https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Ghostyc&feedformat=atomAoPS Wiki - User contributions [en]2024-03-29T15:33:41ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=Resources_for_mathematics_competitions&diff=158246Resources for mathematics competitions2021-07-12T17:24:56Z<p>Ghostyc: /* Problem Sets */</p>
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<div>The [[Art of Problem Solving]] hosts this [[AoPSWiki]] as well as many other online resources for students interested in [[mathematics competitions]]. Look around the AoPSWiki. Individual articles often have sample problems and solutions for many levels of problem solvers. Many also have links to books, websites, and other resources relevant to the topic.<br />
<br />
* [[Math books]]<br />
* [[Mathematics forums]]<br />
* [[Mathematics websites]]<br />
<br />
== Training Websites ==<br />
* AMC Problem Trainer: https://amctrainer.com/play<br />
<br />
==Free eBook of Mathematical Formulas and Strategies==<br />
130+ page free eBook of Mathematical Formulas and Strategies: https://www.omegalearn.org/thebookofformulas<br />
<br />
== Math Courses ==<br />
Introduction To Number Theory: https://thepuzzlr.com/math-courses<br />
<br />
Free AMC 8 Bootcamp covering all the essential concepts: https://thepuzzlr.com/courses/amc-8-bootcamp/<br />
<br />
==List of Resources==<br />
<br />
Elementary: https://www.omegalearn.org/elementary-competition-math <br />
<br />
Middle: https://www.omegalearn.org/middle-competition-math <br />
<br />
High: https://www.omegalearn.org/high-competition-math <br />
<br />
Everything you need to know for the AMC 8: https://thepuzzlr.com/courses/amc-8-bootcamp/<br />
<br />
<br />
AMC 8 Fundamentals Class: https://www.omegalearn.org/amc8-fundamentals<br />
<br />
AMC 8 Video Solutions Playlist: https://www.youtube.com/watch?v=TRGPF3BxujE&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL<br />
<br />
AMC/MATHCOUNTS Class: https://www.omegalearn.org/amc8-advanced<br />
<br />
== Math Competition Classes ==<br />
* [[Art of Problem Solving]] hosts classes that are popular among many of the highest performing students in the United States. [http://www.artofproblemsolving.com/Classes/AoPS_C_PSeries.php AoPS Problem Series].<br />
<br />
== Math Competition Problems ==<br />
=== Problem Books ===<br />
Many mathematics competitions sell books of past competitions and solutions. These books can be great supplementary material for avid students of mathematics.<br />
* [[ARML]] has four problem books covering most ARML as well as some [[NYSML]] competitions. However, they are generally difficult to find. Some can be ordered [http://www.arml2.com/arml_2018/page/index.php?page_type=public&page=books here].<br />
* [[MOEMS]] books are available [https://www.artofproblemsolving.com/store/list/other-products here] at [[AoPS]].<br />
* [[MATHCOUNTS]] books are available [https://www.artofproblemsolving.com/store/list/other-products here] at [[AoPS]].<br />
* [[AMC]] books are available [https://www.artofproblemsolving.com/store/list/other-products here] at [[AoPS]].<br />
* [[Mandelbrot Competition]] books are available [https://www.artofproblemsolving.com/store/list/other-products here] at [[AoPS]].<br />
<br />
=== Problems Online ===<br />
[[Art of Problem Solving]] maintains a very large database of [http://www.artofproblemsolving.com/Forum/resources.php math contest problems]. Many math contest websites include archives of past problems. The [[List of mathematics competitions]] leads to links for many of these competition homepages. Here are a few examples:<br />
==== Introductory Problem Solvers ====<br />
* [[Mu Alpha Theta]].org hosts past [http://www.mualphatheta.org/index.php?chapters/national-convention/past-tests contest problems].<br />
* Noetic Learning [http://www.noetic-learning.com/gifted/index.jsp Challenge Math] - Problem Solving for the Gifted Elementary Students .<br />
* Elias Saab's [[MathCounts]] [http://mathcounts.saab.org/ Drills page].<br />
* [[Alabama Statewide High School Mathematics Contest]] [http://mcis.jsu.edu/mathcontest/ homepage].<br />
* The [[South African Mathematics Olympiad]] [http://www.samf.ac.za/QuestionPapers.aspx here] includes many years of past problems with solutions.<br />
* [http://www.beestar.org/index.jsp?adid=106 Beestar.org] - Weekly problem solving challenges and honor roll ranking, Grade 1 - 8<br />
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==== Intermediate Problem Solvers ====<br />
* [[AoPS]] [http://www.artofproblemsolving.com/Forum/resources.php math contest problems and solutions]<br />
* Past [[United States of America Mathematical Talent Search | USAMTS]] problems can be found at the [http://usamts.org USAMTS homepage].<br />
* [http://www.albany.edu/ialexandrova/HS.html Ivana Alexandrova's Weekly Math Problems for High School Students] contains good problems that will make you think and that will teach you new skills and material<br />
* The [http://www.kalva.demon.co.uk/ Kalva site] is one of the best resources for math problems on the planet. (Currently offline. Mirror can be found at [https://mks.mff.cuni.cz/kalva/ this page])<br />
* Past [[Colorado Mathematical Olympiad]] (CMO) problems can be found at the [http://www.uccs.edu/%7Easoifer/olympiad.html CMO homepage].<br />
* Past [[International Mathematical Talent Search]] (IMTS) problems can be found [http://www.cms.math.ca/Competitions/IMTS/ here]<br />
* [https://brilliant.org/ Brilliant] is a website where one can solve problems to gain points and go to higher levels.<br />
*[https://www.clevermath.org/ Clevermath] Is similar to above<br />
<br />
==== Olympiad Problem Solvers ====<br />
* [[AoPS]] [http://www.artofproblemsolving.com/Forum/resources.php math contest problems and solutions]<br />
* [https://mathcsr.org Math and CS Research] is a math and computer science publication with articles and problem sets on a wide range of topics.<br />
* Past [[United States of America Mathematical Talent Search | USAMTS]] problems can be found at the [http://usamts.org USAMTS homepage].<br />
* The [http://www.kalva.demon.co.uk/ Kalva site] is one of the best resources for math problems on the planet. (Currently offline - but a few mirrors are available, e.g [https://mks.mff.cuni.cz/kalva/ here].)<br />
* [http://www.qbyte.org/puzzles/ Nick's Mathematical Puzzles] -- Challenging problems with hints and solutions.<br />
* [[Canadian Mathematical Olympiad]] are hosted [http://www.cms.math.ca/Competitions/CMO/ here by the Canadian Mathematical Society].<br />
* [http://web.archive.org/web/20120825124642/http://pertselv.tripod.com/RusMath.html Problems of the All-Soviet-Union math competitions 1961-1986] - Many problems, no solutions. [Site no longer exists. Site has been replaced by a web capture]<br />
* Past [[International Mathematical Talent Search]] (IMTS) problems can be found [http://www.cms.math.ca/Competitions/IMTS/ here]<br />
* [http://web.archive.org/web/20091027032345/http://geocities.com/CapeCanaveral/Lab/4661/ Olympiad Math Madness] - Stacks of challenging problems, no solutions. [Site no longer exists. Site has been replaced by a web capture]<br />
<br />
== Articles ==<br />
* [https://thepuzzlr.com/math-kangaroo-time-management Time Management]<br />
* [https://artofproblemsolving.com/news/articles/pros-cons-math-competitions Pros and Cons of Math Competitions] by [[Richard Rusczyk]].<br />
* [https://artofproblemsolving.com/news/articles/establishing-a-positive-culture Establishing a Positive Culture of Expectation in Math Education] by [[Sister Scholastica Award]] winner Darryl Hill.<br />
* [https://artofproblemsolving.com/news/articles/stop-making-stupid-mistakes Stop Making Stupid Mistakes] by [[Richard Rusczyk]].<br />
* [https://artofproblemsolving.com/news/articles/what-are-stupid-questions What Questions Really Are the Stupid Questions?] by [[Richard Rusczyk]].<br />
* [https://artofproblemsolving.com/news/articles/learning-through-teaching Learning Through Teaching]<br />
* [https://artofproblemsolving.com/news/articles/how-to-write-a-solution How to Write a Math Solution] by [[Richard Rusczyk]] and [[user:MCrawford | Mathew Crawford]].<br />
* [https://artofproblemsolving.com/articles/files/KedlayaInequalities.pdf Inequalities] by Dr. Kiran Kedlaya<br />
* [https://artofproblemsolving.com/articles/files/MildorfInequalities.pdf Olympiad Inequalities] by Thomas J. Mildorf<br />
* [https://artofproblemsolving.com/articles/files/MildorfNT.pdf Olympiad Number Theory: An Abstract Perspective] by Thomas J. Mildorf<br />
* [https://artofproblemsolving.com/articles/files/SatoNT.pdf Number Theory] by Naoki Sato<br />
* [https://s3.amazonaws.com/aops-cdn.artofproblemsolving.com/resources/articles/olympiad-number-theory.pdf Olympiad Number Theory Through Challenging Problems] by Justin Stevens<br />
* [https://s3.amazonaws.com/aops-cdn.artofproblemsolving.com/resources/articles/bary.pdf Barycentric Coordinates in Olympiad Geometry] by Max Schindler and Evan Chen<br />
* [https://s3.amazonaws.com/aops-cdn.artofproblemsolving.com/resources/articles/lifting-the-exponent.pdf Lifting the Exponent (LTE)] by Amir Hossein Parvardi<br />
* [https://s3.amazonaws.com/aops-cdn.artofproblemsolving.com/resources/articles/uvw.pdf The uvw Method] by Mathias Bæk Tejs Knudsen<br />
* [https://s3.amazonaws.com/aops-cdn.artofproblemsolving.com/resources/articles/crt.pdf The Chinese Remainder Theorem] by Evan Chen<br />
* [https://s3.amazonaws.com/aops-cdn.artofproblemsolving.com/resources/articles/Recollection.pdf Contest Reflections] by Wanlin Li<br />
<br />
== A Huge List of Links ==<br />
=== AoPS Course Recommendations ===<br />
*[http://www.artofproblemsolving.com/School/recommendations.php Art of Problem Solving Course Recommendations]<br />
*Do you still have trouble deciding which course? Go to the above link and click '''contact us''' at the bottom of the Course Map section to ask for personal recommendations!<br />
<br />
===AMC 8 Preparation===<br />
<br />
Free AMC 8 Classes: https://thepuzzlr.com/courses/amc-8-bootcamp/<br />
====Problems====<br />
Free AMC 8 Classes: omegalearn.org/amc8-fundamentals<br />
omegalearn.org/amc8-advanced<br />
<br />
These classes cover all of the important concepts needed to do well on the AMC 8. <br />
<br />
AMC 8 Video Solutions: https://www.youtube.com/watch?v=TRGPF3BxujE&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL<br />
<br />
[http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=42 AMC 8 Problems in the Resources Section]<br />
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Problem and Solutions: [http://www.artofproblemsolving.com/Wiki/index.php/AMC_8_Problems_and_Solutions AMC 8 Problems in the AoPS wiki]<br />
<br />
===AMC 10/12 Preparation===<br />
AMC 10/12 130+ page Book of Mathematical Formulas and Strategies: https://www.omegalearn.org/thebookofformulas<br />
<br />
Free AMC 10/12 Classes: omegalearn.org/amc10-12<br />
<br />
[http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&t=388108&hilit=preparation How preparing for the AIME will help AMC 10/12 Score] <br />
<br />
[http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&t=396741&hilit=preparation What class to take?]<br />
<br />
[http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&t=387918&hilit=preparation AMC 10 for AMC 12 practice]<br />
<br />
[http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&t=385418&hilit=preparation AMC prep]<br />
<br />
[http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&t=384828&hilit=preparation AMC 10/12 Preparation]<br />
<br />
[http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&t=384747&hilit=preparation AIME/AMC 10 Overlap and Preparation]<br />
<br />
[http://www.artofproblemsolving.com/Forum/viewtopic.php?f=149&t=378851&hilit=preparation How to prepare for amc10 and aime?]<br />
<br />
[http://www.artofproblemsolving.com/Forum/viewtopic.php?f=149&t=369849&hilit=preparation Preparation for AMC 10?]<br />
====Problems====<br />
<br />
[http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=43 AMC 10 Problems in the Resources Section]<br />
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[http://www.artofproblemsolving.com/wiki/index.php/AMC_10_Problems_and_Solutions AMC 10 Problems in the AoPS Wiki]<br />
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[http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=44 AMC 12 Problems in the Resources Section]<br />
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[http://www.artofproblemsolving.com/wiki/index.php/AHSME_Problems_and_Solutions AHSME (Old AMC 12) Problems in the AoPS Wiki]<br />
<br />
[http://www.artofproblemsolving.com/wiki/index.php/AMC_12_Problems_and_Solutions AMC 12 Problems in the AoPS Wiki]<br />
<br />
===AIME Preparation===<br />
[http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&t=397954&hilit=preparation Studying to qualify for USAMO]<br />
<br />
[http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&t=400442&hilit=preparation How to prepare for the AIME]<br />
<br />
[http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&t=399160&hilit=preparation Preparation for the AIME]<br />
<br />
[http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&t=357602&hilit=preparation Using non-AIME questions to prepare for AIME]<br />
<br />
[http://www.artofproblemsolving.com/Forum/viewtopic.php?f=149&t=355918&hilit=preparation Best books to prepare for AIME?]<br />
<br />
[http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&t=344816&hilit=preparation How to improve AIME score to make JMO?]<br />
<br />
[http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&t=341827&hilit=preparation Preparation for AIME and USAMO]<br />
<br />
====Problems====<br />
<br />
[http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45 AIME Problems in the Resources Section]<br />
<br />
[http://www.artofproblemsolving.com/Wiki/index.php/AIME_Problems_and_Solutions AIME Problems in the AoPS Wiki]<br />
<br />
'''[https://drive.google.com/file/d/0B8JbOaFM5Xo_bnc2NUd0dDFLY1U/view?pref=2&pli=AIME AIME problems sorted by difficulty]'''<br />
<br />
===Beginning Olympiad Preparation===<br />
* [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&t=480253 General]<br />
* [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&t=481746&p=2698978 General]<br />
* [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&t=401061&hilit=preparation How to Prepare for USAJMO?]<br />
* [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&t=399023&hilit=preparation USAMO preparation/doing problems]<br />
* [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&t=396736&hilit=preparation Easier Olympiads for USAJMO practice?]<br />
* [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&t=366383&hilit=preparation For the USAMO: ACoPS or Engel?]<br />
* [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&t=360619&hilit=preparation Olympiad problems- how to prepare]<br />
* [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&t=354103&hilit=preparation USAMO/Olympiads Preparation: Where to start?]<br />
* [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&t=344929&hilit=preparation USAJMO prep]<br />
<br />
====Bunch of General links====<br />
* [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&t=31888&hilit=USAMO+prep USAMO Prep]<br />
* [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&t=71008&hilit=USAMO+prep USAMO Prep]<br />
* [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&t=79077&hilit=USAMO+prep USAMO]<br />
* [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&t=81296&hilit=USAMO+prep Usamo prep ]<br />
* [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&t=143168&hilit=USAMO+prep USAMO Prep]<br />
* [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&t=273572&hilit=USAMO+prep Counting down to USAMO]<br />
* [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&t=294132&hilit=USAMO+prep USAMO Preparation]<br />
* [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&t=344929&hilit=olympiad+prep USAJMO prep]<br />
* [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&t=385092&hilit=USAMO+prep USAMO Preparation]<br />
* [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&t=397424&hilit=olympiad+prep USAJMO Prep]<br />
* [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&t=401201&hilit=USAMO+prep How to prepare for the USAMO/Making Red MOP]<br />
* [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&t=401640&hilit=USAMO+prep Preparing in a hardcore way]<br />
* [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&t=406402&hilit=USAMO+prep USAMO and JMO prep]<br />
* [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&t=411476&hilit=USAMO+prep USAMO PREP]<br />
* [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&t=419800&hilit=USAMO+prep Beginner to USAMO]<br />
* [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&t=447454 What should I be doing?]<br />
* [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&t=453638&hilit=USAMO+prep Improve to USAMO and IMO level]<br />
* [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&t=474960&hilit=USAMO+prep Prep for USA(J)MO]<br />
* '''[http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&t=385654 contest math stuff/some advice on how to get good]'''<br />
* [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&t=420845 Olympiad Prep]<br />
* '''[http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2379622#p2379622 Olympiad Prep]'''<br />
* '''[http://www.artofproblemsolving.com/community/c5h619416p3697136 USAJMO Prep]'''<br />
* '''[http://www.artofproblemsolving.com/community/c5h520900 The Right Training]'''<br />
* '''[https://usamo.wordpress.com/2014/07/27/what-leads-to-success-at-math-contests/ What Leads to Success]'''<br />
<br />
====Problems====<br />
* [http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=176 USAJMO Problems in the Resources Section]<br />
* [http://www.artofproblemsolving.com/Wiki/index.php/USAJMO_Problems_and_Solutions USAJMO Problems in the AoPS Wiki]<br />
* [http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=27 USAMO Problems in the Resources Section]<br />
* [http://www.artofproblemsolving.com/Wiki/index.php/USAMO_Problems_and_Solutions USAMO Problems in the AoPS Wiki]<br />
<br />
===Middle/Advanced Olympiad Preparation===<br />
<br />
====Problems====<br />
* [http://www.artofproblemsolving.com/Forum/download/file.php?id=38803 Practice Olympiad 1]<br />
* [http://www.artofproblemsolving.com/Forum/download/file.php?id=38804 Practice Olympiad 2]<br />
* [http://www.artofproblemsolving.com/Forum/download/file.php?id=38805 Practice Olympiad 3]<br />
* [http://www.artofproblemsolving.com/Forum/download/file.php?id=38806 Practice Olympiad Solutions]<br />
* [http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=27 USAMO Problems in the Resources Section] <br />
* [http://www.artofproblemsolving.com/Wiki/index.php/USAMO_Problems_and_Solutions USAMO Problems in the AoPS Wiki]<br />
* [http://www.artofproblemsolving.com/Forum/resources.php?c=1&cid=16 IMO Problems in the Resources Section]<br />
* [http://www.artofproblemsolving.com/Wiki/index.php/IMO_Problems_and_Solutions IMO Problems in the AoPS Wiki]<br />
<br />
===Book Links:===<br />
====Olympiad Level====<br />
=====Free=====<br />
* [https://drive.google.com/file/d/0B3gLVLnxtyRvejlpenRmZGh6SDQ/view?usp=sharing Lemmas in Olympiad Geometry article]<br />
* [http://students.imsa.edu/~tliu/Math/planegeo.pdf Plane Geometry]<br />
* [https://web.evanchen.cc/textbooks/OTIS-Excerpts.pdf Evan Chen OTIS-Excerpts]<br />
* [https://www.dropbox.com/s/0fvyelr8rdh837b/olympiad-number-theory.pdf?dl=0 The Basics of Olympiad Number Theory]<br />
* [http://s3.amazonaws.com/aops-cdn.artofproblemsolving.com/resources/articles/olympiad-number-theory.pdf Olympiad Number Theory Through Challenging Problems]<br />
* [https://drive.google.com/file/d/1BcJTLjQaelZ4w_70oHKyImC2I8zLfyrt/view?usp=sharing Modern Olympiad Number Theory]<br />
<br />
=====Not Free=====<br />
* [http://www.amazon.com/Plane-Euclidean-Geometry-Theory-Problems/dp/0953682366/ref=sr_1_1?s=books&ie=UTF8&qid=1338742080&sr=1-1 Plane Euclidean Geometry: Theory and Problems]<br />
*[https://www.amazon.com/dp/0883858398/ref=cm_sw_r_cp_awdb_t1_WTaBCbBFKWMPK Euclidean Geometry in Mathematical Olympiads]<br />
*[http://www.amazon.com/Complex-Geometry-Mathematical-Association-Textbooks/dp/0883855100/ref=sr_1_1?s=books&ie=UTF8&qid=1338742131&sr=1-1 Complex Numbers and Geometry]<br />
* [http://www.amazon.com/Geometry-Complex-Numbers-Dover-Mathematics/dp/0486638308/ref=sr_1_1?s=books&ie=UTF8&qid=1338742156&sr=1-1 Geometry of Complex Numbers]<br />
* [http://www.amazon.com/Complex-Numbers-Z-Titu-Andreescu/dp/0817643265/ref=sr_1_1?s=books&ie=UTF8&qid=1338741912&sr=1-1 Complex Numbers from A to …Z]<br />
* [http://www.amazon.com/103-Trigonometry-Problems-Training-Team/dp/0817643346/ref=sr_1_1?s=books&ie=UTF8&qid=1338742048&sr=1-1 103 Trigonometry Problems: From the Training of the USA IMO Team]<br />
* [http://www.amazon.com/An-Introduction-Diophantine-Equations-Problem-Based/dp/0817645489/ref=sr_1_1?ie=UTF8&qid=1338741533&sr=8-1 An Introduction to Diophantine Equations: A Problem-Based Approach]<br />
* [http://www.amazon.com/Introductions-Number-Theory-Inequalities-Bradley/dp/0953682382/ref=sr_1_1?s=books&ie=UTF8&qid=1338741653&sr=1-1 Introductions to Number Theory and Inequalities]<br />
* [http://www.amazon.com/104-Number-Theory-Problems-Training/dp/0817645276/ref=sr_1_1?s=books&ie=UTF8&qid=1338741697&sr=1-1 104 Number Theory Problems: From the Training of the USA IMO Team]<br />
* [http://www.amazon.com/102-Combinatorial-Problems-Titu-Andreescu/dp/0817643176/ref=sr_1_1?s=books&ie=UTF8&qid=1338741741&sr=1-1 102 Combinatorial Problems]<br />
* [http://www.amazon.com/Path-Combinatorics-Undergraduates-Counting-Strategies/dp/8181283368/ref=sr_1_2?s=books&ie=UTF8&qid=1338741874&sr=1-2 A Path to Combinatorics for Undergraduates: Counting Strategies]<br />
* [http://www.amazon.com/Mathematical-Olympiads-1972-1986-Problems-Solutions/dp/0883856344/ref=sr_1_fkmr1_1?s=books&ie=UTF8&qid=1338742228&sr=1-1 -fkmr1 USA Mathematical Olympiads 1972-1986 Problems and Solutions]<br />
* [http://www.amazon.com/s/ref=nb_sb_noss_1?url=search-alias%3Daps&field-keywords=art+and+craft+of+problem+solving Art and Craft of Problem Solving]<br />
* [http://www.amazon.com/Problem-Solving-Strategies-Problem-Books-Mathematics/dp/0387982191/ref=sr_1_1?ie=UTF8&qid=1338865322&sr=8-1 Problem Solving Strategies]<br />
<br />
===Problem Sets===<br />
* [http://www.casperyc.club/amc Collection of Past Papers in PDF Format]<br />
* [https://drive.google.com/file/d/0B3gLVLnxtyRvS05vQ0N6aEVqSGs/view?usp=sharing Practice problems from around the world]<br />
* [https://drive.google.com/file/d/0B3gLVLnxtyRvQkkwS0xsVVZ3Z1E/view?usp=sharing General problems in Olympiad Mathematics]<br />
* [https://mathcsr.org/article.html?type=problemsolving&a=cHJvYiBwcm9ibGVt&v=1&n=1&t=Q29uZGl0aW9uYWwgUHJvYmFiaWxpdHkgUHJvYmxlbSBTZXQ= Conditional Probability Problem Set]<br />
* [http://cdn.artofproblemsolving.com/aops20/attachments/probability_problems_306.pdf 31 Olympiad problems about Probabilistic Method]<br />
* [http://cdn.artofproblemsolving.com/aops20/attachments/67223_e31f7415b8debe259534f245ab6f402c 567 Nice and Hard Inequalities]<br />
* [http://cdn.artofproblemsolving.com/aops20/attachments/78444_44261211c725bfc07fe3a40698ce18b0 Inequalities]<br />
* [http://cdn.artofproblemsolving.com/aops20/attachments/67223_e3aaee2729d3d97d945b66e06971e5bc 100 Polynomial Problems]<br />
* [http://cdn.artofproblemsolving.com/aops20/attachments/67223_afdd8bf4469c8aa0dd12406f9d246da1 Trigonometry Problems]<br />
* [http://cdn.artofproblemsolving.com/aops20/attachments/82357_cfd6bc6cd06884730701121c0146d455 General all levels]<br />
* [http://cdn.artofproblemsolving.com/aops20/attachments/85314_e1e907a6c92d64bea241ed35b6414d3a Number Theory]<br />
* [http://cdn.artofproblemsolving.com/aops20/attachments/78444_c808a31bb53413a4d4c90426d66df645 Olympiad Problems]<br />
* [http://cdn.artofproblemsolving.com/aops20/attachments/110524_b361f5fe3bb9b43c6d8a1b516e200bc6 33 Functional Equations]<br />
* [http://cdn.artofproblemsolving.com/aops20/attachments/114639_fbc5d584d5949657f5b7db35d2548f4b Induction Problems]<br />
* [http://cdn.artofproblemsolving.com/aops20/attachments/114639_c6b98fb00f3baca6fefb3978db3dda1d Induction Solutions]<br />
* [http://cdn.artofproblemsolving.com/aops20/attachments/369_675af9219c63516a024366c1d8d28251 260 Geometry Problems]<br />
