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• USA(J)MO
  67 bytes (8 words) - 08:30, 13 July 2010
• J. E. Littlewood
  little-wood
  11 bytes (1 word) - 19:53, 1 July 2020

Page text matches

• Resources for mathematics competitions
  * Challenge Math Online: http://www.noetic-learning.com/gifted ...ration Open]Elementary/middle school contests: https://www.beestar.org/math-competition/
  24 KB (3,269 words) - 22:58, 18 March 2024
• Math books
  ...ing Started with Competition Math], a textbook meant for true beginners (on-target middle school students, or advanced elementary school students). It i ...+maths+kawasaki&qid=1581288606&sprefix=after+school+maths+%2Caps%2C268&sr=8-2 100 Challenging Maths Problems]
  24 KB (3,177 words) - 12:53, 20 February 2024
• Chemistry competitions
  The U.S. National Chemistry Olympiad national exam (USNCO) is a 3-part, 4 hour and 45 minute exam administered in mid or late April by ACS Loc ...www.amazon.com/gp/product/0618221565/ref=pd_lpo_k2_dp_k2a_1_img/102-5655201-2084940?%5Fencoding=UTF8 ''Chemistry''] by Steven S. Zumdahl, Susan A. Zumda
  2 KB (258 words) - 19:31, 8 March 2023
• Inequality
  ...b</math> if <math>a</math> is greater than <math>b</math>, that is, <math>a-b</math> is positive. ...b</math> if <math>a</math> is smaller than <math>b</math>, that is, <math>a-b</math> is negative.
  12 KB (1,798 words) - 16:20, 14 March 2023
• 2015 IMO Problems
  <math>\mathcal{S}</math> is <i>centre-free</i> if for any three points <math>A</math>, <math>B</math>, <math>C</ma <ol style="list-style-type: lower-latin;">
  4 KB (692 words) - 22:33, 15 February 2021
• Proof writing
  ...wider population. Students who spend time studying maths can develop proof-writing skills over time. ...proof|two-column format]], as favored by many geometry teachers. In higher-level mathematics (taken as meaning an advanced undergraduate level of mathe
  3 KB (502 words) - 18:16, 18 January 2016
• Simon's Favorite Factoring Trick
  ...bles in a linear term on the other side. An example would be: <cmath>xy+66x-88y=23333</cmath>where <math>23333</math> is the constant term, <math>xy</ma ...e previous example, <math>xy+66x-88y=23333</math> is the same as: <cmath>(x-88)(y+66)=(23333)+(-88)(66)</cmath>
  7 KB (1,107 words) - 07:35, 26 March 2024
• Pigeonhole Principle
  ...th> where <math>i>j</math>, we may multiply <math>2^jb</math> by <math>2^{i-j}</math> to get <math>2^ib</math>, as desired. <math>\square</math> ...math> and <math>n</math> possible remainders (namely, <math>0, 1, \ldots, n-1</math>) modulo <math>n</math>. Then by the pigeonhole principle, there exi
  11 KB (1,985 words) - 21:03, 5 August 2023
• Cauchy-Schwarz Inequality
  ...ormulations in abstract algebra, calculus, and contest mathematics. In high-school competitions, its applications are limited to elementary and linear a ...intermediate and olympiad competitions. It is particularly crucial in proof-based contests.
  13 KB (2,048 words) - 15:28, 22 February 2024
• Fermat's Little Theorem
  ...ath> is not [[divisibility|divisible]] by <math>{p}</math>, then <math>a^{p-1}\equiv 1 \pmod {p}</math>. ...> denotes [[Euler's totient function]]. In particular, <math>\varphi(p) = p-1</math> for prime numbers <math>p</math>. In turn, this is a special case o
  16 KB (2,675 words) - 10:57, 7 March 2024
• Brute forcing
  1 KB (190 words) - 13:22, 5 May 2023
• Real number
  Let <math>\alpha\subset\mathbb{Q}</math> be non-empty
  3 KB (496 words) - 23:22, 5 January 2022
• Pi
  ...imes(0,1)}\frac {x^ky^\ell}{1-xy}\,dx\,dy</math>. Expanding <math>\frac 1{1-xy}=\sum_{j=0}^\infty x^jy^j</math> and integrating term by term, we get <ma <math>I(k,\ell)=\frac 1{\ell-k}\left[\frac 1{k+1}+\frac 1{k+2}+\dots+\frac 1\ell\right]</math>
  8 KB (1,469 words) - 21:11, 16 September 2022
• Chain Rule
  Here are some more precise statements for the single-variable and multi-variable cases. ===Multi-dimensional Chain Rule===
  12 KB (2,377 words) - 11:48, 22 July 2009
• Muirhead's Inequality
  ...an inequality ''can'' be proved with AM-GM before demonstrating the full AM-GM proof. ...</math> from the [[AM-GM Inequality#Weighted AM-GM Inequality|(weighted) AM-GM inequality]].
  8 KB (1,346 words) - 12:53, 8 October 2023
• Vector
  ...ected line segment. In many situations, a vector is best considered as an n-tuple of numbers (often real or complex). Most generally, but also most abst ...h (or magnitude) and the angle it makes with some fixed line (usually the x-axis) or by describing it as an arrow beginning at the origin and ending at
  7 KB (1,265 words) - 13:22, 14 July 2021
• Area
  ...math>c</math> are the lengths of the sides and <math>s</math> is the [[semi-perimeter]] <math>s=\frac{a+b+c}{2}</math>. ...>, where <math>r</math> is the radius of the [[incircle]] and s is the semi-perimeter.
  6 KB (1,181 words) - 22:37, 22 January 2023
• Improper fractional base
  787 bytes (118 words) - 19:20, 23 October 2010
• Newton's Sums
  <center><math> P(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0</math></center> <cmath>a_nP_1 + a_{n-1} = 0</cmath>
  4 KB (690 words) - 13:11, 20 February 2024
• Fundamental Theorem of Algebra
  <cmath> \lvert P(z) \rvert \ge \lvert z^n \rvert - (R-1) \lvert z^{n-1} \rvert = \lvert z^{n-1} \rvert \cdot \bigl[ \lvert z \rvert - (R-1) \bigr]
  5 KB (832 words) - 14:22, 11 January 2024
• ...bda_a = \lambda_b = 1/2</math>, this is the elementary form of the [[Cauchy-Schwarz Inequality]]. ...he left-hand side of the desired inequality but may contribute to the right-hand side.
  4 KB (774 words) - 12:12, 29 October 2016
• Mock AMC
  Listed below are the [higher-quality] Mock AMCs that have been hosted over AoPS in the past. Feel free to {| class="wikitable" style="text-align:center;width:100%"
  51 KB (6,175 words) - 20:58, 6 December 2023
• 2006 AIME I Problems
  Let set <math> \mathcal{A} </math> be a 90-element subset of <math> \{1,2,3,\ldots,100\}, </math> and let <math> S </ma A collection of 8 cubes consists of one cube with edge-length <math> k </math> for each integer <math> k, 1 \le k \le 8. </math> A
  7 KB (1,173 words) - 03:31, 4 January 2023
• 2006 AIME I Problems/Problem 15
  Given that a sequence satisfies <math> x_0=0 </math> and <math> |x_k|=|x_{k-1}+3| </math> for all integers <math> k\ge 1, </math> find the minimum possi {{AIME box|year=2006|n=I|num-b=14|after=Last Question}}
  6 KB (910 words) - 19:31, 24 October 2023
• 2006 AIME I Problems/Problem 13
  ...For each positive integer <math> n, </math> let <math> S_n=\sum_{k=1}^{2^{n-1}}g(2k). </math> Find the greatest integer <math> n </math> less than 1000 ...math> but not by <math>2^n, \ldots,</math> and <math>2^{n-1}-2^{n-2} = 2^{n-2}</math> elements of <math>S</math> that are divisible by <math>2^1</math>
  10 KB (1,702 words) - 00:45, 16 November 2023
• LaTeX:Symbols
  ....)) maintains a [http://docs.mathjax.org/en/latest/tex.html#supported-latex-commands list of supported commands]. [http://mirrors.ctan.org/info/symbols/comprehensive/symbols-a4.pdf The Comprehensive LaTeX Symbol List].
  16 KB (2,324 words) - 16:50, 19 February 2024
• 2006 AIME I Problems/Problem 10
  ...will not work in general). Denote the intersection of the line and the [[x-axis]] as <math>(x, 0)</math>. ...ezoid]] and a [[rectangle]], and the areas are <math>\frac{1}{2}((1-x) + (2-x))(3)</math> and <math>2 \cdot 1 = 2</math>, totaling <math>\frac{13}{2} -
  4 KB (731 words) - 17:59, 4 January 2022
• 2006 AIME I Problems/Problem 8
  ...ath>44.5^2</math> is also less than <math>2006</math>, so we have numbers 1-44, times two because 0.5 can be added to each of time, plus 1, because 0.5 ...have <math>[GIHJ]=0,</math> implying the minimum possible positive-integer-valued area is <math>1.</math>
  5 KB (730 words) - 15:05, 15 January 2024
• 2021 JMC 10
  <math>\textbf{(A) }\pi-e \qquad\textbf{(B) }2\pi-2e\qquad\textbf{(C) }2e\qquad\textbf{(D) }2\pi \qquad\textbf{(E) }2\pi +e</m ...h>74</math> and <math>83</math> are pretentious. How many pretentious three-digit numbers are odd?
  12 KB (1,784 words) - 16:49, 1 April 2021
• 2006 AMC 12B Problems
  ...1.99</math>, and <math>\textdollar0.99</math>. Mary will pay with a twenty-dollar bill. Which of the following is closest to the percentage of the <ma How many even three-digit integers have the property that their digits, read left to right, are
  13 KB (2,058 words) - 12:36, 4 July 2023
• 2001 AMC 12 Problems
  two-digit number such that <math>N = P(N)+S(N)</math>. What is the units digit o A telephone number has the form <math>\text{ABC-DEF-GHIJ}</math>, where each letter represents
  13 KB (1,957 words) - 12:53, 24 January 2024
• 2004 AMC 12B Problems
  In the expression <math>c\cdot a^b-d</math>, the values of <math>a</math>, <math>b</math>, <math>c</math>, and Minneapolis-St. Paul International Airport is 8 miles southwest of downtown St. Paul and
  13 KB (2,049 words) - 13:03, 19 February 2020
• 2006 AMC 12B Problems/Problem 14
  {{AMC12 box|year=2006|ab=B|num-b=13|num-a=15}}
  1 KB (227 words) - 17:21, 8 December 2013
• 2006 AMC 12B Problems/Problem 25
  A sequence <math>a_1,a_2,\dots</math> of non-negative integers is defined by the rule <math>a_{n+2}=|a_{n+1}-a_n|</math> is a 1-1 correspondence between the odd and even numbers less than and relatively p
  5 KB (924 words) - 12:02, 15 June 2022
• 2005 AMC 12B Problems/Problem 24
  ...ive x- axis, the answer is <math>\dfrac{\tan(BOJ) + \tan(BOJ+60) + \tan(BOJ-60)}{2}</math>. {{AMC12 box|year=2005|ab=B|num-b=23|num-a=25}}
  4 KB (761 words) - 09:10, 1 August 2023
• Root (polynomial)
  ...<math>a</math> in the [[domain]] of the function such that <math>f(a) = (x-a) = 0</math>. Let <cmath>P(x) = c_nx^n + c_{n-1}x^{n-1} + \dots + c_1x + c_0 = \sum_{j=0}^{n} c_jx^j</cmath> with all <math>c_j
  8 KB (1,427 words) - 21:37, 13 March 2022
• Prime ideal
  967 bytes (176 words) - 18:08, 7 April 2012
• 2005 AIME II Problems/Problem 12
  ...0)); pair A=(0,9), B=(9,9), C=(9,0), D=(0,0), E=(2.5-0.5*sqrt(7),9), F=(6.5-0.5*sqrt(7),9), G=(4.5,9), O=(4.5,4.5); draw(A--B--C--D--A);draw(E--O--F);dr ...*sqrt(7),9), F=(6.5-0.5*sqrt(7),9), O=(4.5,4.5), G=O+(E-O)*dir(-90), J=O+(F-O)*dir(-90); draw(A--B--C--D--A);draw(E--O--F);draw(G--O--J);draw(F--G,linet
  13 KB (2,080 words) - 21:20, 11 December 2022
• 2005 AIME I Problems/Problem 13
  ...n range from <math>l = 5</math> to <math>9</math>; notice that <math>d = 10-l</math> gives the number of diagonals. Let <math>R</math> represent a move ...jacent. The <math>D</math>'s split the string into three sections (<math>-D-D-</math>): by the [[Pigeonhole principle]] all of at least one of the two l
  5 KB (897 words) - 00:21, 29 July 2022
• 2005 AIME II Problems/Problem 14
  ...from Pythagorean theorem, <math>SC = \frac{18}{5}</math> and <math>AS = 14-SC = \frac{52}{5}</math>. We can also use the Pythagorean theorem on triangl ...and <math>RE</math> as <math>x</math>. <math>RB</math> then equals <math>13-y</math>. Then, we have two similar triangles.
  13 KB (2,129 words) - 18:56, 1 January 2024
• 2004 AIME I Problems/Problem 7
  where the left-hand sum can be computed from: <center><math>\sum_{i=1}^{15} S_i = S_{15} + \left(\sum_{i=1}^{7} S_{2i-1} + S_{2i}\right) = -15 + 7 = -8</math></center>
  5 KB (833 words) - 19:43, 1 October 2023
• 2004 AIME II Problems/Problem 10
  ...math> is clearly never divisible by 9, so we can just focus on <math>1+2^{y-x}</math>. ...1+2^{y-x} \equiv 0 \pmod 9 \implies 2^{y-x} \equiv -1 \pmod 9 \implies 2^{y-x} \equiv 8 \pmod 9</cmath>
  7 KB (1,091 words) - 18:41, 4 January 2024
• 2019 AIME I Problems/Problem 2
  ...\ne 0</math> because <math>B \ne J</math>, so the probability that <math>B-J < 0</math> is <math>\frac{1}{2}</math> by symmetry. The probability that <math>B-J = 1</math> is <math>\frac{19}{20 \times 19} = \frac{1}{20}</math> because
  5 KB (830 words) - 22:15, 28 December 2023
• 1989 AIME Problems
  <center><math>r+si=a_m(-n+i)^m+a_{m-1}(-n+i)^{m-1}+\cdots +a_1(-n+i)+a_0</math></center> is true for a unique choice of non-negative integer <math>m^{}_{}</math> and digits <math>a_0,a_1^{},\ldots,a_m
  7 KB (1,045 words) - 20:47, 14 December 2023
• 1990 AIME Problems
  Find the value of <math>(52+6\sqrt{43})^{3/2}-(52-6\sqrt{43})^{3/2}</math>. <center><math>\frac 1{x^2-10x-29}+\frac1{x^2-10x-45}-\frac 2{x^2-10x-69}=0</math></center>
  6 KB (870 words) - 10:14, 19 June 2021
• 1998 AIME Problems
  7 KB (1,084 words) - 02:01, 28 November 2023
• 2000 AIME II Problems
  ...ht distinguishable rings, let <math>n</math> be the number of possible five-ring arrangements on the four fingers (not the thumb) of one hand. The order The equation <math>2000x^6+100x^5+10x^3+x-2=0</math> has exactly two real roots, one of which is <math>\frac{m+\sqrt{n
  6 KB (947 words) - 21:11, 19 February 2019
• 2001 AIME II Problems
  ...the largest number of students who could study both languages. Find <math>M-m</math>. x_{n}&=x_{n-1}-x_{n-2}+x_{n-3}-x_{n-4}\ \text{when}\ n\geq5, \end{align*}
  8 KB (1,282 words) - 21:12, 19 February 2019
• 2003 AIME II Problems
  ...th>C</math> is never immediately followed by <math>A</math>. How many seven-letter good words are there? ...ne{EF,}</math> <math>\overline{CD}\parallel \overline{FA},</math> and the y-coordinates of its vertices are distinct elements of the set <math>\{0,2,4,6
  7 KB (1,127 words) - 09:02, 11 July 2023
• 1985 AIME Problems/Problem 10
  ...changes when <math>\{x\}=\frac{m}{n}</math>, where <math>m\in\{1,2,3,...,n-1\}</math>, <math>n\in\{2,4,6,8\}</math>. Using [[Euler's Totient Function]] ...6x\rfloor+\lfloor 8x\rfloor</math>, then <math>k=20a+b</math> for some non-negative integers <math>a</math>, <math>b</math>, where there are <math>12</
  12 KB (1,859 words) - 18:16, 28 March 2022
• 1985 AIME Problems/Problem 5
  {{AIME box|year=1985|num-b=4|num-a=6}}
  2 KB (410 words) - 13:37, 1 May 2022
• 1986 AIME Problems/Problem 11
  The [[polynomial]] <math>1-x+x^2-x^3+\cdots+x^{16}-x^{17}</math> may be written in the form <math>a_0+a_1y+a_ ...<math>1 - x + x^2 + \cdots - x^{17} = \frac {1 - x^{18}}{1 + x} = \frac {1-x^{18}}{y}</math>. Since <math>x = y - 1</math>, this becomes <math>\frac {1
  6 KB (872 words) - 16:51, 9 June 2023
• 1987 AIME Problems/Problem 7
  {{AIME box|year=1987|num-b=6|num-a=8}}
  3 KB (547 words) - 22:54, 4 April 2016
• 1989 AIME Problems/Problem 5
  ...rac{1}{3}</math>. The answer we are looking for is <math>{5\choose3}(h)^3(1-h)^2 = 10\left(\frac{1}{3}\right)^3\left(\frac{2}{3}\right)^2 = \frac{40}{24 {{AIME box|year=1989|num-b=4|num-a=6}}
  2 KB (258 words) - 00:07, 25 June 2023
• 1990 AIME Problems/Problem 9
  ...ing this pattern, we find that there are <math>\sum_{i=6}^{11} {i\choose{11-i}} = {6\choose5} + {7\choose4} + {8\choose3} + {9\choose2} + {{10}\choose1} ...th>n-1</math> flips must fall under one of the configurations of <math>S_{n-1}</math>.
  3 KB (425 words) - 19:31, 30 July 2021
• 1991 AIME Problems/Problem 3
  ...>. Taking logarithms in both sides of this last equation and using the well-known fact <math>\log(a_{}^{}b)=\log a + \log b</math> (valid if <math>a_{}^ ...t[\prod_{j=1}^{k}\frac{(N-j+1)x}{j}\right]=\sum_{j=1}^{k}\log\left[\frac{(N-j+1)x}{j}\right]\, .
  5 KB (865 words) - 12:13, 21 May 2020
• 2006 AMC 10B Problems/Problem 22
  {{AMC10 box|year=2006|ab=B|num-b=21|num-a=23}}
  2 KB (394 words) - 00:51, 25 November 2023
• 1993 AIME Problems/Problem 13
  ...h>, <math>CX=AC\cdot\left(\frac{CD}{AB-CD}\right)=200\cdot\left(\frac{t}{3t-t}\right)=100</math>. ...f this line with <math>AB</math> as <math>S</math>. Then <math>SB=AB-AS=3t-t=2t</math>.
  8 KB (1,231 words) - 20:06, 26 November 2023
• 1994 AIME Problems/Problem 4
  ...th> such that <math>\lfloor\log_2{a}\rfloor=j</math>, and there are <math>n-2^k+1</math> such integers such that <math>\lfloor\log_2{a}\rfloor=k</math>. ...3}\rfloor+\cdots+\lfloor\log_2{n}\rfloor= \sum_{j=0}^{k-1}(j\cdot2^j) + k(n-2^k+1) = 1994</math>.
