Create the page "A- 1" on this wiki! See also the search results found.

Page title matches

• Power of a Point Theorem/Introductory Problem 1
  448 bytes (67 words) - 15:15, 23 March 2020
• If I have four boxes arranged in a $2 \times 2$ grid, in how many distinct ways can I place the digits $1$, $2$, and $3$ in the boxes, using each digit exactly once, such that each box contains at most one digit? (I only have one of each digit, so one box
  ...he third, and one for the last. Thus, there are <math>4\cdot 3\cdot 2\cdot 1 = \boxed{24}</math> distinct ways to fill the boxes.
  328 bytes (60 words) - 17:19, 12 June 2022
• The perfect squares from $1$ through $2500,$ inclusive, are printed in a sequence of digits $1491625\ldots2500.$ How many digits are in the sequence?
  ...<math>3</math> perfect squares that only have <math>1</math> digit, <math>1^{2},</math> <math>2^{2},</math> and <math>3^{2}.</math> So we have a total of <math>1\times3+2\times6+3\times22+4\times19=\boxed{157}</math> digits.
  939 bytes (140 words) - 17:27, 12 June 2022
• The table shows the percent of families in Mathville that have $0, 1, 2, 3$ and $4$ or more children. If there are a total of $10,250$ families, how many are there with no children?
  The table shows the percent of families in Mathville that have <math>0, 1, 2, 3</math> and <math>4</math> or more children. If there are a total of <
  214 bytes (35 words) - 09:55, 1 August 2022
• In triangle ABC, D be a point in BC, ∠BAD = 30 , ∠CAD = 90 , BD=1=AC, DC =
  0 bytes (0 words) - 13:26, 6 February 2024

Page text matches

• Pythagorean Theorem
  === Proof 1 === <math>(a+b)^2=c^2+4\left(\frac{1}{2}ab\right)\implies a^2+2ab+b^2=c^2+2ab\implies a^2 + b^2=c^2</math>. {{Ha
  5 KB (886 words) - 13:51, 15 May 2024
• 2005 AMC 12A Problems/Problem 16
  pair O0=(9,9), O1=(1,1), O2=(3,1), O3=(1,3); draw(Circle(O1,1));
  2 KB (307 words) - 15:30, 30 March 2024
• 2016 AMC 10A Problems/Problem 2
  <math>\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textb ==Solution 1==
  1 KB (190 words) - 10:58, 16 June 2023
• 2016 AMC 10A Problems/Problem 3
  ==Solution 1==
  1 KB (176 words) - 10:58, 16 June 2023
• 2016 AMC 10A Problems/Problem 4
  <math>\textbf{(A) } -\frac{3}{8} \qquad \textbf{(B) } -\frac{1}{40} \qquad \textbf{(C) } 0 \qquad \textbf{(D) } \frac{3}{8} \qquad \textbf ==Solution 1==
  2 KB (257 words) - 10:57, 16 June 2023
• 2016 AMC 10A Problems/Problem 5
  A rectangular box has integer side lengths in the ratio <math>1: 3: 4</math>. Which of the following could be the volume of the box? == Solution 1==
  1 KB (184 words) - 13:58, 22 August 2023
• 2016 AMC 10A Problems/Problem 6
  ..., replacing each occurrence of the digit <math>2</math> by the digit <math>1</math>. Star adds her numbers and Emilio adds his numbers. How much larger ...git and 2 appears 3 times as a units digit, the answer is <math>10\cdot 10+1\cdot 3=\boxed{\textbf{(D) }103.}</math>
  967 bytes (143 words) - 03:18, 27 June 2023
• 2016 AMC 10A Problems/Problem 7
  == Solution 1 ==
  2 KB (268 words) - 18:19, 27 September 2023
• 2016 AMC 10A Problems/Problem 8
  ...eft(\frac{1}{2}\right)=30 , \Rightarrow (+40)=70 , \Rightarrow \left(\frac{1}{2}\right)=\boxed{\textbf{(C) }35}</cmath>
  1 KB (169 words) - 14:59, 8 August 2021
• 2016 AMC 10A Problems/Problem 9
  A triangular array of <math>2016</math> coins has <math>1</math> coin in the first row, <math>2</math> coins in the second row, <math == Solution 1==
  2 KB (315 words) - 15:34, 18 June 2022
• 2016 AMC 10A Problems/Problem 10
  ...er rectangle is one foot wide, and each of the two shaded regions is <math>1</math> foot wide on all four sides. What is the length in feet of the inner filldraw(rectangle((1,1),(6,4)),gray(0.75));
  2 KB (337 words) - 14:56, 25 June 2023
• 2016 AMC 10A Problems/Problem 11
  filldraw((7,0)--(8,0)--(8,1)--(0,4)--(0,5)--(1,5)--cycle,gray(0.8)); label("$1$",(1/2,5),dir(90));
  8 KB (1,016 words) - 00:17, 31 December 2023
• 2016 AMC 10A Problems/Problem 12
  Three distinct integers are selected at random between <math>1</math> and <math>2016</math>, inclusive. Which of the following is a correc ...\dfrac{1}{3}\qquad\textbf{(D)}\ p=\dfrac{1}{3}\qquad\textbf{(E)}\ p>\dfrac{1}{3}</math>
  2 KB (297 words) - 14:54, 25 June 2023
• 2016 AMC 10A Problems/Problem 13
  ...n a movie theater in a row containing <math>5</math> seats, numbered <math>1</math> to <math>5</math> from left to right. (The directions "left" and "ri <math>\textbf{(A) }1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4\qquad
  2 KB (402 words) - 14:54, 25 June 2023
• List of United States middle school mathematics competitions
  ...tp://euclidlab.org/programs/archimedean-challenge/1 Archimedean Challenge #1] * [http://www.mindresearch.org/gameathon/ Game-a-thon] challenges students to design and build math games in this national co
  7 KB (792 words) - 10:14, 23 April 2024
• MathILy (serious mathematics infused with levity)
  ...duate or undergraduate mathematics students. The weeks break down into a 2-1-2 schedule: We start with two weeks of Root Class, which consists of a gall
  5 KB (706 words) - 23:49, 29 January 2024
• MATHCOUNTS
  ...rint</u>: 1-1.5 (School/Chapter), 2-2.5 (State/National)<br><u>Target:</u> 1.5 (School), 2 (Chapter), 2-2.5 (State/National)}}
  10 KB (1,506 words) - 21:31, 14 May 2024
• Georgia mathematics competitions
  * [[Archimedean Challenge #1]] * [[Archimedean Challenge #1]]
  4 KB (565 words) - 13:24, 13 September 2019
• ICTM Math Contest
  ...wo areas of math contests: Grade School (Grades 3, 4, 5, 6, 7, 8 + Algebra 1) and High School (Regional and State Finals). The contest is given in-school May 1-15. Each school sets its own testing date and time within these guidelines.
  8 KB (1,182 words) - 14:26, 3 April 2024
• Oklahoma mathematics competitions
  *[[Math Kangaroo]] is available to students in grades 1-12. The competition occurs annually during March held at ORU. Students do
  2 KB (279 words) - 19:41, 23 March 2017
• Art of Problem Solving
  ...artofproblemsolving.com/store/item/aops-vol1 Art of Problem Solving Volume 1: the Basics]
  5 KB (667 words) - 17:09, 3 July 2023
• Resources for mathematics competitions
  * [https://artofproblemsolving.com/news/articles/establishing-a-positive-culture Establishing a Positive Culture of Expectation in Math Educ * [https://artofproblemsolving.com/news/articles/how-to-write-a-solution How to Write a Math Solution] by [[Richard Rusczyk]] and [[user:MCr
  16 KB (2,152 words) - 21:46, 6 May 2024
• Math books
  ...zon.com/gp/product/B09PMLFHX2/ref=ox_sc_act_title_1?smid=ATVPDKIKX0DER&psc=1 Getting Started with Competition Math], a textbook meant for true beginners ...itu-Andreescu/dp/0817643265/ref=sr_1_1?ie=UTF8&s=books&qid=1204029652&sr=1-1 Complex Numbers from A to... Z] by [[Titu Andreescu]]
  24 KB (3,177 words) - 12:53, 20 February 2024
• Richard Rusczyk
  ...fproblemsolving.com/store/item/aops-vol1 the Art of Problem Solving Volume 1: the Basics]''
  2 KB (254 words) - 09:04, 25 January 2019
• Math Olympiads for Elementary and Middle Schools
  ...Response|difficulty=1|breakdown=<u>Division E</u>: 1<br><u>Division M</u>: 1}}
  2 KB (215 words) - 02:54, 18 November 2020
• William Lowell Putnam Mathematical Competition
  ...me=Putnam|region=USA|type=Proof|difficulty=7 - 9|breakdown=<u>Problem A/B, 1/2</u>: 7<br><u>Problem A/B, 3/4</u>: 8<br><u>Problem A/B, 5/6</u>: 9}} ...if the top three students tie, they are all awarded a rank of <math>\frac{1 + 2 + 3}{3}=2</math>.) Before 2019, schools were required to choose their t
  4 KB (623 words) - 13:11, 20 February 2024
• Physics competitions
  Khan Academy: AP Physics 1/2
  4 KB (506 words) - 11:46, 6 September 2023
• Computer science books
  ...s/ASIN/0201485419/artofproblems-20 The Art of Computer Programming Volumes 1-3 (boxed set)] by [[D. E. Knuth]]
  2 KB (251 words) - 00:45, 17 November 2023
• Physics
  ...Thermodynamics states that the efficiency of heat engines must always be < 1.
  9 KB (1,355 words) - 07:29, 29 September 2021
• 2005 Alabama ARML TST Problems/Problem 7
  ...c{n^2}{2^n}\right)+\sum_{n=1}^{\infty} \left(\frac{2n}{2^n}\right)+\sum_{n=1}^{\infty} \left(\frac{3}{2^n}\right)</math> \sum_{n=1}^{\infty} \left(\frac{3}{2^n}\right)=x\\
  1 KB (193 words) - 21:13, 18 May 2021
• Mathematics
  ==Overview=={{asy image|<math>1\,2\,3\,4\,5\,6\,7\,8\,9\,0</math>|right|The ten [[digit]]s making up <br />
  6 KB (902 words) - 12:53, 3 September 2019
• AMC historical results
  *Distinguished Honor Roll: Awarded to top 1% of scorers on each AMC 8, 10 and 12 respectively. *AIME qualifiers: 960 (1.5%)
  17 KB (1,921 words) - 20:53, 10 May 2024
• 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
• United States of America Mathematical Talent Search
  ...Info|name=USAMTS|region=USA|type=Proof|difficulty=3-6|breakdown=<u>Problem 1-2</u>: 3-4<br><u>Problem 3-5</u>: 5-6}}
  4 KB (613 words) - 13:08, 18 July 2023
• AMC 8
  ...ty=1 - 1.5|breakdown=<u>Problems 1 - 12</u>: 1<br><u>Problems 13 - 25</u>: 1.5}} The AMC 8 is usually administered on the third week of January. There is a 1-week window for students to take the test.