* [http://cdn.artofproblemsolving.com/aops20/attachments/67223_4872d512bdc737f3379ad265e17340d1 150 Geometry Problems]<br />
* [http://cdn.artofproblemsolving.com/aops20/attachments/67223_65dac7d2c6936fe666a991407da6cbeb 50 Diophantine Equation Problems]<br />
* [http://cdn.artofproblemsolving.com/aops20/attachments/67223_d8b71468d1bc454fb4ca732ed7950e19 60 Geometry Problems]<br />
* [http://cdn.artofproblemsolving.com/aops20/attachments/62273_8f432f52f6a8b50ffab885999aacb652 116 Problems]<br />
* [http://cdn.artofproblemsolving.com/aops20/attachments/47577_3609b09820bce0490302aca817405696 Algebraic Inequalities]<br />
* [http://cdn.artofproblemsolving.com/aops20/attachments/67223_f6f1d08183b32b889ee7d4315ef6f620 100 Combinatorics Problems]<br />
* [http://cdn.artofproblemsolving.com/aops20/attachments/67223_55fffd4049ab2308ae770a143f03d7c1 100 Problems]<br />
* [http://cdn.artofproblemsolving.com/aops20/attachments/16383_6600f91864a38bc31764efe5d8b69475 Number Theory]<br />
* [http://cdn.artofproblemsolving.com/aops20/attachments/97235_946d6121c243f09232ea6c390236abfd Geometry]<br />
* [http://cdn.artofproblemsolving.com/aops20/attachments/97235_fd4595a2f0efac349561fae52fa49c2b General]<br />
* [http://cdn.artofproblemsolving.com/aops20/attachments/67223_f0a58877b5326ff79aca7f091559d4e4 100 Number Theory Problems]<br />
* [http://cdn.artofproblemsolving.com/aops20/attachments/67223_cc845a4f181a71ab84cf7af38b79018c 100 Functional Equation Problems]<br />
* [http://cdn.artofproblemsolving.com/aops20/attachments/110524_4dfa3fbf02e9fd55fcfa3ece6a1a1835 Beginning/Intermediate Counting and Probability]<br />
* [http://cdn.artofproblemsolving.com/aops20/attachments/82334_ea1bd0c9afa7d8b36d4b7c9e53048d37 40 Functional Equations]<br />
* [http://cdn.artofproblemsolving.com/aops20/attachments/91148_d2614226d17974c7cae22fd6819e6a35 100 Geometric Inequalities]<br />
* [http://cdn.artofproblemsolving.com/aops20/attachments/139996_1ce94c21ab568094858ede6bdfbec235 10 Fun Unconventional Problems :)]<br />
* [http://cdn.artofproblemsolving.com/aops20/attachments/118_a49397fc1e83d0931c0e72c9139d5a29 169 Functional Equations]<br />
* [http://cdn.artofproblemsolving.com/aops20/attachments/113521_6df2379fee465fac760b788106151ed8 Triangle Geometry]<br />
* [http://cdn.artofproblemsolving.com/aops20/attachments/113521_a090e8492fc3b5e6c9f4ef72e8c25f53 Probability]<br />
* [http://cdn.artofproblemsolving.com/aops20/attachments/113521_3b10e491c0ba9d6421f55aab84c24632 Algebra]<br />
* [http://cdn.artofproblemsolving.com/aops20/attachments/113521_cc6e30d39d41a700d293c1c36869ca6f Number Theory]<br />
* [http://cdn.artofproblemsolving.com/aops20/attachments/113521_c06dc3f7dddb7efe22b5f9c0db0d3a49 Circle Geometry]<br />
* [http://cdn.artofproblemsolving.com/aops20/attachments/113521_a7c63b4b9529686a56abe8298ed1eaa4 Other Geometry]<br />
<br />
'''[http://www.artofproblemsolving.com/wiki/index.php/AoPSWiki:Competition_ratings Ranking of all Olympiads (Difficulty Level)]'''<br />
<br />
== See also ==<br />
<br />
* [[List of mathematics competitions]]<br />
* [[Mathematics scholarships]]<br />
* [[Science competitions]]<br />
* [[Informatics competitions]]<br />
* [http://artofproblemsolving.com/wiki/index.php?title=How_should_I_prepare%3F How should I prepare]</div>Ghostychttps://artofproblemsolving.com/wiki/index.php?title=2007_AMC_8_Problems/Problem_12&diff=1582422007 AMC 8 Problems/Problem 122021-07-12T16:23:03Z<p>Ghostyc: /* Problem */</p>
<hr />
<div>==Problem==<br />
A unit hexagram is composed of a regular hexagon of side length <math>1</math> and its <math>6</math><br />
equilateral triangular extensions, as shown in the diagram. What is the ratio of<br />
the area of the extensions to the area of the original hexagon?<br />
<br />
<asy><br />
defaultpen(linewidth(0.7));<br />
draw(polygon(3));<br />
pair D=origin+1*dir(270), E=origin+1*dir(150), F=1*dir(30);<br />
draw(D--E--F--cycle);<br />
</asy><br />
<br />
<math>\mathrm{(A)}\ 1:1 \qquad \mathrm{(B)}\ 6:5 \qquad \mathrm{(C)}\ 3:2 \qquad \mathrm{(D)}\ 2:1 \qquad \mathrm{(E)}\ 3:1</math><br />
<br />
==Solution==<br />
The six equilateral triangular extensions fit perfectly into the hexagon meaning the answer is <math>\boxed{\textbf{(A) }1:1}</math><br />
<br />
==See Also==<br />
{{AMC8 box|year=2007|num-b=11|num-a=13}}<br />
{{MAA Notice}}</div>Ghostychttps://artofproblemsolving.com/wiki/index.php?title=2007_AMC_8_Problems/Problem_12&diff=1582412007 AMC 8 Problems/Problem 122021-07-12T16:22:49Z<p>Ghostyc: /* Problem */</p>
<hr />
<div>==Problem==<br />
A unit hexagram is composed of a regular hexagon of side length <math>1</math> and its <math>6</math><br />
equilateral triangular extensions, as shown in the diagram. What is the ratio of<br />
the area of the extensions to the area of the original hexagon?<br />
<br />
<asy><br />
defaultpen(linewidth(0.7));<br />
draw(polygon(3));<br />
pair D=origin+1*dir(270), E=origin+1*dir(150), F=1*dir(30);<br />
draw(D--E--F--cycle);<br />
<\asy><br />
<br />
<math>\mathrm{(A)}\ 1:1 \qquad \mathrm{(B)}\ 6:5 \qquad \mathrm{(C)}\ 3:2 \qquad \mathrm{(D)}\ 2:1 \qquad \mathrm{(E)}\ 3:1</math><br />
<br />
==Solution==<br />
The six equilateral triangular extensions fit perfectly into the hexagon meaning the answer is <math>\boxed{\textbf{(A) }1:1}</math><br />
<br />
==See Also==<br />
{{AMC8 box|year=2007|num-b=11|num-a=13}}<br />
{{MAA Notice}}</div>Ghostychttps://artofproblemsolving.com/wiki/index.php?title=2007_AMC_8_Problems/Problem_2&diff=1582402007 AMC 8 Problems/Problem 22021-07-12T16:21:54Z<p>Ghostyc: /* Problem */</p>
<hr />
<div>== Problem ==<br />
<math>650</math> students were surveyed about their pasta preferences. The choices were lasagna, manicotti, ravioli and spaghetti. The results of the survey are displayed in the bar graph. What is the ratio of the number of students who preferred spaghetti to the number of students who preferred manicotti?<br />
<br />
<asy><br />
size(200);<br />
defaultpen(linewidth(0.7));<br />
defaultpen(fontsize(8));<br />
draw(origin--(0,250));<br />
int i;<br />
for(i=0; i<6; i=i+1) {<br />
draw((0,50*i)--(5,50*i));<br />
}<br />
filldraw((25,0)--(75,0)--(75,150)--(25,150)--cycle, gray, black);<br />
filldraw((75,0)--(125,0)--(125,100)--(75,100)--cycle, gray, black);<br />
filldraw((125,0)--(175,0)--(175,150)--(125,150)--cycle, gray, black);<br />
filldraw((225,0)--(175,0)--(175,250)--(225,250)--cycle, gray, black);<br />
label("$50$", (0,50), W);<br />
label("$100$", (0,100), W);<br />
label("$150$", (0,150), W);<br />
label("$200$", (0,200), W);<br />
label("$250$", (0,250), W);<br />
label(rotate(90)*"Lasagna", (50,0), S);<br />
label(rotate(90)*"Manicotti", (100,0), S);<br />
label(rotate(90)*"Ravioli", (150,0), S);<br />
label(rotate(90)*"Spaghetti", (200,0), S);<br />
label(rotate(90)*"$\mbox{Number of People}$", (-40,140), W);<br />
</asy><br />
<br />
<math>\mathrm{(A)} \frac{2}{5} \qquad \mathrm{(B)} \frac{1}{2} \qquad \mathrm{(C)} \frac{5}{4} \qquad \mathrm{(D)} \frac{5}{3} \qquad \mathrm{(E)} \frac{5}{2}</math><br />
<br />
== Solution ==<br />
<br />
The answer is <math>\dfrac{\text{number of students who preferred spaghetti}}{\text{number of students who preferred manicotti}}</math><br />
<br />
So,<br />
<br />
<math>\frac{250}{100}</math> <br />
<br />
Simplify,<br />
<br />
<math>\frac{5}{2}</math> <br />
<br />
The answer is <math> \boxed{\textbf{(E)}\ \dfrac{5}{2}} </math><br />
<br />
==See Also==<br />
{{AMC8 box|year=2007|num-b=1|num-a=3}}<br />
{{MAA Notice}}</div>Ghostychttps://artofproblemsolving.com/wiki/index.php?title=2004_AMC_8_Problems/Problem_23&diff=1582392004 AMC 8 Problems/Problem 232021-07-12T16:15:20Z<p>Ghostyc: /* Problem */</p>
<hr />
<div>==Problem==<br />
Tess runs counterclockwise around rectangular block <math>JKLM</math>. She lives at corner <math>J</math>. Which graph could represent her straight-line distance from home?<br />
<br />
<asy><br />
unitsize(5mm);<br />
pair J=(-3,2); pair K=(-3,-2); pair L=(3,-2); pair M=(3,2); <br />
draw(J--K--L--M--cycle);<br />
label("$J$",J,NW);<br />
label("$K$",K,SW);<br />
label("$L$",L,SE);<br />
label("$M$",M,NE);<br />
</asy><br />
<br />
\textbf{A}<br />
<asy><br />
size(80);defaultpen(linewidth(0.8));<br />
//A<br />
draw((16,0)--origin--(0,16));<br />
draw(origin--(15,15));<br />
label("time", (8,0), S);<br />
label(rotate(90)*"distance", (0,8), W);<br />
</asy><br />
<br />
\textbf{B}<br />
<asy><br />
size(80);defaultpen(linewidth(0.8));<br />
//B<br />
draw((16,0)--origin--(0,16));<br />
draw((0,6)--(1,6)--(1,12)--(2,12)--(2,11)--(3,11)--(3,1)--(12,1)--(12,0));<br />
label("time", (8,0), S);<br />
label(rotate(90)*"distance", (0,8), W);<br />
</asy><br />
<br />
\textbf{C}<br />
<asy><br />
size(80);defaultpen(linewidth(0.8));<br />
//C<br />
draw((16,0)--origin--(0,16));<br />
draw(origin--(2.7,8)--(3,9)^^(11,9)--(14,0));<br />
draw(Arc((4,9), 1, 0, 180));<br />
draw(Arc((10,9), 1, 0, 180));<br />
draw(Arc((7,9), 2, 180,360));<br />
label("time", (8,0), S);<br />
label(rotate(90)*"distance", (0,8), W);<br />
</asy><br />
<br />
\textbf{D}<br />
<asy><br />
size(80);defaultpen(linewidth(0.8));<br />
//D<br />
draw((16,0)--origin--(0,16));<br />
draw(origin--(2,6)--(7,14)--(10,12)--(14,0));<br />
label("time", (8,0), S);<br />
label(rotate(90)*"distance", (0,8), W);<br />
</asy><br />
<br />
\textbf{E}<br />
<asy><br />
size(80);defaultpen(linewidth(0.8));<br />
//E<br />
draw((16,0)--origin--(0,16));<br />
draw(origin--(3,6)--(7,6)--(10,12)--(14,12));<br />
label("time", (8,0), S);<br />
label(rotate(90)*"distance", (0,8), W);<br />
</asy><br />
<br />
==Solution==<br />
For her distance to be represented as a constant horizontal line, Tess would have to be running in a circular shape with her home as the center. Since she is running around a rectangle, this is not possible, ruling out <math>B</math> and <math>E</math> with straight lines. Because <math>JL</math> is the diagonal of the rectangle, and <math>L</math> is at the middle distance around the perimeter, her maximum distance should be in the middle of her journey. The maximum in <math>A</math> is at the end, and <math>C</math> has two maximums, ruling both out. Thus the answer is <math>\boxed{\textbf{(D)}}</math>.<br />
<br />
==See Also==<br />
{{AMC8 box|year=2004|num-b=22|num-a=24}}<br />
{{MAA Notice}}</div>Ghostychttps://artofproblemsolving.com/wiki/index.php?title=2004_AMC_8_Problems/Problem_23&diff=1582382004 AMC 8 Problems/Problem 232021-07-12T16:13:39Z<p>Ghostyc: /* Problem */</p>
<hr />
<div>==Problem==<br />
Tess runs counterclockwise around rectangular block <math>JKLM</math>. She lives at corner <math>J</math>. Which graph could represent her straight-line distance from home?<br />
<br />
<asy><br />
unitsize(5mm);<br />
pair J=(-3,2); pair K=(-3,-2); pair L=(3,-2); pair M=(3,2); <br />
draw(J--K--L--M--cycle);<br />
label("$J$",J,NW);<br />
label("$K$",K,SW);<br />
label("$L$",L,SE);<br />
label("$M$",M,NE);<br />
</asy><br />
<br />
<asy><br />
size(80);defaultpen(linewidth(0.8));<br />
//A<br />
draw((16,0)--origin--(0,16));<br />
draw(origin--(15,15));<br />
label("time", (8,0), S);<br />
label(rotate(90)*"distance", (0,8), W);<br />
</asy><br />
<br />
<br />
<asy><br />
size(80);defaultpen(linewidth(0.8));<br />
//B<br />
draw((16,0)--origin--(0,16));<br />
draw((0,6)--(1,6)--(1,12)--(2,12)--(2,11)--(3,11)--(3,1)--(12,1)--(12,0));<br />
label("time", (8,0), S);<br />
label(rotate(90)*"distance", (0,8), W);<br />
</asy><br />
<br />
<br />
<asy><br />
size(80);defaultpen(linewidth(0.8));<br />
//C<br />
draw((16,0)--origin--(0,16));<br />
draw(origin--(2.7,8)--(3,9)^^(11,9)--(14,0));<br />
draw(Arc((4,9), 1, 0, 180));<br />
draw(Arc((10,9), 1, 0, 180));<br />
draw(Arc((7,9), 2, 180,360));<br />
label("time", (8,0), S);<br />
label(rotate(90)*"distance", (0,8), W);<br />
</asy><br />
<br />
<br />
<asy><br />
size(80);defaultpen(linewidth(0.8));<br />
//D<br />
draw((16,0)--origin--(0,16));<br />
draw(origin--(2,6)--(7,14)--(10,12)--(14,0));<br />
label("time", (8,0), S);<br />
label(rotate(90)*"distance", (0,8), W);<br />
</asy><br />
<br />
<br />
<asy><br />
size(80);defaultpen(linewidth(0.8));<br />
//E<br />
draw((16,0)--origin--(0,16));<br />
draw(origin--(3,6)--(7,6)--(10,12)--(14,12));<br />
label("time", (8,0), S);<br />
label(rotate(90)*"distance", (0,8), W);<br />
</asy><br />
<br />
==Solution==<br />
For her distance to be represented as a constant horizontal line, Tess would have to be running in a circular shape with her home as the center. Since she is running around a rectangle, this is not possible, ruling out <math>B</math> and <math>E</math> with straight lines. Because <math>JL</math> is the diagonal of the rectangle, and <math>L</math> is at the middle distance around the perimeter, her maximum distance should be in the middle of her journey. The maximum in <math>A</math> is at the end, and <math>C</math> has two maximums, ruling both out. Thus the answer is <math>\boxed{\textbf{(D)}}</math>.<br />
<br />
==See Also==<br />
{{AMC8 box|year=2004|num-b=22|num-a=24}}<br />
{{MAA Notice}}</div>Ghostychttps://artofproblemsolving.com/wiki/index.php?title=2004_AMC_8_Problems/Problem_23&diff=1582372004 AMC 8 Problems/Problem 232021-07-12T16:12:53Z<p>Ghostyc: /* Problem */</p>
<hr />
<div>==Problem==<br />
Tess runs counterclockwise around rectangular block <math>JKLM</math>. She lives at corner <math>J</math>. Which graph could represent her straight-line distance from home?<br />
<br />
<asy><br />
unitsize(5mm);<br />
pair J=(-3,2); pair K=(-3,-2); pair L=(3,-2); pair M=(3,2); <br />
draw(J--K--L--M--cycle);<br />
label("$J$",J,NW);<br />
label("$K$",K,SW);<br />
label("$L$",L,SE);<br />
label("$M$",M,NE);<br />
</asy><br />
<br />
<asy><br />
size(80);defaultpen(linewidth(0.8));<br />
//A<br />
draw((16,0)--origin--(0,16));<br />
draw(origin--(15,15));<br />
label("time", (8,0), S);<br />
label(rotate(90)*"distance", (0,8), W);<br />
</asy><br />
<asy><br />
size(80);defaultpen(linewidth(0.8));<br />
//B<br />
draw((16,0)--origin--(0,16));<br />
draw((0,6)--(1,6)--(1,12)--(2,12)--(2,11)--(3,11)--(3,1)--(12,1)--(12,0));<br />
label("time", (8,0), S);<br />
label(rotate(90)*"distance", (0,8), W);<br />
</asy><br />
<asy><br />
size(80);defaultpen(linewidth(0.8));<br />
//C<br />
draw((16,0)--origin--(0,16));<br />
draw(origin--(2.7,8)--(3,9)^^(11,9)--(14,0));<br />
draw(Arc((4,9), 1, 0, 180));<br />
draw(Arc((10,9), 1, 0, 180));<br />
draw(Arc((7,9), 2, 180,360));<br />
label("time", (8,0), S);<br />
label(rotate(90)*"distance", (0,8), W);<br />
</asy><br />
<asy><br />
size(80);defaultpen(linewidth(0.8));<br />
//D<br />
draw((16,0)--origin--(0,16));<br />
draw(origin--(2,6)--(7,14)--(10,12)--(14,0));<br />
label("time", (8,0), S);<br />
label(rotate(90)*"distance", (0,8), W);<br />
</asy><br />
<asy><br />
size(80);defaultpen(linewidth(0.8));<br />
//E<br />
draw((16,0)--origin--(0,16));<br />
draw(origin--(3,6)--(7,6)--(10,12)--(14,12));<br />
label("time", (8,0), S);<br />
label(rotate(90)*"distance", (0,8), W);<br />
</asy><br />
<br />
==Solution==<br />
For her distance to be represented as a constant horizontal line, Tess would have to be running in a circular shape with her home as the center. Since she is running around a rectangle, this is not possible, ruling out <math>B</math> and <math>E</math> with straight lines. Because <math>JL</math> is the diagonal of the rectangle, and <math>L</math> is at the middle distance around the perimeter, her maximum distance should be in the middle of her journey. The maximum in <math>A</math> is at the end, and <math>C</math> has two maximums, ruling both out. Thus the answer is <math>\boxed{\textbf{(D)}}</math>.<br />
<br />
==See Also==<br />
{{AMC8 box|year=2004|num-b=22|num-a=24}}<br />
{{MAA Notice}}</div>Ghostychttps://artofproblemsolving.com/wiki/index.php?title=2004_AMC_8_Problems/Problem_23&diff=1582362004 AMC 8 Problems/Problem 232021-07-12T16:11:39Z<p>Ghostyc: /* Problem */</p>
<hr />
<div>==Problem==<br />
Tess runs counterclockwise around rectangular block <math>JKLM</math>. She lives at corner <math>J</math>. Which graph could represent her straight-line distance from home?<br />
<br />
<asy><br />
unitsize(5mm);<br />
pair J=(-3,2); pair K=(-3,-2); pair L=(3,-2); pair M=(3,2); <br />
draw(J--K--L--M--cycle);<br />
label("$J$",J,NW);<br />
label("$K$",K,SW);<br />
label("$L$",L,SE);<br />
label("$M$",M,NE);<br />
</asy><br />
<br />
<asy><br />
size(80);defaultpen(linewidth(0.8));<br />
//A<br />
draw((16,0)--origin--(0,16));<br />
draw(origin--(15,15));<br />
label("time", (8,0), S);<br />
label(rotate(90)*"distance", (0,8), W);<br />
</asy><br />
size(80);defaultpen(linewidth(0.8));<br />
//B<br />
draw((16,0)--origin--(0,16));<br />
draw((0,6)--(1,6)--(1,12)--(2,12)--(2,11)--(3,11)--(3,1)--(12,1)--(12,0));<br />
label("time", (8,0), S);<br />
label(rotate(90)*"distance", (0,8), W);<br />
</asy><br />
size(80);defaultpen(linewidth(0.8));<br />
//C<br />
draw((16,0)--origin--(0,16));<br />
draw(origin--(2.7,8)--(3,9)^^(11,9)--(14,0));<br />
draw(Arc((4,9), 1, 0, 180));<br />
draw(Arc((10,9), 1, 0, 180));<br />
draw(Arc((7,9), 2, 180,360));<br />
label("time", (8,0), S);<br />
label(rotate(90)*"distance", (0,8), W);<br />
</asy><br />
size(80);defaultpen(linewidth(0.8));<br />
//D<br />
draw((16,0)--origin--(0,16));<br />
draw(origin--(2,6)--(7,14)--(10,12)--(14,0));<br />
label("time", (8,0), S);<br />
label(rotate(90)*"distance", (0,8), W);<br />
</asy><br />
size(80);defaultpen(linewidth(0.8));<br />
//E<br />
draw((16,0)--origin--(0,16));<br />
draw(origin--(3,6)--(7,6)--(10,12)--(14,12));<br />
label("time", (8,0), S);<br />
label(rotate(90)*"distance", (0,8), W);<br />
</asy><br />
<br />
==Solution==<br />
For her distance to be represented as a constant horizontal line, Tess would have to be running in a circular shape with her home as the center. Since she is running around a rectangle, this is not possible, ruling out <math>B</math> and <math>E</math> with straight lines. Because <math>JL</math> is the diagonal of the rectangle, and <math>L</math> is at the middle distance around the perimeter, her maximum distance should be in the middle of her journey. The maximum in <math>A</math> is at the end, and <math>C</math> has two maximums, ruling both out. Thus the answer is <math>\boxed{\textbf{(D)}}</math>.<br />
<br />
==See Also==<br />
{{AMC8 box|year=2004|num-b=22|num-a=24}}<br />
{{MAA Notice}}</div>Ghostychttps://artofproblemsolving.com/wiki/index.php?title=2004_AMC_8_Problems/Problem_23&diff=1582352004 AMC 8 Problems/Problem 232021-07-12T16:10:59Z<p>Ghostyc: /* Problem */</p>
<hr />
<div>==Problem==<br />
Tess runs counterclockwise around rectangular block <math>JKLM</math>. She lives at corner <math>J</math>. Which graph could represent her straight-line distance from home?<br />
<br />
<asy><br />
unitsize(5mm);<br />
pair J=(-3,2); pair K=(-3,-2); pair L=(3,-2); pair M=(3,2); <br />
draw(J--K--L--M--cycle);<br />
label("$J$",J,NW);<br />
label("$K$",K,SW);<br />
label("$L$",L,SE);<br />
label("$M$",M,NE);<br />
</asy><br />
<br />
<asy><br />
size(80);defaultpen(linewidth(0.8));<br />
//A<br />
draw((16,0)--origin--(0,16));<br />
draw(origin--(15,15));<br />
label("time", (8,0), S);<br />
label(rotate(90)*"distance", (0,8), W);<br />
<br />
//B<br />
draw((16,0)--origin--(0,16));<br />
draw((0,6)--(1,6)--(1,12)--(2,12)--(2,11)--(3,11)--(3,1)--(12,1)--(12,0));<br />
label("time", (8,0), S);<br />
label(rotate(90)*"distance", (0,8), W);<br />
<br />
//C<br />
draw((16,0)--origin--(0,16));<br />
draw(origin--(2.7,8)--(3,9)^^(11,9)--(14,0));<br />
draw(Arc((4,9), 1, 0, 180));<br />
draw(Arc((10,9), 1, 0, 180));<br />
draw(Arc((7,9), 2, 180,360));<br />
label("time", (8,0), S);<br />
label(rotate(90)*"distance", (0,8), W);<br />
<br />
//D<br />
draw((16,0)--origin--(0,16));<br />
draw(origin--(2,6)--(7,14)--(10,12)--(14,0));<br />
label("time", (8,0), S);<br />
label(rotate(90)*"distance", (0,8), W);<br />
<br />
//E<br />
draw((16,0)--origin--(0,16));<br />
draw(origin--(3,6)--(7,6)--(10,12)--(14,12));<br />
label("time", (8,0), S);<br />
label(rotate(90)*"distance", (0,8), W);<br />
</asy><br />
<br />
==Solution==<br />
For her distance to be represented as a constant horizontal line, Tess would have to be running in a circular shape with her home as the center. Since she is running around a rectangle, this is not possible, ruling out <math>B</math> and <math>E</math> with straight lines. Because <math>JL</math> is the diagonal of the rectangle, and <math>L</math> is at the middle distance around the perimeter, her maximum distance should be in the middle of her journey. The maximum in <math>A</math> is at the end, and <math>C</math> has two maximums, ruling both out. Thus the answer is <math>\boxed{\textbf{(D)}}</math>.<br />
<br />
==See Also==<br />
{{AMC8 box|year=2004|num-b=22|num-a=24}}<br />
{{MAA Notice}}</div>Ghostychttps://artofproblemsolving.com/wiki/index.php?title=2002_AMC_8_Problems/Problem_6&diff=1582342002 AMC 8 Problems/Problem 62021-07-12T15:50:43Z<p>Ghostyc: /* Problem */</p>
<hr />
<div>==Problem==<br />
<br />
A birdbath is designed to overflow so that it will be self-cleaning. Water flows in at the rate of 20 milliliters per minute and drains at the rate of 18 milliliters per minute. One of these graphs shows the volume of water in the birdbath during the filling time and continuing into the overflow time. Which one is it?<br />
<br />
<asy><br />
size(450);<br />
defaultpen(linewidth(0.8));<br />
path[] p={origin--(8,8)--(14,8), (0,10)--(4,10)--(14,0), origin--(14,14), (0,14)--(14,14), origin--(7,7)--(14,0)};<br />
int i;<br />
for(i=0; i<5; i=i+1) {<br />
draw(shift(21i,0)*((0,16)--origin--(14,0)));<br />
draw(shift(21i,0)*(p[i]));<br />
label("Time", (7+21i,0), S);<br />
label(rotate(90)*"Volume", (21i,8), W);<br />
}<br />
<br />
label("$A$", (0*21 + 7,-5), S);<br />
label("$B$", (1*21 + 7,-5), S);<br />
label("$C$", (2*21 + 7,-5), S);<br />
label("$D$", (3*21 + 7,-5), S);<br />
label("$E$", (4*21 + 7,-5), S);<br />
</asy><br />
<br />
<math>\text{(A)}\ \text{A} \qquad \text{(B)}\ \text{B} \qquad \text{(C)}\ \text{C} \qquad \text{(D)}\ \text{D} \qquad \text{(E)}\ \text{E}</math><br />
<br />
==Solution==<br />
The change in the water volume has a net gain of <math>20-18=2</math> millimeters per minute. The birdbath's volume increases at a constant rate until it reaches its maximum and starts overflowing to keep a constant volume. This is best represented by graph <math>\boxed{\text{(A)}\ A}</math>.<br />
<br />
==See Also==<br />
{{AMC8 box|year=2002|num-b=5|num-a=7}}<br />
{{MAA Notice}}</div>Ghostychttps://artofproblemsolving.com/wiki/index.php?title=2005_AMC_10A_Problems&diff=1582332005 AMC 10A Problems2021-07-12T15:45:28Z<p>Ghostyc: /* Problem 12 */</p>
<hr />
<div>{{AMC10 Problems|year=2005|ab=A}}<br />
== Problem 1 ==<br />
While eating out, Mike and Joe each tipped their server <math>2</math> dollars. Mike tipped <math>10\%</math> of his bill and Joe tipped <math>20\%</math> of his bill. What was the difference, in dollars between their bills? <br />
<br />
<math> \mathrm{(A) \ } 2\qquad \mathrm{(B) \ } 4\qquad \mathrm{(C) \ } 5\qquad \mathrm{(D) \ } 10\qquad \mathrm{(E) \ } 20 </math><br />
<br />
[[2005 AMC 10A Problems/Problem 1|Solution]]<br />
<br />
== Problem 2 ==<br />
For each pair of real numbers <math>a \neq b</math>, define the [[operation]] <math>\star</math> as<br />
<br />
<math> (a \star b) = \frac{a+b}{a-b} </math>.<br />
<br />
What is the value of <math> ((1 \star 2) \star 3)</math>?<br />
<br />
<math> \mathrm{(A) \ } -\frac{2}{3}\qquad \mathrm{(B) \ } -\frac{1}{5}\qquad \mathrm{(C) \ } 0\qquad \mathrm{(D) \ } \frac{1}{2}\qquad \mathrm{(E) \ } \textrm{This\, value\, is\, not\, defined.} </math><br />
<br />
[[2005 AMC 10A Problems/Problem 2|Solution]]<br />
<br />
== Problem 3 ==<br />
The equations <math> 2x + 7 = 3 </math> and <math> bx - 10 = -2 </math> have the same solution <math>x</math>. What is the value of <math>b</math>? <br />
<br />
<math> \mathrm{(A) \ } -8\qquad \mathrm{(B) \ } -4\qquad \mathrm{(C) \ } -2\qquad \mathrm{(D) \ } 4\qquad \mathrm{(E) \ } 8 </math><br />
<br />
[[2005 AMC 10A Problems/Problem 3|Solution]]<br />
<br />
== Problem 4 ==<br />
A rectangle with a diagonal of length <math>x</math> is twice as long as it is wide. What is the area of the rectangle? <br />
<br />
<math> \mathrm{(A) \ } \frac{1}{4}x^2\qquad \mathrm{(B) \ } \frac{2}{5}x^2\qquad \mathrm{(C) \ } \frac{1}{2}x^2\qquad \mathrm{(D) \ } x^2\qquad \mathrm{(E) \ } \frac{3}{2}x^2 </math><br />
<br />
[[2005 AMC 10A Problems/Problem 4|Solution]]<br />
<br />
== Problem 5 ==<br />
