169}{289}[ABC]&lt;/math&gt;. Therefore, by finding the area of triangle &lt;math&gt;CBD&lt;/math&gt;, we see that &lt;cmath&gt;\frac{507\cdot BD}{2}\sin\frac{B}{2}=\frac{169}{289}[ABC].&lt;/cmath&gt;<br /> Solving for &lt;math&gt;BD&lt;/math&gt; yields &lt;cmath&gt;BD=\frac{2[ABC]}{3\cdot289\sin\frac{B}{2}}.&lt;/cmath&gt;<br /> Furthermore, &lt;math&gt;\cos\frac{B}{2}=\frac{BD}{BF}&lt;/math&gt;, so &lt;cmath&gt;BF=\frac{BD}{\cos\frac{B}{2}}=\frac{2[ABC]}{3\cdot289\sin\frac{B}{2}\cos\frac{B}{2}}.&lt;/cmath&gt;<br /> Now by the identity &lt;math&gt;2\sin\frac{B}{2}\cos\frac{B}{2}=\sin B&lt;/math&gt;, we get &lt;cmath&gt;BF=\frac{4[ABC]}{3\cdot289\sin B}.&lt;/cmath&gt;<br /> But then &lt;math&gt;[ABC]=\frac{360\cdot 507}{2}\sin B&lt;/math&gt;, so &lt;math&gt;BF=\frac{240}{289}\cdot 507&lt;/math&gt;. Thus &lt;math&gt;BF:FC=240:49&lt;/math&gt;.<br /> <br /> Now by the Angle Bisector Theorem, &lt;math&gt;CD=\frac{169}{289}\cdot 780&lt;/math&gt;, and we know that &lt;math&gt;MC=\frac{1}{2}\cdot 780&lt;/math&gt; so &lt;math&gt;DM:MC=\frac{169}{289}-\frac{1}{2}:\frac{1}{2}=49:289&lt;/math&gt;.<br /> <br /> We can now use mass points on triangle CBD. Assign a mass of &lt;math&gt;240\cdot 49&lt;/math&gt; to point &lt;math&gt;C&lt;/math&gt;. Then &lt;math&gt;D&lt;/math&gt; must have mass &lt;math&gt;240\cdot 289&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; must have mass &lt;math&gt;49\cdot 49&lt;/math&gt;. This gives &lt;math&gt;F&lt;/math&gt; a mass of &lt;math&gt;240\cdot 49+49\cdot 49=289\cdot 49&lt;/math&gt;. Therefore, &lt;math&gt;DE:EF=\frac{289\cdot 49}{240\cdot 289}=\frac{49}{240}&lt;/math&gt;, giving us an answer of &lt;math&gt;\boxed{289}.&lt;/math&gt;<br /> <br /> ===Solution 3===<br /> Let &lt;math&gt;\angle{DBM}=\theta&lt;/math&gt; and &lt;math&gt;\angle{DBC}=\alpha&lt;/math&gt;. Then because &lt;math&gt;BM&lt;/math&gt; is a median we have &lt;math&gt;360\sin{(\alpha+\theta)}=507\sin{(\alpha-\theta)}&lt;/math&gt;. Now we know &lt;cmath&gt;\sin{(\alpha+\theta)}=\sin{\alpha}\cos{\theta}+\sin{\theta}\cos{\alpha}=\dfrac{DF\cdot BD}{BF\cdot BE}+\dfrac{DE\cdot BD}{BE\cdot BF}=\dfrac{BD(DF+DE)}{BF\cdot BE}&lt;/cmath&gt; Expressing the area of &lt;math&gt;\triangle{BEF}&lt;/math&gt; in two ways we have &lt;cmath&gt;\dfrac{1}{2}BE\cdot BF\sin{(\alpha-\theta)}=\dfrac{1}{2}EF\cdot BD&lt;/cmath&gt; so &lt;cmath&gt;\sin{(\alpha-\theta)}=\dfrac{EF\cdot BD}{BF\cdot BE}&lt;/cmath&gt; Plugging this in we have &lt;cmath&gt;\dfrac{360\cdot BD(DF+DE)}{BF\cdot BE}=\dfrac{507\cdot BD\cdot EF}{BF\cdot BE}&lt;/cmath&gt; so &lt;math&gt;\dfrac{DF+DE}{EF}=\dfrac{507}{360}&lt;/math&gt;. But &lt;math&gt;DF=DE+EF&lt;/math&gt;, so this simplifies to &lt;math&gt;1+\dfrac{2DE}{EF}=\dfrac{507}{360}=\dfrac{169}{120}&lt;/math&gt;, and thus &lt;math&gt;\dfrac{DE}{EF}=\dfrac{49}{240}&lt;/math&gt;, and &lt;math&gt;m+n=\boxed{289}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2003|n=I|num-b=14|after=Last question}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Thedoge https://artofproblemsolving.com/wiki/index.php?title=2017_AIME_II_Problems/Problem_7&diff=85003 2017 AIME II Problems/Problem 7 2017-03-24T20:48:08Z <p>Thedoge: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> Find the number of integer values of &lt;math&gt;k&lt;/math&gt; in the closed interval &lt;math&gt;[-500,500]&lt;/math&gt; for which the equation &lt;math&gt;\log(kx)=2\log(x+2)&lt;/math&gt; has exactly one real solution.<br /> <br /> ==Solution 1==<br /> <br /> &lt;asy&gt;<br /> Label f; <br /> f.p=fontsize(5); <br /> xaxis(-3,3,Ticks(f,1.0));<br /> yaxis(-3,26,Ticks(f,1.0));<br /> real f(real x){return (x+2)^2;}<br /> real g(real x){return x*-1;}<br /> real h(real x){return x*-2;}<br /> real i(real x){return x*-3;}<br /> real j(real x){return x*8;}<br /> draw(graph(f,-2,3),green);<br /> draw(graph(g,-2,2),red);<br /> draw(graph(h,-2,1),red);<br /> draw(graph(i,-2,1/3),red);<br /> draw(graph(j,-0.25,3),red);<br /> &lt;/asy&gt;<br /> Note the equation &lt;math&gt;\log(kx)=2\log(x+2)&lt;/math&gt; is valid for &lt;math&gt;kx&gt;0&lt;/math&gt; and &lt;math&gt;x&gt;-2&lt;/math&gt;. &lt;math&gt;\log(kx)=2\log(x+2)=\log((x+2)^2)&lt;/math&gt;. The equation &lt;math&gt;kx=(x+2)^2&lt;/math&gt; is derived by taking away the outside logs from the previous equation. Because &lt;math&gt;(x+2)^2&lt;/math&gt; is always non-negative, &lt;math&gt;kx&lt;/math&gt; must also be non-negative; therefore this takes care of the &lt;math&gt;kx&gt;0&lt;/math&gt; condition as long as &lt;math&gt;k\neq0&lt;/math&gt;, i.e. &lt;math&gt;k&lt;/math&gt; cannot be &lt;math&gt;0&lt;/math&gt;. Now, we graph both &lt;math&gt;(x+2)^2&lt;/math&gt; (the green graph) and &lt;math&gt;kx&lt;/math&gt; (the red graph for &lt;math&gt;k=-1,k=-2,k=-3,k=8&lt;/math&gt;) for &lt;math&gt;x&gt;-2&lt;/math&gt;. It is easy to see that all negative values of &lt;math&gt;k&lt;/math&gt; make the equation &lt;math&gt;\log(kx)=2\log(x+2)&lt;/math&gt; have only one solution. However, there is also one positive value of &lt;math&gt;k&lt;/math&gt; that makes the equation only have one solution, as shown by the steepest line in the diagram. We can show that the slope of this line is a positive integer by setting the discriminant of the equation &lt;math&gt;(x+2)^2=kx&lt;/math&gt; to be &lt;math&gt;0&lt;/math&gt; and solving for &lt;math&gt;k&lt;/math&gt;. Therefore, there are &lt;math&gt;500&lt;/math&gt; negative solutions and &lt;math&gt;1&lt;/math&gt; positive solution, for a total of &lt;math&gt;\boxed{501}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> We use an algebraic approach. Since &lt;math&gt;\log(kx)=2\log(x+2)&lt;/math&gt;, then &lt;math&gt;kx = (x+2)^2&lt;/math&gt; (the converse isn't necessarily true!), or &lt;math&gt;x^2+(4-k)x+4=0&lt;/math&gt;. Our original equation has exactly one solution if and only if there is only one solution to the above equation, or one of the solutions is extraneous; it involves the computation of the log of a nonpositive number.<br /> <br /> For the first case, we note that this can only occur when it is a perfect square trinomal, or &lt;math&gt;k = 0, 8&lt;/math&gt;. However, &lt;math&gt;k = 0&lt;/math&gt; results in &lt;math&gt;\log(0)&lt;/math&gt; on the LHS, which is invalid. &lt;math&gt;k = 8&lt;/math&gt; yields &lt;math&gt;x = 2&lt;/math&gt;, so that is one solution.<br /> <br /> For the second case, we can use the quadratic formula. We have &lt;cmath&gt;x = \frac{k-4 \pm \sqrt{k^2-8k}}2,&lt;/cmath&gt; so in order for there to be at least one real solution, the discriminant must be nonnegative, or &lt;math&gt;k &lt; 0&lt;/math&gt; or &lt;math&gt;k &gt; 8&lt;/math&gt;. Note that if &lt;math&gt;k &gt; 8&lt;/math&gt;, then both solutions will be positive, and therefore both valid. Therefore, &lt;math&gt;k &lt; 0&lt;/math&gt;.