https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Amackenzie1&feedformat=atom AoPS Wiki - User contributions [en] 2021-10-21T06:31:23Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=AoPS_Wiki:Competition_ratings&diff=82098 AoPS Wiki:Competition ratings 2016-12-31T19:14:37Z <p>Amackenzie1: </p> <hr /> <div>This page contains an approximate estimation of the difficulty level of various [[List of mathematics competitions|competitions]]. It is designed with the intention of introducing contests of similar difficulty levels (but possibly different styles of problems) that readers may like to try to gain more experience.<br /> <br /> Each entry groups the problems into sets of similar difficulty levels and suggests an approximate difficulty rating, on a scale from 1 to 10 (from easiest to hardest). Note that many of these ratings are not directly comparable, because the actual competitions have many different rules; the ratings are generally synchronized with the amount of available time, etc. Also, due to variances within a contest, ranges shown may overlap. A sample problem is provided with each entry, with a link to a solution. <br /> <br /> As you may have guessed with time many competitions got more challenging because many countries got more access to books targeted at olympiad preparation. But especially web site where one can discuss Olympiads such as our very own AoPS!<br /> <br /> If you have some experience with mathematical competitions, we hope that you can help us make the difficulty rankings more accurate. Currently, the system is on a scale from 1 to 10 where 1 is the easiest level, e.g. [url=http://www.mathlinks.ro/resources.php?c=182&amp;cid=44]early AMC problems[/url] and 10 is hardest level, e.g. [http://www.mathlinks.ro/resources.php?c=37&amp;cid=47 China IMO Team Selection Test.] When considering problem difficulty put more emphasis on problem-solving aspects and less so on technical skill requirements.{{ref|1}}<br /> <br /> == Competitions ==<br /> <br /> === [[MOEMS]] ===<br /> *Division E: '''1'''<br /> *: ''The whole number &lt;math&gt;N&lt;/math&gt; is divisible by &lt;math&gt;7&lt;/math&gt;. &lt;math&gt;N&lt;/math&gt; leaves a remainder of &lt;math&gt;1&lt;/math&gt; when divided by &lt;math&gt;2,3,4,&lt;/math&gt; or &lt;math&gt;5&lt;/math&gt;. What is the smallest value that &lt;math&gt;N&lt;/math&gt; can be?'' ([http://www.moems.org/sample_files/SampleE.pdf Solution])<br /> *Division M: '''1'''<br /> *: ''The value of a two-digit number is &lt;math&gt;10&lt;/math&gt; times more than the sum of its digits. The units digit is 1 more than twice the tens digit. Find the two-digit number.'' ([http://www.moems.org/sample_files/SampleM.pdf Solution])<br /> <br /> === [[AMC 8]] ===<br /> <br /> * Problem 1 - Problem 12: '''1''' <br /> *: ''What is the number of degrees in the smaller angle between the hour hand and the minute hand on a clock that reads seven o'clock?'' ([[1989 AJHSME Problems/Problem 10|Solution]])<br /> * Problem 13 - Problem 25: '''1.5'''<br /> *: ''A fifth number, &lt;math&gt;n&lt;/math&gt;, is added to the set &lt;math&gt;\{ 3,6,9,10 \}&lt;/math&gt; to make the mean of the set of five numbers equal to its median. What is the number of possible values of &lt;math&gt;n&lt;/math&gt;? '' ([[1988 AJHSME Problems/Problem 21|Solution]])<br /> <br /> === [[Mathcounts]] ===<br /> <br /> * Countdown: '''0.5''' (School, Chapter), '''1''' (State, National)<br /> * Sprint: '''1-1.5''' (school), '''1.5''' (Chapter),'''1.5-2''' (State), '''2''' (National)<br /> * Target: '''1.5''' (school), '''2''' (Chapter), '''2''' (State), '''2.5''' (National)<br /> <br /> === [[AMC 10]] ===<br /> <br /> * Problem 1 - 5: '''1'''<br /> *: ''The larger of two consecutive odd integers is three times the smaller. What is their sum?'' ([[2007 AMC 10A Problems/Problem 4|Solution]])<br /> * Problem 6 - 20: '''2'''<br /> *: ''How many non-similar triangles have angles whose degree measures are distinct positive integers in arithmetic progression?'' ([[2006 AMC 10A Problems/Problem 19|Solution]])<br /> * Problem 21 - 25: '''3'''<br /> *: ''Mr. Jones has eight children of different ages. On a family trip his oldest child, who is 9, spots a license plate with a 4-digit number in which each of two digits appears two times. &quot;Look, daddy!&quot; she exclaims. &quot;That number is evenly divisible by the age of each of us kids!&quot; &quot;That's right,&quot; replies Mr. Jones, &quot;and the last two digits just happen to be my age.