  4 KB (558 words) - 22:25, 28 April 2024
• AMC 10
  ...gion=USA|type=Multiple Choice|difficulty=1-3|breakdown=<u>Problem 1-5</u>: 1<br><u>Problem 6-20</u>: 2<br><u>Problem 21-25</u>: 3}} ...incorrect questions are worth 0 points, and unanswered questions are worth 1.5 points, to give a total score out of 150 points. From 2002 to 2006, unan
  4 KB (574 words) - 15:28, 22 February 2024
• AMC 12
  ...AMC 12|region=USA|type=Multiple Choice|difficulty=2-4|breakdown=<u>Problem 1-10</u>: 2<br><u>Problem 11-20</u>: 3<br><u>Problem 21-25</u>: 4}} ..., incorrect answers are worth 0 points, and unanswered questions are worth 1.5 points, to give a total score out of 150 points. From 2002 to 2006, the
  4 KB (520 words) - 12:11, 13 March 2024
• 1955 AHSME Problems/Problem 38
  ...c}{3}+d+16=a+b+c+d</math>. This, with some algebra, means that <math>\frac{1}{3}(a+b+c)=8</math>. <math>d</math> must be <math>\boxed{\textbf{(B)} 21}</
  1 KB (200 words) - 23:35, 28 August 2020
• American Invitational Mathematics Examination
  ...ame=AIME|region=USA|type=Free Response|difficulty=3-6|breakdown=<u>Problem 1-5</u>: 3<br><u>Problem 6-10</u>: 4<br><u>Problem 10-12</u>: 5<br><u>Problem The AIME is a 15 question, 3 hour exam<math>^1</math> taken by high scorers on the [[AMC 10]], [[AMC 12]], and [[USAMTS]]
  8 KB (1,057 words) - 12:02, 25 February 2024
• 1954 AHSME Problems/Problem 46
  draw((-2/3,1/2)--(-sqrt(3)/6,1/2)--(0,0)--(sqrt(3)/6,1/2)--(2/3,1/2)); label("$\frac{1}{2}$",(.5,.25));
  3 KB (415 words) - 18:01, 24 May 2020
• 2015 IMO Problems
  ==Problem 1== [[2015 IMO Problems/Problem 1|Solution]]
  4 KB (692 words) - 22:33, 15 February 2021
• United States of America Mathematical Olympiad
  ...Info|name=USAMO|region=USA|type=Proof|difficulty=7-9|breakdown=<u>Problem 1/4</u>: 7<br><u>Problem 2/5</u>: 8<br><u>Problem 3/6</u>: 9}}
  6 KB (869 words) - 12:52, 20 February 2024
• American Regions Mathematics League
  ...dividual</u>: 4 (Problem 6/8), 6 (Problem 10)<br><u>Team</u>: 3.5 (Problem 1-5), 5 (Problem 6-10)}}
  2 KB (267 words) - 17:06, 7 March 2020
• Joining an ARML team
  ...The three teams are the Reals, Naturals, and Primes. Two compete in A and 1 in B... Michigan was also awarded the best shirt design at ARML Iowa 2015.. ...ille, etc.) on the MO ARML team. (eg. the 2008 team fielded 6 from KS, and 1 from IL)
  21 KB (3,500 words) - 18:41, 23 April 2024
• Chemistry scholarships
  ....htm American Electroplaters and Surface Finishers Society Scholarship] of 1,500 dollars for undergraduate and graduate students in chemistry and some t
  1 KB (182 words) - 22:00, 4 February 2017
• Engineering scholarships
  ...ican Electroplaters and Surface Finishers Society Scholarship] of <dollar/>1,500 for undergraduate and graduate students in chemistry and some types of ....com/2020-usa-cargo-trailer-scholarship USA Cargo Trailer Scholarship] of $1,500 for students pursuing engineering.
  4 KB (511 words) - 14:57, 16 July 2020
• Princeton University Mathematics Competition
  ...participate in the Finals round, and they have to solve three problems in 1 hour.
  2 KB (295 words) - 23:19, 5 January 2019
• Mu Alpha Theta
  *[[Algebra]] 1 *-1 for each incorrect question
  4 KB (632 words) - 17:09, 11 October 2020
• AoPS Online School
  * [[AoPS Online School/Prealgebra 1 | Prealgebra 1]] — [https://artofproblemsolving.com/school/course/catalog/prealgebra1 De .../Introduction to Algebra A | Introduction to Algebra A]] (formerly Algebra 1) — [https://artofproblemsolving.com/school/course/catalog/algebra-a Detai
  8 KB (965 words) - 03:41, 17 September 2020
• Power Mean Inequality
  ...th> positive real weights <math>w_i</math> with sum <math>\sum_{i=1}^n w_i=1</math>, the power mean with exponent <math>t</math>, where <math>t\in\mathb \prod_{i=1}^n a_i^{w_i} &\text{if } t=0 \\
  3 KB (606 words) - 23:59, 1 July 2022
• Wilson's Theorem
  ...rem''' states that if [[integer ]]<math>p > 1</math> , then <math>(p-1)! + 1</math> is divisible by <math>p</math> if and only if <math>p</math> is prim ...p-1)! + 1</math>. Therefore <math>p</math> does not divide <math>(p-1)! + 1</math>.
  4 KB (639 words) - 01:53, 2 February 2023
• Trivial Inequality
  ...lutions <math>x,y,z</math> of the equation <math>x^2+5y^2+10z^2=4xy+6yz+2z-1</math>. *Show that <math>\sum_{k=1}^{n}a_k^2 \geq a_1a_2+a_2a_3+\cdots+a_{n-1}a_n+a_na_1</math>. [[Inequality_Introductory_Problem_2|Solution]]
  3 KB (560 words) - 22:51, 13 January 2024
• Arithmetic mean
  For example, if I wanted to find the average of the numbers 3, 1, 4, 1, and 5, I would compute: <center><math> \frac{3+1+4+1+5}{5} = \frac{14}{5}.</math></center>
  699 bytes (110 words) - 12:44, 20 September 2015
• Completing the square
  What is the value of <math>x</math> if <math>x=1+\dfrac{1}{x}</math>
  2 KB (422 words) - 16:20, 5 March 2023
• Heron's Formula
  <cmath>[ABC]=\frac{ab}{2}\sqrt{1-\cos^2 C}.</cmath> <math>[ABC]=\frac{ab}{2}\sqrt{1-\left(\frac{a^2+b^2-c^2}{2ab}\right)^2}.</math>
  4 KB (675 words) - 00:05, 22 January 2024
• Polynomial
  * <math>x^{-1}+2+3x+x^2</math> * <math>x^{1/3}=\sqrt[3]{x}</math>
  6 KB (1,100 words) - 01:44, 17 January 2024
• Simon's Favorite Factoring Trick
  ...h> are integer constants, and the coefficient of xy must be 1(If it is not 1, then divide the coefficient off of the equation.). According to Simon's Fa ...y)</math> that are solutions to the equation <math>\frac{4}{x}+\frac{5}{y}=1</math>. (2021 CEMC Galois #4b)
  7 KB (1,107 words) - 07:35, 26 March 2024
• University of South Carolina High School Math Contest
  The test is scored as 5 points for every correct response, 1 point for a blank response, and 0 points for an incorrect response.
  972 bytes (141 words) - 11:12, 30 September 2018
• Factoring
  <cmath>a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b + \cdots + ab^{n-2} + b^{n-1})</cmath> ...this creates the difference of squares factorization, <cmath>a^2-b^2=(a+b)(a-b)</cmath>
  3 KB (532 words) - 22:00, 13 January 2024
• Proof by contradiction
  ...e|primes]] <math>p_1, p_2,\ldots, p_n</math>. Let <math>N=p_1p_2p_3...p_n+1</math>. <math>N</math> is not divisible by any of the known primes since i
  2 KB (374 words) - 14:01, 21 August 2022
• Mathematical problem solving
  ...e sum of the [[series]] <math>\frac11 + \frac14 + \frac19 + \cdots + \frac{1}{n^2} + \cdots</math><br> ...{\pi}\right)\left(1+\frac{x}{\pi}\right)\left(1-\frac{x}{2\pi}\right)\left(1+\frac{x}{2\pi}\right)\cdots</math><br>
  2 KB (314 words) - 06:45, 1 May 2014
• Pigeonhole Principle
  In [[combinatorics]], the '''pigeonhole principle''' states that if <math>n+1</math> or more pigeons are placed into <math>n</math> holes, one hole must ...ontain at most one ball implies that <math>b_r \leq 1</math> for all <math>1 \leq r \leq n</math>, so <cmath>b_1 + b_1 + \cdots + b_n \geq n.</cmath> Ho
  11 KB (1,985 words) - 21:03, 5 August 2023
• Quadratic equation
  Next, we factor out our common terms to get <math>x(x-1)-2(x-1)=0</math>. ...<math>(x-1)(x-2)=0</math>. By the zero-product property, either <math> (x-1) </math> or <math> (x-2) </math> equals zero.
  2 KB (264 words) - 12:04, 15 July 2021
• Relatively prime
  ...r than 1. That is, their [[greatest common divisor]] is <math>\gcd(m, n) = 1</math>. Equivalently, <math>m</math> and <math>n</math> must have no [[pri ...r difference <math>(n+1)-n = 1</math>, which is impossible since <math>p > 1</math>.
  2 KB (245 words) - 15:51, 25 February 2020
• 1965 AHSME Problems/Problem 40
  \textbf{(D) }\ 1 \qquad We can then try <math>(x^2+3x+1)=x^4+6x^3+11x^2+6x+1</math>.
  3 KB (571 words) - 00:42, 22 October 2021
• Trigonometric substitution
  ...>\int \sqrt{a^2+x^2}\,dx</math>, we make use of the identity <math>\tan^2x+1=\sec^2x</math>. Set <math>x=a\tan\theta</math> and the radical will go awa Making use of the identity <math>\sin^2\theta+\cos^2\theta=1</math>, simply let <math>x=a\sin\theta</math>.