  2 KB (264 words) - 13:33, 11 August 2018
• 1995 AIME Problems/Problem 12
  ...>AP = 1.</math> It follows that <math>\triangle OPA</math> is a <math>45-45-90</math> [[right triangle]], so <math>OP = AP = 1,</math> <math>OB = OA = \ Without loss of generality, place the pyramid in a 3-dimensional coordinate system such that <math>A = (1,0,0),</math> <math>B =
  8 KB (1,172 words) - 21:57, 22 September 2022
• 1996 AIME Problems/Problem 12
  <cmath>|a_1-a_2|+|a_3-a_4|+|a_5-a_6|+|a_7-a_8|+|a_9-a_{10}|.</cmath> ...obtained from these paired sequences are also obtained in another <math>2^5-1</math> ways by permuting the adjacent terms <math>\{a_1,a_2\},\{a_3,a_4\},
  5 KB (879 words) - 11:23, 5 September 2021
• 1996 AIME Problems/Problem 7
  ...l be <br /> three other equivalent boards.</font></td><td><font style="font-size:85%">For those symmetric about the center, <br /> there is only one oth ...re rotationally symmetric about the center square; there are <math>\frac{49-1}{2}=24</math> such pairs. There are then <math>{49 \choose 2}-24</math> pa
  4 KB (551 words) - 11:44, 26 June 2020
• 1996 AIME Problems/Problem 3
  ...itive [[integer]] <math>n</math> for which the expansion of <math>(xy-3x+7y-21)^n</math>, after like terms have been collected, has at least 1996 terms. ...3)^n</math>. Both [[binomial expansion]]s will contain <math>n+1</math> non-like terms; their product will contain <math>(n+1)^2</math> terms, as each t
  3 KB (515 words) - 04:29, 27 November 2023
• 1998 AIME Problems/Problem 15
  ...s a path that retraces no segment. Each time that such a path reaches a non-terminal vertex, it must leave it. ...it can be arranged that <math>n-2</math> segments will emanate from <math>n-2</math> of the vertices and that an odd number of segments will emanate fro
  9 KB (1,671 words) - 22:10, 15 March 2024
• 1999 AIME Problems/Problem 7
  ...math> take on the values <math>0, 1, \ldots, 9</math>. At step i of a 1000-step process, the <math>i</math>-th switch is advanced one step, and so are ...}{d_{i}}= 2^{9-x_{i}}3^{9-y_{i}}5^{9-z_{i}}</math>. In general, the divisor-count of <math>\frac{N}{d}</math> must be a multiple of 4 to ensure that a s
  3 KB (475 words) - 13:33, 4 July 2016
• 2001 AIME I Problems/Problem 15
  .../math>, which can be done in <math>4! = 24</math> ways. Then choose a three-edge path along tetrahedron <math>DBEG</math> which, because it must start a ...faces, and one face adjacent to the three B-faces, which we will call the C-face.
  11 KB (1,837 words) - 18:53, 22 January 2024
• 2001 AIME I Problems/Problem 12
  ...ween a plane and a point <math>I</math> can be calculated as <math>\frac{(I-G) \cdot P}{|P|}</math>, where G is any point on the plane, and P is a vecto ...rpendicular to plane <math>ABC</math> can be found as <math>V=(A-C)\times(B-C)=\langle 8, 12, 24 \rangle</math>
  6 KB (1,050 words) - 18:44, 27 September 2023
• 2003 AIME I Problems/Problem 3
  ...1, 13, 34, 3, 21, 2\}. </math> Susan makes a list as follows: for each two-element subset of <math> \mathcal{S}, </math> she writes on her list the gre Each [[element]] of the [[set]] will appear in <math>7</math> two-element [[subset]]s, once with each other number.
  2 KB (317 words) - 00:09, 9 January 2024
• 2003 AIME II Problems/Problem 15
  ...that <math> \left( x^{23} + x^{22} + \cdots + x^2 + x + 1 \right) \cdot (x-1) = x^{24} - 1 </math>. The five-element sum is just <math>\sin 30^\circ + \sin 60^\circ + \sin 90^\circ + \s
  4 KB (675 words) - 17:23, 30 July 2022
• 2003 AIME II Problems/Problem 13
  ...ing the starting vertex in the next move. Thus <math>P_n=\frac{1}{2}(1-P_{n-1})</math>. ...-1</math> steps plus the number of ways to get to <math>C</math> in <math>n-1</math> steps.
  15 KB (2,406 words) - 23:56, 23 November 2023
• 2001 AIME II Problems/Problem 15
  {{AIME box|year=2001|n=II|num-b=14|after=Last Question}}
  4 KB (518 words) - 15:01, 31 December 2021
• 2001 AIME II Problems/Problem 10
  ...plies that <math>10^{6} - 1 | 10^{6k} - 1</math>, and so any <math>\boxed{j-i \equiv 0 \pmod{6}}</math> will work. ...le 5</math>, and let <math>j-i-a = 6k</math>. Then we can write <math>10^{j-i} - 1 = 10^{a} (10^{6k} - 1) + (10^{a} - 1)</math>, and we can easily verif
  4 KB (549 words) - 23:16, 19 January 2024
• 2001 AIME II Problems/Problem 6
  Let point <math>A</math> be the top-left corner of square <math>ABCD</math> and the rest of the vertices be arra ...factor. We get <math>b = 2/5a</math>, meaning the ratio of areas <math>((a-2b)/a)^2</math> = <math>(1/5)^2</math> = <math>1/25</math> = <math>m/n.</mat
  4 KB (772 words) - 19:31, 6 December 2023
• 2000 AIME II Problems/Problem 14
  ...8!-64!+\cdots+1968!-1984!+2000!</math>, find the value of <math>f_1-f_2+f_3-f_4+\cdots+(-1)^{j+1}f_j</math>. ...{k=1}^{n-1} {((k+1)!- k!)} = 1 + ((2! - 1!) + (3! - 2!) + \cdots + (n! - (n-1)!)) = n!</math>.
  7 KB (1,131 words) - 14:49, 6 April 2023
• 2006 USAMO Problems/Problem 1
  if and only if <math>s </math> is not a divisor of <math>p-1 </math>. ...>s|(p-1)</math>. Then for some positive integer <math>k</math>, <math>sk=p-1</math>. The conditions given are equivalent to stating that <math>sm \bmo
  3 KB (506 words) - 17:54, 22 June 2023
• 2006 USAMO Problems/Problem 3
  Let <math>f(x)</math> be a non-constant polynomial in <math>x</math> of degree <math>d</math> with any non-negative integer roots, so <math>a_i > 1</math> and thus <math>b_i > 0</math
  9 KB (1,699 words) - 13:48, 11 April 2020
• 2006 USAMO Problems/Problem 5
  ...is from <math>2^m(2^{i_0-m} - 1) = 2^{i_0} - 2^m</math> to <math>2^m(2^{i_0-m}+1) = 2^{i_0} + 2^m</math>. ...- 2^{i_0} = 2^p(2t - 2^{i_0-p} + 1)</math> and the number <math>2t + 2^{i_0-p} + 1</math> is odd, a jump of size <math>2^{p+1}</math> can be made from <
  7 KB (1,280 words) - 17:23, 26 March 2016
• Multinomial Theorem
  For a positive integer <math>k</math> and a non-negative integer <math>n</math>, ..._1 + b_2 + b_3 + ... + b_{k-1} + b_k= n}{\binom{n}{b_1, b_2, b_3, ..., b_{k-1}, b_k} \prod_{j=1}^{k}{x_j^{b_j}}}</cmath>
  3 KB (476 words) - 19:37, 4 January 2023
• Dedekind domain
  ...\cdots P_{n-1})\not\subseteq (a)</math>. Take any <math>b\in P_1\cdots P_{n-1}\setminus (a)</math> let <math>\gamma = b/a\in K</math>. We claim that thi ...ot\in R</math>. Now for any <math>x\in J</math>, <math>bx\in P_1\cdots P_{n-1}J\subseteq P_1\cdots P_n\subseteq (a)</math>, and so <math>bx = ar</math>
  9 KB (1,648 words) - 16:36, 14 October 2017
• 2006 AMC 10B Problems
  ...</math> and <math>y</math>, define <math> x \mathop{\spadesuit} y = (x+y)(x-y) </math>. What is <math> 3 \mathop{\spadesuit} (4 \mathop{\spadesuit} 5) < ...rcle is painted blue. What is the ratio of the blue-painted area to the red-painted area?
  14 KB (2,059 words) - 01:17, 30 January 2024
• 2000 AMC 12 Problems/Problem 8
  <cmath>S_{100}=\frac{100[2\cdot 4+(100-1)4]}{2}</cmath> ...se Figure 0 exists) <math>\dbinom{101-1}{0}+4\dbinom{101-1}{1}+4\dbinom{101-1}{2}=20201</math> or <math>\textbf{(C)}</math>
  7 KB (988 words) - 15:14, 10 April 2024
• 2005 AMC 10A Problems/Problem 1
  So the desired difference is <math>m-j=20-10=10 \Rightarrow \boxed{\textbf{(D) }10}</math> {{AMC10 box|year=2005|ab=A|before=First Problem|num-a=2}}
  909 bytes (134 words) - 19:05, 25 December 2022
• Navier-Stokes Equation
  ...interactions and dictate how sticky (viscous) a fluid is. Thus, the Navier-Stokes equations are a dynamical statement of the balance of forces acting a ...cal terms these rates correspond to their [[derivative]]s. Thus, the Navier-Stokes equations for the most simple case of an ideal fluid with zero viscos
  3 KB (553 words) - 22:08, 2 May 2022
• 2015 IMO Problems/Problem 6
  ...th>b</math> and <math>N</math> for which<cmath>\left\vert\sum_{j=m+1}^n(a_j-b)\right\vert\le1007^2</cmath>for all integers <math>m</math> and <math>n</m <u>'''Theorem'''</u> ''Let <math>T</math> be a non-negative integer parameter. If given a sequence <math>a_1,a_2,\dots</math> t
  4 KB (833 words) - 01:33, 31 December 2019
• Volume
  ...t defined by an upper bound of <math>f(x,y,z)</math> in the Cartesian three-space can be found using a triple [[integral]]: <math>\int_{a_z}^{b_z}\int_{
  3 KB (523 words) - 20:24, 17 August 2023
• Divisibility rules/Rule for 3 and 9 proof
  Suppose that the [[base numbers | base-ten]] representation of <math>N</math> is <center><math>N = a_k a_{k-1} \cdots a_2 a_1 a_0</math>,</center>
  2 KB (316 words) - 20:29, 6 March 2014
• AwesomeMath
  ...or bright middle and high school students who wish to sharpen their problem-solving skills and further their mathematics education. Many of our particip ...ve, instructor-led, virtual classes are taught once per weekend for each 12-week semester. Students choose their classes and there is no application req
  1 KB (195 words) - 11:19, 30 November 2023
• Stirling number
  <cmath> S(n,k) = \frac{1}{k!}\sum_{j=0}^{k}(-1)^j{k \choose j} (k-j)^n. </cmath>
  2 KB (253 words) - 00:04, 5 December 2020
• 2007 iTest Problems
  Compute the sum of all twenty-one terms of the geometric series <cmath>1+1+1-1-1+1-1+1-1+1-1-1-1+1+1.</cmath>
  33 KB (5,177 words) - 21:05, 4 February 2023
• Euler line
  ...th>L</math>. It is named after [[Leonhard Euler]]. Its existence is a non-trivial fact of Euclidean [[geometry]]. Certain fixed orders and distance [[ ...riangle CH_AH_B</math> [[concurrence | concur]] at <math>N</math>, the nine-point circle of <math>\triangle ABC</math>.
  59 KB (10,203 words) - 04:47, 30 August 2023
• Pick's Theorem
  2 KB (301 words) - 13:08, 20 February 2024
• Fermat point
  We shall present a standard triangle inequality proof as well as a less-known vector proof: ...\vec{0}</math> and adding, we see that <math>|a|+|b|+|c|\leq |a-x|+|b-x|+|c-x|</math>, or <math>AP+BP+CP\leq AX+BX+CX</math>. Thus, the origin or point
  4 KB (769 words) - 16:07, 29 December 2019
• Chinese Remainder Theorem
  ...or instance, [[Fermat's Little Theorem]] may be generalized to the [[Fermat-Euler Theorem]] in this manner. ...: the proof of the general case follows by induction to the above result (k-1) times.
  6 KB (1,022 words) - 14:57, 6 May 2023
• Matrix
  4 KB (856 words) - 15:29, 30 March 2013
• 1985 IMO Problems/Problem 3
  Now assume that the problem holds for <math>k-1</math>. We now have two cases: <math> i_1 \ge k</math>, and <math>i_1 < k ..._{j=0}^{k-1}a_j x^j + (1+x)^k \sum_{j=0}^{k-1} b_j x^j \equiv \sum_{j=0}^{k-1}\left[ (a_j + b_j)x^j + b_j x^{j+k} \right] \pmod{2}
  2 KB (354 words) - 04:56, 11 March 2023
• 2001 AMC 8 Problems
  Granny Smith has \$63. Elberta has \$2 more than Anjou and Anjou has one-third as much as Granny Smith. How many dollars does Elberta have? ...3, 4 and 9 are each used once to form the smallest possible '''even''' five-digit number. The digit in the tens place is
  13 KB (1,994 words) - 13:04, 18 February 2024
• 2003 AMC 12A Problems/Problem 12
  {{AMC10 box|year=2003|ab=A|num-b=23|num-a=25}} {{AMC12 box|year=2003|ab=A|num-b=11|num-a=13}}
  2 KB (368 words) - 18:04, 28 December 2020
• Minkowski Inequality
  1 KB (240 words) - 16:49, 29 December 2021
• 2001 IMO Shortlist Problems/A2
  ...>b_0, b_1, \ldots</math> of positive integers for which <math>1+b_n\le b_{n-1}\sqrt[n]{2}</math>, it is clear that there will not exist any sequence <ma ...exist such a sequence. Then, define <math>x_0=b_0</math> and <math>x_n=x_{n-1}\sqrt[n]{2} -1</math>. It is clear that <math>x_n\ge b_n</math> for all <m
  2 KB (411 words) - 04:38, 17 June 2019
• Quaternion
  1,007 bytes (155 words) - 20:47, 14 October 2013
• 2020 AMC 10A Problems
  <cmath>\frac{a-3}{5-c} \cdot \frac{b-4}{3-a} \cdot \frac{c-5}{4-b}</cmath> What is the sum of all real numbers <math>x</math> for which <math>|x^2-12x+34|=2?</math>
  13 KB (1,968 words) - 18:32, 29 February 2024
• Asymptote: Logical Operators and Loops
  if (floor((i-j)/2)==((i-j)/2)) if (floor((i-j)/2)==((i-j)/2))
  2 KB (324 words) - 16:50, 2 October 2016
• Pigeonhole Principle/Solutions
  ...riends, then the remaining friends must have from <math>1</math> to <math>n-2</math> friends for the remaining friends not to also have no friends. By p ...same remainder when divided by 5. Since there are 5 possible remainders (0-4), by the pigeonhole principle, at least two of the integers must share the
  10 KB (1,617 words) - 01:34, 26 October 2021
• 2002 USAMO Problems/Problem 3
  ...it is necessary and sufficient that <math> P(x) = Q(x) + \prod_{i=1}^{n}(x-x_i) </math>. ...s. But a polynomial with real coefficients must have an even number of non-real roots, so <math>P(x) </math> must have <math>n </math> real roots. Sim
  4 KB (688 words) - 13:38, 4 July 2013
• 2007 AIME I Problems/Problem 10
  [[Image:AIME I 2007-10.png]] ...ath> (<math>j < k</math>) here. We can now use the [[Principle of Inclusion-Exclusion]] based on the stipulation that <math>j\ne k</math> to solve the p
  13 KB (2,328 words) - 00:12, 29 November 2023
• 2004 USAMO Problems
  ...egers <math> x, y \in S </math>, if <math> x+y \in S </math>, then <math> x-y \in S </math>. * [http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2004-ua/04usamo-test.shtml 2004 USAMO Problems]
  3 KB (474 words) - 09:20, 14 May 2021
• 2007 AIME II Problems/Problem 3
  ...ince <math>\triangle ABE \cong \triangle CDF</math> and are both <math>5-12-13</math> [[right triangle]]s, we can come to the conclusion that the two ne A slightly more analytic/brute-force approach:
  6 KB (933 words) - 00:05, 8 July 2023
• 2007 AIME II Problems/Problem 13
  {{AIME box|year=2007|n=II|num-b=12|num-a=14}}
  3 KB (600 words) - 11:10, 22 January 2023
• 2007 AIME II Problems
  <div style="text-align:center;">[[Image:2007 AIME II-3.png]]</div> ..._{i+1}</math> if <math>a_{i}</math> is [[even]]. How many four-digit parity-monotonic integers are there?
  9 KB (1,435 words) - 01:45, 6 December 2021
• 2003 USAMO Problems/Problem 1
  ...-1} + a </math> is divisible by 5, which is true when <math> k \equiv -3^{n-1}a \pmod{5} </math>. Since there is an odd digit in each of the residue cl ...s after it, and all multiples of 5 end in 5. Therefore, <math> a*10^x*5^{n-x}</math> always contains a 5 as its <math> (x+1)^{st}</math> digit, and we
  4 KB (736 words) - 22:17, 3 March 2023
• 2003 USAMO Problems/Problem 3
  ...{} c_j = \cdots = c_{j+i} = \cdots = c_k = j </math> (<math> 0 \le i \le k-j</math>). ''Proof.'' Since the <math>j </math> terms <math>c_0, \ldots, c_{j-1} </math> are all less than <math>j </math>, no other terms that precede <m
  3 KB (636 words) - 13:39, 4 July 2013
• Pentagon
  1 KB (191 words) - 09:59, 6 June 2022
• 2004 IMO Shortlist Problems/N2
  ...e-i}) = p^e + \sum_{i=0}^{e-1}p^i(p^{e-i} - p^{e-i-1}) = p^e + e(p^e - p^{e-1}) </math>. </center> ...^{e_i} - e_i \cdot p_i^{e_i}]}{\prod p_i^{e_i}} = \prod \left(e_i \frac{p_i-1}{p_i} + 1 \right) </math>.
  6 KB (1,007 words) - 09:10, 29 August 2011
• Parallelepiped
  2 KB (248 words) - 20:08, 17 August 2023
• Chicken McNugget Theorem
  ...am + bn</math> for [[nonnegative]] integers <math>a, b</math> is <math>mn-m-n</math>. ...e proof is based on the fact that in each pair of the form <math>(k, mn-m-n-k)</math>, exactly one element is expressible.
  17 KB (2,748 words) - 19:22, 24 February 2024
• Dodecagon
  A '''dodecagon''' is a 12-sided [[polygon]]. The sum of its internal [[angle]]s is <math>1800^{\circ}<
  1 KB (219 words) - 13:08, 15 June 2018
• Triangular number
  <math>1 +2+3 + 4............. +(n-1)+(n)</math>. ...math>T_{n} = 1 + 2 + \ldots + (n-1) + n = (1 + 2 + \ldots + n-1) + n = T_{n-1} + n</math>.
  2 KB (275 words) - 08:39, 7 July 2021
• 2007 AMC 12A Problems/Problem 24
  ...mod{4}</math> is that <math>v_2(2j+1)=0,</math> so we must have <math>v_2(n-1)>v_2(n+1)</math> since <math>v_2(2k) \geq 1 >v_2(2j+1).</math> Therefore, {{AMC12 box|year=2007|num-b=23|num-a=25|ab=A}}
  4 KB (588 words) - 14:40, 23 August 2023
• 2006 AIME II Problems/Problem 13
  ...math>, <math>n\geq 0</math>. Then the final odd integer is <math>2n+1 + 2(j-1) = 2(n+j) - 1</math>. The odd integers form an [[arithmetic sequence]] wit ...ual <math>0</math>. We then perform casework based on the parity of <math>p-q</math>.