&quot; Which of the following is not the age of one of Mr. Jones's children?'' ([[2006 AMC 10B Problems/Problem 25|Solution]])<br /> === Austrian MO ===<br /> <br /> * Regional Competition for Advanced Students, Problems 1-4: '''2''' <br /> * Federal Competition for Advanced Students, Part 1. Problems 1-4: '''3''' <br /> * Federal Competition for Advanced Students, Part 2, Problems 1-6: '''4'''<br /> <br /> === [[AMC 12]] ===<br /> <br /> * Problem 1-10: '''2'''<br /> *: ''A solid box is &lt;math&gt;15&lt;/math&gt; cm by &lt;math&gt;10&lt;/math&gt; cm by &lt;math&gt;8&lt;/math&gt; cm. A new solid is formed by removing a cube &lt;math&gt;3&lt;/math&gt; cm on a side from each corner of this box. What percent of the original volume is removed?'' ([[2003 AMC 12A Problems/Problem 3|Solution]])<br /> * Problem 11-20: '''3'''<br /> *: ''An object in the plane moves from one lattice point to another. At each step, the object may move one unit to the right, one unit to the left, one unit up, or one unit down. If the object starts at the origin and takes a ten-step path, how many different points could be the final point?'' ([[2006 AMC 12B Problems/Problem 18|Solution]])<br /> * Problem 21-25: '''4'''<br /> *: ''Functions &lt;math&gt;f&lt;/math&gt; and &lt;math&gt;g&lt;/math&gt; are quadratic, &lt;math&gt;g(x) = - f(100 - x)&lt;/math&gt;, and the graph of &lt;math&gt;g&lt;/math&gt; contains the vertex of the graph of &lt;math&gt;f&lt;/math&gt;. The four &lt;math&gt;x&lt;/math&gt;-intercepts on the two graphs have &lt;math&gt;x&lt;/math&gt;-coordinates &lt;math&gt;x_1&lt;/math&gt;, &lt;math&gt;x_2&lt;/math&gt;, &lt;math&gt;x_3&lt;/math&gt;, and &lt;math&gt;x_4&lt;/math&gt;, in increasing order, and &lt;math&gt;x_3 - x_2 = 150&lt;/math&gt;. The value of &lt;math&gt;x_4 - x_1&lt;/math&gt; is &lt;math&gt;m + n\sqrt p&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt;, &lt;math&gt;n&lt;/math&gt;, and &lt;math&gt;p&lt;/math&gt; are positive integers, and &lt;math&gt;p&lt;/math&gt; is not divisible by the square of any prime. What is &lt;math&gt;m + n + p&lt;/math&gt;?'' ([[2009 AMC 12A Problems/Problem 23|Solution]])<br /> <br /> === [[AIME]] ===<br /> <br /> * Problem 1 - 5: '''3'''<br /> *: Starting at &lt;math&gt;(0,0),&lt;/math&gt; an object moves in the coordinate plane via a sequence of steps, each of length one. Each step is left, right, up, or down, all four equally likely. Let &lt;math&gt;p&lt;/math&gt; be the probability that the object reaches &lt;math&gt;(2,2)&lt;/math&gt; in six or fewer steps. Given that &lt;math&gt;p&lt;/math&gt; can be written in the form &lt;math&gt;m/n,&lt;/math&gt; where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers, find &lt;math&gt;m+n.&lt;/math&gt; ([[1995 AIME Problems/Problem 3|Solution]])<br /> * Problem 6 - 10: '''4''' <br /> *: ''Triangle &lt;math&gt;ABC&lt;/math&gt; has &lt;math&gt;AB=21&lt;/math&gt;, &lt;math&gt;AC=22&lt;/math&gt; and &lt;math&gt;BC=20&lt;/math&gt;. Points &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; are located on &lt;math&gt;\overline{AB}&lt;/math&gt; and &lt;math&gt;\overline{AC}&lt;/math&gt;, respectively, such that &lt;math&gt;\overline{DE}&lt;/math&gt; is [[parallel]] to &lt;math&gt;\overline{BC}&lt;/math&gt; and contains the center of the inscribed circle of triangle &lt;math&gt;ABC&lt;/math&gt;. Then &lt;math&gt;DE=m/n&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m+n&lt;/math&gt;.'' ([[2001 AIME I Problems/Problem 7|Solution]])<br /> * Problem 10 - 12: '''5'''<br /> *: Let &lt;math&gt;z&lt;/math&gt; be a complex number with &lt;math&gt;|z|=2014&lt;/math&gt;. Let &lt;math&gt;P&lt;/math&gt; be the polygon in the complex plane whose vertices are &lt;math&gt;z&lt;/math&gt; and every &lt;math&gt;w&lt;/math&gt; such that &lt;math&gt;\frac{1}{z+w}=\frac{1}{z}+\frac{1}{w}&lt;/math&gt;. Then the area enclosed by &lt;math&gt;P&lt;/math&gt; can be written in the form &lt;math&gt;n\sqrt{3}&lt;/math&gt;, where &lt;math&gt;n&lt;/math&gt; is an integer. Find the remainder when &lt;math&gt;n&lt;/math&gt; is divided by &lt;math&gt;1000&lt;/math&gt;. ([[2014 AIME II Problems/Problem 10|Solution]])<br /> * Problem 12 - 15: '''6'''<br /> *: ''Let<br /> <br /> &lt;cmath&gt;P(x) = 24x^{24} + \sum_{j = 1}^{23}(24 - j)(x^{24 - j} + x^{24 + j}).&lt;/cmath&gt;<br /> Let &lt;math&gt;z_{1},z_{2},\ldots,z_{r}&lt;/math&gt; be the distinct zeros of &lt;math&gt;P(x),&lt;/math&gt; and let &lt;math&gt;z_{k}^{2} = a_{k} + b_{k}i&lt;/math&gt; for &lt;math&gt;k = 1,2,\ldots,r,&lt;/math&gt; where &lt;math&gt;i = \sqrt { - 1},&lt;/math&gt; and &lt;math&gt;a_{k}&lt;/math&gt; and &lt;math&gt;b_{k}&lt;/math&gt; are real numbers. Let<br /> <br /> &lt;cmath&gt;\sum_{k = 1}^{r}|b_{k}| = m + n\sqrt {p},&lt;/cmath&gt;<br /> where &lt;math&gt;m,&lt;/math&gt; &lt;math&gt;n,&lt;/math&gt; and &lt;math&gt;p&lt;/math&gt; are integers and &lt;math&gt;p&lt;/math&gt; is not divisible by the square of any prime. Find &lt;math&gt;m + n + p.