  1 KB (173 words) - 18:42, 30 May 2021
• Geometric mean
  ...eometric mean of the numbers 6, 4, 1 and 2 is <math>\sqrt[4]{6\cdot 4\cdot 1 \cdot 2} = \sqrt[4]{48} = 2\sqrt[4]{3}</math>. MC("\sqrt{ab}",D(A--M,orange+linewidth(1)),W);
  2 KB (282 words) - 22:04, 11 July 2008
• Principle of Inclusion-Exclusion
  ...gain, <math>|A\cap C|</math> would be putting five guys in order, so <math>1!\binom{6}{5}=6</math>. <math>|A\cap D|</math> is just choosing <math>3</mat ...>, <math>|A\cap C\cap D|</math> is again ordering everybody which is <math>1</math>, and <math>|B\cap C\cap D|</math> is the same as <math>|A\cap B\cap
  9 KB (1,703 words) - 07:25, 24 March 2024
• Googol
  Googol is a huge number. It has a 1 followed by 100 zeroes or <math>10^{100}.</math>
  84 bytes (14 words) - 10:46, 25 October 2020
• Combination
  * <math>\binom{n-1}{r-1}+\binom{n-1}{r}=\binom{n}{r}</math> * <math>\binom{n}{r}=\frac{n}{r}\binom{n-1}{r-1}</math>
  4 KB (615 words) - 11:43, 21 May 2021
• 2020 CIME II Problems/Problem 10
  ...row (a-1)(b-1)=2</math> from whence we have <math>(a,b,c)\in\{(2,3,1),(3,2,1)\}</math>. ...b\geq 6c</math> we have <math>6c\leq 2c+6\Rightarrow c\leq3/2\Rightarrow c=1</math>; a contradiction since <math>c\geq 2</math>.
  2 KB (332 words) - 09:37, 30 December 2021
• Cauchy-Schwarz Inequality
  ...a constant <math>t</math> such that <math>a_i = t b_i</math> for all <math>1 \leq i \leq n</math>, or if one list consists of only zeroes. Along with th ...a^2b^2</math>. The inequality then follows from <math> |\cos\theta | \le 1 </math>, with equality when one of <math> \mathbf{a,b} </math> is a multipl
  13 KB (2,048 words) - 15:28, 22 February 2024
• Factorial
  ...[recursion|recursive definition]] for the factorial is <math>n!=n \cdot (n-1)!</math>. * <math>0! = 1</math> (remember! this is 1, not 0! (the '!' was an exclamation mark, not a factorial sign))
  10 KB (809 words) - 16:40, 17 March 2024
• Discriminant
  ...ng about the roots of a given [[polynomial]] <math>p(x)=a_nx^n+a_{n-1}x^{n-1}+...+a_0</math> of degree <math>n</math> with all the coefficients being re ...integer such that <math>k\geq\frac{n-2}{4}</math>, then there are <math>2k+1</math> pairs of complex conjugate roots and <math>n-4k+2</math> real roots.
  4 KB (734 words) - 19:19, 10 October 2023
• Menelaus' theorem
  <cmath>\frac{PB}{CP} \cdot \frac{QC}{QA} \cdot \frac{AR}{RB} = 1.</cmath> ...;label("C",C,(1,0));label("P",P,(0,-1));label("Q",Q,(1,0));label("R",R,(-1,1));
  5 KB (804 words) - 03:01, 12 June 2023
• ARML historical results
  | [[South Carolina ARML]] (South Carolina #1)
  19 KB (2,632 words) - 14:31, 12 June 2022
• Fermat's Little Theorem
  ...t [[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 of
  16 KB (2,658 words) - 16:02, 8 May 2024
• Parabola
  xaxis(-9,9,Ticks(f, 1.0)); yaxis(-9,9,Ticks(f, 1.0));
  3 KB (551 words) - 16:22, 13 September 2023
• Chebyshev's Inequality
  ...\sum_{i=1}^{n}a_ib_i\right)\geq\left(\sum_{i=1}^{n}a_i\right)\left(\sum_{i=1}^{n}b_i\right)</math>. <math> b_n\geq b_{n-1}\geq ... \geq b_1 </math> then:
  1 KB (214 words) - 20:32, 13 March 2022
• Euler's Totient Theorem
  ...er [[relatively prime]] to <math>a</math>, then <math>{a}^{\phi (m)}\equiv 1 \pmod {m}</math>. ...hi(m)} \pmod{m} </math> <math> \implies </math> <math> a^{\phi (m)} \equiv 1 \pmod{m}</math> as desired. Note that dividing by <math> n_1 n_2 ... n_{\ph
  3 KB (542 words) - 17:45, 21 March 2023
• Geometric inequality
  .../math> and [[perimeter]] <math>P</math>, then <math>\frac{4\pi A}{P^2} \le 1</math>. This means that given a perimeter <math>P</math> for a plane figure ...o <math>2\sin A \cos B<2\sin A</math>, which is equivalent to <math>\cos B<1</math>. Since this is always true for <math>0<B<180</math>, this inequality
  7 KB (1,296 words) - 14:22, 22 October 2023
• Brahmagupta's Formula
  Substituting <math>\sin^2B=1-\cos^2B</math> results in <cmath>4[ABCD]^2=(1-\cos^2B)(ab+cd)^2=(ab+cd)^2-\cos^2B(ab+cd)^2</cmath>
  3 KB (465 words) - 18:31, 3 July 2023
• Ptolemy's theorem
  == Proof 1 == ...<math> ABCDEFG </math>, prove that: <math> \frac{1}{AB}=\frac{1}{AC}+\frac{1}{AE} </math>.
  7 KB (1,198 words) - 20:39, 9 March 2024
• Elementary symmetric sum
  ...mbers is the sum of all products of <math>k</math> of those numbers (<math>1 \leq k \leq n</math>). For example, if <math>n = 4</math>, and our set of ...)^nS_n</math>, and the coefficient of the <math>x^k</math> term is <math>(-1)^{n-k}S_{n-k}</math>, where the symmetric sums are taken over the roots of
  2 KB (275 words) - 12:51, 26 July 2023
• Number theory/Advanced
  <cmath>z'=\frac{az+b}{cz+d},\quad a,b,c,d\in\mathbb{Z}, ad-bc=1.</cmath>
  5 KB (849 words) - 16:14, 18 May 2021
• 1959 IMO Problems/Problem 2
  <math> \sqrt{x+\sqrt{2x-1}} + \sqrt{x-\sqrt{2x-1}} = A,</math> given (a) <math>A=\sqrt{2}</math>, (b) <math>A=1</math>, (c) <math>A=2</math>, where only non-negative real numbers are admi
  3 KB (466 words) - 12:04, 12 April 2024
• Harmonic mean
  ...> is defined to be: <math> \frac{n} {\frac{1}{x_1}+\frac{1}{x_2}+...+\frac{1}{x_n}}</math>. ..._1\cdot x_2 \cdots x_n}\ge \frac{n} {\frac{1}{x_1}+\frac{1}{x_2}+...+\frac{1}{x_n}} </math>
  1 KB (196 words) - 00:49, 6 January 2021
• Euler's totient function
  ...(1-\frac{1}{p_1} \right) \left(1-\frac{1}{p_2} \right)\cdots \left(1-\frac{1}{p_m}\right).</cmath> ...fine the [[prime factorization]] of <math> n </math> as <math> n =\prod_{i=1}^{m}p_i^{e_i} =p_1^{e_1}p_2^{e_2}\cdots p_m^{e_m} </math> where the <math>p
  5 KB (898 words) - 19:12, 28 January 2024
• Probability
  ..., <math>(\Omega, \mathfrak{a})</math>. <math>\mathit{P}:\mathfrak{a}\to [0,1]</math> is an assignment with certain properties (it is a special kind of [ <math>\mathit{P}(\{H, T\})=1</math>.
  4 KB (588 words) - 12:47, 2 October 2022
• LaTeX
  <cmath>1+2+3+4+5+\sin \pi = \frac{5\cdot 6}{2}+0=15.</cmath>
  1 KB (164 words) - 19:09, 14 February 2024
• Divisibility
  * The positive divisors of <math>35</math> are <math>1</math>, <math>5</math>, <math>7</math>, and <math>35</math>.
  2 KB (277 words) - 16:21, 29 April 2023
• Rearrangement Inequality
  ...in increasing order, then the maximum sum is just <math>-a_1b_k - a_2b_{k-1} + \ldots</math>. Thus, by negating all values the inequality follows.
  5 KB (804 words) - 13:54, 26 January 2023
• Real number
  ...umber line. It can have any value. Some examples of real numbers are:<math>1, 2, -23.25, 0, \frac{\pi}{\phi}</math>, and so on. Numbers that are not rea (1) <math>\alpha\neq\mathbb{Q}</math> and <math>\alpha</math> is [[bounded]] a
  3 KB (496 words) - 23:22, 5 January 2022
• Divisibility rules
  Rule 1: Partition <math>N</math> into 3 digit numbers from the right (<math>d_3d_ [[Divisibility rules/Rule 1 for 7 proof | Proof]]
  8 KB (1,315 words) - 18:18, 2 March 2024
• Induction
  ...works for <math>n=1+1=2</math>, which in turn means it works for <math>n=2+1=3</math>, and so on. ...+1)}{2}</math> (the <math>n</math>th triangular number is defined as <math>1+2+\cdots +n</math>; imagine an [[equilateral polygon | equilateral]] [[tria
  5 KB (768 words) - 20:45, 1 September 2022
• Incenter/excenter lemma
  ...wever, <cmath>\angle BIL = \angle BAI + \angle ABI = \frac{1}{2} A + \frac{1}{2} B.</cmath> Hence, <math>\triangle BIL</math> is isosceles, so <math>LB
  2 KB (291 words) - 16:31, 18 May 2021
• Orthic triangle
  [[Image:Acute_orthic_triangle.png|thumb|right|300px|Case 1: <math>\triangle ABC</math> is acute.]] ...enter of <math>\triangle DEF</math>, <math>\angle EDC = 90^{\circ} - \frac{1}{2} \angle D</math>. Thus, <math>\angle ADC = 90^{\circ}</math>, and becaus
  8 KB (1,408 words) - 11:54, 8 December 2021
• Triangle
  {{asy image|<asy>draw((0,1)--(2,0)--(3,2)--cycle);</asy>|right|A triangle.}} {{asy image|<asy>draw((0,0)--(1,0)--(0.5,0.5)--cycle);</asy>|right|An isosceles triangle.}}
  4 KB (628 words) - 17:17, 17 May 2018
• Greatest common divisor
  ...use the recursive formula <math>GCD(a_1,\dots,a_n)=GCD(GCD(a_1,\dots,a_{n-1}),a_n)</math>.
  2 KB (288 words) - 22:40, 26 January 2021
• Overcounting
  Here is MATHCOUNTS 2008 National Target #1: Try to solve this. ...According to the formula we get 2 handshakes, but wait, we will have only 1 handshake between two persons. That means we have overcounted somewhere.
  4 KB (635 words) - 12:19, 2 January 2022
• Constructive counting
  === Example 1 === ...first digit is, we know that it removes one option, so there are <math>8 - 1 = 7</math> options for the second digit.