  4 KB (675 words) - 10:40, 14 July 2022
• 2006 AIME II Problems
  When rolling a certain unfair six-sided die with faces numbered <math>1, 2, 3, 4, 5</math>, and <math>6</math> ...h> and <math>t_2</math> intersect at <math>(x,y),</math> and that <math>x=p-q\sqrt{r},</math> where <math>p, q,</math> and <math>r</math> are positive i
  8 KB (1,350 words) - 12:00, 4 December 2022
• 1998 PMWC Problems
  for(int j = n-i; j > 0; --j){ ...ad a total of <math>30</math> balls, how many balls were there in the right-hand box?
  11 KB (1,738 words) - 19:25, 10 March 2015
• 2001 PMWC Problems
  F=O+(G-O)+(E-O); draw((xstart,A.y)--(xstart,A.y-len));
  4 KB (641 words) - 21:24, 21 April 2014
• 2002 PMWC Problems
  label("$" + string(i) + "^\textrm{" + ord[i-1] + "}$ Row:", (-5,(-i+1)*sqrt(3)/2+correction)); draw( (1-cos(pi/3)*correction,-sin(pi/3)*correction)--(0+cos(pi/3)*correction,-sqrt(3
  5 KB (725 words) - 16:07, 23 April 2014
• 2003 PMWC Problems
  filldraw((p-a-b)--(p+a-b)--(p+a+b)--(p-a+b)--cycle,black); fill((p-a-b)--(p+a-b)--(p+a+b)--(p-a+b)--cycle,white);
  7 KB (918 words) - 16:15, 22 April 2014
• 1989 USAMO Problems/Problem 3
  Let <math>P(z)= z^n + c_1 z^{n-1} + c_2 z^{n-2} + \cdots + c_n</math> be a [[polynomial]] in the complex variable <math>z <cmath> \lvert z_i-i \rvert \cdot \lvert z_j-i \rvert < 1. </cmath>
  2 KB (340 words) - 19:11, 18 July 2016
• Cramer's Rule
  2 KB (352 words) - 18:22, 11 October 2023
• Lucas' Theorem
  ...n of <math>n</math> in [[base]] <math>p</math> and <math>(\overline{i_mi_{m-1}\cdots i_0})_p</math> is the representation of <math>i</math> in base <mat For all <math>1\leq k \leq p-1</math>, <math>\binom{p}{k}\equiv 0 \pmod{p}</math>. Then we have
  1 KB (251 words) - 15:13, 11 August 2020
• Summation
  ...r algebraic structure), <math>\sum_{i=a}^{b}c_i=c_a+c_{a+1}+c_{a+2}...+c_{b-1}+c_{b}</math>. Here <math>i</math> refers to the index of summation, <math ...i= \frac{n(n+1)}{2}</math>, and in general <math>\sum_{i=a}^{b} i= \frac{(b-a+1)(a+b)}{2}</math>
  3 KB (482 words) - 16:39, 8 October 2023
• Zorn's Lemma
  ...intermediate results, viz., Bourbaki's Theorem (also known as the Bourbaki-Witt theorem). ...h>x \in A</math> does <math>x = f(x)</math>, which contradicts the Bourbaki-Witt Theorem. {{halmos}}
  9 KB (1,669 words) - 19:02, 1 August 2018
• Ideal
  ...ls in rings are the [[kernel]]s of ring [[homomorphism]]s; in this way, two-sided ideals of rings are similar to [[normal subgroup]]s of [[group]]s. ...frak{a}</math> is both a left ideal and a right ideal, it is called a ''two-sided ideal''. In a [[commutative ring]], all three kinds of ideals are the
  8 KB (1,389 words) - 23:44, 17 February 2020
• Work
  1 KB (192 words) - 00:46, 16 July 2018
• Celsius
  4 KB (357 words) - 22:40, 17 April 2024
• 1999 USAMO Problems/Problem 4
  <cmath> \sum_{i=k+1}^n -a_i \le 2k-n . </cmath> <cmath> \sum_{i=k+1}^n a_i^2 \le \left( \sum_{i=k+1}^n-a_i \right)^2 \le (2k-n)^2 . </cmath>
  3 KB (499 words) - 09:47, 20 July 2016
• 2001 IMO Shortlist Problems/A3
  By the [[Cauchy-Schwarz Inequality]], By the [[Cauchy-Schwarz Inequality]]
  3 KB (441 words) - 00:54, 24 October 2023
• 1999 IMO Problems/Problem 2
  ...ath> and the right-hand side becomes <math>(x_a + x_b)^4/8</math>. By [[AM-GM]], ...ase the left-hand side of the desired inequality without changing the right-hand side; and then to use the inductive hypothesis.
  4 KB (838 words) - 01:04, 17 November 2023
• Proofs of AM-GM
  This pages lists some proofs of the weighted [[AM-GM Inequality]]. The inequality's statement is as follows: for all nonnegati ...and <math>\lambda_i \neq 0</math>, for some <math>i</math>, then the right-hand side of the inequality is zero and the left hand of the inequality is g
  12 KB (2,171 words) - 07:55, 11 May 2023
• Carleman's Inequality
  Define <math>c_n = n \left( 1 + \frac{1}{n} \right)^n = \frac{(n+1)^n}{n^{n-1}}</math>. Then for all positive integers <math>k</math>, Now, by [[AM-GM]],
  2 KB (278 words) - 16:39, 29 December 2021
• 2008 AMC 12B Problems/Problem 23
  ...square. When these are multiplied, they equal <math>2^{a+n-a} \times 5^{b+n-b} = 10^n</math>. <math>\log 10^n=n</math> so the number of factors divided .../math> satisfies <math>ab = 10^n</math> -- ex. <math>(1, 10^n), (2, 5*10^{n-1})</math>, etc. Then the sum of the base-<math>10</math> logarithms is <mat
  5 KB (814 words) - 18:02, 17 January 2023
• Vieta's formulas
  Let <math>P(x) = a_n x^n + a_{n-1}x^{n-1} + \cdots + a_1 x + a_0</math> be any polynomial with [[Complex number | c ....</cmath> This can be compactly summarized as <math>s_j = (-1)^j \frac{a_{n-j}}{a_n}</math> for some <math>j</math> such that <math>1 \leq j \leq n</mat
  3 KB (512 words) - 15:31, 22 February 2024
• 2000 AMC 12 Problems/Problem 16
  <math>f(i, j) = 17(i-1) + j = 17i + j - 17</math> {{AMC12 box|year=2000|num-b=15|num-a=17}}
  2 KB (310 words) - 11:28, 3 August 2021
• 2007 AMC 10A Problems
  draw((-h-1,-h-1)--(-h-1,h+1)--(h+1,h+1)--(h+1,-h-1)--cycle); ...at random from 0 to 2007, inclusive. What is the probability that <math>ad-bc</math> is even?
  13 KB (2,058 words) - 17:54, 29 March 2024
• 1990 USAMO Problems/Problem 4
  '''Lemma.''' Let <math>a</math> be an integer. Then there are <math>2^{k-1}</math> <math>k</math>-good sequences starting on <math>a</math>, and furt ...M</math>. Then the only possibilities for <math>a_{k+1}</math> are <math>m-1</math> and <math>M+1</math>; either way, <math>\{ a_i \}_{i=1}^{k+1}</math
  3 KB (529 words) - 19:15, 18 July 2016
• 2007 AMC 10A Problems/Problem 18
  ...\cdot 12 = \frac {36}5 </math>. (An alternate way is by seeing that the set-up AHGCM is similar to the 2 pole problem (http://www.artofproblemsolving.co {{AMC10 box|year=2007|ab=A|num-b=17|num-a=19}}
  6 KB (867 words) - 00:17, 20 May 2023
• 1974 USAMO Problems/Problem 1
  <cmath> a-b \mid P(a) - P(b) = b-c \mid P(b)-P(c) = c-a \mid P(c) - P(a) = a-b , </cmath> ...absolute value. In fact, two of them, say <math>(a-b)</math> and <math>(b-c)</math>, must be equal. Then
  7 KB (1,291 words) - 20:30, 27 April 2020
• 1994 USAMO Problems
  The sides of a 99-gon are initially colored so that consecutive sides are red, blue, red, blue ...sual, let <math>\, \binom{n}{k} \,</math> denote <math>\, n! \over k! \, (n-k)!</math>. Prove that <cmath> \sum_{U \subseteq S} (-1)^{|U|} \binom{m - \s
  2 KB (391 words) - 07:58, 19 July 2016
• Isogonal conjugate
  Let <math>I'</math> and <math>J'</math> be the A-excenters of triangles <math>\triangle ABC</math> and <math>\triangle ADC,</ ...oordinates <cmath>(p' : q' : r'), p' = \frac {|B - C|^2}{p}, q' = \frac {|A-C|^2}{q}, r' = \frac {|A - B|^2}{r}.</cmath>
  54 KB (9,416 words) - 08:40, 18 April 2024
• 1976 IMO Problems/Problem 2
  ...inct real solutions, where <math>2^{n-1}</math> are positive and <math>2^{n-1}</math> are negative. Also, for ever root <math>r</math>, <math>|r|<2</mat ...values less than 2, where <math>2^{k-1}</math> are positive and <math>2^{k-1}</math> are negative.
  3 KB (596 words) - 16:19, 28 July 2015
• 1976 IMO Problems/Problem 5
  {{IMO box|year=1976|num-b=4|num-a=6}}
  2 KB (377 words) - 16:28, 29 January 2021
• Partition (combinatorics)
  ...iven by <math>F(x)= \sum_{n \geq 0}P(n) x^n = \prod_{n=1}^\infty \frac{1}{1-x^n}</math>. Partitions can also be studied by using the [[Jacobi theta func ...epresents a different addend in the partition. The rows are ordered in non-increasing order so that that the row with the most dots is on the top and t
  10 KB (1,508 words) - 14:24, 17 September 2017
• 2001 AMC 12 Problems/Problem 6
  A telephone number has the form <math>\text{ABC-DEF-GHIJ}</math>, where each letter represents {{AMC12 box|year=2001|num-b=5|num-a=7}}
  2 KB (340 words) - 03:02, 28 June 2023
• 2008 AIME II Problems
  ...ch is three-quarters the rate that Rudolph bikes, but Jennifer takes a five-minute break at the end of every two miles. Jennifer and Rudolph begin bikin a^2 + y^2 = b^2 + x^2 = (a-x)^2 + (b-y)^2
  7 KB (1,167 words) - 21:33, 12 August 2020
• 2008 AIME I Problems/Problem 3
  <math>(13,1)</math> and <math>(3,9)</math> give non-integral <math>b</math>, but <math>(8,5)</math> gives <math>b = 15</math>. T ...<math>2y+3z+4x=91</math>. Subtracting the two expressions we get <math>y+z-2x=-17</math>. Note that <math>-17</math> is odd, so one of <math>x,y,z</mat
  3 KB (564 words) - 22:15, 28 November 2023
• Mock AIME 5 Pre 2005 Problems
  Two 5-digit numbers are called "responsible" if they are:
  6 KB (909 words) - 07:27, 12 October 2022
• 2008 AIME II Problems/Problem 10
  ...our points using coordinates <math>0 \le x,y \le 3</math>, with the bottom-left point being <math>(0,0)</math>. By the [[Pythagorean Theorem]], the dis {{AIME box|year=2008|n=II|num-b=9|num-a=11}}
  4 KB (569 words) - 09:44, 25 November 2019
• 2005 USAMO Problems/Problem 2
  (1) <math>(x^3+y)(x^3+1) = (x^3+y)(x+1)(x^2-x+1) = 147^{157} = 7^{314}3^{152}</math>, <math>x^2-x+1 = (x+1)^2-3(x+1)+3 \rightarrow \gcd(x+1, x^2-x+1) \le 3</math>.
  7 KB (1,053 words) - 10:38, 12 August 2015
• 2005 USAMO Problems/Problem 1
  ...t with the circular arrangement of <math>n,p_{1}p_{2},p_{2}p_{3}\ldots,p_{k-1}p_{k}</math> as shown. ...by placing <math>p_k,p^{2}_{k},\ldots,p^{e_k}_{k}</math> between <math>p_{k-1}p_k</math> and <math>n</math>. It is easy to see that each element of <mat
  4 KB (650 words) - 13:40, 4 July 2013
• 2006 IMO Problems/Problem 5
  a_1 - a_0 &\mid P(a_1)- P(a_0) = a_2-a_1 \\ &\mid P(a_r)- P(a_{r-1}) = a_1 - a_0 .
  3 KB (704 words) - 14:42, 7 September 2021
• Schreier's Theorem
  ...iscovered it in 1928, and used it to give an improved proof of the [[Jordan-Hölder Theorem]]. Six years later, Hans Zassenhaus published his [[Zassenh ...ies in question. For integers <math>j \in [1,m-1]</math>, <math>i \in [0,n-1]</math>, let <math>H'_{im+j} = H_{i+1} \cdot (H_i \cap K_j)</math>, and fo
  2 KB (337 words) - 12:13, 9 April 2019
• 2008 Mock ARML 2 Problems
  ...th>f(x)</math> equal the number of zeroes to the right of the rightmost non-zero digit in the decimal form of <math>x!</math>, and let <math>n = \frac { ...sum_{i = 0}^{n}a_ia_{n - i} = 1</math> and <math>a_n > 0</math> for all non-negative integers <math>n</math>, evaluate <math>\sum_{j = 0}^{\infty}\frac
  4 KB (582 words) - 21:57, 8 May 2019
• 2008 Mock ARML 2 Problems/Problem 8
  ...sum_{i = 0}^{n}a_ia_{n - i} = 1</math> and <math>a_n > 0</math> for all non-negative integers <math>n</math>, evaluate <math>\sum_{j = 0}^{\infty}\frac By the given, the coefficients on the right-hand side are all equal to <math>1</math>, yielding the [[geometric series]]
  1 KB (190 words) - 00:57, 31 May 2016
• 2003 AMC 12A Problems/Problem 17
  <math>x^2 + (y-4)^2 = 4^2</math> <math>(x-2)^2 + y^2 = 2^2</math>
  6 KB (1,026 words) - 22:35, 29 March 2023
• 2003 AMC 12B Problems/Problem 18
  ...s 7 and 11 in order to keep <math>x</math> at a minimum. Moving all the non-y terms to the left hand side of the equation, we end up with: <cmath>7^{5b+1}11^{5d-1}=y^{13}</cmath>
  6 KB (914 words) - 11:07, 7 September 2023
• Jean-Victor Poncelet
  '''Jean-Victor Poncelet''' (July 1, 1788 – December 22, 1867) was a French mathema ...son of Claude de Poncelet, a lawyer in the local Parliament, and a well-to-do landowner. Poncelet attended local schools in his home town before going
  2 KB (253 words) - 11:41, 19 December 2018
• KGS math club/solution 6 1
  ...know that <math>\forall i: 2 | (W-w_i)</math>, that is, each remaining ball-mass is divisible by two. Combining these, we get <math>\forall i,j: 2 | (w_i-w_j)</math> by subtracting the case i from the case j.
  4 KB (759 words) - 06:27, 18 July 2009
• Legendre's Formula
  ...i=1}^{\infty} \left\lfloor \dfrac{n}{p^i}\right\rfloor =\frac{n-S_{p}(n)}{p-1}</cmath> <cmath>e_2(27!)=\frac{27-S_2(27)}{2-1}=27-S_2(27)</cmath>
  4 KB (699 words) - 17:55, 5 August 2023
• 2001 IMO Shortlist Problems/C2
  ...that <math>S(x)\equiv y\bmod{n}</math> for all integers <math>0\leq y\leq n-1</math>. This means that the sum of every <math>S(a)</math> is congruent to ...(c_i\sum_{j=1}^{n} (n-1)!*a_j\right)=\sum_{i=1}^{n} c_i\cdot \frac{n(n+1)(n-1)!}{2}=\frac{c_i(n+1)!}{2}\bmod{n!}</cmath>
  2 KB (364 words) - 08:55, 31 August 2011
• 2001 IMO Shortlist Problems/C5
  519 bytes (73 words) - 20:07, 29 May 2020
• 2001 IMO Shortlist Problems/N6
  ...ath>(c+d-b)^2+b^2\equiv c^2+d^2\pmod p</math>. Factorizing, <cmath>2(b^2-bc-bd+cd)\equiv0\pmod p</cmath> <cmath>\implies(b-c)(b-d)\equiv0\pmod p</cmath>
  2 KB (443 words) - 13:08, 17 August 2011
• Millennium Problems
  ...], the [[Poincaré Conjecture]], the [[Riemann Hypothesis]], and the [[Yang-Mills Theory]]. In 2003, the Poincaré Conjecture was proven by Russian math ...ent. He gave three lectures, titled "Ricci Flow and Geometrization of Three-Manifolds", on April 7, 9, and 11. These were his first public presentation
  13 KB (1,969 words) - 17:57, 22 February 2024
• 1989 AHSME Problems/Problem 29
  {{AHSME box|year=1989|num-b=28|num-a=30}}
  1 KB (173 words) - 19:52, 30 December 2020
• 2008 IMO Problems/Problem 5
  a_i-n, & \mbox{ if } a_i \in B .../math> elements from <math>\cal{N}</math>, which would imply <math>N = 2^{k-n}M</math> and solve the problem.
  4 KB (677 words) - 01:10, 19 November 2023
• 2000 USAMO Problems
  Call a real-valued function <math>f</math> very convex if
  3 KB (495 words) - 19:02, 18 April 2014
• 2000 USAMO Problems/Problem 4
  if((i == 0 || j == 9) && !(j-i == 9)) fill(shift(i,j)*unitsquare,rgb(0.3,0.3,0.3)); ...r of the rows containing the colored squares. Hence, each of the <math>1999-m</math> colored squares must be placed in different rows, but as there are
  2 KB (382 words) - 13:37, 4 July 2013
• 2000 USAMO Problems/Problem 5
  ...e externally tangent at the point <math>A_k</math>, so <math>O_k, A_k, O_{k-1}</math> are [[collinear]]. ...ath>\theta_k = 180 - \theta_{k-1}</math>, and so <math>\theta_k = \theta_{k-2}</math>. Thus, <math>\theta_{1} = \theta_{7}</math>.
  3 KB (609 words) - 09:52, 20 July 2016
• 2001 USAMO Problems/Problem 1
  ...olors in 3 boxes and <math>3n-48</math> colors in 2. Thus <math>n\geq 20+48-2n,</math> so <math>3n\geq 68</math>, and <math>n\geq23</math> and we are do {{USAMO newbox|year=2001|before=First question|num-a=2}}
  5 KB (841 words) - 17:19, 5 May 2022
• Karamata's Inequality
  ...a\leq x_1\leq x_2 \leq b</math> and <math>\Gamma(x, y):=\frac{f(y)-f(x)}{y-x}</math>, <math>\Gamma(x_1, x)\leq \Gamma (x_2, x)</math> <cmath>\Gamma(x_1, x)=\frac{f(x_1)-f(x)}{x_1-x}\leq \frac{f(x_2)-f(x)}{x_2-x}=\Gamma(x_2, x)</cmath>
  2 KB (370 words) - 03:39, 28 March 2024
• 1973 USAMO Problems/Problem 2
  <center> <math>X_0=1</math>, <math>X_1=1</math>, <math>X_{n+1}=X_n+2X_{n-1}</math> <math>(n=1,2,3,\dots),</math></center> <center><math>Y_0=1</math>, <math>Y_1=7</math>, <math>Y_{n+1}=2Y_n+3Y_{n-1}</math> <math>(n=1,2,3,\dots)</math>.</center>
  2 KB (342 words) - 18:54, 3 July 2013
• 2004 IMO Problems/Problem 4
  ...math>t_3 = t_2 + t_1 + k</math> for <math>k \ge 0</math>; then (by [[Cauchy-Schwarz Inequality]]) ...} \right) + t_n \sum_{i=1}^{n-1} \frac 1{t_i} + \frac{1}{t_n} \sum_{i=1}^{n-1} t_i + 1\end{align*}</cmath>
  2 KB (322 words) - 00:54, 19 November 2023
• 2002 AMC 10A Problems
  Given that <math>a, b,</math> and <math>c</math> are non-zero real numbers, define <math>(a, b, c) = \frac{a}{b} + \frac{b}{c} + \fra ..., and each of those circles is tangent to the large circle and to its small-circle neighbors. Find the area of the shaded region.