&lt;/math&gt;.'' ([[2003 AIME II Problems/Problem 15|Solution]])<br /> <br /> === [[Indonesian MO]] ===<br /> * Problem 1/5: '''3.5'''<br /> *: '' In a drawer, there are at most &lt;math&gt;2009&lt;/math&gt; balls, some of them are white, the rest are blue, which are randomly distributed. If two balls were taken at the same time, then the probability that the balls are both blue or both white is &lt;math&gt;\frac12&lt;/math&gt;. Determine the maximum amount of white balls in the drawer, such that the probability statement is true?'' &lt;url&gt;viewtopic.php?t=294065 (Solution)&lt;/url&gt;<br /> * Problem 2/6: '''4.5'''<br /> *: ''Find the lowest possible values from the function<br /> &lt;math&gt;f(x) = x^{2008} - 2x^{2007} + 3x^{2006} - 4x^{2005} + 5x^{2004} - \cdots - 2006x^3 + 2007x^2 - 2008x + 2009&lt;/math&gt;<br /> <br /> for any real numbers &lt;math&gt;x&lt;/math&gt;.''&lt;url&gt;viewtopic.php?t=294067 (Solution)&lt;/url&gt;<br /> * Problem 3/7: '''5'''<br /> *: ''A pair of integers &lt;math&gt;(m,n)&lt;/math&gt; is called ''good'' if<br /> &lt;math&gt;m\mid n^2 + n \ \text{and} \ n\mid m^2 + m&lt;/math&gt;<br /> <br /> Given 2 positive integers &lt;math&gt;a,b &gt; 1&lt;/math&gt; which are relatively prime, prove that there exists a ''good'' pair &lt;math&gt;(m,n)&lt;/math&gt; with &lt;math&gt;a\mid m&lt;/math&gt; and &lt;math&gt;b\mid n&lt;/math&gt;, but &lt;math&gt;a\nmid n&lt;/math&gt; and &lt;math&gt;b\nmid m&lt;/math&gt;.'' &lt;url&gt;viewtopic.php?t=294068 (Solution)&lt;/url&gt;<br /> * Problem 4/8: '''6'''<br /> *: ''Given an acute triangle &lt;math&gt;ABC&lt;/math&gt;. The incircle of triangle &lt;math&gt;ABC&lt;/math&gt; touches &lt;math&gt;BC,CA,AB&lt;/math&gt; respectively at &lt;math&gt;D,E,F&lt;/math&gt;. The angle bisector of &lt;math&gt;\angle A&lt;/math&gt; cuts &lt;math&gt;DE&lt;/math&gt; and &lt;math&gt;DF&lt;/math&gt; respectively at &lt;math&gt;K&lt;/math&gt; and &lt;math&gt;L&lt;/math&gt;. Suppose &lt;math&gt;AA_1&lt;/math&gt; is one of the altitudes of triangle &lt;math&gt;ABC&lt;/math&gt;, and &lt;math&gt;M&lt;/math&gt; be the midpoint of &lt;math&gt;BC&lt;/math&gt;.<br /> <br /> (a) Prove that &lt;math&gt;BK&lt;/math&gt; and &lt;math&gt;CL&lt;/math&gt; are perpendicular with the angle bisector of &lt;math&gt;\angle BAC&lt;/math&gt;.<br /> <br /> (b) Show that &lt;math&gt;A_1KML&lt;/math&gt; is a cyclic quadrilateral.'' &lt;url&gt;viewtopic.php?t=294069 (Solution)&lt;/url&gt;<br /> <br /> === [[ARML]] ===<br /> <br /> * Individuals, Problem 6 and 8: '''4''' <br /> <br /> * Individuals, Problem 10: '''6'''<br /> <br /> * Team/power, Problem 1-5: '''3.5''' <br /> <br /> * Team/power, Problem 6-10: '''5'''<br /> <br /> === [[Central American Olympiad]] ===<br /> * Problem 1: '''4'''<br /> *: ''Find all three-digit numbers &lt;math&gt;abc&lt;/math&gt; (with &lt;math&gt;a \neq 0&lt;/math&gt;) such that &lt;math&gt;a^{2} + b^{2} + c^{2}&lt;/math&gt; is a divisor of 26.'' (&lt;url&gt;viewtopic.php?p=903856#903856 Solution&lt;/url&gt;)<br /> * Problem 2,4,5: '''5-6'''<br /> *: ''Show that the equation &lt;math&gt;a^{2}b^{2} + b^{2}c^{2} + 3b^{2} - c^{2} - a^{2} = 2005&lt;/math&gt; has no integer solutions.'' (&lt;url&gt;viewtopic.php?p=291301#291301 Solution&lt;/url&gt;)<br /> * Problem 3/6: '''6.5''' <br /> *: ''Let &lt;math&gt;ABCD&lt;/math&gt; be a convex quadrilateral. &lt;math&gt;I = AC\cap BD&lt;/math&gt;, and &lt;math&gt;E&lt;/math&gt;, &lt;math&gt;H&lt;/math&gt;, &lt;math&gt;F&lt;/math&gt; and &lt;math&gt;G&lt;/math&gt; are points on &lt;math&gt;AB&lt;/math&gt;, &lt;math&gt;BC&lt;/math&gt;, &lt;math&gt;CD&lt;/math&gt; and &lt;math&gt;DA&lt;/math&gt; respectively, such that &lt;math&gt;EF \cap GH = I&lt;/math&gt;. If &lt;math&gt;M = EG \cap AC&lt;/math&gt;, &lt;math&gt;N = HF \cap AC&lt;/math&gt;, show that &lt;math&gt;\frac {AM}{IM}\cdot \frac {IN}{CN} = \frac {IA}{IC}.&lt;/math&gt;'' (&lt;url&gt;viewtopic.php?p=828841#p828841 Solution&lt;/url&gt;<br /> <br /> === [[Canadian MO]] ===<br /> <br /> * Problem 1: '''5'''<br /> * Problem 2: '''6'''<br /> * Problem 3: '''6.5''' <br /> * Problem 4: '''7-7.5'''<br /> * Problem 5: '''7.5-8'''<br /> <br /> === [[APMO]] ===<br /> *Problem 1: '''6'''<br /> *Problem 2: '''7'''<br /> *Problem 3: '''7'''<br /> *Problem 4: '''7.5'''<br /> *Problem 5: '''8'''<br /> <br /> === Balkan MO ===<br /> <br /> * Problem 1: '''6'''<br /> *: '' Solve the equation &lt;math&gt;3^x - 5^y = z^2&lt;/math&gt; in positive integers. '' <br /> * Problem 2: '''6.5'''<br /> *: '' Let &lt;math&gt;MN&lt;/math&gt; be a line parallel to the side &lt;math&gt;BC&lt;/math&gt; of a triangle &lt;math&gt;ABC&lt;/math&gt;, with &lt;math&gt;M&lt;/math&gt; on the side &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;N&lt;/math&gt; on the side &lt;math&gt;AC&lt;/math&gt;. The lines &lt;math&gt;BN&lt;/math&gt; and &lt;math&gt;CM&lt;/math&gt; meet at point &lt;math&gt;P&lt;/math&gt;. The circumcircles of triangles &lt;math&gt;BMP&lt;/math&gt; and &lt;math&gt;CNP&lt;/math&gt; meet at two distinct points &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt;. Prove that &lt;math&gt;\angle BAQ = \angle CAP&lt;/math&gt;. ''<br /> * Problem 3: '''7.5'''<br /> *: '' A &lt;math&gt;9 \times 12&lt;/math&gt; rectangle is partitioned into unit squares. The centers of all the unit squares, except for the four corner squares and eight squares sharing a common side with one of them, are coloured red. Is it possible to label these red centres &lt;math&gt;C_1,C_2...,C_{96}&lt;/math&gt; in such way that the following to conditions are both fulfilled<br /> <br /> &lt;math&gt;(i)&lt;/math&gt; the distances &lt;math&gt;C_1C_2,...C_{95}C_{96}, C_{96}C_{1}&lt;/math&gt; are all equal to &lt;math&gt;\sqrt {13}&lt;/math&gt;<br /> <br /> &lt;math&gt;(ii)&lt;/math&gt; the closed broken line &lt;math&gt;C_1C_2...C_{96}C_1&lt;/math&gt; has a centre of symmetry? ''<br /> * Problem 4: '''8'''<br /> *: '' Denote by &lt;math&gt;S&lt;/math&gt; the set of all positive integers. Find all functions &lt;math&gt;f: S \rightarrow S&lt;/math&gt; such that<br /> <br /> &lt;math&gt;f \bigg(f^2(m) + 2f^2(n)\bigg) = m^2 + 2 n^2&lt;/math&gt; for all &lt;math&gt;m,n \in S&lt;/math&gt;. '<br /> <br /> === [[HMMT]] ===<br /> '''February Contest'''<br /> * Individual Round, Problem 1-5: '''5'''<br /> * Individual Round, Problem 6-10: '''6'''<br /> * Team Round: '''7.5'''<br /> * HMIC: '''8'''<br /> <br /> === [[Ibero American Olympiad]] ===<br /> <br /> * Problem 1/4: '''5.5'''<br /> * Problem 2/5: '''6.5'''<br /> * Problem 3/6: '''7.5'''<br /> <br /> === [[IMO]] ===<br /> <br /> * Problem 1/4: '''6.5'''<br /> *: ''Find all functions &lt;math&gt;f: (0, \infty) \mapsto (0, \infty)&lt;/math&gt; (so that &lt;math&gt;f&lt;/math&gt; is a function from the positive real numbers) such that<br /> &lt;center&gt;&lt;math&gt;\frac {\left( f(w) \right)^2 + \left( f(x) \right)^2}{f(y^2) + f(z^2) } = \frac {w^2 + x^2}{y^2 + z^2}&lt;/math&gt;&lt;/center&gt; for all positive real numbers &lt;math&gt;w,x,y,z,&lt;/math&gt; satisfying &lt;math&gt;wx = yz.&lt;/math&gt; ([[2008 IMO Problems/Problem 4|Solution]])<br /> ''<br /> * Problem 2/5: '''7.5-8'''<br /> *: ''Let &lt;math&gt;P(x)&lt;/math&gt; be a polynomial of degree &lt;math&gt;n&gt;1&lt;/math&gt; with integer coefficients, and let &lt;math&gt;k&lt;/math&gt; be a positive integer. Consider the polynomial &lt;math&gt;Q(x) = P( P ( \ldots P(P(x)) \ldots ))&lt;/math&gt;, where &lt;math&gt;P&lt;/math&gt; occurs &lt;math&gt;k&lt;/math&gt; times. Prove that there are at most &lt;math&gt;n&lt;/math&gt; integers &lt;math&gt;t&lt;/math&gt; such that &lt;math&gt;Q(t)=t&lt;/math&gt;.'' ([[2006 IMO Problems/Problem 5|Solution]])<br /> * Problem 3/6: '''9.5'''<br /> *: ''Assign to each side &lt;math&gt;b&lt;/math&gt; of a convex polygon &lt;math&gt;P&lt;/math&gt; the maximum area of a triangle that has &lt;math&gt;b&lt;/math&gt; as a side and is contained in &lt;math&gt;P&lt;/math&gt;. Show that the sum of the areas assigned to the sides of &lt;math&gt;P&lt;/math&gt; is at least twice the area of &lt;math&gt;P&lt;/math&gt;.'' (&lt;url&gt;viewtopic.php?p=572824#572824 Solution&lt;/url&gt;)<br /> <br /> === [[IMO Shortlist]] ===<br /> <br /> * Problem 1-2: '''5.5-7'''<br /> * Problem 3-4: '''7-8'''<br /> * Problem 5+: '''8-10'''<br /> <br /> === [[JBMO]] ===<br /> <br /> * Problem 1: '''4'''<br /> *: ''Find all real numbers &lt;math&gt;a,b,c,d&lt;/math&gt; such that <br /> &lt;cmath&gt; \left\{\begin{array}{cc}a+b+c+d = 20,\\ ab+ac+ad+bc+bd+cd = 150.\end{array}\right. &lt;/cmath&gt;''<br /> * Problem 2: '''5'''<br /> *: ''Let &lt;math&gt;ABCD&lt;/math&gt; be a convex quadrilateral with &lt;math&gt;\angle DAC=\angle BDC=36^\circ&lt;/math&gt;, &lt;math&gt;\angle CBD=18^\circ&lt;/math&gt; and &lt;math&gt;\angle BAC=72^\circ&lt;/math&gt;. The diagonals intersect at point &lt;math&gt;P&lt;/math&gt;. Determine the measure of &lt;math&gt;\angle APD&lt;/math&gt;.''<br /> * Problem 3: '''5'''<br /> *: ''Find all prime numbers &lt;math&gt;p,q,r&lt;/math&gt;, such that &lt;math&gt;\frac pq-\frac4{r+1}=1&lt;/math&gt;.''<br /> * Problem 4: '''6'''<br /> *: ''A &lt;math&gt;4\times4&lt;/math&gt; table is divided into &lt;math&gt;16&lt;/math&gt; white unit square cells. Two cells are called neighbors if they share a common side. A '''move''' consists in choosing a cell and changing the colors of neighbors from white to black or from black to white. After exactly &lt;math&gt;n&lt;/math&gt; moves all the &lt;math&gt;16&lt;/math&gt; cells were black. Find all possible values of &lt;math&gt;n&lt;/math&gt;.''