https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Problemsolver2026&feedformat=atom AoPS Wiki - User contributions [en] 2022-08-07T16:29:26Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10A_Problems/Problem_1&diff=159231 2019 AMC 10A Problems/Problem 1 2021-07-28T19:56:14Z <p>Problemsolver2026: /* Video Solution 2 */</p> <hr /> <div>== Problem ==<br /> What is the value of &lt;cmath&gt;2^{\left(0^{\left(1^9\right)}\right)}+\left(\left(2^0\right)^1\right)^9?&lt;/cmath&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 0 \qquad\textbf{(B) } 1 \qquad\textbf{(C) } 2 \qquad\textbf{(D) } 3 \qquad\textbf{(E) } 4&lt;/math&gt;<br /> <br /> == Solution == <br /> &lt;math&gt;2^{\left(0^{\left(1^9\right)}\right)}+\left(\left(2^0\right)^1\right)^9&lt;/math&gt;<br /> <br /> &lt;math&gt;= 1+1 = \boxed{2}&lt;/math&gt; which corresponds to &lt;math&gt;\boxed{\text{C}}&lt;/math&gt;.<br /> <br /> == Video Solution 1 ==<br /> <br /> https://youtu.be/K8je0WYBHFc<br /> <br /> Education, The Study Of Everything<br /> <br /> <br /> == Video Solution 2==<br /> https://youtu.be/Ad8WKcwZcTA<br /> <br /> ~savannahsolver<br /> <br /> == Video Solution 3==<br /> https://www.youtube.com/watch?v=OhCy9c2RTFo&amp;list=PLbhMrFqoXXwnSCZShTPinX9ZHWjzIlQr_&amp;index=1<br /> <br /> == See Also ==<br /> {{AMC10 box|year=2019|ab=A|before=First Problem|num-a=2}}<br /> {{MAA Notice}}</div> Problemsolver2026 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_22&diff=159230 2019 AMC 8 Problems/Problem 22 2021-07-28T19:53:42Z <p>Problemsolver2026: /* Video explaining solution */</p> <hr /> <div>==Problem 22==<br /> A store increased the original price of a shirt by a certain percent and then lowered the new price by the same amount. Given that the resulting price was &lt;math&gt;84\%&lt;/math&gt; of the original price, by what percent was the price increased and decreased&lt;math&gt;?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }16\qquad\textbf{(B) }20\qquad\textbf{(C) }28\qquad\textbf{(D) }36\qquad\textbf{(E) }40&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Suppose the fraction of discount is &lt;math&gt;x&lt;/math&gt;. That means &lt;math&gt;(1-x)(1+x)=0.84&lt;/math&gt;; so &lt;math&gt;1-x^{2}=0.84&lt;/math&gt;, and &lt;math&gt;(x^{2})=0.16&lt;/math&gt;, obtaining &lt;math&gt;x=0.4&lt;/math&gt;. Therefore, the price was increased and decreased by &lt;math&gt;40&lt;/math&gt;%, or &lt;math&gt;\boxed{\textbf{(E)}\ 40}&lt;/math&gt;.<br /> <br /> ==Solution 2 (Answer options)==<br /> We can try out every option and see which one works out. By this method, we get &lt;math&gt;\boxed{\textbf{(E)}\ 40}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> Let x be the discount. We can also work in reverse such as (&lt;math&gt;84&lt;/math&gt;)&lt;math&gt;(\frac{100}{100-x})&lt;/math&gt;&lt;math&gt;(\frac{100}{100+x})&lt;/math&gt; = &lt;math&gt;100&lt;/math&gt;.<br /> <br /> Thus &lt;math&gt;8400&lt;/math&gt; = &lt;math&gt;(100+x)(100-x)&lt;/math&gt;. Solving for &lt;math&gt;x&lt;/math&gt; gives us &lt;math&gt;x = 40, -40&lt;/math&gt;. But &lt;math&gt;x&lt;/math&gt; has to be positive. Thus &lt;math&gt;x&lt;/math&gt; = &lt;math&gt;40&lt;/math&gt;.<br /> <br /> ==Solution 4 ~ using the answer choices==<br /> <br /> Let our original cost be &lt;math&gt;\$100.&lt;/math&gt; We are looking for a result of &lt;math&gt;\$ 84,&lt;/math&gt; then. We try 16% and see it gets us higher than 84. We try 20% and see it gets us lower than 16 but still higher than 84. We know that the higher the percent, the less the value. We try 36, as we are not progressing much, and we are close! We try &lt;math&gt;\boxed{40\%}&lt;/math&gt;, and we have the answer; it worked.<br /> <br /> ==Video explaining solution== <br /> <br /> https://www.youtube.com/watch?v=_TheVi-6LWE - Happytwin<br /> <br /> Associated video - https://www.youtube.com/watch?v=aJX27Cxvwlc<br /> <br /> https://youtu.be/gX_l0PGsQao<br /> <br /> https://www.youtube.com/watch?v=RcBDdB35Whk&amp;list=PLLCzevlMcsWNBsdpItBT4r7Pa8cZb6Viu&amp;index=4 ~ MathEx<br /> <br /> https://youtu.be/h2GlK7itc2g<br /> <br /> ~savannahsolver<br /> <br /> https://www.youtube.com/watch?v=DaF8uD8V8u0&amp;list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&amp;index=24<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2019|num-b=21|num-a=23}}<br /> <br /> [[Category:Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Problemsolver2026 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_21&diff=159229 2019 AMC 8 Problems/Problem 21 2021-07-28T19:53:24Z <p>Problemsolver2026: /* Video Solutions */</p> <hr /> <div><br /> ==Problem 21==<br /> What is the area of the triangle formed by the lines &lt;math&gt;y=5&lt;/math&gt;, &lt;math&gt;y=1+x&lt;/math&gt;, and &lt;math&gt;y=1-x&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }4\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }12\qquad\textbf{(E) }16&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> First we need to find the coordinates where the graphs intersect. <br /> <br /> We want the points x and y to be the same. Thus, we set &lt;math&gt;5=x+1,&lt;/math&gt; and get &lt;math&gt;x=4.&lt;/math&gt; Plugging this into the equation, &lt;math&gt;y=1-x,&lt;/math&gt; <br /> &lt;math&gt;y=5&lt;/math&gt;, and &lt;math&gt;y=1+x&lt;/math&gt; intersect at &lt;math&gt;(4,5)&lt;/math&gt;, we call this line x.<br /> <br /> Doing the same thing, we get &lt;math&gt;x=-4.&lt;/math&gt; Thus &lt;math&gt;y=5&lt;/math&gt; also.<br /> &lt;math&gt;y=5&lt;/math&gt;, and &lt;math&gt;y=1-x&lt;/math&gt; intersect at &lt;math&gt;(-4,5)&lt;/math&gt;, we call this line y.<br /> <br /> It's apparent the only solution to &lt;math&gt;1-x=1+x&lt;/math&gt; is &lt;math&gt;0.&lt;/math&gt; Thus, &lt;math&gt;y=1.&lt;/math&gt;<br /> &lt;math&gt;y=1-x&lt;/math&gt; and &lt;math&gt;y=1+x&lt;/math&gt; intersect at &lt;math&gt;(0,1)&lt;/math&gt;, we call this line z.<br /> <br /> Using the [[Shoelace Theorem]] we get: &lt;cmath&gt;\left(\frac{(20-4)-(-20+4)}{2}\right)=\frac{32}{2}&lt;/cmath&gt; &lt;math&gt;=&lt;/math&gt; So our answer is &lt;math&gt;\boxed{\textbf{(E)}\ 16.}&lt;/math&gt;<br /> <br /> We might also see that the lines &lt;math&gt;y&lt;/math&gt; and &lt;math&gt;x&lt;/math&gt; are mirror images of each other. This is because, when rewritten, their slopes can be multiplied by &lt;math&gt;-1&lt;/math&gt; to get the other. As the base is horizontal, this is a isosceles triangle with base 8, as the intersection points have distance 8. The height is &lt;math&gt;5-1=4,&lt;/math&gt; so &lt;math&gt;\frac{4\cdot 8}{2} = \boxed{\textbf{(E)} 16.}&lt;/math&gt;<br /> <br /> Warning: Do not use the distance formula for the base then use Heron's formula. It will take you half of the time you have left!<br /> <br /> ==Solution 2==<br /> Graphing the lines, using the intersection points we found in Solution 1, we can see that the height of the triangle is 4, and the base is 8. Using the formula for the area of a triangle, we get &lt;math&gt;\frac{4\cdot8}{2}&lt;/math&gt; which is equal to &lt;math&gt;\boxed{\textbf{(E)}\ 16}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> <br /> &lt;math&gt;y = x + 1&lt;/math&gt; and &lt;math&gt;y = -x + 1&lt;/math&gt; have &lt;math&gt;y&lt;/math&gt;-intercepts at &lt;math&gt;(1, 0)&lt;/math&gt; and slopes of &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;-1&lt;/math&gt;, respectively. Since the product of these slopes is &lt;math&gt;-1&lt;/math&gt;, the two lines are perpendicular. From &lt;math&gt;y = 5&lt;/math&gt;, we see that &lt;math&gt;(-4, 5)&lt;/math&gt; and &lt;math&gt;(4, 5)&lt;/math&gt; are the other two intersection points, and they are &lt;math&gt;8&lt;/math&gt; units apart. By symmetry, this triangle is a &lt;math&gt;45-45-90&lt;/math&gt; triangle, so the legs are &lt;math&gt;4\sqrt{2}&lt;/math&gt; each and the area is &lt;math&gt;\frac{(4\sqrt{2})^2}{2} = \boxed{\textbf{(E)}\ 16}&lt;/math&gt;. <br /> ==Video Solutions==<br /> https://www.youtube.com/watch?v=mz3DY1rc5ao - Happytwin<br /> <br /> Associated Video - https://www.youtube.com/watch?v=ie3tlSNyiaY<br /> <br /> https://www.youtube.com/watch?v=9nlX9VCisQc<br /> <br /> https://www.youtube.com/watch?v=Z27G0xy5AgA&amp;list=PLLCzevlMcsWNBsdpItBT4r7Pa8cZb6Viu&amp;index=3 ~ MathEx<br /> <br /> https://youtu.be/RvtOX17DemY<br /> <br /> ~savannahsolver<br /> <br /> https://www.youtube.com/watch?v=SBGcumVOroI&amp;list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&amp;index=23<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2019|num-b=20|num-a=22}}<br /> <br /> {{MAA Notice}}</div> Problemsolver2026 https://artofproblemsolving.com/wiki/index.php?title=Resources_for_mathematics_competitions&diff=152585 Resources for mathematics competitions 2021-04-24T20:39:03Z <p>Problemsolver2026: /* AMC 8 Preparation */</p> <hr /> <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 /> ==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&amp;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&amp;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 /> <br /> ==== 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&amp;list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL<br /> <br /> [http://www.artofproblemsolving.com/Forum/resources.php?c=182&amp;cid=42 AMC 8 Problems in the Resources Section]<br /> <br /> 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&amp;t=388108&amp;hilit=preparation How preparing for the AIME will help AMC 10/12 Score] <br /> <br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=396741&amp;hilit=preparation What class to take?]<br /> <br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=387918&amp;hilit=preparation AMC 10 for AMC 12 practice]<br /> <br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=385418&amp;hilit=preparation AMC prep]<br /> <br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=384828&amp;hilit=preparation AMC 10/12 Preparation]<br /> <br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=384747&amp;hilit=preparation AIME/AMC 10 Overlap and Preparation]<br /> <br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=149&amp;t=378851&amp;hilit=preparation How to prepare for amc10 and aime?]<br /> <br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=149&amp;t=369849&amp;hilit=preparation Preparation for AMC 10?]<br /> ====Problems====<br /> <br /> [http://www.artofproblemsolving.com/Forum/resources.php?c=182&amp;cid=43 AMC 10 Problems in the Resources Section]<br /> <br /> [http://www.artofproblemsolving.com/wiki/index.php/AMC_10_Problems_and_Solutions AMC 10 Problems in the AoPS Wiki]<br /> <br /> [http://www.artofproblemsolving.com/Forum/resources.php?c=182&amp;cid=44 AMC 12 Problems in the Resources Section]<br /> <br /> [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&amp;t=397954&amp;hilit=preparation Studying to qualify for USAMO]<br /> <br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=400442&amp;hilit=preparation How to prepare for the AIME]<br /> <br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=399160&amp;hilit=preparation Preparation for the AIME]<br /> <br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=357602&amp;hilit=preparation Using non-AIME questions to prepare for AIME]<br /> <br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=149&amp;t=355918&amp;hilit=preparation Best books to prepare for AIME?]<br /> <br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=344816&amp;hilit=preparation How to improve AIME score to make JMO?]