  11 KB (1,733 words) - 11:04, 12 October 2021
• 2002 AMC 10A Problems/Problem 20
  {{AMC10 box|year=2002|ab=A|num-b=19|num-a=21}}
  3 KB (439 words) - 22:15, 9 June 2023
• 1973 USAMO Problems/Problem 1
  http://www.cut-the-knot.org/wiki-math/index.php?n=MathematicalOlympiads.USA1973Problem1 {{USAMO box|year=1973|before=First Question|num-a=2}}
  2 KB (276 words) - 08:10, 4 January 2022
• 1975 USAMO Problems/Problem 3
  ...], we can write <math>Q(x)</math> as <cmath>Q(x) = c(x)(x-1)(x-2) \cdots (x-n)</cmath> where <math>c</math> is a constant. Thus, <cmath>(x+1)P(x) - x = c(x)(x-1)(x-2) \cdots (x-n).</cmath>
  3 KB (465 words) - 17:29, 8 January 2024
• 1962 IMO Problems/Problem 7
  <math>M_{12}=\frac{V_1-V_2}{2}=\left( \frac{\sqrt{2}}{6}r_c ,\;\frac{\sqrt{6}}{6}r_c,\;-\frac{r_c}{ <math>M_{13}=\frac{V_1-V_3}{2}=\left( \frac{\sqrt{2}}{6}r_c ,\;-\frac{\sqrt{6}}{6}r_c,\;-\frac{r_c}
  11 KB (1,928 words) - 20:52, 21 November 2023
• 1963 IMO Problems/Problem 3
  ...textrm{-coordinate} = \sum_{k = 1}^{\frac{n}{2}}a_k \cdot \sin \frac{2\pi(k-1)}{n}.\textbf{ (1)}</cmath> ...xtrm{-coordinate} = \sum_{k = \frac{n}{2}+1}^{n}a_k \cdot \sin \frac{2\pi(k-1)}{n}</cmath>
  4 KB (836 words) - 17:58, 7 December 2022
• 2001 AMC 10 Problems
  two-digit number such that <math>N = P(N)+S(N)</math>. What is the units digit o ...t), and the rest sells for half price. How much money is raised by the full-price tickets?
  14 KB (1,983 words) - 16:25, 2 June 2022
• 2000 AMC 10 Problems/Problem 16
  ..., hence we can compute the slope, <math>m_1</math> to be <math>\frac{0-3}{6-0}=-\frac{1}{2}</math>, so <math>l_1</math> is the line <math>y=\frac{-1}{2} ...th> gives us <math>b_2=-2</math>, so <math>l_2</math> is the line <math>y=x-2</math>.
  7 KB (966 words) - 23:22, 31 July 2023
• 2007 IMO Shortlist Problems/A1
  <center><math>\max\{|x_i-a_i|:1\leq i\leq n\}\geq \frac{d}{2}</math>.</center> ..._j:i\le j\le n\}</math>, all <math>d_i</math> can be expressed as <math>a_p-a_q</math>, where <math>1\le p\le i\le q \le n</math>.
  3 KB (734 words) - 05:11, 26 March 2024
• 2004 AMC 12B Problems/Problem 14
  pair J=intersectionpoint(A--B, M--(M+rotate(90)*(B-A)) ); pair K=intersectionpoint(A--B, N--(N+rotate(90)*(B-A)) );
  7 KB (1,083 words) - 22:41, 23 November 2020
• 1976 IMO Problems
  For all positive integral <math>n</math>, <math>u_{n+1}=u_n(u_{n-1}^2-2)-u_1</math>, <math>u_0=2</math>, and <math>u_1=2\frac12</math>. Prove that
  2 KB (409 words) - 16:25, 29 January 2021
• 1971 IMO Problems/Problem 6
  ...})(i, j = 1, 2, \cdots, n)</math> be a square matrix whose elements are non-negative integers. Suppose that whenever an element <math>a_{ij} = 0</math>, ...h>. We thus infer that<cmath>s(\ell_j)+s(c_{\sigma(j)})\ge (n-t)+(n-k+t)=2n-k\ge n</cmath>so <math>s(\ell_j)+s(c_{\sigma(j)})\ge n,\ \forall j\ge k+1</m
  6 KB (1,192 words) - 14:14, 29 January 2021
• 2001 AMC 12 Problems/Problem 10
  {{AMC12 box|year=2001|num-b=9|num-a=11}} {{AMC10 box|year=2001|num-b=17|num-a=19}}
  2 KB (252 words) - 00:11, 15 August 2022
• 2001 AMC 12 Problems/Problem 22
  Therefore <math>[EHJ]=[ACD]-[AHD]-[DEH]-[EJC]=35-14-\frac {21}2-\frac{15}2 = \boxed{3}</math>. {{AMC12 box|year=2001|num-b=21|num-a=23}}
  7 KB (1,120 words) - 02:27, 29 August 2023
• 2009 AMC 12B Problems
  Each morning of her five-day workweek, Jane bought either a <math>50</math>-cent muffin or a <math>75 Twenty percent less than <math>60</math> is one-third more than what number?
  13 KB (2,030 words) - 03:04, 5 September 2021
• 2009 AMC 12B Problems/Problem 20
  ...rtices, implying each vertex owns 4 edge ends. There are twice as many edge-ends as there are edges, and <math>2 \cdot 100 = 200</math>. {{AMC12 box|year=2009|ab=B|num-b=19|num-a=21}}
  9 KB (1,567 words) - 13:43, 19 August 2023
• 2009 AMC 12B Problems/Problem 25
  Given any square, we can circumscribe another axes-parallel square around it. In the picture below, the original square is red ...n a clever bijection given in [http://web2.slc.qc.ca/sh/Contest/AMC12_2009B-S.pdf this page].
  15 KB (2,229 words) - 03:36, 4 September 2021
• Maximal ideal
  968 bytes (183 words) - 19:50, 23 August 2009
• Polynomial ring
  ...biggr)\cdot \biggl(\sum_j b_jx^j\biggr) = \sum_k\biggl(\sum_{i=0}^k a_ib_{k-i}\biggr)x^k ...nto <math>R[x]/(x-a)</math> and the canonical homomorphism of <math>R[x]/(x-a)</math> into <math>R</math>.)
  12 KB (2,010 words) - 00:10, 3 August 2020
• 1989 AJHSME Problems
  Which of the five "T-like shapes" would be symmetric to the one shown with respect to the dashed ...per is then cut in half along the dashed line. Three rectangles are formed-a large one and two small ones. What is the ratio of the perimeter of one o
  13 KB (1,765 words) - 11:55, 22 November 2023
• Hilbert's Basis Theorem
  ascending chain of (two-sided) [[ideal]]s of <math>A[x]</math>. <math>\mathfrak{c}_{i,j}</math> is a two-sided ideal of <math>A</math>; furthermore,
  4 KB (617 words) - 19:59, 23 April 2023
• Determinant
  ...n\times n</math> determinant can be written in terms of <math>(n-1)\times(n-1)</math> determinants: where <math>m_i</math> is the <math>(n-1)\times(n-1)</math> matrix formed by removing the <math>1</math>st row and <math>i</ma
  8 KB (1,345 words) - 00:31, 9 May 2020
• 2004 USAMO Problems/Problem 2
  By Schur's theorem (a generalization of the more well-known Chicken McNugget theorem), every integer greater than some integer <ma {{USAMO newbox|year=2004|num-b=1|num-a=3}}
  3 KB (508 words) - 14:23, 17 July 2014
• 1968 AHSME Problems/Problem 35
  {{AHSME 35p box|year=1968|num-b=34|after=Last Problem}}
  3 KB (597 words) - 01:52, 16 August 2023
• 1989 AJHSME Problems/Problem 19
  <cmath>4.8-2.2=2.6\approx 2.5 \rightarrow \boxed{\text{B}}</cmath> {{AJHSME box|year=1989|num-b=18|num-a=20}}
  2 KB (255 words) - 03:28, 28 November 2019
• 1989 AJHSME Problems/Problem 22
  {{AJHSME box|year=1989|num-b=21|num-a=23}}
  1 KB (177 words) - 16:03, 19 April 2021
• 1965 IMO Problems/Problem 4
  ...= 1-P \Longleftrightarrow x_i = 1 \pm t</math> where <math>t = \pm \sqrt{1-P} \in \mathbb{R}</math>. Which has no non-zero solutions for <math>t</math>.
  2 KB (382 words) - 12:51, 29 January 2021
• 2009 USAMO Problems
  ...ile a chessboard polygon by dominoes is to exactly cover the polygon by non-overlapping <math>1 \times 2</math> rectangles. Finally, a ''tasteful tiling
  4 KB (718 words) - 18:16, 17 September 2012
• 2009 USAMO Problems/Problem 6
  {{USAMO newbox|year=2009|num-b=5|after=Last question}}
  2 KB (463 words) - 12:19, 21 August 2020
• 2009 IMO Problems/Problem 6
  ...h> be distinct positive integers and let <math>M</math> be a set of <math>n-1</math> positive integers not containing <math>s=a_1+a_2+\ldots+a_n</math>. ...ts are identical, we can treat them as one and use induction on <math>n = k-1</math>. We now consider four cases:
  5 KB (1,055 words) - 01:18, 19 November 2023
• 1998 USAMO Problems/Problem 3
  By AM-GM, {{USAMO newbox|year=1998|num-b=2|num-a=4}}
  2 KB (322 words) - 13:31, 23 August 2023
• Semisimple module
  Let us take a well-ordering on <math>I</math> (such an ordering exists by the [[well-ordering theorem]], a consequence of the [[axiom of choice]]),
  6 KB (1,183 words) - 15:02, 18 August 2009
• Tower law
  3 KB (567 words) - 08:42, 21 August 2009
• 1994 IMO Problems/Problem 3
  For space-time saving we say that a positive integer is a T-number or simply is T if it has exactly three 1s in his base two representat * <math>f(2^n) = \tbinom n2 = \sum_{i=1}^{n-1} i</math> whence
  7 KB (1,298 words) - 19:59, 9 February 2024
• 1989 IMO Problems/Problem 6
  ...\sum_{x \in \mathbb{U}} (1-I_{(1,3)}(x))(1-I_{(3,1)}(x))(1-I_{(2,4)}(x))(1-I_{(4,2)}(x))</math> So there are <math>2 \times 2 = 4</math> non-zero pairs. Each pair however has 2 ways to rearrange. So the third sum equa
  11 KB (1,928 words) - 12:26, 26 July 2023
• Mock AIME 3 2006-2007/Problem 3
  *[[Mock AIME 3 2006-2007/Problem 2 | Next Problem]] *[[Mock AIME 3 2006-2007]]
  862 bytes (151 words) - 01:35, 30 December 2009
• Characteristic polynomial
  In the context of problem-solving, the characteristic polynomial is often used to find closed forms fo ...{2n} \\ \vdots & \vdots & \ddots & \vdots \\ -a_{n1} & -a_{n2} & \cdots & t-a_{nn}\end{vmatrix}</cmath></center>
  19 KB (3,412 words) - 14:57, 21 September 2022
• KGS math club/solution 9 1
  629 bytes (125 words) - 06:42, 4 March 2010
• Proofs without words
  http://mathoverflow.net/questions/8846/proofs-without-words http://gurmeet.net/computer-science/mathematical-recreations-proofs-without-words/
  55 KB (7,986 words) - 17:04, 20 December 2018
• Pell equation
  ...al number]] <math>D</math>. Generally, <math>D</math> is taken to be square-free, since otherwise we can "absorb" the largest square factor <math>d^2 | ...using [[difference of squares]]. We would have <math>x^2 - Dy^2 = (x+dy)(x-dy) = 1</math>, from which we can use casework to quickly determine the solu
  6 KB (1,076 words) - 10:20, 29 October 2023
• 2010 AIME I Problems/Problem 11
  pair A = (8,10), B = (4.5,6.5), C= (9.75,8.25), F=foot(A,B,C), G=2*F-A; fill(A--B--C--cycle,rgb(0.9,0.9,0.9)); label("$y \le 10 - |x-8|$",(max,Q(max)),E,fontsize(10));
  4 KB (636 words) - 16:46, 25 November 2023
• Mock AIME 2 2010 Answer Key
  ...to 5 less. We can repeat this step again, and we will end up at <math>2010-15=1995</math>. This process can be repeated every for every 15 lemmings, so ...0})) = (b_1, b_2, \ldots, b_{10})</math>, with <math>0 \le b_i \le 2, 3|a_i-b_i</math> for <math>1 \le i \le 10</math>. Given that <math>f</math> can ta
  36 KB (6,214 words) - 20:22, 13 July 2023
• 2007 IMO Problems
  <cmath>\max\{|x_i-a_i|:1\le i\le n\}\ge \dfrac{d}{2} \qquad (*)</cmath> ...be positive integers. Show that if <math>4ab-1</math> divides <math>(4a^2-1)^2</math>, then <math>a=b</math>.
  3 KB (505 words) - 09:24, 10 September 2020
• 2007 IMO Problems/Problem 1
  <cmath>\max\{|x_i-a_i|:1\le i\le n\}\ge \dfrac{d}{2} (*)</cmath> ..._j:i\le j\le n\}</math>, all <math>d_i</math> can be expressed as <math>a_p-a_q</math>, where <math>1\le p\le i\le q \le n</math>.
  4 KB (768 words) - 23:15, 31 March 2010
• 2010 AIME II Problems
  Let <math>K</math> be the product of all factors <math>(b-a)</math> (not necessarily distinct) where <math>a</math> and <math>b</math> ...of <math>P(z)</math> are <math>w+3i</math>, <math>w+9i</math>, and <math>2w-4</math>, where <math>i^2=-1</math>. Find <math>|a+b+c|</math>.
  8 KB (1,246 words) - 21:58, 10 August 2020
• 2010 AIME II Problems/Problem 9
  <cmath>MN=BI-NI-BM=BI-(BM+MH).</cmath> ...he [[Law of cosines]] on <math>\triangle BCI</math>, <math>BI=\sqrt{2^2+1^2-2(2)(1)(\cos(120^\circ))}=\sqrt{7}</math>
  5 KB (857 words) - 22:22, 27 August 2023
• LaTeX:LaTeX on AoPS
  ==== In-line Math Mode ==== ...e>$</code> signs to enclose the math we want to display, and it displays in-line with our text. For example, typing <code>$\sqrt{x} = 5$</code> gives us
  8 KB (1,356 words) - 22:35, 26 June 2020
• 2010 USAMO Problems
  ...ath>h_k</math> is standing directly behind a student with height <math>h_{k-2}</math> or less, the two students are permitted to switch places. Prove th Let <math>q = \dfrac{3p-5}{2}</math> where <math>p</math> is an odd prime, and let
  3 KB (525 words) - 13:44, 4 July 2013
• 2010 USAMO Problems/Problem 3
  <cmath> \prod_{i=1}^{1005}(4i-1) = 3\times 7 \times \ldots \times 4019. </cmath> <math>(a_{2i-1},a_{2i}), 1\le i \le 1005</math>, the product is at most
  11 KB (1,889 words) - 13:45, 4 July 2013
• 2010 USAJMO Problems
  ...be an integer. Find, with proof, all sequences <math>x_1, x_2, \ldots, x_{n-1}</math> of positive integers with the following three properties: <ol style="list-style-type:lower-alpha">
  3 KB (538 words) - 13:55, 16 June 2020
• 2010 USAJMO Problems/Problem 1
  ...closed under multiplication and a non-square times a square is always a non-square. ...lement <math>g_m</math> of the equivalence class <math>C_m</math> is square-free, since if it were divisible by the square of a prime, the quotient woul
  12 KB (2,338 words) - 20:30, 13 February 2024
• 2010 USAJMO Problems/Problem 2
  <math>x_1, x_2, \ldots, x_{n-1}</math> of positive integers with the following <ol style="list-style-type:lower-alpha">
  6 KB (1,177 words) - 23:09, 20 November 2022
• 2010 USAMO Problems/Problem 1
  // Semi-circle: centre O, radius r, diameter A--B. pair P = foot(Y, A, X); dot(P); label("$P$", P, unit(P-Y)); dot(P);
  13 KB (2,178 words) - 14:14, 11 September 2021
• 2010 USAMO Problems/Problem 2
  with height <math>h_{k-2}</math> or less, the two students are permitted to \binom{0}{0} = 1, \quad \binom{n}{k} = \binom{n-1}{k} + \binom{n-1}{k-1}
  5 KB (823 words) - 19:20, 3 October 2017
• Scholarships by State
  ...s primarily for students in the U.S. Some of these scholarships are subject-specific. ...fornia [http://www.cagovernorsscholars.org/ website] (NOTE: granted in 2000-2002)
  9 KB (1,107 words) - 15:32, 16 July 2010
• 2008 iTest Problems/Problem 77
  ...<math>4</math> triangles of size <math>2^{n-1}</math> which include the all-zero "inner" triangle with vertices <math>\Bigg\{\binom{2^{n-1}}{1},\binom{2^{n-1}}{2^{n-1}-1},\binom{2^n-2}{2^{n-1}-1}\Bigg\}</math>
  5 KB (788 words) - 05:15, 28 January 2019
• 2010 IMO Problems/Problem 2
  Let <math>A'</math> be A-excenter <math>\triangle ABC \implies</math> {{IMO box|year=2010|num-b=1|num-a=3}}
  3 KB (525 words) - 14:52, 16 July 2023
• 2010 IMO Problems/Problem 5
  Type 1) Choose a non-empty box <math>B_j</math>, <math>1\leq j \leq 5</math>, remove one coin fro Type 2) Choose a non-empty box <math>B_k</math>, <math>1\leq k \leq 4</math>, remove one coin fro
  3 KB (411 words) - 06:25, 25 June 2018
• 2010 IMO Problems/Problem 6
  <cmath>a_n = \max \{ a_k + a_{n-k} \mid 1 \leq k \leq n-1 \} \ \textrm{ for all } \ n > s.</cmath> :<math>\text{To this end, let’s define an n-type to be a vector T = ⟨t1,...,ts⟩ of nonnegative integers such that}</
  4 KB (786 words) - 08:46, 12 March 2024
• 1971 IMO Problems
  All the faces of tetrahedron <math>ABCD</math> are acute-angled triangles. We consider all closed polygonal paths of the form <math>X ...})(i, j = 1, 2, \cdots, n)</math> be a square matrix whose elements are non-negative integers. Suppose that whenever an element <math>a_{ij} = 0</math>,
  3 KB (495 words) - 13:52, 29 January 2021
• 2010 IMO Problems
  Type 1) Choose a non-empty box <math>B_j</math>, <math>1\leq j \leq 5</math>, remove one coin fro Type 2) Choose a non-empty box <math>B_k</math>, <math>1\leq k \leq 4</math>, remove one coin fro
  4 KB (603 words) - 09:22, 10 September 2020
• 2004 AMC 8 Problems
  How many different four-digit numbers can be formed by rearranging the four digits in <math>2004</ma ...lton’s eighth-grade class wants to participate in the annual three-person-team basketball tournament. Lance, Sally, Joy, and Fred are chosen for the t
  13 KB (1,860 words) - 19:58, 8 May 2023
• Fractional ideal
  2 KB (288 words) - 20:05, 23 January 2017
• 1997 AJHSME Problems
  Ahn chooses a two-digit integer, subtracts it from 200, and doubles the result. What is the l ...speech for her class. Her speech must last between one-half hour and three-quarters of an hour. The ideal rate of speech is 150 words per minute. If
  12 KB (1,702 words) - 12:35, 6 November 2022
• Hölder's Inequality
  ...bda_a = \lambda_b = 1/2</math>, this is the elementary form of the [[Cauchy-Schwarz Inequality]]. ...he left-hand side of the desired inequality but may contribute to the right-hand side.