<br /> We now wish to show that if &lt;math&gt;k &lt; 0&lt;/math&gt;, then there is exactly one solution that works. Note that whenever &lt;math&gt;k &lt; 0&lt;/math&gt;, both &quot;solutions&quot; in &lt;math&gt;x&lt;/math&gt; are negative. One of the solutions to the equation is &lt;math&gt;x = \frac{k-4 + \sqrt{k^2-8k}}2&lt;/math&gt;. We wish to prove that &lt;math&gt;x + 2 &gt; 0&lt;/math&gt;, or &lt;math&gt;x &gt; -2&lt;/math&gt; (therefore the RHS in the original equation will be defined). Substituting, we have &lt;math&gt;\frac{k-4 + \sqrt{k^2-8k}}2 &gt; -2&lt;/math&gt;, or &lt;math&gt;\sqrt{k^2 - 8k} &gt; -k&lt;/math&gt;. Since both sides are positive, we can square both sides (if &lt;math&gt;k &lt; 0&lt;/math&gt;, then &lt;math&gt;-k &gt; 0&lt;/math&gt;) to get &lt;math&gt;k^2-8k &gt; k^2&lt;/math&gt;, or &lt;math&gt;8k &lt; 0 \implies k &lt; 0&lt;/math&gt;, which was our original assumption, so this solution satisfies the original equation. The other case is when &lt;math&gt;x = \frac{k-4 - \sqrt{k^2-8k}}2&lt;/math&gt;, which we wish to show is less that &lt;math&gt;-2&lt;/math&gt;, or &lt;math&gt;\frac{k-4 - \sqrt{k^2-8k}}2 &lt; -2 \implies k &lt; \sqrt{k^2-8k}&lt;/math&gt;. However, since the square root is defined to be positive, then this is always true, which implies that whenever &lt;math&gt;k &lt; 0&lt;/math&gt;, there is exactly one real solution that satisfies the original equation. Combining this with &lt;math&gt;k \in [-500, 500]&lt;/math&gt;, we find that the answer is &lt;math&gt;500 + 1 = \boxed{501}&lt;/math&gt;.<br /> <br /> =See Also=<br /> {{AIME box|year=2017|n=II|num-b=6|num-a=8}}<br /> {{MAA Notice}}</div> Thedoge https://artofproblemsolving.com/wiki/index.php?title=2017_AIME_II_Problems/Problem_4&diff=85002 2017 AIME II Problems/Problem 4 2017-03-24T19:41:19Z <p>Thedoge: /* Solution */</p> <hr /> <div>==Problem==<br /> Find the number of positive integers less than or equal to &lt;math&gt;2017&lt;/math&gt; whose base-three representation contains no digit equal to &lt;math&gt;0&lt;/math&gt;.<br /> <br /> ==Solution==<br /> ===Solution 1===<br /> The base-&lt;math&gt;3&lt;/math&gt; representation of &lt;math&gt;2017_{10}&lt;/math&gt; is &lt;math&gt;2202201_3&lt;/math&gt;. Because any &lt;math&gt;7&lt;/math&gt;-digit base-&lt;math&gt;3&lt;/math&gt; number that starts with &lt;math&gt;22&lt;/math&gt; and has no digit equal to &lt;math&gt;0&lt;/math&gt; must be greater than &lt;math&gt;2017_{10}&lt;/math&gt;, all &lt;math&gt;7&lt;/math&gt;-digit numbers that have no digit equal to &lt;math&gt;0&lt;/math&gt; must start with &lt;math&gt;21&lt;/math&gt; or &lt;math&gt;1&lt;/math&gt; in base &lt;math&gt;3&lt;/math&gt;. Of the base-&lt;math&gt;3&lt;/math&gt; numbers that have no digit equal to &lt;math&gt;0&lt;/math&gt;, there are &lt;math&gt;2^5&lt;/math&gt; &lt;math&gt;7&lt;/math&gt;-digit numbers that start with &lt;math&gt;21&lt;/math&gt;, &lt;math&gt;2^6&lt;/math&gt; &lt;math&gt;7&lt;/math&gt;-digit numbers that start with &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;2^6&lt;/math&gt; &lt;math&gt;6&lt;/math&gt;-digit numbers, &lt;math&gt;2^5&lt;/math&gt; &lt;math&gt;5&lt;/math&gt;-digit numbers, &lt;math&gt;2^4&lt;/math&gt; &lt;math&gt;4&lt;/math&gt;-digit numbers, &lt;math&gt;2^3&lt;/math&gt; &lt;math&gt;3&lt;/math&gt;-digit numbers, &lt;math&gt;2^2&lt;/math&gt; &lt;math&gt;2&lt;/math&gt;-digit numbers, and &lt;math&gt;2^1&lt;/math&gt; &lt;math&gt;1&lt;/math&gt;-digit numbers. Summing these up, we find that the answer is &lt;math&gt;2^5+2^6+2^6+2^5+2^4+2^3+2^2+2^1=\boxed{222}&lt;/math&gt;.<br /> <br /> <br /> ===Solution 2===<br /> <br /> Note that &lt;math&gt;2017=220221_{3}&lt;/math&gt;, and &lt;math&gt;2187=3^7=10000000_{3}&lt;/math&gt;. There can be a &lt;math&gt;1,2,...,7&lt;/math&gt; digit number less than &lt;math&gt;2187&lt;/math&gt;, and each digit can either be &lt;math&gt;1&lt;/math&gt; or &lt;math&gt;2&lt;/math&gt;. So &lt;math&gt;2^1&lt;/math&gt; one digit numbers and so on up to &lt;math&gt;2^7&lt;/math&gt; &lt;math&gt;7&lt;/math&gt; digit.<br /> <br /> <br /> Now we have to subtract out numbers from &lt;math&gt;2018&lt;/math&gt; to &lt;math&gt;2187&lt;/math&gt;<br /> <br /> Then either the number must begin &lt;math&gt;221...&lt;/math&gt; or &lt;math&gt;222...&lt;/math&gt; with four more digits at the end<br /> <br /> Using &lt;math&gt;1&lt;/math&gt;s and &lt;math&gt;2&lt;/math&gt;s there are &lt;math&gt;2^4&lt;/math&gt; options for each so:<br /> <br /> &lt;math&gt;2+4+8+16+32+64+128-2*16=256-2-32=\boxed{222}&lt;/math&gt;<br /> <br /> =See Also=<br /> {{AIME box|year=2017|n=II|num-b=3|num-a=5}}<br /> {{MAA Notice}}</div> Thedoge https://artofproblemsolving.com/wiki/index.php?title=2017_AIME_II_Problems/Problem_7&diff=85001 2017 AIME II Problems/Problem 7 2017-03-24T19:40:19Z <p>Thedoge: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> Find the number of integer values of &lt;math&gt;k&lt;/math&gt; in the closed interval &lt;math&gt;[-500,500]&lt;/math&gt; for which the equation &lt;math&gt;\log(kx)=2\log(x+2)&lt;/math&gt; has exactly one real solution.<br /> <br /> ==Solution 1==<br /> <br /> &lt;asy&gt;<br /> Label f; <br /> f.p=fontsize(5); <br /> xaxis(-3,3,Ticks(f,1.0));<br /> yaxis(-3,26,Ticks(f,1.0));<br /> real f(real x){return (x+2)^2;}<br /> real g(real x){return x*-1;}<br /> real h(real x){return x*-2;}<br /> real i(real x){return x*-3;}<br /> real j(real x){return x*8;}<br /> draw(graph(f,-2,3),green);<br /> draw(graph(g,-2,2),red);<br /> draw(graph(h,-2,1),red);<br /> draw(graph(i,-2,1/3),red);<br /> draw(graph(j,-0.25,3),red);<br /> &lt;/asy&gt;<br /> Note the equation &lt;math&gt;\log(kx)=2\log(x+2)&lt;/math&gt; is valid for &lt;math&gt;kx&gt;0&lt;/math&gt; and &lt;math&gt;x&gt;-2&lt;/math&gt;. &lt;math&gt;\log(kx)=2\log(x+2)=\log((x+2)^2)&lt;/math&gt;. The equation &lt;math&gt;kx=(x+2)^2&lt;/math&gt; is derived by taking away the outside logs from the previous equation. Because &lt;math&gt;(x+2)^2&lt;/math&gt; is always non-negative, &lt;math&gt;kx&lt;/math&gt; must also be non-negative; therefore this takes care of the &lt;math&gt;kx&gt;0&lt;/math&gt; condition as long as &lt;math&gt;k\neq0&lt;/math&gt;, i.e. &lt;math&gt;k&lt;/math&gt; cannot be &lt;math&gt;0&lt;/math&gt;. Now, we graph both &lt;math&gt;(x+2)^2&lt;/math&gt; (the green graph) and &lt;math&gt;kx&lt;/math&gt; (the red graph for &lt;math&gt;k=-1,k=-2,k=-3,k=8&lt;/math&gt;) for &lt;math&gt;x&gt;-2&lt;/math&gt;. It is easy to see that all negative values of &lt;math&gt;k&lt;/math&gt; make the equation &lt;math&gt;\log(kx)=2\log(x+2)&lt;/math&gt; have only one solution. However, there is also one positive value of &lt;math&gt;k&lt;/math&gt; that makes the equation only have one solution, as shown by the steepest line in the diagram. We can show that the slope of this line is a positive integer by setting the discriminant of the equation &lt;math&gt;(x+2)^2=kx&lt;/math&gt; to be &lt;math&gt;0&lt;/math&gt; and solving for &lt;math&gt;k&lt;/math&gt;. Therefore, there are &lt;math&gt;500&lt;/math&gt; negative solutions and &lt;math&gt;1&lt;/math&gt; positive solution, for a total of &lt;math&gt;\boxed{501}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> We use an algebraic approach. Since &lt;math&gt;\log(kx)=2\log(x+2)&lt;/math&gt;, then &lt;math&gt;kx = (x+2)^2&lt;/math&gt; (the converse isn't necessarily true!), or &lt;math&gt;x^2+(4-k)x+4=0&lt;/math&gt;. Our original equation has exactly one solution if and only if there is only one solution to the above equation, or one of the solutions is extraneous; it involves the computation of the log of a nonpositive number.