<br /> <br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=341827&amp;hilit=preparation Preparation for AIME and USAMO]<br /> <br /> ====Problems====<br /> <br /> [http://www.artofproblemsolving.com/Forum/resources.php?c=182&amp;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&amp;pli=AIME AIME problems sorted by difficulty]'''<br /> <br /> ===Beginning Olympiad Preparation===<br /> * [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=480253 General]<br /> * [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=481746&amp;p=2698978 General]<br /> * [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=401061&amp;hilit=preparation How to Prepare for USAJMO?]<br /> * [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=399023&amp;hilit=preparation USAMO preparation/doing problems]<br /> * [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=396736&amp;hilit=preparation Easier Olympiads for USAJMO practice?]<br /> * [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=366383&amp;hilit=preparation For the USAMO: ACoPS or Engel?]<br /> * [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=360619&amp;hilit=preparation Olympiad problems- how to prepare]<br /> * [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=354103&amp;hilit=preparation USAMO/Olympiads Preparation: Where to start?]<br /> * [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=344929&amp;hilit=preparation USAJMO prep]<br /> <br /> ====Bunch of General links====<br /> * [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=31888&amp;hilit=USAMO+prep USAMO Prep]<br /> * [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=71008&amp;hilit=USAMO+prep USAMO Prep]<br /> * [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=79077&amp;hilit=USAMO+prep USAMO]<br /> * [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=81296&amp;hilit=USAMO+prep Usamo prep ]<br /> * [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=143168&amp;hilit=USAMO+prep USAMO Prep]<br /> * [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=273572&amp;hilit=USAMO+prep Counting down to USAMO]<br /> * [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=294132&amp;hilit=USAMO+prep USAMO Preparation]<br /> * [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=344929&amp;hilit=olympiad+prep USAJMO prep]<br /> * [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=385092&amp;hilit=USAMO+prep USAMO Preparation]<br /> * [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=397424&amp;hilit=olympiad+prep USAJMO Prep]<br /> * [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=401201&amp;hilit=USAMO+prep How to prepare for the USAMO/Making Red MOP]<br /> * [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=401640&amp;hilit=USAMO+prep Preparing in a hardcore way]<br /> * [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=406402&amp;hilit=USAMO+prep USAMO and JMO prep]<br /> * [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=411476&amp;hilit=USAMO+prep USAMO PREP]<br /> * [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=419800&amp;hilit=USAMO+prep Beginner to USAMO]<br /> * [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=447454 What should I be doing?]<br /> * [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=453638&amp;hilit=USAMO+prep Improve to USAMO and IMO level]<br /> * [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=474960&amp;hilit=USAMO+prep Prep for USA(J)MO]<br /> * '''[http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=385654 contest math stuff/some advice on how to get good]'''<br /> * [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;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&amp;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&amp;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&amp;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&amp;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 /> <br /> =====Not Free=====<br /> * [http://www.amazon.com/Plane-Euclidean-Geometry-Theory-Problems/dp/0953682366/ref=sr_1_1?s=books&amp;ie=UTF8&amp;qid=1338742080&amp;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&amp;ie=UTF8&amp;qid=1338742131&amp;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&amp;ie=UTF8&amp;qid=1338742156&amp;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&amp;ie=UTF8&amp;qid=1338741912&amp;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&amp;ie=UTF8&amp;qid=1338742048&amp;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&amp;qid=1338741533&amp;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&amp;ie=UTF8&amp;qid=1338741653&amp;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&amp;ie=UTF8&amp;qid=1338741697&amp;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&amp;ie=UTF8&amp;qid=1338741741&amp;sr=1-1 102 Combinatorial Problems]<br /> * [http://www.amazon.com/Path-Combinatorics-Undergraduates-Counting-Strategies/dp/8181283368/ref=sr_1_2?s=books&amp;ie=UTF8&amp;qid=1338741874&amp;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&amp;ie=UTF8&amp;qid=1338742228&amp;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&amp;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&amp;qid=1338865322&amp;sr=8-1 Problem Solving Strategies]<br /> <br /> ===Problem Sets===<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&amp;a=cHJvYiBwcm9ibGVt&amp;v=1&amp;n=1&amp;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> Problemsolver2026 https://artofproblemsolving.com/wiki/index.php?title=Resources_for_mathematics_competitions&diff=152584 Resources for mathematics competitions 2021-04-24T20:38:02Z <p>Problemsolver2026: </p> <hr /> <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 /> ==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&amp;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&amp;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 /> <br /> ==== 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 /> [http://www.artofproblemsolving.com/Forum/resources.php?c=182&amp;cid=42 AMC 8 Problems in the Resources Section]<br /> <br /> 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&amp;t=388108&amp;hilit=preparation How preparing for the AIME will help AMC 10/12 Score] <br /> <br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=396741&amp;hilit=preparation What class to take?]<br /> <br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=387918&amp;hilit=preparation AMC 10 for AMC 12 practice]<br /> <br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=385418&amp;hilit=preparation AMC prep]<br /> <br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=384828&amp;hilit=preparation AMC 10/12 Preparation]<br /> <br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=384747&amp;hilit=preparation AIME/AMC 10 Overlap and Preparation]<br /> <br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=149&amp;t=378851&amp;hilit=preparation How to prepare for amc10 and aime?]<br /> <br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=149&amp;t=369849&amp;hilit=preparation Preparation for AMC 10?]<br /> ====Problems====<br /> <br /> [http://www.artofproblemsolving.com/Forum/resources.php?c=182&amp;cid=43 AMC 10 Problems in the Resources Section]<br /> <br /> [http://www.artofproblemsolving.com/wiki/index.php/AMC_10_Problems_and_Solutions AMC 10 Problems in the AoPS Wiki]<br /> <br /> [http://www.artofproblemsolving.com/Forum/resources.php?c=182&amp;cid=44 AMC 12 Problems in the Resources Section]<br /> <br /> [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&amp;t=397954&amp;hilit=preparation Studying to qualify for USAMO]<br /> <br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=400442&amp;hilit=preparation How to prepare for the AIME]<br /> <br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=399160&amp;hilit=preparation Preparation for the AIME]<br /> <br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=357602&amp;hilit=preparation Using non-AIME questions to prepare for AIME]<br /> <br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=149&amp;t=355918&amp;hilit=preparation Best books to prepare for AIME?]<br /> <br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=344816&amp;hilit=preparation How to improve AIME score to make JMO?]<br /> <br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=341827&amp;hilit=preparation Preparation for AIME and USAMO]<br /> <br /> ====Problems====<br /> <br /> [http://www.artofproblemsolving.com/Forum/resources.php?c=182&amp;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&amp;pli=AIME AIME problems sorted by difficulty]'''<br /> <br /> ===Beginning Olympiad Preparation===<br /> * [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=480253 General]<br /> * [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=481746&amp;p=2698978 General]<br /> * [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=401061&amp;hilit=preparation How to Prepare for USAJMO?]<br /> * [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=399023&amp;hilit=preparation USAMO preparation/doing problems]<br /> * [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=396736&amp;hilit=preparation Easier Olympiads for USAJMO practice?]<br /> * [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=366383&amp;hilit=preparation For the USAMO: ACoPS or Engel?]<br /> * [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=360619&amp;hilit=preparation Olympiad problems- how to prepare]<br /> * [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=354103&amp;hilit=preparation USAMO/Olympiads Preparation: Where to start?]<br /> * [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=344929&amp;hilit=preparation USAJMO prep]<br /> <br /> ====Bunch of General links====<br /> * [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=31888&amp;hilit=USAMO+prep USAMO Prep]<br /> * [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=71008&amp;hilit=USAMO+prep USAMO Prep]<br /> * [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=79077&amp;hilit=USAMO+prep USAMO]<br /> * [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=81296&amp;hilit=USAMO+prep Usamo prep ]<br /> * [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=143168&amp;hilit=USAMO+prep USAMO Prep]<br /> * [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=273572&amp;hilit=USAMO+prep Counting down to USAMO]<br /> * [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=294132&amp;hilit=USAMO+prep USAMO Preparation]<br /> * [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=344929&amp;hilit=olympiad+prep USAJMO prep]<br /> * [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=385092&amp;hilit=USAMO+prep USAMO Preparation]<br /> * [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=397424&amp;hilit=olympiad+prep USAJMO Prep]<br /> * [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=401201&amp;hilit=USAMO+prep How to prepare for the USAMO/Making Red MOP]<br /> * [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=401640&amp;hilit=USAMO+prep Preparing in a hardcore way]<br /> * [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=406402&amp;hilit=USAMO+prep USAMO and JMO prep]<br /> * [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=411476&amp;hilit=USAMO+prep USAMO PREP]<br /> * [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=419800&amp;hilit=USAMO+prep Beginner to USAMO]<br /> * [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=447454 What should I be doing?]