A store normally sells windows at \$100 each. This week the store is offering one free window for each purchase of four. Dave needs seven windows and Doug needs eight windows. How many dollars will they save if they purchase the windows together rather than separately?<br />
<br />
<math> \mathrm{(A) \ } 100\qquad \mathrm{(B) \ } 200\qquad \mathrm{(C) \ } 300\qquad \mathrm{(D) \ } 400\qquad \mathrm{(E) \ } 500 </math><br />
<br />
[[2005 AMC 10A Problems/Problem 5|Solution]]<br />
<br />
== Problem 6 ==<br />
The average (mean) of <math>20</math> numbers is <math>30</math>, and the average of <math>30</math> other numbers is <math>20</math>. What is the average of all <math>50</math> numbers?<br />
<br />
<math> \mathrm{(A) \ } 23\qquad \mathrm{(B) \ } 24\qquad \mathrm{(C) \ } 25\qquad \mathrm{(D) \ } 26\qquad \mathrm{(E) \ } 27 </math><br />
<br />
[[2005 AMC 10A Problems/Problem 6|Solution]]<br />
<br />
== Problem 7 ==<br />
Josh and Mike live <math>13</math> miles apart. Yesterday Josh started to ride his bicycle toward Mike's house. A little later Mike started to ride his bicycle toward Josh's house. When they met, Josh had ridden for twice the length of time as Mike and at four-fifths of Mike's rate. How many miles had Mike ridden when they met? <br />
<br />
<math> \mathrm{(A) \ } 4\qquad \mathrm{(B) \ } 5\qquad \mathrm{(C) \ } 6\qquad \mathrm{(D) \ } 7\qquad \mathrm{(E) \ } 8 </math><br />
<br />
[[2005 AMC 10A Problems/Problem 7|Solution]]<br />
<br />
== Problem 8 ==<br />
In the figure, the length of side <math>AB</math> of square <math>ABCD</math> is <math>\sqrt{50}</math> and <math>BE=1</math>. What is the area of the inner square <math>EFGH</math>?<br />
<br />
<asy><br />
unitsize(4cm);<br />
defaultpen(linewidth(.8pt)+fontsize(10pt));<br />
<br />
pair D=(0,0), C=(1,0), B=(1,1), A=(0,1);<br />
pair F=intersectionpoints(Circle(D,2/sqrt(5)),Circle(A,1))[0];<br />
pair G=foot(A,D,F), H=foot(B,A,G), E=foot(C,B,H);<br />
<br />
draw(A--B--C--D--cycle);<br />
draw(D--F);<br />
draw(C--E);<br />
draw(B--H);<br />
draw(A--G);<br />
<br />
label("$A$",A,NW);<br />
label("$B$",B,NE);<br />
label("$C$",C,SE);<br />
label("$D$",D,SW);<br />
label("$E$",E,NNW);<br />
label("$F$",F,ENE);<br />
label("$G$",G,SSE);<br />
label("$H$",H,WSW);<br />
</asy><br />
<br />
<math> \textbf{(A)}\ 25\qquad\textbf{(B)}\ 32\qquad\textbf{(C)}\ 36\qquad\textbf{(D)}\ 40\qquad\textbf{(E)}\ 42 </math><br />
<br />
[[2005 AMC 10A Problems/Problem 8|Solution]]<br />
<br />
== Problem 9 ==<br />
Three tiles are marked <math>X</math> and two other tiles are marked <math>O</math>. The five tiles are randomly arranged in a row. What is the probability that the arrangement reads <math>XOXOX</math>?<br />
<br />
<math> \mathrm{(A) \ } \frac{1}{12}\qquad \mathrm{(B) \ } \frac{1}{10}\qquad \mathrm{(C) \ } \frac{1}{6}\qquad \mathrm{(D) \ } \frac{1}{4}\qquad \mathrm{(E) \ } \frac{1}{3} </math><br />
<br />
[[2005 AMC 10A Problems/Problem 9|Solution]]<br />
<br />
== Problem 10 ==<br />
There are two values of <math>a</math> for which the equation <math> 4x^2 + ax + 8x + 9 = 0 </math> has only one solution for <math>x</math>. What is the sum of those values of <math>a</math>?<br />
<br />
<math> \mathrm{(A) \ } -16\qquad \mathrm{(B) \ } -8\qquad \mathrm{(C) \ } 0\qquad \mathrm{(D) \ } 8\qquad \mathrm{(E) \ } 20 </math><br />
<br />
[[2005 AMC 10A Problems/Problem 10|Solution]]<br />
<br />
== Problem 11 ==<br />
A wooden cube <math>n</math> units on a side is painted red on all six faces and then cut into <math>n^3</math> unit cubes. Exactly one-fourth of the total number of faces of the unit cubes are red. What is <math>n</math>?<br />
<br />
<math> \mathrm{(A) \ } 3\qquad \mathrm{(B) \ } 4\qquad \mathrm{(C) \ } 5\qquad \mathrm{(D) \ } 6\qquad \mathrm{(E) \ } 7 </math><br />
<br />
[[2005 AMC 10A Problems/Problem 11|Solution]]<br />
<br />
== Problem 12 ==<br />
The figure shown is called a ''trefoil'' and is constructed by drawing circular sectors about the sides of the congruent equilateral triangles. What is the area of a trefoil whose horizontal base has length <math>2</math>?<br />
<br />
<asy><br />
unitsize(1.5cm);<br />
defaultpen(linewidth(.8pt)+fontsize(12pt));<br />
<br />
pair O=(0,0), A=dir(0), B=dir(60), C=dir(120), D=dir(180);<br />
pair E=B+C;<br />
<br />
draw(D--E--B--O--C--B--A,linetype("4 4"));<br />
draw(Arc(O,1,0,60),linewidth(1.2pt));<br />
draw(Arc(O,1,120,180),linewidth(1.2pt));<br />
draw(Arc(C,1,0,60),linewidth(1.2pt));<br />
draw(Arc(B,1,120,180),linewidth(1.2pt));<br />
draw(A--D,linewidth(1.2pt));<br />
draw(O--dir(40),EndArrow(HookHead,4));<br />
draw(O--dir(140),EndArrow(HookHead,4));<br />
draw(C--C+dir(40),EndArrow(HookHead,4));<br />
draw(B--B+dir(140),EndArrow(HookHead,4));<br />
<br />
label("2",O,S);<br />
draw((0.1,-0.12)--(1,-0.12),EndArrow(HookHead,4),EndBar);<br />
draw((-0.1,-0.12)--(-1,-0.12),EndArrow(HookHead,4),EndBar);<br />
</asy><br />
<br />
<math> \mathrm{(A) \ } \frac{1}{3}\pi+\frac{\sqrt{3}}{2}\qquad \mathrm{(B) \ } \frac{2}{3}\pi\qquad \mathrm{(C) \ } \frac{2}{3}\pi+\frac{\sqrt{3}}{4}\qquad \mathrm{(D) \ } \frac{2}{3}\pi+\frac{\sqrt{3}}{3}\qquad \mathrm{(E) \ } \frac{2}{3}\pi+\frac{\sqrt{3}}{2} </math><br />
<br />
[[2005 AMC 10A Problems/Problem 12|Solution]]<br />
<br />
== Problem 13 ==<br />
How many positive integers <math>n</math> satisfy the following condition:<br />
<br />
<math> (130n)^{50} > n^{100} > 2^{200} </math>?<br />
<br />
<math> \mathrm{(A) \ } 0\qquad \mathrm{(B) \ } 7\qquad \mathrm{(C) \ } 12\qquad \mathrm{(D) \ } 65\qquad \mathrm{(E) \ } 125 </math><br />
<br />
[[2005 AMC 10A Problems/Problem 13|Solution]]<br />
<br />
== Problem 14 ==<br />
How many three-digit numbers satisfy the property that the middle digit is the average of the first and the last digits? <br />
<br />
<math> \mathrm{(A) \ } 41\qquad \mathrm{(B) \ } 42\qquad \mathrm{(C) \ } 43\qquad \mathrm{(D) \ } 44\qquad \mathrm{(E) \ } 45 </math><br />
<br />
[[2005 AMC 10A Problems/Problem 14|Solution]]<br />
<br />
== Problem 15 ==<br />
How many positive cubes divide <math> 3! \cdot 5! \cdot 7! </math> ?<br />
<br />
<math> \mathrm{(A) \ } 2\qquad \mathrm{(B) \ } 3\qquad \mathrm{(C) \ } 4\qquad \mathrm{(D) \ } 5\qquad \mathrm{(E) \ } 6 </math><br />
<br />
[[2005 AMC 10A Problems/Problem 15|Solution]]<br />
<br />
== Problem 16 ==<br />
The sum of the digits of a two-digit number is subtracted from the number. The units digit of the result is <math>6</math>. How many two-digit numbers have this property? <br />
<br />
<math> \mathrm{(A) \ } 5\qquad \mathrm{(B) \ } 7\qquad \mathrm{(C) \ } 9\qquad \mathrm{(D) \ } 10\qquad \mathrm{(E) \ } 19 </math><br />
<br />
[[2005 AMC 10A Problems/Problem 16|Solution]]<br />
<br />
== Problem 17 ==<br />
In the five-sided star shown, the letters <math>A</math>, <math>B</math>, <math>C</math>, <math>D</math>, and <math>E</math> are replaced by the numbers <math>3</math>, <math>5</math>, <math>6</math>, <math>7</math>, and <math>9</math>, although not necessarily in this order. The sums of the numbers at the ends of the line segments <math>AB</math>, <math>BC</math>, <math>CD</math>, <math>DE</math>, and <math>EA</math> form an arithmetic sequence, although not necessarily in this order. What is the middle term of the sequence? <br />
<br />
<asy><br />
size(150);<br />
defaultpen(linewidth(0.8));<br />
string[] strng = {'A','D','B','E','C'};<br />
pair A=dir(90),B=dir(306),C=dir(162),D=dir(18),E=dir(234);<br />
draw(A--B--C--D--E--cycle);<br />
for(int i=0;i<=4;i=i+1)<br />
{<br />
path circ=circle(dir(90-72*i),0.125);<br />
unfill(circ);<br />
draw(circ);<br />
label("$"+strng[i]+"$",dir(90-72*i));<br />
}<br />
</asy><br />
<br />
<math> \mathrm{(A) \ } 9\qquad \mathrm{(B) \ } 10\qquad \mathrm{(C) \ } 11\qquad \mathrm{(D) \ } 12\qquad \mathrm{(E) \ } 13 </math><br />
<br />
[[2005 AMC 10A Problems/Problem 17|Solution]]<br />
<br />
== Problem 18 ==<br />
Team A and team B play a series. The first team to win three games wins the series. Each team is equally likely to win each game, there are no ties, and the outcomes of the individual games are independent. If team B wins the second game and team A wins the series, what is the probability that team B wins the first game? <br />
<br />
<math> \mathrm{(A) \ } \frac{1}{5}\qquad \mathrm{(B) \ } \frac{1}{4}\qquad \mathrm{(C) \ } \frac{1}{3}\qquad \mathrm{(D) \ } \frac{1}{2}\qquad \mathrm{(E) \ } \frac{2}{3} </math><br />
<br />
[[2005 AMC 10A Problems/Problem 18|Solution]]<br />
<br />
== Problem 19 ==<br />
Three one-inch squares are placed with their bases on a line. The center square is lifted out and rotated 45 degrees, as shown. Then it is centered and lowered into its original location until it touches both of the adjoining squares. How many inches is the point <math>B</math> from the line on which the bases of the original squares were placed?<br />
<br />
<asy><br />
unitsize(1inch);<br />
defaultpen(linewidth(.8pt)+fontsize(8pt));<br />
draw((0,0)--((1/3) + 3*(1/2),0));<br />
fill(((1/6) + (1/2),0)--((1/6) + (1/2),(1/2))--((1/6) + 1,(1/2))--((1/6) + 1,0)--cycle, rgb(.7,.7,.7));<br />
draw(((1/6),0)--((1/6) + (1/2),0)--((1/6) + (1/2),(1/2))--((1/6),(1/2))--cycle);<br />
draw(((1/6) + (1/2),0)--((1/6) + (1/2),(1/2))--((1/6) + 1,(1/2))--((1/6) + 1,0)--cycle);<br />
draw(((1/6) + 1,0)--((1/6) + 1,(1/2))--((1/6) + (3/2),(1/2))--((1/6) + (3/2),0)--cycle);<br />
draw((2,0)--(2 + (1/3) + (3/2),0));<br />
draw(((2/3) + (3/2),0)--((2/3) + 2,0)--((2/3) + 2,(1/2))--((2/3) + (3/2),(1/2))--cycle);<br />
draw(((2/3) + (5/2),0)--((2/3) + (5/2),(1/2))--((2/3) + 3,(1/2))--((2/3) + 3,0)--cycle);<br />
label("$B$",((1/6) + (1/2),(1/2)),NW);<br />
label("$B$",((2/3) + 2 + (1/4),(29/30)),NNE);<br />
draw(((1/6) + (1/2),(1/2)+0.05)..(1,.8)..((2/3) + 2 + (1/4)-.05,(29/30)),EndArrow(HookHead,3));<br />
fill(((2/3) + 2 + (1/4),(1/4))--((2/3) + (5/2) + (1/10),(1/2) + (1/9))--((2/3) + 2 + (1/4),(29/30))--((2/3) + 2 - (1/10),(1/2) + (1/9))--cycle, rgb(.7,.7,.7));<br />
draw(((2/3) + 2 + (1/4),(1/4))--((2/3) + (5/2) + (1/10),(1/2) + (1/9))--((2/3) + 2 + (1/4),(29/30))--((2/3) + 2 - (1/10),(1/2) + (1/9))--cycle);</asy><br />
<br />
<math> \textbf{(A)}\ 1\qquad\textbf{(B)}\ \sqrt{2}\qquad\textbf{(C)}\ \frac{3}{2}\qquad\textbf{(D)}\ \sqrt{2}+\frac{1}{2}\qquad\textbf{(E)}\ 2 </math><br />
<br />
[[2005 AMC 10A Problems/Problem 19|Solution]]<br />
<br />
== Problem 20 ==<br />
An equiangular octagon has four sides of length 1 and four sides of length <math>\frac{\sqrt{2}}{2}</math>, arranged so that no two consecutive sides have the same length. What is the area of the octagon?<br />
<br />
<math> \mathrm{(A) \ } \frac72\qquad \mathrm{(B) \ } \frac{7\sqrt2}{2}\qquad \mathrm{(C) \ } \frac{5+4\sqrt2}{2}\qquad \mathrm{(D) \ } \frac{4+5\sqrt2}{2}\qquad \mathrm{(E) \ } 7 </math><br />
<br />
[[2005 AMC 10A Problems/Problem 20|Solution]]<br />
<br />
== Problem 21 ==<br />
For how many positive integers <math>n</math> does <math> 1+2+...+n </math> evenly divide <math>6n</math>? <br />
<br />
<math> \mathrm{(A) \ } 3\qquad \mathrm{(B) \ } 5\qquad \mathrm{(C) \ } 7\qquad \mathrm{(D) \ } 9\qquad \mathrm{(E) \ } 11 </math><br />
<br />
[[2005 AMC 10A Problems/Problem 21|Solution]]<br />
<br />
== Problem 22 ==<br />
Let <math>S</math> be the set of the <math>2005</math> smallest positive multiples of <math>4</math>, and let <math>T</math> be the set of the <math>2005</math> smallest positive multiples of <math>6</math>. How many elements are common to <math>S</math> and <math>T</math>?<br />
<br />
<math> \mathrm{(A) \ } 166\qquad \mathrm{(B) \ } 333\qquad \mathrm{(C) \ } 500\qquad \mathrm{(D) \ } 668\qquad \mathrm{(E) \ } 1001 </math><br />
<br />
[[2005 AMC 10A Problems/Problem 22|Solution]]<br />
<br />
== Problem 23 ==<br />
Let <math>AB</math> be a diameter of a circle and let <math>C</math> be a point on <math>AB</math> with <math>2\cdot AC=BC</math>. Let <math>D</math> and <math>E</math> be points on the circle such that <math>DC \perp AB</math> and <math>DE</math> is a second diameter. What is the ratio of the area of <math>\triangle DCE</math> to the area of <math>\triangle ABD</math>?<br />
<br />
<asy><br />
unitsize(2.5cm);<br />
defaultpen(fontsize(10pt)+linewidth(.8pt));<br />
dotfactor=3;<br />
pair O=(0,0), C=(-1/3.0), B=(1,0), A=(-1,0);<br />
pair D=dir(aCos(C.x)), E=(-D.x,-D.y);<br />
draw(A--B--D--cycle);<br />
draw(D--E--C);<br />
draw(unitcircle,white);<br />
drawline(D,C);<br />
dot(O);<br />
clip(unitcircle);<br />
draw(unitcircle);<br />
label("$E$",E,SSE);<br />
label("$B$",B,E);<br />
label("$A$",A,W);<br />
label("$D$",D,NNW);<br />
label("$C$",C,SW);<br />
draw(rightanglemark(D,C,B,2));</asy><br />
<br />
<math> \mathrm{(A) \ } \frac{1}{6} \qquad \mathrm{(B) \ } \frac{1}{4} \qquad \mathrm{(C) \ } \frac{1}{3} \qquad \mathrm{(D) \ } \frac{1}{2} \qquad \mathrm{(E) \ } \frac{2}{3} </math><br />
<br />
[[2005 AMC 10A Problems/Problem 23|Solution]]<br />
<br />
== Problem 24 ==<br />
For each positive integer <math> n > 1 </math>, let <math>P(n)</math> denote the greatest prime factor of <math>n</math>. For how many positive integers <math>n</math> is it true that both <math> P(n) = \sqrt{n} </math> and <math> P(n+48) = \sqrt{n+48} </math>?<br />
<br />
<math> \mathrm{(A) \ } 0\qquad \mathrm{(B) \ } 1\qquad \mathrm{(C) \ } 3\qquad \mathrm{(D) \ } 4\qquad \mathrm{(E) \ } 5 </math><br />
<br />
[[2005 AMC 10A Problems/Problem 24|Solution]]<br />
<br />
== Problem 25 ==<br />
In <math>ABC</math> we have <math> AB = 25 </math>, <math> BC = 39 </math>, and <math>AC=42</math>. Points <math>D</math> and <math>E</math> are on <math>AB</math> and <math>AC</math> respectively, with <math> AD = 19 </math> and <math> AE = 14 </math>. What is the ratio of the area of triangle <math>ADE</math> to the area of the quadrilateral <math>BCED</math>?<br />
<br />
<math> \mathrm{(A) \ } \frac{266}{1521}\qquad \mathrm{(B) \ } \frac{19}{75}\qquad \mathrm{(C) \ } \frac{1}{3}\qquad \mathrm{(D) \ } \frac{19}{56}\qquad \mathrm{(E) \ } 1 </math><br />
<br />
[[2005 AMC 10A Problems/Problem 25|Solution]]<br />
<br />
== See also ==<br />
{{AMC10 box|year=2005|ab=A|before=[[2004 AMC 10B Problems]]|after=[[2005 AMC 10B Problems]]}}<br />
*[[AMC 10 Problems and Solutions]]<br />
* [[AMC Problems and Solutions]]<br />
{{MAA Notice}}</div>Ghostychttps://artofproblemsolving.com/wiki/index.php?title=2005_AMC_10A_Problems&diff=1582322005 AMC 10A Problems2021-07-12T15:44:58Z<p>Ghostyc: /* Problem 17 */</p>
<hr />
<div>{{AMC10 Problems|year=2005|ab=A}}<br />
== Problem 1 ==<br />
While eating out, Mike and Joe each tipped their server <math>2</math> dollars. Mike tipped <math>10\%</math> of his bill and Joe tipped <math>20\%</math> of his bill. What was the difference, in dollars between their bills? <br />
<br />
<math> \mathrm{(A) \ } 2\qquad \mathrm{(B) \ } 4\qquad \mathrm{(C) \ } 5\qquad \mathrm{(D) \ } 10\qquad \mathrm{(E) \ } 20 </math><br />
<br />
[[2005 AMC 10A Problems/Problem 1|Solution]]<br />
<br />
== Problem 2 ==<br />
For each pair of real numbers <math>a \neq b</math>, define the [[operation]] <math>\star</math> as<br />
<br />
<math> (a \star b) = \frac{a+b}{a-b} </math>.<br />
<br />
What is the value of <math> ((1 \star 2) \star 3)</math>?<br />
<br />
<math> \mathrm{(A) \ } -\frac{2}{3}\qquad \mathrm{(B) \ } -\frac{1}{5}\qquad \mathrm{(C) \ } 0\qquad \mathrm{(D) \ } \frac{1}{2}\qquad \mathrm{(E) \ } \textrm{This\, value\, is\, not\, defined.} </math><br />
<br />
[[2005 AMC 10A Problems/Problem 2|Solution]]<br />
<br />
== Problem 3 ==<br />
The equations <math> 2x + 7 = 3 </math> and <math> bx - 10 = -2 </math> have the same solution <math>x</math>. What is the value of <math>b</math>? <br />
<br />
<math> \mathrm{(A) \ } -8\qquad \mathrm{(B) \ } -4\qquad \mathrm{(C) \ } -2\qquad \mathrm{(D) \ } 4\qquad \mathrm{(E) \ } 8 </math><br />
<br />
[[2005 AMC 10A Problems/Problem 3|Solution]]<br />
<br />
== Problem 4 ==<br />
A rectangle with a diagonal of length <math>x</math> is twice as long as it is wide. What is the area of the rectangle? <br />
<br />
<math> \mathrm{(A) \ } \frac{1}{4}x^2\qquad \mathrm{(B) \ } \frac{2}{5}x^2\qquad \mathrm{(C) \ } \frac{1}{2}x^2\qquad \mathrm{(D) \ } x^2\qquad \mathrm{(E) \ } \frac{3}{2}x^2 </math><br />
<br />
[[2005 AMC 10A Problems/Problem 4|Solution]]<br />
<br />
== Problem 5 ==<br />
A store normally sells windows at \$100 each. This week the store is offering one free window for each purchase of four. Dave needs seven windows and Doug needs eight windows. How many dollars will they save if they purchase the windows together rather than separately?<br />
<br />
<math> \mathrm{(A) \ } 100\qquad \mathrm{(B) \ } 200\qquad \mathrm{(C) \ } 300\qquad \mathrm{(D) \ } 400\qquad \mathrm{(E) \ } 500 </math><br />
<br />
[[2005 AMC 10A Problems/Problem 5|Solution]]<br />
<br />
== Problem 6 ==<br />
The average (mean) of <math>20</math> numbers is <math>30</math>, and the average of <math>30</math> other numbers is <math>20</math>. What is the average of all <math>50</math> numbers?<br />
<br />
<math> \mathrm{(A) \ } 23\qquad \mathrm{(B) \ } 24\qquad \mathrm{(C) \ } 25\qquad \mathrm{(D) \ } 26\qquad \mathrm{(E) \ } 27 </math><br />
<br />
[[2005 AMC 10A Problems/Problem 6|Solution]]<br />
<br />
== Problem 7 ==<br />
Josh and Mike live <math>13</math> miles apart. Yesterday Josh started to ride his bicycle toward Mike's house. A little later Mike started to ride his bicycle toward Josh's house. When they met, Josh had ridden for twice the length of time as Mike and at four-fifths of Mike's rate. How many miles had Mike ridden when they met? <br />
<br />
<math> \mathrm{(A) \ } 4\qquad \mathrm{(B) \ } 5\qquad \mathrm{(C) \ } 6\qquad \mathrm{(D) \ } 7\qquad \mathrm{(E) \ } 8 </math><br />
<br />
[[2005 AMC 10A Problems/Problem 7|Solution]]<br />
<br />
== Problem 8 ==<br />
In the figure, the length of side <math>AB</math> of square <math>ABCD</math> is <math>\sqrt{50}</math> and <math>BE=1</math>. What is the area of the inner square <math>EFGH</math>?<br />
<br />
<asy><br />
unitsize(4cm);<br />
defaultpen(linewidth(.8pt)+fontsize(10pt));<br />
<br />
pair D=(0,0), C=(1,0), B=(1,1), A=(0,1);<br />
pair F=intersectionpoints(Circle(D,2/sqrt(5)),Circle(A,1))[0];<br />
pair G=foot(A,D,F), H=foot(B,A,G), E=foot(C,B,H);<br />
<br />
draw(A--B--C--D--cycle);<br />
draw(D--F);<br />
draw(C--E);<br />
draw(B--H);<br />
draw(A--G);<br />
<br />
label("$A$",A,NW);<br />
label("$B$",B,NE);<br />
label("$C$",C,SE);<br />
label("$D$",D,SW);<br />
label("$E$",E,NNW);<br />
label("$F$",F,ENE);<br />
label("$G$",G,SSE);<br />
label("$H$",H,WSW);<br />
</asy><br />
<br />
<math> \textbf{(A)}\ 25\qquad\textbf{(B)}\ 32\qquad\textbf{(C)}\ 36\qquad\textbf{(D)}\ 40\qquad\textbf{(E)}\ 42 </math><br />
<br />
[[2005 AMC 10A Problems/Problem 8|Solution]]<br />
<br />
== Problem 9 ==<br />
Three tiles are marked <math>X</math> and two other tiles are marked <math>O</math>. The five tiles are randomly arranged in a row. What is the probability that the arrangement reads <math>XOXOX</math>?<br />
<br />
<math> \mathrm{(A) \ } \frac{1}{12}\qquad \mathrm{(B) \ } \frac{1}{10}\qquad \mathrm{(C) \ } \frac{1}{6}\qquad \mathrm{(D) \ } \frac{1}{4}\qquad \mathrm{(E) \ } \frac{1}{3} </math><br />
<br />
[[2005 AMC 10A Problems/Problem 9|Solution]]<br />
<br />
== Problem 10 ==<br />
There are two values of <math>a</math> for which the equation <math> 4x^2 + ax + 8x + 9 = 0 </math> has only one solution for <math>x</math>. What is the sum of those values of <math>a</math>?<br />
<br />
<math> \mathrm{(A) \ } -16\qquad \mathrm{(B) \ } -8\qquad \mathrm{(C) \ } 0\qquad \mathrm{(D) \ } 8\qquad \mathrm{(E) \ } 20 </math><br />
<br />
[[2005 AMC 10A Problems/Problem 10|Solution]]<br />
<br />
== Problem 11 ==<br />
A wooden cube <math>n</math> units on a side is painted red on all six faces and then cut into <math>n^3</math> unit cubes. Exactly one-fourth of the total number of faces of the unit cubes are red. What is <math>n</math>?<br />
<br />
<math> \mathrm{(A) \ } 3\qquad \mathrm{(B) \ } 4\qquad \mathrm{(C) \ } 5\qquad \mathrm{(D) \ } 6\qquad \mathrm{(E) \ } 7 </math><br />
<br />
[[2005 AMC 10A Problems/Problem 11|Solution]]<br />
<br />
== Problem 12 ==<br />
The figure shown is called a ''trefoil'' and is constructed by drawing circular sectors about the sides of the congruent equilateral triangles. What is the area of a trefoil whose horizontal base has length <math>2</math>?<br />
<br />
[[Image:2005amc10a12.gif]]<br />
<br />
<math> \mathrm{(A) \ } \frac{1}{3}\pi+\frac{\sqrt{3}}{2}\qquad \mathrm{(B) \ } \frac{2}{3}\pi\qquad \mathrm{(C) \ } \frac{2}{3}\pi+\frac{\sqrt{3}}{4}\qquad \mathrm{(D) \ } \frac{2}{3}\pi+\frac{\sqrt{3}}{3}\qquad \mathrm{(E) \ } \frac{2}{3}\pi+\frac{\sqrt{3}}{2} </math><br />
<br />
[[2005 AMC 10A Problems/Problem 12|Solution]]<br />
<br />
== Problem 13 ==<br />
How many positive integers <math>n</math> satisfy the following condition:<br />
<br />
<math> (130n)^{50} > n^{100} > 2^{200} </math>?<br />
<br />
<math> \mathrm{(A) \ } 0\qquad \mathrm{(B) \ } 7\qquad \mathrm{(C) \ } 12\qquad \mathrm{(D) \ } 65\qquad \mathrm{(E) \ } 125 </math><br />
<br />
[[2005 AMC 10A Problems/Problem 13|Solution]]<br />
<br />
== Problem 14 ==<br />
How many three-digit numbers satisfy the property that the middle digit is the average of the first and the last digits? <br />
<br />
<math> \mathrm{(A) \ } 41\qquad \mathrm{(B) \ } 42\qquad \mathrm{(C) \ } 43\qquad \mathrm{(D) \ } 44\qquad \mathrm{(E) \ } 45 </math><br />
<br />
[[2005 AMC 10A Problems/Problem 14|Solution]]<br />
<br />
== Problem 15 ==<br />
How many positive cubes divide <math> 3! \cdot 5! \cdot 7! </math> ?<br />
<br />
<math> \mathrm{(A) \ } 2\qquad \mathrm{(B) \ } 3\qquad \mathrm{(C) \ } 4\qquad \mathrm{(D) \ } 5\qquad \mathrm{(E) \ } 6 </math><br />
<br />
[[2005 AMC 10A Problems/Problem 15|Solution]]<br />
<br />
== Problem 16 ==<br />
The sum of the digits of a two-digit number is subtracted from the number. The units digit of the result is <math>6</math>. How many two-digit numbers have this property? <br />
<br />
<math> \mathrm{(A) \ } 5\qquad \mathrm{(B) \ } 7\qquad \mathrm{(C) \ } 9\qquad \mathrm{(D) \ } 10\qquad \mathrm{(E) \ } 19 </math><br />
<br />
[[2005 AMC 10A Problems/Problem 16|Solution]]<br />
<br />
== Problem 17 ==<br />
In the five-sided star shown, the letters <math>A</math>, <math>B</math>, <math>C</math>, <math>D</math>, and <math>E</math> are replaced by the numbers <math>3</math>, <math>5</math>, <math>6</math>, <math>7</math>, and <math>9</math>, although not necessarily in this order. The sums of the numbers at the ends of the line segments <math>AB</math>, <math>BC</math>, <math>CD</math>, <math>DE</math>, and <math>EA</math> form an arithmetic sequence, although not necessarily in this order. What is the middle term of the sequence? <br />
<br />
<asy><br />
size(150);<br />
defaultpen(linewidth(0.8));<br />
string[] strng = {'A','D','B','E','C'};<br />
pair A=dir(90),B=dir(306),C=dir(162),D=dir(18),E=dir(234);<br />
draw(A--B--C--D--E--cycle);<br />
for(int i=0;i<=4;i=i+1)<br />
{<br />
path circ=circle(dir(90-72*i),0.125);<br />
unfill(circ);<br />
draw(circ);<br />
label("$"+strng[i]+"$",dir(90-72*i));<br />
}<br />
</asy><br />
<br />
<math> \mathrm{(A) \ } 9\qquad \mathrm{(B) \ } 10\qquad \mathrm{(C) \ } 11\qquad \mathrm{(D) \ } 12\qquad \mathrm{(E) \ } 13 </math><br />
<br />
[[2005 AMC 10A Problems/Problem 17|Solution]]<br />
<br />
== Problem 18 ==<br />
Team A and team B play a series. The first team to win three games wins the series. Each team is equally likely to win each game, there are no ties, and the outcomes of the individual games are independent. If team B wins the second game and team A wins the series, what is the probability that team B wins the first game? <br />
<br />
<math> \mathrm{(A) \ } \frac{1}{5}\qquad \mathrm{(B) \ } \frac{1}{4}\qquad \mathrm{(C) \ } \frac{1}{3}\qquad \mathrm{(D) \ } \frac{1}{2}\qquad \mathrm{(E) \ } \frac{2}{3} </math><br />
<br />
[[2005 AMC 10A Problems/Problem 18|Solution]]<br />
<br />
== Problem 19 ==<br />
Three one-inch squares are placed with their bases on a line. The center square is lifted out and rotated 45 degrees, as shown. Then it is centered and lowered into its original location until it touches both of the adjoining squares. How many inches is the point <math>B</math> from the line on which the bases of the original squares were placed?<br />
<br />
<asy><br />
unitsize(1inch);<br />
defaultpen(linewidth(.8pt)+fontsize(8pt));<br />
draw((0,0)--((1/3) + 3*(1/2),0));<br />
fill(((1/6) + (1/2),0)--((1/6) + (1/2),(1/2))--((1/6) + 1,(1/2))--((1/6) + 1,0)--cycle, rgb(.7,.7,.7));<br />