  4 KB (762 words) - 00:11, 18 June 2023
• Mobius function
  ...>b</math>. Via a combinatorial argument using the [[Principle of Inclusion-Exclusion]] (PIE), it can be seen that the number of integers less than or e ...u{p_i}+\sum_{i,j}\mu(p_ip_j)+\ldots=1-k+\binom{k}{2}-\binom{k}{3}-\ldots=(1-1)^k=0^k=0</math>.</center>
  5 KB (910 words) - 02:58, 1 March 2022
• 2012 IMO Problems/Problem 1
  Let <math>R</math> be the radius of the A-excircle. Because <math>CM = CL</math>, we have <math>CML</math> isosceles a ...w of Sines on triangle <math>BFM</math> and <math>ABC</math> and the double-angle formula for sine, we have
  7 KB (1,189 words) - 01:22, 19 November 2023
• Energy
  636 bytes (99 words) - 21:44, 12 April 2021
• 2002 AMC 8 Problems
  A birdbath is designed to overflow so that it will be self-cleaning. Water flows in at the rate of 20 milliliters per minute and drains Right isosceles triangles are constructed on the sides of a 3-4-5 right triangle, as shown. A capital letter represents the area of each tri
  15 KB (2,102 words) - 11:35, 19 February 2024
• 1994 USAMO Problems/Problem 4
  ...timal equation is <math>a_n = \sqrt{n} - \sqrt{n-1} = 1/(\sqrt{n} + \sqrt{n-1})</math>. Note that this is strictly greater than <math>1/(2\sqrt{n})</mat {{USAMO box|year=1994|num-b=3|num-a=5}}
  1 KB (210 words) - 13:30, 4 July 2013
• 2011 AIME I Problems/Problem 11
  ...ed by <math>d = \text{ord}_{125}(2).</math> To begin with, we'll use a well-known property of the order to get a bound on <math>d.</math> <cmath>2^b\left(2^{a-b} - 1\right) \equiv 0 \mod 1000</cmath>
  11 KB (1,668 words) - 22:10, 24 February 2023
• 2011 AIME I Problems/Problem 7
  ...h> to be anything we want (including 0), all we care about is that <math>(m-1)\text{ }|\text{ } 2010</math> which happens <math>\boxed{016}</math> times ...<math>(m-1)Q(m) = 2010</math>. Thus, if <math>P(m) = 0</math>, then <math>m-1 | 2010</math> .
  10 KB (1,673 words) - 19:27, 8 April 2024
• 2011 AIME II Problems/Problem 8
  <center>[[File:2011_AIME_II_-8.png‎]]</center> <cmath>(z^6-2^{18})(z^6+2^{18})=0</cmath>
  5 KB (805 words) - 18:46, 27 January 2024
• 2011 AIME II Problems
  The degree measures of the angles in a convex 18-sided polygon form an increasing arithmetic sequence with integer values. Fi
  8 KB (1,301 words) - 08:43, 11 October 2020
• 2011 AIME II Problems/Problem 14
  Each POSITION in the 30-position permutation is uniquely defined by an ordered triple <math>(i, j, k ...or the number 2, there are (2-1) choices for i, (3-1) choices for j, and (5-1) choices for k. Thus, there are 1*2*4=8 possible placements for the number
  10 KB (1,581 words) - 22:09, 27 August 2023
• 2011 AIME II Problems/Problem 9
  Let <math>x_1, x_2, ... , x_6</math> be non-negative real numbers such that <math>x_1 +x_2 +x_3 +x_4 +x_5 +x_6 =1</math> ...n is the product of three non-negative terms whose sum is equal to 1. By AM-GM this product is at most <math>\frac1{27}</math>. Since we have added at l
  3 KB (466 words) - 15:06, 16 January 2023
• 1998 USAMO Problems
  ...> (<math>1\leq i\leq 999</math>) so that for all <math>i</math>, <math>|a_i-b_i|</math> equals <math>1</math> or <math>6</math>. Prove that the sum <cmath> |a_1-b_1|+|a_2-b_2|+\cdots +|a_{999}-b_{999}| </cmath>
  3 KB (486 words) - 06:11, 24 November 2020
• 1999 AMC 8 Problems
  ...is at milepost 160. There is a service center on the highway located three-fourths of the way from the third exit to the tenth exit. At what milepost w Walter can buy eggs by the half-dozen. How many half-dozens should he buy to make enough cookies? (Some eggs and some cookies may
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• KGS math club/solution 11 5
  ...<math>\frac{1}{y}, \frac{2}{y}, \ldots, \frac{y}{y}=1</math>, i.e. <math>y-x</math> needs to divide <math>y</math> perfectly. This condition is also en ...now that for every pair <math>i</math> and <math>j</math> we know <math>s_j-s_i</math> divides <math>s_j</math> exactly an integer number of times. We w
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• 2011 USAJMO Problems/Problem 3
  ..._1), l(P_2), l(P_3)</math> are <cmath>y=2a_1x-a_1^2, y=2a_2x-a_2^2, y=2a_3x-a_3^2,</cmath> respectively. It is easy to deduce that the three points of i &= \frac{\Im(1+6a_ni-12a_n^2-8a_n^3i)}{\Re(1+6a_ni-12a_n^2-8a_n^3i)}\\
  15 KB (2,593 words) - 13:37, 29 January 2021
• 2011 USAMO Problems/Problem 6
  ...ach intersection of three of the 11 sets. Since each pair of sets is in 9 3-way intersections&mdash;one with each of the 9 remaining sets&mdash;any two ...he number of <math>b_k>0</math>, where <math>1\le k\le225</math>. By the QM-AM inequality, we know <math>\frac{\sum_{k=1}^{225}b_k^2}n\ge\Bigg(\frac{\su
  7 KB (1,209 words) - 12:50, 25 August 2023
• 2011 AMC 10B Problems/Problem 16
  ...}} = \frac{1}{2+2\sqrt{2}} \times \frac{2-2\sqrt{2}}{2-2\sqrt{2}} = \frac{2-2\sqrt{2}}{-4} = \boxed{\textbf{(A)} \frac{\sqrt{2}-1}{2}}</cmath> ...he denominator reduces to <math>-4</math> and the fraction is <math>\frac{2-2\sqrt{2}}{-4}=\frac{\sqrt{2}-1}{2}</math> which is <math>\boxed{\textbf{(A)
  4 KB (606 words) - 06:14, 25 June 2022
• 2001 AMC 8 Problems/Problem 11
  Using the diagram above, the figure can be divided along the x-axis into two familiar regions that do not overlap: a right triangle and a r {{AMC8 box|year=2001|num-b=10|num-a=12}}
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• 2002 AMC 10B Problems/Problem 23
  ...ore, <math>a_m = a_{m-1} + m, a_{m-1}=a_{m-2}+(m-1), a_{m-2} = a_{m-3} + (m-2)</math>, and so on until <math>a_2 = a_1 + 2</math>. ...Sides of all of these equations gives <math>a_m + a_{m-1} + a_{m-2} + a_{m-3} + \cdots + a_2</math>.
  5 KB (781 words) - 06:38, 23 October 2023
• Olympiad books
  Here is a list of '''Olympiad Books''' that have Olympiad-level problems used to train students for future [[mathematics]] competition ...ric/dp/1794193928/ref=tmm_pap_swatch_0?_encoding=UTF8&qid=1556093238&sr=1-1-catcorr Advanced Olympiad Inequalities: Algebraic & Geometric Olympiad Inequ
  17 KB (2,251 words) - 00:47, 28 March 2024
• 2007 AMC 10B Problems/Problem 16
  {{AMC12 box|year=2007|ab=B|num-b=11|num-a=13}} {{AMC10 box|year=2007|ab=B|num-b=15|num-a=17}}
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• Block walking
  9 KB (1,228 words) - 21:47, 26 February 2022
• Mock Amc 12
  {| class="wikitable" style="text-align:center;width:100%" | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=9572 1-5]
  18 KB (2,206 words) - 19:41, 24 December 2020
• 2003 AMC 8 Problems
  Blake and Jenny each took four 100-point tests. Blake averaged 78 on the four tests. Jenny scored 10 points hig When a fair six-sided die is tossed on a table top, the bottom face cannot be seen. What is
  16 KB (2,236 words) - 12:02, 19 February 2024
• 1997 USAMO Problems
  ...\text{ if }x_{k-1}=0\\ \left\{\frac{p_{k}}{x_{k-1}}\right\}&\text{ if }x_{k-1}\ne0\end{cases} </math>
  3 KB (507 words) - 14:31, 12 April 2023
• 1997 USAMO Problems/Problem 6
  <cmath>a_n\ge a_{n-1}+a_1\ge (a_{n-2}+a_1)+a_1\ge \cdots \ge na_1.</cmath> <cmath>a_k\le a_{k-1}+(a_1+1)\le (a_{k-2}+(a_1+1))+(a_1+1)\le\cdots\le ka_1+(k-1).</cmath>
  6 KB (1,107 words) - 14:12, 12 April 2023
• 2011 IMO Problems
  Let <math>f : \mathbb{R} \rightarrow \mathbb{R}</math> be a real-valued function defined on the set of real numbers that satisfies ...a balance and <math>n</math> weights of weight <math>2^0, 2^1,\ldots, 2^{n-1}</math> . We are to place each of the <math>n</math> weights on the balanc
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• 2011 IMO Problems/Problem 1
  .../math> isn't a good couple. We have in total 6 couples. So <math>n_A \leq 6-2=4</math>. ...So we have <math>2a+x | 2a+2b</math> and <math>a+b-x|2a+2b \Rightarrow a+b-x|2x</math>.
  9 KB (1,718 words) - 23:08, 26 June 2014
• 1987 USAMO Problems
  ...olutions to <math>(m^2+n)(m + n^2)= (m - n)^3</math>, where m and n are non-zero integers. ...ctors of <math>\Delta ABC</math> form a right-angled triangle. If the right-angle is at <math>X</math>, where <math>AX</math> is the bisector of <math>\
  2 KB (356 words) - 18:42, 18 July 2016
• 2002 AMC 8 Problems/Problem 15
  {{AMC8 box|year=2002|num-b=14|num-a=16}}
  1 KB (197 words) - 04:47, 25 November 2019
• 1999 AMC 8 Problems/Problem 25
  ...ake an underestimate by using the current diagram. All triangles are right-isosceles triangles. {{AMC8 box|year=1999|num-b=24|after=Last Question}}
  4 KB (655 words) - 05:21, 12 March 2024
• 1997 AJHSME Problems/Problem 12
  {{AJHSME box|year=1997|num-b=11|num-a=13}}
  1 KB (188 words) - 12:37, 2 May 2021
• 1979 USAMO Problems/Problem 5
  ...do not contain <math>k, m</math>. Because the latter subset has <math>(n-t-2) < (n+1)-t - 1</math> elements, we may use infinite descent to contradict {{USAMO box|year=1979|num-b=4|after=Last Question}}
  2 KB (410 words) - 21:04, 11 November 2015
• 2003 AMC 8 Problems/Problem 18
  {{AMC8 box|year=2003|num-b=17|num-a=19}}
  2 KB (250 words) - 22:17, 5 January 2024
• 1984 AHSME Problems/Problem 30
  Now we multiply <math>S</math> by <math>1-w</math>: <cmath>S(1-w)=\sum_{i=1}^{9}\sum_{j=i}^{9} w^j(1-w)</cmath>
  3 KB (447 words) - 21:21, 17 July 2020
• 1985 AHSME Problems
  ...ame as <math>a(b-c)</math> in ordinary algebraic notation. If <math>a\div b-c+d</math> is evaluated in such a language, the result in ordinary algebraic ...\qquad \mathrm{(D) \ } \frac{a}{b-c+d} \qquad \mathrm{(E) \ }\frac{a}{b-c-d} </math>
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• 1985 AHSME Problems/Problem 9
  label(string(16*i+15-2*j), (2*j,-2*i-1)); label(string(15-2*i), (2*(i-1),-2*1.35));
  1 KB (213 words) - 21:08, 19 March 2024
• 1985 AHSME Problems/Problem 13
  {{AHSME box|year=1985|num-b=12|num-a=14}}
  2 KB (287 words) - 21:05, 19 March 2024
• 2010 AMC 8 Problems
  <math>\textbf{(B)}\ \text{non-square rhombus} </math> <math>\textbf{(C)}\ \text{non-square rectangle}</math>
  18 KB (2,768 words) - 21:05, 9 January 2024
• 2011 AMC 8 Problems
  ...h> apples at a cost of <math> 50 </math> cents per apple. She paid with a 5-dollar bill. How much change did Margie receive? A fair 6-sided die is rolled twice. What is the probability that the first number tha
  16 KB (2,371 words) - 17:34, 9 January 2024
• 2011 AMC 8 Problems/Problem 19
  {{AMC8 box|year=2011|num-b=18|num-a=20}}
  1 KB (158 words) - 15:06, 17 December 2023
• 1973 Canadian MO Problems
  Find all real numbers that satisfy the equation <math>|x+3|-|x-1|=x+1</math>. (Note: <math>|a| = a</math> if <math>a\ge 0; |a|=-a</math> if Prove that <math>n+h(1)+h(2)+h(3)+\cdots+h(n-1) = nh(n)\qquad</math> for <math>n=2,3,4,\ldots</math>
  4 KB (540 words) - 18:23, 8 October 2014
• 1973 Canadian MO Problems/Problem 7
  {{CanadaMO box|year=1973||num-b=6||num-a=1}}
  2 KB (351 words) - 18:56, 27 May 2022
• Euler's Four-Square Identity
  ...)^2</cmath> <cmath>+(x_1y_3-x_3y_1+x_4y_2-x_2y_4)^2</cmath> <cmath>+(x_1y_4-x_4y_1 + x_2y_3 - x_3y_2)^2.</cmath> Then Euler's Four-Square Identity simply reads <math>|XY|^2 = |X|^2 |Y|^2</math>; i.e. the qua
  1 KB (179 words) - 23:24, 5 November 2019
• 2011 AMC 8 Problems/Problem 7
  {{AMC8 box|year=2011|num-b=6|num-a=8}}
  4 KB (557 words) - 00:30, 4 January 2023
• 1971 Canadian MO Problems/Problem 9
  {{Old CanadaMO box|num-b=8|num-a=10|year=1971}}
  1 KB (204 words) - 18:09, 19 September 2012
• 1989 AHSME Problems
  ...th>n</math> between <math>1</math> and <math>100</math> does <math> x^{2}+x-n </math> factor into the product of two linear factors with integer coeffic ...e following is closest to the number of westbound vehicles present in a 100-mile section of highway?
  15 KB (2,247 words) - 13:44, 19 February 2020
• 2012 AMC 10A Problems/Problem 25
  ...ints satisfying <math>|x-y|\geq1</math>, <math>|y-z|\geq1</math>, <math>|z-x|\geq1</math>. ...n\geq x \geq y \geq z \geq 0</math>, we have <math>x-y\geq1</math>, <math>y-z\geq1</math>.
  13 KB (2,133 words) - 01:22, 6 February 2024
• 2012 AMC 12A Problems
  a^{2}+3b^{2}+3c^{2}-3ab-2ac-2bc&=-1997 (0.\underbrace{20101\cdots 0101}_{k+2\text{ digits}})^{a_{k-1}} & \text{if }k\text{ is odd,}\\
  14 KB (2,197 words) - 13:34, 12 August 2020
• 2012 AMC 12A Problems/Problem 22
  ...ath>k</math> is <math>7</math>. The answer to this problem is then <math>27-7=20</math> ... <math>\framebox{C}</math>. {{AMC12 box|year=2012|ab=A|num-b=21|num-a=23}}
  5 KB (940 words) - 17:13, 4 April 2020
• 2014 USAMO Problems
  Let <math>a,b,c,d</math> be real numbers such that <math>b-d \ge 5</math> and all zeros <math>x_1, x_2, x_3,</math> and <math>x_4</math ...athbb{Z} \rightarrow \mathbb{Z}</math> such that <cmath>xf(2f(y)-x)+y^2f(2x-f(y))=\frac{(f(x))^2}{x}+f(yf(y))</cmath> for all <math>x, y \in \mathbb{Z}<
  3 KB (496 words) - 02:14, 14 February 2024
• 2012 AMC 12A Problems/Problem 23
  {{AMC12 box|year=2012|ab=A|num-b=22|num-a=24}}
  4 KB (658 words) - 20:37, 19 January 2021
• 2011 IMO Shortlist Problems/A1
  500 bytes (91 words) - 06:44, 28 October 2013
• 1989 AHSME Problems/Problem 13
  {{AHSME box|year=1989|num-b=12|num-a=14}}
  2 KB (242 words) - 07:51, 22 October 2014
• 2012 AMC 12B Problems
  ...abbits. How many more students than rabbits are there in all 4 of the third-grade classrooms? In order to estimate the value of <math>x-y</math> where <math>x</math> and <math>y</math> are real numbers with <math
  20 KB (2,681 words) - 09:47, 29 June 2023
• 2012 AMC 12B Problems/Problem 19
  [[File:2012_AMC-12B-19‎.jpg]] .... This will occur when the distance from <math>(x,0,0)</math> to <math>(1,1-x, 1)</math> is <math>(x)(\sqrt{2}</math>).
  5 KB (815 words) - 21:59, 19 September 2023
• 2015 AIME I Problems
  ...ates a sequence according to the rule <math>a_n=a_{n-1}\cdot | a_{n-2}-a_{n-3} |</math> for all <math>n\ge 4</math>. Find the number of such sequences f Let <math>f(x)</math> be a third-degree polynomial with real coefficients satisfying
  10 KB (1,615 words) - 21:48, 13 January 2024
• 2012 AIME I Problems
  ...eaches the first of the hitching posts that are conveniently located at one-mile intervals along their route, he ties Sparky to the post and begins walk ...math>\mathcal{T}</math> be the set of all numbers of the form <math>\frac{x-256}{1000}</math>, where <math>x</math> is in <math>\mathcal{S}</math>. In o
  10 KB (1,617 words) - 14:49, 2 June 2023
• 2012 AIME I Problems/Problem 7
  {{AIME box|year=2012|n=I|num-b=6|num-a=8}}
  6 KB (1,058 words) - 01:49, 25 November 2023
• 2007 iTest Problems/Problem 15
  {{iTest box|year=2007|num-b=14|num-a=16}}
  3 KB (539 words) - 21:01, 29 July 2018
• Mock AIME II 2012 Problems/Problem 8
  .... We can use the law of cosines to find <math>EC</math>, <math>EC^2=9^2+9^2-2(\cos 135)\cdot 9\cdot 9=2(81)+81\sqrt{2}=162+81\sqrt{2}</math>.
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• Pascal (unit)
  309 bytes (52 words) - 16:59, 28 February 2020
• 1979 USAMO Problems
  Determine all non-negative integral solutions <math>(n_1,n_2,\dots , n_{14})</math> if any, ap
  2 KB (321 words) - 13:04, 24 December 2015
• 2000 USAMO Problems/Problem 6
  '''Lemma 1:''' If <math>r_1, r_2, \ldots , r_n</math> are non-negative reals and <math>x_1, x_2, \ldots x_n</math> are reals, then <cmath>\sum_{i}(r_{i}-r_{i-1})\biggl(\sum_{j=i}^{n}x_{j}\biggr)^{2}\, .</cmath>
  3 KB (556 words) - 15:08, 15 July 2021
• 2000 USAMO Problems/Problem 3
  <cmath>f(B,W,R) = \min(W+f(B-1,W,R),2R+f(B,W-1,R),3B+f(B,W,R-1)) </cmath> ...o <math>W+3R(B-1) < BW</math>, whence <math>3R < W</math>. But <math>W+3R(B-1) < 3RB</math>, so that <math>W < 3R</math>, a contradiction.