  12 KB (1,896 words) - 23:55, 27 December 2023
• Jensen's Inequality
  ...math> and let <math>a_1,\dots, a_n\ge 0</math> satisfy <math>a_1+\dots+a_n=1</math>. Then Let <math>\bar{x}=\sum_{i=1}^n a_ix_i</math>.
  3 KB (623 words) - 13:10, 20 February 2024
• Casework
  === Example 1 === '''Case 1''': The word is one letter long. Clearly, there are <math>5</math> of these
  5 KB (709 words) - 10:28, 19 February 2024
• Euclidean algorithm
  <math>r_{n-1} \pmod {r_n} \equiv 0</math><br> for <math>r_{k+1} < r_k < r_{k-1}</math><br>
  6 KB (924 words) - 21:50, 8 May 2022
• Conic section
  ...here the distance from a line (the directrix) is some number <math>0 < e < 1</math> times the distance to some fixed point (the focus). ...nce away from a point (focus) and a line (called the directrix) (<math>e = 1</math>).
  5 KB (891 words) - 01:14, 9 January 2023
• Generating function
  ...ath>A(k)={n \choose k}</math>, then we have <math>{n \choose 0}+{n \choose 1}x + {n \choose 2}x^2+\cdots+</math><math>{n \choose n}x^n</math>. .....+{n \choose n}=2^n</math>(let <math>{x}=1</math>), also <math>{n \choose 1}+{n \choose 3}+\cdots={n \choose 0}+{n \choose 2}+\cdots</math>.
  4 KB (659 words) - 12:54, 7 March 2022
• Binomial Theorem
  ...ath>, with coefficients <math>1 = \binom{5}{0}</math>, <math>5 = \binom{5}{1}</math>, <math>10 = \binom{5}{2}</math>, etc. ...(a+b)^{n} = \sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^{k}</math>, <cmath>(a+b)^{n+1} = (\sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^{k})(a+b)</cmath>
  5 KB (935 words) - 13:11, 20 February 2024
• Prime number
  ...integer]] <math>p>1</math> whose only positive [[divisor | divisors]] are 1 and itself. Note that <math>1</math> is usually defined as being neither prime nor [[composite number|com
  6 KB (985 words) - 12:38, 25 February 2024
• Recursion
  ...early values of the sequence in terms of previous values: <math>F_0=1, F_1=1, F_2=2, F_3=3, F_4=5, F_5=8</math>, and so on. ...defined recursively by <math>a_0 = 1</math> and <math>a_n = 2\cdot a_{n - 1}</math> for <math>n > 0</math> also has the closed-form definition <math>a_
  2 KB (316 words) - 16:03, 1 January 2024
• Function
  ...exactly one value in the second. For instance, one function may map 1 to 1, 2 to 4, 3 to 9, 4 to 16, and so on. This function has the rule that it ta ...'.) Often the inverse of a function <math>f</math> is denoted by <math>f^{-1}</math>.
  10 KB (1,761 words) - 03:16, 12 May 2023
• Complementary counting
  ...bility it does not happen is one. Thus, we have the identity <cmath>P(A) = 1 - P(A^c).</cmath> Like its counting analog, complementary probability often === Example 1 ===
  8 KB (1,192 words) - 17:20, 16 June 2023
• Combinatorial identity
  <math>{n \choose k}={n-1\choose k-1}+{n-1\choose k}</math> ...s added to the number of ways to choose <math>k</math> things from <math>n-1</math> things.
  12 KB (1,996 words) - 12:01, 18 May 2024
• Integer
  ...ath> are integers, then their sum <math>a+b</math>, their difference <math>a-b</math>, and their product <math>ab</math> are all integers (that is, the i ...</math> bits), which limits their maximum value (typically to <math>2^{31}-1</math> for signed <math>32</math>-bit integers). Integers in Python can be
  2 KB (296 words) - 15:04, 5 August 2022
• Prime factorization
  * [[2000 AIME I Problems/Problem 1]]
  3 KB (496 words) - 22:14, 5 January 2024
• Composite number
  ...''' is a [[positive integer]] with at least one [[divisor]] different from 1 and itself. Some composite numbers are <math>4=2^2</math> and <math>12=2\t Every positive integer either is prime, composite, or 1.
  6 KB (350 words) - 12:58, 26 September 2023
• Natural number
  A '''natural number''' is any positive [[integer]]: <math>\text{1, 2, 3, 4, 5, 6, 7,\dots}</math>. The set of '''natural numbers''', denoted
  1 KB (162 words) - 21:44, 13 March 2022
• Circle
  '''Case 1:''' The circle's area is greater than the triangle's area. ...han the radius so <math>a<r</math>. Therefore <math> P=\frac{1}{2}ap<\frac{1}{2}r\cdot 2\pi r=T</math>. However it cannot be both <math>P>T</math> and <
  9 KB (1,581 words) - 18:59, 9 May 2024
• Ellipse
  MC(90,"\mbox{semiminor axis}",7,D((0,0)--(0,3),green+linewidth(1)),E); MC("\mbox{semimajor axis}",7,D((0,0)--(5,0),red+linewidth(1)),S);
  5 KB (892 words) - 21:52, 1 May 2021
• Divisor
  ...ath>d(n)=(\alpha_{1} + 1)\cdot(\alpha_{2} + 1)\cdot\dots\cdot(\alpha_{m} + 1)</math>. It is often useful to know that this expression grows slower than * <math>{\sum_{n=1}^N d(n)=\left\lfloor\frac N1\right\rfloor+\left\lfloor\frac N2\right\rfloor
  1 KB (274 words) - 19:50, 29 August 2023
• Base numbers
  ...number can be rewritten as <math>2746_{10}=2\cdot10^3+7\cdot10^2+4\cdot10^1+6\cdot10^0.</math> ...six <math>10^0</math>'s, the second digit tells us there are four <math>10^1</math>'s, the third digit tells us there are seven <math>10^2</math>'s, and
  4 KB (547 words) - 17:23, 30 December 2020
• Base numbers/Common bases
  ...ong computer programmers. It has just two digits: <math>0</math> and <math>1</math>. [[Hexadecimal]] is base 16. The digits in hexadecimal are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, and F. One of its common uses is f
  2 KB (351 words) - 10:39, 1 October 2015
• Partition of an interval
  ...thcal{P}</math> is defined as <math>\|\mathcal{P}\|=\sup\{x_i-x_{i-1}\}_{i=1}^n</math> ...he set of ordered pairs <math>\mathcal{\dot{P}}=\{([x_{i-1},x_i],t_i)\}_{i=1}^n</math>.
  1 KB (178 words) - 20:34, 6 March 2022
• Henstock-Kurzweil integral
  Consider the function <math>f:[0,1]\rightarrow\mathbb{R}</math> <math>f\left( \frac{1}{n}\right) =n\forall n\in\mathbb{N}</math>
  2 KB (401 words) - 09:46, 31 January 2018
• 2001 AMC 12 Problems/Problem 1
  {{duplicate|[[2001 AMC 12 Problems|2001 AMC 12 #1]] and [[2001 AMC 10 Problems|2001 AMC 10 #3]]}}
  788 bytes (120 words) - 10:32, 8 November 2021
• 2001 AMC 12 Problems/Problem 2
  {{AMC12 box|year=2001|num-b=1|num-a=3}}
  1,007 bytes (165 words) - 00:28, 30 December 2023
• Square (geometry)
  A = (-1, 1); B = (1, 1);
  1 KB (169 words) - 01:12, 13 June 2022
• Mathematical Kangaroo
  ...ngaroo is an international mathematical competition for students in grades 1 through 12. The competition consists of a single round that is taken on th Any student that is currently enrolled in grades 1 through 12 (or homeschooled equivalent) is eligible to participate. They m
  6 KB (936 words) - 15:38, 22 February 2024
• American Scholastic Mathematics Association
  Contest #1 - October 10, 2019
  1 KB (153 words) - 13:11, 14 May 2019
• Least common multiple
  <math>12 = 2^2\cdot 3^1</math> <math>15 = 3^1\cdot 5^1</math>
  2 KB (383 words) - 10:49, 4 September 2022
• Common factorizations
  ...llet}&&x^{2n+1}+y^{2n+1}&=(x+y)(x^{2n}-x^{2n-1}y+x^{2n-2}y^2-\ldots-xy^{2n-1}+y^{2n})\\ \text{\textbullet}&&x^{n}-y^{n}&=(x-y)(x^{n-1}+x^{n-2}y+\cdots +xy^{n-2}+y^{n-1})
  2 KB (327 words) - 02:06, 28 April 2024
• Math Bee
  ...evel I</u>: 0.5 - 1<br><u>Level II</u>: 0.5 - 1.5<br><u>Level III</u>: 1 - 1.5}} * Level I: For grades 1 and 2.
  1 KB (197 words) - 10:59, 14 April 2024
• Math Day at the Beach
  == Individual Round - Part 1 == * 1 point is awarded for each question left blank
  4 KB (644 words) - 12:56, 29 March 2017
• Ptolemy's Inequality
  BD = \frac{BA \cdot DC }{AP} \; (1) ...both sides of the inequality by <math>BD</math> and using equations <math>(1) </math> and <math>(2) </math> gives us
  3 KB (602 words) - 09:01, 7 June 2023
• Median of a triangle
  A = (1, 2); draw(B--M, StickIntervalMarker(1));
  1 KB (185 words) - 20:24, 6 March 2024
• Pi
  ...e most surprising places, such as in the sum <math>\sum_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6}</math>. Some common [[fraction]]al approximations fo ...rmula for pi is <math>4\left( \sum_{i = 0}^\infty (-1)^i \left(\frac{1}{2n+1}\right)\right) </math>. This can be computed to the desired degree of accur
  8 KB (1,469 words) - 21:11, 16 September 2022
• Phi
  ...ently in mathematical writing, often to represent the constant <math>\frac{1+\sqrt{5}}{2}</math>. (The Greek letter [[Tau]] (<math>\tau</math>) was also ...as well as the positive solution of the [[quadratic equation]] <math>x^2-x-1=0</math>.
  2 KB (302 words) - 14:04, 1 January 2024
• Fibonacci sequence
  ...term is the sum of the two preceding it. The first few terms are <math>1, 1, 2, 3, 5, 8, 13, 21, 34, 55,...</math>. ...[[recursion|recursively]] as <math>F_1 = F_2 = 1</math> and <math>F_n=F_{n-1}+F_{n-2}</math> for <math>n \geq 3</math>. This is the simplest nontrivial
  6 KB (957 words) - 23:49, 7 March 2024
• Muirhead's Inequality
  ...sequence <math>(5,1)</math> majorizes <math>(4,2)</math> (as <math>5>4, 5+1=4+2</math>), Muirhead's inequality states that for any positive <math>x,y</ <cmath>x^5y^1+y^5x^1\geq x^4y^2+y^4x^2</cmath>
  8 KB (1,346 words) - 12:53, 8 October 2023
• Cardinality
  ...y of <math>\{3, 4\}</math> is 2, the cardinality of <math>\{1, \{2, 3\}, \{1, 2, 3\}\}</math> is 3, and the cardinality of the [[empty set]] is 0.