<br /> <br /> === [[Putnam]] ===<br /> <br /> * Problem A/B,1-2: '''7'''<br /> *: ''Find the least possible area of a convex set in the plane that intersects both branches of the hyperbola &lt;math&gt;xy = 1&lt;/math&gt; and both branches of the hyperbola &lt;math&gt;xy = - 1.&lt;/math&gt; (A set &lt;math&gt;S&lt;/math&gt; in the plane is called ''convex'' if for any two points in &lt;math&gt;S&lt;/math&gt; the line segment connecting them is contained in &lt;math&gt;S.&lt;/math&gt;)'' (&lt;url&gt;viewtopic.php?p=978383#p978383 Solution&lt;/url&gt;)<br /> * Problem A/B,3-4: '''8'''<br /> *: ''Let &lt;math&gt;H&lt;/math&gt; be an &lt;math&gt;n\times n&lt;/math&gt; matrix all of whose entries are &lt;math&gt;\pm1&lt;/math&gt; and whose rows are mutually orthogonal. Suppose &lt;math&gt;H&lt;/math&gt; has an &lt;math&gt;a\times b&lt;/math&gt; submatrix whose entries are all &lt;math&gt;1.&lt;/math&gt; Show that &lt;math&gt;ab\le n&lt;/math&gt;.'' (&lt;url&gt;viewtopic.php?p=383280#383280 Solution&lt;/url&gt;)<br /> * Problem A/B,5-6: '''9'''<br /> *: ''For any &lt;math&gt;a &gt; 0&lt;/math&gt;, define the set &lt;math&gt;S(a) = \{[an]|n = 1,2,3,...\}&lt;/math&gt;. Show that there are no three positive reals &lt;math&gt;a,b,c&lt;/math&gt; such that &lt;math&gt;S(a)\cap S(b) = S(b)\cap S(c) = S(c)\cap S(a) = \emptyset,S(a)\cup S(b)\cup S(c) = \{1,2,3,...\}&lt;/math&gt;.'' (&lt;url&gt;viewtopic.php?t=127810 Solution&lt;/url&gt;)<br /> <br /> === [[USAJMO]] ===<br /> * Problem 1/4: '''6'''<br /> * Problem 2/5: '''6.5'''<br /> * Problem 3/6: '''7'''<br /> <br /> === [[USAMO]] ===<br /> * Problem 1/4: '''7'''<br /> *: ''Let &lt;math&gt;\mathcal{P}&lt;/math&gt; be a convex polygon with &lt;math&gt;n&lt;/math&gt; sides, &lt;math&gt;n\ge3&lt;/math&gt;. Any set of &lt;math&gt;n - 3&lt;/math&gt; diagonals of &lt;math&gt;\mathcal{P}&lt;/math&gt; that do not intersect in the interior of the polygon determine a ''triangulation'' of &lt;math&gt;\mathcal{P}&lt;/math&gt; into &lt;math&gt;n - 2&lt;/math&gt; triangles. If &lt;math&gt;\mathcal{P}&lt;/math&gt; is regular and there is a triangulation of &lt;math&gt;\mathcal{P}&lt;/math&gt; consisting of only isosceles triangles, find all the possible values of &lt;math&gt;n&lt;/math&gt;.'' ([[2008 USAMO Problems/Problem 4|Solution]]) <br /> * Problem 2/5: '''8'''<br /> *: ''Three nonnegative real numbers &lt;math&gt;r_1&lt;/math&gt;, &lt;math&gt;r_2&lt;/math&gt;, &lt;math&gt;r_3&lt;/math&gt; are written on a blackboard. These numbers have the property that there exist integers &lt;math&gt;a_1&lt;/math&gt;, &lt;math&gt;a_2&lt;/math&gt;, &lt;math&gt;a_3&lt;/math&gt;, not all zero, satisfying &lt;math&gt;a_1r_1 + a_2r_2 + a_3r_3 = 0&lt;/math&gt;. We are permitted to perform the following operation: find two numbers &lt;math&gt;x&lt;/math&gt;, &lt;math&gt;y&lt;/math&gt; on the blackboard with &lt;math&gt;x \le y&lt;/math&gt;, then erase &lt;math&gt;y&lt;/math&gt; and write &lt;math&gt;y - x&lt;/math&gt; in its place. Prove that after a finite number of such operations, we can end up with at least one &lt;math&gt;0&lt;/math&gt; on the blackboard.'' ([[2008 USAMO Problems/Problem 5|Solution]])<br /> * Problem 3/6: '''9'''<br /> *: ''Prove that any monic polynomial (a polynomial with leading coefficient 1) of degree &lt;math&gt;n &lt;/math&gt; with real coefficients is the average of two monic polynomials of degree &lt;math&gt;n &lt;/math&gt; with &lt;math&gt;n &lt;/math&gt; real roots.'' ([[2002 USAMO Problems/Problem 3|Solution]])<br /> <br /> === [[USAMTS]] ===<br /> USAMTS generally has a different feel to it than olympiads, and is mainly for proofwriting practice instead of olympiad practice depending on how one takes the test. USAMTS allows an entire month to solve problems, with internet resources and books being allowed. However, the ultimate gap is that it permits computer programs to be used, and that Problem 1 is not a proof problem. However, it can still be roughly put to this rating scale:<br /> * Problem 1-2: '''3-4'''<br /> *: ''Find three isosceles triangles, no two of which are congruent, with integer sides, such that each triangle’s area is numerically equal to 6 times its perimeter.'' ([http://usamts.org/Solutions/Solution2_3_16.pdf Solution])<br /> * Problem 3-5: '''5-6'''<br /> *: ''Call a positive real number groovy if it can be written in the form &lt;math&gt;\sqrt{n} + \sqrt{n + 1}&lt;/math&gt; for some positive integer &lt;math&gt;n&lt;/math&gt;. Show