<br /> * [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=453638&amp;hilit=USAMO+prep Improve to USAMO and IMO level]<br /> * [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=474960&amp;hilit=USAMO+prep Prep for USA(J)MO]<br /> * '''[http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=385654 contest math stuff/some advice on how to get good]'''<br /> * [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;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&amp;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&amp;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&amp;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&amp;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 /> <br /> =====Not Free=====<br /> * [http://www.amazon.com/Plane-Euclidean-Geometry-Theory-Problems/dp/0953682366/ref=sr_1_1?s=books&amp;ie=UTF8&amp;qid=1338742080&amp;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&amp;ie=UTF8&amp;qid=1338742131&amp;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&amp;ie=UTF8&amp;qid=1338742156&amp;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&amp;ie=UTF8&amp;qid=1338741912&amp;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&amp;ie=UTF8&amp;qid=1338742048&amp;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&amp;qid=1338741533&amp;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&amp;ie=UTF8&amp;qid=1338741653&amp;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&amp;ie=UTF8&amp;qid=1338741697&amp;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&amp;ie=UTF8&amp;qid=1338741741&amp;sr=1-1 102 Combinatorial Problems]<br /> * [http://www.amazon.com/Path-Combinatorics-Undergraduates-Counting-Strategies/dp/8181283368/ref=sr_1_2?s=books&amp;ie=UTF8&amp;qid=1338741874&amp;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&amp;ie=UTF8&amp;qid=1338742228&amp;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&amp;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&amp;qid=1338865322&amp;sr=8-1 Problem Solving Strategies]<br /> <br /> ===Problem Sets===<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&amp;a=cHJvYiBwcm9ibGVt&amp;v=1&amp;n=1&amp;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> Problemsolver2026 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_20&diff=152521 2019 AMC 8 Problems/Problem 20 2021-04-23T18:35:13Z <p>Problemsolver2026: </p> <hr /> <div>==Problem 20==<br /> How many different real numbers &lt;math&gt;x&lt;/math&gt; satisfy the equation &lt;cmath&gt;(x^{2}-5)^{2}=16?&lt;/cmath&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }4\qquad\textbf{(E) }8&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> We have that &lt;math&gt;(x^2-5)^2 = 16&lt;/math&gt; if and only if &lt;math&gt;x^2-5 = \pm 4&lt;/math&gt;. If &lt;math&gt;x^2-5 = 4&lt;/math&gt;, then &lt;math&gt;x^2 = 9 \implies x = \pm 3&lt;/math&gt;, giving 2 solutions. If &lt;math&gt;x^2-5 = -4&lt;/math&gt;, then &lt;math&gt;x^2 = 1 \implies x = \pm 1&lt;/math&gt;, giving 2 more solutions. All four of these solutions work, so the answer is &lt;math&gt;\boxed{\textbf{(D)} 4}&lt;/math&gt;. Further, the equation is a [[quartic Equation|quartic]] in &lt;math&gt;x&lt;/math&gt;, so by the [https://artofproblemsolving.com/wiki/index.php/Fundamental_Theorem_of_Algebra Fundamental Theorem of Algebra], there can be at most four real solutions.<br /> <br /> ==Solution 2==<br /> We can expand &lt;math&gt;(x^2-5)^2&lt;/math&gt; to get &lt;math&gt;x^4-10x^2+25&lt;/math&gt;, so now our equation is &lt;math&gt;x^4-10x^2+25=16&lt;/math&gt;. Subtracting &lt;math&gt;16&lt;/math&gt; from both sides gives us &lt;math&gt;x^4-10x^2+9=0&lt;/math&gt;. Now, we can factor the left hand side to get &lt;math&gt;(x^2-9)(x^2-1)=0&lt;/math&gt;. If &lt;math&gt;x^2-9&lt;/math&gt; and/or &lt;math&gt;x^2-1&lt;/math&gt; equals &lt;math&gt;0&lt;/math&gt;, then the whole left side will equal &lt;math&gt;0&lt;/math&gt;. Since the solutions can be both positive and negative, we have &lt;math&gt;4&lt;/math&gt; solutions: &lt;math&gt;-3,3,-1,1&lt;/math&gt; (we can find these solutions by setting &lt;math&gt;x^2-9&lt;/math&gt; and &lt;math&gt;x^2-1&lt;/math&gt; equal to &lt;math&gt;0&lt;/math&gt; and solving for &lt;math&gt;x&lt;/math&gt;). So the answer is &lt;math&gt;\boxed{\textbf{(D)} 4}&lt;/math&gt;. <br /> <br /> ~UnstoppableGoddess<br /> <br /> ==Solution 3==<br /> Associated Video - https://www.youtube.com/watch?v=Q5yfodutpsw<br /> <br /> https://youtu.be/0AY1klX3gBo<br /> <br /> ==Solution 4==<br /> https://youtu.be/5BXh0JY4klM (Uses a difference of squares &amp; factoring method, different from above solutions)<br /> <br /> ==Solution 5 (Fast)==<br /> https://www.youtube.com/watch?v=44vrsk_CbF8&amp;list=PLLCzevlMcsWNBsdpItBT4r7Pa8cZb6Viu&amp;index=2 ~ MathEx<br /> <br /> ==Video Solution==<br /> https://youtu.be/V3HxkJhSn08<br /> <br /> -Happytwin<br /> <br /> == Video Solution ==<br /> <br /> Solution detailing how to solve the problem: https://www.youtube.com/watch?v=eXJnG96Xqp4&amp;list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&amp;index=21<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2019|num-b=19|num-a=21}}<br /> <br /> {{MAA Notice}}</div> Problemsolver2026 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_19&diff=152520 2019 AMC 8 Problems/Problem 19 2021-04-23T18:34:48Z <p>Problemsolver2026: </p> <hr /> <div>==Problem 19==<br /> In a tournament there are six teams that play each other twice. A team earns &lt;math&gt;3&lt;/math&gt; points for a win, &lt;math&gt;1&lt;/math&gt; point for a draw, and &lt;math&gt;0&lt;/math&gt; points for a loss. After all the games have been played it turns out that the top three teams earned the same number of total points. What is the greatest possible number of total points for each of the top three teams?<br /> <br /> &lt;math&gt;\textbf{(A) }22\qquad\textbf{(B) }23\qquad\textbf{(C) }24\qquad\textbf{(D) }26\qquad\textbf{(E) }30&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> After fully understanding the problem, we immediately know that the three top teams, say team &lt;math&gt;A&lt;/math&gt;, team &lt;math&gt;B&lt;/math&gt;, and team &lt;math&gt;C&lt;/math&gt;, must beat the other three teams &lt;math&gt;D&lt;/math&gt;, &lt;math&gt;E&lt;/math&gt;, &lt;math&gt;F&lt;/math&gt;. Therefore, &lt;math&gt;A&lt;/math&gt;,&lt;math&gt;B&lt;/math&gt;,&lt;math&gt;C&lt;/math&gt; must each obtain &lt;math&gt;(3+3+3)=9&lt;/math&gt; points. However, they play against each team twice, for a total of &lt;math&gt;18&lt;/math&gt; points against &lt;math&gt;D&lt;/math&gt;, &lt;math&gt;E&lt;/math&gt;, and &lt;math&gt;F&lt;/math&gt;. For games between &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, &lt;math&gt;C&lt;/math&gt;, we have 2 cases. In both cases, there is an equality of points between &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;C&lt;/math&gt;.<br /> <br /> Case 1: A team ties the two other teams. For a tie, we have 1 point, so we have &lt;math&gt;(1+1)*2=4&lt;/math&gt; points (they play twice). Therefore, this case brings a total of &lt;math&gt;4+18=22&lt;/math&gt; points.<br /> <br /> Case 2: A team beats one team while losing to another. This gives equality, as each team wins once and loses once as well. For a win, we have &lt;math&gt;3&lt;/math&gt; points, so a team gets &lt;math&gt;3\times2=6&lt;/math&gt; points if they each win a game and lose a game. This case brings a total of &lt;math&gt;18+6=24&lt;/math&gt; points. <br /> <br /> Therefore, we use Case 2 since it brings the greater amount of points, or &lt;math&gt;\boxed{(C)24}&lt;/math&gt;.<br /> <br /> <br /> --------------------------<br /> Note that case 2 can be easily seen to be better as follows. Let &lt;math&gt;x_A&lt;/math&gt; be the number of points &lt;math&gt;A&lt;/math&gt; gets, &lt;math&gt;x_B&lt;/math&gt; be the number of points &lt;math&gt;B&lt;/math&gt; gets, and &lt;math&gt;x_C&lt;/math&gt; be the number of points &lt;math&gt;C&lt;/math&gt; gets. Since &lt;math&gt;x_A = x_B = x_C&lt;/math&gt;, to maximize &lt;math&gt;x_A&lt;/math&gt;, we can just maximize &lt;math&gt;x_A + x_B + x_C&lt;/math&gt;. But in each match, if one team wins then the total sum increases by &lt;math&gt;3&lt;/math&gt; points, whereas if they tie, the total sum increases by &lt;math&gt;2&lt;/math&gt; points. So it is best if there are the fewest ties possible.<br /> <br /> ==Solution 2==<br /> <br /> (1st match(3) + 2nd match(1)) * number of teams(6) = 24, &lt;math&gt;\boxed{C}&lt;/math&gt;.<br /> <br /> Explanation: So after reading the problem we see that there are 6 teams and each team versus each other twice. This means one of the two matches has to be a win, so 3 points so far. Now if we say that the team won again and make it 6 points, that would mean that team would be dominating the leader-board and the problem says that all the top 3 people have the same score. So that means the maximum amount of points we could get is 1 so that each team gets the same amount of matches won &amp; drawn so that adds up to 4. 4 * the number of teams(6) = 24 so the answer is &lt;math&gt;\boxed{C}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> <br /> <br /> We can name the top three teams as &lt;math&gt;A, B,&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt;. We can see that &lt;math&gt;A=B=C&lt;/math&gt;, because these teams have the same points. If we look at the matches that involve the top three teams, we see that there are some duplicates: &lt;math&gt;AB, BC,&lt;/math&gt; and &lt;math&gt;AC&lt;/math&gt; come twice. In order to even out the scores and get the maximum score, we can say that in match &lt;math&gt;AB, A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; each win once out of the two games that they play. We can say the same thing for &lt;math&gt;AC&lt;/math&gt; and &lt;math&gt;BC&lt;/math&gt;. This tells us that each team &lt;math&gt;A, B,&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; win and lose twice. This gives each team a total of 3 + 3 + 0 + 0 = 6 points. Now, we need to include the other three teams. We can label these teams as &lt;math&gt;D, E,&lt;/math&gt; and &lt;math&gt;F&lt;/math&gt;. We can write down every match that &lt;math&gt;A, B,&lt;/math&gt; or &lt;math&gt;C&lt;/math&gt; plays in that we haven't counted yet: &lt;math&gt;AD, AD, AE, AE, AF, AF, BD, BD, BE, BE, BF, BF, CD, CD, CE, CE, CF,&lt;/math&gt; and &lt;math&gt;CF&lt;/math&gt;. We can say &lt;math&gt;A, B,&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; win each of these in order to obtain the maximum score that &lt;math&gt;A, B,&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; can have. If &lt;math&gt;A, B,&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; win all six of their matches, &lt;math&gt;A, B,&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; will have a score of &lt;math&gt;18&lt;/math&gt;. &lt;math&gt;18 + 6&lt;/math&gt; results in a maximum score of &lt;math&gt;\boxed{24}&lt;/math&gt;. This tells us that the correct answer choice is &lt;math&gt;\boxed{C}&lt;/math&gt;.<br /> <br /> == Video Solutions ==<br /> <br /> Associated Video - https://youtu.be/s0O3_uXZrOI<br /> <br /> https://youtu.be/hM4sHJSMNDs<br /> <br /> -Happpytwin<br /> <br /> == Video Solution ==<br /> https://youtu.be/HISL2-N5NVg?t=4616<br /> <br /> ~ pi_is_3.14<br /> <br /> == Video Solution ==<br /> <br /> Solution detailing how to solve the problem: https://www.youtube.com/watch?v=k_AuB_bzidc&amp;list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&amp;index=20<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2019|num-b=18|num-a=20}}<br /> <br /> {{MAA Notice}}</div> Problemsolver2026 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_18&diff=152519 2019 AMC 8 Problems/Problem 18 2021-04-23T18:33:56Z <p>Problemsolver2026: </p> <hr /> <div>==Problem 18==<br /> The faces of each of two fair dice are numbered &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;5&lt;/math&gt;, &lt;math&gt;7&lt;/math&gt;, and &lt;math&gt;8&lt;/math&gt;. When the two dice are tossed, what is the probability that their sum will be an even number?