draw(((1/6),0)--((1/6) + (1/2),0)--((1/6) + (1/2),(1/2))--((1/6),(1/2))--cycle);<br />
draw(((1/6) + (1/2),0)--((1/6) + (1/2),(1/2))--((1/6) + 1,(1/2))--((1/6) + 1,0)--cycle);<br />
draw(((1/6) + 1,0)--((1/6) + 1,(1/2))--((1/6) + (3/2),(1/2))--((1/6) + (3/2),0)--cycle);<br />
draw((2,0)--(2 + (1/3) + (3/2),0));<br />
draw(((2/3) + (3/2),0)--((2/3) + 2,0)--((2/3) + 2,(1/2))--((2/3) + (3/2),(1/2))--cycle);<br />
draw(((2/3) + (5/2),0)--((2/3) + (5/2),(1/2))--((2/3) + 3,(1/2))--((2/3) + 3,0)--cycle);<br />
label("$B$",((1/6) + (1/2),(1/2)),NW);<br />
label("$B$",((2/3) + 2 + (1/4),(29/30)),NNE);<br />
draw(((1/6) + (1/2),(1/2)+0.05)..(1,.8)..((2/3) + 2 + (1/4)-.05,(29/30)),EndArrow(HookHead,3));<br />
fill(((2/3) + 2 + (1/4),(1/4))--((2/3) + (5/2) + (1/10),(1/2) + (1/9))--((2/3) + 2 + (1/4),(29/30))--((2/3) + 2 - (1/10),(1/2) + (1/9))--cycle, rgb(.7,.7,.7));<br />
draw(((2/3) + 2 + (1/4),(1/4))--((2/3) + (5/2) + (1/10),(1/2) + (1/9))--((2/3) + 2 + (1/4),(29/30))--((2/3) + 2 - (1/10),(1/2) + (1/9))--cycle);</asy><br />
<br />
<math> \textbf{(A)}\ 1\qquad\textbf{(B)}\ \sqrt{2}\qquad\textbf{(C)}\ \frac{3}{2}\qquad\textbf{(D)}\ \sqrt{2}+\frac{1}{2}\qquad\textbf{(E)}\ 2 </math><br />
<br />
[[2005 AMC 10A Problems/Problem 19|Solution]]<br />
<br />
== Problem 20 ==<br />
An equiangular octagon has four sides of length 1 and four sides of length <math>\frac{\sqrt{2}}{2}</math>, arranged so that no two consecutive sides have the same length. What is the area of the octagon?<br />
<br />
<math> \mathrm{(A) \ } \frac72\qquad \mathrm{(B) \ } \frac{7\sqrt2}{2}\qquad \mathrm{(C) \ } \frac{5+4\sqrt2}{2}\qquad \mathrm{(D) \ } \frac{4+5\sqrt2}{2}\qquad \mathrm{(E) \ } 7 </math><br />
<br />
[[2005 AMC 10A Problems/Problem 20|Solution]]<br />
<br />
== Problem 21 ==<br />
For how many positive integers <math>n</math> does <math> 1+2+...+n </math> evenly divide <math>6n</math>? <br />
<br />
<math> \mathrm{(A) \ } 3\qquad \mathrm{(B) \ } 5\qquad \mathrm{(C) \ } 7\qquad \mathrm{(D) \ } 9\qquad \mathrm{(E) \ } 11 </math><br />
<br />
[[2005 AMC 10A Problems/Problem 21|Solution]]<br />
<br />
== Problem 22 ==<br />
Let <math>S</math> be the set of the <math>2005</math> smallest positive multiples of <math>4</math>, and let <math>T</math> be the set of the <math>2005</math> smallest positive multiples of <math>6</math>. How many elements are common to <math>S</math> and <math>T</math>?<br />
<br />
<math> \mathrm{(A) \ } 166\qquad \mathrm{(B) \ } 333\qquad \mathrm{(C) \ } 500\qquad \mathrm{(D) \ } 668\qquad \mathrm{(E) \ } 1001 </math><br />
<br />
[[2005 AMC 10A Problems/Problem 22|Solution]]<br />
<br />
== Problem 23 ==<br />
Let <math>AB</math> be a diameter of a circle and let <math>C</math> be a point on <math>AB</math> with <math>2\cdot AC=BC</math>. Let <math>D</math> and <math>E</math> be points on the circle such that <math>DC \perp AB</math> and <math>DE</math> is a second diameter. What is the ratio of the area of <math>\triangle DCE</math> to the area of <math>\triangle ABD</math>?<br />
<br />
<asy><br />
unitsize(2.5cm);<br />
defaultpen(fontsize(10pt)+linewidth(.8pt));<br />
dotfactor=3;<br />
pair O=(0,0), C=(-1/3.0), B=(1,0), A=(-1,0);<br />
pair D=dir(aCos(C.x)), E=(-D.x,-D.y);<br />
draw(A--B--D--cycle);<br />
draw(D--E--C);<br />
draw(unitcircle,white);<br />
drawline(D,C);<br />
dot(O);<br />
clip(unitcircle);<br />
draw(unitcircle);<br />
label("$E$",E,SSE);<br />
label("$B$",B,E);<br />
label("$A$",A,W);<br />
label("$D$",D,NNW);<br />
label("$C$",C,SW);<br />
draw(rightanglemark(D,C,B,2));</asy><br />
<br />
<math> \mathrm{(A) \ } \frac{1}{6} \qquad \mathrm{(B) \ } \frac{1}{4} \qquad \mathrm{(C) \ } \frac{1}{3} \qquad \mathrm{(D) \ } \frac{1}{2} \qquad \mathrm{(E) \ } \frac{2}{3} </math><br />
<br />
[[2005 AMC 10A Problems/Problem 23|Solution]]<br />
<br />
== Problem 24 ==<br />
For each positive integer <math> n > 1 </math>, let <math>P(n)</math> denote the greatest prime factor of <math>n</math>. For how many positive integers <math>n</math> is it true that both <math> P(n) = \sqrt{n} </math> and <math> P(n+48) = \sqrt{n+48} </math>?<br />
<br />
<math> \mathrm{(A) \ } 0\qquad \mathrm{(B) \ } 1\qquad \mathrm{(C) \ } 3\qquad \mathrm{(D) \ } 4\qquad \mathrm{(E) \ } 5 </math><br />
<br />
[[2005 AMC 10A Problems/Problem 24|Solution]]<br />
<br />
== Problem 25 ==<br />
In <math>ABC</math> we have <math> AB = 25 </math>, <math> BC = 39 </math>, and <math>AC=42</math>. Points <math>D</math> and <math>E</math> are on <math>AB</math> and <math>AC</math> respectively, with <math> AD = 19 </math> and <math> AE = 14 </math>. What is the ratio of the area of triangle <math>ADE</math> to the area of the quadrilateral <math>BCED</math>?<br />
<br />
<math> \mathrm{(A) \ } \frac{266}{1521}\qquad \mathrm{(B) \ } \frac{19}{75}\qquad \mathrm{(C) \ } \frac{1}{3}\qquad \mathrm{(D) \ } \frac{19}{56}\qquad \mathrm{(E) \ } 1 </math><br />
<br />
[[2005 AMC 10A Problems/Problem 25|Solution]]<br />
<br />
== See also ==<br />
{{AMC10 box|year=2005|ab=A|before=[[2004 AMC 10B Problems]]|after=[[2005 AMC 10B Problems]]}}<br />
*[[AMC 10 Problems and Solutions]]<br />
* [[AMC Problems and Solutions]]<br />
{{MAA Notice}}</div>Ghostychttps://artofproblemsolving.com/wiki/index.php?title=2005_AMC_10A_Problems&diff=1582312005 AMC 10A Problems2021-07-12T15:41:10Z<p>Ghostyc: /* Problem 8 */</p>
<hr />
<div>{{AMC10 Problems|year=2005|ab=A}}<br />
== Problem 1 ==<br />
While eating out, Mike and Joe each tipped their server <math>2</math> dollars. Mike tipped <math>10\%</math> of his bill and Joe tipped <math>20\%</math> of his bill. What was the difference, in dollars between their bills? <br />
<br />
<math> \mathrm{(A) \ } 2\qquad \mathrm{(B) \ } 4\qquad \mathrm{(C) \ } 5\qquad \mathrm{(D) \ } 10\qquad \mathrm{(E) \ } 20 </math><br />
<br />
[[2005 AMC 10A Problems/Problem 1|Solution]]<br />
<br />
== Problem 2 ==<br />
For each pair of real numbers <math>a \neq b</math>, define the [[operation]] <math>\star</math> as<br />
<br />
<math> (a \star b) = \frac{a+b}{a-b} </math>.<br />
<br />
What is the value of <math> ((1 \star 2) \star 3)</math>?<br />
<br />
<math> \mathrm{(A) \ } -\frac{2}{3}\qquad \mathrm{(B) \ } -\frac{1}{5}\qquad \mathrm{(C) \ } 0\qquad \mathrm{(D) \ } \frac{1}{2}\qquad \mathrm{(E) \ } \textrm{This\, value\, is\, not\, defined.} </math><br />
<br />
[[2005 AMC 10A Problems/Problem 2|Solution]]<br />
<br />
== Problem 3 ==<br />
The equations <math> 2x + 7 = 3 </math> and <math> bx - 10 = -2 </math> have the same solution <math>x</math>. What is the value of <math>b</math>? <br />
<br />
<math> \mathrm{(A) \ } -8\qquad \mathrm{(B) \ } -4\qquad \mathrm{(C) \ } -2\qquad \mathrm{(D) \ } 4\qquad \mathrm{(E) \ } 8 </math><br />
<br />
[[2005 AMC 10A Problems/Problem 3|Solution]]<br />
<br />
== Problem 4 ==<br />
A rectangle with a diagonal of length <math>x</math> is twice as long as it is wide. What is the area of the rectangle? <br />
<br />
<math> \mathrm{(A) \ } \frac{1}{4}x^2\qquad \mathrm{(B) \ } \frac{2}{5}x^2\qquad \mathrm{(C) \ } \frac{1}{2}x^2\qquad \mathrm{(D) \ } x^2\qquad \mathrm{(E) \ } \frac{3}{2}x^2 </math><br />
<br />
[[2005 AMC 10A Problems/Problem 4|Solution]]<br />
<br />
== Problem 5 ==<br />
A store normally sells windows at \$100 each. This week the store is offering one free window for each purchase of four. Dave needs seven windows and Doug needs eight windows. How many dollars will they save if they purchase the windows together rather than separately?<br />
<br />
<math> \mathrm{(A) \ } 100\qquad \mathrm{(B) \ } 200\qquad \mathrm{(C) \ } 300\qquad \mathrm{(D) \ } 400\qquad \mathrm{(E) \ } 500 </math><br />
<br />
[[2005 AMC 10A Problems/Problem 5|Solution]]<br />
<br />
== Problem 6 ==<br />
The average (mean) of <math>20</math> numbers is <math>30</math>, and the average of <math>30</math> other numbers is <math>20</math>. What is the average of all <math>50</math> numbers?<br />
<br />
<math> \mathrm{(A) \ } 23\qquad \mathrm{(B) \ } 24\qquad \mathrm{(C) \ } 25\qquad \mathrm{(D) \ } 26\qquad \mathrm{(E) \ } 27 </math><br />
<br />
[[2005 AMC 10A Problems/Problem 6|Solution]]<br />
<br />
== Problem 7 ==<br />
Josh and Mike live <math>13</math> miles apart. Yesterday Josh started to ride his bicycle toward Mike's house. A little later Mike started to ride his bicycle toward Josh's house. When they met, Josh had ridden for twice the length of time as Mike and at four-fifths of Mike's rate. How many miles had Mike ridden when they met? <br />
<br />
<math> \mathrm{(A) \ } 4\qquad \mathrm{(B) \ } 5\qquad \mathrm{(C) \ } 6\qquad \mathrm{(D) \ } 7\qquad \mathrm{(E) \ } 8 </math><br />
<br />
[[2005 AMC 10A Problems/Problem 7|Solution]]<br />
<br />
== Problem 8 ==<br />
In the figure, the length of side <math>AB</math> of square <math>ABCD</math> is <math>\sqrt{50}</math> and <math>BE=1</math>. What is the area of the inner square <math>EFGH</math>?<br />
<br />
<asy><br />
unitsize(4cm);<br />
defaultpen(linewidth(.8pt)+fontsize(10pt));<br />
<br />
pair D=(0,0), C=(1,0), B=(1,1), A=(0,1);<br />
pair F=intersectionpoints(Circle(D,2/sqrt(5)),Circle(A,1))[0];<br />
pair G=foot(A,D,F), H=foot(B,A,G), E=foot(C,B,H);<br />
<br />
draw(A--B--C--D--cycle);<br />
draw(D--F);<br />
draw(C--E);<br />
draw(B--H);<br />
draw(A--G);<br />
<br />
label("$A$",A,NW);<br />
label("$B$",B,NE);<br />
label("$C$",C,SE);<br />
label("$D$",D,SW);<br />
label("$E$",E,NNW);<br />
label("$F$",F,ENE);<br />
label("$G$",G,SSE);<br />
label("$H$",H,WSW);<br />
</asy><br />
<br />
<math> \textbf{(A)}\ 25\qquad\textbf{(B)}\ 32\qquad\textbf{(C)}\ 36\qquad\textbf{(D)}\ 40\qquad\textbf{(E)}\ 42 </math><br />
<br />
[[2005 AMC 10A Problems/Problem 8|Solution]]<br />
<br />
== Problem 9 ==<br />
Three tiles are marked <math>X</math> and two other tiles are marked <math>O</math>. The five tiles are randomly arranged in a row. What is the probability that the arrangement reads <math>XOXOX</math>?<br />
<br />
<math> \mathrm{(A) \ } \frac{1}{12}\qquad \mathrm{(B) \ } \frac{1}{10}\qquad \mathrm{(C) \ } \frac{1}{6}\qquad \mathrm{(D) \ } \frac{1}{4}\qquad \mathrm{(E) \ } \frac{1}{3} </math><br />
<br />
[[2005 AMC 10A Problems/Problem 9|Solution]]<br />
<br />
== Problem 10 ==<br />
There are two values of <math>a</math> for which the equation <math> 4x^2 + ax + 8x + 9 = 0 </math> has only one solution for <math>x</math>. What is the sum of those values of <math>a</math>?<br />
<br />
<math> \mathrm{(A) \ } -16\qquad \mathrm{(B) \ } -8\qquad \mathrm{(C) \ } 0\qquad \mathrm{(D) \ } 8\qquad \mathrm{(E) \ } 20 </math><br />
<br />
[[2005 AMC 10A Problems/Problem 10|Solution]]<br />
<br />
== Problem 11 ==<br />
A wooden cube <math>n</math> units on a side is painted red on all six faces and then cut into <math>n^3</math> unit cubes. Exactly one-fourth of the total number of faces of the unit cubes are red. What is <math>n</math>?<br />
<br />
<math> \mathrm{(A) \ } 3\qquad \mathrm{(B) \ } 4\qquad \mathrm{(C) \ } 5\qquad \mathrm{(D) \ } 6\qquad \mathrm{(E) \ } 7 </math><br />
<br />
[[2005 AMC 10A Problems/Problem 11|Solution]]<br />
<br />
== Problem 12 ==<br />
The figure shown is called a ''trefoil'' and is constructed by drawing circular sectors about the sides of the congruent equilateral triangles. What is the area of a trefoil whose horizontal base has length <math>2</math>?<br />
<br />
[[Image:2005amc10a12.gif]]<br />
<br />
<math> \mathrm{(A) \ } \frac{1}{3}\pi+\frac{\sqrt{3}}{2}\qquad \mathrm{(B) \ } \frac{2}{3}\pi\qquad \mathrm{(C) \ } \frac{2}{3}\pi+\frac{\sqrt{3}}{4}\qquad \mathrm{(D) \ } \frac{2}{3}\pi+\frac{\sqrt{3}}{3}\qquad \mathrm{(E) \ } \frac{2}{3}\pi+\frac{\sqrt{3}}{2} </math><br />
<br />
[[2005 AMC 10A Problems/Problem 12|Solution]]<br />
<br />
== Problem 13 ==<br />
How many positive integers <math>n</math> satisfy the following condition:<br />
<br />
<math> (130n)^{50} > n^{100} > 2^{200} </math>?<br />
<br />
<math> \mathrm{(A) \ } 0\qquad \mathrm{(B) \ } 7\qquad \mathrm{(C) \ } 12\qquad \mathrm{(D) \ } 65\qquad \mathrm{(E) \ } 125 </math><br />
<br />
[[2005 AMC 10A Problems/Problem 13|Solution]]<br />
<br />
== Problem 14 ==<br />
How many three-digit numbers satisfy the property that the middle digit is the average of the first and the last digits? <br />
<br />
<math> \mathrm{(A) \ } 41\qquad \mathrm{(B) \ } 42\qquad \mathrm{(C) \ } 43\qquad \mathrm{(D) \ } 44\qquad \mathrm{(E) \ } 45 </math><br />
<br />
[[2005 AMC 10A Problems/Problem 14|Solution]]<br />
<br />
== Problem 15 ==<br />
How many positive cubes divide <math> 3! \cdot 5! \cdot 7! </math> ?<br />
<br />
<math> \mathrm{(A) \ } 2\qquad \mathrm{(B) \ } 3\qquad \mathrm{(C) \ } 4\qquad \mathrm{(D) \ } 5\qquad \mathrm{(E) \ } 6 </math><br />
<br />
[[2005 AMC 10A Problems/Problem 15|Solution]]<br />
<br />
== Problem 16 ==<br />
The sum of the digits of a two-digit number is subtracted from the number. The units digit of the result is <math>6</math>. How many two-digit numbers have this property? <br />
<br />
<math> \mathrm{(A) \ } 5\qquad \mathrm{(B) \ } 7\qquad \mathrm{(C) \ } 9\qquad \mathrm{(D) \ } 10\qquad \mathrm{(E) \ } 19 </math><br />
<br />
[[2005 AMC 10A Problems/Problem 16|Solution]]<br />
<br />
== Problem 17 ==<br />
In the five-sided star shown, the letters <math>A</math>, <math>B</math>, <math>C</math>, <math>D</math>, and <math>E</math> are replaced by the numbers <math>3</math>, <math>5</math>, <math>6</math>, <math>7</math>, and <math>9</math>, although not necessarily in this order. The sums of the numbers at the ends of the line segments <math>AB</math>, <math>BC</math>, <math>CD</math>, <math>DE</math>, and <math>EA</math> form an arithmetic sequence, although not necessarily in this order. What is the middle term of the sequence? <br />
<br />
[[Image:2005amc10a17.gif]]<br />
<br />
<math> \mathrm{(A) \ } 9\qquad \mathrm{(B) \ } 10\qquad \mathrm{(C) \ } 11\qquad \mathrm{(D) \ } 12\qquad \mathrm{(E) \ } 13 </math><br />
<br />
[[2005 AMC 10A Problems/Problem 17|Solution]]<br />
<br />
== Problem 18 ==<br />
Team A and team B play a series. The first team to win three games wins the series. Each team is equally likely to win each game, there are no ties, and the outcomes of the individual games are independent. If team B wins the second game and team A wins the series, what is the probability that team B wins the first game? <br />
<br />
<math> \mathrm{(A) \ } \frac{1}{5}\qquad \mathrm{(B) \ } \frac{1}{4}\qquad \mathrm{(C) \ } \frac{1}{3}\qquad \mathrm{(D) \ } \frac{1}{2}\qquad \mathrm{(E) \ } \frac{2}{3} </math><br />
<br />
[[2005 AMC 10A Problems/Problem 18|Solution]]<br />
<br />
== Problem 19 ==<br />
Three one-inch squares are placed with their bases on a line. The center square is lifted out and rotated 45 degrees, as shown. Then it is centered and lowered into its original location until it touches both of the adjoining squares. How many inches is the point <math>B</math> from the line on which the bases of the original squares were placed?<br />
<br />
<asy><br />
unitsize(1inch);<br />
defaultpen(linewidth(.8pt)+fontsize(8pt));<br />
draw((0,0)--((1/3) + 3*(1/2),0));<br />
fill(((1/6) + (1/2),0)--((1/6) + (1/2),(1/2))--((1/6) + 1,(1/2))--((1/6) + 1,0)--cycle, rgb(.7,.7,.7));<br />
draw(((1/6),0)--((1/6) + (1/2),0)--((1/6) + (1/2),(1/2))--((1/6),(1/2))--cycle);<br />
draw(((1/6) + (1/2),0)--((1/6) + (1/2),(1/2))--((1/6) + 1,(1/2))--((1/6) + 1,0)--cycle);<br />
draw(((1/6) + 1,0)--((1/6) + 1,(1/2))--((1/6) + (3/2),(1/2))--((1/6) + (3/2),0)--cycle);<br />
draw((2,0)--(2 + (1/3) + (3/2),0));<br />
draw(((2/3) + (3/2),0)--((2/3) + 2,0)--((2/3) + 2,(1/2))--((2/3) + (3/2),(1/2))--cycle);<br />
draw(((2/3) + (5/2),0)--((2/3) + (5/2),(1/2))--((2/3) + 3,(1/2))--((2/3) + 3,0)--cycle);<br />
label("$B$",((1/6) + (1/2),(1/2)),NW);<br />
label("$B$",((2/3) + 2 + (1/4),(29/30)),NNE);<br />
draw(((1/6) + (1/2),(1/2)+0.05)..(1,.8)..((2/3) + 2 + (1/4)-.05,(29/30)),EndArrow(HookHead,3));<br />
fill(((2/3) + 2 + (1/4),(1/4))--((2/3) + (5/2) + (1/10),(1/2) + (1/9))--((2/3) + 2 + (1/4),(29/30))--((2/3) + 2 - (1/10),(1/2) + (1/9))--cycle, rgb(.7,.7,.7));<br />
draw(((2/3) + 2 + (1/4),(1/4))--((2/3) + (5/2) + (1/10),(1/2) + (1/9))--((2/3) + 2 + (1/4),(29/30))--((2/3) + 2 - (1/10),(1/2) + (1/9))--cycle);</asy><br />
<br />
<math> \textbf{(A)}\ 1\qquad\textbf{(B)}\ \sqrt{2}\qquad\textbf{(C)}\ \frac{3}{2}\qquad\textbf{(D)}\ \sqrt{2}+\frac{1}{2}\qquad\textbf{(E)}\ 2 </math><br />
<br />
[[2005 AMC 10A Problems/Problem 19|Solution]]<br />
<br />
== Problem 20 ==<br />
An equiangular octagon has four sides of length 1 and four sides of length <math>\frac{\sqrt{2}}{2}</math>, arranged so that no two consecutive sides have the same length. What is the area of the octagon?<br />
<br />
<math> \mathrm{(A) \ } \frac72\qquad \mathrm{(B) \ } \frac{7\sqrt2}{2}\qquad \mathrm{(C) \ } \frac{5+4\sqrt2}{2}\qquad \mathrm{(D) \ } \frac{4+5\sqrt2}{2}\qquad \mathrm{(E) \ } 7 </math><br />
<br />
[[2005 AMC 10A Problems/Problem 20|Solution]]<br />
<br />
== Problem 21 ==<br />
For how many positive integers <math>n</math> does <math> 1+2+...+n </math> evenly divide <math>6n</math>? <br />
<br />
<math> \mathrm{(A) \ } 3\qquad \mathrm{(B) \ } 5\qquad \mathrm{(C) \ } 7\qquad \mathrm{(D) \ } 9\qquad \mathrm{(E) \ } 11 </math><br />
<br />
[[2005 AMC 10A Problems/Problem 21|Solution]]<br />
<br />
== Problem 22 ==<br />
Let <math>S</math> be the set of the <math>2005</math> smallest positive multiples of <math>4</math>, and let <math>T</math> be the set of the <math>2005</math> smallest positive multiples of <math>6</math>. How many elements are common to <math>S</math> and <math>T</math>?<br />
<br />
<math> \mathrm{(A) \ } 166\qquad \mathrm{(B) \ } 333\qquad \mathrm{(C) \ } 500\qquad \mathrm{(D) \ } 668\qquad \mathrm{(E) \ } 1001 </math><br />
<br />
[[2005 AMC 10A Problems/Problem 22|Solution]]<br />
<br />
== Problem 23 ==<br />
Let <math>AB</math> be a diameter of a circle and let <math>C</math> be a point on <math>AB</math> with <math>2\cdot AC=BC</math>. Let <math>D</math> and <math>E</math> be points on the circle such that <math>DC \perp AB</math> and <math>DE</math> is a second diameter. What is the ratio of the area of <math>\triangle DCE</math> to the area of <math>\triangle ABD</math>?<br />
<br />
<asy><br />
unitsize(2.5cm);<br />
defaultpen(fontsize(10pt)+linewidth(.8pt));<br />
dotfactor=3;<br />
pair O=(0,0), C=(-1/3.0), B=(1,0), A=(-1,0);<br />
pair D=dir(aCos(C.x)), E=(-D.x,-D.y);<br />
draw(A--B--D--cycle);<br />
draw(D--E--C);<br />
draw(unitcircle,white);<br />
drawline(D,C);<br />
dot(O);<br />
clip(unitcircle);<br />
draw(unitcircle);<br />
label("$E$",E,SSE);<br />
label("$B$",B,E);<br />
label("$A$",A,W);<br />
label("$D$",D,NNW);<br />
label("$C$",C,SW);<br />
draw(rightanglemark(D,C,B,2));</asy><br />
<br />
<math> \mathrm{(A) \ } \frac{1}{6} \qquad \mathrm{(B) \ } \frac{1}{4} \qquad \mathrm{(C) \ } \frac{1}{3} \qquad \mathrm{(D) \ } \frac{1}{2} \qquad \mathrm{(E) \ } \frac{2}{3} </math><br />
<br />
[[2005 AMC 10A Problems/Problem 23|Solution]]<br />
<br />
== Problem 24 ==<br />
For each positive integer <math> n > 1 </math>, let <math>P(n)</math> denote the greatest prime factor of <math>n</math>. For how many positive integers <math>n</math> is it true that both <math> P(n) = \sqrt{n} </math> and <math> P(n+48) = \sqrt{n+48} </math>?<br />
<br />
<math> \mathrm{(A) \ } 0\qquad \mathrm{(B) \ } 1\qquad \mathrm{(C) \ } 3\qquad \mathrm{(D) \ } 4\qquad \mathrm{(E) \ } 5 </math><br />
<br />
[[2005 AMC 10A Problems/Problem 24|Solution]]<br />
<br />
== Problem 25 ==<br />
In <math>ABC</math> we have <math> AB = 25 </math>, <math> BC = 39 </math>, and <math>AC=42</math>. Points <math>D</math> and <math>E</math> are on <math>AB</math> and <math>AC</math> respectively, with <math> AD = 19 </math> and <math> AE = 14 </math>. What is the ratio of the area of triangle <math>ADE</math> to the area of the quadrilateral <math>BCED</math>?<br />
<br />
<math> \mathrm{(A) \ } \frac{266}{1521}\qquad \mathrm{(B) \ } \frac{19}{75}\qquad \mathrm{(C) \ } \frac{1}{3}\qquad \mathrm{(D) \ } \frac{19}{56}\qquad \mathrm{(E) \ } 1 </math><br />
<br />
[[2005 AMC 10A Problems/Problem 25|Solution]]<br />
<br />
== See also ==<br />
{{AMC10 box|year=2005|ab=A|before=[[2004 AMC 10B Problems]]|after=[[2005 AMC 10B Problems]]}}<br />
*[[AMC 10 Problems and Solutions]]<br />
* [[AMC Problems and Solutions]]<br />
{{MAA Notice}}</div>Ghostychttps://artofproblemsolving.com/wiki/index.php?title=2005_AMC_10A_Problems/Problem_12&diff=1582302005 AMC 10A Problems/Problem 122021-07-12T15:39:23Z<p>Ghostyc: /* Problem */</p>
<hr />
<div>==Problem==<br />
The figure shown is called a ''trefoil'' and is constructed by drawing circular sectors about the sides of the congruent equilateral triangles. What is the area of a trefoil whose horizontal base has length <math>2</math>?<br />
<br />
<asy><br />
unitsize(1.5cm);<br />
defaultpen(linewidth(.8pt)+fontsize(12pt));<br />
<br />
pair O=(0,0), A=dir(0), B=dir(60), C=dir(120), D=dir(180);<br />
pair E=B+C;<br />
<br />
draw(D--E--B--O--C--B--A,linetype("4 4"));<br />
draw(Arc(O,1,0,60),linewidth(1.2pt));<br />
draw(Arc(O,1,120,180),linewidth(1.2pt));<br />
draw(Arc(C,1,0,60),linewidth(1.2pt));<br />
draw(Arc(B,1,120,180),linewidth(1.2pt));<br />
draw(A--D,linewidth(1.2pt));<br />
draw(O--dir(40),EndArrow(HookHead,4));<br />
draw(O--dir(140),EndArrow(HookHead,4));<br />
draw(C--C+dir(40),EndArrow(HookHead,4));<br />
draw(B--B+dir(140),EndArrow(HookHead,4));<br />
<br />
label("2",O,S);<br />
draw((0.1,-0.12)--(1,-0.12),EndArrow(HookHead,4),EndBar);<br />
draw((-0.1,-0.12)--(-1,-0.12),EndArrow(HookHead,4),EndBar);<br />
</asy><br />
<br />
<math> \mathrm{(A) \ } \frac{1}{3}\pi+\frac{\sqrt{3}}{2}\qquad \mathrm{(B) \ } \frac{2}{3}\pi\qquad \mathrm{(C) \ } \frac{2}{3}\pi+\frac{\sqrt{3}}{4}\qquad \mathrm{(D) \ } \frac{2}{3}\pi+\frac{\sqrt{3}}{3}\qquad \mathrm{(E) \ } \frac{2}{3}\pi+\frac{\sqrt{3}}{2} </math><br />
<br />
==Solution==<br />
The area of the ''trefoil'' is equal to the area of a small equilateral triangle plus the area of four <math>60^\circ</math> sectors with a radius of <math>\frac{2}{2}=1</math> minus the area of a small equilateral triangle. <br />
<br />
This is equivalent to the area of four <math>60^\circ</math> sectors with a radius of <math>1</math>. <br />
<br />
So the answer is: <br />
<br />
<math>4\cdot\frac{60}{360}\cdot\pi\cdot1^2 = \frac{4}{6}\cdot\pi = \frac{2}{3}\pi \Rightarrow B </math><br />
<br />
==See also==<br />
{{AMC10 box|year=2005|ab=A|num-b=11|num-a=13}}<br />
<br />
{{MAA Notice}}</div>Ghostychttps://artofproblemsolving.com/wiki/index.php?title=2005_AMC_10A_Problems/Problem_12&diff=1582292005 AMC 10A Problems/Problem 122021-07-12T15:38:22Z<p>Ghostyc: /* Problem */</p>
<hr />
<div>==Problem==<br />
The figure shown is called a ''trefoil'' and is constructed by drawing circular sectors about the sides of the congruent equilateral triangles. What is the area of a trefoil whose horizontal base has length <math>2</math>?<br />
<br />
[asy]<br />
unitsize(1.5cm);<br />
defaultpen(linewidth(.8pt)+fontsize(12pt));<br />
<br />
pair O=(0,0), A=dir(0), B=dir(60), C=dir(120), D=dir(180);<br />
pair E=B+C;<br />
<br />
draw(D--E--B--O--C--B--A,linetype("4 4"));<br />
draw(Arc(O,1,0,60),linewidth(1.2pt));<br />
draw(Arc(O,1,120,180),linewidth(1.2pt));<br />
draw(Arc(C,1,0,60),linewidth(1.2pt));<br />
draw(Arc(B,1,120,180),linewidth(1.2pt));<br />
draw(A--D,linewidth(1.2pt));<br />
draw(O--dir(40),EndArrow(HookHead,4));<br />
draw(O--dir(140),EndArrow(HookHead,4));<br />
draw(C--C+dir(40),EndArrow(HookHead,4));<br />
draw(B--B+dir(140),EndArrow(HookHead,4));<br />
<br />
label("2",O,S);<br />
draw((0.1,-0.12)--(1,-0.12),EndArrow(HookHead,4),EndBar);<br />
draw((-0.1,-0.12)--(-1,-0.12),EndArrow(HookHead,4),EndBar);<br />
[/asy]<br />
<br />
<math> \mathrm{(A) \ } \frac{1}{3}\pi+\frac{\sqrt{3}}{2}\qquad \mathrm{(B) \ } \frac{2}{3}\pi\qquad \mathrm{(C) \ } \frac{2}{3}\pi+\frac{\sqrt{3}}{4}\qquad \mathrm{(D) \ } \frac{2}{3}\pi+\frac{\sqrt{3}}{3}\qquad \mathrm{(E) \ } \frac{2}{3}\pi+\frac{\sqrt{3}}{2} </math><br />
<br />
==Solution==<br />
The area of the ''trefoil'' is equal to the area of a small equilateral triangle plus the area of four <math>60^\circ</math> sectors with a radius of <math>\frac{2}{2}=1</math> minus the area of a small equilateral triangle. <br />
<br />
This is equivalent to the area of four <math>60^\circ</math> sectors with a radius of <math>1</math>. <br />
<br />
So the answer is: <br />
<br />
<math>4\cdot\frac{60}{360}\cdot\pi\cdot1^2 = \frac{4}{6}\cdot\pi = \frac{2}{3}\pi \Rightarrow B </math><br />
<br />
==See also==<br />