<br /> <br /> For the first case, we note that this can only occur when it is a perfect square trinomal, or &lt;math&gt;k = 0, 8&lt;/math&gt;. However, &lt;math&gt;k = 0&lt;/math&gt; results in &lt;math&gt;\log(0)&lt;/math&gt; on the LHS, which is invalid. &lt;math&gt;k = 8&lt;/math&gt; yields &lt;math&gt;x = 2&lt;/math&gt;, so that is one solution.<br /> <br /> For the second case, we can use the quadratic formula. We have &lt;cmath&gt;x = \frac{k-4 \pm \sqrt{k^2-8k}}2,&lt;/cmath&gt; so in order for there to be at least one real solution, the discriminant must be nonnegative, or &lt;math&gt;k &lt; 0&lt;/math&gt; or &lt;math&gt;k &gt; 8&lt;/math&gt;. Note that if &lt;math&gt;k &gt; 8&lt;/math&gt;, then both solutions will be positive, and therefore both valid. Therefore, &lt;math&gt;k &lt; 0&lt;/math&gt;.<br /> We now wish to show that if &lt;math&gt;k &lt; 0&lt;/math&gt;, then there is exactly one solution that works. Note that whenever &lt;math&gt;k &lt; 0&lt;/math&gt;, both &quot;solutions&quot; in &lt;math&gt;x&lt;/math&gt; are negative. One of the solutions to the equation is &lt;math&gt;x = \frac{k-4 + \sqrt{k^2-8k}}2&lt;/math&gt;. We wish to prove that &lt;math&gt;x + 2 &gt; 0&lt;/math&gt;, or &lt;math&gt;x &gt; -2&lt;/math&gt; (therefore the RHS in the original equation will be defined). Substituting, we have &lt;math&gt;\frac{k-4 + \sqrt{k^2-8k}}2 &gt; -2&lt;/math&gt;, or &lt;math&gt;\sqrt{k^2 - 8k} &gt; -k&lt;/math&gt;. Since both sides are positive we can square both sides (if &lt;math&gt;k &lt; 0&lt;/math&gt;, then &lt;math&gt;-k &gt; 0&lt;/math&gt;) to get &lt;math&gt;k^2-8k &gt; k^2&lt;/math&gt;, or &lt;math&gt;8k &lt; 0 \implies k &lt; 0&lt;/math&gt;, which was our original assumption, so this solution satisfies the original equation. The other case is when &lt;math&gt;x = \frac{k-4 - \sqrt{k^2-8k}}2&lt;/math&gt;, which we wish to show is less that &lt;math&gt;-2&lt;/math&gt;, or &lt;math&gt;\frac{k-4 - \sqrt{k^2-8k}}2 &lt; -2 \implies k &lt; \sqrt{k^2-8k}&lt;/math&gt;. However, since the square root is defined to be positive, then this is always true, which implies that whenever &lt;math&gt;k &lt; 0&lt;/math&gt;, there is exactly one real solution that satisfies the original equation. Combining this with &lt;math&gt;k \in [-500, 500]&lt;/math&gt;, we find that the answer is &lt;math&gt;500 + 1 = \boxed{501}&lt;/math&gt;.<br /> <br /> =See Also=<br /> {{AIME box|year=2017|n=II|num-b=6|num-a=8}}<br /> {{MAA Notice}}</div> Thedoge https://artofproblemsolving.com/wiki/index.php?title=2017_AIME_II_Problems/Problem_15&diff=84998 2017 AIME II Problems/Problem 15 2017-03-24T03:02:25Z <p>Thedoge: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> Tetrahedron &lt;math&gt;ABCD&lt;/math&gt; has &lt;math&gt;AD=BC=28&lt;/math&gt;, &lt;math&gt;AC=BD=44&lt;/math&gt;, and &lt;math&gt;AB=CD=52&lt;/math&gt;. For any point &lt;math&gt;X&lt;/math&gt; in space, define &lt;math&gt;f(X)=AX+BX+CX+DX&lt;/math&gt;. The least possible value of &lt;math&gt;f(X)&lt;/math&gt; can be expressed as &lt;math&gt;m\sqrt{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are positive integers, and &lt;math&gt;n&lt;/math&gt; is not divisible by the square of any prime. Find &lt;math&gt;m+n&lt;/math&gt;.<br /> <br /> ==Solution==<br /> ===Solution 1===<br /> Let &lt;math&gt;M&lt;/math&gt; and &lt;math&gt;N&lt;/math&gt; be midpoints of &lt;math&gt;\overline{AB}&lt;/math&gt; and &lt;math&gt;\overline{CD}&lt;/math&gt;. The given conditions imply that &lt;math&gt;\triangle ABD\cong\triangle BAC&lt;/math&gt; and &lt;math&gt;\triangle CDA\cong\triangle DCB&lt;/math&gt;, and therefore &lt;math&gt;MC=MD&lt;/math&gt; and &lt;math&gt;NA=NB&lt;/math&gt;. It follows that &lt;math&gt;M&lt;/math&gt; and &lt;math&gt;N&lt;/math&gt; both lie on the common perpendicular bisector of &lt;math&gt;\overline{AB}&lt;/math&gt; and &lt;math&gt;\overline{CD}&lt;/math&gt;, and thus line &lt;math&gt;MN&lt;/math&gt; is that common perpendicular bisector. Points &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; are symmetric to &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt; with respect to line &lt;math&gt;MN&lt;/math&gt;. If &lt;math&gt;X&lt;/math&gt; is a point in space and &lt;math&gt;X'&lt;/math&gt; is the point symmetric to &lt;math&gt;X&lt;/math&gt; with respect to line &lt;math&gt;MN&lt;/math&gt;, then &lt;math&gt;BX=AX'&lt;/math&gt; and &lt;math&gt;CX=DX'&lt;/math&gt;, so &lt;math&gt;f(X) = AX+AX'+DX+DX'&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;Q&lt;/math&gt; be the intersection of &lt;math&gt;\overline{XX'}&lt;/math&gt; and &lt;math&gt;\overline{MN}&lt;/math&gt;. Then &lt;math&gt;AX+AX'\geq 2AQ&lt;/math&gt;, from which it follows that &lt;math&gt;f(X) \geq 2(AQ+DQ) = f(Q)&lt;/math&gt;. It remains to minimize &lt;math&gt;f(Q)&lt;/math&gt; as &lt;math&gt;Q&lt;/math&gt; moves along &lt;math&gt;\overline{MN}&lt;/math&gt;.<br /> <br /> Allow &lt;math&gt;D&lt;/math&gt; to rotate about &lt;math&gt;\overline{MN}&lt;/math&gt; to point &lt;math&gt;D'&lt;/math&gt; in the plane &lt;math&gt;AMN&lt;/math&gt; on the side of &lt;math&gt;\overline{MN}&lt;/math&gt; opposite &lt;math&gt;A&lt;/math&gt;. Because &lt;math&gt;\angle DNM&lt;/math&gt; is a right angle, &lt;math&gt;D'N=DN&lt;/math&gt;. It then follows that &lt;math&gt;f(Q) = 2(AQ+D'Q)\geq 2AD'&lt;/math&gt;, and equality occurs when &lt;math&gt;Q&lt;/math&gt; is the intersection of &lt;math&gt;\overline{AD'}&lt;/math&gt; and &lt;math&gt;\overline{MN}&lt;/math&gt;. Thus &lt;math&gt;\min f(Q) = 2AD'&lt;/math&gt;. Because &lt;math&gt;\overline{MD}&lt;/math&gt; is the median of &lt;math&gt;\triangle ADB&lt;/math&gt;, the Length of Median Formula shows that &lt;math&gt;4MD^2 = 2AD^2 + 2BD^2 - AB^2 = 2\cdot 28^2 + 2 \cdot 44^2 - 52^2&lt;/math&gt; and &lt;math&gt;MD^2 = 684&lt;/math&gt;. By the Pythagorean Theorem &lt;math&gt;MN^2 = MD^2 - ND^2 = 8&lt;/math&gt;.<br /> <br /> Because &lt;math&gt;\angle AMN&lt;/math&gt; and &lt;math&gt;\angle D'NM&lt;/math&gt; are right angles, &lt;cmath&gt;(AD')^2 = (AM+D'N)^2 + MN^2 = (2AM)^2 + MN^2 = 52^2 + 8 = 4\cdot 678.&lt;/cmath&gt;It follows that &lt;math&gt;\min f(Q) = 2AD' = 4\sqrt{678}&lt;/math&gt;. The requested sum is &lt;math&gt;4+678=\boxed{682}&lt;/math&gt;.<br /> <br /> ===Solution 2===<br /> Set &lt;math&gt;a=BC=28&lt;/math&gt;, &lt;math&gt;b=CA=44&lt;/math&gt;, &lt;math&gt;c=AB=52&lt;/math&gt;. Let &lt;math&gt;O&lt;/math&gt; be the point which minimizes &lt;math&gt;f(X)&lt;/math&gt;.<br /> <br /> Claim: &lt;math&gt;O&lt;/math&gt; is the gravity center &lt;math&gt;\tfrac14(\vec A + \vec B + \vec C + \vec D)&lt;/math&gt;.<br /> Proof. Let &lt;math&gt;M&lt;/math&gt; and &lt;math&gt;N&lt;/math&gt; denote the midpoints of &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;CD&lt;/math&gt;. From &lt;math&gt;\triangle ABD \cong \triangle BAC&lt;/math&gt; and &lt;math&gt;\triangle CDA \cong \triangle DCB&lt;/math&gt;, we have &lt;math&gt;MC=MD&lt;/math&gt;, &lt;math&gt;NA=NB&lt;/math&gt; an hence &lt;math&gt;MN&lt;/math&gt; is a perpendicular bisector of both segments &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;CD&lt;/math&gt;. Then if &lt;math&gt;X&lt;/math&gt; is any point inside tetrahedron &lt;math&gt;ABCD&lt;/math&gt;, its orthogonal projection onto line &lt;math&gt;MN&lt;/math&gt; will have smaller &lt;math&gt;f&lt;/math&gt;-value; hence we conclude that &lt;math&gt;O&lt;/math&gt; must lie on &lt;math&gt;MN&lt;/math&gt;. Similarly, &lt;math&gt;O&lt;/math&gt; must lie on the line joining the midpoints of &lt;math&gt;AC&lt;/math&gt; and &lt;math&gt;BD&lt;/math&gt;. &lt;math&gt;\blacksquare&lt;/math&gt;<br /> <br /> Claim: The gravity center &lt;math&gt;O&lt;/math&gt; coincides with the circumcenter.