  4 KB (800 words) - 21:24, 28 August 2019
• 2012 USAMO Problems/Problem 6
  ...=1}^{n} x_i^2\right)+2^{n-1}\left(\sum_{1\leq i<j\leq n} x_ix_j\right)=2^{n-2}.</cmath> Since <math>S_A^2\geq 0</math>, it follows that at most <math>2^{n-2}/\lambda^2</math> sets <math>A\subseteq N</math> have <math>|S_A|\geq \lam
  6 KB (1,081 words) - 13:37, 21 June 2023
• 2015 IMO Problems/Problem 1
  <math>\mathcal{S}</math> is <i>centre-free</i> if for any three points <math>A</math>, <math>B</math>, <math>C</ma <ol style="list-style-type: lower-latin;">
  4 KB (773 words) - 08:14, 19 July 2016
• 1966 AHSME Problems/Problem 14
  {{AHSME box|year=1966|num-b=13|num-a=15}}
  3 KB (434 words) - 23:00, 12 December 2019
• 2006 SMT/Advanced Topics Problems
  Ten teams of five runners each compete in a cross-country race. A runner finishing in <math> n\text{th} </math> place contribu Evaluate: <math> \sum_{n=1}^{\infty}\arctan\left(\frac{1}{n^2-n+1}\right) </math>
  4 KB (596 words) - 20:09, 27 May 2012
• 2006 SMT/Team Problems
  ...th>, for terms <math> n\ge3 </math>, <math> S_n=\sum_{i=1}^{n-1}i\cdot S_{n-i} </math>. For example, if the first two elements are <math> 2 </math> and ...igin tangent to the graph of the upper part of the hyperbola <math> y^2=x^2-x+1 </math> in the first quadrant. This ray makes an angle of <math> \theta
  6 KB (910 words) - 17:32, 27 May 2012
• 2006 SMT/Advanced Topics Problems/Problem 3
  434 bytes (58 words) - 20:20, 27 May 2012
• 2010 AMC 8 Problems/Problem 3
  ...$10. 17 is <math>\frac{17}{10}\cdot100=170\%</math> of 10, and is <math>170-100=70\%</math> more than 10. {{AMC8 box|year=2010|num-b=2|num-a=4}}
  5 KB (847 words) - 19:59, 5 February 2023
• 2012 AMC 12B Problems/Problem 16
  ...But this overcounts the case 2 with 36 ways. So total ways are <math>168-36=132</math>. {{AMC10 box|year=2012|ab=B|num-b=23|num-a=25}}
  8 KB (1,464 words) - 00:31, 6 September 2023
• 2005 IMO Shortlist Problems/N2
  ...a_1,a_2,\ldots a_{n}</math> are all in the closed interval <math>[a_k,a_k+n-1]</math>, and hence <math>a_1,a_2,\ldots a_n</math> is a permutation of <ma
  2 KB (333 words) - 10:43, 1 September 2012
• 1998 USAMO Problems/Problem 6
  C = (sqrt(2)/2-1,sqrt(2)/2); J = (1-sqrt(2)/2, sqrt(2)/2);
  5 KB (871 words) - 18:59, 10 May 2023
• 1989 AHSME Problems/Problem 30
  {{AHSME box|year=1989|num-b=29|after=Last Question}}
  2 KB (293 words) - 17:13, 24 August 2020
• 1989 AHSME Problems/Problem 5
  ...the frame can be dissected into <math>20\cdot10</math> L shapes, the right-hand border, and the bottom border: p = (i+0.75,j)--(i,j)--(i,j-0.75);
  2 KB (295 words) - 07:43, 22 October 2014
• 2012 IMO Problems
  Find all positive integers n for which there exist non-negative integers <math>a_1</math>, <math>a_2</math>, <math>\ldots</math> ,
  5 KB (834 words) - 11:00, 24 February 2021
• 2020 Mock Combo AMC 10
  ...ths from the origin <math>(0, 0)</math> to a point on the line <math>y=2020-2x</math> such that each step is from <math>(x, y)</math> to either <math>(x <cmath>r_1 = 1</cmath><cmath>r_n = nr_{n-1} + n</cmath>Let <math>S</math> be the sum of all <math>n</math> such that
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• 2012 AMC 8 Problems
  Peter's family ordered a 12-slice pizza for dinner. Peter ate one slice and shared another slice equally Isabella must take four 100-point tests in her math class. Her goal is to achieve an average grade of 95
  13 KB (1,835 words) - 08:51, 8 March 2024
• 2012 AMC 8 Problems/Problem 5
  https://youtu.be/m4g-Nmot-c8 ~savannahsolver {{AMC8 box|year=2012|num-b=4|num-a=6}}
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• 2004 AMC 8 Problems/Problem 23
  ...he lives at corner <math>J</math>. Which graph could represent her straight-line distance from home? {{AMC8 box|year=2004|num-b=22|num-a=24}}
  2 KB (308 words) - 17:26, 25 July 2021
• 2013 AMC 10A Problems/Problem 25
  ...ed to subtract <math>\dbinom{4}{2} -1 = 5</math> from this count, <math>70-5 = 65</math>. Note that diagonals like <math>\overline{AD}</math>, <math>\ ..., not double-counted -which the calculation by Lcz accounts for- but triple-counted. Thus, taking away one for each of these points is more than enough
  6 KB (1,054 words) - 09:21, 28 December 2021
• 2013 AMC 10B Problems/Problem 22
  {{AMC10 box|year=2013|ab=B|num-b=21|num-a=23}}
  4 KB (746 words) - 17:29, 30 September 2023
• 2013 AMC 12B Problems/Problem 20
  <math>\textbf{(A)} \ 2-2\sqrt{2}\qquad\textbf{(B)}\ 3\sqrt{3}-6\qquad\textbf{(C)}\ 3\sqrt{2}-5\qqua ...points to be parallel, so that <math>\frac{g^2-f^2}{g-f} = \frac{j^2-h^2}{j-h}</math>, or <math>g+f = j+h</math>.
  2 KB (375 words) - 00:54, 28 September 2021
• 2013 AMC 10B Problems
  ...ir parents is 33. What is the average age of all of these parents and fifth-graders? ...y attempted 50% more two-point shots than three-point shots. How many three-point shots did they attempt?
  12 KB (1,926 words) - 21:54, 6 October 2022
• 1982 USAMO Problems/Problem 5
  {{USAMO box|year=1982|num-b=4|after=Last Question}}
  3 KB (524 words) - 16:08, 11 July 2018
• 1991 AHSME Problems
  ...>, <math>b</math>, and <math>c</math> we define <math>f(a,b,c)=\frac{c+a}{c-b}</math>, then <math>f(1,-2,-3)</math> is ...\qquad \textbf{(C) }3-\pi \qquad \textbf{(D) }3+\pi \qquad \textbf{(E) }\pi-3 </math>
  16 KB (2,451 words) - 04:27, 6 September 2021
• 2013 AIME II Problems/Problem 8
  A=expi(pi); B=expi(pi-a); C=expi(a); D=expi(0); E=expi(-a); F=expi(pi+a); ...means <cmath>\sin{\theta}=\frac{BY}{r}=\frac{11}{r}</cmath> <cmath>\sin{(90-2\theta)}=\frac{BX}{r}=\frac{10}{r}</cmath>
  8 KB (1,357 words) - 09:23, 11 March 2024
• 2013 USAMO Problems/Problem 2
  ...ending at <math>A</math>, without repeating a move. Prove that <math>a_{n-1}+a_n=2^n</math> for all <math>n\geq 4</math>. label("$\dots$",(8-1,1));
  14 KB (2,076 words) - 20:29, 10 July 2023
• 1989 AHSME Problems/Problem 27
  ...)^3\cdot\left(\frac{1}{1+k}\right)^2\ =\ k^5\cdot(1+k)\cdot\left(\frac{1}{1-k^2}\right)^3</math> ...k^6+...)</math>, the last part having general term <math>\tbinom{j}{2}k^{2j-4}</math>
  2 KB (285 words) - 12:39, 5 July 2021
• 2014 AIME II Problems/Problem 15
  ...ath>x_{n+2^{k-1}} = P_k \cdot x_{n}</math> for all <math>0 \leq n \leq 2^{k-1} - 1.</math> .../math> is now the least <math>n</math> such that <cmath>P_{k+1} = p(x_{2^{n-1}}), </cmath> in which <math>P_a = p(x_n)</math> where <math>a > k</math> i
  4 KB (770 words) - 17:44, 11 October 2023
• 1986 USAMO Problems/Problem 3
  ...smallest integer <math>n</math>, greater than one, for which the root-mean-square of the first <math>n</math> positive integers is an integer? <math>\mathbf{Note.}</math> The root-mean-square of <math>n</math> numbers <math>a_1, a_2, \cdots, a_n</math> is defin
  3 KB (554 words) - 13:19, 19 July 2023
• 2013 IMO Problems/Problem 1
  <math>1+\frac{2^k-1}{n}=(1+\frac{1}{m_1})(1+\frac{1}{m_2})...(1+\frac{1}{m_k})</math>. '''Base case:''' If <math>k = 1</math> then <math>1 +\frac{2^1-1}{n} = 1 + \frac{1}{n}</math>, so the claim is true for all positive intege
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• 1971 IMO Problems/Problem 5
  ...l)</math> iff either <math>v_i=v_k,|w_j-w_l|=1</math> or <math>w_j=w_l,|v_i-v_k|=1</math>. clearly if the condition holds then the adjacency is satisfie ...>. since we can form the graph <math>K_2</math>, we can form <math>Q_n=Q_{n-1}\times K_2</math>(the unit cube) and that solves the problem.
  4 KB (749 words) - 14:09, 29 January 2021
• 2013 AMC 8 Problems
  A sign at the fish market says, "50% off, today only: half-pound packages for just \$3 per package." What is the regular price for a fu What is the value of <math>4 \cdot (-1+2-3+4-5+6-7+\cdots+1000)</math>?
  15 KB (2,162 words) - 20:05, 8 May 2023
• 2013 AMC 8 Problems/Problem 18
  Isabella uses one-foot cubical blocks to build a rectangular fort that is <math>12</math> feet ...</math>. Therefore, the number of blocks contained in the fort is <math>600-320=\boxed{\textbf{(B)}\ 280}</math>
  3 KB (412 words) - 22:30, 18 December 2023
• 2013 AMC 8 Problems/Problem 21
  https://youtu.be/OOdK-nOzaII?t=2679 ==Video Solution for Problems 21-25 By Rohan Dhillon==
  2 KB (263 words) - 19:21, 19 December 2023
• 2013 AMC 8 Problems/Problem 22
  for (int i=0; i<3; ++i){for (int j=0; j>-2; --j){if ((i-j)<3){add(corner,(50i,50j));}}} ==Video Solution for Problems 21-25==
  2 KB (256 words) - 19:06, 22 January 2024
• 2013 AMC 8 Problems/Problem 24
  ==Video Solution for Problems 21-25= Shaded area = 12-8=4 <br>
  6 KB (1,076 words) - 22:15, 20 December 2023
• 1958 AHSME Problems/Problem 49
  <cmath>\sum\limits_{k=0}^{10} \binom{10}{k} (a+b)^k c^{10-k}</cmath> {{AHSME 50p box|year=1958|num-b=48|num-a=50}}
  1 KB (191 words) - 00:59, 1 January 2024
• 1981 AHSME Problems
  If <math> \dfrac{x}{x-1}=\dfrac{y^2+2y-1}{y^2+2y-2} </math>, then <math>x</math> equals ...tbf{(C)}\ y^2+2y+2 \qquad \\ \textbf{(D)}\ y^2+2y+1\qquad\textbf{(E)}\ -y^2-2y+1 </math>
  17 KB (2,633 words) - 15:44, 16 September 2023
• 1968 AHSME Problems
  ...lue of <math>x</math> such that <math>64^{x-1}</math> divided by <math>4^{x-1}</math> equals <math>256^{2x}</math> is: ...hrough the point <math>(0,4)</math> is perpendicular to the line <math>x-3y-7=0</math>. Its equation is:
  16 KB (2,571 words) - 14:13, 20 February 2020
• 2014 AMC 10A Problems/Problem 13
  Since <math>\angle C=360</math>, <math>\angle GCH=360-90-90-60=120</math>. ...itude from <math>C</math> to <math>GH</math> allows to create a <math>30-60-90</math> triangle since <math>\triangle GCH</math> is isosceles. This means
  7 KB (1,040 words) - 02:07, 8 October 2022
• 2014 AMC 10A Problems/Problem 14
  draw((-3,0)--(9,0),linewidth(1),Arrows); //x-axis draw((0,-13)--(0,13),linewidth(1),Arrows); //y-axis
  6 KB (1,001 words) - 13:07, 25 July 2022
• 2014 AMC 10A Problems/Problem 16
  ...math>, <math>IJ = 2x</math>. So each perpendicular is length <math>\dfrac{1-2x}{2}</math>. So taking our numbers and plugging them into <math> [ABCD] = {{AMC10 box|year=2014|ab=A|num-b=15|num-a=17}}
  9 KB (1,411 words) - 19:51, 25 July 2023
• 2013 Mock AIME I Problems/Problem 14
  ...{2011}+16x^{2010}+\cdots + 4052169x + 4056196 = \sum_{j=1}^{2014}j^2x^{2014-j}.</math> If <math>a_1, a_2, \cdots a_{2013}</math> are its roots, then com
  783 bytes (105 words) - 19:04, 15 December 2020
• 1998 PMWC Problems/Problem I2
  for(int j = n-i; j > 0; --j){ We want <math>n^2-(n-1)^2=11</math> or <math>2n-1=11</math> for <math>n=6</math>. Therefore the two square numbers are <math
  1 KB (231 words) - 15:06, 23 July 2014
• 2014 USAJMO Problems/Problem 2
  Let <math>\triangle{ABC}</math> be a non-equilateral, acute triangle with <math>\angle A=60^{\circ}</math>, and let < i1=2*O-H;
  7 KB (1,197 words) - 19:40, 28 February 2023
• 2014 USAMO Problems/Problem 6
  2 KB (361 words) - 11:55, 25 June 2020
• 2007 iTest Problems/Problem 12
  This problem is phrased oddly, leading to the non-standard result of <math>\frac{3}{2}</math> instead of the standard <math>\f {{iTest box|year=2007|num-b=11|num-a=13}}
  3 KB (500 words) - 04:44, 10 June 2018
• 2006 IMO Problems/Problem 1
  ...ath>\triangle ABC\, </math> at point <math>J</math>. Then, by the Incenter-Excenter Lemma, <math>JB=JC=JI=JP</math>. {{IMO box|year=2006|before=First Problem|num-a=2}}
  1 KB (247 words) - 01:02, 19 November 2023
• 2014 AMC 12B Problems
  ...l road, the next <math> 20 </math> miles on pavement, and the remaining one-fifth on a dirt road. In miles, how long was Randy's trip? Doug constructs a square window using <math> 8 </math> equal-size panes of glass, as shown. The ratio of the height to width for each pan
  13 KB (2,066 words) - 14:08, 1 November 2022
• 2014 AMC 10B Problems
  ...vel road, the next <math>20</math> miles on pavement, and the remaining one-fifth on a dirt road. In miles how long was Peter's trip? Camden constructs a square window using <math> 8 </math> equal-size panes of glass, as shown. The ratio of the height to width for each pan
  13 KB (2,011 words) - 21:54, 8 November 2022
• 2014 AMC 10B Problems/Problem 18
  {{AMC10 box|year=2014|ab=B|num-b=17|num-a=19}}
  2 KB (369 words) - 17:44, 30 January 2021
• 2014 AMC 12B Problems/Problem 11
  {{AMC12 box|year=2014|ab=B|num-b=10|num-a=12}}
  2 KB (277 words) - 10:13, 3 March 2015
• 2014 AMC 12B Problems/Problem 22
  ...is on pad <math>N</math>, <math>0<N<10</math>, it will jump to pad <math>N-1</math> with probability <math>\frac{N}{10}</math> and to pad <math>N+1</ma ...es b=\frac{10a-\emptyset}{9},c=\frac{5b-a}{4},d=\frac{10c-3b}{7},e=\frac{5d-2c}{3}=1/2</cmath>
  7 KB (1,082 words) - 22:35, 3 April 2024
• 2014 AMC 12B Problems/Problem 21
  ...(2-sqrt(3),1), G=(1,sqrt(3)/2), H=(2.5-sqrt(3),1), J=(.5,0), K=(2-sqrt(3),1-sqrt(3)/2); ...so we must have <math>\triangle HGJ\cong\triangle JLH</math> by Hypotenuse-Leg congruence. From this congruence we have <math>LJ=HG=BE</math>.
  8 KB (1,307 words) - 02:07, 1 January 2023
• 2014 AIME I Problems
  Jon and Steve ride their bicycles along a path that parallels two side-by-side train tracks running the east/west direction. Jon rides east at <math>2 ...<math>2014</math>, respectively, and each graph has two positive integer x-intercepts. Find <math>h</math>.
  9 KB (1,472 words) - 13:59, 30 November 2021
• 2014 AIME I Problems/Problem 6
  ...<math>2014</math>, respectively, and each graph has two positive integer x-intercepts. Find <math>h</math>. ...respectively. Notice that because the two parabolas have to have positive x-intercepts, <math>h\ge32</math>.
  7 KB (1,158 words) - 20:50, 8 December 2021
• 2014 AIME I Problems/Problem 13
  ...(A,D,O,dir(90+theta)); J=extension(B,C,O,dir(-90+theta)); H=(A+I)/2; F=H+(J-I); R=midpoint(H--F); S=midpoint(C--D); T=(R.x, A.y); draw(A--B--C--D--cycle Because these areas are in the ratio <math>411:405=(408+3):(408-3)</math>, it follows that <cmath>\frac{\frac 1{400}\sin 2\theta}{\frac 3{10
  9 KB (1,404 words) - 21:07, 13 October 2023
• 2014 AIME I Problems/Problem 15
  ...rc</math>. But <math>DF=FE</math>, so <math>\triangle DEF</math> is a 45-45-90 triangle. Letting <math>DG=3x</math>, we have that <math>EG=4x</math>, <m ...while the segment <math>CJ</math> is <math>2x\sqrt{2}</math> since its 3-4-5 again. Now adding all those segments together we can find that <math>AC=5=
  10 KB (1,643 words) - 22:30, 28 January 2024
• 1993 AHSME Problems/Problem 29
  {{AHSME box|year=1993|num-b=28|num-a=30}}
  2 KB (322 words) - 15:07, 6 July 2022
• 1991 AHSME Problems/Problem 9
  {{AHSME box|year=1991|num-b=8|num-a=10}}
  1,012 bytes (158 words) - 13:29, 14 March 2023
• 1988 AHSME Problems
  <math>{a \choose k} = \frac{a(a-1)(a-2)\cdots(a-(k-1))}{k(k-1)(k-2)\cdots(2)(1)}</math> ...e <math>x_{0}</math>, consider the sequence defined by <math>x_{n} = f(x_{n-1})</math> for all <math>n \ge 1</math>.