{{AMC10 box|year=2005|ab=A|num-b=11|num-a=13}}<br />
<br />
{{MAA Notice}}</div>Ghostychttps://artofproblemsolving.com/wiki/index.php?title=2021_AMC_10A_Problems&diff=1554132021 AMC 10A Problems2021-06-07T03:22:40Z<p>Ghostyc: /* Problem 3 */</p>
<hr />
<div>{{AMC10 Problems|year=2021|ab=A}}<br />
<br />
==Problem 1==<br />
What is the value of<cmath>(2^2-2)-(3^2-3)+(4^2-4)?</cmath><math>\textbf{(A)} ~1 \qquad\textbf{(B)} ~2 \qquad\textbf{(C)} ~5 \qquad\textbf{(D)} ~8 \qquad\textbf{(E)} ~12 </math><br />
<br />
[[2021 AMC 10A Problems/Problem 1|Solution]]<br />
<br />
==Problem 2==<br />
Portia's high school has <math>3</math> times as many students as Lara's high school. The two high schools have a total of <math>2600</math> students. How many students does Portia's high school have?<br />
<br />
<math>\textbf{(A)} ~600 \qquad\textbf{(B)} ~650 \qquad\textbf{(C)} ~1950 \qquad\textbf{(D)} ~2000\qquad\textbf{(E)} ~2050</math><br />
<br />
[[2021 AMC 10A Problems/Problem 2|Solution]]<br />
<br />
==Problem 3==<br />
The sum of two natural numbers is <math>17{,}402</math>. One of the two numbers is divisible by <math>10</math>. If the units digit of that number is erased, the other number is obtained. What is the difference of these two numbers?<br />
<br />
<math>\textbf{(A)} ~10{,}272\qquad\textbf{(B)} ~11{,}700\qquad\textbf{(C)} ~13{,}362\qquad\textbf{(D)} ~14{,}238\qquad\textbf{(E)} ~15{,}426</math><br />
<br />
[[2021 AMC 10A Problems/Problem 3|Solution]]<br />
<br />
==Problem 4==<br />
A cart rolls down a hill, travelling <math>5</math> inches the first second and accelerating so that during each successive <math>1</math>-second time interval, it travels <math>7</math> inches more than during the previous <math>1</math>-second interval. The cart takes <math>30</math> seconds to reach the bottom of the hill. How far, in inches, does it travel?<br />
<br />
<math>\textbf{(A)} ~215 \qquad\textbf{(B)} ~360\qquad\textbf{(C)} ~2992\qquad\textbf{(D)} ~3195\qquad\textbf{(E)} ~3242</math><br />
<br />
[[2021 AMC 10A Problems/Problem 4|Solution]]<br />
<br />
==Problem 5==<br />
The quiz scores of a class with <math>k > 12</math> students have a mean of <math>8</math>. The mean of a collection of <math>12</math> of these quiz scores is <math>14</math>. What is the mean of the remaining quiz scores in terms of <math>k</math>?<br />
<br />
<math>\textbf{(A)} ~\frac{14-8}{k-12} \qquad\textbf{(B)} ~\frac{8k-168}{k-12} \qquad\textbf{(C)} ~\frac{14}{12} - \frac{8}{k} \qquad\textbf{(D)} ~\frac{14(k-12)}{k^2} \qquad\textbf{(E)} ~\frac{14(k-12)}{8k}</math><br />
<br />
[[2021 AMC 10A Problems/Problem 5|Solution]]<br />
<br />
==Problem 6==<br />
Chantal and Jean start hiking from a trailhead toward a fire tower. Jean is wearing a heavy backpack and walks slower. Chantal starts walking at <math>4</math> miles per hour. Halfway to the tower, the trail becomes really steep, and Chantal slows down to <math>2</math> miles per hour. After reaching the tower, she immediately turns around and descends the steep part of the trail at <math>3</math> miles per hour. She meets Jean at the halfway point. What was Jean's average speed, in miles per hour, until they meet?<br />
<br />
<math>\textbf{(A)} ~\frac{12}{13} \qquad\textbf{(B)} ~1 \qquad\textbf{(C)} ~\frac{13}{12} \qquad\textbf{(D)} <br />
~\frac{24}{13} \qquad\textbf{(E)} ~2</math><br />
<br />
[[2021 AMC 10A Problems/Problem 6|Solution]]<br />
==Problem 7==<br />
Tom has a collection of <math>13</math> snakes, <math>4</math> of which are purple and <math>5</math> of which are happy. He observes that<br />
<br />
* all of his happy snakes can add,<br />
<br />
* none of his purple snakes can subtract, and<br />
<br />
* all of his snakes that can't subtract also can't add.<br />
<br />
Which of these conclusions can be drawn about Tom's snakes?<br />
<br />
<math>\textbf{(A) }</math> Purple snakes can add.<br />
<br />
<math>\textbf{(B) }</math> Purple snakes are happy.<br />
<br />
<math>\textbf{(C) }</math> Snakes that can add are purple.<br />
<br />
<math>\textbf{(D) }</math> Happy snakes are not purple.<br />
<br />
<math>\textbf{(E) }</math> Happy snakes can't subtract.<br />
<br />
[[2021 AMC 10A Problems/Problem 7|Solution]]<br />
<br />
==Problem 8==<br />
When a student multiplied the number <math>66</math> by the repeating decimal<br />
<cmath>\underline{1}.\underline{a} \underline{b} \underline{a} \underline{b} \cdots = <br />
\underline{1}.\overline{\underline{ab}}</cmath>Where <math>a</math> and <math>b</math> are digits. He did not notice the notation and just multiplied <math>66</math> times <math>\underline{1}.\underline{a}\underline{b}</math>. Later he found that his answer is <math>0.5</math> less than the correct answer. What is the <math>2</math>-digit integer <math>\underline{ab}</math>?<br />
<br />
<math>\textbf{(A)} ~15\qquad\textbf{(B)} ~30\qquad\textbf{(C)} ~45\qquad\textbf{(D)} ~60\qquad\textbf{(E)} ~75</math><br />
<br />
[[2021 AMC 10A Problems/Problem 8|Solution]]<br />
==Problem 9==<br />
What is the least possible value of <math>(xy-1)^2 + (x+y)^2</math> for real numbers <math>x</math> and <math>y</math>?<br />
<br />
<math>\textbf{(A)} ~0\qquad\textbf{(B)} ~\frac{1}{4}\qquad\textbf{(C)} ~\frac{1}{2}\qquad\textbf{(D)} ~1\qquad\textbf{(E)} ~2</math><br />
<br />
[[2021 AMC 10A Problems/Problem 9|Solution]]<br />
<br />
==Problem 10==<br />
Which of the following is equivalent to<br />
<cmath>(2+3)(2^2+3^2)(2^4+3^4)(2^8+3^8)(2^{16}+3^{16})(2^{32}+3^{32})(2^{64}+3^{64})?</cmath><br />
<math>\textbf{(A)} ~3^{127} + 2^{127} \qquad\textbf{(B)} ~3^{127} + 2^{127} + 2 \cdot 3^{63} + 3 \cdot 2^{63} \qquad\textbf{(C)} ~3^{128}-2^{128} \qquad\textbf{(D)} ~3^{128} + 3^{128} \qquad\textbf{(E)} ~5^{127}</math><br />
<br />
[[2021 AMC 10A Problems/Problem 10|Solution]]<br />
<br />
==Problem 11==<br />
For which of the following integers <math>b</math> is the base-<math>b</math> number <math>2021_b - 221_b</math> not divisible by <math>3</math>?<br />
<br />
<math>\textbf{(A)} ~3 \qquad\textbf{(B)} ~4\qquad\textbf{(C)} ~6\qquad\textbf{(D)} ~7\qquad\textbf{(E)} ~8</math><br />
<br />
[[2021 AMC 10A Problems/Problem 11|Solution]]<br />
<br />
==Problem 12==<br />
<br />
Two right circular cones with vertices facing down as shown in the figure below contain the same amount of liquid. The radii of the tops of the liquid surfaces are <math>3 \text{ cm}</math> and <math>6 \text{ cm}</math>. Into each cone is dropped a spherical marble of radius <math>1 \text{ cm}</math>, which sinks to the bottom and is completely submerged without spilling any liquid. What is the ratio of the rise of the liquid level in the narrow cone to the rise of the liquid level in the wide cone? <br />
<br />
<asy><br />
size(350);<br />
defaultpen(linewidth(0.8));<br />
real h1 = 10, r = 3.1, s=0.75;<br />
pair P = (r,h1), Q = (-r,h1), Pp = s * P, Qp = s * Q;<br />
path e = ellipse((0,h1),r,0.9), ep = ellipse((0,h1*s),r*s,0.9);<br />
draw(ellipse(origin,r*(s-0.1),0.8));<br />
fill(ep,gray(0.8));<br />
fill(origin--Pp--Qp--cycle,gray(0.8));<br />
draw((-r,h1)--(0,0)--(r,h1)^^e);<br />
draw(subpath(ep,0,reltime(ep,0.5)),linetype("4 4"));<br />
draw(subpath(ep,reltime(ep,0.5),reltime(ep,1)));<br />
draw(Qp--(0,Qp.y),Arrows(size=8));<br />
draw(origin--(0,12),linetype("4 4"));<br />
draw(origin--(r*(s-0.1),0));<br />
label("$3$",(-0.9,h1*s),N,fontsize(10));<br />
<br />
real h2 = 7.5, r = 6, s=0.6, d = 14;<br />
pair P = (d+r-0.05,h2-0.15), Q = (d-r+0.05,h2-0.15), Pp = s * P + (1-s)*(d,0), Qp = s * Q + (1-s)*(d,0);<br />
path e = ellipse((d,h2),r,1), ep = ellipse((d,h2*s+0.09),r*s,1);<br />
draw(ellipse((d,0),r*(s-0.1),0.8));<br />
fill(ep,gray(0.8));<br />
fill((d,0)--Pp--Qp--cycle,gray(0.8));<br />
draw(P--(d,0)--Q^^e);<br />
draw(subpath(ep,0,reltime(ep,0.5)),linetype("4 4"));<br />
draw(subpath(ep,reltime(ep,0.5),reltime(ep,1)));<br />
draw(Qp--(d,Qp.y),Arrows(size=8));<br />
draw((d,0)--(d,10),linetype("4 4"));<br />
draw((d,0)--(d+r*(s-0.1),0));<br />
label("$6$",(d-r/4,h2*s-0.06),N,fontsize(10));<br />
</asy><br />
<br />
<math>\textbf{(A) }1:1 \qquad \textbf{(B) }47:43 \qquad \textbf{(C) }2:1 \qquad \textbf{(D) }40:13 \qquad \textbf{(E) }4:1</math><br />
<br />
[[2021 AMC 10A Problems/Problem 12|Solution]]<br />
<br />
==Problem 13==<br />
What is the volume of tetrahedron <math>ABCD</math> with edge lengths <math>AB = 2</math>, <math>AC = 3</math>, <math>AD = 4</math>, <math>BC = \sqrt{13}</math>, <math>BD = 2\sqrt{5}</math>, and <math>CD = 5</math> ?<br />
<br />
<math>\textbf{(A)} ~3 \qquad\textbf{(B)} ~2\sqrt{3} \qquad\textbf{(C)} ~4\qquad\textbf{(D)} ~3\sqrt{3}\qquad\textbf{(E)} ~6</math><br />
<br />
[[2021 AMC 10A Problems/Problem 13|Solution]]<br />
==Problem 14==<br />
All the roots of the polynomial <math>z^6-10z^5+Az^4+Bz^3+Cz^2+Dz+16</math> are positive integers, possibly repeated. What is the value of <math>B</math>?<br />
<br />
<math>\textbf{(A)} ~-88\qquad\textbf{(B)} ~-80\qquad\textbf{(C)} ~-64\qquad\textbf{(D)} ~-41\qquad\textbf{(E)} ~-40</math><br />
<br />
[[2021 AMC 10A Problems/Problem 14|Solution]]<br />
<br />
==Problem 15==<br />
Values for <math>A,B,C,</math> and <math>D</math> are to be selected from <math>\{1, 2, 3, 4, 5, 6\}</math> without replacement (i.e. no two letters have the same value). How many ways are there to make such choices so that the two curves <math>y=Ax^2+B</math> and <math>y=Cx^2+D</math> intersect? (The order in which the curves are listed does not matter; for example, the choices <math>A=3, B=2, C=4, D=1</math> is considered the same as the choices <math>A=4, B=1, C=3, D=2.</math>)<br />
<br />
<math>\textbf{(A) }30 \qquad \textbf{(B) }60 \qquad \textbf{(C) }90 \qquad \textbf{(D) }180 \qquad \textbf{(E) }360</math><br />
<br />
[[2021 AMC 10A Problems/Problem 15|Solution]]<br />
<br />
==Problem 16==<br />
In the following list of numbers, the integer <math>n</math> appears <math>n</math> times in the list for <math>1 \leq n \leq 200</math>.<cmath>1, 2, 2, 3, 3, 3, 4, 4, 4, 4, \dots, 200, 200, \dots , 200</cmath>What is the median of the numbers in this list?<br />
<br />
<math>\textbf{(A)} ~100.5\qquad\textbf{(B)} ~134\qquad\textbf{(C)} ~142\qquad\textbf{(D)} ~150.5\qquad\textbf{(E)} ~167</math><br />
<br />
[[2021 AMC 10A Problems/Problem 16|Solution]]<br />
<br />
==Problem 17==<br />
Trapezoid <math>ABCD</math> has <math>\overline{AB} \parallel \overline{CD}</math>, <math>BC = CD = 43</math>, and <math>\overline{AD} \perp \overline{BD}</math>. Let <math>O</math> be the intersection of the diagonals <math>\overline{AC}</math> and <math>\overline{BD}</math>, and let <math>P</math> be the midpoint of <math>\overline{BD}</math>. GIven that <math>OP = 11</math>, the length <math>AD</math> can be written in the form <math>m\sqrt{n}</math>, where <math>m</math> and <math>n</math> are positive integers and <math>n</math> is not divisible by the square of any prime. What is <math>m + n</math>?<br />
<br />
<math>\textbf{(A)} ~65\qquad\textbf{(B)} ~132\qquad\textbf{(C)} ~157\qquad\textbf{(D)} ~194\qquad\textbf{(E)} ~215</math><br />
<br />
[[2021 AMC 10A Problems/Problem 17|Solution]]<br />
<br />
==Problem 18==<br />
Let <math>f</math> be a function defined on the set of positive rational numbers with the property that <math>f(a\cdot b)=f(a)+f(b)</math> for all positive rational numbers <math>a</math> and <math>b</math>. Furthermore, suppose that <math>f</math> also has the property that <math>f(p)=p</math> for every prime number <math>p</math>. For which of the following numbers <math>x</math> is <math>f(x)<0</math>?<br />
<br />
<math>\textbf{(A)} ~\frac{17}{32}\qquad\textbf{(B)} ~\frac{11}{16}\qquad\textbf{(C)} ~\frac{7}{9}\qquad\textbf{(D)} ~\frac{7}{6} \qquad\textbf{(E)} ~\frac{25}{11}</math><br />
<br />
[[2021 AMC 10A Problems/Problem 18|Solution]]<br />
<br />
==Problem 19==<br />
The area of the region bounded by the graph of <cmath>x^2+y^2 = 3|x-y| + 3|x+y|</cmath>is <math>m+n\pi</math>, where <math>m</math> and <math>n</math> are integers. What is <math>m + n</math>?<br />
<br />
<math>\textbf{(A)} ~18\qquad\textbf{(B)} ~27\qquad\textbf{(C)} ~36\qquad\textbf{(D)} ~45\qquad\textbf{(E)} ~54</math><br />
<br />
[[2021 AMC 10A Problems/Problem 19|Solution]]<br />
<br />
==Problem 20==<br />
In how many ways can the sequence <math>1, 2, 3, 4, 5</math> be rearranged so that no three consecutive terms are increasing and no three consecutive terms are decreasing?<br />
<br />
<math>\textbf{(A)} ~10\qquad\textbf{(B)} ~18\qquad\textbf{(C)} ~24\qquad\textbf{(D)} ~32\qquad\textbf{(E)} ~44</math><br />
<br />
[[2021 AMC 10A Problems/Problem 20|Solution]]<br />
<br />
==Problem 21==<br />
Let <math>ABCDEF</math> be an equiangular hexagon. The lines <math>AB, CD,</math> and <math>EF</math> determine a triangle with area <math>192\sqrt{3}</math>, and the lines <math>BC, DE,</math> and <math>FA</math> determine a triangle with area <math>324\sqrt{3}</math>. The perimeter of hexagon <math>ABCDEF</math> can be expressed as <math>m +n\sqrt{p}</math>, where <math>m, n,</math> and <math>p</math> are positive integers and <math>p</math> is not divisible by the square of any prime. What is <math>m + n + p</math>?<br />
<br />
<math>\textbf{(A)} ~47\qquad\textbf{(B)} ~52\qquad\textbf{(C)} ~55\qquad\textbf{(D)} ~58\qquad\textbf{(E)} ~63</math><br />
<br />
[[2021 AMC 10A Problems/Problem 21|Solution]]<br />
<br />
==Problem 22==<br />
Hiram's algebra notes are <math>50</math> pages long and are printed on <math>25</math> sheets of paper; the first sheet contains pages <math>1</math> and <math>2</math>, the second sheet contains pages <math>3</math> and <math>4</math>, and so on. One day he leaves his notes on the table before leaving for lunch, and his roommate decides to borrow some pages from the middle of the notes. When Hiram comes back, he discovers that his roommate has taken a consecutive set of sheets from the notes and that the average (mean) of the page numbers on all remaining sheets is exactly <math>19</math>. How many sheets were borrowed?<br />
<br />
<math>\textbf{(A)} ~10\qquad\textbf{(B)} ~13\qquad\textbf{(C)} ~15\qquad\textbf{(D)} ~17\qquad\textbf{(E)} ~20</math><br />
<br />
[[2021 AMC 10A Problems/Problem 22|Solution]]<br />
<br />
==Problem 23==<br />
Frieda the frog begins a sequence of hops on a <math>3 \times 3</math> grid of squares, moving one square on each hop and choosing at random the direction of each hop-up, down, left, or right. She does not hop diagonally. When the direction of a hop would take Frieda off the grid, she "wraps around" and jumps to the opposite edge. For example if Frieda begins in the center square and makes two hops "up", the first hop would place her in the top row middle square, and the second hop would cause Frieda to jump to the opposite edge, landing in the bottom row middle square. Suppose Frieda starts from the center square, makes at most four hops at random, and stops hopping if she lands on a corner square. What is the probability that she reaches a corner square on one of the four hops?<br />
<br />
<math>\textbf{(A)} ~\frac{9}{16}\qquad\textbf{(B)} ~\frac{5}{8}\qquad\textbf{(C)} ~\frac{3}{4}\qquad\textbf{(D)} ~\frac{25}{32}\qquad\textbf{(E)} ~\frac{13}{16}</math><br />
<br />
[[2021 AMC 10A Problems/Problem 23|Solution]]<br />
<br />
==Problem 24==<br />
The interior of a quadrilateral is bounded by the graphs of <math>(x+ay)^2 = 4a^2</math> and <math>(ax-y)^2 = a^2</math>, where <math>a</math> a positive real number. What is the area of this region in terms of <math>a</math>, valid for all <math>a > 0</math>?<br />
<br />
<math>\textbf{(A)} ~\frac{8a^2}{(a+1)^2}\qquad\textbf{(B)} ~\frac{4a}{a+1}\qquad\textbf{(C)} ~\frac{8a}{a+1}\qquad\textbf{(D)} ~\frac{8a^2}{a^2+1}\qquad\textbf{(E)} ~\frac{8a}{a^2+1}</math><br />
<br />
[[2021 AMC 10A Problems/Problem 24|Solution]]<br />
<br />
==Problem 25==<br />
How many ways are there to place <math>3</math> indistinguishable red chips, <math>3</math> indistinguishable blue chips, and <math>3</math> indistinguishable green chips in the squares of a <math>3 \times 3</math> grid so that no two chips of the same color are directly adjacent to each other, either vertically or horizontally?<br />
<br />
<math>\textbf{(A)} ~12\qquad\textbf{(B)} ~18\qquad\textbf{(C)} ~24\qquad\textbf{(D)} ~30\qquad\textbf{(E)} ~36</math><br />
<br />
[[2021 AMC 10A Problems/Problem 25|Solution]]<br />
<br />
==See also==<br />
{{AMC10 box|year=2021|ab=A|before=[[2020 AMC 10B]]|after=[[2021 AMC 10B]]}}<br />
* [[AMC 10]]<br />
* [[AMC 10 Problems and Solutions]]<br />
* [[Mathematics competitions]]<br />
* [[Mathematics competition resources]]<br />
{{MAA Notice}}</div>Ghostychttps://artofproblemsolving.com/wiki/index.php?title=1980_AHSME_Problems&diff=1529871980 AHSME Problems2021-04-30T14:47:06Z<p>Ghostyc: /* Problem 5 */</p>
<hr />
<div>{{AHSME Problems<br />
|year = 1980<br />
}}<br />
== Problem 1 ==<br />
The largest whole number such that seven times the number is less than <math>100</math> is<br />
<br />
<math>\text{(A)} \ 12 \qquad \text{(B)} \ 13 \qquad \text{(C)} \ 14 \qquad \text{(D)} \ 15 \qquad \text{(E)} \ 16</math><br />
<br />
[[1980 AHSME Problems/Problem 1|Solution]]<br />
<br />
== Problem 2 ==<br />
The degree of <math>(x^2+1)^4 (x^3+1)^3</math> as a polynomial in <math>x</math> is<br />
<br />
<math>\text{(A)} \ 5 \qquad \text{(B)} \ 7 \qquad \text{(C)} \ 12 \qquad \text{(D)} \ 17 \qquad \text{(E)} \ 72</math><br />
<br />
[[1980 AHSME Problems/Problem 2|Solution]]<br />
<br />
== Problem 3 ==<br />
If the ratio of <math>2x-y</math> to <math>x+y</math> is <math>\frac{2}{3}</math>, what is the ratio of <math>x</math> to <math>y</math>?<br />
<br />
<math>\text{(A)} \ \frac{1}{5} \qquad \text{(B)} \ \frac{4}{5} \qquad \text{(C)} \ 1 \qquad \text{(D)} \ \frac{6}{5} \qquad \text{(E)} \ \frac{5}{4}</math><br />
<br />
[[1980 AHSME Problems/Problem 3|Solution]]<br />
<br />
== Problem 4 ==<br />
In the adjoining figure, CDE is an equilateral triangle and ABCD and DEFG are squares. The measure of <math>\angle GDA</math> is<br />
<br />
<math>\text{(A)} \ 90^\circ \qquad \text{(B)} \ 105^\circ \qquad \text{(C)} \ 120^\circ \qquad \text{(D)} \ 135^\circ \qquad \text{(E)} \ 150^\circ</math><br />
<br />
<asy><br />
defaultpen(linewidth(0.7)+fontsize(10));<br />
pair D=origin, C=D+dir(240), E=D+dir(300), F=E+dir(30), G=D+dir(30), A=D+dir(150), B=C+dir(150);<br />
draw(E--D--G--F--E--C--D--A--B--C);<br />
pair point=(0,0.5);<br />
label("$A$", A, dir(point--A));<br />
label("$B$", B, dir(point--B));<br />
label("$C$", C, dir(point--C));<br />
label("$D$", D, dir(-15));<br />
label("$E$", E, dir(point--E));<br />
label("$F$", F, dir(point--F));<br />
label("$G$", G, dir(point--G));</asy><br />
<br />
[[1980 AHSME Problems/Problem 4|Solution]]<br />
<br />
== Problem 5 ==<br />
If <math>AB</math> and <math>CD</math> are perpendicular diameters of circle <math>Q</math>, <math>P</math> in <math>\overline{AQ}</math>, and <math>\measuredangle QPC = 60^\circ</math>, then the length of <math>PQ</math> divided by the length of <math>AQ</math> is <br />
<br />
<asy><br />
defaultpen(linewidth(0.7)+fontsize(10));<br />
pair A=(-1,0), B=(1,0), C=(0,1), D=(0,-1), Q=origin, P=(-0.5,0);<br />
draw(P--C--D^^A--B^^Circle(Q,1));<br />
label("$A$", A, W);<br />
label("$B$", B, E);<br />
label("$C$", C, N);<br />
label("$D$", D, S);<br />
label("$P$", P, S);<br />
label("$Q$", Q, SE);<br />
label("$60^\circ$", P+0.05*dir(30), dir(30));</asy><br />
<br />
<math> \text{(A)} \ \frac{\sqrt{3}}{2} \qquad \text{(B)} \ \frac{\sqrt{3}}{3} \qquad \text{(C)} \ \frac{\sqrt{2}}{2} \qquad \text{(D)} \ \frac12 \qquad \text{(E)} \ \frac23 </math><br />
<br />
[[1980 AHSME Problems/Problem 5|Solution]]<br />
<br />
== Problem 6 ==<br />
A positive number <math>x</math> satisfies the inequality <math>\sqrt{x} < 2x</math> if and only if<br />
<br />
<math>\text{(A)} \ x > \frac{1}{4} \qquad \text{(B)} \ x > 2 \qquad \text{(C)} \ x > 4 \qquad \text{(D)} \ x < \frac{1}{4}\qquad \text{(E)} \ x < 4</math><br />
<br />
[[1980 AHSME Problems/Problem 6|Solution]]<br />
<br />
== Problem 7 ==<br />
Sides <math>AB,BC,CD</math> and <math>DA</math> of convex polygon <math>ABCD</math> have lengths 3,4,12, and 13, respectively, and <math>\measuredangle CBA</math> is a right angle. The area of the quadrilateral is<br />
<br />
<asy><br />
defaultpen(linewidth(0.7)+fontsize(10));<br />
real r=degrees((12,5)), s=degrees((3,4));<br />
pair D=origin, A=(13,0), C=D+12*dir(r), B=A+3*dir(180-(90-r+s));<br />
draw(A--B--C--D--cycle);<br />
markscalefactor=0.05;<br />
draw(rightanglemark(A,B,C));<br />
pair point=incenter(A,C,D);<br />
label("$A$", A, dir(point--A));<br />
label("$B$", B, dir(point--B));<br />
label("$C$", C, dir(point--C));<br />
label("$D$", D, dir(point--D));<br />
label("$3$", A--B, dir(A--B)*dir(-90));<br />
label("$4$", B--C, dir(B--C)*dir(-90));<br />
label("$12$", C--D, dir(C--D)*dir(-90));<br />
label("$13$", D--A, dir(D--A)*dir(-90));</asy><br />
<br />
<math>\text{(A)} \ 32 \qquad \text{(B)} \ 36 \qquad \text{(C)} \ 39 \qquad \text{(D)} \ 42 \qquad \text{(E)} \ 48</math><br />
<br />
[[1980 AHSME Problems/Problem 7|Solution]]<br />
<br />
== Problem 8 ==<br />
<br />
How many pairs <math>(a,b)</math> of non-zero real numbers satisfy the equation<br />
<br />
<cmath> \frac{1}{a} + \frac{1}{b} = \frac{1}{a+b} </cmath><br />
<math>\text{(A)} \ \text{none} \qquad \text{(B)} \ 1 \qquad \text{(C)} \ 2 \qquad \text{(D)} \ \text{one pair for each} ~b \neq 0</math><br />
<math>\text{(E)} \ \text{two pairs for each} ~b \neq 0</math><br />
<br />
[[1980 AHSME Problems/Problem 8|Solution]]<br />
<br />
== Problem 9 ==<br />
A man walks <math>x</math> miles due west, turns <math>150^\circ</math> to his left and walks 3 miles in the new direction. If he finishes a a point <math>\sqrt{3}</math> from his starting point, then <math>x</math> is<br />
<br />
<math>\text{(A)} \ \sqrt 3 \qquad \text{(B)} \ 2\sqrt{5} \qquad \text{(C)} \ \frac 32 \qquad \text{(D)} \ 3 \qquad \text{(E)} \ \text{not uniquely determined}</math><br />
<br />
[[1980 AHSME Problems/Problem 9|Solution]]<br />
<br />
== Problem 10 ==<br />
<br />
The number of teeth in three meshed gears <math>A</math>, <math>B</math>, and <math>C</math> are <math>x</math>, <math>y</math>, and <math>z</math>, respectively. (The teeth on all gears are the same size and regularly spaced.) The angular speeds, in revolutions per minutes of <math>A</math>, <math>B</math>, and <math>C</math> are in the proportion<br />
<br />
<math>\text{(A)} \ x: y: z ~~\text{(B)} \ z: y: x ~~ \text{(C)} \ y: z: x~~ \text{(D)} \ yz: xz: xy ~~ \text{(E)} \ xz: yx: zy</math><br />
<br />
[[1980 AHSME Problems/Problem 10|Solution]]<br />
<br />
== Problem 11 ==<br />
<br />
If the sum of the first <math>10</math> terms and the sum of the first <math>100</math> terms of a given arithmetic progression are <math>100</math> and <math>10</math>, <br />
respectively, then the sum of first <math>110</math> terms is:<br />
<br />
<math>\text{(A)} \ 90 \qquad \text{(B)} \ -90 \qquad \text{(C)} \ 110 \qquad \text{(D)} \ -110 \qquad \text{(E)} \ -100</math><br />
<br />
[[1980 AHSME Problems/Problem 11|Solution]]<br />
<br />
== Problem 12 ==<br />
<br />
The equations of <math>L_1</math> and <math>L_2</math> are <math>y=mx</math> and <math>y=nx</math>, respectively. Suppose <math>L_1</math> makes twice as large of an angle with the horizontal (measured counterclockwise from the positive x-axis ) as does <math>L_2</math>, and that <math>L_1</math> has 4 times the slope of <math>L_2</math>. If <math>L_1</math> is not horizontal, then <math>mn</math> is<br />
<br />
<math>\text{(A)} \ \frac{\sqrt{2}}{2} \qquad \text{(B)} \ -\frac{\sqrt{2}}{2} \qquad \text{(C)} \ 2 \qquad \text{(D)} \ -2 \qquad \text{(E)} \ \text{not uniquely determined}</math><br />
<br />
[[1980 AHSME Problems/Problem 12|Solution]]<br />
<br />
== Problem 13 ==<br />
<br />
A bug (of negligible size) starts at the origin on the coordinate plane. First, it moves one unit right to <math>(1,0)</math>. Then it makes a <math>90^\circ</math> counterclockwise and travels <math>\frac 12</math> a unit to <math>\left(1, \frac 12 \right)</math>. If it continues in this fashion, each time making a <math>90^\circ</math> degree turn counterclockwise and traveling half as far as the previous move, to which of the following points will it come closest? <br />
<br />
<math>\text{(A)} \ \left(\frac 23, \frac 23 \right) \qquad \text{(B)} \ \left( \frac 45, \frac 25 \right) \qquad \text{(C)} \ \left( \frac 23, \frac 45 \right) \qquad \text{(D)} \ \left(\frac 23, \frac 13 \right) \qquad \text{(E)} \ \left(\frac 25, \frac 45 \right)</math><br />