<br /> Proof. Let &lt;math&gt;G_D&lt;/math&gt; be the centroid of triangle &lt;math&gt;ABC&lt;/math&gt;; then &lt;math&gt;DO = \tfrac 34 DG_D&lt;/math&gt; (by vectors). If we define &lt;math&gt;G_A&lt;/math&gt;, &lt;math&gt;G_B&lt;/math&gt;, &lt;math&gt;G_C&lt;/math&gt; similarly, we get &lt;math&gt;AO = \tfrac 34 AG_A&lt;/math&gt; and so on. But from symmetry we have &lt;math&gt;AG_A = BG_B = CG_C = DG_D&lt;/math&gt;, hence &lt;math&gt;AO = BO = CO = DO&lt;/math&gt;. &lt;math&gt;\blacksquare&lt;/math&gt;<br /> <br /> Now we use the fact that an isosceles tetrahedron has circumradius &lt;math&gt;R = \sqrt{\frac18(a^2+b^2+c^2)}&lt;/math&gt;. Here &lt;math&gt;R = \sqrt{678}&lt;/math&gt; so &lt;math&gt;f(O) = 4R = 4\sqrt{678}&lt;/math&gt;. Therefore, the answer is &lt;math&gt;4 + 678 = \boxed{682}&lt;/math&gt;.<br /> <br /> =See Also=<br /> {{AIME box|year=2017|n=II|num-b=14|after=Last Question}}<br /> {{MAA Notice}}</div> Thedoge https://artofproblemsolving.com/wiki/index.php?title=2017_AIME_II_Problems/Problem_12&diff=84997 2017 AIME II Problems/Problem 12 2017-03-24T00:29:37Z <p>Thedoge: /* Solution */</p> <hr /> <div>==Problem==<br /> Circle &lt;math&gt;C_0&lt;/math&gt; has radius &lt;math&gt;1&lt;/math&gt;, and the point &lt;math&gt;A_0&lt;/math&gt; is a point on the circle. Circle &lt;math&gt;C_1&lt;/math&gt; has radius &lt;math&gt;r&lt;1&lt;/math&gt; and is internally tangent to &lt;math&gt;C_0&lt;/math&gt; at point &lt;math&gt;A_0&lt;/math&gt;. Point &lt;math&gt;A_1&lt;/math&gt; lies on circle &lt;math&gt;C_1&lt;/math&gt; so that &lt;math&gt;A_1&lt;/math&gt; is located &lt;math&gt;90^{\circ}&lt;/math&gt; counterclockwise from &lt;math&gt;A_0&lt;/math&gt; on &lt;math&gt;C_1&lt;/math&gt;. Circle &lt;math&gt;C_2&lt;/math&gt; has radius &lt;math&gt;r^2&lt;/math&gt; and is internally tangent to &lt;math&gt;C_1&lt;/math&gt; at point &lt;math&gt;A_1&lt;/math&gt;. In this way a sequence of circles &lt;math&gt;C_1,C_2,C_3,\ldots&lt;/math&gt; and a sequence of points on the circles &lt;math&gt;A_1,A_2,A_3,\ldots&lt;/math&gt; are constructed, where circle &lt;math&gt;C_n&lt;/math&gt; has radius &lt;math&gt;r^n&lt;/math&gt; and is internally tangent to circle &lt;math&gt;C_{n-1}&lt;/math&gt; at point &lt;math&gt;A_{n-1}&lt;/math&gt;, and point &lt;math&gt;A_n&lt;/math&gt; lies on &lt;math&gt;C_n&lt;/math&gt; &lt;math&gt;90^{\circ}&lt;/math&gt; counterclockwise from point &lt;math&gt;A_{n-1}&lt;/math&gt;, as shown in the figure below. There is one point &lt;math&gt;B&lt;/math&gt; inside all of these circles. When &lt;math&gt;r = \frac{11}{60}&lt;/math&gt;, the distance from the center &lt;math&gt;C_0&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; is &lt;math&gt;\frac{m}{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m+n&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> draw(Circle((0,0),125));<br /> draw(Circle((25,0),100));<br /> draw(Circle((25,20),80));<br /> draw(Circle((9,20),64));<br /> dot((125,0));<br /> label(&quot;$A_0$&quot;,(125,0),E);<br /> dot((25,100));<br /> label(&quot;$A_1$&quot;,(25,100),SE);<br /> dot((-55,20));<br /> label(&quot;$A_2$&quot;,(-55,20),E);<br /> &lt;/asy&gt;<br /> <br /> ==Solution==<br /> Impose a coordinate system and let the center of &lt;math&gt;C_0&lt;/math&gt; be &lt;math&gt;(0,0)&lt;/math&gt; and &lt;math&gt;A_0&lt;/math&gt; be &lt;math&gt;(1,0)&lt;/math&gt;. Therefore &lt;math&gt;A_1=(1-r,r)&lt;/math&gt;, &lt;math&gt;A_2=(1-r-r^2,r-r^2)&lt;/math&gt;, &lt;math&gt;A_3=(1-r-r^2+r^3,r-r^2-r^3)&lt;/math&gt;, &lt;math&gt;A_4=(1-r-r^2+r^3+r^4,r-r^2-r^3+r^4)&lt;/math&gt;, and so on, where the signs alternate in groups of &lt;math&gt;2&lt;/math&gt;. The limit of all these points is point &lt;math&gt;B&lt;/math&gt;. Using the geometric series formula on &lt;math&gt;B&lt;/math&gt; and reducing the expression, we get &lt;math&gt;B=\left(\frac{1-r}{r^2+1},\frac{r-r^2}{r^2+1}\right)&lt;/math&gt;. The distance from &lt;math&gt;B&lt;/math&gt; to the origin is &lt;math&gt;\sqrt{\left(\frac{1-r}{r^2+1}\right)^2+\left(\frac{r-r^2}{r^2+1}\right)^2}=\frac{1-r}{\sqrt{r^2+1}}.&lt;/math&gt; Let &lt;math&gt;r=\frac{11}{60}&lt;/math&gt;, and the distance from the origin is &lt;math&gt;\frac{49}{61}&lt;/math&gt;. &lt;math&gt;49+61=\boxed{110}&lt;/math&gt;.<br /> <br /> =See Also=<br /> {{AIME box|year=2017|n=II|num-b=11|num-a=13}}<br /> {{MAA Notice}}</div> Thedoge https://artofproblemsolving.com/wiki/index.php?title=2017_AIME_II_Problems/Problem_15&diff=84986 2017 AIME II Problems/Problem 15 2017-03-23T22:55:06Z <p>Thedoge: /* Solution */</p> <hr /> <div>==Problem==<br /> Tetrahedron &lt;math&gt;ABCD&lt;/math&gt; has &lt;math&gt;AD=BC=28&lt;/math&gt;, &lt;math&gt;AC=BD=44&lt;/math&gt;, and &lt;math&gt;AB=CD=52&lt;/math&gt;. For any point &lt;math&gt;X&lt;/math&gt; in space, define &lt;math&gt;f(X)=AX+BX+CX+DX&lt;/math&gt;. The least possible value of &lt;math&gt;f(X)&lt;/math&gt; can be expressed as &lt;math&gt;m\sqrt{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are positive integers, and &lt;math&gt;n&lt;/math&gt; is not divisible by the square of any prime. Find &lt;math&gt;m+n&lt;/math&gt;.<br /> <br /> ==Solution==<br /> ===Solution 1===<br /> Let &lt;math&gt;M&lt;/math&gt; and &lt;math&gt;N&lt;/math&gt; be midpoints of &lt;math&gt;\overline{AB}&lt;/math&gt; and &lt;math&gt;\overline{CD}&lt;/math&gt;. The given conditions imply that &lt;math&gt;\triangle ABD\cong\triangle BAC&lt;/math&gt; and &lt;math&gt;\triangle CDA\cong\triangle DCB&lt;/math&gt;, and therefore &lt;math&gt;MC=MD&lt;/math&gt; and &lt;math&gt;NA=NB&lt;/math&gt;. It follows that &lt;math&gt;M&lt;/math&gt; and &lt;math&gt;N&lt;/math&gt; both lie on the common perpendicular bisector of &lt;math&gt;\overline{AB}&lt;/math&gt; and &lt;math&gt;\overline{CD}&lt;/math&gt;, and thus line &lt;math&gt;MN&lt;/math&gt; is that common perpendicular bisector. Points &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; are symmetric to &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt; with respect to line &lt;math&gt;MN&lt;/math&gt;. If &lt;math&gt;X&lt;/math&gt; is a point in space and &lt;math&gt;X'&lt;/math&gt; is the point symmetric to &lt;math&gt;X&lt;/math&gt; with respect to line &lt;math&gt;MN&lt;/math&gt;, then &lt;math&gt;BX=AX'&lt;/math&gt; and &lt;math&gt;CX=DX'&lt;/math&gt;, so &lt;math&gt;f(X) = AX+AX'+DX+DX'&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;Q&lt;/math&gt; be the intersection of &lt;math&gt;\overline{XX'}&lt;/math&gt; and &lt;math&gt;\overline{MN}&lt;/math&gt;. Then &lt;math&gt;AX+AX'\geq 2AQ&lt;/math&gt;, from which it follows that &lt;math&gt;f(X) \geq 2(AQ+DQ) = f(Q)&lt;/math&gt;. It remains to minimize &lt;math&gt;f(Q)&lt;/math&gt; as &lt;math&gt;Q&lt;/math&gt; moves along &lt;math&gt;\overline{MN}&lt;/math&gt;.