  2 KB (263 words) - 00:54, 17 November 2019
• Injection
  ...thbb R</math>, <math>f(x)=x^2</math> is not an injection (<math>f(-1)=f(1)=1</math>), the function <math>g:[0,\infty)\to\mathbb R</math>, <math>g(x)=x^2
  1 KB (228 words) - 01:01, 17 November 2019
• Fundamental Theorem of Calculus
  ...ve.) Interpret the distance that the object travels between times <math>t=1</math> and <math>t=2</math> geometrically, as an area under a curve. ...is the amount that <math>F</math> changes on the interval <math>[x_i, x_{i+1}]</math>, then <math>\Delta F_i \approx F'(x_i)\Delta x</math>.
  11 KB (2,082 words) - 15:23, 2 January 2022
• Polygon
  ...-- it will have "degenerated" from an <math>n</math>-gon to an <math>(n - 1)</math>-gon. (In the case of triangles, this will result in either a line
  2 KB (372 words) - 19:04, 30 May 2015
• Isoperimetric Inequalities
  ...h>A</math> and perimeter <math>P</math> then <math>\frac{4\pi A}{P^2} \leq 1</math>. This means that given a perimeter <math>P</math> for a plane figure
  789 bytes (115 words) - 17:08, 29 December 2021
• Writing scholarships
  ...s://seasonsbeachcottage.com/mental-health-education-scholarship-2023-round-1 Mental health Education Annual Scholarship 2023] Applicants must be current ...g Scholarship] Zampi, Inc. A small business marketing platform is offering 1,000 dollar scholarship. To enter, simple create a small business marketing
  7 KB (1,039 words) - 18:45, 18 January 2024
• National Scholarships
  * [[Best Buy @15 Scholarship]] of <dollar/>1,000 for students in grades 9-12. [https://www.at15.com/contests_scholarship
  4 KB (538 words) - 00:48, 28 January 2024
• 2011 AMC 10B Problems/Problem 22
  A pyramid has a square base with sides of length <math>1</math> and has lateral faces that are equilateral triangles. A cube is plac == Solution 1 ==
  4 KB (691 words) - 18:38, 19 September 2021
• 2011 AMC 10B Problems/Problem 4
  == Solution 1==
  1 KB (249 words) - 13:05, 24 January 2024
• Bretschneider's formula
  <math>[ABCD]=\frac{1}{4} \cdot \sqrt{4p^2q^2-(b^2+d^2-a^2-c^2)^2}</math>. ...vec{c} - \vec{d}</math>. The area of any such quadrilateral is <math>\frac{1}{2} |\vec{p} \times \vec{q}|</math>.
  3 KB (566 words) - 03:51, 12 February 2021
• Vector
  pen p=linewidth(1); MA("\theta",(5,-1),(2,3),(4,6),0.3,9,yellow);
  7 KB (1,265 words) - 13:22, 14 July 2021
• Logarithm
  ...bers. ex. :<cmath>((((((3^5)^6)^7)^8)^9)^{10})^{11}=\underbrace{1177\ldots 1}_{\text{793549 digits}}</cmath> would be a pain to have to calculate any ti <cmath>\log_4(5)\approx 1.160964047443681173935159715\ldots</cmath>
  4 KB (680 words) - 12:54, 16 October 2023
• Law of Cosines
  ==Proof 1== label("A",A,(1,0));
  6 KB (1,003 words) - 09:11, 7 June 2023
• Schur's Inequality
  <cmath>a^r(a-b)(a-c)+b^r(b-a)(b-c)+c^r(c-a)(c-b) \geq 0</cmath> The <math>r=1</math> case yields the well-known inequality:
  2 KB (398 words) - 16:57, 29 December 2021
• Trigonometric identities
  * <math>\sin^2x + \cos^2x = 1</math> * <math>1 + \cot^2x = \csc^2x</math>
  8 KB (1,397 words) - 21:55, 20 January 2024
• Irrational number
  ...>\sqrt{2}</math> as the length of a diagonal of a square with side lengths 1 in the sixth century <math>B.C</math>. The Pythagoreans lived by the doctri <math>3\frac{10}{71}</math> <math>to</math> <math>3\frac{1}{7}</math>.
  3 KB (368 words) - 19:26, 6 June 2015
• Complex number
  ...numbers''' arise when we try to solve [[equation]]s such as <math> x^2 = -1 </math>. ...this addition, we are not only able to find the solutions of <math> x^2 = -1 </math> but we can now find ''all'' solutions to ''every'' polynomial. (Se
  5 KB (860 words) - 15:36, 10 December 2023
• Rational number
  ...ecause every integer <math>a</math> can be represented as <math>a=\frac{a}{1}</math>
  1 KB (207 words) - 15:51, 25 August 2022
• Roots of unity
  ...the [[complex number|complex]] [[root]]s of the [[polynomial]] <math> x^n=1 </math>. ...ik}</math>. The magnitude of the RHS is 1, making <math>r^n=1\Rightarrow r=1</math> (magnitude is always expressed as a positive real number). This lea
  3 KB (558 words) - 21:36, 11 December 2011
• Angle bisector
  markangle(n=1,radius=20,D,A,F,green); markangle(n=1,radius=22,F,A,B,green);
  3 KB (575 words) - 15:27, 19 March 2023
• Without loss of generality
  * [[2005_Austrian_Mathematical_Olympiad_Final_Round-Part 1/Problem 5]]
  2 KB (280 words) - 15:30, 22 February 2024
• Law of Sines
  === Method 1 === real r = 1;
  4 KB (658 words) - 16:19, 28 April 2024
• Sequence
  ...<math>\mathbb{N}</math> corresponds to the sequence <math>X = (x_n) = (0, 1, 4, 9, 16, \ldots)</math>. A classic example of convergence would be to show that <math>1/n\to 0</math> as <math>n\to \infty</math>.
  2 KB (413 words) - 21:18, 13 November 2022
• Geometric sequence
  ...ratio <math>-1/2</math>; however, <math>1, 3, 9, -27</math> and <math>-3, 1, 5, 9, \ldots</math> are not geometric sequences, as the ratio between cons ...ogression if and only if <math>a_2 / a_1 = a_3 / a_2 = \cdots = a_n / a_{n-1}</math>. A similar definition holds for infinite geometric sequences. It ap
  4 KB (644 words) - 12:55, 7 March 2022
• Arithmetic sequence
  ...h>1, 2, 3, 4</math> is an arithmetic sequence with common difference <math>1</math> and <math>99, 91, 83, 75</math> is an arithmetic sequence with commo ...ogression if and only if <math>a_2 - a_1 = a_3 - a_2 = \cdots = a_n - a_{n-1}</math>. A similar definition holds for infinite arithmetic sequences. It a
  4 KB (736 words) - 02:00, 7 March 2024
• Fermat's Last Theorem
  ...n unique factorization in [[ring]]s of the form <math>\mathbb{Z}[\sqrt[n]{-1}]</math>. Unfortunately, this is not often the case. In fact, it has now
  3 KB (453 words) - 11:13, 9 June 2023
• Homothety
  * A homothety with factor <math>-1</math> is a <math>180^\circ</math> rotation about the center.
  3 KB (532 words) - 01:11, 11 January 2021
• 2005 USAMO Problems/Problem 4
  ...a piece of length <math>k_i</math> from the end of leg <math>L_i \; (i = 1,2,3,4)</math> and still have a stable table? == Solution 1 ==
  7 KB (1,276 words) - 20:51, 6 January 2024
• 2000 AMC 12 Problems/Problem 1
  ...0 AMC 12 Problems|2000 AMC 12 #1]] and [[2000 AMC 10 Problems|2000 AMC 10 #1]]}} == Solution 1 (Verifying the Statement)==
  2 KB (276 words) - 05:25, 9 December 2023
• Diophantine equation
  ...there will always be an infinite number of solutions when <math>\gcd(a,b)=1</math>. If <math>\gcd(a,b)\nmid c</math>, then there are no solutions to t .../math> is an [[odd]] number, then <math>m, \frac {m^2 -1}{2}, \frac {m^2 + 1}{2}</math> is a Pythagorean triple.
  9 KB (1,434 words) - 13:10, 20 February 2024
• Area
  ...ices to the [[circumcenter]]. This creates a triangle that is <math>\frac{1}{n},</math> of the total area (consider the regular [[octagon]] below as an ..._b^{-1})(H-h_c^{-1})}</math>, where <math>H=\frac{(h_a^{-1}+h_b^{-1}+h_c^{-1})}{2}</math> and the triangle has altitudes <math>h_a</math>, <math>h_b</ma
  6 KB (1,181 words) - 22:37, 22 January 2023
• Fraction
  ...the denominator such as <math>\frac{3}{3}</math> is always equal to <math>1</math>. ...d denominator by the same quantity (since we're essentially multiplying by 1).
  3 KB (432 words) - 19:34, 11 June 2020
• Functional equation
  ...'.) Often the inverse of a function <math>f</math> is denoted by <math>f^{-1}</math>. ...le of such a function is <math>f(x) = 1/x</math> because <math>f(f(x)) = f(1/x) = x</math>. Cyclic functions can significantly help in solving function
  2 KB (361 words) - 14:40, 24 August 2021
• Exponential form
  A=(0,1); C=(1,0);
  1 KB (238 words) - 22:51, 20 February 2022
• Floor function
  ...loor+\left\lfloor a+\frac{1}{n}\right\rfloor+\ldots+\left\lfloor a+\frac{n-1}{n}\right\rfloor</cmath> ...te the largest integer not exceeding <math>x</math>. For example, <math>[2.1]=2</math>, <math>[4]=4</math> and <math>[5.7]=5</math>. How many positive i
  3 KB (508 words) - 21:05, 26 February 2024
• Pascal's triangle
  ...sum of the two numbers appearing above it. For example, <math>{5 \choose 1}+{5 \choose 2} = 5 + 10 = 15 = {6 \choose 2}</math>. This property allows ...math>{6 \choose 0}+{5 \choose 1}+{4 \choose 2}+{3 \choose 3} = 1 + 5 + 6 + 1 = 13 = F(7)</math>. A "shallow diagonal" is plotted in the diagram.