<br />
[[1980 AHSME Problems/Problem 13|Solution]]<br />
<br />
== Problem 14 ==<br />
<br />
If the function <math>f</math> is defined by <br />
<cmath> f(x)=\frac{cx}{2x+3} ,\quad x\neq -\frac{3}{2} , </cmath> <br />
satisfies <math>x=f(f(x))</math> for all real numbers <math>x</math> except <math>-\frac{3}{2}</math>, then <math>c</math> is<br />
<br />
<math>\text{(A)} \ -3 \qquad <br />
\text{(B)} \ - \frac{3}{2} \qquad <br />
\text{(C)} \ \frac{3}{2} \qquad <br />
\text{(D)} \ 3 \qquad <br />
\text{(E)} \ \text{not uniquely determined}</math><br />
<br />
[[1980 AHSME Problems/Problem 14|Solution]]<br />
<br />
== Problem 15 ==<br />
A store prices an item in dollars and cents so that when 4% sales tax is added, no rounding is necessary because the result is exactly <math>n</math> dollars where <math>n</math> is a positive integer. The smallest value of <math>n</math> is<br />
<br />
<math>\text{(A)} \ 1 \qquad \text{(B)} \ 13 \qquad \text{(C)} \ 25 \qquad \text{(D)} \ 26 \qquad \text{(E)} \ 100</math><br />
<br />
[[1980 AHSME Problems/Problem 15|Solution]]<br />
<br />
== Problem 16 ==<br />
Four of the eight vertices of a cube are the vertices of a regular tetrahedron. Find the ratio of the surface area of the cube to the surface area of the tetrahedron.<br />
<br />
<math>\text{(A)} \ \sqrt 2 \qquad \text{(B)} \ \sqrt 3 \qquad \text{(C)} \ \sqrt{\frac{3}{2}} \qquad \text{(D)} \ \frac{2}{\sqrt{3}} \qquad \text{(E)} \ 2</math><br />
<br />
[[1980 AHSME Problems/Problem 16|Solution]]<br />
<br />
== Problem 17 ==<br />
Given that <math>i^2=-1</math>, for how many integers <math>n</math> is <math>(n+i)^4</math> an integer?<br />
<br />
<math>\text{(A)} \ \text{none} \qquad \text{(B)} \ 1 \qquad \text{(C)} \ 2 \qquad \text{(D)} \ 3 \qquad \text{(E)} \ 4</math><br />
<br />
[[1980 AHSME Problems/Problem 17|Solution]]<br />
<br />
== Problem 18 ==<br />
If <math>b>1</math>, <math>\sin x>0</math>, <math>\cos x>0</math>, and <math>\log_b \sin x = a</math>, then <math>\log_b \cos x</math> equals<br />
<br />
<math>\text{(A)} \ 2\log_b(1-b^{a/2}) ~~\text{(B)} \ \sqrt{1-a^2} ~~\text{(C)} \ b^{a^2} ~~\text{(D)} \ \frac 12 \log_b(1-b^{2a}) ~~\text{(E)} \ \text{none of these}</math><br />
<br />
[[1980 AHSME Problems/Problem 18|Solution]]<br />
<br />
==Problem 19==<br />
<br />
Let <math>C_1, C_2</math> and <math>C_3</math> be three parallel chords of a circle on the same side of the center. <br />
The distance between <math>C_1</math> and <math>C_2</math> is the same as the distance between <math>C_2</math> and <math>C_3</math>. <br />
The lengths of the chords are <math>20, 16</math>, and <math>8</math>. The radius of the circle is <br />
<br />
<math>\text{(A)} \ 12 \qquad <br />
\text{(B)} \ 4\sqrt{7} \qquad <br />
\text{(C)} \ \frac{5\sqrt{65}}{3} \qquad <br />
\text{(D)}\ \frac{5\sqrt{22}}{2}\qquad<br />
\text{(E)}\ \text{not uniquely determined} </math> <br />
<br />
[[1980 AHSME Problems/Problem 19|Solution]]<br />
<br />
==Problem 20==<br />
<br />
A box contains <math>2</math> pennies, <math>4</math> nickels, and <math>6</math> dimes. Six coins are drawn without replacement, <br />
with each coin having an equal probability of being chosen. What is the probability that the value of coins drawn is at least <math>50</math> cents? <br />
<br />
<math>\text{(A)} \ \frac{37}{924} \qquad <br />
\text{(B)} \ \frac{91}{924} \qquad <br />
\text{(C)} \ \frac{127}{924} \qquad <br />
\text{(D)}\ \frac{132}{924}\qquad<br />
\text{(E)}\ \text{none of these} </math> <br />
<br />
[[1980 AHSME Problems/Problem 20|Solution]]<br />
<br />
==Problem 21==<br />
<br />
<asy><br />
defaultpen(linewidth(0.7)+fontsize(10));<br />
pair B=origin, C=(15,3), D=(5,1), A=7*dir(72)*dir(B--C), E=midpoint(A--C), F=intersectionpoint(A--D, B--E);<br />
draw(E--B--A--C--B^^A--D);<br />
label("$A$", A, dir(D--A));<br />
label("$B$", B, dir(E--B));<br />
label("$C$", C, dir(0));<br />
label("$D$", D, SE);<br />
label("$E$", E, N);<br />
label("$F$", F, dir(80));</asy><br />
<br />
In triangle <math>ABC</math>, <math>\measuredangle CBA=72^\circ</math>, <math>E</math> is the midpoint of side <math>AC</math>, <br />
and <math>D</math> is a point on side <math>BC</math> such that <math>2BD=DC</math>; <math>AD</math> and <math>BE</math> intersect at <math>F</math>. <br />
The ratio of the area of triangle <math>BDF</math> to the area of quadrilateral <math>FDCE</math> is<br />
<br />
<math>\text{(A)} \ \frac 15 \qquad <br />
\text{(B)} \ \frac 14 \qquad <br />
\text{(C)} \ \frac 13 \qquad <br />
\text{(D)}\ \frac{2}{5}\qquad<br />
\text{(E)}\ \text{none of these}</math><br />
<br />
[[1980 AHSME Problems/Problem 21|Solution]]<br />
<br />
==Problem 22==<br />
<br />
For each real number <math>x</math>, let <math>f(x)</math> be the minimum of the numbers <math>4x+1, x+2</math>, and <math>-2x+4</math>. Then the maximum value of <math>f(x)</math> is <br />
<br />
<math>\text{(A)} \ \frac{1}{3} \qquad <br />
\text{(B)} \ \frac{1}{2} \qquad <br />
\text{(C)} \ \frac{2}{3} \qquad <br />
\text{(D)} \ \frac{5}{2} \qquad <br />
\text{(E)}\ \frac{8}{3} </math> <br />
<br />
[[1980 AHSME Problems/Problem 22|Solution]]<br />
<br />
==Problem 23==<br />
<br />
Line segments drawn from the vertex opposite the hypotenuse of a right triangle to the points trisecting the hypotenuse have lengths <math>\sin x</math> and <math>\cos x</math>, where <math>x</math> is a real number such that <math>0<x<\frac{\pi}{2}</math>. The length of the hypotenuse is <br />
<br />
<math>\text{(A)} \ \frac{4}{3} \qquad <br />
\text{(B)} \ \frac{3}{2} \qquad <br />
\text{(C)} \ \frac{3\sqrt{5}}{5} \qquad <br />
\text{(D)}\ \frac{2\sqrt{5}}{3}\qquad<br />
\text{(E)}\ \text{not uniquely determined}</math> <br />
<br />
[[1980 AHSME Problems/Problem 23|Solution]]<br />
<br />
==Problem 24==<br />
<br />
For some real number <math>r</math>, the polynomial <math>8x^3-4x^2-42x+45</math> is divisible by <math>(x-r)^2</math>. Which of the following numbers is closest to <math>r</math>? <br />
<br />
<math>\text{(A)} \ 1.22 \qquad <br />
\text{(B)} \ 1.32 \qquad <br />
\text{(C)} \ 1.42 \qquad <br />
\text{(D)} \ 1.52 \qquad <br />
\text{(E)} \ 1.62 </math> <br />
<br />
[[1980 AHSME Problems/Problem 24|Solution]]<br />
<br />
==Problem 25==<br />
<br />
In the non-decreasing sequence of odd integers <math>\{a_1,a_2,a_3,\ldots \}=\{1,3,3,3,5,5,5,5,5,\ldots \}</math> each odd positive integer <math>k</math><br />
appears <math>k</math> times. It is a fact that there are integers <math>b, c</math>, and <math>d</math> such that for all positive integers <math>n</math>, <br />
<math>a_n=b\lfloor \sqrt{n+c} \rfloor +d</math>, <br />
where <math>\lfloor x \rfloor</math> denotes the largest integer not exceeding <math>x</math>. The sum <math>b+c+d</math> equals <br />
<br />
<math>\text{(A)} \ 0 \qquad <br />
\text{(B)} \ 1 \qquad <br />
\text{(C)} \ 2 \qquad <br />
\text{(D)} \ 3 \qquad <br />
\text{(E)} \ 4 </math> <br />
<br />
[[1980 AHSME Problems/Problem 25|Solution]]<br />
<br />
==Problem 26==<br />
<br />
Four balls of radius <math>1</math> are mutually tangent, three resting on the floor and the fourth resting on the others. <br />
A tetrahedron, each of whose edges have length <math>s</math>, is circumscribed around the balls. Then <math>s</math> equals <br />
<br />
<math>\text{(A)} \ 4\sqrt 2 \qquad <br />
\text{(B)} \ 4\sqrt 3 \qquad <br />
\text{(C)} \ 2\sqrt 6 \qquad <br />
\text{(D)}\ 1+2\sqrt 6\qquad<br />
\text{(E)}\ 2+2\sqrt 6</math> <br />
<br />
[[1980 AHSME Problems/Problem 26|Solution]]<br />
<br />
==Problem 27==<br />
<br />
The sum <math>\sqrt[3] {5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}</math> equals <br />
<br />
<math>\text{(A)} \ \frac 32 \qquad <br />
\text{(B)} \ \frac{\sqrt[3]{65}}{4} \qquad <br />
\text{(C)} \ \frac{1+\sqrt[6]{13}}{2} \qquad <br />
\text{(D)}\ \sqrt[3]{2}\qquad<br />
\text{(E)}\ \text{none of these} </math><br />
<br />
[[1980 AHSME Problems/Problem 27|Solution]]<br />
<br />
==Problem 28==<br />
The polynomial <math>x^{2n}+1+(x+1)^{2n}</math> is not divisible by <math>x^2+x+1</math> if <math>n</math> equals <br />
<br />
<math>\text{(A)} \ 17 \qquad <br />
\text{(B)} \ 20 \qquad <br />
\text{(C)} \ 21 \qquad <br />
\text{(D)} \ 64 \qquad <br />
\text{(E)} \ 65 </math> <br />
<br />
[[1980 AHSME Problems/Problem 28|Solution]]<br />
<br />
==Problem 29==<br />
<br />
How many ordered triples (x,y,z) of integers satisfy the system of equations below? <br />
<br />
<cmath>\begin{array}{l} x^2-3xy+2y^2-z^2=31 \\ -x^2+6yz+2z^2=44 \\ x^2+xy+8z^2=100\\ \end{array} </cmath><br />
<br />
<math>\text{(A)} \ 0 \qquad <br />
\text{(B)} \ 1 \qquad <br />
\text{(C)} \ 2 \qquad <br />
\text{(D)}\ \text{a finite number greater than 2}\qquad<br />
\text{(E)}\ \text{infinitely many} </math> <br />
<br />
[[1980 AHSME Problems/Problem 29|Solution]]<br />
<br />
==Problem 30==<br />
<br />
A six digit number (base 10) is squarish if it satisfies the following conditions: <br />
<br />
(i) none of its digits are zero; <br />
<br />
(ii) it is a perfect square; and <br />
<br />
(iii) the first of two digits, the middle two digits and the last two digits of the number are all perfect squares when considered as two digit numbers. <br />
<br />
How many squarish numbers are there? <br />
<br />
<math>\text{(A)} \ 0 \qquad <br />
\text{(B)} \ 2 \qquad <br />
\text{(C)} \ 3 \qquad <br />
\text{(D)} \ 8 \qquad <br />
\text{(E)} \ 9 </math><br />
<br />
[[1980 AHSME Problems/Problem 30|Solution]]<br />
<br />
== See also ==<br />
<br />
* [[AMC 12 Problems and Solutions]]<br />
* [[Mathematics competition resources]]<br />
<br />
{{AHSME box|year=1980|before=[[1979 AHSME]]|after=[[1981 AHSME]]}} <br />
<br />
{{MAA Notice}}</div>Ghostychttps://artofproblemsolving.com/wiki/index.php?title=1959_AHSME_Problems/Problem_46&diff=1528851959 AHSME Problems/Problem 462021-04-28T09:59:25Z<p>Ghostyc: Created page with "== Problem 46== A student on vacation for <math>d</math> days observed that (1) it rained <math>7</math> times, morning or afternoon (2) when it rained in the afternoon, it w..."</p>
<hr />
<div>== Problem 46==<br />
A student on vacation for <math>d</math> days observed that (1) it rained <math>7</math> times, morning or afternoon (2) when it rained in the afternoon, <br />
it was clear in the morning (3) there were five clear afternoons (4) there were six clear mornings. Then <math>d</math> equals:<br />
<math>\textbf{(A)}\ 7\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ 11\qquad\textbf{(E)}\ 12 </math> <br />
<br />
[[1959 AHSME Problems/Problem 46|Solution]]</div>Ghostychttps://artofproblemsolving.com/wiki/index.php?title=2007_AMC_8_Problems&diff=1410202007 AMC 8 Problems2020-12-30T09:01:46Z<p>Ghostyc: /* Problem 8 */</p>
<hr />
<div>==Problem 1==<br />
Theresa's parents have agreed to buy her tickets to see her favorite band if she spends an average of <math>10</math> hours per week helping around the house for <math>6</math> weeks. For the first <math>5</math> weeks she helps around the house for <math>8</math>, <math>11</math>, <math>7</math>, <math>12</math> and <math>10</math> hours. How many hours must she work for the final week to earn the tickets?<br />
<br />
<math>\mathrm{(A)}\ 9 \qquad\mathrm{(B)}\ 10 \qquad\mathrm{(C)}\ 11 \qquad\mathrm{(D)}\ 12 \qquad\mathrm{(E)}\ 13</math><br />
<br />
[[2007 AMC 8 Problems/Problem 1|Solution]]<br />
<br />
==Problem 2==<br />
<math>650</math> students were surveyed about their pasta preferences. The choices were lasagna, manicotti, ravioli and spaghetti. The results of the survey are displayed in the bar graph. What is the ratio of the number of students who preferred spaghetti to the number of students who preferred manicotti?<br />
<br />
<center>[[Image:AMC8_2007_2.png]]</center><br />
<br />
<math>\mathrm{(A)} \frac{2}{5} \qquad \mathrm{(B)} \frac{1}{2} \qquad \mathrm{(C)} \frac{5}{4} \qquad \mathrm{(D)} \frac{5}{3} \qquad \mathrm{(E)} \frac{5}{2}</math><br />
<br />
[[2007 AMC 8 Problems/Problem 2|Solution]]<br />
<br />
==Problem 3==<br />
<br />
What is the sum of the two smallest prime factors of <math>250</math>?<br />
<br />
<math>\mathrm{(A)}\ 2 \qquad\mathrm{(B)}\ 5 \qquad\mathrm{(C)}\ 7 \qquad\mathrm{(D)}\ 10 \qquad\mathrm{(E)}\ 12</math><br />
<br />
[[2007 AMC 8 Problems/Problem 3|Solution]]<br />
<br />
==Problem 4==<br />
<br />
A haunted house has six windows. In how many ways can<br />
Georgie the Ghost enter the house by one window and leave<br />
by a different window?<br />
<br />
<math>\mathrm{(A)}\ 12 \qquad\mathrm{(B)}\ 15 \qquad\mathrm{(C)}\ 18 \qquad\mathrm{(D)}\ 30 \qquad\mathrm{(E)}\ 36</math><br />
<br />
[[2007 AMC 8 Problems/Problem 4|Solution]]<br />
<br />
==Problem 5==<br />
<br />
Chandler wants to buy a <math>\textdollar 500</math> mountain bike. For his birthday, his grandparents<br />
send him <math>\textdollar 50</math>, his aunt sends him <math>\textdollar 35</math> and his cousin gives him <math>\textdollar 15</math>. He earns<br />
<math>\textdollar 16</math> per week for his paper route. He will use all of his birthday money and all<br />
of the money he earns from his paper route. In how many weeks will he be able<br />
to buy the mountain bike?<br />
<br />
<math>\mathrm{(A)}\ 24 \qquad\mathrm{(B)}\ 25 \qquad\mathrm{(C)}\ 26 \qquad\mathrm{(D)}\ 27 \qquad\mathrm{(E)}\ 28</math><br />
<br />
[[2007 AMC 8 Problems/Problem 5|Solution]]<br />
<br />
==Problem 6==<br />
<br />
The average cost of a long-distance call in the USA in <math>1985</math> was<br />
<math>41</math> cents per minute, and the average cost of a long-distance<br />
call in the USA in <math>2005</math> was <math>7</math> cents per minute. Find the<br />
approximate percent decrease in the cost per minute of a long-<br />
distance call.<br />
<br />
<math>\mathrm{(A)}\ 7 \qquad\mathrm{(B)}\ 17 \qquad\mathrm{(C)}\ 34 \qquad\mathrm{(D)}\ 41 \qquad\mathrm{(E)}\ 80</math><br />
<br />
[[2007 AMC 8 Problems/Problem 6|Solution]]<br />
<br />
==Problem 7==<br />
<br />
The average age of <math>5</math> people in a room is <math>30</math> years. An <math>18</math>-year-old person leaves<br />
the room. What is the average age of the four remaining people?<br />
<br />
<math>\mathrm{(A)}\ 25 \qquad\mathrm{(B)}\ 26 \qquad\mathrm{(C)}\ 29 \qquad\mathrm{(D)}\ 33 \qquad\mathrm{(E)}\ 36</math><br />
<br />
[[2007 AMC 8 Problems/Problem 7|Solution]]<br />
<br />
==Problem 8==<br />
<br />
In trapezoid <math>ABCD</math>, <math>AD</math> is perpendicular to <math>DC</math>,<br />
<math>AD</math> = <math>AB</math> = <math>3</math>, and <math>DC</math> = <math>6</math>. In addition, <math>E</math> is on<br />
<math>DC</math>, and <math>BE</math> is parallel to <math>AD</math>. Find the area of<br />
<math>\triangle BEC</math>.<br />
<br />
<center><br />
<asy><br />
defaultpen(linewidth(0.7));<br />
pair A=(0,3), B=(3,3), C=(6,0), D=origin, E=(3,0);<br />
draw(E--B--C--D--A--B);<br />
draw(rightanglemark(A, D, C));<br />
label("$A$", A, NW);<br />
label("$B$", B, NW);<br />
label("$C$", C, SE);<br />
label("$D$", D, SW);<br />
label("$E$", E, NW);<br />
label("$3$", A--D, W);<br />
label("$3$", A--B, N);<br />
label("$6$", E, S);<br />
</asy><br />
</center><br />
<br />
<math>\text{(A)}\ 3 \qquad \text{(B)}\ 4.5 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 9 \qquad \text{(E)}\ 18</math><br />
<br />
[[2007 AMC 8 Problems/Problem 8|Solution]]<br />
<br />
==Problem 9==<br />
<br />
To complete the grid below, each of the digits 1 through 4 must occur once<br />
in each row and once in each column. What number will occupy the lower<br />
right-hand square?<br />
<br />
<center>[[Image:AMC8_2007_9.png]]</center><br />
<br />
<math>\mathrm{(A)}\ 1 \qquad \mathrm{(B)}\ 2 \qquad \mathrm{(C)}\ 3 \qquad \mathrm{(D)}\ 4 \qquad \mathrm{(E)}</math> cannot be determined<br />
<br />
[[2007 AMC 8 Problems/Problem 9|Solution]]<br />
<br />
==Problem 10==<br />
<br />
For any positive integer <math>n</math>, define <math>\boxed{n}</math> to be the sum of the positive factors of <math>n</math>.<br />
For example, <math>\boxed{6} = 1 + 2 + 3 + 6 = 12</math>. Find <math>\boxed{\boxed{11}}</math> .<br />
<br />
<math>\mathrm{(A)}\ 13 \qquad \mathrm{(B)}\ 20 \qquad \mathrm{(C)}\ 24 \qquad \mathrm{(D)}\ 28 \qquad \mathrm{(E)}\ 30</math><br />
<br />
[[2007 AMC 8 Problems/Problem 10|Solution]]<br />
<br />
==Problem 11==<br />
<br />
Tiles <math>I, II, III</math> and <math>IV</math> are translated so one tile coincides with each of the rectangles <math>A, B, C</math> and <math>D</math>. In the final arrangement, the two numbers on any side common to two adjacent tiles must be the same. Which of the tiles is translated to Rectangle <math>C</math>?<br />
<br />
<center>[[Image:AMC8_2007_11.png]]</center><br />
<br />
<math>\mathrm{(A)}\ I \qquad \mathrm{(B)}\ II \qquad \mathrm{(C)}\ III \qquad \mathrm{(D)}\ IV \qquad \mathrm{(E)}</math> cannot be determined<br />
<br />
[[2007 AMC 8 Problems/Problem 11|Solution]]<br />
<br />
==Problem 12==<br />
<br />
A unit hexagram is composed of a regular hexagon of side length <math>1</math> and its <math>6</math><br />
equilateral triangular extensions, as shown in the diagram. What is the ratio of<br />
the area of the extensions to the area of the original hexagon?<br />
<br />
<center>[[Image:AMC8_2007_12.png]]</center><br />
<br />
<math>\mathrm{(A)}\ 1:1 \qquad \mathrm{(B)}\ 6:5 \qquad \mathrm{(C)}\ 3:2 \qquad \mathrm{(D)}\ 2:1 \qquad \mathrm{(E)}\ 3:1</math> <br />
<br />
[[2007 AMC 8 Problems/Problem 12|Solution]]<br />
<br />
==Problem 13==<br />
<br />
Sets <math>A</math> and <math>B</math>, shown in the Venn diagram, have the same number of elements.<br />
Their union has <math>2007</math> elements and their intersection has <math>1001</math> elements. Find<br />
the number of elements in <math>A</math>.<br />
<br />
<center>[[Image:AMC8_2007_13.png]]</center><br />
<br />
<math>\mathrm{(A)}\ 503 \qquad \mathrm{(B)}\ 1006 \qquad \mathrm{(C)}\ 1504 \qquad \mathrm{(D)}\ 1507 \qquad \mathrm{(E)}\ 1510</math><br />
<br />
[[2007 AMC 8 Problems/Problem 13|Solution]]<br />
<br />
==Problem 14==<br />
<br />
The base of isosceles <math>\triangle ABC</math> is <math>24</math> and its area is <math>60</math>. What is the length of one<br />
of the congruent sides?<br />
<br />
<math>\mathrm{(A)}\ 5 \qquad \mathrm{(B)}\ 8 \qquad \mathrm{(C)}\ 13 \qquad \mathrm{(D)}\ 14 \qquad \mathrm{(E)}\ 18</math><br />
<br />
[[2007 AMC 8 Problems/Problem 14|Solution]]<br />
<br />
==Problem 15==<br />
<br />
Let <math>a, b</math> and <math>c</math> be numbers with <math>0 < a < b < c</math>. Which of the following is<br />
impossible?<br />
<br />
<math>\mathrm{(A)} \ a + c < b \qquad \mathrm{(B)} \ a \cdot b < c \qquad \mathrm{(C)} \ a + b < c \qquad \mathrm{(D)} \ a \cdot c < b \qquad \mathrm{(E)}\frac{b}{c} = a</math><br />
<br />
[[2007 AMC 8 Problems/Problem 15|Solution]]<br />
<br />
==Problem 16==<br />
<br />
Amanda draws five circles with radii <math>1, 2, 3,<br />
4</math> and <math>5</math>. Then for each circle she plots the point <math>(C,A)</math>,<br />
where <math>C</math> is its circumference and <math>A</math> is its area. Which of the<br />
following could be her graph?<br />
<br />
<center>[[Image:AMC8_2007_16.png]]</center><br />
<br />
[[2007 AMC 8 Problems/Problem 16|Solution]]<br />
<br />
==Problem 17==<br />
<br />
A mixture of <math>30</math> liters of paint is <math>25\%</math> red tint, <math>30\%</math> yellow<br />
tint and <math>45\%</math> water. Five liters of yellow tint are added to<br />
the original mixture. What is the percent of yellow tint<br />
in the new mixture?<br />
<br />
<math>\mathrm{(A)}\ 25 \qquad \mathrm{(B)}\ 35 \qquad \mathrm{(C)}\ 40 \qquad \mathrm{(D)}\ 45 \qquad \mathrm{(E)}\ 50</math><br />
<br />
[[2007 AMC 8 Problems/Problem 17|Solution]]<br />
<br />
==Problem 18==<br />
<br />
The product of the two <math>99</math>-digit numbers<br />
<br />
<math>303,\!030,\!303,\!...,\!030,\!303</math> and <math>505,\!050,\!505,\!...,\!050,\!505</math><br />
<br />
has thousands digit <math>A</math> and units digit <math>B</math>. What is the sum of <math>A</math> and <math>B</math>?<br />
<br />
<math>\mathrm{(A)}\ 3 \qquad \mathrm{(B)}\ 5 \qquad \mathrm{(C)}\ 6 \qquad \mathrm{(D)}\ 8 \qquad \mathrm{(E)}\ 10</math><br />
<br />
[[2007 AMC 8 Problems/Problem 18|Solution]]<br />
<br />
==Problem 19==<br />
<br />
Pick two consecutive positive integers whose sum is less than <math>100</math>. Square both<br />
of those integers and then find the difference of the squares. Which of the<br />
following could be the difference?<br />
<br />
<math>\mathrm{(A)}\ 2 \qquad \mathrm{(B)}\ 64 \qquad \mathrm{(C)}\ 79 \qquad \mathrm{(D)}\ 96 \qquad \mathrm{(E)}\ 131</math><br />
<br />
[[2007 AMC 8 Problems/Problem 19|Solution]]<br />
<br />
==Problem 20==<br />
<br />
Before district play, the Unicorns had won <math>45\%</math> of their<br />
basketball games. During district play, they won six more<br />
games and lost two, to finish the season having won half<br />
their games. How many games did the Unicorns play in<br />
all?<br />
<br />
<math>\mathrm{(A)}\ 48 \qquad \mathrm{(B)}\ 50 \qquad \mathrm{(C)}\ 52 \qquad \mathrm{(D)}\ 54 \qquad \mathrm{(E)}\ 60</math><br />
<br />
[[2007 AMC 8 Problems/Problem 20|Solution]]<br />
<br />
==Problem 21==<br />
<br />
Two cards are dealt from a deck of four red cards labeled <math>A, B, C, D</math> and four<br />
green cards labeled <math>A, B, C, D</math>. A winning pair is two of the same color or two<br />
of the same letter. What is the probability of drawing a winning pair?<br />
<br />
<math>\mathrm{(A)} \frac{2}{7} \qquad \mathrm{(B)} \frac{3}{8} \qquad \mathrm{(C)} \frac{1}{2} \qquad \mathrm{(D)} \frac{4}{7} \qquad \mathrm{(E)} \frac{5}{8}</math><br />
<br />
[[2007 AMC 8 Problems/Problem 21|Solution]]<br />
<br />
==Problem 22==<br />
<br />
A lemming sits at a corner of a square with side length <math>10</math> meters. The lemming<br />
runs <math>6.2</math> meters along a diagonal toward the opposite corner. It stops, makes<br />
a <math>90</math> degree right turn and runs <math>2</math> more meters. A scientist measures the shortest<br />
distance between the lemming and each side of the square. What is the average<br />
of these four distances in meters?<br />
<br />
<math>\mathrm{(A)}\ 2 \qquad \mathrm{(B)}\ 4.5 \qquad \mathrm{(C)}\ 5 \qquad \mathrm{(D)}\ 6.2 \qquad \mathrm{(E)}\ 7</math><br />
<br />
[[2007 AMC 8 Problems/Problem 22|Solution]]<br />
<br />
==Problem 23==<br />
<br />
What is the area of the shaded part shown in the <math>5</math> x <math>5</math> grid?<br />
<br />
<center>[[Image:AMC8_2007_23.png]]</center><br />
<br />
<math>\mathrm{(A)}\ 4 \qquad\mathrm{(B)}\ 6 \qquad\mathrm{(C)}\ 8 \qquad\mathrm{(D)}\ 10 \qquad\mathrm{(E)}\ 12</math><br />
<br />
[[2007 AMC 8 Problems/Problem 23|Solution]]<br />
<br />
==Problem 24==<br />
<br />
A bag contains four pieces of paper, each labeled with one of the digits "1, 2, 3"<br />
or "4", with no repeats. Three of these pieces are drawn, one at a time without<br />
replacement, to construct a three-digit number. What is the probability that<br />
the three-digit number is a multiple of 3?<br />
<br />
<math>\mathrm{(A)} \frac{1}{4} \qquad \mathrm{(B)} \frac{1}{3} \qquad \mathrm{(C)} \frac{1}{2} \qquad \mathrm{(D)} \frac{2}{3} \qquad \mathrm{(E)} \frac{3}{4}</math><br />
<br />
[[2007 AMC 8 Problems/Problem 24|Solution]]<br />
<br />
==Problem 25==<br />
<br />
On the dart board shown in the Figure, the outer circle has radius <math>6</math> and the inner circle has a radius of 3.<br />
Three radii divide each circle into three congruent<br />
regions, with point values shown. The probability that a dart will hit a given<br />
region is proportional to the area of the region. When two darts hit this board,<br />
the score is the sum of the point values in the regions. What is the probability<br />
that the score is odd?<br />
<br />
<asy><br />
draw(Circle(origin, 2));<br />
draw(Circle(origin, 1));<br />
draw(origin--2*dir(90));<br />
draw(origin--2*dir(210));<br />
draw(origin--2*dir(330));<br />
label("$1$", 0.35*dir(150), dir(150));<br />
label("$1$", 1.3*dir(30), dir(30));<br />
label("$1$", (0,-1.3), dir(270));<br />
label("$2$", 1.3*dir(150), dir(150));<br />
label("$2$", 0.35*dir(30), dir(30));<br />
label("$2$", (0,-0.35), dir(270));<br />
</asy><br />
<br />
<math>\mathrm{(A)} \frac{17}{36} \qquad \mathrm{(B)} \frac{35}{72} \qquad \mathrm{(C)} \frac{1}{2} \qquad \mathrm{(D)} \frac{37}{72} \qquad \mathrm{(E)} \frac{19}{36}</math><br />
<br />
[[2007 AMC 8 Problems/Problem 25|Solution]]<br />