<br /> <br /> Allow &lt;math&gt;D&lt;/math&gt; to rotate about &lt;math&gt;\overline{MN}&lt;/math&gt; to point &lt;math&gt;D'&lt;/math&gt; in the plane &lt;math&gt;AMN&lt;/math&gt; on the side of &lt;math&gt;\overline{MN}&lt;/math&gt; opposite &lt;math&gt;A&lt;/math&gt;. Because &lt;math&gt;\angle DNM&lt;/math&gt; is a right angle, &lt;math&gt;D'N=DN&lt;/math&gt;. It then follows that &lt;math&gt;f(Q) = 2(AQ+D'Q)\geq 2AD'&lt;/math&gt;, and equality occurs when &lt;math&gt;Q&lt;/math&gt; is the intersection of &lt;math&gt;\overline{AD'}&lt;/math&gt; and &lt;math&gt;\overline{MN}&lt;/math&gt;. Thus &lt;math&gt;\min f(Q) = 2AD'&lt;/math&gt;. Because &lt;math&gt;\overline{MD}&lt;/math&gt; is the median of &lt;math&gt;\triangle ADB&lt;/math&gt;, the Length of Median Formula shows that &lt;math&gt;4MD^2 = 2AD^2 + 2BD^2 - AB^2 = 2\cdot 28^2 + 2 \cdot 44^2 - 52^2&lt;/math&gt; and &lt;math&gt;MD^2 = 684&lt;/math&gt;. By the Pythagorean Theorem &lt;math&gt;MN^2 = MD^2 - ND^2 = 8&lt;/math&gt;.<br /> <br /> Because &lt;math&gt;\angle AMN&lt;/math&gt; and &lt;math&gt;\angle D'NM&lt;/math&gt; are right angles, &lt;cmath&gt;(AD')^2 = (AM+D'N)^2 + MN^2 = (2AM)^2 + MN^2 = 52^2 + 8 = 4\cdot 678.&lt;/cmath&gt;It follows that &lt;math&gt;\min f(Q) = 2AD' = 4\sqrt{678}&lt;/math&gt;. The requested sum is &lt;math&gt;4+678=\boxed{682}&lt;/math&gt;.<br /> <br /> ===Solution 2===<br /> Set &lt;math&gt;a=BC=28&lt;/math&gt;, &lt;math&gt;b=CA=44&lt;/math&gt;, &lt;math&gt;c=AB=52&lt;/math&gt;. Let &lt;math&gt;O&lt;/math&gt; be the point which minimizes &lt;math&gt;f(X)&lt;/math&gt;.<br /> <br /> Claim: &lt;math&gt;O&lt;/math&gt; is the gravity center &lt;math&gt;\tfrac14(\vec A + \vec B + \vec C + \vec D)&lt;/math&gt;.<br /> Proof. Let &lt;math&gt;M&lt;/math&gt; and &lt;math&gt;N&lt;/math&gt; denote the midpoints of &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;CD&lt;/math&gt;. From &lt;math&gt;\triangle ABD \cong \triangle BAC&lt;/math&gt; and &lt;math&gt;\triangle CDA \cong \triangle DCB&lt;/math&gt;, we have &lt;math&gt;MC=MD&lt;/math&gt;, &lt;math&gt;NA=NB&lt;/math&gt; an hence &lt;math&gt;MN&lt;/math&gt; is a perpendicular bisector of both segments &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;CD&lt;/math&gt;. Then if &lt;math&gt;X&lt;/math&gt; is any point inside tetrahedron &lt;math&gt;ABCD&lt;/math&gt;, its orthogonal projection onto line &lt;math&gt;MN&lt;/math&gt; will have smaller &lt;math&gt;f&lt;/math&gt;-value; hence we conclude that &lt;math&gt;O&lt;/math&gt; must lie on &lt;math&gt;MN&lt;/math&gt;. Similarly, &lt;math&gt;O&lt;/math&gt; must lie on the line joining the midpoints of &lt;math&gt;AC&lt;/math&gt; and &lt;math&gt;BD&lt;/math&gt;. &lt;math&gt;\blacksquare&lt;/math&gt;<br /> <br /> Claim: The gravity center &lt;math&gt;O&lt;/math&gt; coincides with the circumcenter.<br /> Proof. Let &lt;math&gt;G_D&lt;/math&gt; be the centroid of triangle &lt;math&gt;ABC&lt;/math&gt;; then &lt;math&gt;DO = \tfrac 34 DG_D&lt;/math&gt; (by vectors). If we define &lt;math&gt;G_A&lt;/math&gt;, &lt;math&gt;G_B&lt;/math&gt;, &lt;math&gt;G_C&lt;/math&gt; similarly, we get &lt;math&gt;AO = \tfrac 34 AG_A&lt;/math&gt; and so on. But from symmetry we have &lt;math&gt;AG_A = BG_B = CG_C = DG_D&lt;/math&gt;, hence &lt;math&gt;AO = BO = CO = DO&lt;/math&gt;. &lt;math&gt;\blacksquare&lt;/math&gt;<br /> <br /> Now we use the fact that an isosceles tetrahedron has circumradius &lt;math&gt;R = \sqrt{\frac18(a^2+b^2+c^2)}&lt;/math&gt;. Here &lt;math&gt;R = \sqrt{678}&lt;/math&gt; so &lt;math&gt;f(O) = 4R = 4\sqrt{678}&lt;/math&gt;.<br /> <br /> =See Also=<br /> {{AIME box|year=2017|n=II|num-b=14|after=Last Question}}<br /> {{MAA Notice}}</div> Thedoge https://artofproblemsolving.com/wiki/index.php?title=2017_AIME_II_Problems/Problem_13&diff=84985 2017 AIME II Problems/Problem 13 2017-03-23T22:49:26Z <p>Thedoge: /* Solution */</p> <hr /> <div>==Problem==<br /> For each integer &lt;math&gt;n\geq3&lt;/math&gt;, let &lt;math&gt;f(n)&lt;/math&gt; be the number of &lt;math&gt;3&lt;/math&gt;-element subsets of the vertices of the regular &lt;math&gt;n&lt;/math&gt;-gon that are the vertices of an isosceles triangle (including equilateral triangles). Find the sum of all values of &lt;math&gt;n&lt;/math&gt; such that &lt;math&gt;f(n+1)=f(n)+78&lt;/math&gt;.<br /> <br /> ==Solution==<br /> Considering &lt;math&gt;n \pmod{6}&lt;/math&gt;, we have the following formulas:<br /> <br /> Even and a multiple of 3: &lt;math&gt;\frac{n(n-4)}{2} + \frac{n}{3}&lt;/math&gt;<br /> <br /> Even and not a multiple of 3: &lt;math&gt;\frac{n(n-2)}{2}&lt;/math&gt;<br /> <br /> Odd and a multiple of 3: &lt;math&gt;\frac{n(n-3)}{2} + \frac{n}{3}&lt;/math&gt;<br /> <br /> Odd and not a multiple of 3: &lt;math&gt;\frac{n(n-1)}{2}&lt;/math&gt;<br /> <br /> To derive these formulas, we note the following:<br /> Any isosceles triangle formed by the vertices of our regular &lt;math&gt;n&lt;/math&gt;-sided polygon &lt;math&gt;P&lt;/math&gt; has its sides from the set of edges and diagonals of &lt;math&gt;P&lt;/math&gt;. Notably, as two sides of an isosceles triangle must be equal, it is important to use the property that same-lengthed edges and diagonals come in groups of &lt;math&gt;n&lt;/math&gt;, unless &lt;math&gt;n&lt;/math&gt; is even when one set of diagonals (those which bisect the polygon) comes in a group of &lt;math&gt;\frac{n}{2}&lt;/math&gt;. Three properties hold true of &lt;math&gt;f(n)&lt;/math&gt;:<br /> <br /> When &lt;math&gt;n&lt;/math&gt; is odd there are &lt;math&gt;\frac{n(n-1)}{2}&lt;/math&gt; satisfactory subsets (This can be chosen with &lt;math&gt;n&lt;/math&gt; choices for the not-base vertex, and &lt;math&gt;\frac{n-1}{2}&lt;/math&gt; for the pair of equal sides as we have &lt;math&gt;n-1&lt;/math&gt; edges to choose from, and we must divide by 2 for over-count).<br /> <br /> When &lt;math&gt;n&lt;/math&gt; is even there are &lt;math&gt;\frac{n(n-2)}{2}&lt;/math&gt; satisfactory subsets (This can be chosen with &lt;math&gt;n&lt;/math&gt; choices for the not-base vertex, and &lt;math&gt;\frac{n-2}{2}&lt;/math&gt; for the pair of equal sides as we have &lt;math&gt;n-1&lt;/math&gt; edges to choose from, one of them which is not satisfactory (the bisecting edge), and we must divide by 2 for over-count).<br /> <br /> When &lt;math&gt;n&lt;/math&gt; is a multiple of three we additionally over-count equilateral triangles, of which there are &lt;math&gt;\frac{n}{3}&lt;/math&gt;. As we count them three times, we are two times over, so we subtract &lt;math&gt;\frac{2n}{3}&lt;/math&gt;.<br /> <br /> Considering the six possibilities &lt;math&gt;n \equiv 0,1,2,3,4,5 \pmod{6}&lt;/math&gt; and solving, we find that the only valid solutions are &lt;math&gt;n = 36, 52, 157&lt;/math&gt;, from which the answer is &lt;math&gt;36 + 52 + 157 = \boxed{245}&lt;/math&gt;.<br /> <br /> =See Also=<br /> {{AIME box|year=2017|n=II|num-b=12|num-a=14}}<br /> {{MAA Notice}}</div> Thedoge https://artofproblemsolving.com/wiki/index.php?title=2017_AIME_II_Problems/Problem_13&diff=84984 2017 AIME II Problems/Problem 13 2017-03-23T22:47:55Z <p>Thedoge: /* Solution */</p> <hr /> <div>==Problem==<br /> For each integer &lt;math&gt;n\geq3&lt;/math&gt;, let &lt;math&gt;f(n)&lt;/math&gt; be the number of &lt;math&gt;3&lt;/math&gt;-element subsets of the vertices of the regular &lt;math&gt;n&lt;/math&gt;-gon that are the vertices of an isosceles triangle (including equilateral triangles). Find the sum of all values of &lt;math&gt;n&lt;/math&gt; such that &lt;math&gt;f(n+1)=f(n)+78&lt;/math&gt;.