  5 KB (838 words) - 17:20, 3 January 2023
• 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
• Exponentiation
  ...words, we want to divide by 2 four times. Therefore, <math> 2^{-4}=\frac 1{2^4}.</math> * <math>b^0 = 1</math> if <math>b \neq 0</math>
  5 KB (803 words) - 16:25, 10 August 2020
• Rational approximation
  ...en denominator <math>q\ge 1</math> with an error not exceeding <math>\frac 1{2q}</math>. ...interval <math>\left[qx-\frac12,qx+\frac12\right]</math> has length <math>1</math> and, therefore, contains at least one integer. Choosing <math>p</mat
  7 KB (1,290 words) - 12:18, 30 May 2019
• Trigonometry
  ...d as the reciprocal of the sine of <math>A</math>. <cmath>\csc (A) = \frac{1}{\sin (x)} = \frac{\textrm{hypotenuse}}{\textrm{opposite}} = \frac{c}{a}.</ ...as the reciprocal of the cosine of <math>A</math>. <cmath>\sec (A) = \frac{1}{\cos (x)} = \frac{\textrm{hypotenuse}}{\textrm{adjacent}} = \frac{c}{b}.</
  8 KB (1,217 words) - 20:15, 7 September 2023
• Euler's number
  ...>e</math> such that <math>\ln e=1</math>, where <math>\ln x=\int_1^x \frac{1}{t} \, dt</math>. It has been shown to be both [[irrational]] and [[transce ...tion such that <math>\frac{d}{dx} \exp(x)=\exp(x)</math> and <math>\exp(0)=1</math> is equal to <math>e^x</math>.
  4 KB (764 words) - 21:09, 13 March 2022
• Exponential function
  :<math>f\left( x \right) = a\left( {{1 \over 2}} \right)^{{x \over h}} :If <math>b > 1</math>, then the function will show growth.
  2 KB (312 words) - 15:57, 6 March 2022
• Mathematical Olympiad Summer Program
  ...three instructional sessions: 8:30 AM - 10:00 AM, 10:15 AM - 11:45 AM, and 1:15 PM - 2:45 PM. Classes usually consist of a lecture followed by a problem ...ered in place of the afternoon lecture, and is graded with comments within 1-2 days.
  6 KB (936 words) - 10:37, 27 November 2023
• Complex analysis
  ...h> f^{(n)}(z_0)=\frac{n!}{2\pi i} \oint_\Gamma \frac{f(z)\; dz}{(z-z_0)^{n+1}}. </cmath>
  2 KB (271 words) - 22:06, 12 April 2022
• Root-Mean Square-Arithmetic Mean-Geometric Mean-Harmonic mean Inequality
  ...cdots+x_a^{n_4}}{a}}</cmath> where <math>n_1>1,~~0<n_2<1,~~-1<n_3<0,~~n_4<-1</math>, and <math>n</math> is the root mean power. ...s greater than or equal to 1. This creates the indeterminate form of <math>1^{\infty}</math>. Then, we can say that the limit as x goes to 0, the result
  5 KB (912 words) - 20:06, 14 March 2023
• Harmonic series
  <math>\sum_{i=1}^{\infty}\frac{1}{i}=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots</math> <math>\sum_{i=1}^{\infty}\frac{(-1)^{i+1}}{i}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots</math> , though, approaches <math>\ln 2</math>.
  2 KB (334 words) - 20:52, 13 March 2022
• Goldbach Conjecture
  He considered 1 to be a prime number, a [[mathematical convention|convention]] subsequently ...m</math> and <math>n-m</math> simultaneously being prime to be <math>\frac{1}{\ln m \cdot \ln (n-m)}</math>. This heuristic is non-rigorous for a numbe
  7 KB (1,201 words) - 16:59, 19 February 2024
• Twin Prime Conjecture
  ...1}{5}+\frac{1}{5}+\frac{1}{7}+\frac{1}{11}+\frac{1}{13}+\frac{1}{17}+\frac{1}{19}+\cdots</math></center> ...y if one of variables is 0 will the factorization be trivial (contain only 1 and itself).
  2 KB (308 words) - 02:27, 1 May 2024
• Fallacy
  === 2 = 1 === <div style="text-align:center"><math> 2 = 1 </math> (dividing by <math>a^2-ab</math>)</div>
  2 KB (429 words) - 08:27, 5 June 2013
• Riemann Hypothesis
  ...[[root |zero]]s of the [[Riemann zeta function]] have [[real part]] <math>1/2</math>. From the [[functional equation]] for the zeta function, it is eas ...atement of the Riemann hypothesis is that <math>\pi(x)=\mathrm{Li}(x)+O(x^{1/2}\ln(x))</math>.
  2 KB (425 words) - 12:01, 20 October 2016
• Quadratic residues
  ...\mathrm{ and }\ a\ \mathrm{\ is\ a\ quadratic\ residue\ modulo\ }\ p, \\ -1 & \mathrm{if }\ p\nmid a\ \mathrm{ and }\ a\ \mathrm{\ is\ a\ quadratic\ no ...left(\frac{p}{q}\right)\left(\frac{q}{p}\right)=(-1)^{\frac{p-1}{2}\frac{q-1}{2}}</math>. This is known as the [[Quadratic Reciprocity Theorem]].
  5 KB (778 words) - 13:10, 29 November 2017
• Power of a Point Theorem
  ''Proof.'' We have already proven the theorem for a <math>1</math>-sphere (i.e., a circle), so it only remains to prove the theorem for ==== Problem 1 ====
  5 KB (827 words) - 17:30, 21 February 2024
• Fundamental Theorem of Algebra
  ...loss of generality, suppose that <math>P</math> is [[monic]]. Then <math>1/P(z)</math> is an [[entire]] function; we wish to show that it is bounded. ...absolute values of the coefficients of <math>P</math>, so that <math>R \ge 1</math>. Then for <math>\lvert z \rvert \ge R</math>,
  5 KB (832 words) - 14:22, 11 January 2024
• Collatz Problem
  ...>, the sequence <math>\{n,f(n),f(f(n)),f(f(f(n))),\ldots\}</math> contains 1. This conjecture is still open. Some people have described it as the easies Observing that if <math>n=2m+1</math> then <math>3n+1=6m+4</math>, as well as: <cmath>{6m+4\over 2}=3m+2</cmath> we can then obse
  1 KB (231 words) - 19:45, 24 February 2020
• Algebraic number
  ...ynomial]] with [[rational]] [[coefficient]]s. Examples include <math>\frac{1}{3}</math>, <math>\sqrt{2}+\sqrt{3}</math>, <math>i</math>, and <math>\frac
  1,006 bytes (151 words) - 21:56, 22 April 2022
• Incenter
  D=intersectionpoint(A+0.1*expi((angle(B-A)+angle(C-A))/2)--A+20*expi((angle(B-A)+angle(C-A))/2), circ dot(A^^B^^C^^D^^I);label("A",A,(0,1));label("B",B,(-1,0));label("C",C,(1,0));
  2 KB (381 words) - 19:38, 24 November 2011
• Triangle Inequality
  ...+ \ldots + a_n = \left(\sum_{j=1}^n a_j\right) - a_i</math> for <math>i = 1, 2 \ldots, n</math>. Expressing the inequality in this form leads to <math
  2 KB (268 words) - 03:02, 3 January 2021
• Majorization
  ...for all <math> 1 \le k \le n </math>, <math> \sum_{i=1}^{k}a_i \ge \sum_{i=1}^{k}b_i </math>, with equality when <math>k = n </math>. If <math>\{a_i\} ...h>\sum_{i=k}^n a_i \le \sum_{i=k}^n b_i</math>, with equality when <math>k=1 </math>. An interesting consequence of this is that the finite sequence <m
  2 KB (288 words) - 22:48, 5 July 2023
• Surjection
  ...function <math>f: \mathbb N \to\mathbb N</math> defined by <math>f(x) = x+1</math> is not surjective because there exists a [[natural number]] which is
  794 bytes (131 words) - 22:39, 13 May 2020
• International Mathematical Olympiad
  ...|region=International|type=Proof|difficulty=5.5 - 10|breakdown=<u>Problem 1/4</u>: 6.5<br><u>Problem 2/5</u>: 7.5-8<br><u>Problem 3/6</u>: 9.5<br><u>Pr * Gold - the top 1/12 of individual scores.
  3 KB (490 words) - 03:32, 23 July 2023
• Prime Number Theorem
  <cmath> \lim_{x\to \infty} \frac{\pi(x) \log x}{x} = 1 . </cmath> with <math>A(x)</math> tending to some constant number around 1.08366.
  10 KB (1,729 words) - 19:52, 21 October 2023
• ...a_a, \lambda_b, \dotsc, \lambda_z</math> are nonnegative reals with sum of 1, then ...f{a}</math> and <math>\mathbf{b}</math>, and <math>\lambda_a = \lambda_b = 1/2</math>, this is the elementary form of the [[Cauchy-Schwarz Inequality]].
  4 KB (774 words) - 12:12, 29 October 2016
• Infinite
  ...ot empty and cannot be put into bijection with any set of the form <math>\{1, 2, \ldots, n\}</math> for a [[positive integer]] <math>n</math>. <math>\sum_{i = 3}^{\infty}{(2i - 1)}</math>
  1 KB (186 words) - 23:19, 16 August 2013
• Isosceles trapezoid
  {{asy image|<asy>draw((0,1)--(1,5)--(3,5)--(4,1)--cycle);</asy>|right|An isosceles trapezoid.}}
  577 bytes (81 words) - 10:33, 18 April 2019
• Mock AMC
  ! scope="row" | '''Mock AMC #1''' | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=9572 1-5]
  51 KB (6,175 words) - 20:58, 6 December 2023
• Mock AIME
  ...Y2QwOTc3NWZiYjY0LnBkZg==&rn=TWlsZG9yZiBNb2NrIEFJTUUucGRm Mildorf Mock AIME 1] ** [[Mock AIME 1 2006-2007]]
  8 KB (906 words) - 17:30, 26 April 2024
• Commutative property
  ...n <math>AB</math> and <math>BA</math> are matrices of sizes <math>1 \times 1</math> and <math>2 \times 2</math>, respectively.
  2 KB (301 words) - 17:46, 16 March 2012
• Permutation
  ...ath>. We often drop the brackets and commas, so the permutation <math>\{2,1,3\}</math> would just be represented by <math>213</math>. ...ent objects we can choose from. For the second element, there are <math>n-1</math> objects we can choose, <math>n-2</math> for the third, and so on. I
  3 KB (422 words) - 11:01, 25 December 2020
• Algebraic topology
  ...ith <math>g(a,0)=p(a)</math>, <math>g(a,1)=q(a)</math>, and <math>g(0,b)=g(1,b)=x</math>. We call <math>g</math> a [[homotopy]]. Now define <math>\pi_1( ...>g\cdot h(a)=\begin{cases} g(2a) & 0\le a\le 1/2, \\ h(2a-1) & 1/2\le a\le 1.\end{cases}</math> One can check that this is indeed [[well-defined]].