<br />
==See Also==<br />
{{AMC8 box|year=2007|before=[[2006 AMC 8 Problems|2006 AMC 8]]|after=[[2008 AMC 8 Problems|2008 AMC 8]]}}<br />
* [[AMC 8]]<br />
* [[AMC 8 Problems and Solutions]]<br />
* [[Mathematics competition resources]]<br />
<br />
<br />
{{MAA Notice}}</div>Ghostychttps://artofproblemsolving.com/wiki/index.php?title=2005_AMC_8_Problems&diff=1409072005 AMC 8 Problems2020-12-29T12:35:48Z<p>Ghostyc: /* Problem 3 */</p>
<hr />
<div>==Problem 1==<br />
Connie multiplies a number by 2 and gets 60 as her answer. However, she should<br />
have divided the number by 2 to get the correct answer. What is the correct<br />
answer?<br />
<br />
<math> \textbf{(A)}\ 7.5\qquad\textbf{(B)}\ 15\qquad\textbf{(C)}\ 30\qquad\textbf{(D)}\ 120\qquad\textbf{(E)}\ 240 </math><br />
<br />
[[2005 AMC 8 Problems/Problem 1|Solution]]<br />
<br />
==Problem 2==<br />
Karl bought five folders from Pay-A-Lot at a cost of <math> \textdollar 2.50 </math> each.<br />
Pay-A-Lot had a 20%-off sale the following day. How much could<br />
Karl have saved on the purchase by waiting a day?<br />
<br />
<math> \textbf{(A)}\ \textdollar 1.00 \qquad\textbf{(B)}\ \textdollar 2.00 \qquad\textbf{(C)}\ \textdollar 2.50\qquad\textbf{(D)}\ \textdollar 2.75 \qquad\textbf{(E)}\ \textdollar 5.00 </math><br />
<br />
[[2005 AMC 8 Problems/Problem 2|Solution]]<br />
<br />
==Problem 3==<br />
What is the minimum number of small squares that must be colored black so that a line of symmetry lies on the diagonal <math> \overline{BD}</math> of square <math> ABCD</math>?<br />
<asy>defaultpen(linewidth(1));<br />
for ( int x = 0; x < 5; ++x )<br />
{<br />
draw((0,x)--(4,x));<br />
draw((x,0)--(x,4));<br />
}<br />
<br />
fill((1,0)--(2,0)--(2,1)--(1,1)--cycle);<br />
fill((0,3)--(1,3)--(1,4)--(0,4)--cycle);<br />
fill((2,3)--(4,3)--(4,4)--(2,4)--cycle);<br />
fill((3,1)--(4,1)--(4,2)--(3,2)--cycle);<br />
label("$A$", (0, 4), NW);<br />
label("$B$", (4, 4), NE);<br />
label("$C$", (4, 0), SE);<br />
label("$D$", (0, 0), SW);</asy><br />
<br />
<math> \textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5 </math><br />
<br />
[[2005 AMC 8 Problems/Problem 3|Solution]]<br />
<br />
==Problem 4==<br />
A square and a triangle have equal perimeters. The lengths of the three sides of the triangle are 6.1 cm, 8.2 cm and 9.7 cm. What is the area of the square in square centimeters?<br />
<br />
<math> \textbf{(A)}\ 24\qquad\textbf{(B)}\ 25\qquad\textbf{(C)}\ 36\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 64 </math><br />
<br />
[[2005 AMC 8 Problems/Problem 4|Solution]]<br />
<br />
==Problem 5==<br />
Soda is sold in packs of 6, 12 and 24 cans. What is the minimum number of packs needed to buy exactly 90 cans of soda?<br />
<br />
<math> \textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 15 </math><br />
<br />
[[2005 AMC 8 Problems/Problem 5|Solution]]<br />
<br />
==Problem 6==<br />
Suppose <math>d</math> is a digit. For how many values of <math>d</math> is <math>2.00d5 > 2.005</math>?<br />
<br />
<math> \textbf{(A)}\ 0\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 10 </math><br />
<br />
[[2005 AMC 8 Problems/Problem 6|Solution]]<br />
<br />
==Problem 7==<br />
Bill walks <math>\tfrac12</math> mile south, then <math>\tfrac34</math> mile east, and finally <math>\tfrac12</math> mile south. How many miles is he, in a direct line, from his starting point?<br />
<br />
<math> \textbf{(A)}\ 1\qquad\textbf{(B)}\ 1\tfrac14\qquad\textbf{(C)}\ 1\tfrac12\qquad\textbf{(D)}\ 1\tfrac34\qquad\textbf{(E)}\ 2 </math><br />
<br />
[[2005 AMC 8 Problems/Problem 7|Solution]]<br />
<br />
==Problem 8==<br />
Suppose m and n are positive odd integers. Which of the following must also be an odd integer?<br />
<br />
<math> \textbf{(A)}\ m+3n\qquad\textbf{(B)}\ 3m-n\qquad\textbf{(C)}\ 3m^2 + 3n^2\qquad\textbf{(D)}\ (nm + 3)^2\qquad\textbf{(E)}\ 3mn </math><br />
<br />
[[2005 AMC 8 Problems/Problem 8|Solution]]<br />
<br />
==Problem 9==<br />
In quadrilateral <math> ABCD</math>, sides <math> \overline{AB}</math> and <math> \overline{BC}</math> both have length 10, sides <math> \overline{CD}</math> and <math> \overline{DA}</math> both have length 17, and the measure of angle <math> ADC</math> is <math> 60^\circ</math>. What is the length of diagonal <math> \overline{AC}</math>?<br />
<asy>draw((0,0)--(17,0));<br />
draw(rotate(301, (17,0))*(0,0)--(17,0));<br />
picture p;<br />
draw(p, (0,0)--(0,10));<br />
draw(p, rotate(115, (0,10))*(0,0)--(0,10));<br />
add(rotate(3)*p);<br />
<br />
draw((0,0)--(8.25,14.5), linetype("8 8"));<br />
<br />
label("$A$", (8.25, 14.5), N);<br />
label("$B$", (-0.25, 10), W);<br />
label("$C$", (0,0), SW);<br />
label("$D$", (17, 0), E);</asy><br />
<br />
<math> \textbf{(A)}\ 13.5\qquad\textbf{(B)}\ 14\qquad\textbf{(C)}\ 15.5\qquad\textbf{(D)}\ 17\qquad\textbf{(E)}\ 18.5 </math><br />
<br />
[[2005 AMC 8 Problems/Problem 9|Solution]]<br />
<br />
==Problem 10==<br />
Joe had walked half way from home to school when he realized he was late. He ran the rest of the way to school. He ran 3 times as fast as he walked. Joe took 6 minutes to walk half way to school. How many minutes did it take Joe to get from home to school?<br />
<br />
<math> \textbf{(A)}\ 7\qquad\textbf{(B)}\ 7.3\qquad\textbf{(C)}\ 7.7\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 8.3 </math><br />
<br />
[[2005 AMC 8 Problems/Problem 10|Solution]]<br />
<br />
==Problem 11==<br />
The sales tax rate in Bergville is 6%. During a sale at the Bergville Coat Closet, the price of a coat is discounted 20% from its &#36;90.00 price. Two clerks, Jack and Jill, calculate the bill independently. Jack brings up &#36;90.00 and adds 6% sales tax, then subtracts 20% from this total. Jill rings up &#36;90.00, subtracts 20% of the price, then adds 6% of the discounted price for sales tax. What is Jack's total minus Jill's total?<br />
<br />
<math> \textbf{(A)}\ - \textdollar1.06\qquad\textbf{(B)}\ - \textdollar 0.53\qquad\textbf{(C)}\ 0\qquad\textbf{(D)}\ \textdollar 0.53\qquad\textbf{(E)}\ \textdollar 1.06 </math><br />
<br />
[[2005 AMC 8 Problems/Problem 11|Solution]]<br />
<br />
==Problem 12==<br />
Big Al the ape ate 100 delicious yellow bananas from May 1 through May 5. Each day he ate six more bananas than on the previous day. How many delicious bananas did Big Al eat on May 5?<br />
<br />
<math> \textbf{(A)}\ 20\qquad\textbf{(B)}\ 22\qquad\textbf{(C)}\ 30\qquad\textbf{(D)}\ 32\qquad\textbf{(E)}\ 34 </math><br />
<br />
[[2005 AMC 8 Problems/Problem 12|Solution]]<br />
<br />
==Problem 13==<br />
The area of polygon <math> ABCDEF</math> is 52 with <math> AB=8</math>, <math> BC=9</math> and <math> FA=5</math>. What is <math> DE+EF</math>?<br />
<asy>pair a=(0,9), b=(8,9), c=(8,0), d=(4,0), e=(4,4), f=(0,4);<br />
draw(a--b--c--d--e--f--cycle);<br />
draw(shift(0,-.25)*a--shift(.25,-.25)*a--shift(.25,0)*a);<br />
draw(shift(-.25,0)*b--shift(-.25,-.25)*b--shift(0,-.25)*b);<br />
draw(shift(-.25,0)*c--shift(-.25,.25)*c--shift(0,.25)*c);<br />
draw(shift(.25,0)*d--shift(.25,.25)*d--shift(0,.25)*d);<br />
draw(shift(.25,0)*f--shift(.25,.25)*f--shift(0,.25)*f);<br />
label("$A$", a, NW);<br />
label("$B$", b, NE);<br />
label("$C$", c, SE);<br />
label("$D$", d, SW);<br />
label("$E$", e, SW);<br />
label("$F$", f, SW);<br />
label("5", (0,6.5), W);<br />
label("8", (4,9), N);<br />
label("9", (8, 4.5), E);</asy><br />
<br />
<math> \textbf{(A)}\ 7\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 11 </math><br />
<br />
[[2005 AMC 8 Problems/Problem 13|Solution]]<br />
<br />
==Problem 14==<br />
The Little Twelve Basketball League has two divisions, with six teams in each division. Each team plays each of the other teams in its own division twice and every team in the other division once. How many games are scheduled?<br />
<br />
<math> \textbf{(A)}\ 80\qquad\textbf{(B)}\ 96\qquad\textbf{(C)}\ 100\qquad\textbf{(D)}\ 108\qquad\textbf{(E)}\ 192 </math><br />
<br />
[[2005 AMC 8 Problems/Problem 14|Solution]]<br />
<br />
==Problem 15==<br />
How many different isosceles triangles have integer side lengths and perimeter 23?<br />
<br />
<math> \textbf{(A)}\ 2\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 11</math><br />
<br />
[[2005 AMC 8 Problems/Problem 15|Solution]]<br />
<br />
==Problem 16==<br />
A five-legged Martian has a drawer full of socks, each of which is red, white or blue, and there are at least five socks of each color. The Martian pulls out one sock at a time without looking. How many socks must the Martian remove from the drawer to be certain there will be 5 socks of the same color?<br />
<br />
<math> \textbf{(A)}\ 6\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 13\qquad\textbf{(E)}\ 15 </math><br />
<br />
[[2005 AMC 8 Problems/Problem 16|Solution]]<br />
<br />
==Problem 17==<br />
The results of a cross-country team's training run are graphed below. Which student has the greatest average speed?<br />
<asy><br />
for ( int i = 1; i <= 7; ++i )<br />
{<br />
draw((i,0)--(i,6));<br />
}<br />
<br />
for ( int i = 1; i <= 5; ++i )<br />
{<br />
draw((0,i)--(8,i));<br />
}<br />
draw((-0.5,0)--(8,0), linewidth(1));<br />
draw((0,-0.5)--(0,6), linewidth(1));<br />
label("$O$", (0,0), SW);<br />
label(scale(.85)*rotate(90)*"distance", (0, 3), W);<br />
label(scale(.85)*"time", (4, 0), S);<br />
dot((1.25, 4.5));<br />
label(scale(.85)*"Evelyn", (1.25, 4.8), N);<br />
dot((2.5, 2.2));<br />
label(scale(.85)*"Briana", (2.5, 2.2), S);<br />
dot((4.25,5.2));<br />
label(scale(.85)*"Carla", (4.25, 5.2), SE);<br />
dot((5.6, 2.8));<br />
label(scale(.85)*"Debra", (5.6, 2.8), N);<br />
dot((6.8, 1.4));<br />
label(scale(.85)*"Angela", (6.8, 1.4), E);<br />
</asy><br />
<br />
<math> \textbf{(A)}\ \text{Angela}\qquad\textbf{(B)}\ \text{Briana}\qquad\textbf{(C)}\ \text{Carla}\qquad\textbf{(D)}\ \text{Debra}\qquad\textbf{(E)}\ \text{Evelyn} </math><br />
<br />
[[2005 AMC 8 Problems/Problem 17|Solution]]<br />
<br />
==Problem 18==<br />
How many three-digit numbers are divisible by 13?<br />
<br />
<math> \textbf{(A)}\ 7\qquad\textbf{(B)}\ 67\qquad\textbf{(C)}\ 69\qquad\textbf{(D)}\ 76\qquad\textbf{(E)}\ 77</math><br />
<br />
[[2005 AMC 8 Problems/Problem 18|Solution]]<br />
<br />
==Problem 19==<br />
What is the perimeter of trapezoid <math> ABCD</math>?<br />
<br />
<asy>size(3inch, 1.5inch);<br />
pair a=(0,0), b=(18,24), c=(68,24), d=(75,0), f=(68,0), e=(18,0);<br />
draw(a--b--c--d--cycle);<br />
draw(b--e);<br />
draw(shift(0,2)*e--shift(2,2)*e--shift(2,0)*e);<br />
label("30", (9,12), W);<br />
label("50", (43,24), N);<br />
label("25", (71.5, 12), E);<br />
label("24", (18, 12), E);<br />
label("$A$", a, SW);<br />
label("$B$", b, N);<br />
label("$C$", c, N);<br />
label("$D$", d, SE);<br />
label("$E$", e, S);</asy><br />
<br />
<math> \textbf{(A)}\ 180\qquad\textbf{(B)}\ 188\qquad\textbf{(C)}\ 196\qquad\textbf{(D)}\ 200\qquad\textbf{(E)}\ 204 </math><br />
<br />
[[2005 AMC 8 Problems/Problem 19|Solution]]<br />
<br />
==Problem 20==<br />
Alice and Bob play a game involving a circle whose circumference is divided by 12 equally-spaced points. The points are numbered clockwise, from 1 to 12. Both start on point 12. Alice moves clockwise and Bob, counterclockwise.<br />
In a turn of the game, Alice moves 5 points clockwise and Bob moves 9 points counterclockwise. The game ends when they stop on the same point. How many turns will this take?<br />
<br />
<math> \textbf{(A)}\ 6\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 14\qquad\textbf{(E)}\ 24 </math><br />
<br />
[[2005 AMC 8 Problems/Problem 20|Solution]]<br />
<br />
==Problem 21==<br />
How many distinct triangles can be drawn using three of the dots below as vertices?<br />
<br />
<asy>dot(origin^^(1,0)^^(2,0)^^(0,1)^^(1,1)^^(2,1));</asy><br />
<br />
<math> \textbf{(A)}\ 9\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 20\qquad\textbf{(E)}\ 24 </math><br />
<br />
[[2005 AMC 8 Problems/Problem 21|Solution]]<br />
<br />
==Problem 22==<br />
A company sells detergent in three different sized boxes: small (S), medium (M) and large (L). The medium size costs 50% more than the small size and contains 20% less detergent than the large size. The large size contains twice as much detergent as the small size and costs 30% more than the medium size. Rank the three sizes from best to worst buy.<br />
<br />
<math> \textbf{(A)}\ \text{SML}\qquad\textbf{(B)}\ \text{LMS}\qquad\textbf{(C)}\ \text{MSL}\qquad\textbf{(D)}\ \text{LSM}\qquad\textbf{(E)}\ \text{MLS} </math><br />
<br />
[[2005 AMC 8 Problems/Problem 22|Solution]]<br />
<br />
==Problem 23==<br />
Isosceles right triangle <math> ABC</math> encloses a semicircle of area <math> 2\pi</math>. The circle has its center <math> O</math> on hypotenuse <math> \overline{AB}</math> and is tangent to sides <math> \overline{AC}</math> and <math> \overline{BC}</math>. What is the area of triangle <math> ABC</math>?<br />
<br />
<asy>pair a=(4,4), b=(0,0), c=(0,4), d=(4,0), o=(2,2);<br />
draw(circle(o, 2));<br />
clip(a--b--c--cycle);<br />
draw(a--b--c--cycle);<br />
dot(o);<br />
label("$C$", c, NW);<br />
label("$A$", a, NE);<br />
label("$B$", b, SW);</asy><br />
<br />
<math> \textbf{(A)}\ 6\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 3\pi\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 4\pi </math><br />
<br />
[[2005 AMC 8 Problems/Problem 23|Solution]]<br />
<br />
==Problem 24==<br />
A certain calculator has only two keys [+1] and [x2]. When you press one of the keys, the calculator automatically displays the result. For instance, if the calculator originally displayed "9" and you pressed [+1], it would display "10." If you then pressed [x2], it would display "20." Starting with the display "1," what is the fewest number of keystrokes you would need to reach "200"?<br />
<br />
<math> \textbf{(A)}\ 8\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ 11\qquad\textbf{(E)}\ 12</math><br />
<br />
[[2005 AMC 8 Problems/Problem 24|Solution]]<br />
<br />
==Problem 25==<br />
A square with side length 2 and a circle share the same center. The total area of the regions that are inside the circle and outside the square is equal to the total area of the regions that are outside the circle and inside the square. What is the radius of the circle?<br />
<br />
<asy>pair a=(4,4), b=(0,0), c=(0,4), d=(4,0), o=(2,2);<br />
draw(a--d--b--c--cycle);<br />
draw(circle(o, 2.25));</asy><br />
<math> \textbf{(A)}\ \frac{2}{\sqrt{\pi}} \qquad \textbf{(B)}\ \frac{1+\sqrt{2}}{2} \qquad \textbf{(C)}\ \frac{3}{2} \qquad \textbf{(D)}\ \sqrt{3} \qquad \textbf{(E)}\ \sqrt{\pi}</math><br />
<br />
[[2005 AMC 8 Problems/Problem 25|Solution]]<br />
<br />
==See Also==<br />
{{AMC8 box|year=2005|before=[[2004 AMC 8 Problems|2004 AMC 8]]|after=[[2006 AMC 8 Problems|2006 AMC 8]]}}<br />
* [[AMC 8]]<br />
* [[AMC 8 Problems and Solutions]]<br />
* [[Mathematics competition resources]]<br />
<br />
<br />
{{MAA Notice}}</div>Ghostychttps://artofproblemsolving.com/wiki/index.php?title=2006_AMC_12A_Problems&diff=1408362006 AMC 12A Problems2020-12-28T17:43:43Z<p>Ghostyc: /* Problem 19 */</p>
<hr />
<div>{{AMC12 Problems|year=2006|ab=A}}<br />
== Problem 1 ==<br />
Sandwiches at Joe's Fast Food cost <math>3</math> dollars each and sodas cost <math>2</math> dollars each. How many dollars will it cost to purchase <math>5</math> sandwiches and <math>8</math> sodas?<br />
<br />
<math> \mathrm{(A) \ } 31\qquad \mathrm{(B) \ } 32\qquad \mathrm{(C) \ } 33\qquad \mathrm{(D) \ } 34\qquad \mathrm{(E) \ } 35 </math><br />
<br />
[[2006 AMC 12A Problems/Problem 1|Solution]]<br />
<br />
== Problem 2 ==<br />
Define <math>x\otimes y=x^3-y</math>. What is <math>h\otimes (h\otimes h)</math>?<br />
<br />
<math> \mathrm{(A) \ } -h\qquad \mathrm{(B) \ } 0\qquad \mathrm{(C) \ } h\qquad \mathrm{(D) \ } 2h\qquad \mathrm{(E) \ } h^3</math><br />
<br />
[[2006 AMC 12A Problems/Problem 2|Solution]]<br />
<br />
== Problem 3 ==<br />
The ratio of Mary's age to Alice's age is <math>3:5</math>. Alice is <math>30</math> years old. How old is Mary?<br />
<br />
<math> \mathrm{(A) \ } 15\qquad \mathrm{(B) \ } 18\qquad \mathrm{(C) \ } 20\qquad \mathrm{(D) \ } 24\qquad \mathrm{(E) \ } 50</math><br />
<br />
[[2006 AMC 12A Problems/Problem 3|Solution]]<br />
<br />
== Problem 4 ==<br />
A digital watch displays hours and minutes with AM and PM. What is the largest possible sum of the digits in the display?<br />
<br />
<math> \mathrm{(A) \ } 17\qquad \mathrm{(B) \ } 19\qquad \mathrm{(C) \ } 21\qquad \mathrm{(D) \ } 22\qquad \mathrm{(E) \ } 23</math><br />
<br />
[[2006 AMC 12A Problems/Problem 4|Solution]]<br />
<br />
== Problem 5 ==<br />
Doug and Dave shared a pizza with <math>8</math> equally-sized slices. Doug wanted a plain pizza, but Dave wanted anchovies on half the pizza. The cost of a plain pizza was <math>8</math> dollars, and there was an additional cost of <math>2</math> dollars for putting anchovies on one half. Dave ate all the slices of anchovy pizza and one plain slice. Doug ate the remainder. Each paid for what he had eaten. How many more dollars did Dave pay than Doug?<br />
<br />
<math> \mathrm{(A) \ } 1\qquad \mathrm{(B) \ } 2\qquad \mathrm{(C) \ } 3\qquad \mathrm{(D) \ } 4\qquad \mathrm{(E) \ } 5</math><br />
<br />
[[2006 AMC 12A Problems/Problem 5|Solution]]<br />
<br />
== Problem 6 ==<br />
The <math>8\times 18</math> rectangle <math>ABCD</math> is cut into two congruent hexagons, as shown, in such a way that the two hexagons can be repositioned without overlap to form a square. What is <math>y</math>?<br />
<asy><br />
unitsize(3mm);<br />
defaultpen(fontsize(10pt)+linewidth(.8pt));<br />
dotfactor=4;<br />
draw((0,4)--(18,4)--(18,-4)--(0,-4)--cycle);<br />
draw((6,4)--(6,0)--(12,0)--(12,-4));<br />
label("$A$",(0,4),NW);<br />
label("$B$",(18,4),NE);<br />
label("$C$",(18,-4),SE);<br />
label("$D$",(0,-4),SW);<br />
label("$y$",(3,4),S);<br />
label("$y$",(15,-4),N);<br />
label("$18$",(9,4),N);<br />
label("$18$",(9,-4),S);<br />
label("$8$",(0,0),W);<br />
label("$8$",(18,0),E);<br />
dot((0,4));<br />
dot((18,4));<br />
dot((18,-4));<br />
dot((0,-4));</asy><br />
<math> \mathrm{(A) \ } 6\qquad \mathrm{(B) \ } 7\qquad \mathrm{(C) \ } 8\qquad \mathrm{(D) \ } 9\qquad \mathrm{(E) \ } 10</math><br />
<br />
[[2006 AMC 12A Problems/Problem 6|Solution]]<br />
<br />
== Problem 7 ==<br />
Mary is <math>20\%</math> older than Sally, and Sally is <math>40\%</math> younger than Danielle. The sum of their ages is <math>23.2</math> years. How old will Mary be on her next birthday?<br />
<br />
<math> \mathrm{(A) \ } 7\qquad \mathrm{(B) \ } 8\qquad \mathrm{(C) \ } 9\qquad \mathrm{(D) \ } 10\qquad \mathrm{(E) \ } 11</math><br />
<br />
[[2006 AMC 12A Problems/Problem 7|Solution]]<br />
<br />
== Problem 8 ==<br />
How many sets of two or more consecutive positive integers have a sum of <math>15</math>?<br />
<br />
<math> \mathrm{(A) \ } 1\qquad \mathrm{(B) \ } 2\qquad \mathrm{(C) \ } 3\qquad \mathrm{(D) \ } 4\qquad \mathrm{(E) \ } 5</math><br />
<br />
[[2006 AMC 12A Problems/Problem 8|Solution]]<br />
<br />
== Problem 9 ==<br />
Oscar buys <math>13</math> pencils and <math>3</math> erasers for <math>\textdollar 1.00</math>. A pencil costs more than an eraser, and both items cost a whole number of cents. What is the total cost, in cents, of one pencil and one eraser?<br />
<br />
<math> \mathrm{(A) \ } 10\qquad \mathrm{(B) \ } 12\qquad \mathrm{(C) \ } 15\qquad \mathrm{(D) \ } 18\qquad \mathrm{(E) \ } 20</math><br />
<br />
[[2006 AMC 12A Problems/Problem 9|Solution]]<br />
<br />
== Problem 10 ==<br />
For how many real values of <math>x</math> is <math>\sqrt{120-\sqrt{x}}</math> an integer?<br />
<br />
<math> \mathrm{(A) \ } 3\qquad \mathrm{(B) \ } 6\qquad \mathrm{(C) \ } 9\qquad \mathrm{(D) \ } 10\qquad \mathrm{(E) \ } 11</math><br />
<br />
[[2006 AMC 12A Problems/Problem 10|Solution]]<br />
<br />
== Problem 11 ==<br />
Which of the following describes the graph of the equation <math>(x+y)^2=x^2+y^2</math>?<br />
<br />
<math>\mathrm{(A)}\ \text{the empty set}\qquad\mathrm{(B)}\ \text{one point}\qquad\mathrm{(C)}\ \text{two lines}\qquad\mathrm{(D)}\ \text{a circle}\qquad\mathrm{(E)}\ \text{the entire plane}</math><br />
<br />
[[2006 AMC 12A Problems/Problem 11|Solution]]<br />
<br />
== Problem 12 ==<br />
<br />
A number of linked rings, each 1 cm thick, are hanging on a peg. The top ring has an outside diameter of 20 cm. The outside diameter of each of the outer rings is 1 cm less than that of the ring above it. The bottom ring has an outside diameter of 3 cm. What is the distance, in cm, from the top of the top ring to the bottom of the bottom ring?<br />
<!-- <center>[[Image:2006_AMC10A-14.png]]</center> --><br />
<asy>size(7cm); pointpen = black; pathpen = linewidth(0.7);<br />
D(CR((0,0),10));<br />
D(CR((0,0),9.5));<br />
D(CR((0,-18.5),9.5));<br />
D(CR((0,-18.5),9));<br />
MP("$\vdots$",(0,-31),(0,0));<br />
D(CR((0,-39),3));<br />
D(CR((0,-39),2.5));<br />
D(CR((0,-43.5),2.5));<br />
D(CR((0,-43.5),2));<br />
D(CR((0,-47),2));<br />
D(CR((0,-47),1.5));<br />
D(CR((0,-49.5),1.5));<br />
D(CR((0,-49.5),1.0));<br />
<br />
D((12,-10)--(12,10)); MP('20',(12,0),E);<br />
D((12,-51)--(12,-48)); MP('3',(12,-49.5),E);<br />
</asy><br />
<math> \mathrm{(A) \ } 171\qquad \mathrm{(B) \ } 173\qquad \mathrm{(C) \ } 182\qquad \mathrm{(D) \ } 188\qquad \mathrm{(E) \ } 210</math><br />
<br />
[[2006 AMC 12A Problems/Problem 12|Solution]]<br />
<br />
== Problem 13 ==<br />
<!-- <center>[[Image:2006_AMC_12A_Problem_13.gif]]</center> --><br />
The vertices of a <math>3-4-5</math> right triangle are the centers of three mutually externally tangent circles, as shown. What is the sum of the areas of the three circles?<br />
<asy><br />
unitsize(5mm);<br />
defaultpen(fontsize(10pt)+linewidth(.8pt));<br />
pair B=(0,0), C=(5,0);<br />
pair A=intersectionpoints(Circle(B,3),Circle(C,4))[0];<br />
draw(A--B--C--cycle);<br />
draw(Circle(C,3));<br />
draw(Circle(A,1));<br />
draw(Circle(B,2));<br />
label("$A$",A,N);<br />
label("$B$",B,W);<br />
label("$C$",C,E);<br />
label("3",midpoint(B--A),NW);<br />
label("4",midpoint(A--C),NE);<br />
label("5",midpoint(B--C),S);</asy><br />
<br />
<math> \mathrm{(A) \ } 12\pi\qquad \mathrm{(B) \ } \frac{25\pi}{2}\qquad \mathrm{(C) \ } 13\pi\qquad \mathrm{(D) \ } \frac{27\pi}{2}\qquad \mathrm{(E) \ } 14\pi</math><br />
<br />
[[2006 AMC 12A Problems/Problem 13|Solution]]<br />
<br />
== Problem 14 ==<br />
Two farmers agree that pigs are worth <math>300</math> dollars and that goats are worth <math>210</math> dollars. When one farmer owes the other money, he pays the debt in pigs or goats, with "change" received in the form of goats or pigs as necessary. (For example, a <math>390</math> dollar debt could be paid with two pigs, with one goat received in change.) What is the amount of the smallest positive debt that can be resolved in this way?<br />
<br />
<math> \mathrm{(A) \ } \textdollar5 \qquad \mathrm{(B) \ } \textdollar 10 \qquad \mathrm{(C) \ } \textdollar 30 \qquad \mathrm{(D) \ } \textdollar 90 \qquad \mathrm{(E) \ } \textdollar 210</math><br />
<br />
[[2006 AMC 12A Problems/Problem 14|Solution]]<br />
<br />
== Problem 15 ==<br />
Suppose <math>\cos x=0</math> and <math>\cos (x+z)=1/2</math>. What is the smallest possible positive value of <math>z</math>?<br />
<br />
<math> \mathrm{(A) \ } \frac{\pi}{6}\qquad \mathrm{(B) \ } \frac{\pi}{3}\qquad \mathrm{(C) \ } \frac{\pi}{2}\qquad \mathrm{(D) \ } \frac{5\pi}{6}\qquad \mathrm{(E) \ } \frac{7\pi}{6}</math><br />
<br />
[[2006 AMC 12A Problems/Problem 15|Solution]]<br />
<br />
== Problem 16 ==<br />
Circles with centers <math>A</math> and <math>B</math> have radii <math>3</math> and <math>8</math>, respectively. A common internal tangent intersects the circles at <math>C</math> and <math>D</math>, respectively. Lines <math>AB</math> and <math>CD</math> intersect at <math>E</math>, and <math>AE=5</math>. What is <math>CD</math>?<br />
<!-- [[Image:2006_AMC12A-16.png|center]] --><br />
<asy>unitsize(2.5mm);<br />
defaultpen(fontsize(10pt)+linewidth(.8pt));<br />
dotfactor=3;<br />
pair A=(0,0), Ep=(5,0), B=(5+40/3,0);<br />
pair M=midpoint(A--Ep);<br />
pair C=intersectionpoints(Circle(M,2.5),Circle(A,3))[1];<br />
pair D=B+8*dir(180+degrees(C));<br />
dot(A);<br />
dot(C);<br />
dot(B);<br />
dot(D);<br />
draw(C--D);<br />
draw(A--B);<br />
draw(Circle(A,3));<br />
draw(Circle(B,8));<br />
label("$A$",A,W);<br />
label("$B$",B,E);<br />
label("$C$",C,SE);<br />
label("$E$",Ep,SSE);<br />
label("$D$",D,NW);</asy><br />