that if &lt;math&gt;x&lt;/math&gt; is groovy, then for any positive integer &lt;math&gt;r&lt;/math&gt;, the number &lt;math&gt;x^r&lt;/math&gt; is groovy as well.'' ([http://usamts.org/Solutions/Solutions_20_1.pdf Solution])<br /> <br /> === [[USA TST]] ===<br /> <br /> (seems to vary more than other contests; estimates based on 08 and 09)<br /> <br /> * Problem 1/4/7: '''7'''<br /> * Problem 2/5/8: '''8'''<br /> * Problem 3/6/9: '''9.5'''<br /> <br /> === [[China TST]] ===<br /> <br /> * Problem 1/4: '''7''' <br /> *: ''Given an integer &lt;math&gt;m,&lt;/math&gt; prove that there exist odd integers &lt;math&gt;a,b&lt;/math&gt; and a positive integer &lt;math&gt;k&lt;/math&gt; such that &lt;cmath&gt;2m=a^{19}+b^{99}+k*2^{1000}.&lt;/cmath&gt;''<br /> * Problem 2/5: '''8.5''' <br /> *: ''Given a positive integer &lt;math&gt;n&gt;1&lt;/math&gt; and real numbers &lt;math&gt;a_1 &lt; a_2 &lt; \ldots &lt; a_n,&lt;/math&gt; such that &lt;math&gt;\dfrac{1}{a_1} + \dfrac{1}{a_2} + \ldots + \dfrac{1}{a_n} \le 1,&lt;/math&gt; prove that for any positive real number &lt;math&gt;x,&lt;/math&gt; &lt;cmath&gt;\left(\dfrac{1}{a_1^2+x} + \dfrac{1}{a_2^2+x} + \ldots + \dfrac{1}{a_n^2+x}\right)^2 \ge \dfrac{1}{2a_1(a_1-1)+2x}.&lt;/cmath&gt;''<br /> * Problem 3/6: '''10'''<br /> *: ''Let &lt;math&gt;n&gt;1&lt;/math&gt; be an integer and let &lt;math&gt;a_0,a_1,\ldots,a_n&lt;/math&gt; be non-negative real numbers. Define &lt;math&gt;S_k=\sum_{i=0}^k \binom{k}{i}a_i&lt;/math&gt; for &lt;math&gt;k=0,1,\ldots,n&lt;/math&gt;. Prove that&lt;cmath&gt;\frac{1}{n} \sum_{k=0}^{n-1} S_k^2-\frac{1}{n^2}\left(\sum_{k=0}^{n} S_k\right)^2\le \frac{4}{45} (S_n-S_0)^2.&lt;/cmath&gt;''<br /> <br /> == Scale ==<br /> {{ref|1}} All levels estimated and refer to ''averages''. The following is a rough standard based on the USA tier system AMC 8 – AMC 10 – AMC 12 – AIME – USAMO/USAJMO, representing Middle School – Junior High – High School – Challenging High School – Olympiad levels. Other contests can be interpolated against this. <br /> # Problems strictly for beginners, on the easiest elementary school or middle school levels. Examples would be MOEMS, easy Mathcounts questions, #1-20 on AMC 8s, #1-5 AMC 10s, and others that involve standard techniques introduced up to the middle school level<br /> # For motivated beginners, harder questions from the previous categories (#21-25 on AMC 8, Challenging Mathcounts questions, #5-20 on AMC 10, #5-10 on AMC 12, the easiest AIME questions, etc).<br /> # For those not too familiar with standard techniques, MathCounts National, #21-25 on AMC 10, #11-20ish on AMC 12, #1-5 on AIMEs, etc.<br /> # Intermediate-leveled problem solvers, the most difficult questions on AMC 12s (#21-25s), more difficult AIME-styled questions #6-10<br /> # Difficult AIME problems (#10-13), others, simple proof-based problems (JBMO etc)<br /> # High-leveled AIME-styled questions, not requiring proofs (#12-15). Introductory-leveled Olympiad-level questions (#1,4s).<br /> # Intermediate-leveled Olympiad-level questions, #1,4s that require more technical knowledge than new students to Olympiad-type questions have, easier #2,5s, etc. <br /> # High-level difficult Olympiad-level questions, eg #2,5s on difficult Olympiad contest and easier #3,6s, etc. <br /> # Difficult Olympiad-level questions, eg #3,6s on difficult Olympiad contests. <br /> # Problems occasionally even unsuitable for very hard competitions (like the IMO) due to being exceedingly tedious/long/difficult (eg very few students are capable of solving, even on a worldwide basis), or involving techniques beyond high school level mathematics.<br /> <br /> == See also ==<br /> [[Category:Mathematics competitions]]</div> Amackenzie1 https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_25&diff=81577 2016 AMC 8 Problems/Problem 25 2016-11-24T18:47:44Z <p>Amackenzie1: /* Solution 3 */</p> <hr /> <div>A semicircle is inscribed in an isosceles triangle with base &lt;math&gt;16&lt;/math&gt; and height &lt;math&gt;15&lt;/math&gt; so that the diameter of the semicircle is contained in the base of the triangle as shown. What is the radius of the semicircle?