<br /> <br /> &lt;math&gt;\textbf{(A) }\frac{4}{9}\qquad\textbf{(B) }\frac{1}{2}\qquad\textbf{(C) }\frac{5}{9}\qquad\textbf{(D) }\frac{3}{5}\qquad\textbf{(E) }\frac{2}{3}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> The approach to this problem:<br /> There are two cases in which the sum can be an even number: both numbers are even and both numbers are odd. This results in only one case where the sum of the numbers are odd (one odd and one even in any order) . We can solve for how many ways the &lt;math&gt;2&lt;/math&gt; numbers add up to an odd number and subtract the answer from &lt;math&gt;1&lt;/math&gt;. <br /> <br /> How to solve the problem:<br /> The probability of getting an odd number first is &lt;math&gt;\frac{4}{6}=\frac{2}{3}&lt;/math&gt;. In order to make the sum odd, we must select an even number next. The probability of getting an even number is &lt;math&gt;\frac{2}{6}=\frac{1}{3}&lt;/math&gt;. Now we multiply the two fractions: &lt;math&gt;\frac{2}{3}\times\frac{1}{3}=2/9&lt;/math&gt;. However, this is not the answer because we could pick an even number first then an odd number. The equation is the same except backward and by the Communitive Property of Multiplication, the equations are it does not matter is the equation is backward or not. Thus we do &lt;math&gt;\frac{2}{9}\times2=\frac{4}{9}&lt;/math&gt;. This is the probability of getting an odd number. In order to get the probability of getting an even number we do &lt;math&gt;1-\frac{4}{9}=\boxed{(\textbf{C})\frac{5}{9}}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> We have a &lt;math&gt;2&lt;/math&gt; die with &lt;math&gt;2&lt;/math&gt; evens and &lt;math&gt;4&lt;/math&gt; odds on both dies. For the sum to be even, the rolls must consist of &lt;math&gt;2&lt;/math&gt; odds or &lt;math&gt;2&lt;/math&gt; evens. <br /> <br /> Ways to roll &lt;math&gt;2&lt;/math&gt; odds (Case &lt;math&gt;1&lt;/math&gt;): The total number of ways to roll &lt;math&gt;2&lt;/math&gt; odds is &lt;math&gt;4*4=16&lt;/math&gt;, as there are &lt;math&gt;4&lt;/math&gt; choices for the first odd on the first roll and &lt;math&gt;4&lt;/math&gt; choices for the second odd on the second roll.<br /> <br /> Ways to roll &lt;math&gt;2&lt;/math&gt; evens (Case &lt;math&gt;2&lt;/math&gt;): Similarly, we have &lt;math&gt;2*2=4&lt;/math&gt; ways to roll &lt;math&gt;{36}=\frac{20}{36}=\frac{5}{9}&lt;/math&gt;, or &lt;math&gt;\framebox{C}&lt;/math&gt;.<br /> <br /> ==Solution 3 (Complementary Counting)==<br /> We count the ways to get an odd. If the sum is odd, then we must have an even and an odd. The probability of an even is &lt;math&gt;\frac{1}{3}&lt;/math&gt;, and the probability of an odd is &lt;math&gt;\frac{2}{3}&lt;/math&gt;. We have to multiply by &lt;math&gt;2!&lt;/math&gt; because the even and odd can be in any order. This gets us &lt;math&gt;\frac{4}{9}&lt;/math&gt;, so the answer is &lt;math&gt;1 - \frac{4}{9} = \frac{5}{9} = \boxed{(\textbf{C})}&lt;/math&gt;.<br /> <br /> ==Solution 4==<br /> To get an even, you must get either 2 odds or 2 evens. The probability of getting 2 odds is &lt;math&gt;\frac{4}{6} * \frac{4}{6}&lt;/math&gt;. The probability of getting 2 evens is &lt;math&gt;\frac{2}{6} * \frac{2}{6}&lt;/math&gt;. If you add them together, you get &lt;math&gt;\frac{16}{36} + \frac{4}{36}&lt;/math&gt; = &lt;math&gt;\boxed{(\textbf{C}) \frac{5}{9}}&lt;/math&gt;.<br /> <br /> ==Solution 5 (Casework)==<br /> To get an even number, we must either have two odds or two evens. We will solve this through casework. The probability of rolling a 1 is &lt;math&gt;\frac{1}{6}&lt;/math&gt;, and the probability of rolling another odd number after this is 4/6=2/3, so the probability of getting a sum of an even number is (1/6)(2/3)=1/9. The probability of rolling a &lt;math&gt;2&lt;/math&gt; is &lt;math&gt;\frac{1}{6}&lt;/math&gt;, and the probability of rolling another even number after this is 2/6=1/3, so the probability of rolling a sum of an even number is &lt;math&gt;\frac{1}{6}\cdot\frac{1}{3}=\frac{1}{18}&lt;/math&gt;. Now, notice that the probability of getting an even sum with two odd numbers is identical for all odd numbers. This is because the probability of probability of getting an even number is identical for all even numbers, so the probability of getting an even sum with only even numbers is (2)(1/18)=1/9. Adding these two up, we get our desired &lt;math&gt;\boxed{(\textbf{C}) \frac{5}{9}}&lt;/math&gt;<br /> <br /> ==Video Solution==<br /> https://youtu.be/8fF55uF64mE - Happytwin<br /> <br /> <br /> Associated video - https://www.youtube.com/watch?v=EoBZy_WYWEw<br /> <br /> https://www.youtube.com/watch?v=H52AqAl4nt4&amp;t=2s<br /> <br /> == Video Solution ==<br /> <br /> Solution detailing how to solve the problem: https://www.youtube.com/watch?v=94D1UnH7seo&amp;list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&amp;index=19<br /> ==See Also==<br /> {{AMC8 box|year=2019|num-b=17|num-a=19}}<br /> <br /> {{MAA Notice}}</div> Problemsolver2026 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_17&diff=152518 2019 AMC 8 Problems/Problem 17 2021-04-23T18:33:29Z <p>Problemsolver2026: </p> <hr /> <div>==Problem==<br /> What is the value of the product <br /> <br /> &lt;cmath&gt;\left(\frac{1\cdot3}{2\cdot2}\right)\left(\frac{2\cdot4}{3\cdot3}\right)\left(\frac{3\cdot5}{4\cdot4}\right)\cdots\left(\frac{97\cdot99}{98\cdot98}\right)\left(\frac{98\cdot100}{99\cdot99}\right)?&lt;/cmath&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }\frac{1}{2}\qquad\textbf{(B) }\frac{50}{99}\qquad\textbf{(C) }\frac{9800}{9801}\qquad\textbf{(D) }\frac{100}{99}\qquad\textbf{(E) }50&lt;/math&gt;<br /> <br /> ==Solution 1(Telescoping)==<br /> We rewrite: &lt;cmath&gt;\frac{1}{2}\cdot\left(\frac{3\cdot2}{2\cdot3}\right)\left(\frac{4\cdot3}{3\cdot4}\right)\cdots\left(\frac{99\cdot98}{98\cdot99}\right)\cdot\frac{100}{99}&lt;/cmath&gt;<br /> <br /> The middle terms cancel, leaving us with<br /> <br /> &lt;cmath&gt;\left(\frac{1\cdot100}{2\cdot99}\right)= \boxed{\textbf{(B)}\frac{50}{99}}&lt;/cmath&gt;<br /> <br /> ==Solution 2==<br /> If you calculate the first few values of the equation, all of the values tend to &lt;math&gt;\frac{1}{2}&lt;/math&gt;, but are not equal to it. The answer closest to &lt;math&gt;\frac{1}{2}&lt;/math&gt; but not equal to it is &lt;math&gt;\boxed{\textbf{(B)}\frac{50}{99}}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> Rewriting the numerator and the denominator, we get &lt;math&gt;\frac{\frac{100! \cdot 98!}{2}}{\left(99!\right)^2}&lt;/math&gt;. We can simplify by canceling 99! on both sides, leaving us with: &lt;math&gt;\frac{100 \cdot 98!}{2 \cdot 99!}&lt;/math&gt; We rewrite &lt;math&gt;99!&lt;/math&gt; as &lt;math&gt;99 \cdot 98!&lt;/math&gt; and cancel &lt;math&gt;98!&lt;/math&gt;, which gets &lt;math&gt;\boxed{(B)\frac{50}{99}}&lt;/math&gt;.<br /> <br /> ==Video Solution==<br /> <br /> Associated video - https://www.youtube.com/watch?v=yPQmvyVyvaM<br /> <br /> https://www.youtube.com/watch?v=ffHl1dAjs7g&amp;list=PLLCzevlMcsWNBsdpItBT4r7Pa8cZb6Viu&amp;index=1 ~ MathEx<br /> <br /> == Video Solution ==<br /> <br /> Solution detailing how to solve the problem: https://www.youtube.com/watch?v=VezsRMJvGPs&amp;list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&amp;index=18<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2019|num-b=16|num-a=18}}<br /> <br /> {{MAA Notice}}</div> Problemsolver2026 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_16&diff=152517 2019 AMC 8 Problems/Problem 16 2021-04-23T18:33:02Z <p>Problemsolver2026: </p> <hr /> <div>==Problem 16==<br /> Qiang drives &lt;math&gt;15&lt;/math&gt; miles at an average speed of &lt;math&gt;30&lt;/math&gt; miles per hour. How many additional miles will he have to drive at &lt;math&gt;55&lt;/math&gt; miles per hour to average &lt;math&gt;50&lt;/math&gt; miles per hour for the entire trip?<br /> <br /> &lt;math&gt;\textbf{(A) }45\qquad\textbf{(B) }62\qquad\textbf{(C) }90\qquad\textbf{(D) }110\qquad\textbf{(E) }135&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> The only option that is easily divisible by &lt;math&gt;55&lt;/math&gt; is &lt;math&gt;110&lt;/math&gt;. Which gives 2 hours of travel. And by the formula &lt;math&gt;\frac{15}{30} + \frac{110}{55} = \frac{5}{2}&lt;/math&gt;<br /> <br /> And &lt;math&gt;\text{Average Speed}&lt;/math&gt; = &lt;math&gt;\frac{\text{Total Distance}}{\text{Total Time}}&lt;/math&gt;<br /> <br /> Thus &lt;math&gt;\frac{125}{50} = \frac{5}{2}&lt;/math&gt;<br /> <br /> Both are equal and thus our answer is &lt;math&gt;\boxed{\textbf{(D)}\ 110}.&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> Note that the average speed is simply the total distance over the total time. Let the number of additional miles he has to drive be &lt;math&gt;x.&lt;/math&gt; Therefore, the total distance is &lt;math&gt;15+x&lt;/math&gt; and the total time (in hours) is &lt;cmath&gt;\frac{15}{30}+\frac{x}{55}=\frac{1}{2}+\frac{x}{55}.&lt;/cmath&gt; We can set up the following equation: &lt;cmath&gt;\frac{15+x}{\frac{1}{2}+\frac{x}{55}}=50.&lt;/cmath&gt; Simplifying the equation, we get &lt;cmath&gt;15+x=25+\frac{10x}{11}.&lt;/cmath&gt; Solving the equation yields &lt;math&gt;x=110,&lt;/math&gt; so our answer is &lt;math&gt;\boxed{\textbf{(D)}\ 110}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> If he travels &lt;math&gt;15&lt;/math&gt; miles at a speed of &lt;math&gt;30&lt;/math&gt; miles per hour, he travels for 30 min. Average rate is total distance over total time so &lt;math&gt;(15+d)/(0.5 + t) = 50&lt;/math&gt;, where d is the distance left to travel and t is the time to travel that distance. solve for &lt;math&gt;d&lt;/math&gt; to get &lt;math&gt;d = 10+50t&lt;/math&gt;. you also know that he has to travel &lt;math&gt;55&lt;/math&gt; miles per hour for some time, so &lt;math&gt;d=55t&lt;/math&gt; plug that in for d to get &lt;math&gt;55t = 10+50t&lt;/math&gt; and &lt;math&gt;t=2&lt;/math&gt; and since &lt;math&gt;d=55t&lt;/math&gt;, &lt;math&gt;d = 2\cdot55 =110&lt;/math&gt; the answer is &lt;math&gt;\boxed{\textbf{(D)}\ 110}&lt;/math&gt;.<br /> <br /> ==Video Solution==<br /> <br /> Associated Video - https://www.youtube.com/watch?v=OC1KdFeZFeE<br /> <br /> https://youtu.be/5K1AgeZ8rUQ - happytwin<br /> <br /> == Video Solution ==<br /> <br /> Solution detailing how to solve the problem: https://www.youtube.com/watch?v=sEZ0sM-d1FA&amp;list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&amp;index=17<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2019|num-b=15|num-a=17}}<br /> <br /> {{MAA Notice}}</div> Problemsolver2026 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_15&diff=152516 2019 AMC 8 Problems/Problem 15 2021-04-23T18:32:39Z <p>Problemsolver2026: </p> <hr /> <div>==Problem 15==<br /> On a beach &lt;math&gt;50&lt;/math&gt; people are wearing sunglasses and &lt;math&gt;35&lt;/math&gt; people are wearing caps. Some people are wearing both sunglasses and caps. If one of the people wearing a cap is selected at random, the probability that this person is also wearing sunglasses is &lt;math&gt;\frac{2}{5}&lt;/math&gt;. If instead, someone wearing sunglasses is selected at random, what is the probability that this person is also wearing a cap?