<br /> <br /> ==Solution==<br /> Considering &lt;math&gt;n \pmod{6}&lt;/math&gt;, we have the following formulas:<br /> <br /> Even and a multiple of 3: &lt;math&gt;\frac{n(n-4)}{2} + \frac{n}{3}&lt;/math&gt;<br /> Even and not a multiple of 3: &lt;math&gt;\frac{n(n-2)}{2}&lt;/math&gt;<br /> Odd and a multiple of 3: &lt;math&gt;\frac{n(n-3)}{2} + \frac{n}{3}&lt;/math&gt;<br /> Odd and not a multiple of 3: &lt;math&gt;\frac{n(n-1)}{2}&lt;/math&gt;<br /> <br /> Considering the six possibilities &lt;math&gt;n \equiv 0,1,2,3,4,5 \pmod{6}&lt;/math&gt; and solving, we find that the only valid solutions are &lt;math&gt;n = 36, 52, 157&lt;/math&gt;, from which the answer is &lt;math&gt;36 + 52 + 157 = \boxed{245}&lt;/math&gt;.<br /> <br /> =See Also=<br /> {{AIME box|year=2017|n=II|num-b=12|num-a=14}}<br /> {{MAA Notice}}</div> Thedoge https://artofproblemsolving.com/wiki/index.php?title=2017_AIME_II_Problems/Problem_14&diff=84983 2017 AIME II Problems/Problem 14 2017-03-23T22:45:47Z <p>Thedoge: /* Solution */</p> <hr /> <div>==Problem==<br /> A &lt;math&gt;10\times10\times10&lt;/math&gt; grid of points consists of all points in space of the form &lt;math&gt;(i,j,k)&lt;/math&gt;, where &lt;math&gt;i&lt;/math&gt;, &lt;math&gt;j&lt;/math&gt;, and &lt;math&gt;k&lt;/math&gt; are integers between &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;10&lt;/math&gt;, inclusive. Find the number of different lines that contain exactly &lt;math&gt;8&lt;/math&gt; of these points.<br /> <br /> ==Solution==<br /> The easiest way to see the case where the lines are not parallel to the faces, is that a line through the point &lt;math&gt;(a,b,c)&lt;/math&gt; must contain &lt;math&gt;(a \pm 1, b \pm 1, c \pm 1)&lt;/math&gt; on it as well, as otherwise the line would not pass through more than 5 points. This corresponds to the 4 diagonals of the cube.<br /> <br /> We look at the one from &lt;math&gt;(0,0,0)&lt;/math&gt; to &lt;math&gt;(10,10,10)&lt;/math&gt;. The lower endpoint of the desired lines must contain both a 0 and a 2 (if min &gt; 0 then the point &lt;math&gt;(a-1,b-1,c-1)&lt;/math&gt; will also be on the line for example, 2 applies to the other end), so it can be &lt;math&gt;(0,0,2), (0,1,2), (0,2,2)&lt;/math&gt;. Accounting for permutations, there are &lt;math&gt;12&lt;/math&gt; ways, so there are &lt;math&gt;12 \cdot 4 = 48&lt;/math&gt; different lines for this case. The answer is, therefore, &lt;math&gt;120 + 48 = \boxed{168}&lt;/math&gt;<br /> <br /> =See Also=<br /> {{AIME box|year=2017|n=II|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Thedoge https://artofproblemsolving.com/wiki/index.php?title=2017_AIME_II_Problems/Problem_14&diff=84982 2017 AIME II Problems/Problem 14 2017-03-23T22:44:56Z <p>Thedoge: /* Solution */</p> <hr /> <div>==Problem==<br /> A &lt;math&gt;10\times10\times10&lt;/math&gt; grid of points consists of all points in space of the form &lt;math&gt;(i,j,k)&lt;/math&gt;, where &lt;math&gt;i&lt;/math&gt;, &lt;math&gt;j&lt;/math&gt;, and &lt;math&gt;k&lt;/math&gt; are integers between &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;10&lt;/math&gt;, inclusive. Find the number of different lines that contain exactly &lt;math&gt;8&lt;/math&gt; of these points.<br /> <br /> ==Solution==<br /> The easiest way to see the case where the lines are not parallel to the faces, is that a line through the point &lt;math&gt;(a,b,c)&lt;/math&gt; must contain &lt;math&gt;(a \pm 1, b \pm 1, c \pm 1)&lt;/math&gt; on it as well, as otherwise the line would not pass through more than 5 points. This corresponds to the 4 diagonals of the cube.<br /> <br /> We look at the one from &lt;math&gt;(0,0,0)&lt;/math&gt; to &lt;math&gt;(10,10,10)&lt;/math&gt;. The lower endpoint of the desired lines must contain both a 0 and a 2 (if min &gt; 0 then the point &lt;math&gt;(a-1,b-1,c-1)&lt;/math&gt; will also be on the line for example, 2 applies to the other end), so it can be &lt;math&gt;(0,0,2), (0,1,2), (0,2,2)&lt;/math&gt;. Accounting for permutations this is 12 ways, so 12*4 = 48 for this case. The answer is, therefore, &lt;math&gt;120 + 48 = \boxed{168}&lt;/math&gt;<br /> <br /> =See Also=<br /> {{AIME box|year=2017|n=II|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Thedoge https://artofproblemsolving.com/wiki/index.php?title=2017_AIME_II_Problems/Problem_12&diff=84981 2017 AIME II Problems/Problem 12 2017-03-23T22:39:50Z <p>Thedoge: /* Problem */</p> <hr /> <div>==Problem==<br /> Circle &lt;math&gt;C_0&lt;/math&gt; has radius &lt;math&gt;1&lt;/math&gt;, and the point &lt;math&gt;A_0&lt;/math&gt; is a point on the circle. Circle &lt;math&gt;C_1&lt;/math&gt; has radius &lt;math&gt;r&lt;1&lt;/math&gt; and is internally tangent to &lt;math&gt;C_0&lt;/math&gt; at point &lt;math&gt;A_0&lt;/math&gt;. Point &lt;math&gt;A_1&lt;/math&gt; lies on circle &lt;math&gt;C_1&lt;/math&gt; so that &lt;math&gt;A_1&lt;/math&gt; is located &lt;math&gt;90^{\circ}&lt;/math&gt; counterclockwise from &lt;math&gt;A_0&lt;/math&gt; on &lt;math&gt;C_1&lt;/math&gt;. Circle &lt;math&gt;C_2&lt;/math&gt; has radius &lt;math&gt;r^2&lt;/math&gt; and is internally tangent to &lt;math&gt;C_1&lt;/math&gt; at point &lt;math&gt;A_1&lt;/math&gt;. In this way a sequence of circles &lt;math&gt;C_1,C_2,C_3,\ldots&lt;/math&gt; and a sequence of points on the circles &lt;math&gt;A_1,A_2,A_3,\ldots&lt;/math&gt; are constructed, where circle &lt;math&gt;C_n&lt;/math&gt; has radius &lt;math&gt;r^n&lt;/math&gt; and is internally tangent to circle &lt;math&gt;C_{n-1}&lt;/math&gt; at point &lt;math&gt;A_{n-1}&lt;/math&gt;, and point &lt;math&gt;A_n&lt;/math&gt; lies on &lt;math&gt;C_n&lt;/math&gt; &lt;math&gt;90^{\circ}&lt;/math&gt; counterclockwise from point &lt;math&gt;A_{n-1}&lt;/math&gt;, as shown in the figure below. There is one point &lt;math&gt;B&lt;/math&gt; inside all of these circles. When &lt;math&gt;r = \frac{11}{60}&lt;/math&gt;, the distance from the center &lt;math&gt;C_0&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; is &lt;math&gt;\frac{m}{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m+n&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> draw(Circle((0,0),125));<br /> draw(Circle((25,0),100));<br /> draw(Circle((25,20),80));<br /> draw(Circle((9,20),64));<br /> dot((125,0));<br /> label(&quot;$A_0$&quot;,(125,0),E);<br /> dot((25,100));<br /> label(&quot;$A_1$&quot;,(25,100),SE);<br /> dot((-55,20));<br /> label(&quot;$A_2$&quot;,(-55,20),E);<br /> &lt;/asy&gt;<br /> <br /> ==Solution==<br /> Impose a coordinate system and let the center of &lt;math&gt;C_0&lt;/math&gt; be &lt;math&gt;(0,0)&lt;/math&gt; and &lt;math&gt;A_0&lt;/math&gt; be &lt;math&gt;(1,0)&lt;/math&gt;. Therefore &lt;math&gt;A_1=(1-r,r)&lt;/math&gt;, &lt;math&gt;A_2=(1-r-r^2,r-r^2)&lt;/math&gt;, &lt;math&gt;A_3=(1-r-r^2+r^3,r-r^2-r^3)&lt;/math&gt;, &lt;math&gt;A_4=(1-r-r^2+r^3+r^4,r-r^2-r^3+r^4)&lt;/math&gt;, and so on, where the signs alternate in groups of &lt;math&gt;2&lt;/math&gt;. The limit of all these points is point &lt;math&gt;B&lt;/math&gt;. Using the geometric series formula on &lt;math&gt;B&lt;/math&gt; and reducing the expression, we get &lt;math&gt;B=(\frac{1-r}{r^2+1},\frac{r-r^2}{r^2+1})&lt;/math&gt;. The distance from &lt;math&gt;B&lt;/math&gt; to the origin is &lt;math&gt;\sqrt{(\frac{1-r}{r^2+1})^2+(\frac{r-r^2}{r^2+1}))^2}=\frac{1-r}{\sqrt{r^2+1}}.