<br /> <br /> &lt;asy&gt;draw((0,0)--(8,15)--(16,0)--(0,0));<br /> draw(arc((8,0),7.0588,0,180));&lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }4 \sqrt{3}\qquad\textbf{(B) } \dfrac{120}{17}\qquad\textbf{(C) }10\qquad\textbf{(D) }\dfrac{17\sqrt{2}}{2}\qquad \textbf{(E)} \dfrac{17\sqrt{3}}{2}&lt;/math&gt;<br /> <br /> <br /> ==Solution 1==<br /> Draw the altitude from the top of the triangle to its base, dividing the isosceles triangle into two right triangles with height &lt;math&gt;15&lt;/math&gt; and base &lt;math&gt;\frac{16}{2} = 8&lt;/math&gt;. The Pythagorean triple &lt;math&gt;8&lt;/math&gt;-&lt;math&gt;15&lt;/math&gt;-&lt;math&gt;17&lt;/math&gt; tells us that these triangles have hypotenuses of &lt;math&gt;17&lt;/math&gt;. <br /> <br /> Now draw an altitude of one of the smaller right triangles, starting from the foot of the first altitude we drew (which is also the center of the circle that contains the semicircle) and going to the hypotenuse of the right triangle. This segment is both an altitude of the right triangle as well as the radius of the semicircle (this is because tangent lines to circles, such as the hypotenuse touching the semicircle, are always perpendicular to the radii of the circles drawn to the point of tangency). Let this segment's length be &lt;math&gt;r&lt;/math&gt;.<br /> <br /> The area of the entire isosceles triangle is &lt;math&gt;\frac{(16)(15)}{2} = 120&lt;/math&gt;, so the area of each of the two congruent right triangles it gets split into is &lt;math&gt;\frac{120}{2} = 60&lt;/math&gt;. We can also find the area of one of the two congruent right triangles by using its hypotenuse as its base and the radius of the semicircle, the altitude we drew, as its height. Then the area of the triangle is &lt;math&gt;\frac{17r}{2}&lt;/math&gt;. Thus we can write the equation &lt;math&gt;\frac{17r}{2} = 60&lt;/math&gt;, so &lt;math&gt;17r = 120&lt;/math&gt;, so &lt;math&gt;r = \boxed{\textbf{(B) }\frac{120}{17}}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> &lt;math&gt;\frac{8}{17} \cdot 15&lt;/math&gt; =<br /> &lt;math&gt; \boxed{\textbf{(B) }\frac{120}{17}}&lt;/math&gt;. (Look at Solution 4 for more details)<br /> <br /> ==Solution 3==<br /> First, we draw a line perpendicular to the base of the triangle and cut it in half. The base of the resulting right triangle would be 8, and the height would be 15. Using the Pythagorean theorem, we can find the length of the hypotenuse, which would be 17. Using the two legs of the right angle, we can find the area of the right triangle, &lt;math&gt;60&lt;/math&gt;. &lt;math&gt;\frac{60}{17}&lt;/math&gt; times &lt;math&gt;2&lt;/math&gt; get you the radius, which is the height of the right triangle when using the hypotenuse as the base. The answer is &lt;math&gt; \boxed{\textbf{(B) }\frac{120}{17}}&lt;/math&gt;.<br /> <br /> ==Solution 4: Similar Triangles==<br /> &lt;asy&gt; pair A, B, C, D, E; A=(0,0); B=(16,0); C=(8,15); D=B/2; E=(64/17*8/17, 64/17*15/17); draw(A--B--C--cycle); draw(C--D); draw(D--E); draw(arc(D,120/17,0,180)); draw(rightanglemark(B,D,C,25)); draw(rightanglemark(A,E,D,25)); label(&quot;$A$&quot;,A,SW); label(&quot;$B$&quot;,B,SE); label(&quot;$C$&quot;,C,N); label(&quot;$D$&quot;,D,S); label(&quot;$E$&quot;,E,NW);&lt;/asy&gt;<br /> Let's call the triangle &lt;math&gt;\triangle ABC,&lt;/math&gt; where &lt;math&gt;AB=16&lt;/math&gt; and &lt;math&gt; AC=BC.&lt;/math&gt; Let's say that &lt;math&gt;D&lt;/math&gt; is the midpoint of &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; is the point where &lt;math&gt;AC&lt;/math&gt; is tangent to the semicircle. We could also use &lt;math&gt;BC&lt;/math&gt; instead of &lt;math&gt;AC&lt;/math&gt; because of symmetry.<br /> <br /> We notice that &lt;math&gt;\triangle ACD \cong \triangle BCD,&lt;/math&gt; and are both 8-15-17 right triangles. We also know that we create a right angle with the intersection of the radius and a tangent line of a circle (or part of a circle). So, by &lt;math&gt;AA&lt;/math&gt; similarity, &lt;math&gt;\triangle AED \sim \triangle ADC,&lt;/math&gt; with &lt;math&gt;\angle EAD \cong \angle DAC&lt;/math&gt; and &lt;math&gt; \angle CDA \cong \angle DEA.&lt;/math&gt; This similarity means that we can create a proportion: &lt;math&gt;\frac{AD}{AB}=\frac{DE}{CD}.&lt;/math&gt; We plug in &lt;math&gt;AD=\frac{AB}{2}=8, AC=17,&lt;/math&gt; and &lt;math&gt;CD=15.&lt;/math&gt; After we multiply both sides by &lt;math&gt;15,&lt;/math&gt; we get &lt;math&gt;DE=\frac{8}{17} \cdot 15= \boxed{\textbf{(B) }\frac{120}{17}}.&lt;/math&gt;<br /> <br /> (By the way, we could also use &lt;math&gt;\triangle DEC \sim \triangle ADC.&lt;/math&gt;)<br /> <br /> {{AMC8 box|year=2016|num-b=24|after=Last Problem}}<br /> {{MAA Notice}}</div> Amackenzie1 https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_25&diff=81576 2016 AMC 8 Problems/Problem 25 2016-11-24T18:47:17Z <p>Amackenzie1: LaTeX</p> <hr /> <div>A semicircle is inscribed in an isosceles triangle with base &lt;math&gt;16&lt;/math&gt; and height &lt;math&gt;15&lt;/math&gt; so that the diameter of the semicircle is contained in the base of the triangle as shown. What is the radius of the semicircle?