  3 KB (479 words) - 15:35, 1 December 2015
• Bertrand's Postulate
  <math>\frac{2^{2n}}{(2n+1)}\le{\binom{2n}{n}}\le <math>\left(\prod_{n<p\le{2n}}p\right)\ge \frac{4^{\frac n3}}{(2n+1)(2n)^{\sqrt {2n}}}</math>
  2 KB (309 words) - 21:43, 11 January 2010
• Riemann zeta function
  <cmath>\zeta (s)=\sum_{n=1}^{\infty}\frac{1}{n^s}= 1+\frac{1}{2^s}+\frac{1}{3^s}+\frac{1}{4^s}+\cdots</cmath>
  9 KB (1,547 words) - 03:04, 13 January 2021
• Mock USAMO
  ...ww.artofproblemsolving.com/Forum/viewtopic.php?p=423926#p423926 Mock USAMO 1 2006] ...http://www.artofproblemsolving.com/Forum/viewtopic.php?t=200572 Mock USAMO 1 2008]
  2 KB (205 words) - 19:56, 4 March 2020
• Modular arithmetic/Introduction
  <math>1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, \ldots </math> This is the way in which we count in '''modulo 12'''. When we add <math>1</math> to <math>11</math>, we arrive back at <math>0</math>. The same is t
  15 KB (2,396 words) - 20:24, 21 February 2024
• Perfect square
  ...th> so that <math>m^2=n</math>. The first few perfect squares are <math>0, 1, 4, 9, 16, 25, 36</math>. ...th> square numbers (starting with <math>1</math>) is <math>\frac{n(n+1)(2n+1)}{6}</math>
  954 bytes (155 words) - 01:14, 29 November 2023
• Convex function
  ...Equivalently, <math>f </math> is convex if for every <math> \lambda \in [0,1] </math> and every <math> x,y \in I</math>, <center><math>\lambda f(x) + (1-\lambda)f(y) \ge f\left( \lambda x + (1-\lambda) y \right) </math>.</center>
  2 KB (417 words) - 00:10, 20 February 2016
• Uncountable
  Suppose that the set <math>A=\{x\in\mathbb{R}:0<x< 1\}</math> is countable. Let <math>\{\omega_1, \omega_2, \omega_3, ...\}</mat ...ath>. However, <math>\omega</math> is clearly a real number between 0 and 1, a [[contradiction]]. Thus our assumption that <math>A</math> is countable
  2 KB (403 words) - 20:53, 13 October 2019
• Gamma function
  ...<math>\Re(z)>0</math>, we define <cmath>\Gamma(z)=\int_0^\infty e^{-t}t^{z-1}\; dt</cmath> ...f by one. Since <math>\Gamma(1)=1</math>, we therefore have <math>\Gamma(n+1)=n!</math> for nonnegative integers <math>n</math>. But with the integral,
  840 bytes (137 words) - 22:26, 22 June 2009
• Modular arithmetic/Intermediate
  * Subtraction: <math>a-c\equiv b-d\pmod {n}</math>. <math>\{ \ldots, -5, -2, 1, 4, 7, \ldots \}</math>
  14 KB (2,317 words) - 19:01, 29 October 2021
• Right triangle
  A = (0, 1); B = (1, 0);
  3 KB (499 words) - 23:41, 11 June 2022
• Subset
  ...thbb{C} \subset \mathbb{C}\cup\{\textrm{Groucho, Harpo, Chico}\} \supset \{1, 2, i, \textrm{Groucho}\}</math>
  1 KB (217 words) - 09:32, 13 August 2011
• 2006 AMC 10A
  * [[2006 AMC 10A Problems/Problem 1]]
  2 KB (180 words) - 18:06, 6 October 2014
• Limit
  ...the value of the function <math>f(x)</math> at <math>x = 0</math> is <math>1</math>, the limit <math>\lim_{x\rightarrow 0} f(x)</math> is, in fact, zero ...ons <math>d_A(a,b)</math> and <math>d_B(a,b)</math> are both simply <math>|a-b|</math>. We then obtain the following definition commonly found in calculu
  7 KB (1,325 words) - 13:51, 1 June 2015
• 2006 AIME I Problems
  == Problem 1 == [[2006 AIME I Problems/Problem 1|Solution]]
  7 KB (1,173 words) - 03:31, 4 January 2023
• 2006 AIME I Problems/Problem 15
  ...</math> and <math> |x_k|=|x_{k-1}+3| </math> for all integers <math> k\ge 1, </math> find the minimum possible value of <math> |x_1+x_2+\cdots+x_{2006} === Solution 1 ===
  6 KB (910 words) - 19:31, 24 October 2023
• 2006 AIME I Problems/Problem 14
  ...s <math>4</math> feet from the ground. In setting up the tripod, the lower 1 foot of one leg breaks off. Let <math> h </math> be the height in feet of t == Solution 1 ==
  6 KB (980 words) - 21:45, 31 March 2020
• 2006 AIME I Problems/Problem 13
  ...or 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 s == Solution 1 ==
  10 KB (1,702 words) - 00:45, 16 November 2023
• 2006 AIME I Problems/Problem 11
  ...e with [[edge]]-[[length]] <math> k </math> for each [[integer]] <math> k, 1 \le k \le 8. </math> A tower is to be built using all 8 cubes according to ...th>, so there are <math>2\cdot 3^6 = 1458</math> towers using blocks <math>1, 2, \ldots, 8</math>, so the answer is <math>\boxed{458}</math>.
  3 KB (436 words) - 05:40, 4 November 2022
• 2006 AIME I Problems/Problem 10
  ...math> c </math> are positive integers whose [[greatest common divisor]] is 1. Find <math> a^2+b^2+c^2. </math> for(i=0; i<3; i=i+1) {
  4 KB (731 words) - 17:59, 4 January 2022
• 2006 AIME I Problems/Problem 9
  == Solution 1 == ...s <math>1 + \frac{1001-11}{22}=1 + \frac{990}{22}=46</math>. (We must add 1 because both endpoints are being included.) So the answer is <math>\boxed{0
  4 KB (651 words) - 18:27, 22 May 2021
• 2006 AIME I Problems/Problem 8
  ...1.65,-1.6), G=(0.45,-1.6), H=(3.75,-1.6), I=(2.1,0), J=(2.1,-3.2), K=(2.1,-1.6); draw(F--G--(2.1,0));
  5 KB (730 words) - 15:05, 15 January 2024
• 2006 AIME I Problems/Problem 7
  for(int i=0; i<4; i=i+1) { fill((2*i,0)--(2*i+1,0)--(2*i+1,6)--(2*i,6)--cycle, mediumgray);
  4 KB (709 words) - 01:50, 10 January 2022
• 2006 AIME I Problems/Problem 6
  == Solution 1 == ...ple, <math>.\overline{123}+.\overline{876}=.\overline{999} \Longrightarrow 1</math>.
  2 KB (237 words) - 19:14, 20 November 2023
• 2006 AIME I Problems/Problem 5
  == Solution 1 ==
  3 KB (439 words) - 18:24, 10 March 2015
• 2006 AIME I Problems/Problem 4
  ...th>'s at the right end of the decimal representation of the product <math> 1!2!3!4!\cdots99!100!. </math> Find the remainder when <math> N </math> is di ...rom <math>100</math>. Thus, our final total is <math>970 + 76 + 51 + 26 + 1 = 1124</math>, and the answer is <math>\boxed{124}</math>.
  2 KB (278 words) - 08:33, 4 November 2022
• 2006 AIME I Problems/Problem 3
  ...en its leftmost [[digit]] is deleted, the resulting integer is <math>\frac{1}{29}</math> of the original integer. === Solution 1 ===
  4 KB (622 words) - 03:53, 10 December 2022
• 2006 AIME I Problems/Problem 2
  ...et]] <math> \mathcal{A} </math> be a 90-[[element]] [[subset]] of <math> \{1,2,3,\ldots,100\}, </math> and let <math> S </math> be the sum of the elemen ...ible values of S, so the number of possible values of S is <math>4995-4095+1=901</math>.
  1 KB (189 words) - 20:05, 4 July 2013
• 2022 AIME I Problems/Problem 1
  ==Solution 1 (Linear Polynomials)== P(16) &= &512 + 16a + b &= 54, \hspace{20mm}&&(1) \\
  4 KB (670 words) - 13:03, 13 November 2023
• 2021 JMC 10
  ==Problem 1== What is the value of <cmath>\dfrac{20}{2\cdot1} - \dfrac{2+0}{2/1}?</cmath>
  12 KB (1,784 words) - 16:49, 1 April 2021
• 2006 AMC 12B Problems
  == Problem 1 == What is <math>( - 1)^1 + ( - 1)^2 + \cdots + ( - 1)^{2006}</math>?
  13 KB (2,058 words) - 12:36, 4 July 2023
• 2006 AMC 12A
  * [[2006 AMC 12A Problems/Problem 1]]
  1 KB (168 words) - 21:51, 6 October 2014
• 2004 AMC 12A
  * [[2004 AMC 12A Problems/Problem 1]]
  2 KB (186 words) - 17:35, 16 December 2019
• 2005 AMC 12A
  ** [[2005 AMC 12A Problems/Problem 1|Problem 1]]
  2 KB (202 words) - 21:30, 6 October 2014
• 2002 AMC 12A
  * [[2002 AMC 12A Problems/Problem 1]]
  1 KB (158 words) - 21:33, 6 October 2014
• 2003 AMC 12A
  * [[2003 AMC 12A Problems/Problem 1]]
  1 KB (162 words) - 21:52, 6 October 2014
• 2006 AMC 12A Problems
  == Problem 1 == [[2006 AMC 12A Problems/Problem 1|Solution]]
  15 KB (2,223 words) - 13:43, 28 December 2020
• 2005 AMC 12A Problems
  == Problem 1 == (\mathrm {A}) \ 1 \qquad (\mathrm {B}) \ 2 \qquad (\mathrm {C})\ 5 \qquad (\mathrm {D}) \ 10
  13 KB (1,971 words) - 13:03, 19 February 2020
• 2004 AMC 12A Problems
  == Problem 1 == Alicia earns <math> 20</math> dollars per hour, of which <math>1.45\%</math> is deducted to pay local taxes. How many cents per hour of Ali
  13 KB (1,953 words) - 00:31, 26 January 2023
• 2003 AMC 12A Problems
  == Problem 1 == <math> \mathrm{(A) \ } 0\qquad \mathrm{(B) \ } 1\qquad \mathrm{(C) \ } 2\qquad \mathrm{(D) \ } 2003\qquad \mathrm{(E) \ } 40
  13 KB (1,955 words) - 21:06, 19 August 2023
• 2002 AMC 12A Problems
  == Problem 1 == [[2002 AMC 12A Problems/Problem 1|Solution]]
  12 KB (1,792 words) - 13:06, 19 February 2020
• 2000 AMC 12 Problems
  == Problem 1 == [[2000 AMC 12 Problems/Problem 1|Solution]]
  13 KB (1,948 words) - 12:26, 1 April 2022
• 2001 AMC 12 Problems
  == Problem 1 == [[2001 AMC 12 Problems/Problem 1|Solution]]
  13 KB (1,957 words) - 12:53, 24 January 2024
• 2002 AMC 12B Problems
  == Problem 1 == [[2002 AMC 12B Problems/Problem 1|Solution]]
  10 KB (1,547 words) - 04:20, 9 October 2022
• 2003 AMC 12B Problems
  == Problem 1 == ...xt {(B) } -\frac{2}{3} \qquad \text {(C) } \frac{2}{3} \qquad \text {(D) } 1 \qquad \text {(E) } \frac{14}{3}
  13 KB (1,987 words) - 18:53, 10 December 2022
• 2004 AMC 12B Problems
  == Problem 1 == [[2004 AMC 12B Problems/Problem 1|Solution]]
  13 KB (2,049 words) - 13:03, 19 February 2020
• 2005 AMC 12B Problems
  == Problem 1 == .../math> dollars. They sell all the candy bars at the price of two for <math>1</math> dollar. What was their profit, in dollars?