  17 KB (2,535 words) - 13:45, 19 February 2020
• 2015 AMC 10B Problems/Problem 22
  <math>\textbf{(A) } 3 \qquad\textbf{(B) } 12-4\sqrt5 \qquad\textbf{(C) } \dfrac{5+2\sqrt5}{3} \qquad\textbf{(D) } 1+\sqrt <cmath>FG^2+FG-1 = 0</cmath>
  7 KB (1,146 words) - 14:09, 13 April 2024
• 1982 AHSME Problems
  When the polynomial <math>x^3-2</math> is divided by the polynomial <math>x^2-2</math>, the remainder is \textbf{(C)} \ -2x-2 \qquad
  17 KB (2,500 words) - 19:05, 11 September 2023
• 2007 iTest Problems/Problem 10
  {{iTest box|year=2007|num-b=9|num-a=11}}
  2 KB (308 words) - 06:13, 14 August 2021
• 2007 iTest Problems/Problem 11
  {{iTest box|year=2007|num-b=10|num-a=12}}
  931 bytes (130 words) - 04:31, 10 June 2018
• 2007 iTest Problems/Problem 13
  {{iTest box|year=2007|num-b=12|num-a=14}}
  1 KB (212 words) - 14:35, 4 August 2018
• 2007 iTest Problems/Problem 14
  ...p_s^{k_s}</math> then <math>\phi(n)=p_1^{k_1-1}\cdots p_s^{k_s-1}\cdot (p_1-1)\cdots (p_s-1)</math>. Hence every prime factor of <math>n</math> is contained in <math>
  2 KB (265 words) - 21:23, 4 February 2023
• 2007 iTest Problems/Problem 19
  {{iTest box|year=2007|num-b=18|num-a=20}}
  3 KB (397 words) - 01:18, 29 October 2023
• 2007 iTest Problems/Problem 18
  <cmath>(x-a)^2 (x-b)</cmath> <cmath>(x^2 - 2ax + a^2)(x-b)</cmath>
  2 KB (393 words) - 17:01, 10 June 2018
• 2007 iTest Problems/Problem 17
  {{iTest box|year=2007|num-b=16|num-a=18}}
  2 KB (266 words) - 21:30, 4 February 2023
• 2007 iTest Problems/Problem 16
  {{iTest box|year=2007|num-b=15|num-a=17}}
  3 KB (420 words) - 20:28, 17 June 2018
• 2007 iTest Problems/Problem 20
  {{iTest box|year=2007|num-b=19|num-a=21}}
  2 KB (220 words) - 21:33, 4 February 2023
• 2007 iTest Problems/Problem 21
  <cmath>1+1+1-1-1+1-1+1-1+1-1-1-1+1+1.</cmath> {{iTest box|year=2007|num-b=20|num-a=22}}
  2 KB (247 words) - 03:34, 14 June 2018
• 2007 iTest Problems/Problem 22
  <cmath> |x+y|=2007 \\ |x-y|=c</cmath> <math>2007|x-y|=c</math> must be one line; otherwise, the system would have four solution
  929 bytes (137 words) - 22:05, 10 January 2019
• 2007 iTest Problems/Problem 25
  We know that <math>\binom{2007}{i} = \binom{2007}{2007-i}</math>, so ...7} (i \cdot \binom{2007}{i}) + \sum_{j=1}^{2007} (j \cdot \binom{2007}{2007-j})</cmath>
  2 KB (270 words) - 14:35, 29 July 2018
• 2007 iTest Problems/Problem 47
  <cmath>\begin{align*}X_{n+1}&=X_n+2X_{n-1}\qquad(n=1,2,3\ldots),\\ Y_{n+1}&=3Y_n+4Y_{n-1}\qquad(n=1,2,3\ldots).\end{align*}</cmath>
  2 KB (415 words) - 11:46, 18 April 2024
• 2007 iTest Problems/Problem 58
  ...red the answer of a previous problem. When the problem is rewritten, the T-value is substituted.'' ...3^{n+1}+1}{3^{n-1}+1}\right\rfloor+\cdots+\left\lfloor\frac{k^{n+1}+1}{k^{n-1}+1}\right\rfloor</cmath>
  2 KB (307 words) - 17:29, 22 July 2018
• 2013 Canadian MO Problems
  <math>(x+1)P(x-1)-(x-1)P(x)</math> Let <math>G</math> be the centroid of a right-angled triangle <math>ABC</math> with <math>\angle BCA = 90^\circ</math>. Le
  2 KB (381 words) - 13:45, 8 October 2014
• 2007 iTest Problems/Problem 23
  Find the product of the non-real roots of the equation <cmath>(x^2-3x)^2+5(x^2-3x)+6=0</cmath>
  1 KB (202 words) - 16:48, 24 November 2018
• 2007 iTest Problems/Problem 24
  {{iTest box|year=2007|num-b=23|num-a=25}}
  2 KB (340 words) - 19:49, 30 June 2018
• Mock AIME 5 2005-2006 Problems/Problem 15
  ...math> and positive integer <math>n < 1003</math> such that <math>a_m = a_{m-n} = a_{m+n}</math>, where all subscripts are taken <math>\pmod{2006}</math> {{Mock AIME box|year=2005-2006|n=5|source=76847|num-b=14|num-a=15}}
  955 bytes (157 words) - 21:20, 8 October 2014
• 1956 AHSME Problems
  \textbf{(B)}\ \frac{1}{x-1}\qquad \textbf{(C)}\ \frac{-1}{x-1}\qquad
  20 KB (3,039 words) - 22:44, 12 February 2021
• 2014 UMO Problems
  Completely describe the set of all right triangles with positive integer-valued legs such that when four draw((10-sqrt(3)/2,-1/2)--(10,1.5-1/2),arrow=Arrow());
  6 KB (982 words) - 15:11, 22 October 2018
• 2014 UMO Problems/Problem 6
  draw((10-sqrt(3)/2,-1/2)--(10,1.5-1/2),arrow=Arrow()); draw((12.5-sqrt(3)/2,-1/2)--(12.5+sqrt(3)/2,-1/2),arrow=Arrow());
  3 KB (628 words) - 19:22, 20 May 2021
• Asymptote: Crash Course
  Suppose you are floating in a three-dimensional universe that extends infinitely in each direction. A bright sou
  3 KB (410 words) - 20:22, 15 December 2020
• 2006 UNCO Math Contest II Problems
  For Colorado Students Grades 7-12. A=dir(90-(j-1)*(360/14));
  5 KB (725 words) - 22:15, 7 November 2014
• 2006 UNCO Math Contest II Problems/Problem 3
  A=dir(90-(j-1)*(360/14)); {{UNCO Math Contest box|n=II|year=2006|num-b=2|num-a=4}}
  713 bytes (103 words) - 19:47, 4 December 2016
• 2006 UNCO Math Contest II Problems/Problem 11
  draw((j,max(0,j-1))--(j,4),black); {{UNCO Math Contest box|n=II|year=2006|num-b=10|after=Last Question}}
  863 bytes (128 words) - 02:35, 13 January 2019
• 1988 AHSME Problems/Problem 9
  {{AHSME box|year=1988|num-b=8|num-a=10}}
  2 KB (367 words) - 08:24, 12 July 2022
• 1988 AHSME Problems/Problem 11
  {{AHSME box|year=1988|num-b=10|num-a=12}}
  2 KB (306 words) - 18:25, 21 November 2017
• 2014 UNM-PNM Statewide High School Mathematics Contest II Problems
  [[UNM-PNM Statewide High School Mathematics Contest II Problems/Problem 1|Solution [[2014 UNM-PNM Statewide High School Mathematics Contest II Problems/Problem 1|Solution
  5 KB (774 words) - 06:07, 29 January 2019
• 2014 UNM-PNM Statewide High School Mathematics Contest II Problems/Problem 7
  {{UNM-PNM Math Contest box|year=2014|n=II|num-b=6|num-a=8}}
  700 bytes (112 words) - 13:23, 20 December 2018
• 2014 AMC 8 Problems
  ...wer is <math>H</math> and Terry's answer is <math>T</math>, what is <math>H-T</math>? ...Paula 35 cents and has a pocket full of 5-cent coins, 10-cent coins, and 25-cent coins that he can use to pay her. What is the difference between the la
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• 2014 AMC 8 Problems/Problem 15
  circum[i]=dir(120-30*i); https://youtu.be/3QHH9xV-QDw
  2 KB (280 words) - 11:10, 2 July 2023
• 2015 AMC 10A Problems
  What is the value of <math>(2^0-1+5^2+0)^{-1}\times5?</math> ...wn in the figure. How many toothpicks does she need to add to complete a 5-step staircase?
  12 KB (1,897 words) - 22:45, 18 March 2024
• 2015 AMC 12A Problems
  What is the value of <math>(2^0-1+5^2-0)^{-1}\times5?</math> A league with 12 teams holds a round-robin tournament, with each team playing every other team exactly once. Game
  13 KB (2,117 words) - 12:33, 24 August 2023
• 2015 AMC 10A Problems/Problem 3
  for(int j=0;j<=3-i;j=j+1) ...have <math>2[2(5)+4+3+2+1]=40</math> toothpicks. So our answer is <math>40-18=22</math> or <math>D</math>.
  2 KB (280 words) - 23:07, 26 June 2023
• 2015 AMC 12A Problems/Problem 25
  ...two circles in the pair. Layer <math>L_k</math> consists of the <math>2^{k-1}</math> circles constructed in this way. Let <math>S=\bigcup_{j=0}^{6}L_j< fill((coord[0].x-910,coord[0].y)--arc1--cycle,gray(0.75));
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• Mock AIME 3 2006-2007 Problems/Problem 3
  769 bytes (139 words) - 23:38, 15 February 2015
• 2015 AMC 10B Problems
  Marie does three equally time-consuming tasks in a row without taking breaks. She begins the first task at ...imum value of the sum of the roots of the equation <math>(x-a)(x-b)+(x-b)(x-c)=0</math>?
  14 KB (2,247 words) - 11:38, 26 March 2024
• 2015 AMC 12B Problems
  Marie does three equally time-consuming tasks in a row without taking breaks. She begins the first task at David, Hikmet, Jack, Marta, Rand, and Todd were in a 12-person race with 6 other people. Rand finished 6 places ahead of Hikmet. Mar
  13 KB (2,064 words) - 13:39, 1 October 2022
• 2015 AMC 12B Problems/Problem 20
  f(i-1,1) & \text{ if } i \ge 1 \text{ and } j = 0 \text{, and} \\ f(i-1, f(i,j-1)) & \text{ if } i \ge 1 \text{ and } 1 \le j \le 4.
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• 2015 AMC 12B Problems/Problem 21
  <cmath>5s+5a-2s-2b=190</cmath> <cmath>3s = 190+2b-5a</cmath>
  6 KB (1,007 words) - 13:54, 12 October 2022
• 2015 AMC 12B Problems/Problem 25
  ...ude of <math>P_{2015}</math> on the complex plane. This is an [[arithmetico-geometric series]]. (1-x)S &=1+x+x^2+\cdots+x^{2014}-2015x^{2015} \\
  13 KB (2,117 words) - 11:30, 8 April 2023
• 2015 AMC 10B Problems/Problem 21
  ...least <math>3 \cdot 19=57</math> steps high. We then start using guess-and-check: {{AMC10 box|year=2015|ab=B|num-b=20|num-a=22}}
  6 KB (983 words) - 01:51, 16 June 2022
• 2015 AIME I Problems/Problem 7
  ...ta JFC</math> are similar to <math>\Delta EDC</math> since they are <math>1-2-\sqrt{5}</math> triangles. Thus, we can rewrite <math>BC</math> in terms o ...m \triangle JFC</math> (and similarly the rest of the triangles are <math>1-2-\sqrt5</math> triangles). We let the sidelength of <math>FGHJ</math> be <m
  6 KB (1,105 words) - 21:02, 9 November 2023
• Hawaii AMC
  356 bytes (44 words) - 15:47, 7 August 2017
• 2015 AIME II Problems/Problem 15
  ..._C) = \left(\frac{24}{15}, \frac{-48}{15}\right)</math>. Using the point-to-line distance formula and the condition <math>n = 4p</math>, we have <cmath> ...g a perpendicular from <math>P</math> to <math>Q</math> creates a <math>3-4-5</math> right triangle, from which <math>BC = 4</math> and, if <math>\alpha
  31 KB (5,086 words) - 19:15, 20 December 2023
• 2016 AIME I Problems
  ...55,-0.85),G=B-(0,1.1),H=F-(0,0.6),I=E-(0,0.6),J=D-(0,1.1),K=C-(0,1.4),L=C+K-A; ...that for every positive integer <math>k</math>, the subsequence <math>a_{2k-1}</math>, <math>a_{2k}</math>, <math>a_{2k+1}</math> is geometric and the s
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• 2015 USAMO Problems/Problem 4
  How many different non-equivalent ways can Steve pile the stones on the grid? ...\forall i</math>. By stars and bars, the number of ways is <math>\binom{n+m-1}{m}^{2}</math>.
  4 KB (770 words) - 20:09, 10 March 2016
• 2015 USAMO Problems
  How many different non-equivalent ways can Steve pile the stones on the grid? ...math> is at most <math>\frac{n(n+1)}{2}\lambda</math>. (A multiset is a set-like collection of elements in which order is ignored, but repetition of ele
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• 2015 USAJMO Problems
  How many different non-equivalent ways can Steve pile the stones on the grid?
  3 KB (565 words) - 16:42, 5 August 2023
• 2015 USAJMO Problems/Problem 1
  ...11}, m_{111})</math> for all the eight elements. Since the sum of the eight-group is <math>0</math>, <math>m_{111}</math> must also be <math>0</math>. T ..., namely the mean. Then before that, if we could always choose <math>M\ge N-2^k</math> members to form pairs, each yielding the average of the total gro
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• Aops font
  For example, to type the symbol for the re-sort button <span class="aops-font">|</span>, you would type: :<code> <nowiki><span class="aops-font">text here!</span></nowiki></code>
  6 KB (924 words) - 21:31, 22 February 2024
• 1993 IMO Problems/Problem 1
  ...<math>f\left(x\right)</math> cannot be expressed as the product of two non-constant polynomials with integer coefficients. ...gives that <math>f\left(x\right)=\sum_{i=0}^{n}\left(\sum_{j=0}^{i}b_jc_{i-j}\right)x^i</math>.
  3 KB (536 words) - 11:26, 21 November 2023
• 2015 AMC 8 Problems/Problem 24
  A baseball league consists of two four-team divisions. Each team plays every other team in its division <math>N</ma ...tart by solving this Diophantine equation. In other words, <math>N=\frac{76-4M}{3}</math>.
  4 KB (669 words) - 20:41, 17 January 2024
• 2015 AMC 8 Problems
  ...\text{mean} \qquad \textbf{(D) } \text{mode} \qquad \textbf{(E) } \text{mid-range}</math> ...econd and third must be two different letters among the <math>21</math> non-vowels, and the fourth must be a digit (<math>0</math> through <math>9</math
  16 KB (2,322 words) - 14:04, 2 February 2024
• 2015 AMC 8 Problems/Problem 21
  draw((-8,0)--(-4,-6*sqrt(2))--(-4-6*sqrt(2),-4-6*sqrt(2))--(-8-6*sqrt(2),-4)--cycle); label("$H$",(-4-6*sqrt(2),-4-6*sqrt(2)),S);
  3 KB (471 words) - 20:36, 17 January 2024
• 2015 AMC 8 Problems/Problem 18
  ...th term <math>25</math>, so we have the common difference is <math>\frac{25-1}4=6</math>. This means we can fill in the first row of the table: ...a fifth term of <math>81</math>, so the common difference is <math>\frac{81-17}4=16</math>. We can fill in the fifth row of the table as shown:
  5 KB (641 words) - 10:28, 13 January 2024
• 2016 AMC 12A Problems/Problem 25
  ...t squares needs to be at least <math>100</math>. Since <math>100\le (x+1)^2-x^2=2x+1</math>, this first happens at <math>x\ge \lfloor 99/2\rfloor = 50</ ...btracted from <math>f(k)</math>. In other words, <math>f(k) = 25\cdot 10^{k-2}+1\cdot 10^{k/2}</math>.
  8 KB (1,224 words) - 18:28, 16 November 2022
• 2016 AMC 10B Problems/Problem 25
  ...k \{ x \} \rfloor</math> can equal integers from <math>0</math> to <math>k-1</math>. ...equal to any of the fractions <math>\frac{1}{k}, \frac{2}{k} \dots \frac{k-1}{k}</math>.
  12 KB (1,951 words) - 18:23, 4 April 2023
• 2016 AMC 10B Problems/Problem 15
  with the numbers <math>1-8</math> around the outsides and <math>9</math> in the middle. We see that t {{AMC10 box|year=2016|ab=B|num-b=14|num-a=16}}
  2 KB (378 words) - 21:13, 18 June 2022
• 2016 AMC 10B Problems/Problem 14
  ...le 5</math>. Then to include all other corners, we need <math>1\le y\le 3(x-1)</math>. ...le x\le 5</math>. To include all other corners, we need <math>2\le y\le 3(x-2)</math>.
  3 KB (497 words) - 08:50, 9 March 2024
• 2016 AIME I Problems/Problem 14
  https://youtu.be/diLCwN-358s {{AIME box|year=2016|n=I|num-b=13|num-a=15}}
  6 KB (967 words) - 21:08, 22 November 2022
• 2016 AIME I Problems/Problem 3
  ...55,-0.85),G=B-(0,1.1),H=F-(0,0.6),I=E-(0,0.6),J=D-(0,1.1),K=C-(0,1.4),L=C+K-A; {{AIME box|year=2016|n=I|num-b=2|num-a=4}}
  4 KB (636 words) - 13:00, 22 December 2020
• 2016 AIME II Problems/Problem 7
  ...th>IB=b</math>, we have <cmath>IJ^{2}=a^{2}+b^{2} \geq 1008</cmath> by [[AM-GM inequality]]. Also, since <math>EFGH||ABCD</math>, the angles that each s ...quare whose sides its vertices touch), so the desired answer is <cmath>1848-1008=\boxed{840}</cmath>
  2 KB (316 words) - 00:10, 2 September 2021
• 2016 AIME II Problems/Problem 9
  ...c_n=a_n+b_n</math>. There is an integer <math>k</math> such that <math>c_{k-1}=100</math> and <math>c_{k+1}=1000</math>. Find <math>c_k</math>. ...h>a_k=r^{k-1}</math> and <math>b_k=(k-1)d</math>. First, <math>b_{k-1}<c_{k-1}=100</math> implies <math>d<100</math>, so <math>b_{k+1}<300</math>.
  6 KB (983 words) - 01:18, 2 February 2023
• 2016 AIME II Problems/Problem 12
  <cmath>f(n,C_i)=\sum_{j\ne i} f(n-1,C_j),</cmath> i.e., <math>f(n,C_i)</math> is equal to the number of colorings of <math>n-1</math> sections that end in any color other than <math>C_i</math>. Using t
  20 KB (3,328 words) - 21:13, 5 April 2024
• 2016 AIME II Problems/Problem 13
  ...however we like, so by PIE we have <math>54 \cdot 4-18 \cdot 6 + 3! \cdot 4-2!=130</math> ways to get a score of <math>5</math>. ...e number of ways to get a sum of <math>6</math> is <math>6!-120-216-222-130-1=31</math>.
  9 KB (1,564 words) - 16:53, 1 December 2021
• Gmaas
  ...lem. EDIT: That theorem was proved by GMAAS. EDIT: GMAAS's Theorem has real-world applications: because GMAAS knows the answer to any math problem, you ...respected. If you want to respect him the most, you should NOT have a high-quality sculpture of Gmaas.
  69 KB (11,805 words) - 20:49, 18 December 2019
• 2006 iTest Problems
  ...a right angle to the line <math>2x + y = 3</math>. Find the x-intercept of the line. Questions 19 and 20 are Sudoku-related questions. Sudoku is a puzzle game that has one and only
  31 KB (4,811 words) - 00:02, 4 November 2023
• 2015 USAJMO Problems/Problem 6
  How many different non-equivalent ways can Steve pile the stones on the grid? ...\forall i</math>. By stars and bars, the number of ways is <math>\binom{n+m-1}{m}^{2}</math>.