<br /> &lt;math&gt;\textbf{(A) }\frac{14}{85}\qquad\textbf{(B) }\frac{7}{25}\qquad\textbf{(C) }\frac{2}{5}\qquad\textbf{(D) }\frac{4}{7}\qquad\textbf{(E) }\frac{7}{10}&lt;/math&gt;<br /> <br /> ==Video Solution==<br /> https://youtu.be/6xNkyDgIhEE?t=252<br /> <br /> ==Solution 1==<br /> The number of people wearing caps and sunglasses is <br /> &lt;math&gt;\frac{2}{5}\cdot35=14&lt;/math&gt;. So then 14 people out of the 50 people wearing sunglasses also have caps. <br /> &lt;math&gt;\frac{14}{50}=\boxed{\textbf{(B)}\frac{7}{25}}&lt;/math&gt;<br /> <br /> ==Video Solution==<br /> https://www.youtube.com/watch?v=gKlYlAiBzrs ~ MathEx<br /> <br /> Another video - https://www.youtube.com/watch?v=afMsUqER13c<br /> <br /> https://youtu.be/37UWNaltvQo -Happytwin<br /> <br /> == Video Solution ==<br /> <br /> Solution detailing how to solve the problem: https://www.youtube.com/watch?v=omRgmX7KXOg&amp;list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&amp;index=16<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2019|num-b=14|num-a=16}}<br /> <br /> {{MAA Notice}}</div> Problemsolver2026 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_14&diff=152513 2019 AMC 8 Problems/Problem 14 2021-04-23T17:53:13Z <p>Problemsolver2026: </p> <hr /> <div>==Problem 14==<br /> Isabella has &lt;math&gt;6&lt;/math&gt; coupons that can be redeemed for free ice cream cones at Pete's Sweet Treats. In order to make the coupons last, she decides that she will redeem one every &lt;math&gt;10&lt;/math&gt; days until she has used them all. She knows that Pete's is closed on Sundays, but as she circles the &lt;math&gt;6&lt;/math&gt; dates on her calendar, she realizes that no circled date falls on a Sunday. On what day of the week does Isabella redeem her first coupon?<br /> <br /> &lt;math&gt;\textbf{(A) }\text{Monday}\qquad\textbf{(B) }\text{Tuesday}\qquad\textbf{(C) }\text{Wednesday}\qquad\textbf{(D) }\text{Thursday}\qquad\textbf{(E) }\text{Friday}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Let &lt;math&gt;\text{Day }1&lt;/math&gt; to &lt;math&gt;\text{Day }2&lt;/math&gt; denote a day where one coupon is redeemed and the day when the second coupon is redeemed. <br /> <br /> If she starts on a &lt;math&gt;\text{Monday}&lt;/math&gt; she redeems her next coupon on &lt;math&gt;\text{Thursday}&lt;/math&gt;. <br /> <br /> &lt;math&gt;\text{Thursday}&lt;/math&gt; to &lt;math&gt;\text{Sunday}&lt;/math&gt;.<br /> <br /> Thus &lt;math&gt;\textbf{(A)}\ \text{Monday}&lt;/math&gt; is incorrect.<br /> <br /> <br /> If she starts on a &lt;math&gt;\text{Tuesday}&lt;/math&gt; she redeems her next coupon on &lt;math&gt;\text{Friday}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\text{Friday}&lt;/math&gt; to &lt;math&gt;\text{Monday}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\text{Monday}&lt;/math&gt; to &lt;math&gt;\text{Thursday}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\text{Thursday}&lt;/math&gt; to &lt;math&gt;\text{Sunday}&lt;/math&gt;.<br /> <br /> Thus &lt;math&gt;\textbf{(B)}\ \text{Tuesday}&lt;/math&gt; is incorrect.<br /> <br /> <br /> If she starts on a &lt;math&gt;\text{Wednesday}&lt;/math&gt; she redeems her next coupon on &lt;math&gt;\text{Saturday}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\text{Saturday}&lt;/math&gt; to &lt;math&gt;\text{Tuesday}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\text{Tuesday}&lt;/math&gt; to &lt;math&gt;\text{Friday}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\text{Friday}&lt;/math&gt; to &lt;math&gt;\text{Monday}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\text{Monday}&lt;/math&gt; to &lt;math&gt;\text{Thursday}&lt;/math&gt;.<br /> <br /> And on &lt;math&gt;\text{Thursday}&lt;/math&gt; she redeems her last coupon. <br /> <br /> <br /> No sunday occured thus &lt;math&gt;\boxed{\textbf{(C)}\ \text{Wednesday}}&lt;/math&gt; is correct. <br /> <br /> <br /> Checking for the other options, <br /> <br /> <br /> If she starts on a &lt;math&gt;\text{Thursday}&lt;/math&gt; she redeems her next coupon on &lt;math&gt;\text{Sunday}&lt;/math&gt;.<br /> <br /> Thus &lt;math&gt;\textbf{(D)}\ \text{Thursday}&lt;/math&gt; is incorrect.<br /> <br /> <br /> If she starts on a &lt;math&gt;\text{Friday}&lt;/math&gt; she redeems her next coupon on &lt;math&gt;\text{Monday}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\text{Monday}&lt;/math&gt; to &lt;math&gt;\text{Thursday}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\text{Thursday}&lt;/math&gt; to &lt;math&gt;\text{Sunday}&lt;/math&gt;.<br /> <br /> <br /> Checking for the other options gave us negative results, thus the answer is &lt;math&gt;\boxed{\textbf{(C)}\ \text{Wednesday}}&lt;/math&gt;.<br /> <br /> == Solution 2==<br /> Let <br /> <br /> &lt;math&gt;Sunday \equiv 0 \pmod{7}&lt;/math&gt;<br /> <br /> &lt;math&gt;Monday \equiv 1 \pmod{7}&lt;/math&gt;<br /> <br /> &lt;math&gt;Tuesday \equiv 2 \pmod{7}&lt;/math&gt;<br /> <br /> &lt;math&gt;Wednesday \equiv 3 \pmod{7}&lt;/math&gt;<br /> <br /> &lt;math&gt;Thursday \equiv 4 \pmod{7}&lt;/math&gt;<br /> <br /> &lt;math&gt;Friday \equiv 5 \pmod{7}&lt;/math&gt;<br /> <br /> &lt;math&gt;Saturday \equiv 6 \pmod{7}&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;10 \equiv 3 \pmod{7}&lt;/math&gt;<br /> <br /> &lt;math&gt;20 \equiv 6 \pmod{7}&lt;/math&gt;<br /> <br /> &lt;math&gt;30 \equiv 2 \pmod{7}&lt;/math&gt;<br /> <br /> &lt;math&gt;40 \equiv 5 \pmod{7}&lt;/math&gt;<br /> <br /> &lt;math&gt;50 \equiv 1 \pmod{7}&lt;/math&gt;<br /> <br /> &lt;math&gt;60 \equiv 4 \pmod{7}&lt;/math&gt;<br /> <br /> <br /> Which clearly indicates if you start form a &lt;math&gt;x \equiv 3 \pmod{7}&lt;/math&gt; you will not get a &lt;math&gt;y \equiv 0 \pmod{7}&lt;/math&gt;.<br /> <br /> Any other starting value may lead to a &lt;math&gt;y \equiv 0 \pmod{7}&lt;/math&gt;.<br /> <br /> Which means our answer is &lt;math&gt;\boxed{\textbf{(C)}\ Wednesday}&lt;/math&gt;.<br /> <br /> ~phoenixfire<br /> <br /> == Solution 3 ==<br /> Like Solution 2, let the days of the week be numbers&lt;math&gt;\pmod 7&lt;/math&gt;. &lt;math&gt;3&lt;/math&gt; and &lt;math&gt;7&lt;/math&gt; are coprime, so continuously adding &lt;math&gt;3&lt;/math&gt; to a number&lt;math&gt;\pmod 7&lt;/math&gt; will cycle through all numbers from &lt;math&gt;0&lt;/math&gt; to &lt;math&gt;6&lt;/math&gt;. If a string of 6 numbers in this cycle does not contain &lt;math&gt;0&lt;/math&gt;, then if you minus 3 from the first number of this cycle, it will always be &lt;math&gt;0&lt;/math&gt;. So, the answer is &lt;math&gt;\boxed{\textbf{(C)}\ Wednesday}&lt;/math&gt;. ~~SmileKat32<br /> <br /> == Solution 4 ==<br /> Since Sunday is the only day that has not been counted yet. We can just add the 3 days as it will become &lt;math&gt;\boxed{\textbf{(C)}\ Wednesday}&lt;/math&gt;. <br /> ~~ gorefeebuddie<br /> Note: This only works when 7 and 3 are relatively prime.<br /> <br /> == Solution 5 ==<br /> Let Sunday be Day 0, Monday be Day 1, Tuesday be Day 2, and so forth. We see that Sundays fall on Day &lt;math&gt;n&lt;/math&gt;, where n is a multiple of seven. If Isabella starts using her coupons on Monday (Day 1), she will fall on a Day that is a multiple of seven, a Sunday (her third coupon will be &quot;used&quot; on Day 21). Similarly, if she starts using her coupons on Tuesday (Day 2), Isabella will fall on a Day that is a multiple of seven (Day 42). Repeating this process, if she starts on Wednesday (Day 3), Isabella will first fall on a Day that is a multiple of seven, Day 63 (13, 23, 33, 43, 53 are not multiples of seven), but on her seventh coupon, of which she only has six. So, the answer is &lt;math&gt;\boxed{\textbf{(C)}\text{ Wednesday}}&lt;/math&gt;.<br /> <br /> == Solution 6 ==<br /> Associated video - https://www.youtube.com/watch?v=LktgMtgb_8E<br /> <br /> == Video Solution ==<br /> <br /> Solution detailing how to solve the problem: https://www.youtube.com/watch?v=MOQj1zxH2gY&amp;list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&amp;index=15<br /> <br /> ==See also==<br /> {{AMC8 box|year=2019|num-b=13|num-a=15}}<br /> <br /> {{MAA Notice}}</div> Problemsolver2026 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_13&diff=152512 2019 AMC 8 Problems/Problem 13 2021-04-23T17:52:15Z <p>Problemsolver2026: </p> <hr /> <div>==Problem 13==<br /> A ''palindrome'' is a number that has the same value when read from left to right or from right to left. (For example, 12321 is a palindrome.) Let &lt;math&gt;N&lt;/math&gt; be the least three-digit integer which is not a palindrome but which is the sum of three distinct two-digit palindromes. What is the sum of the digits of &lt;math&gt;N&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }2\qquad\textbf{(B) }3\qquad\textbf{(C) }4\qquad\textbf{(D) }5\qquad\textbf{(E) }6&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Note that the only positive 2-digit palindromes are multiples of 11, namely &lt;math&gt;11, 22, \ldots, 99&lt;/math&gt;. Since &lt;math&gt;N&lt;/math&gt; is the sum of 2-digit palindromes, &lt;math&gt;N&lt;/math&gt; is necessarily a multiple of 11. The smallest 3-digit multiple of 11 which is not a palindrome is 110, so &lt;math&gt;N=110&lt;/math&gt; is a candidate solution. We must check that 110 can be written as the sum of three distinct 2-digit palindromes; this suffices as &lt;math&gt;110=77+22+11&lt;/math&gt;. Then &lt;math&gt;N = 110&lt;/math&gt;, and the sum of the digits of &lt;math&gt;N&lt;/math&gt; is &lt;math&gt;1+1+0 = \boxed{\textbf{(A) }2}&lt;/math&gt;.<br /> <br /> *Another set of 2-digit numbers is &lt;math&gt;110 = 11+33+66&lt;/math&gt;<br /> <br /> ==Video Solution==<br /> https://www.youtube.com/watch?v=bOnNFeZs7S8<br /> <br /> == Video Solution ==<br /> <br /> Solution detailing how to solve the problem: https://www.youtube.com/watch?v=PJpDJ23sOJM&amp;list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&amp;index=14<br /> <br /> ==See also==<br /> {{AMC8 box|year=2019|num-b=12|num-a=14}}<br /> <br /> {{MAA Notice}}</div> Problemsolver2026 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_12&diff=152511 2019 AMC 8 Problems/Problem 12 2021-04-23T17:50:56Z <p>Problemsolver2026: </p> <hr /> <div>==Problem==<br /> The faces of a cube are painted in six different colors: red &lt;math&gt;(R)&lt;/math&gt;, white &lt;math&gt;(W)&lt;/math&gt;, green &lt;math&gt;(G)&lt;/math&gt;, brown &lt;math&gt;(B)&lt;/math&gt;, aqua &lt;math&gt;(A)&lt;/math&gt;, and purple &lt;math&gt;(P)&lt;/math&gt;. Three views of the cube are shown below. What is the color of the face opposite the aqua face?<br /> <br /> [[File:2019AMC8Prob12.png]]<br /> <br /> ==Solution 1==<br /> &lt;math&gt;B&lt;/math&gt; is on the top, and &lt;math&gt;R&lt;/math&gt; is on the side, and &lt;math&gt;G&lt;/math&gt; is on the right side. That means that (image &lt;math&gt;2&lt;/math&gt;) &lt;math&gt;W&lt;/math&gt; is on the left side. From the third image, you know that &lt;math&gt;P&lt;/math&gt; must be on the bottom since &lt;math&gt;G&lt;/math&gt; is sideways. That leaves us with the back, so the back must be &lt;math&gt;A&lt;/math&gt;. The front is opposite of the back, so the answer is &lt;math&gt;\boxed{\textbf{(A)}\ R}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> Looking closely we can see that all faces are connected with &lt;math&gt;R&lt;/math&gt; except for &lt;math&gt;A&lt;/math&gt;. Thus the answer is &lt;math&gt;\boxed{\textbf{(A)}\ R}&lt;/math&gt;.<br /> <br /> It is A, just draw it out!<br /> ~phoenixfire<br /> <br /> ==Solution 3==<br /> Associated video - https://www.youtube.com/watch?v=K5vaX_EzjEM<br /> <br /> == Video Solution ==<br /> <br /> Solution detailing how to solve the problem: https://www.youtube.com/watch?v=VXBqE-jh2WA&amp;list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&amp;index=13<br /> <br /> ==See also==<br /> {{AMC8 box|year=2019|num-b=11|num-a=13}}<br /> Only two of the cubes are required to solve the problem.