&lt;/math&gt; Let &lt;math&gt;r=\frac{11}{60}&lt;/math&gt;, we find that the distance from the origin is &lt;math&gt;\frac{49}{61} \implies 49+61=\boxed{110}&lt;/math&gt;.<br /> <br /> =See Also=<br /> {{AIME box|year=2017|n=II|num-b=11|num-a=13}}<br /> {{MAA Notice}}</div> Thedoge https://artofproblemsolving.com/wiki/index.php?title=2017_AIME_II_Problems/Problem_12&diff=84980 2017 AIME II Problems/Problem 12 2017-03-23T22:39:27Z <p>Thedoge: /* Solution */</p> <hr /> <div>==Problem==<br /> Impose a coordinate system and let the center of &lt;math&gt;C_0&lt;/math&gt; be &lt;math&gt;(0,0)&lt;/math&gt; and &lt;math&gt;A_0&lt;/math&gt; be &lt;math&gt;(1,0)&lt;/math&gt;. Therefore &lt;math&gt;A_1=(1-r,r)&lt;/math&gt;, &lt;math&gt;A_2=(1-r-r^2,r-r^2)&lt;/math&gt;, &lt;math&gt;A_3=(1-r-r^2+r^3,r-r^2-r^3)&lt;/math&gt;, &lt;math&gt;A_4=(1-r-r^2+r^3+r^4,r-r^2-r^3+r^4)&lt;/math&gt;, and so on, where the signs alternate in groups of &lt;math&gt;2&lt;/math&gt;. The limit of all these points is point &lt;math&gt;B&lt;/math&gt;. Using the geometric series formula on &lt;math&gt;B&lt;/math&gt; and reducing the expression, we get &lt;math&gt;B=\left(\frac{1-r}{r^2+1},\frac{r-r^2}{r^2+1}\right)&lt;/math&gt;. The distance from &lt;math&gt;B&lt;/math&gt; to the origin is &lt;math&gt;\sqrt{\left(\frac{1-r}{r^2+1}\right)^2+\left(\frac{r-r^2}{r^2+1}\right)^2}=\frac{1-r}{\sqrt{r^2+1}}.&lt;/math&gt; Let &lt;math&gt;r=\frac{11}{60}&lt;/math&gt;, and the distance from the origin is &lt;math&gt;\frac{49}{61}&lt;/math&gt;. &lt;math&gt;49+61=\boxed{110}&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> draw(Circle((0,0),125));<br /> draw(Circle((25,0),100));<br /> draw(Circle((25,20),80));<br /> draw(Circle((9,20),64));<br /> dot((125,0));<br /> label(&quot;$A_0$&quot;,(125,0),E);<br /> dot((25,100));<br /> label(&quot;$A_1$&quot;,(25,100),SE);<br /> dot((-55,20));<br /> label(&quot;$A_2$&quot;,(-55,20),E);<br /> &lt;/asy&gt;<br /> <br /> ==Solution==<br /> Impose a coordinate system and let the center of &lt;math&gt;C_0&lt;/math&gt; be &lt;math&gt;(0,0)&lt;/math&gt; and &lt;math&gt;A_0&lt;/math&gt; be &lt;math&gt;(1,0)&lt;/math&gt;. Therefore &lt;math&gt;A_1=(1-r,r)&lt;/math&gt;, &lt;math&gt;A_2=(1-r-r^2,r-r^2)&lt;/math&gt;, &lt;math&gt;A_3=(1-r-r^2+r^3,r-r^2-r^3)&lt;/math&gt;, &lt;math&gt;A_4=(1-r-r^2+r^3+r^4,r-r^2-r^3+r^4)&lt;/math&gt;, and so on, where the signs alternate in groups of &lt;math&gt;2&lt;/math&gt;. The limit of all these points is point &lt;math&gt;B&lt;/math&gt;. Using the geometric series formula on &lt;math&gt;B&lt;/math&gt; and reducing the expression, we get &lt;math&gt;B=(\frac{1-r}{r^2+1},\frac{r-r^2}{r^2+1})&lt;/math&gt;. The distance from &lt;math&gt;B&lt;/math&gt; to the origin is &lt;math&gt;\sqrt{(\frac{1-r}{r^2+1})^2+(\frac{r-r^2}{r^2+1}))^2}=\frac{1-r}{\sqrt{r^2+1}}.&lt;/math&gt; Let &lt;math&gt;r=\frac{11}{60}&lt;/math&gt;, we find that the distance from the origin is &lt;math&gt;\frac{49}{61} \implies 49+61=\boxed{110}&lt;/math&gt;.<br /> <br /> =See Also=<br /> {{AIME box|year=2017|n=II|num-b=11|num-a=13}}<br /> {{MAA Notice}}</div> Thedoge https://artofproblemsolving.com/wiki/index.php?title=2017_AIME_II_Problems/Problem_11&diff=84979 2017 AIME II Problems/Problem 11 2017-03-23T22:38:31Z <p>Thedoge: /* Solution */</p> <hr /> <div>==Problem==<br /> Five towns are connected by a system of roads. There is exactly one road connecting each pair of towns. Find the number of ways there are to make all the roads one-way in such a way that it is still possible to get from any town to any other town using the roads (possibly passing through other towns on the way).<br /> <br /> ==Solution==<br /> It is obvious that any configuration of one-way roads which contains a town whose roads all lead into it or lead out of it cannot satisfy the given. We claim that any configuration which does not have a town whose roads all lead into it or lead out of it does satisfy the given conditions. Now we show that a loop of &lt;math&gt;3&lt;/math&gt; or more towns exist. Pick a town, then choose a neighboring town to travel to &lt;math&gt;5&lt;/math&gt; times. Of these &lt;math&gt;6&lt;/math&gt; towns visited, at least two of them must be the same; therefore there must exist a loop of &lt;math&gt;3&lt;/math&gt; or more towns because a loop of &lt;math&gt;2&lt;/math&gt; towns cannot exist. We want to show that the loop can be reached from any town, and any town can be reached from the loop.<br /> <br /> &lt;math&gt;\textbf{Case 1}&lt;/math&gt;. The loop has &lt;math&gt;5&lt;/math&gt; towns.<br /> Clearly every town can be reached by going around the loop.<br /> <br /> &lt;math&gt;\textbf{Case 2}&lt;/math&gt;. The loop has &lt;math&gt;4&lt;/math&gt; towns.<br /> The town not on the loop must have a road leading to it. This road comes from a town on the loop. Therefore this town can be reached from the loop. This town not on the loop must also have a road leading out of it. This road leads to a town on the loop. Therefore the loop can be reached from the town.<br /> <br /> &lt;math&gt;\textbf{Case 3}&lt;/math&gt;. The loop has &lt;math&gt;3&lt;/math&gt; towns.<br /> There are two towns not on the loop; call them Town &lt;math&gt;A&lt;/math&gt; and Town &lt;math&gt;B&lt;/math&gt;. Without loss of generality assume &lt;math&gt;A&lt;/math&gt; leads to &lt;math&gt;B&lt;/math&gt;. Because a road must lead to &lt;math&gt;A&lt;/math&gt;, the town where this road comes from must be on the loop. Therefore &lt;math&gt;A&lt;/math&gt; and therefore &lt;math&gt;B&lt;/math&gt; can be reached from the loop. Because a road must lead out of &lt;math&gt;B&lt;/math&gt;, the town it leads to must be on the loop. Therefore the loop can be reached from &lt;math&gt;B&lt;/math&gt; and also &lt;math&gt;A&lt;/math&gt;.<br /> <br /> The number of good configurations is the total number of configurations minus the number of bad configurations. There are &lt;math&gt;2^{{5\choose2}}&lt;/math&gt; total configurations. To find the number of bad configurations in which a town exists such that all roads lead to it, there are &lt;math&gt;5&lt;/math&gt; ways to choose this town and &lt;math&gt;2^6&lt;/math&gt; ways to assign the six other roads that do not connect to this town. The same logic is used to find the number of bad configurations in which a town exists such that all roads lead out of it. It might be tempting to conclude that there are &lt;math&gt;5 \cdot 2^6+5 \cdot 2^6&lt;/math&gt; bad configurations, but the configurations in which there exists a town such that all roads lead to it and a town such that all roads lead out of it are overcounted. There are &lt;math&gt;5&lt;/math&gt; ways to choose the town for which all roads lead to it, &lt;math&gt;4&lt;/math&gt; ways to choose the town for which all roads lead out of it, and &lt;math&gt;2^3&lt;/math&gt; ways to assign the remaining &lt;math&gt;3&lt;/math&gt; roads not connected to either of these towns. Therefore, the answer is &lt;math&gt;2^{{5\choose2}}-(5 \cdot 2^6+5 \cdot 2^6-5\cdot 4 \cdot 2^3)=\boxed{544}&lt;/math&gt;.