<br /> <br /> &lt;asy&gt;draw((0,0)--(8,15)--(16,0)--(0,0));<br /> draw(arc((8,0),7.0588,0,180));&lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }4 \sqrt{3}\qquad\textbf{(B) } \dfrac{120}{17}\qquad\textbf{(C) }10\qquad\textbf{(D) }\dfrac{17\sqrt{2}}{2}\qquad \textbf{(E)} \dfrac{17\sqrt{3}}{2}&lt;/math&gt;<br /> <br /> <br /> ==Solution 1==<br /> Draw the altitude from the top of the triangle to its base, dividing the isosceles triangle into two right triangles with height &lt;math&gt;15&lt;/math&gt; and base &lt;math&gt;\frac{16}{2} = 8&lt;/math&gt;. The Pythagorean triple &lt;math&gt;8&lt;/math&gt;-&lt;math&gt;15&lt;/math&gt;-&lt;math&gt;17&lt;/math&gt; tells us that these triangles have hypotenuses of &lt;math&gt;17&lt;/math&gt;. <br /> <br /> Now draw an altitude of one of the smaller right triangles, starting from the foot of the first altitude we drew (which is also the center of the circle that contains the semicircle) and going to the hypotenuse of the right triangle. This segment is both an altitude of the right triangle as well as the radius of the semicircle (this is because tangent lines to circles, such as the hypotenuse touching the semicircle, are always perpendicular to the radii of the circles drawn to the point of tangency). Let this segment's length be &lt;math&gt;r&lt;/math&gt;.<br /> <br /> The area of the entire isosceles triangle is &lt;math&gt;\frac{(16)(15)}{2} = 120&lt;/math&gt;, so the area of each of the two congruent right triangles it gets split into is &lt;math&gt;\frac{120}{2} = 60&lt;/math&gt;. We can also find the area of one of the two congruent right triangles by using its hypotenuse as its base and the radius of the semicircle, the altitude we drew, as its height. Then the area of the triangle is &lt;math&gt;\frac{17r}{2}&lt;/math&gt;. Thus we can write the equation &lt;math&gt;\frac{17r}{2} = 60&lt;/math&gt;, so &lt;math&gt;17r = 120&lt;/math&gt;, so &lt;math&gt;r = \boxed{\textbf{(B) }\frac{120}{17}}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> &lt;math&gt;\frac{8}{17} \cdot 15&lt;/math&gt; =<br /> &lt;math&gt; \boxed{\textbf{(B) }\frac{120}{17}}&lt;/math&gt;. (Look at Solution 4 for more details)<br /> <br /> ==Solution 3==<br /> First, we draw a line perpendicular to the base of the triangle and cut it in half. The base of the resulting right triangle would be 8, and the height would be 15. Using the Pythagorean theorem, we can find the length of the hypotenuse, which would be 17. Using the two legs of the right angle, we can find the area of the right triangle, &lt;math&gt;60&lt;/math&gt;. &lt;math&gt;\frac{60}{17}&lt;/math&gt; times 2 get you the radius, which is the height of the right triangle when using the hypotenuse as the base. The answer is &lt;math&gt; \boxed{\textbf{(B) }\frac{120}{17}}&lt;/math&gt;.<br /> <br /> ==Solution 4: Similar Triangles==<br /> &lt;asy&gt; pair A, B, C, D, E; A=(0,0); B=(16,0); C=(8,15); D=B/2; E=(64/17*8/17, 64/17*15/17); draw(A--B--C--cycle); draw(C--D); draw(D--E); draw(arc(D,120/17,0,180)); draw(rightanglemark(B,D,C,25)); draw(rightanglemark(A,E,D,25)); label(&quot;$A$&quot;,A,SW); label(&quot;$B$&quot;,B,SE); label(&quot;$C$&quot;,C,N); label(&quot;$D$&quot;,D,S); label(&quot;$E$&quot;,E,NW);&lt;/asy&gt;<br /> Let's call the triangle &lt;math&gt;\triangle ABC,&lt;/math&gt; where &lt;math&gt;AB=16&lt;/math&gt; and &lt;math&gt; AC=BC.&lt;/math&gt; Let's say that &lt;math&gt;D&lt;/math&gt; is the midpoint of &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; is the point where &lt;math&gt;AC&lt;/math&gt; is tangent to the semicircle. We could also use &lt;math&gt;BC&lt;/math&gt; instead of &lt;math&gt;AC&lt;/math&gt; because of symmetry.<br /> <br /> We notice that &lt;math&gt;\triangle ACD \cong \triangle BCD,&lt;/math&gt; and are both 8-15-17 right triangles. We also know that we create a right angle with the intersection of the radius and a tangent line of a circle (or part of a circle). So, by &lt;math&gt;AA&lt;/math&gt; similarity, &lt;math&gt;\triangle AED \sim \triangle ADC,&lt;/math&gt; with &lt;math&gt;\angle EAD \cong \angle DAC&lt;/math&gt; and &lt;math&gt; \angle CDA \cong \angle DEA.&lt;/math&gt; This similarity means that we can create a proportion: &lt;math&gt;\frac{AD}{AB}=\frac{DE}{CD}.&lt;/math&gt; We plug in &lt;math&gt;AD=\frac{AB}{2}=8, AC=17,&lt;/math&gt; and &lt;math&gt;CD=15.&lt;/math&gt; After we multiply both sides by &lt;math&gt;15,&lt;/math&gt; we get &lt;math&gt;DE=\frac{8}{17} \cdot 15= \boxed{\textbf{(B) }\frac{120}{17}}.&lt;/math&gt;<br /> <br /> (By the way, we could also use &lt;math&gt;\triangle DEC \sim \triangle ADC.&lt;/math&gt;)<br /> <br /> {{AMC8 box|year=2016|num-b=24|after=Last Problem}}<br /> {{MAA Notice}}</div> Amackenzie1