</div> Problemsolver2026 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_11&diff=152510 2019 AMC 8 Problems/Problem 11 2021-04-23T17:50:32Z <p>Problemsolver2026: </p> <hr /> <div>==Problem 11==<br /> The eighth grade class at Lincoln Middle School has &lt;math&gt;93&lt;/math&gt; students. Each student takes a math class or a foreign language class or both. There are &lt;math&gt;70&lt;/math&gt; eighth graders taking a math class, and there are &lt;math&gt;54&lt;/math&gt; eighth graders taking a foreign language class. How many eighth graders take ''only'' a math class and ''not'' a foreign language class?<br /> <br /> &lt;math&gt;\textbf{(A) }16\qquad\textbf{(B) }23\qquad\textbf{(C) }31\qquad\textbf{(D) }39\qquad\textbf{(E) }70&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Let &lt;math&gt;x&lt;/math&gt; be the number of students taking both a math and a foreign language class.<br /> <br /> By P-I-E, we get &lt;math&gt;70 + 54 - x&lt;/math&gt; = &lt;math&gt;93&lt;/math&gt;. <br /> <br /> Solving gives us &lt;math&gt;x = 31&lt;/math&gt;.<br /> <br /> But we want the number of students taking only a math class.<br /> <br /> Which is &lt;math&gt;70 - 31 = 39&lt;/math&gt;.<br /> <br /> &lt;math&gt;\boxed{\textbf{(D)}\ 39}&lt;/math&gt;<br /> <br /> ~phoenixfire<br /> <br /> ==Solution 2==<br /> We have &lt;math&gt;70 + 54 = 124&lt;/math&gt; people taking classes. However we over-counted the number of people who take both classes. If we subtract the original amount of people who take classes we get that &lt;math&gt;31&lt;/math&gt; people took the two classes. To find the amount of people who took only math class web subtract the people who didn't take only one math class, so we get &lt;math&gt;70 - 31 = \boxed{\textbf{D} \, 39}&lt;/math&gt; -fath2012<br /> <br /> ==Solution 3==<br /> &lt;asy&gt;<br /> draw(circle((-0.5,0),1));<br /> draw(circle((0.5,0),1));<br /> label(&quot;$\huge{x}$&quot;, (0, 0));<br /> label(&quot;$70-x$&quot;, (-1, 0));<br /> label(&quot;$54-x$&quot;, (1, 0));<br /> &lt;/asy&gt;<br /> <br /> We know that the sum of all three areas is &lt;math&gt;93&lt;/math&gt;<br /> So, we have: <br /> &lt;cmath&gt;93 = 70-x+x+54-x&lt;/cmath&gt;<br /> &lt;cmath&gt;93 = 70+54-x&lt;/cmath&gt;<br /> &lt;cmath&gt;93 = 124 - x&lt;/cmath&gt;<br /> &lt;cmath&gt;-31=-x&lt;/cmath&gt;<br /> &lt;cmath&gt;x=31&lt;/cmath&gt;<br /> <br /> We are looking for the number of students in only math. This is &lt;math&gt;70-x&lt;/math&gt;. Substituting &lt;math&gt;x&lt;/math&gt; with &lt;math&gt;31&lt;/math&gt;, our answer is &lt;math&gt;\boxed{39}&lt;/math&gt;.<br /> <br /> -mathnerdnair<br /> <br /> ==Solution 4==<br /> Associated video - https://www.youtube.com/watch?v=onPaMTO3dSA<br /> <br /> == Video Solution ==<br /> <br /> Solution detailing how to solve the problem: https://www.youtube.com/watch?v=Kanl4ni2y0o&amp;list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&amp;index=12<br /> <br /> ==See also==<br /> {{AMC8 box|year=2019|num-b=10|num-a=12}}<br /> <br /> {{MAA Notice}}</div> Problemsolver2026 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_10&diff=152509 2019 AMC 8 Problems/Problem 10 2021-04-23T17:50:08Z <p>Problemsolver2026: </p> <hr /> <div>==Problem 10==<br /> The diagram shows the number of students at soccer practice each weekday during last week. After computing the mean and median values, Coach discovers that there were actually &lt;math&gt;21&lt;/math&gt; participants on Wednesday. Which of the following statements describes the change in the mean and median after the correction is made?<br /> <br /> &lt;center&gt;&lt;asy&gt;<br /> unitsize(2mm);<br /> defaultpen(fontsize(8bp));<br /> real d = 5;<br /> real t = 0.7;<br /> real r;<br /> int[] num = {20,26,16,22,16};<br /> string[] days = {&quot;Monday&quot;,&quot;Tuesday&quot;,&quot;Wednesday&quot;,&quot;Thursday&quot;,&quot;Friday&quot;};<br /> for (int i=0; i&lt;30;<br /> i=i+2) { draw((i,0)--(i,-5*d),gray);<br /> }for (int i=0;<br /> i&lt;5;<br /> ++i) { r = -1*(i+0.5)*d;<br /> fill((0,r-t)--(0,r+t)--(num[i],r+t)--(num[i],r-t)--cycle,gray);<br /> label(days[i],(-1,r),W);<br /> }for(int i=0;<br /> i&lt;32;<br /> i=i+4) { label(string(i),(i,1));<br /> }label(&quot;Number of students at soccer practice&quot;,(14,3.5));<br /> &lt;/asy&gt;&lt;/center&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }&lt;/math&gt;The mean increases by &lt;math&gt;1&lt;/math&gt; and the median does not change.<br /> <br /> &lt;math&gt;\textbf{(B) }&lt;/math&gt;The mean increases by &lt;math&gt;1&lt;/math&gt; and the median increases by &lt;math&gt;1&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{(C) }&lt;/math&gt;The mean increases by &lt;math&gt;1&lt;/math&gt; and the median increases by &lt;math&gt;5&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{(D) }&lt;/math&gt;The mean increases by &lt;math&gt;5&lt;/math&gt; and the median increases by &lt;math&gt;1&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{(E) }&lt;/math&gt;The mean increases by &lt;math&gt;5&lt;/math&gt; and the median increases by &lt;math&gt;5&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> On Monday, &lt;math&gt;20&lt;/math&gt; people come. On Tuesday, &lt;math&gt;26&lt;/math&gt; people come. On Wednesday, &lt;math&gt;16&lt;/math&gt; people come. On Thursday, &lt;math&gt;22&lt;/math&gt; people come. Finally, on Friday, &lt;math&gt;16&lt;/math&gt; people come. &lt;math&gt;20+26+16+22+16=100&lt;/math&gt;, so the mean is &lt;math&gt;20&lt;/math&gt;. The median is &lt;math&gt;(16, 16, 20, 22, 26) 20&lt;/math&gt;. The coach figures out that actually &lt;math&gt;21&lt;/math&gt; people come on Wednesday. The new mean is &lt;math&gt;21&lt;/math&gt;, while the new median is &lt;math&gt;(16, 20, 21, 22, 26) 21&lt;/math&gt;. The median and mean both change, so the answer is &lt;math&gt;\boxed{\textbf{(B)}}&lt;/math&gt;<br /> Another way to compute the change in mean is to notice that the sum increased by &lt;math&gt;5&lt;/math&gt; with the correction. So the average increased by &lt;math&gt;5/5 = 1&lt;/math&gt;.<br /> <br /> == Video Solution ==<br /> <br /> Solution detailing how to solve the problem: https://www.youtube.com/watch?v=vkm1ZXuuQcc&amp;list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&amp;index=11<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2019|num-b=9|num-a=11}}<br /> <br /> {{MAA Notice}}</div> Problemsolver2026 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_9&diff=152508 2019 AMC 8 Problems/Problem 9 2021-04-23T17:49:30Z <p>Problemsolver2026: </p> <hr /> <div>==Problem 9==<br /> Alex and Felicia each have cats as pets. Alex buys cat food in cylindrical cans that are &lt;math&gt;6&lt;/math&gt; cm in diameter and &lt;math&gt;12&lt;/math&gt; cm high. Felicia buys cat food in cylindrical cans that are &lt;math&gt;12&lt;/math&gt; cm in diameter and &lt;math&gt;6&lt;/math&gt; cm high. What is the ratio of the volume one of Alex's cans to the volume one of Felicia's cans?<br /> <br /> &lt;math&gt;\textbf{(A) }1:4\qquad\textbf{(B) }1:2\qquad\textbf{(C) }1:1\qquad\textbf{(D) }2:1\qquad\textbf{(E) }4:1&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> Using the formula for the volume of a cylinder, we get Alex, &lt;math&gt;\pi108&lt;/math&gt;, and Felicia, &lt;math&gt;\pi216&lt;/math&gt;. We can quickly notice that &lt;math&gt;\pi&lt;/math&gt; cancels out on both sides, and that Alex's volume is &lt;math&gt;1/2&lt;/math&gt; of Felicia's leaving &lt;math&gt;1/2 = \boxed{1:2}&lt;/math&gt; as the answer. <br /> <br /> ~aopsav<br /> <br /> ==Solution 2==<br /> <br /> Using the formula for the volume of a cylinder, we get that the volume of Alex's can is &lt;math&gt;3^2\cdot12\cdot\pi&lt;/math&gt;, and that the volume of Felicia's can is &lt;math&gt;6^2\cdot6\cdot\pi&lt;/math&gt;. Now we divide the volume of Alex's can by the volume of Felicia's can, so we get &lt;math&gt;\frac{1}{2}&lt;/math&gt;, which is &lt;math&gt;\boxed{\textbf{(B)}\ 1:2}&lt;/math&gt; <br /> <br /> lol this is something no one should be able to do.-(Algebruh123)2020<br /> <br /> ==Solution 3==<br /> <br /> The ratio of the numbers is &lt;math&gt;1/2&lt;/math&gt;. Looking closely at the formula &lt;math&gt;r^2 * h * \pi&lt;/math&gt;, we see that the &lt;math&gt;r * h * \pi&lt;/math&gt; will cancel, meaning that the ratio of them will be &lt;math&gt;\frac{1(2)}{2(2)}&lt;/math&gt; = &lt;math&gt;\boxed{\textbf{(B)}\ 1:2}&lt;/math&gt; <br /> <br /> -Lcz<br /> <br /> == Video Solution ==<br /> https://youtu.be/FDgcLW4frg8?t=2440<br /> <br /> ~ pi_is_3.14<br /> <br /> == Video Solution ==<br /> <br /> Solution detailing how to solve the problem: https://www.youtube.com/watch?v=G-gEdWP0S9M&amp;list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&amp;index=10<br /> <br /> ==See also==<br /> {{AMC8 box|year=2019|num-b=8|num-a=10}}<br /> <br /> {{MAA Notice}}</div> Problemsolver2026 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_8&diff=152507 2019 AMC 8 Problems/Problem 8 2021-04-23T17:48:52Z <p>Problemsolver2026: </p> <hr /> <div>==Problem 8==<br /> Gilda has a bag of marbles. She gives &lt;math&gt;20\%&lt;/math&gt; of them to her friend Pedro. Then Gilda gives &lt;math&gt;10\%&lt;/math&gt; of what is left to another friend, Ebony. Finally, Gilda gives &lt;math&gt;25\%&lt;/math&gt; of what is now left in the bag to her brother Jimmy. What percentage of her original bag of marbles does Gilda have left for herself?<br /> <br /> &lt;math&gt;\textbf{(A) }20\qquad\textbf{(B) }33\frac{1}{3}\qquad\textbf{(C) }38\qquad\textbf{(D) }45\qquad\textbf{(E) }54&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> After Gilda gives &lt;math&gt;20&lt;/math&gt;% of the marbles to Pedro, she has &lt;math&gt;80&lt;/math&gt;% of the marbles left. If she then gives &lt;math&gt;10&lt;/math&gt;% of what's left to Ebony, she has &lt;math&gt;(0.8*0.9)&lt;/math&gt; = &lt;math&gt;72&lt;/math&gt;% of what she had at the beginning. Finally, she gives &lt;math&gt;25&lt;/math&gt;% of what's left to her brother, so she has &lt;math&gt;(0.75*0.72)&lt;/math&gt; &lt;math&gt;\boxed{\textbf{(E)}\ 54}&lt;/math&gt;. of what she had in the beginning left.<br /> <br /> ==Solution 2==<br /> Suppose Gilda has 100 marbles. <br /> <br /> Then she gives Pedro 20% of 100 = 20, she remains with 80 marbles.<br /> <br /> Out of 80 marbles she gives 10% of 80 = 8 to Ebony. <br /> <br /> Thus she remains with 72 marbles. <br /> <br /> Then she gives 25% of 72 = 18 to Jimmy, finally leaving her with 54. <br /> <br /> And &lt;math&gt;\frac{54}{100}&lt;/math&gt;=54%=&lt;math&gt;\boxed{\textbf{(E)}\ 54}&lt;/math&gt;<br /> <br /> ~phoenixfire<br /> <br /> ==Solution 3==<br /> (Only if you have lots of time do it this way)<br /> Since she gave away 20% and 10% of what is left and then another 25% of what is actually left, we can do 20+10+25 or 55%. But it is actually going to be a bit more than 55% because 10% of what is left is not 10% of the total amount. So the only option that is greater than 100% - 55% is &lt;math&gt;\boxed{\textbf{(E)}\ 54}&lt;/math&gt;.<br /> <br /> == Video Solution ==<br /> <br /> Solution detailing how to solve the problem: https://www.youtube.com/watch?v=WAmvVQzuzfc&amp;list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&amp;index=9<br /> <br /> ==See also== <br /> {{AMC8 box|year=2019|num-b=7|num-a=9}}<br /> <br /> {{MAA Notice}}</div> Problemsolver2026 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_7&diff=152506 2019 AMC 8 Problems/Problem 7 2021-04-23T17:48:25Z <p>Problemsolver2026: </p> <hr /> <div>==Problem 7==<br /> Shauna takes five tests, each worth a maximum of &lt;math&gt;100&lt;/math&gt; points. Her scores on the first three tests are &lt;math&gt;76&lt;/math&gt;, &lt;math&gt;94&lt;/math&gt;, and &lt;math&gt;87&lt;/math&gt;. In order to average &lt;math&gt;81&lt;/math&gt; for all five tests, what is the lowest score she could earn on one of the other two tests?