<br /> <br /> =See Also=<br /> {{AIME box|year=2017|n=II|num-b=10|num-a=12}}<br /> {{MAA Notice}}</div> Thedoge https://artofproblemsolving.com/wiki/index.php?title=2017_AIME_II_Problems/Problem_10&diff=84977 2017 AIME II Problems/Problem 10 2017-03-23T22:37:08Z <p>Thedoge: /* Solution */</p> <hr /> <div>==Problem==<br /> Rectangle &lt;math&gt;ABCD&lt;/math&gt; has side lengths &lt;math&gt;AB=84&lt;/math&gt; and &lt;math&gt;AD=42&lt;/math&gt;. Point &lt;math&gt;M&lt;/math&gt; is the midpoint of &lt;math&gt;\overline{AD}&lt;/math&gt;, point &lt;math&gt;N&lt;/math&gt; is the trisection point of &lt;math&gt;\overline{AB}&lt;/math&gt; closer to &lt;math&gt;A&lt;/math&gt;, and point &lt;math&gt;O&lt;/math&gt; is the intersection of &lt;math&gt;\overline{CM}&lt;/math&gt; and &lt;math&gt;\overline{DN}&lt;/math&gt;. Point &lt;math&gt;P&lt;/math&gt; lies on the quadrilateral &lt;math&gt;BCON&lt;/math&gt;, and &lt;math&gt;\overline{BP}&lt;/math&gt; bisects the area of &lt;math&gt;BCON&lt;/math&gt;. Find the area of &lt;math&gt;\triangle CDP&lt;/math&gt;.<br /> <br /> ==Solution==<br /> &lt;asy&gt;<br /> pair A,B,C,D,M,n,O,P;<br /> A=(0,42);B=(84,42);C=(84,0);D=(0,0);M=(0,21);n=(28,42);O=(12,18);P=(32,13);<br /> fill(C--D--P--cycle,lightgray);<br /> draw(A--B--C--D--cycle);<br /> draw(C--M);<br /> draw(D--n);<br /> draw(B--P);<br /> draw(D--P);<br /> label(&quot;$A$&quot;,A,NW);<br /> label(&quot;$B$&quot;,B,NE);<br /> label(&quot;$C$&quot;,C,SE);<br /> label(&quot;$D$&quot;,D,SW);<br /> label(&quot;$M$&quot;,M,W);<br /> label(&quot;$N$&quot;,n,N);<br /> label(&quot;$O$&quot;,O,(-0.5,1));<br /> label(&quot;$P$&quot;,P,N);<br /> dot(A);<br /> dot(B);<br /> dot(C);<br /> dot(D);<br /> dot(M);<br /> dot(n);<br /> dot(O);<br /> dot(P);<br /> label(&quot;28&quot;,(0,42)--(28,42),N);<br /> label(&quot;56&quot;,(28,42)--(84,42),N);<br /> &lt;/asy&gt;<br /> Impose a coordinate system on the diagram where point &lt;math&gt;D&lt;/math&gt; is the origin. Therefore &lt;math&gt;A=(0,42)&lt;/math&gt;, &lt;math&gt;B=(84,42)&lt;/math&gt;, &lt;math&gt;C=(84,0)&lt;/math&gt;, and &lt;math&gt;D=(0,0)&lt;/math&gt;. Because &lt;math&gt;M&lt;/math&gt; is a midpoint and &lt;math&gt;M&lt;/math&gt; is a trisection point, &lt;math&gt;M=(0,21)&lt;/math&gt; and &lt;math&gt;N=(28,42)&lt;/math&gt;. The equation for line &lt;math&gt;DN&lt;/math&gt; is &lt;math&gt;y=\frac{3}{2}x&lt;/math&gt; and the equation for line &lt;math&gt;CM&lt;/math&gt; is &lt;math&gt;\frac{1}{84}x+\frac{1}{21}y=1&lt;/math&gt;, so their intersection, point &lt;math&gt;O&lt;/math&gt;, is &lt;math&gt;(12,18)&lt;/math&gt;. Using the shoelace formula on quadrilateral &lt;math&gt;BCON&lt;/math&gt;, or or drawing diagonal &lt;math&gt;\overline{BO}&lt;/math&gt; and using &lt;math&gt;\frac 12 bh&lt;/math&gt;, we find that its area is &lt;math&gt;2184&lt;/math&gt;. Therefore the area of triangle &lt;math&gt;BCP&lt;/math&gt; is &lt;math&gt;\frac{2184}{2}&lt;/math&gt; and the distance from &lt;math&gt;P&lt;/math&gt; to line &lt;math&gt;PC&lt;/math&gt; is &lt;math&gt;52&lt;/math&gt; and its &lt;math&gt;x&lt;/math&gt;-coordinate is &lt;math&gt;32&lt;/math&gt;. Because &lt;math&gt;P&lt;/math&gt; lies on the equation &lt;math&gt;\frac{1}{84}x+\frac{1}{21}y=1&lt;/math&gt;, &lt;math&gt;P&lt;/math&gt;'s &lt;math&gt;y&lt;/math&gt;-coordinate is &lt;math&gt;13&lt;/math&gt;, which is also the height of &lt;math&gt;CDP&lt;/math&gt;. Therefore the area of &lt;math&gt;CDP&lt;/math&gt; is &lt;math&gt;\frac{1}{2} \cdot 13 \cdot 84=\boxed{546}&lt;/math&gt;.<br /> <br /> =See Also=<br /> {{AIME box|year=2017|n=II|num-b=9|num-a=11}}<br /> {{MAA Notice}}</div> Thedoge https://artofproblemsolving.com/wiki/index.php?title=2017_AIME_II_Problems/Problem_10&diff=84976 2017 AIME II Problems/Problem 10 2017-03-23T22:36:39Z <p>Thedoge: /* Solution */</p> <hr /> <div>==Problem==<br /> Rectangle &lt;math&gt;ABCD&lt;/math&gt; has side lengths &lt;math&gt;AB=84&lt;/math&gt; and &lt;math&gt;AD=42&lt;/math&gt;. Point &lt;math&gt;M&lt;/math&gt; is the midpoint of &lt;math&gt;\overline{AD}&lt;/math&gt;, point &lt;math&gt;N&lt;/math&gt; is the trisection point of &lt;math&gt;\overline{AB}&lt;/math&gt; closer to &lt;math&gt;A&lt;/math&gt;, and point &lt;math&gt;O&lt;/math&gt; is the intersection of &lt;math&gt;\overline{CM}&lt;/math&gt; and &lt;math&gt;\overline{DN}&lt;/math&gt;. Point &lt;math&gt;P&lt;/math&gt; lies on the quadrilateral &lt;math&gt;BCON&lt;/math&gt;, and &lt;math&gt;\overline{BP}&lt;/math&gt; bisects the area of &lt;math&gt;BCON&lt;/math&gt;. Find the area of &lt;math&gt;\triangle CDP&lt;/math&gt;.<br /> <br /> ==Solution==<br /> &lt;asy&gt;<br /> pair A,B,C,D,M,n,O,P;<br /> A=(0,42);B=(84,42);C=(84,0);D=(0,0);M=(0,21);n=(28,42);O=(12,18);P=(32,13);<br /> fill(C--D--P--cycle,lightgray);<br /> draw(A--B--C--D--cycle);<br /> draw(C--M);<br /> draw(D--n);<br /> draw(B--P);<br /> draw(D--P);<br /> label(&quot;$A$&quot;,A,NW);<br /> label(&quot;$B$&quot;,B,NE);<br /> label(&quot;$C$&quot;,C,SE);<br /> label(&quot;$D$&quot;,D,SW);<br /> label(&quot;$M$&quot;,M,W);<br /> label(&quot;$N$&quot;,n,N);<br /> label(&quot;$O$&quot;,O,(-0.5,1));<br /> label(&quot;$P$&quot;,P,N);<br /> dot(A);<br /> dot(B);<br /> dot(C);<br /> dot(D);<br /> dot(M);<br /> dot(n);<br /> dot(O);<br /> dot(P);<br /> label(&quot;28&quot;,(0,42)--(28,42),N);<br /> label(&quot;56&quot;,(28,42)--(84,42),N);<br /> &lt;/asy&gt;<br /> Impose a coordinate system on the diagram where point &lt;math&gt;D&lt;/math&gt; is the origin. Therefore &lt;math&gt;A=(0,42)&lt;/math&gt;, &lt;math&gt;B=(84,42)&lt;/math&gt;, &lt;math&gt;C=(84,0)&lt;/math&gt;, and &lt;math&gt;D=(0,0)&lt;/math&gt;. Because &lt;math&gt;M&lt;/math&gt; is a midpoint and &lt;math&gt;M&lt;/math&gt; is a trisection point, &lt;math&gt;M=(0,21)&lt;/math&gt; and &lt;math&gt;N=(28,42)&lt;/math&gt;. The equation for line &lt;math&gt;DN&lt;/math&gt; is &lt;math&gt;y=\frac{3}{2}x&lt;/math&gt; and the equation for line &lt;math&gt;CM&lt;/math&gt; is &lt;math&gt;\frac{1}{84}x+\frac{1}{21}y=1&lt;/math&gt;, so their intersection, point &lt;math&gt;O&lt;/math&gt;, is &lt;math&gt;(12,18)&lt;/math&gt;. Using the shoelace formula on quadrilateral &lt;math&gt;BCON&lt;/math&gt;, or or drawing diagonal &lt;math&gt;\overline{CO}&lt;/math&gt; and using &lt;math&gt;\frac 12 bh&lt;/math&gt;, we find that its area is &lt;math&gt;2184&lt;/math&gt;. Therefore the area of triangle &lt;math&gt;BCP&lt;/math&gt; is &lt;math&gt;\frac{2184}{2}&lt;/math&gt; and the distance from &lt;math&gt;P&lt;/math&gt; to line &lt;math&gt;PC&lt;/math&gt; is &lt;math&gt;52&lt;/math&gt; and its &lt;math&gt;x&lt;/math&gt;-coordinate is &lt;math&gt;32&lt;/math&gt;. Because &lt;math&gt;P&lt;/math&gt; lies on the equation &lt;math&gt;\frac{1}{84}x+\frac{1}{21}y=1&lt;/math&gt;, &lt;math&gt;P&lt;/math&gt;'s &lt;math&gt;y&lt;/math&gt;-coordinate is &lt;math&gt;13&lt;/math&gt;, which is also the height of &lt;math&gt;CDP&lt;/math&gt;. Therefore the area of &lt;math&gt;CDP&lt;/math&gt; is &lt;math&gt;\frac{1}{2} \cdot 13 \cdot 84=\boxed{546}&lt;/math&gt;.<br /> <br /> =See Also=<br /> {{AIME box|year=2017|n=II|num-b=9|num-a=11}}<br /> {{MAA Notice}}</div> Thedoge