  4 KB (769 words) - 20:24, 16 March 2019
• 1982 AHSME Problems/Problem 26
  If <math>k=8j+5</math>, then <math>(8j+5)\equiv 64j^2+80j-1\equiv 16j-1\equiv 24+c \implies 16j\equiv 25+c</math>, which clearly can only have the {{AHSME box|year=1982|num-b=25|num-a=27}}
  2 KB (403 words) - 15:50, 17 June 2021
• 2016 USAMO Problems
  <cmath>\left(k^2\right)!\cdot\prod_{j=0}^{k-1}\frac{j!}{\left(j+k\right)!}</cmath> ...math> and <math>y</math>, <cmath>(f(x)+xy)\cdot f(x-3y)+(f(y)+xy)\cdot f(3x-y)=(f(x+y))^2.</cmath>
  4 KB (608 words) - 13:49, 22 November 2023
• 2016 USAMO Problems/Problem 2
  <cmath>\left(k^2\right)!\cdot\prod_{j=0}^{k-1}\frac{j!}{\left(j+k\right)!}</cmath> ...{\frac{k^{2}}{m}}\right\rfloor-\frac{\sum_{i=0}^{k-1} r(i,m)-\sum_{i=k}^{2k-1} r(i,m)}{m} \geq \sum_{m=p^{i}, i\in \mathbb{N}} \left\lceil -\frac{k^{2}}
  4 KB (732 words) - 20:59, 8 April 2021
• 2016 USAMO Problems/Problem 5
  ...lane with <math>A</math> at the origin and <math>C</math> on the positive x-axis. Assume without loss of generality that C is acute. ...ta</math> and <math>\gamma</math> as the angles made between the positive x-axis and <math>\overrightarrow{MN}</math> and <math>\overrightarrow{QP}</mat
  8 KB (1,449 words) - 00:09, 12 October 2023
• 2016 USAMO Problems/Problem 6
  ...bitrarily permutes the <math>k</math> chosen cards and turns them back face-down. Then, it is your turn again. ...hat you do not win on any earlier move, repeat this for <math>1\le i \le 2n-k+1.</math>
  8 KB (1,516 words) - 10:11, 8 April 2023
• 1954 AHSME Problems/Problem 3
  {{AHSME 50p box|year=1954|num-b=2|num-a=4}}
  627 bytes (105 words) - 20:34, 17 February 2020
• 1990 IMO Problems/Problem 6
  ...en if we can find a sequence <math>\{b_n\}</math> such that <math>b_i=a_p^2-a_q^2, (p\not=q)</math> and <math>\sum_{i=1}^{995}b_ie^{i\theta}=0</math>, t Note that <math>2^2-1^2=3,4^2-3^2=7,\cdots,1990^2-1989^2=3979</math>, then <math>3,7,\cdots,3979</math> can be written as a su
  3 KB (522 words) - 13:54, 30 January 2021
• 1985 USAMO Problems/Problem 1
  {{USAMO box|year=1985|before=First<br>Problem|num-a=2}}
  2 KB (261 words) - 04:18, 23 October 2022
• 1985 USAMO Problems/Problem 5
  Let <math>a_1,a_2,a_3,\cdots</math> be a non-decreasing sequence of positive integers. For <math>m\ge1</math>, define <ma ...ed cells in the first 19 columns of row <math>j</math> is equal to <math>20-b_j</math>.
  2 KB (295 words) - 13:03, 18 July 2016
• 1987 USAMO Problems/Problem 5
  {{USAMO box|year=1987|num-b=4|after=Last Question}}
  662 bytes (128 words) - 18:44, 18 July 2016
• 2016 AMC 8 Problems/Problem 22
  ...t, we find the height of each triangle to calculate their area. The two non-colored isosceles triangles are similar, and are in a <math>3:1</math> ratio ...dot 4 + 1\cdot 0)-(\dfrac{3}{2}\cdot 0 + 1\cdot 3 + 4\cdot 0)}{2} = \frac{6-3}{2} = \frac{3}{2}</math>
  8 KB (1,157 words) - 16:12, 18 January 2024
• Mock AIME 3 2010 Problems
  Let x be the integer two below one hundred, and let y be the number of two-digit primes. Find x − y. How many three-digit numbers (with no leading zeroes) with three distinct digits satisfy th
  14 KB (2,904 words) - 18:24, 16 May 2017
• 2017 AMC 10A Problems/Problem 23
  ...l the ones in a line from <math>2300</math>, so we have <math>2300-120-16-4-12=\boxed{(\textbf{B})\ 2148}</math> ...cases where the chosen 3 points make a line. The answer would be <math>2300-152=\boxed{(\textbf{B})\ 2148}</math>
  3 KB (456 words) - 23:08, 7 October 2023
• 2017 AMC 12A Problems/Problem 5
  https://youtu.be/TCxjwO-I8kQ {{AMC10 box|year=2017|ab=A|num-b=7|num-a=9}}
  3 KB (495 words) - 15:40, 9 June 2023
• 2017 AMC 12A Problems/Problem 4
  <math>\frac{j-s}{j} \approx \frac{2 - 1.4}{2} = 0.3 = 30\%</math> {{AMC10 box|year=2017|ab=A|num-b=6|num-a=8}}
  1 KB (220 words) - 15:41, 9 June 2023
• 2017 AMC 10A Problems/Problem 7
  <math>\frac{j-s}{j} \approx \frac{2 - 1.4}{2}=0.3 \implies \boxed{\textbf{(A)}\ 30\%}</mat https://youtu.be/MUxbCp-2OYQ
  1 KB (213 words) - 21:40, 1 November 2023
• 2017 AMC 10A Problems
  ...r garden. The beds are separated and also surrounded by <math>1</math>-foot-wide walkways, as shown on the diagram. What is the total area of the walkwa The region consisting of all points in three-dimensional space within <math>3</math> units of line segment <math>\overlin
  15 KB (2,285 words) - 18:02, 28 October 2023
• 2017 AMC 12A Problems/Problem 25
  ...1+i),\frac{1}{\sqrt{8}}(-1+i),\frac{1}{\sqrt{8}}(1-i),\frac{1}{\sqrt{8}}(-1-i) \right\}.</cmath> ...we draw these 6 complex numbers out on the complex plane, we get a crystal-looking thing. Note that the total number of ways to choose 12 complex numbe
  18 KB (2,878 words) - 01:47, 16 December 2023
• 2017 AMC 12A Problems
  ...store sells single popsicles for \$1 each, 3-popsicle boxes for \$2, and 5-popsicle boxes for \$3. What is the greatest number of popsicles that Pablo ...th>f(n) = f(n-1) + 1</math> if <math>n</math> is even, and <math>f(n) = f(n-2) + 2</math> if <math>n</math> is odd and greater than <math>1</math>. What
  15 KB (2,418 words) - 16:58, 7 November 2022
• Mock AIME I 2015 Problems/Problem 11
  .... Define a polynomial <math>P(x)</math> such that <math>P(x) = (x-a)(x-b)(x-c) = x^3 - (a+b+c)x^2 + (ab+ac+bc)x - abc</math>. Let <math>j = ab + ac + bc
  3 KB (435 words) - 11:29, 29 October 2019
• 2017 AMC 10B Problems/Problem 21
  {{AMC10 box|year=2017|ab=B|num-b=20|num-a=22}}
  4 KB (601 words) - 00:34, 8 August 2023
• Areteem Institute's Math Zoom Winter Boot Camp (Locations Vary each year including one in Irvine, CA
  ...ely practice sessions with challenging written and oral problems and a good-natured spirit of competition. Locations vary each year with one in Irvine, ...place Dec 19-23 and the Irvine Winter Boot Camp always taking place Dec 26-30. We will have updated information on the 2017 Winter Math Boot Camps in L
  1 KB (166 words) - 17:04, 9 March 2017
• 2017 AIME II Problems
  ...ber of positive integers less than or equal to <math>2017</math> whose base-three representation contains no digit equal to <math>0</math>. ...</math> to <math>7</math> and colored with one of seven colors. Each number-color combination appears on exactly one card. Sharon will select a set of e
  7 KB (1,200 words) - 15:02, 8 September 2020
• 2017 AIME II Problems/Problem 7
  ...ath>(x+2)^2</math> is always non-negative, <math>kx</math> must also be non-negative; therefore this takes care of the <math>kx>0</math> condition as lo ...kx = (x+2)^2</math> (the converse isn't necessarily true!), or <math>x^2+(4-k)x+4=0</math>. Our original equation has exactly one solution if and only i
  6 KB (983 words) - 23:03, 5 January 2024
• 2017 AIME II Problems/Problem 2
  {{AIME box|year=2017|n=II|num-b=1|num-a=3}}
  2 KB (358 words) - 18:57, 24 March 2017
• 2017 AIME II Problems/Problem 14
  ...3)</math>. If <math>\textrm{min} > 0</math> then the point <math>(a-1,b-1,c-1)</math> will also be on the line for example, 3 applies to the other end. So the answer is <math>192-24=\boxed{168}</math>.
  6 KB (1,015 words) - 18:02, 31 December 2023
• 2017 USAJMO Problems/Problem 6
  *Here is a counter-case. <math>BRR_{1}B_{1}RR</math> : j = 1, k = 2 => <math>BRRR</math> : j = ...e, by assumption, the total amount of arcs is <math>1+\text{max}(0,k-1)=1+k-1=k=\text{max}(j,k)</math>.
  8 KB (1,465 words) - 15:30, 12 June 2020
• 2017 UNCO Math Contest II Problems
  Twenty-fifth Annual UNC Math Contest Final Round January 21, 2017 Harold writes an integer; its right-most digit is 4. When Curious George
  7 KB (1,192 words) - 15:14, 20 August 2020
• 2017 UNCO Math Contest II Problems/Problem 2
  {{UNCO Math Contest box|year=2017|n=II|num-b=1|num-a=3}}
  1 KB (195 words) - 18:22, 16 January 2023
• Bill's Triangle
  draw((0,8-j)--(j,8)); draw((8-j,0)--(8,j));
  2 KB (307 words) - 20:25, 12 December 2017
• 2020 AMC 12A Problems
  Jason rolls three fair standard six-sided dice. Then he looks at the rolls and chooses a subset of the dice (pos
  14 KB (2,073 words) - 15:15, 21 October 2021
• 2020 AMC 12B Problems
  <cmath>\frac{100^2-7^2}{70^2-11^2} \cdot \frac{(70-11)(70+11)}{(100-7)(100+7)}</cmath><math>\textbf{(A) } 1 \qquad \textbf{(B) } \frac{9951}{995 A three-quarter sector of a circle of radius <math>4</math> inches together with its
  16 KB (2,417 words) - 01:03, 28 April 2022
• 2017 AMC 8 Problems/Problem 24
  ==Solution 1 (Principle of Inclusion-Exclusion)== We use Principle of Inclusion-Exclusion. There are <math>365</math> days in the year, and we subtract the
  6 KB (907 words) - 21:11, 20 January 2024
• The Pot Method
  The Pot Method is a double-counting method used to evaluate series with binomial coefficients, especial The other way to count this expected value is case-by-case based on the number of blue marbles.
  3 KB (414 words) - 19:10, 4 April 2019
• MIE 2015
  ...ree sets <math>F</math>, <math>G</math> and <math>H</math>. The set <math>G-H</math> is equal to the set: (a) <math>(G\cup F)-(F-H)</math>
  7 KB (1,127 words) - 18:23, 11 January 2018
• 2018 AMC 10A Problems/Problem 20
  ...y it in particular. Each of these sections has <math>10</math> distinct sub-squares, whether partially or in full. So since each can be colored either w .... Since we must subtract off <math>2</math> cases for the all-black and all-white cases, the answer is <math>2^{10}-2=\boxed{\textbf{(B) } 1022.}</math>
  5 KB (827 words) - 21:31, 24 October 2023
• 2018 AMC 10A Problems/Problem 22
  == Video Solution by OmegaLearn (Meta-Solving Technique) == {{AMC10 box|year=2018|ab=A|num-b=21|num-a=23}}
  10 KB (1,599 words) - 04:51, 6 August 2023
• 2018 AMC 10A Problems/Problem 21
  ...math>a</math> for which the curves <math>x^2+y^2=a^2</math> and <math>y=x^2-a</math> in the real <math>xy</math>-plane intersect at exactly <math>3</mat Substituting <math>y=x^2-a</math> into <math>x^2+y^2=a^2</math>, we get
  9 KB (1,502 words) - 23:31, 19 August 2023
• 2018 AMC 12A Problems/Problem 19
  {{AMC12 box|year=2018|ab=A|num-b=18|num-a=20}}
  5 KB (751 words) - 03:12, 15 December 2022
• 2018 AMC 12B Problems
  ...sqrt2 \qquad \textbf{(D) } \frac{1}{2}\sqrt{2} \qquad \textbf{(E) } \sqrt 3-1</math> ...ac{4}{7}</cmath>and <math>q</math> is as small as possible. What is <math>q-p</math>?
  14 KB (2,118 words) - 15:36, 28 October 2021
• 2018 AMC 10B Problems/Problem 19
  ...>p^2q^2.</math> Since <math>c-1<100,</math> the only possibility is <math>c-1=2^2\cdot3^2=36,</math> from which <math>c=37.</math> We conclude that Joey ...</math>. Now, since <math>j-1=37</math>, by similar logic, <math>37=(1+k)(a-1)</math>, so <math>k=36</math> and Joey will be <math>38+36=74</math> and t
  5 KB (888 words) - 05:20, 12 April 2024
• 2018 AMC 10B Problems/Problem 16
  ...ath> consecutive numbers must be a multiple of <math>k</math>, so <math>a^3-a</math> is both divisible by <math>2</math> and <math>3</math>. This provid {{AMC10 box|year=2018|ab=B|num-b=15|num-a=17}}
  5 KB (814 words) - 19:27, 5 March 2024
• 2018 AMC 10B Problems
  A three-dimensional rectangular box with dimensions <math>X</math>, <math>Y</math>, for(int j=0;j<=3-i;j=j+1)
  15 KB (2,237 words) - 23:08, 15 November 2023
• 2018 AMC 12B Problems/Problem 9
  <ol style="margin-left: 1.5em;"> More generally, the term <math>k+101</math> occurs <math>100-k</math> times for <math>k\in\{1,2,3,\ldots,99\}.</math><p></li>
  5 KB (594 words) - 22:03, 27 May 2023
• 2018 AMC 10B Problems/Problem 8
  for(int j=0;j<=3-i;j=j+1) This is a 3-step staircase and uses 18 toothpicks. How many steps would be in a staircas
  4 KB (609 words) - 19:49, 27 August 2023
• 2018 USAJMO Problems/Problem 5
  ...th>\{1, 2, ..., p\},</math> so they have a total of exactly <math>\frac{p(p-1)}{2}</math> edges. Therefore, there exists at least one graph <math>G_k</m {{USAJMO newbox|year=2018|num-b=4|num-a=6}}
  2 KB (455 words) - 13:53, 21 April 2018
• 2018 USAMO Problems/Problem 4
  ...th>\{1, 2, ..., p\},</math> so they have a total of exactly <math>\frac{p(p-1)}{2}</math> edges. Therefore, there exists at least one graph <math>G_k</m The first equality is given by a well known theorem, which can be proven by C-S or Jensen's.
  4 KB (794 words) - 21:46, 20 March 2023
• 2018 USAMO Problems/Problem 3
  s+1. Therefore, the desired sum of k-th powers is equal is a complete reduced residue set modulo nq. The new sum of k-th powers is equal to
  5 KB (939 words) - 22:00, 11 February 2024
• 2018 USAMO Problems/Problem 6
  Now, consider what happens when the two rows are swapped (and the top-bottom pairs are reordered so that the top reads (1,2,3,...,n)). This will ...ds <math>b_n=n\cdot b_{n-1}</math>* which has the same parity as <math>b_{n-1}</math>, so we need only consider the parity of <math>b_n</math> for odd <
  4 KB (700 words) - 04:36, 5 March 2023
• Lagrange's Identity
  ...\biggr )}-{\biggl (}\sum _{k=1}^{n}a_{k}b_{k}{\biggr )}^{2}&=\sum _{i=1}^{n-1}\sum _{j=i+1}^{n}(a_{i}b_{j}-a_{j}b_{i})^{2}\\&{\biggl (}={\frac {1}{2}}\s The vector form follows from the Binet-Cauchy identity by setting <math>ci = ai</math> and <math>di = bi</math>. Th
  4 KB (914 words) - 17:27, 31 August 2022
• 1960 AHSME Problems/Problem 34
  ...answer. From the graph (where the x-axis is the time in seconds and the y-axis is distance from one side of the pool), there are five meeting points, {{AHSME 40p box|year=1960|num-b=33|num-a=35}}
  2 KB (262 words) - 19:18, 17 May 2018
• 2018 AIME I Problems
  Find the number of four-element subsets of <math>\{1,2,3,4,\dots, 20\}</math> with the property that ...at when <math>3^n</math> is written in base <math>143</math>, its two right-most digits in base <math>143</math> are <math>01</math>.
  8 KB (1,284 words) - 14:35, 9 August 2021
• 2018 AIME II Problems
  ...math> recursively to be the remainder when <math>4(a_{n-1} + a_{n-2} + a_{n-3})</math> is divided by <math>11</math>. Find <math>a_{2018} \cdot a_{2020} ...math>. The self-intersecting octagon <math>CORNELIA</math> encloses six non-overlapping triangular regions. Let <math>K</math> be the area enclosed by <
  9 KB (1,385 words) - 00:26, 21 January 2024
• 2018 AIME I Problems/Problem 10
  label("$A$", (1-r) * dir( 90), -dir( 90)); label("$B$", (1-r) * dir(162), -dir(162));
  11 KB (1,934 words) - 12:18, 29 March 2024
• 2018 AIME I Problems/Problem 11
  ...at when <math>3^n</math> is written in base <math>143</math>, its two right-most digits in base <math>143</math> are <math>01</math>. ...iv \binom{p}{0}26^p+\binom{p}{1}26^{p-1}....+\binom{p}{p-2}26^2+\binom{p}{p-1}26+\binom{p}{p}\equiv 26p+1\equiv 1\pmod{169}</math>, so <math>26p\equiv 0
  10 KB (1,448 words) - 19:22, 4 January 2024
• 2018 AIME II Problems/Problem 9
  ...10</math> and <math>BC = DE = FG = HA = 11</math> is formed by removing 6-8-10 triangles from the corners of a <math>23</math> <math>\times</math> <math <math>=\left|\frac{297}{9}-\frac{690}{6}-102\right|=\left| 33-115-102\right|=\left|-184\right|=\boxed{184}</math>
  5 KB (782 words) - 02:30, 5 January 2024
• 2017 IMO Problems
  ...bbit is at point <math>A_{n-1}</math> and the hunter is at point <math>B_{n-1}</math>. In the nth round of the game, three things occur in order. ...sibly to a point <math>A_n</math> such that the distance between <math>A_{n-1}</math> and <math>A_n</math> is exactly 1.
  4 KB (720 words) - 12:12, 5 August 2021
• 1964 AHSME Problems/Problem 13
  ...implies <math>AJ=x</math>. Thus <math>KB=BL=17-x</math> and <math>JC=LC=13-x</math>. ...+LC=(17-x)+(13-x)=8</math>, <math>x=11</math>. Thus <math>r:s=CL:BL=13-x:17-x=2:6=1:3</math>, hence our answer is <math>\fbox{A}</math>.
  2 KB (306 words) - 18:26, 23 July 2019
• 2018 IMO Problems/Problem 3
  ...the two numbers immediately below it. For example, the following is an anti-Pascal triangle with four rows which contains every integer from <math>1</ma Does there exist an anti-Pascal triangle with <math>2018</math> rows which contains every integer fro
  4 KB (626 words) - 01:45, 19 November 2023
• 2003 Indonesia MO Problems
  ...o <math>3</math> congruent triangles (refer to the figure). Every triangle-region is colored with a certain color so that each tile has <math>3</math> ...if <math>k-n</math> divides <math>k^m - n^{m-1}</math>, then <math>k \le 2n-1</math>.
  3 KB (495 words) - 19:16, 7 August 2018
• 2003 Indonesia MO Problems/Problem 4
  |num-b=3 |num-a=5
  2 KB (296 words) - 18:00, 11 August 2018

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