<br /> <br /> &lt;math&gt;\textbf{(A) }48\qquad\textbf{(B) }52\qquad\textbf{(C) }66\qquad\textbf{(D) }70\qquad\textbf{(E) }74&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> We should notice that we can turn the information we are given into a linear equation and just solve for our set variables. I'll use the variables &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; for the scores on the last two tests. &lt;cmath&gt;\frac{76+94+87+x+y}{5} = 81,&lt;/cmath&gt; &lt;cmath&gt;\frac{257+x+y}{5} = 81.&lt;/cmath&gt; We can now cross multiply to get rid of the denominator. &lt;cmath&gt;257+x+y = 405,&lt;/cmath&gt; &lt;cmath&gt;x+y = 148.&lt;/cmath&gt; Now that we have this equation, we will assign &lt;math&gt;y&lt;/math&gt; as the lowest score of the two other tests, and so: &lt;cmath&gt;x = 100,&lt;/cmath&gt; &lt;cmath&gt;y=48.&lt;/cmath&gt; Now we know that the lowest score on the two other tests is &lt;math&gt;\boxed{48}&lt;/math&gt;.<br /> <br /> ~ aopsav<br /> <br /> ==Solution 2==<br /> Right now, she scored &lt;math&gt;76, 94,&lt;/math&gt; and &lt;math&gt;87&lt;/math&gt; points, for a total of &lt;math&gt;257&lt;/math&gt; points. She wants her average to be &lt;math&gt;81&lt;/math&gt; for her &lt;math&gt;5&lt;/math&gt; tests, so she needs to score &lt;math&gt;405&lt;/math&gt; points in total. This means she needs to score a total of &lt;math&gt;405-257= <br /> 148&lt;/math&gt; points in her next &lt;math&gt;2&lt;/math&gt; tests. Since the maximum score she can get on one of her &lt;math&gt;2&lt;/math&gt; tests is &lt;math&gt;100&lt;/math&gt;, the least possible score she can get is &lt;math&gt;\boxed{\textbf{(A)}\ 48}&lt;/math&gt;.<br /> <br /> Note: You can verify that &lt;math&gt;\boxed{48}&lt;/math&gt; is the right answer because it is the lowest answer out of the 5. Since it is possible to get 48, we are guaranteed that that is the right answer.<br /> <br /> ==Solution 3==<br /> We can compare each of the scores with the average of &lt;math&gt;81&lt;/math&gt;:<br /> &lt;math&gt;76&lt;/math&gt; &lt;math&gt;\rightarrow&lt;/math&gt; &lt;math&gt;-5&lt;/math&gt;,<br /> &lt;math&gt;94&lt;/math&gt; &lt;math&gt;\rightarrow&lt;/math&gt; &lt;math&gt;+13&lt;/math&gt;,<br /> &lt;math&gt;87&lt;/math&gt; &lt;math&gt;\rightarrow&lt;/math&gt; &lt;math&gt;+6&lt;/math&gt;,<br /> &lt;math&gt;100&lt;/math&gt; &lt;math&gt;\rightarrow&lt;/math&gt; &lt;math&gt;+19&lt;/math&gt;;<br /> <br /> So the last one has to be &lt;math&gt;-33&lt;/math&gt; (since all the differences have to sum to &lt;math&gt;0&lt;/math&gt;), which corresponds to &lt;math&gt;81-33 = \boxed{48}&lt;/math&gt;.<br /> <br /> == Video Solution ==<br /> <br /> Solution detailing how to solve the problem: https://www.youtube.com/watch?v=mwHrUESo2_A&amp;list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&amp;index=8<br /> <br /> ==See also==<br /> {{AMC8 box|year=2019|num-b=6|num-a=8}}<br /> <br /> {{MAA Notice}}</div> Problemsolver2026 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_6&diff=152499 2019 AMC 8 Problems/Problem 6 2021-04-23T16:39:19Z <p>Problemsolver2026: </p> <hr /> <div>== Problem 6 ==<br /> <br /> There are &lt;math&gt;81&lt;/math&gt; grid points (uniformly spaced) in the square shown in the diagram below, including the points on the edges. Point &lt;math&gt;P&lt;/math&gt; is in the center of the square. Given that point &lt;math&gt;Q&lt;/math&gt; is randomly chosen among the other &lt;math&gt;80&lt;/math&gt; points, what is the probability that the line &lt;math&gt;PQ&lt;/math&gt; is a line of symmetry for the square?<br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(0,8));<br /> draw((0,8)--(8,8));<br /> draw((8,8)--(8,0));<br /> draw((8,0)--(0,0));<br /> dot((0,0));<br /> dot((0,1));<br /> dot((0,2));<br /> dot((0,3));<br /> dot((0,4));<br /> dot((0,5));<br /> dot((0,6));<br /> dot((0,7));<br /> dot((0,8));<br /> <br /> dot((1,0));<br /> dot((1,1));<br /> dot((1,2));<br /> dot((1,3));<br /> dot((1,4));<br /> dot((1,5));<br /> dot((1,6));<br /> dot((1,7));<br /> dot((1,8));<br /> <br /> dot((2,0));<br /> dot((2,1));<br /> dot((2,2));<br /> dot((2,3));<br /> dot((2,4));<br /> dot((2,5));<br /> dot((2,6));<br /> dot((2,7));<br /> dot((2,8));<br /> <br /> dot((3,0));<br /> dot((3,1));<br /> dot((3,2));<br /> dot((3,3));<br /> dot((3,4));<br /> dot((3,5));<br /> dot((3,6));<br /> dot((3,7));<br /> dot((3,8));<br /> <br /> dot((4,0));<br /> dot((4,1));<br /> dot((4,2));<br /> dot((4,3));<br /> dot((4,4));<br /> dot((4,5));<br /> dot((4,6));<br /> dot((4,7));<br /> dot((4,8));<br /> <br /> dot((5,0));<br /> dot((5,1));<br /> dot((5,2));<br /> dot((5,3));<br /> dot((5,4));<br /> dot((5,5));<br /> dot((5,6));<br /> dot((5,7));<br /> dot((5,8));<br /> <br /> dot((6,0));<br /> dot((6,1));<br /> dot((6,2));<br /> dot((6,3));<br /> dot((6,4));<br /> dot((6,5));<br /> dot((6,6));<br /> dot((6,7));<br /> dot((6,8));<br /> <br /> dot((7,0));<br /> dot((7,1));<br /> dot((7,2));<br /> dot((7,3));<br /> dot((7,4));<br /> dot((7,5));<br /> dot((7,6));<br /> dot((7,7));<br /> dot((7,8));<br /> <br /> dot((8,0));<br /> dot((8,1));<br /> dot((8,2));<br /> dot((8,3));<br /> dot((8,4));<br /> dot((8,5));<br /> dot((8,6));<br /> dot((8,7));<br /> dot((8,8));<br /> label(&quot;P&quot;,(4,4),NE);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }\frac{1}{5}\qquad\textbf{(B) }\frac{1}{4} \qquad\textbf{(C) }\frac{2}{5} \qquad\textbf{(D) }\frac{9}{20} \qquad\textbf{(E) }\frac{1}{2}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> &lt;asy&gt;<br /> draw((0,0)--(0,8));<br /> draw((0,8)--(8,8));<br /> draw((8,8)--(8,0));<br /> draw((8,0)--(0,0));<br /> dot((0,0));<br /> dot((0,1));<br /> dot((0,2));<br /> dot((0,3));<br /> dot((0,4));<br /> dot((0,5));<br /> dot((0,6));<br /> dot((0,7));<br /> dot((0,8));<br /> <br /> dot((1,0));<br /> dot((1,1));<br /> dot((1,2));<br /> dot((1,3));<br /> dot((1,4));<br /> dot((1,5));<br /> dot((1,6));<br /> dot((1,7));<br /> dot((1,8));<br /> <br /> dot((2,0));<br /> dot((2,1));<br /> dot((2,2));<br /> dot((2,3));<br /> dot((2,4));<br /> dot((2,5));<br /> dot((2,6));<br /> dot((2,7));<br /> dot((2,8));<br /> <br /> dot((3,0));<br /> dot((3,1));<br /> dot((3,2));<br /> dot((3,3));<br /> dot((3,4));<br /> dot((3,5));<br /> dot((3,6));<br /> dot((3,7));<br /> dot((3,8));<br /> <br /> dot((4,0));<br /> dot((4,1));<br /> dot((4,2));<br /> dot((4,3));<br /> dot((4,4));<br /> dot((4,5));<br /> dot((4,6));<br /> dot((4,7));<br /> dot((4,8));<br /> <br /> dot((5,0));<br /> dot((5,1));<br /> dot((5,2));<br /> dot((5,3));<br /> dot((5,4));<br /> dot((5,5));<br /> dot((5,6));<br /> dot((5,7));<br /> dot((5,8));<br /> <br /> dot((6,0));<br /> dot((6,1));<br /> dot((6,2));<br /> dot((6,3));<br /> dot((6,4));<br /> dot((6,5));<br /> dot((6,6));<br /> dot((6,7));<br /> dot((6,8));<br /> <br /> dot((7,0));<br /> dot((7,1));<br /> dot((7,2));<br /> dot((7,3));<br /> dot((7,4));<br /> dot((7,5));<br /> dot((7,6));<br /> dot((7,7));<br /> dot((7,8));<br /> <br /> dot((8,0));<br /> dot((8,1));<br /> dot((8,2));<br /> dot((8,3));<br /> dot((8,4));<br /> dot((8,5));<br /> dot((8,6));<br /> dot((8,7));<br /> dot((8,8));<br /> label(&quot;P&quot;,(4,4),NE);<br /> draw((0,4)--(3,4));<br /> draw((0,8)--(3,5));<br /> draw((8,8)--(5,5));<br /> draw((4,8)--(4,5));<br /> draw((4,0)--(4,3));<br /> draw((0,0)--(3,3));<br /> draw((8,0)--(5,3));<br /> draw((5,4)--(8,4));<br /> &lt;/asy&gt;<br /> Lines of symmetry go through point &lt;math&gt;P&lt;/math&gt;, and there are &lt;math&gt;8&lt;/math&gt; directions the lines could go, and there are &lt;math&gt;4&lt;/math&gt; dots at each direction.&lt;math&gt;\frac{4\times8}{80}=\boxed{\textbf{(C)} \frac{2}{5}}&lt;/math&gt;.<br /> <br /> == Video Solution ==<br /> <br /> Solution detailing how to solve the problem: https://www.youtube.com/watch?v=4L95z9DwlhI&amp;list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&amp;index=7<br /> <br /> ==See also==<br /> {{AMC8 box|year=2019|num-b=5|num-a=7}}<br /> <br /> {{MAA Notice}}</div> Problemsolver2026 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_5&diff=152492 2019 AMC 8 Problems/Problem 5 2021-04-23T15:19:58Z <p>Problemsolver2026: </p> <hr /> <div>== Problem 5 ==<br /> A tortoise challenges a hare to a race. The hare eagerly agrees and quickly runs ahead, leaving the slow-moving tortoise behind. Confident that he will win, the hare stops to take a nap. Meanwhile, the tortoise walks at a slow steady pace for the entire race. The hare awakes and runs to the finish line, only to find the tortoise already there. Which of the following graphs matches the description of the race, showing the distance &lt;math&gt;d&lt;/math&gt; traveled by the two animals over time &lt;math&gt;t&lt;/math&gt; from start to finish?<br /> <br /> [[File:2019_AMC_8_-4_Image_1.png|900px]]<br /> <br /> [[File:2019_AMC_8_-4_Image_2.png|600px]]<br /> <br /> ==Solution 1==<br /> First, the tortoise walks at a constant rate, ruling out &lt;math&gt;(D)&lt;/math&gt;<br /> Second, when the hare is resting, the distance will stay the same, ruling out &lt;math&gt;(E)&lt;/math&gt; and &lt;math&gt;(C)&lt;/math&gt;.<br /> Third, the tortoise wins the race, meaning that the non-constant one should go off the graph last, ruling out &lt;math&gt;(A)&lt;/math&gt;.<br /> Therefore, the answer &lt;math&gt;\boxed{\textbf{(B)}}&lt;/math&gt; is the only one left.<br /> <br /> &lt;math&gt;\phantom{Note to the original author of this solution: &quot;we shouldn't be able to edit&quot; is incorrect (if its definition is what I think it is), because I was able to edit. Also, I deleted that, (but did nothing else) }&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> First, we know that the rabbit beats the tortoise in the first half of the race. So he is going to be ahead of the tortoise. We also know, while he rested, he didn't move. The only graph portraying that is going to be &lt;math&gt;\boxed{\textbf{(B)}}&lt;/math&gt;. This is our answer. ~bobthefam<br /> <br /> == Video Solution ==<br /> <br /> Solution detailing how to solve the problem: https://www.youtube.com/watch?v=uQTM8Kmh9B8&amp;list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&amp;index=6<br /> <br /> ==See also==<br /> {{AMC8 box|year=2019|num-b=4|num-a=6}}<br /> <br /> {{MAA Notice}}</div> Problemsolver2026 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_4&diff=152490 2019 AMC 8 Problems/Problem 4 2021-04-23T14:45:28Z <p>Problemsolver2026: </p> <hr /> <div>== Problem 4 ==<br /> <br /> Quadrilateral &lt;math&gt;ABCD&lt;/math&gt; is a rhombus with perimeter &lt;math&gt;52&lt;/math&gt; meters. The length of diagonal &lt;math&gt;\overline{AC}&lt;/math&gt; is &lt;math&gt;24&lt;/math&gt; meters. What is the area in square meters of rhombus &lt;math&gt;ABCD&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> draw((-13,0)--(0,5));<br /> draw((0,5)--(13,0));<br /> draw((13,0)--(0,-5));<br /> draw((0,-5)--(-13,0));<br /> dot((-13,0));<br /> dot((0,5));<br /> dot((13,0));<br /> dot((0,-5));<br /> label(&quot;A&quot;,(-13,0),W);<br /> label(&quot;B&quot;,(0,5),N);<br /> label(&quot;C&quot;,(13,0),E);<br /> label(&quot;D&quot;,(0,-5),S);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }60\qquad\textbf{(B) }90\qquad\textbf{(C) }105\qquad\textbf{(D) }120\qquad\textbf{(E) }144&lt;/math&gt;<br /> <br /> <br /> == Solution 1 ==<br /> &lt;asy&gt;<br /> draw((-12,0)--(0,5));<br /> draw((0,5)--(12,0));<br /> draw((12,0)--(0,-5));<br /> draw((0,-5)--(-12,0));<br /> draw((0,0)--(12,0));<br /> draw((0,0)--(0,5));<br /> draw((0,0)--(-12,0));<br /> draw((0,0)--(0,-5));<br /> dot((-12,0));<br /> dot((0,5));<br /> dot((12,0));<br /> dot((0,-5));<br /> label(&quot;A&quot;,(-12,0),W);<br /> label(&quot;B&quot;,(0,5),N);<br /> label(&quot;C&quot;,(12,0),E);<br /> label(&quot;D&quot;,(0,-5),S);<br /> label(&quot;E&quot;,(0,0),SW);<br /> &lt;/asy&gt;<br /> <br /> A rhombus has sides of equal length. Because the perimeter of the rhombus is &lt;math&gt;52&lt;/math&gt;, each side is &lt;math&gt;\frac{52}{4}=13&lt;/math&gt;. In a rhombus, diagonals are perpendicular and bisect each other, which means &lt;math&gt;\overline{AE}&lt;/math&gt; = &lt;math&gt;12&lt;/math&gt; = &lt;math&gt;\overline{EC}&lt;/math&gt;.