<math>\mathrm{(A)}\ 13\qquad\mathrm{(B)}\ \frac{44}{3}\qquad\mathrm{(C)}\ \sqrt{221}\qquad\mathrm{(D)}\ \sqrt{255}\qquad\mathrm{(E)}\ \frac{55}{3}</math><br />
<br />
[[2006 AMC 12A Problems/Problem 16|Solution]]<br />
<br />
== Problem 17 ==<br />
Square <math>ABCD</math> has side length <math>s</math>, a circle centered at <math>E</math> has radius <math>r</math>, and <math>r</math> and <math>s</math> are both rational. The circle passes through <math>D</math>, and <math>D</math> lies on <math>\overline{BE}</math>. Point <math>F</math> lies on the circle, on the same side of <math>\overline{BE}</math> as <math>A</math>. Segment <math>AF</math> is tangent to the circle, and <math>AF=\sqrt{9+5\sqrt{2}}</math>. What is <math>r/s</math>?<br />
<!-- [[Image:AMC12_2006A_17.png|center]] --><br />
<asy>unitsize(6mm);<br />
defaultpen(linewidth(.8pt)+fontsize(10pt));<br />
dotfactor=3;<br />
pair B=(0,0), C=(3,0), D=(3,3), A=(0,3);<br />
pair Ep=(3+5*sqrt(2)/6,3+5*sqrt(2)/6);<br />
pair F=intersectionpoints(Circle(A,sqrt(9+5*sqrt(2))),Circle(Ep,5/3))[0];<br />
pair[] dots={A,B,C,D,Ep,F};<br />
draw(A--F);<br />
draw(Circle(Ep,5/3));<br />
draw(A--B--C--D--cycle);<br />
dot(dots);<br />
label("$A$",A,NW);<br />
label("$B$",B,SW);<br />
label("$C$",C,SE);<br />
label("$D$",D,SW);<br />
label("$E$",Ep,E);<br />
label("$F$",F,NW);<br />
</asy><br />
<math> \mathrm{(A) \ } \frac{1}{2}\qquad \mathrm{(B) \ } \frac{5}{9}\qquad \mathrm{(C) \ } \frac{3}{5}\qquad \mathrm{(D) \ } \frac{5}{3}\qquad \mathrm{(E) \ } \frac{9}{5}</math><br />
<br />
[[2006 AMC 12A Problems/Problem 17|Solution]]<br />
<br />
== Problem 18 ==<br />
The function <math>f</math> has the property that for each real number <math>x</math> in its domain, <math>1/x</math> is also in its domain and <br />
<br />
<math>f(x)+f\left(\frac{1}{x}\right)=x</math><br />
<br />
What is the largest set of real numbers that can be in the domain of <math>f</math>?<br />
<br />
<math> \mathrm{(A) \ } \{x|x\ne 0\}\qquad \mathrm{(B) \ } \{x|x<0\}\qquad \mathrm{(C) \ } \{x|x>0\}\qquad \mathrm{(D) \ } \{x|x\ne -1\;</math> <math>\mathrm{and}\; x\ne 0\;\mathrm{and}\; x\ne 1\}\qquad \mathrm{(E) \ } \{-1,1\}</math><br />
<br />
[[2006 AMC 12A Problems/Problem 18|Solution]]<br />
<br />
== Problem 19 ==<br />
Circles with centers <math>(2,4)</math> and <math>(14,9)</math> have radii <math>4</math> and <math>9</math>, respectively. The equation of a common external tangent to the circles can be written in the form <math>y=mx+b</math> with <math>m>0</math>. What is <math>b</math>?<br />
<br />
<!-- [[Image:AMC12_2006A_19.png|center]] --><br />
<asy><br />
size(150);<br />
defaultpen(linewidth(0.7)+fontsize(8));<br />
draw(circle((2,4),4));draw(circle((14,9),9));<br />
draw((0,-2)--(0,20));draw((-6,0)--(25,0));<br />
draw((2,4)--(2,4)+4*expi(pi*4.5/11));<br />
draw((14,9)--(14,9)+9*expi(pi*6/7));<br />
label("4",(2,4)+2*expi(pi*4.5/11),(-1,0));<br />
label("9",(14,9)+4.5*expi(pi*6/7),(1,1));<br />
label("(2,4)",(2,4),(0.5,-1.5));label("(14,9)",(14,9),(1,-1));<br />
draw((-4,120*-4/119+912/119)--(11,120*11/119+912/119));<br />
dot((2,4)^^(14,9));<br />
</asy><br />
<br />
<math> \mathrm{(A) \ } \frac{908}{119}\qquad \mathrm{(B) \ } \frac{909}{119}\qquad \mathrm{(C) \ } \frac{130}{17}\qquad \mathrm{(D) \ } \frac{911}{119}\qquad \mathrm{(E) \ } \frac{912}{119}</math><br />
<br />
[[2006 AMC 12A Problems/Problem 19|Solution]]<br />
<br />
== Problem 20 ==<br />
A bug starts at one vertex of a cube and moves along the edges of the cube according to the following rule. At each vertex the bug will choose to travel along one of the three edges emanating from that vertex. Each edge has equal probability of being chosen, and all choices are independent. What is the probability that after seven moves the bug will have visited every vertex exactly once?<br />
<br />
<math> \mathrm{(A) \ } \frac{1}{2187}\qquad \mathrm{(B) \ } \frac{1}{729}\qquad \mathrm{(C) \ } \frac{2}{243}\qquad \mathrm{(D) \ } \frac{1}{81}\qquad \mathrm{(E) \ } \frac{5}{243}</math><br />
<br />
[[2006 AMC 12A Problems/Problem 20|Solution]]<br />
<br />
== Problem 21 ==<br />
Let <br />
<br />
<math>S_1=\{(x,y)|\log_{10}(1+x^2+y^2)\le 1+\log_{10}(x+y)\}</math><br />
<br />
and <br />
<br />
<math>S_2=\{(x,y)|\log_{10}(2+x^2+y^2)\le 2+\log_{10}(x+y)\}</math>.<br />
<br />
What is the ratio of the area of <math>S_2</math> to the area of <math>S_1</math>?<br />
<br />
<math> \mathrm{(A) \ } 98\qquad \mathrm{(B) \ } 99\qquad \mathrm{(C) \ } 100\qquad \mathrm{(D) \ } 101\qquad \mathrm{(E) \ } 102</math><br />
<br />
[[2006 AMC 12A Problems/Problem 21|Solution]]<br />
<br />
== Problem 22 ==<br />
A circle of radius <math>r</math> is concentric with and outside a regular hexagon of side length <math>2</math>. The probability that three entire sides of hexagon are visible from a randomly chosen point on the circle is <math>1/2</math>. What is <math>r</math>?<br />
<br />
<math> \mathrm{(A) \ } 2\sqrt{2}+2\sqrt{3}\qquad \mathrm{(B) \ } 3\sqrt{3}+\sqrt{2}\qquad \mathrm{(C) \ } 2\sqrt{6}+\sqrt{3} \qquad \mathrm{(D) \ } 3\sqrt{2}+\sqrt{6}\qquad \mathrm{(E) \ } 6\sqrt{2}-\sqrt{3}</math><br />
<br />
[[2006 AMC 12A Problems/Problem 22|Solution]]<br />
<br />
== Problem 23 ==<br />
Given a finite sequence <math>S=(a_1,a_2,\ldots ,a_n)</math> of <math>n</math> real numbers, let <math>A(S)</math> be the sequence <br />
<br />
<math>\left(\frac{a_1+a_2}{2},\frac{a_2+a_3}{2},\ldots ,\frac{a_{n-1}+a_n}{2}\right)</math><br />
<br />
of <math>n-1</math> real numbers. Define <math>A^1(S)=A(S)</math> and, for each integer <math>m</math>, <math>2\le m\le n-1</math>, define <math>A^m(S)=A(A^{m-1}(S))</math>. Suppose <math>x>0</math>, and let <math>S=(1,x,x^2,\ldots ,x^{100})</math>. If <math>A^{100}(S)=(1/2^{50})</math>, then what is <math>x</math>?<br />
<br />
<math> \mathrm{(A) \ } 1-\frac{\sqrt{2}}{2}\qquad \mathrm{(B) \ } \sqrt{2}-1\qquad \mathrm{(C) \ } \frac{1}{2}\qquad \mathrm{(D) \ } 2-\sqrt{2}\qquad \mathrm{(E) \ } \frac{\sqrt{2}}{2}</math><br />
<br />
[[2006 AMC 12A Problems/Problem 23|Solution]]<br />
<br />
== Problem 24 ==<br />
The expression <br />
<br />
<math>(x+y+z)^{2006}+(x-y-z)^{2006}</math><br />
<br />
is simplified by expanding it and combining like terms. How many terms are in the simplified expression?<br />
<br />
<math> \mathrm{(A) \ } 6018\qquad \mathrm{(B) \ } 671,676\qquad \mathrm{(C) \ } 1,007,514\qquad \mathrm{(D) \ } 1,008,016\qquad \mathrm{(E) \ } 2,015,028</math><br />
<br />
[[2006 AMC 12A Problems/Problem 24|Solution]]<br />
<br />
== Problem 25 ==<br />
How many non-empty subsets <math>S</math> of <math>\lbrace 1,2,3,\ldots ,15\rbrace</math> have the following two properties? <br />
<br />
<math>(1)</math> No two consecutive integers belong to <math>S</math>.<br />
<br />
<math>(2)</math> If <math>S</math> contains <math>k</math> elements, then <math>S</math> contains no number less than <math>k</math>.<br />
<br />
<math> \mathrm{(A) \ } 277\qquad \mathrm{(B) \ } 311\qquad \mathrm{(C) \ } 376\qquad \mathrm{(D) \ } 377\qquad \mathrm{(E) \ } 405</math><br />
<br />
[[2006 AMC 12A Problems/Problem 25|Solution]]<br />
<br />
== See also ==<br />
{{AMC12 box|year=2006|ab=A|before=[[2005 AMC 12B Problems]]|after=[[2006 AMC 12B Problems]]}}<br />
* [[AMC 12]]<br />
* [[AMC 12 Problems and Solutions]]<br />
* [http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=142 2006 AMC A Math Jam Transcript]<br />
* [[Mathematics competition resources]]<br />
{{MAA Notice}}</div>Ghostychttps://artofproblemsolving.com/wiki/index.php?title=1983_AIME_Problems/Problem_4&diff=1302261983 AIME Problems/Problem 42020-08-02T08:33:11Z<p>Ghostyc: /* Problem */</p>
<hr />
<div>==Problem==<br />
A machine-shop cutting tool has the shape of a notched circle, as shown. The radius of the circle is <math>\sqrt{50}</math> cm, the length of <math>AB</math> is <math>6</math> cm and that of <math>BC</math> is <math>2</math> cm. The angle <math>ABC</math> is a right angle. Find the square of the distance (in centimeters) from <math>B</math> to the center of the circle.<br />
<br />
<asy><br />
size(150); <br />
defaultpen(linewidth(0.6)+fontsize(11));<br />
real r=10;<br />
pair O=(0,0),<br />
A=r*dir(45),B=(A.x,A.y-r),C;<br />
path P=circle(O,r);<br />
C=intersectionpoint(B--(B.x+r,B.y),P);<br />
draw(P);<br />
draw(C--B--A--B);<br />
dot(A); dot(B); dot(C);<br />
label("$A$",A,NE);<br />
label("$B$",B,S);<br />
label("$C$",C,SE);<br />
</asy><br />
[[File:pdfresizer.com-pdf-convert-aimeq4.png]]<br />
<br />
==Solution==<br />
===Solution 1===<br />
Because we are given a right angle, we look for ways to apply the [[Pythagorean Theorem]]. Let the foot of the [[perpendicular]] from <math>O</math> to <math>AB</math> be <math>D</math> and let the foot of the perpendicular from <math>O</math> to the [[line]] <math>BC</math> be <math>E</math>. Let <math>OE=x</math> and <math>OD=y</math>. We're trying to find <math>x^2+y^2</math>.<br />
<br />
<center><asy><br />
size(150); defaultpen(linewidth(0.6)+fontsize(11));<br />
real r=10;<br />
pair O=(0,0),A=r*dir(45),B=(A.x,A.y-r),C;<br />
pair D=(A.x,0),F=(0,B.y);<br />
path P=circle(O,r);<br />
C=intersectionpoint(B--(B.x+r,B.y),P);<br />
draw(P);<br />
draw(C--B--O--A--B);<br />
draw(D--O--F--B,dashed);<br />
dot(O); dot(A); dot(B); dot(C);<br />
label("$O$",O,SW);<br />
label("$A$",A,NE);<br />
label("$B$",B,S);<br />
label("$C$",C,SE);<br />
label("$D$",D,NE);<br />
label("$E$",F,SW);<br />
</asy></center><!-- Asymptote replacement for Image:AIME_83_-4_Modified.JPG by bpms --><br />
<br />
Applying the Pythagorean Theorem, <math>OA^2 = OD^2 + AD^2</math> and <math>OC^2 = EC^2 + EO^2</math>.<br />
<br />
Thus, <math>\left(\sqrt{50}\right)^2 = y^2 + (6-x)^2</math>, and <math>\left(\sqrt{50}\right)^2 = x^2 + (y+2)^2</math>. We solve this system to get <math>x = 1</math> and <math>y = 5</math>, such that the answer is <math>1^2 + 5^2 = \boxed{026}</math>.<br />
<br />
===Solution 2===<br />
Drop perpendiculars from <math>O</math> to <math>AB</math> (with foot <math>T_1</math>), <math>M</math> to <math>OT_1</math> (with foot <math>T_2</math>), and <math>M</math> to <math>AB</math> (with foot <math>T_3</math>).<br />
Also, mark the midpoint <math>M</math> of <math>AC</math>.<br />
<br />
Then the problem is trivialized. Why?<br />
<center><asy><br />
size(200);<br />
pair dl(string name, pair loc, pair offset) {<br />
dot(loc);<br />
label(name,loc,offset);<br />
return loc;<br />
};<br />
pair a[] = {(0,0),(0,5),(1,5),(1,7),(-2,6),(-5,5),(-2,5),(-2,6),(0,6)};<br />
string n[] = {"O","$T_1$","B","C","M","A","$T_3$","M","$T_2$"};<br />
for(int i=0;i<a.length;++i) {<br />
dl(n[i],a[i],dir(degrees(a[i],false) ) );<br />
draw(a[(i-1)%a.length]--a[i]);<br />
};<br />
dot(a);<br />
draw(a[5]--a[1]);<br />
draw(a[0]--a[3]);<br />
draw(a[0]--a[4]);<br />
draw(a[0]--a[2]);<br />
draw(a[0]--a[5]);<br />
<br />
draw(a[5]--a[2]--a[3]--cycle,blue+linewidth(0.7));<br />
draw(a[0]--a[8]--a[7]--cycle,blue+linewidth(0.7));<br />
</asy></center><br />
First notice that by computation, <math>OAC</math> is a <math>\sqrt {50} - \sqrt {40} - \sqrt {50}</math> isosceles triangle, so <math>AC = MO</math>.<br />
Then, notice that <math>\angle MOT_2 = \angle T_3MO = \angle BAC</math>. Therefore, the two blue triangles are congruent, from which we deduce <math>MT_2 = 2</math> and <math>OT_2 = 6</math>. As <math>T_3B = 3</math> and <math>MT_3 = 1</math>, we subtract and get <math>OT_1 = 5,T_1B = 1</math>. Then the Pythagorean Theorem tells us that <math>OB^2 = \boxed{026}</math>.<br />
<br />
===Solution 3===<br />
Draw segment <math>OB</math> with length <math>x</math>, and draw radius <math>OQ</math> such that <math>OQ</math> bisects chord <math>AC</math> at point <math>M</math>. This also means that <math>OQ</math> is perpendicular to <math>AC</math>. By the Pythagorean Theorem, we get that <math>AC=\sqrt{(BC)^2+(AB)^2}=2\sqrt{10}</math>, and therefore <math>AM=\sqrt{10}</math>. Also by the Pythagorean theorem, we can find that <math>OM=\sqrt{50-10}=2\sqrt{10}</math>.<br />
<br />
Next, find <math>\angle BAC=\arctan{\left(\frac{2}{6}\right)}</math> and <math>\angle OAM=\arctan{\left(\frac{2\sqrt{10}}{\sqrt{10}}\right)}</math>. Since <math>\angle OAB=\angle OAM-\angle BAC</math>, we get <cmath>\angle OAB=\arctan{2}-\arctan{\frac{1}{3}}</cmath><cmath>\tan{(\angle OAB)}=\tan{(\arctan{2}-\arctan{\frac{1}{3}})}</cmath>By the subtraction formula for <math>\tan</math>, we get<cmath>\tan{(\angle OAB)}=\frac{2-\frac{1}{3}}{1+2\cdot \frac{1}{3}}</cmath><cmath>\tan{(\angle OAB)}=1</cmath><cmath>\cos{(\angle OAB)}=\frac{1}{\sqrt{2}}</cmath>Finally, by the Law of Cosines on <math>\triangle OAB</math>, we get <cmath>x^2=50+36-2(6)\sqrt{50}\frac{1}{\sqrt{2}}</cmath><cmath>x^2=\boxed{026}.</cmath><br />
<br />
== See Also ==<br />
{{AIME box|year=1983|num-b=3|num-a=5}}<br />
<br />
[[Category:Intermediate Geometry Problems]]</div>Ghostychttps://artofproblemsolving.com/wiki/index.php?title=1983_AIME_Problems/Problem_4&diff=1302251983 AIME Problems/Problem 42020-08-02T08:31:16Z<p>Ghostyc: /* Problem */</p>
<hr />
<div>==Problem==<br />
A machine-shop cutting tool has the shape of a notched circle, as shown. The radius of the circle is <math>\sqrt{50}</math> cm, the length of <math>AB</math> is <math>6</math> cm and that of <math>BC</math> is <math>2</math> cm. The angle <math>ABC</math> is a right angle. Find the square of the distance (in centimeters) from <math>B</math> to the center of the circle.<br />
<br />
<asy><br />
size(180);<br />
import three; pathpen = black+linewidth(0.65); pointpen = black;<br />
currentprojection = perspective(30,-20,10);<br />
real s = 6 * 2^.5;<br />
triple A=(0,0,0),B=(s,0,0),C=(s,s,0),D=(0,s,0),E=(-s/2,s/2,6),F=(3*s/2,s/2,6);<br />
draw(A--B--C--D--A--E--D);<br />
draw(B--F--C);<br />
draw(E--F);<br />
label("A",A,W);<br />
label("B",B,S);<br />
label("C",C,SE);<br />
label("D",D,NE);<br />
label("E",E,N);<br />
label("F",F,N);<br />
</asy><br />
[[File:pdfresizer.com-pdf-convert-aimeq4.png]]<br />
<br />
==Solution==<br />
===Solution 1===<br />
Because we are given a right angle, we look for ways to apply the [[Pythagorean Theorem]]. Let the foot of the [[perpendicular]] from <math>O</math> to <math>AB</math> be <math>D</math> and let the foot of the perpendicular from <math>O</math> to the [[line]] <math>BC</math> be <math>E</math>. Let <math>OE=x</math> and <math>OD=y</math>. We're trying to find <math>x^2+y^2</math>.<br />
<br />
<center><asy><br />
size(150); defaultpen(linewidth(0.6)+fontsize(11));<br />
real r=10;<br />
pair O=(0,0),A=r*dir(45),B=(A.x,A.y-r),C;<br />
pair D=(A.x,0),F=(0,B.y);<br />
path P=circle(O,r);<br />
C=intersectionpoint(B--(B.x+r,B.y),P);<br />
draw(P);<br />
draw(C--B--O--A--B);<br />
draw(D--O--F--B,dashed);<br />
dot(O); dot(A); dot(B); dot(C);<br />
label("$O$",O,SW);<br />
label("$A$",A,NE);<br />
label("$B$",B,S);<br />
label("$C$",C,SE);<br />
label("$D$",D,NE);<br />
label("$E$",F,SW);<br />
</asy></center><!-- Asymptote replacement for Image:AIME_83_-4_Modified.JPG by bpms --><br />
<br />
Applying the Pythagorean Theorem, <math>OA^2 = OD^2 + AD^2</math> and <math>OC^2 = EC^2 + EO^2</math>.<br />
<br />
Thus, <math>\left(\sqrt{50}\right)^2 = y^2 + (6-x)^2</math>, and <math>\left(\sqrt{50}\right)^2 = x^2 + (y+2)^2</math>. We solve this system to get <math>x = 1</math> and <math>y = 5</math>, such that the answer is <math>1^2 + 5^2 = \boxed{026}</math>.<br />
<br />
===Solution 2===<br />
Drop perpendiculars from <math>O</math> to <math>AB</math> (with foot <math>T_1</math>), <math>M</math> to <math>OT_1</math> (with foot <math>T_2</math>), and <math>M</math> to <math>AB</math> (with foot <math>T_3</math>).<br />
Also, mark the midpoint <math>M</math> of <math>AC</math>.<br />
<br />
Then the problem is trivialized. Why?<br />
<center><asy><br />
size(200);<br />
pair dl(string name, pair loc, pair offset) {<br />
dot(loc);<br />
label(name,loc,offset);<br />
return loc;<br />
};<br />
pair a[] = {(0,0),(0,5),(1,5),(1,7),(-2,6),(-5,5),(-2,5),(-2,6),(0,6)};<br />
string n[] = {"O","$T_1$","B","C","M","A","$T_3$","M","$T_2$"};<br />
for(int i=0;i<a.length;++i) {<br />
dl(n[i],a[i],dir(degrees(a[i],false) ) );<br />
draw(a[(i-1)%a.length]--a[i]);<br />
};<br />
dot(a);<br />
draw(a[5]--a[1]);<br />
draw(a[0]--a[3]);<br />
draw(a[0]--a[4]);<br />
draw(a[0]--a[2]);<br />
draw(a[0]--a[5]);<br />
<br />
draw(a[5]--a[2]--a[3]--cycle,blue+linewidth(0.7));<br />
draw(a[0]--a[8]--a[7]--cycle,blue+linewidth(0.7));<br />
</asy></center><br />
First notice that by computation, <math>OAC</math> is a <math>\sqrt {50} - \sqrt {40} - \sqrt {50}</math> isosceles triangle, so <math>AC = MO</math>.<br />
Then, notice that <math>\angle MOT_2 = \angle T_3MO = \angle BAC</math>. Therefore, the two blue triangles are congruent, from which we deduce <math>MT_2 = 2</math> and <math>OT_2 = 6</math>. As <math>T_3B = 3</math> and <math>MT_3 = 1</math>, we subtract and get <math>OT_1 = 5,T_1B = 1</math>. Then the Pythagorean Theorem tells us that <math>OB^2 = \boxed{026}</math>.<br />
<br />
===Solution 3===<br />
Draw segment <math>OB</math> with length <math>x</math>, and draw radius <math>OQ</math> such that <math>OQ</math> bisects chord <math>AC</math> at point <math>M</math>. This also means that <math>OQ</math> is perpendicular to <math>AC</math>. By the Pythagorean Theorem, we get that <math>AC=\sqrt{(BC)^2+(AB)^2}=2\sqrt{10}</math>, and therefore <math>AM=\sqrt{10}</math>. Also by the Pythagorean theorem, we can find that <math>OM=\sqrt{50-10}=2\sqrt{10}</math>.<br />
<br />
Next, find <math>\angle BAC=\arctan{\left(\frac{2}{6}\right)}</math> and <math>\angle OAM=\arctan{\left(\frac{2\sqrt{10}}{\sqrt{10}}\right)}</math>. Since <math>\angle OAB=\angle OAM-\angle BAC</math>, we get <cmath>\angle OAB=\arctan{2}-\arctan{\frac{1}{3}}</cmath><cmath>\tan{(\angle OAB)}=\tan{(\arctan{2}-\arctan{\frac{1}{3}})}</cmath>By the subtraction formula for <math>\tan</math>, we get<cmath>\tan{(\angle OAB)}=\frac{2-\frac{1}{3}}{1+2\cdot \frac{1}{3}}</cmath><cmath>\tan{(\angle OAB)}=1</cmath><cmath>\cos{(\angle OAB)}=\frac{1}{\sqrt{2}}</cmath>Finally, by the Law of Cosines on <math>\triangle OAB</math>, we get <cmath>x^2=50+36-2(6)\sqrt{50}\frac{1}{\sqrt{2}}</cmath><cmath>x^2=\boxed{026}.</cmath><br />
<br />
== See Also ==<br />
{{AIME box|year=1983|num-b=3|num-a=5}}<br />
<br />
[[Category:Intermediate Geometry Problems]]</div>Ghostychttps://artofproblemsolving.com/wiki/index.php?title=2011_AMC_8_Problems/Problem_25&diff=1293242011 AMC 8 Problems/Problem 252020-07-26T09:26:20Z<p>Ghostyc: /* Solution 3 */</p>
<hr />
<div>==Problem==<br />
A circle with radius <math>1</math> is inscribed in a square and circumscribed about another square as shown. Which fraction is closest to the ratio of the circle's shaded area to the area between the two squares?<br />
<br />
<asy><br />
filldraw((-1,-1)--(-1,1)--(1,1)--(1,-1)--cycle,gray,black);<br />
filldraw(Circle((0,0),1), mediumgray,black);<br />
filldraw((-1,0)--(0,1)--(1,0)--(0,-1)--cycle,white,black);</asy><br />
<br />
<math> \textbf{(A)}\ \frac{1}2\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ \frac{3}2\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ \frac{5}2 </math><br />
<br />
==Solution 1==<br />
The area of the smaller square is the one half of the product of its diagonals. Note that the distance from a corner of the smaller square to the center is equivalent to the circle's radius so the diagonal is equal to the diameter: <math>2*2*1/2=2.</math><br />
<br />
The circle's shaded area is the area of the smaller square subtracted from the area of the circle: <math>\pi - 2.</math><br />
<br />
If you draw the diagonals of the smaller square, you will see that the larger square is split <math>4</math> congruent half-shaded squares. The area between the squares is equal to the area of the smaller square: <math>2.</math><br />
<br />
Approximating <math>\pi</math> to <math>3.14,</math> the ratio of the circle's shaded area to the area between the two squares is about<br />
<br />
<cmath>\frac{\pi-2}{2} \approx \frac{3.14-2}{2} = \frac{1.14}{2} \approx \boxed{\textbf{(A)}\ \frac12}</cmath><br />
<br />
==Solution 2==<br />
For the ratio of the circle's shaded area to the area between the squares to be <math>1,</math> they would have to be approximately the same size. For any ratio larger than that, the circle's shaded area must be greater. However, we can clearly see that the circle's shaded area is part of the area between the squares, and is approximately <math>\boxed{\textbf{(A)}\ \frac12}</math>.<br />
<br />
Note that this solution is not rigorous, because we still should show that the ratio is less than <math>\frac{3}{4}</math>.<br />
<br />
<br />
==Solution 3==<br />
<br />
Set the side length of the bigger square to be <math>8</math>. <br />
Then the area of the big square is <math>8^2 =64</math> and <br />
the area of the small square <math>(4\sqrt{2})^2 = 32</math>. <br />
The difference is <math>32</math>. <br />
The area of the circle is <math>4^2</math> times <math>\pi</math> which is <math>16 \pi</math> or about <math>48</math>. <br />
Knowing the area of the small square is <math>32</math>. <math>48-32</math> is <math>16</math>. <br />
The area of the big square is <math>64</math>. <br />
So <math>32/64</math> is <math>1/2</math>, or <math>\boxed{\textbf{(A)}\ \frac12}</math>.<br />
<br />
<br />
- by goldenn<br />
<br />
- (tex) updated by CasperYC on 26th July 2020<br />
<br />
==See Also==<br />
{{AMC8 box|year=2011|num-b=24|after=Last Problem}}<br />
{{MAA Notice}}</div>Ghostychttps://artofproblemsolving.com/wiki/index.php?title=2011_AMC_8_Problems/Problem_25&diff=1293232011 AMC 8 Problems/Problem 252020-07-26T09:25:48Z<p>Ghostyc: /* Solution 3 */</p>
<hr />
<div>==Problem==<br />
A circle with radius <math>1</math> is inscribed in a square and circumscribed about another square as shown. Which fraction is closest to the ratio of the circle's shaded area to the area between the two squares?<br />
<br />
<asy><br />
filldraw((-1,-1)--(-1,1)--(1,1)--(1,-1)--cycle,gray,black);<br />
filldraw(Circle((0,0),1), mediumgray,black);<br />
filldraw((-1,0)--(0,1)--(1,0)--(0,-1)--cycle,white,black);</asy><br />
<br />
<math> \textbf{(A)}\ \frac{1}2\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ \frac{3}2\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ \frac{5}2 </math><br />
<br />
==Solution 1==<br />
The area of the smaller square is the one half of the product of its diagonals. Note that the distance from a corner of the smaller square to the center is equivalent to the circle's radius so the diagonal is equal to the diameter: <math>2*2*1/2=2.</math><br />
<br />
The circle's shaded area is the area of the smaller square subtracted from the area of the circle: <math>\pi - 2.</math><br />
<br />
If you draw the diagonals of the smaller square, you will see that the larger square is split <math>4</math> congruent half-shaded squares. The area between the squares is equal to the area of the smaller square: <math>2.</math><br />
<br />
Approximating <math>\pi</math> to <math>3.14,</math> the ratio of the circle's shaded area to the area between the two squares is about<br />
<br />
<cmath>\frac{\pi-2}{2} \approx \frac{3.14-2}{2} = \frac{1.14}{2} \approx \boxed{\textbf{(A)}\ \frac12}</cmath><br />
<br />
==Solution 2==<br />
For the ratio of the circle's shaded area to the area between the squares to be <math>1,</math> they would have to be approximately the same size. For any ratio larger than that, the circle's shaded area must be greater. However, we can clearly see that the circle's shaded area is part of the area between the squares, and is approximately <math>\boxed{\textbf{(A)}\ \frac12}</math>.<br />
<br />
Note that this solution is not rigorous, because we still should show that the ratio is less than <math>\frac{3}{4}</math>.<br />
<br />
<br />
==Solution 3==<br />
<br />
Set the side length of the bigger square to be <math>8</math>. <br />
Then the area of the big square is <math>8^2 =64</math> and <br />
the area of the small square <math>(4\sqrt{2})^2 = 32</math>. <br />
The difference is <math>32</math>. <br />
The area of the circle is <math>4^2</math> times <math>\pi</math> which is <math>16 \pi</math> or about <math>48</math>. <br />
Knowing the area of the small square is <math>32</math>. <math>48-32</math> is <math>16</math>. <br />
The area of the big square is <math>64</math>. <br />
So <math>32/64</math> is <math>1/2</math>, or <math>\boxed{\textbf{(A)}\ \frac12}</math>.<br />
<br />
- by goldenn<br />
- (tex) updated by CasperYC<br />
<br />
==See Also==<br />
{{AMC8 box|year=2011|num-b=24|after=Last Problem}}<br />
{{MAA Notice}}</div>Ghostyc