<br /> <br /> Consider one of the right triangles:<br /> <br /> &lt;asy&gt;<br /> draw((-12,0)--(0,5));<br /> draw((0,0)--(-12,0));<br /> draw((0,0)--(0,5));<br /> dot((-12,0));<br /> dot((0,5));<br /> label(&quot;A&quot;,(-12,0),W);<br /> label(&quot;B&quot;,(0,5),N);<br /> label(&quot;E&quot;,(0,0),SE);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\overline{AB}&lt;/math&gt; = &lt;math&gt;13&lt;/math&gt;, and &lt;math&gt;\overline{AE}&lt;/math&gt; = &lt;math&gt;12&lt;/math&gt;. Using Pythagorean theorem, we find that &lt;math&gt;\overline{BE}&lt;/math&gt; = &lt;math&gt;5&lt;/math&gt;.<br /> You may recall the famous Pythagorean triple, (5, 12, 13).<br /> <br /> Thus the values of the two diagonals are &lt;math&gt;\overline{AC}&lt;/math&gt; = &lt;math&gt;24&lt;/math&gt; and &lt;math&gt;\overline{BD}&lt;/math&gt; = &lt;math&gt;10&lt;/math&gt;.<br /> The area of a rhombus is = &lt;math&gt;\frac{d_1\cdot{d_2}}{2}&lt;/math&gt; = &lt;math&gt;\frac{24\cdot{10}}{2}&lt;/math&gt; = &lt;math&gt;120&lt;/math&gt;<br /> <br /> &lt;math&gt;\boxed{\textbf{(D)}\ 120}&lt;/math&gt;<br /> <br /> == Video Solution ==<br /> <br /> Solution detailing how to solve the problem:https://www.youtube.com/watch?v=-yHfOUapg7I&amp;list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&amp;index=5<br /> <br /> ==See also==<br /> {{AMC8 box|year=2019|num-b=3|num-a=5}}<br /> <br /> {{MAA Notice}}</div> Problemsolver2026 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_3&diff=152489 2019 AMC 8 Problems/Problem 3 2021-04-23T14:44:41Z <p>Problemsolver2026: </p> <hr /> <div>==Problem 3==<br /> Which of the following is the correct order of the fractions &lt;math&gt;\frac{15}{11},\frac{19}{15},&lt;/math&gt; and &lt;math&gt;\frac{17}{13},&lt;/math&gt; from least to greatest? <br /> <br /> &lt;math&gt;\textbf{(A) }\frac{15}{11}&lt; \frac{17}{13}&lt; \frac{19}{15} \qquad\textbf{(B) }\frac{15}{11}&lt; \frac{19}{15}&lt;\frac{17}{13} \qquad\textbf{(C) }\frac{17}{13}&lt;\frac{19}{15}&lt;\frac{15}{11} \qquad\textbf{(D) } \frac{19}{15}&lt;\frac{15}{11}&lt;\frac{17}{13} \qquad\textbf{(E) } \frac{19}{15}&lt;\frac{17}{13}&lt;\frac{15}{11}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> We take a common denominator:<br /> &lt;cmath&gt;\frac{15}{11},\frac{19}{15}, \frac{17}{13} = \frac{15\cdot 15 \cdot 13}{11\cdot 15 \cdot 13},\frac{19 \cdot 11 \cdot 13}{15\cdot 11 \cdot 13}, \frac{17 \cdot 11 \cdot 15}{13\cdot 11 \cdot 15} = \frac{2925}{2145},\frac{2717}{2145},\frac{2805}{2145}.&lt;/cmath&gt;<br /> <br /> Since &lt;math&gt;2717&lt;2805&lt;2925&lt;/math&gt; it follows that the answer is &lt;math&gt;\boxed{\textbf{(E)}\frac{19}{15}&lt;\frac{17}{13}&lt;\frac{15}{11}}&lt;/math&gt;.<br /> <br /> -xMidnightFirex<br /> <br /> ~ dolphin7 - I took your idea and made it an explanation.<br /> <br /> ==Solution 2==<br /> When &lt;math&gt;\frac{x}{y}&gt;1&lt;/math&gt; and &lt;math&gt;z&gt;0&lt;/math&gt;, &lt;math&gt;\frac{x+z}{y+z}&lt;\frac{x}{y}&lt;/math&gt;. Hence, the answer is &lt;math&gt;\boxed{\textbf{(E)}\frac{19}{15}&lt;\frac{17}{13}&lt;\frac{15}{11}}&lt;/math&gt;.<br /> ~ ryjs<br /> <br /> This is also similar to Problem 20 on the AMC 2012.<br /> <br /> ==Solution 3 (probably won't use this solution)==<br /> We use our insane mental calculator to find out that &lt;math&gt;\frac{15}{11} \approx 1.36&lt;/math&gt;, &lt;math&gt;\frac{19}{15} \approx 1.27&lt;/math&gt;, and &lt;math&gt;\frac{17}{13} \approx 1.31&lt;/math&gt;. Thus, our answer is &lt;math&gt;\boxed{\textbf{(E)}\frac{19}{15}&lt;\frac{17}{13}&lt;\frac{15}{11}}&lt;/math&gt;.<br /> <br /> ~~ by an insane math guy.<br /> <br /> ==Solution 4==<br /> Suppose each fraction is expressed with denominator &lt;math&gt;2145&lt;/math&gt;: &lt;math&gt;\frac{2925}{2145}, \frac{2717}{2145}, \frac{2805}{2145}&lt;/math&gt;. Clearly &lt;math&gt;2717&lt;2805&lt;2925&lt;/math&gt; so the answer is &lt;math&gt;\boxed{\textbf{(E)}}&lt;/math&gt;.<br /> <br /> ==Solution 5 -SweetMango77==<br /> We notice that each of these fraction's numerator &lt;math&gt;-&lt;/math&gt; denominator &lt;math&gt;=4&lt;/math&gt;. If we take each of the fractions, and subtract &lt;math&gt;1&lt;/math&gt; from each, we get &lt;math&gt;\frac{4}{11}&lt;/math&gt;, &lt;math&gt;\frac{4}{15}&lt;/math&gt;, and &lt;math&gt;\frac{4}{19}&lt;/math&gt;. These are easy to order because the numerators are the same, we get &lt;math&gt;\frac{4}{15}&lt;\frac{4}{13}&lt;\frac{4}{11}&lt;/math&gt;. Because it is a subtraction by a constant, in order to order them, we keep the inequality signs to get &lt;math&gt;\boxed{\textbf{(E)}\;\frac{19}{15}&lt;\frac{17}{13}&lt;\frac{15}{11}}&lt;/math&gt;.<br /> <br /> == Video Solution ==<br /> <br /> Solution detailing how to solve the problem:https://www.youtube.com/watch?v=q27qEcr7TbQ&amp;list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&amp;index=4<br /> <br /> ==See also==<br /> {{AMC8 box|year=2019|num-b=2|num-a=4}}<br /> <br /> {{MAA Notice}} The butterfly method is a method when you multiply the denominator of the second fraction and multiply it by the numerator from the first fraction.</div> Problemsolver2026 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_2&diff=152488 2019 AMC 8 Problems/Problem 2 2021-04-23T14:43:57Z <p>Problemsolver2026: </p> <hr /> <div>==Problem==<br /> Three identical rectangles are put together to form rectangle &lt;math&gt;ABCD&lt;/math&gt;, as shown in the figure below. Given that the length of the shorter side of each of the smaller rectangles is 5 feet, what is the area in square feet of rectangle &lt;math&gt;ABCD&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(3,0));<br /> draw((0,0)--(0,2));<br /> draw((0,2)--(3,2));<br /> draw((3,2)--(3,0));<br /> dot((0,0));<br /> dot((0,2));<br /> dot((3,0));<br /> dot((3,2));<br /> draw((2,0)--(2,2));<br /> draw((0,1)--(2,1));<br /> label(&quot;A&quot;,(0,0),S);<br /> label(&quot;B&quot;,(3,0),S);<br /> label(&quot;C&quot;,(3,2),N);<br /> label(&quot;D&quot;,(0,2),N);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }45\qquad\textbf{(B) }75\qquad\textbf{(C) }100\qquad\textbf{(D) }125\qquad\textbf{(E) }150&lt;/math&gt;<br /> <br /> ==Solutions==<br /> <br /> ===Solution 1===<br /> <br /> We can see that there are &lt;math&gt;2&lt;/math&gt; rectangles lying on top of the other and that is the same as the length of one rectangle. Now we know that the shorter side is &lt;math&gt;5&lt;/math&gt;, if we take the information of the problem, and the bigger side is &lt;math&gt;10&lt;/math&gt;, if we do &lt;math&gt;5 \cdot 2 = 10&lt;/math&gt;. Now we get the sides of the big rectangles being &lt;math&gt;15&lt;/math&gt; and &lt;math&gt;10&lt;/math&gt;, so the area is &lt;math&gt;\boxed{\textbf{(E)}\ 150}&lt;/math&gt;. ~avamarora<br /> <br /> ===Solution 2===<br /> Using the diagram we find that the larger side of the small rectangle is &lt;math&gt;2&lt;/math&gt; times the length of the smaller side. Therefore, the longer side is &lt;math&gt;5 \cdot 2 = 10&lt;/math&gt;. So the area of the identical rectangles is &lt;math&gt;5 \cdot 10 = 50&lt;/math&gt;. We have 3 identical rectangles that form the large rectangle. Therefore the area of the large rectangle is &lt;math&gt;50 \cdot 3 = \boxed{\textbf{(E)}\ 150}&lt;/math&gt;. ~~fath2012<br /> <br /> ===Solution 3===<br /> We see that if the short sides are 5, the long side has to be &lt;math&gt;5\cdot2=10&lt;/math&gt; because the long side is equal to the 2 short sides and because the rectangles are congruent. If that is to be, then the long side of the BIG rectangle(rectangle &lt;math&gt;ABCD&lt;/math&gt;)<br /> is &lt;math&gt;10+5=15&lt;/math&gt; because long side + short side of the small rectangle is &lt;math&gt;15&lt;/math&gt;. The short side of rectangle &lt;math&gt;ABCD&lt;/math&gt; is &lt;math&gt;10&lt;/math&gt; because it is the long side of the short rectangle. Multiplying &lt;math&gt;15&lt;/math&gt; and &lt;math&gt;10&lt;/math&gt; together gets us &lt;math&gt;15\cdot10&lt;/math&gt; which is &lt;math&gt;\boxed{\textbf{(E)}\ 150}&lt;/math&gt;.<br /> ~~mathboy282<br /> <br /> == Video Solution ==<br /> <br /> Solution detailing how to solve the problem:https://www.youtube.com/watch?v=eEqtoI8BQKE&amp;list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&amp;index=3<br /> <br /> ==See also==<br /> {{AMC8 box|year=2019|num-b=1|num-a=3}}<br /> <br /> {{MAA Notice}}</div> Problemsolver2026 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_1&diff=152487 2019 AMC 8 Problems/Problem 1 2021-04-23T14:42:12Z <p>Problemsolver2026: </p> <hr /> <div>== Problem 1 ==<br /> Ike and Mike go into a sandwich shop with a total of &lt;math&gt;\$30.00&lt;/math&gt; to spend. Sandwiches cost &lt;math&gt;\$4.50&lt;/math&gt; each and soft drinks cost &lt;math&gt;\\$1.00&lt;/math&gt; each. Ike and Mike plan to buy as many sandwiches as they can,<br /> and use any remaining money to buy soft drinks. Counting both sandwiches and soft drinks, how<br /> many items will they buy?<br /> <br /> &lt;math&gt;\textbf{(A) }6\qquad\textbf{(B) }7\qquad\textbf{(C) }8\qquad\textbf{(D) }9\qquad\textbf{(E) }10&lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> We know that the sandwiches cost &lt;math&gt;4.50&lt;/math&gt; dollars. Guessing will bring us to multiplying &lt;math&gt;4.50&lt;/math&gt; by 6, which gives us &lt;math&gt;27.00&lt;/math&gt;. Since they can spend &lt;math&gt;30.00&lt;/math&gt; they have &lt;math&gt;3&lt;/math&gt; dollars left. Since sodas cost &lt;math&gt;1.00&lt;/math&gt; dollar each, they can buy 3 sodas, which makes them spend &lt;math&gt;30.00&lt;/math&gt; Since they bought 6 sandwiches and 3 sodas, they bought a total of &lt;math&gt;9&lt;/math&gt; items. Therefore, the answer is &lt;math&gt;\boxed{D = 9 }&lt;/math&gt;<br /> <br /> - SBose<br /> <br /> == Solution 2 (Using Algebra) ==<br /> Let &lt;math&gt;s&lt;/math&gt; be the number of sandwiches and &lt;math&gt;d&lt;/math&gt; be the number of sodas. We have to satisfy the equation of<br /> &lt;cmath&gt;4.50s+d=30&lt;/cmath&gt;<br /> In the question, it states that Ike and Mike buys as many sandwiches as possible. <br /> So, we drop the number of sodas for a while.<br /> We have: <br /> &lt;cmath&gt;4.50s=30&lt;/cmath&gt;<br /> &lt;cmath&gt;s=\frac{30}{4.5}&lt;/cmath&gt;<br /> &lt;cmath&gt;s=6R30&lt;/cmath&gt;<br /> We don't want a remainder so the maximum number of sandwiches is &lt;math&gt;6&lt;/math&gt;.<br /> The total money spent is &lt;math&gt;6\cdot 4.50=27&lt;/math&gt;.<br /> The number of dollar left to spent on sodas is &lt;math&gt;30-27=3&lt;/math&gt; dollars.<br /> &lt;math&gt;3&lt;/math&gt; dollars can buy &lt;math&gt;3&lt;/math&gt; sodas leading us to a total of <br /> &lt;math&gt;6+3=9&lt;/math&gt; items. <br /> Hence, the answer is &lt;math&gt;\boxed{(D) = 9}&lt;/math&gt;<br /> <br /> - by interactivemath<br /> <br /> == Video Solution ==<br /> <br /> Solution detailing how to solve the problem: https://www.youtube.com/watch?v=Puzy1HAlAKk&amp;list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&amp;index=2<br /> <br /> ==See also==<br /> {{AMC8 box|year=2019|before = First Problem|num-a=2}}<br /> <br /> {{MAA Notice}}</div> Problemsolver2026