  12 KB (1,781 words) - 12:38, 14 July 2022
• 2006 AMC 12B Problems/Problem 1
  What is <math>( - 1)^1 + ( - 1)^2 + \cdots + ( - 1)^{2006}</math>? ...} - 2006 \qquad \text {(B) } - 1 \qquad \text {(C) } 0 \qquad \text {(D) } 1 \qquad \text {(E) } 2006
  468 bytes (70 words) - 10:40, 15 September 2017
• 2006 AMC 12B Problems/Problem 2
  {{AMC12 box|year=2006|ab=B|num-b=1|num-a=3}}
  473 bytes (71 words) - 10:44, 4 July 2013
• 2006 AMC 12B Problems/Problem 3
  ===Solution 1===
  910 bytes (136 words) - 13:39, 13 February 2016
• 2006 AMC 12B Problems/Problem 4
  ...e items are <math>7.99</math>, <math>4.99</math>, <math>2.99</math>, <math>1.99</math>, and <math>0.99</math>. Mary will pay with a twenty-dollar bill. The total price of the items is <math>(8-.01)+(5-.01)+(3-.01)+(2-.01)+(1-.01)=19-.05=18.95</math>
  1 KB (152 words) - 16:11, 8 December 2013
• 2006 AMC 12B Problems/Problem 5
  ...b is also walking east, but at a speed of 5 miles per hour. If Bob is now 1 mile west of John, how many minutes will it take for Bob to catch up to Joh ...iles per hour. Since Bob is one mile behind John, it will take <math>\frac{1}{2} \Rightarrow \text{(A)}</math> of an hour to catch up to John.
  654 bytes (115 words) - 21:47, 1 August 2020
• 2006 AMC 12B Problems/Problem 7
  == Solution 1 == Now there are <math>2 \times 1</math> ways to seat the adults.
  1 KB (213 words) - 15:33, 9 April 2024
• 2006 AMC 12B Problems/Problem 8
  ...a</math> and <math>y = \frac 14x + b</math> intersect at the point <math>(1,2)</math>. What is <math>a + b</math>?<!-- don't remove the following tag, \text {(A) } 0 \qquad \text {(B) } \frac 34 \qquad \text {(C) } 1 \qquad \text {(D) } 2 \qquad \text {(E) } \frac 94
  1 KB (235 words) - 00:46, 6 January 2022
• 2006 AMC 12B Problems/Problem 9
  ==Solution 1 (Alcumus Edition)== ...then <math>a</math> and <math>b</math> must each be chosen from the digits 1, 2, and 3. Therefore there are <math>\binom{3}{2}=3</math> choices for <mat
  3 KB (409 words) - 17:10, 30 April 2024
• 2006 AMC 12B Problems/Problem 11
  \text {(A) } \frac 67 \qquad \text {(B) } \frac {13}{14} \qquad \text {(C) } 1 \qquad \text {(D) } \frac {14}{13} \qquad \text {(E) } \frac 76 ...and drank <math>\frac{1}{7}</math> of it. Therefore, she drank <math>\frac{1}{7}</math> of her cream, leaving her <math>2*\frac{6}{7}</math>.
  927 bytes (137 words) - 10:45, 4 July 2013
• 2006 AMC 12B Problems/Problem 12
  == Solution 1==
  1 KB (239 words) - 22:20, 17 October 2020
• 2006 AMC 12B Problems/Problem 13
  pair A=origin, B=(2,0), C=(3, sqrt(3)), D=(1, sqrt(3)), E=(1, 1/sqrt(3)), F=(2, 2/sqrt(3)); == Solution 1==
  3 KB (447 words) - 03:49, 16 January 2021
• 2006 AMC 12B Problems/Problem 14
  ..., <math>J</math> and <math>N</math> are all positive integers with <math>N>1</math>. What is the cost of the jam Elmo uses to make the sandwiches? \mathrm{(A)}\ 1.05
  1 KB (227 words) - 17:21, 8 December 2013
• 2006 AMC 12B Problems/Problem 15
  O[1] = (6,0); draw(Circle(O[1],2));
  3 KB (458 words) - 16:40, 6 October 2019
• 2006 AMC 12B Problems/Problem 16
  ...tices <math>A</math> and <math>C</math> at <math>(0,0)</math> and <math>(7,1)</math>, respectively. What is its area? ...ore, the side length of the hexagon is <math>\frac{5\sqrt{2}}{2}\cdot\frac{1}{\sqrt{3}}\cdot2 = \frac{5\sqrt{6}}{3}</math>.
  1 KB (203 words) - 16:36, 18 September 2023
• 2006 AMC 12B Problems/Problem 17
  ...ath>, <math>5</math> and <math>6</math> on each die are in the ratio <math>1:2:3:4:5:6</math>. What is the probability of rolling a total of <math>7</ma ...>1</math> on the first and <math>6</math> on the second die is <math>\frac 1{21}\cdot\frac 6{21}</math>. Similarly we can express the probabilities for
  1 KB (188 words) - 22:10, 9 June 2016
• 2006 AMC 12B Problems/Problem 18
  == Solution 1 == If the object only makes <math>1</math> move, it is obvious that there are only 4 possible points that the o
  2 KB (354 words) - 16:57, 28 December 2020
• 2006 AMC 12B Problems/Problem 19
  If <math>b=1</math>, the number is not divisible by <math>2</math> (unless it's <math>18 ...e is even, because you can only discard one number from the integers <math>1</math> through <math>9</math>, inclusive (<math>8</math> children, oldest i
  4 KB (696 words) - 09:47, 10 August 2015
• 2006 AMC 12B Problems/Problem 20
  Let <math>x</math> be chosen at random from the interval <math>(0,1)</math>. What is the probability that ...both hold at the same time if and only if <math>10^k \leq x < \frac{10^{k+1}}4</math>.
  3 KB (485 words) - 14:09, 21 May 2021
• 2006 AMC 12B Problems/Problem 22
  == Solution 1== Consider <math>n! = 1\cdot 2 \cdot \dots \cdot n</math>. Each fifth term is divisible by <math>5<
  5 KB (881 words) - 15:52, 23 June 2021
• 2006 AMC 12B Problems/Problem 23
  Now we use <math>\sin^2(\alpha) + \cos^2(\alpha) = 1</math>. ...Rightarrow \frac{s^4-26s^2+169}{144s^2} + \frac{s^4-170s^2+7225}{144s^2} = 1 \\
  7 KB (1,169 words) - 14:04, 10 June 2022
• 2006 AMC 12B Problems/Problem 24
  D((1,-0.1)--(1,0.1)); D((2,-0.1)--(2,0.1));
  3 KB (563 words) - 22:45, 24 October 2021
• 2006 AMC 12B Problems/Problem 25
  ...1</math>. If <math>a_1=999</math>, <math>a_2<999</math> and <math>a_{2006}=1</math>, how many different values of <math>a_2</math> are possible? == Solution 1==
  5 KB (924 words) - 12:02, 15 June 2022
• 2006 AMC 12A Problems/Problem 10
  ...fect squares that are less than or equal to <math>120</math> are <math>\{0,1,4,9,16,25,36,49,64,81,100\}</math>, so there are <math>11</math> values for
  1 KB (167 words) - 23:23, 16 December 2021
• 2006 AMC 12A Problems/Problem 13
  ...1</math>, <math>r_B = 2</math>, and <math>r_C = 3</math> Then we get <math>1^2 \pi + 2^2 \pi + 3^2 \pi = 14 \pi \iff\mathrm{(E)}</math>
  1 KB (184 words) - 13:57, 19 January 2021
• 2006 AMC 12A Problems/Problem 14
  ==Solution 1 (Diophantine Equation)== ...et the smallest positive difference - <math>5\cdot 10 - 7\cdot 7 = 50-49 = 1</math>, since we can't make a non-integer with a linear combination of inte
  3 KB (442 words) - 03:13, 8 August 2022
• 2006 AMC 12A Problems/Problem 15
  Suppose <math>\cos x=0</math> and <math>\cos (x+z)=1/2</math>. What is the smallest possible positive value of <math>z</math>? *For <math>\cos (x+z) = 1 / 2</math>, <math>x + z = \frac{\pi}{3} +2\pi n, \frac{5\pi}{3} + 2\pi n</m
  919 bytes (138 words) - 12:45, 4 August 2017
• 2006 AMC 12A Problems/Problem 16
  pair C=intersectionpoints(Circle(M,2.5),Circle(A,3))[1];
  2 KB (286 words) - 10:16, 19 December 2021
• 2006 AMC 12A Problems/Problem 17
  <math> \mathrm{(A) \ } \frac{1}{2}\qquad \mathrm{(B) \ } \frac{5}{9}\qquad \mathrm{(C) \ } \frac{3}{5}\qqu === Solution 1 ===
  6 KB (958 words) - 23:29, 28 September 2023
• 2006 AMC 12A Problems/Problem 18
  ...the property that for each real number <math>x</math> in its domain, <math>1/x</math> is also in its domain and <math>f(x)+f\left(\frac{1}{x}\right)=x</math>
  2 KB (334 words) - 18:34, 18 September 2020
• 2006 AMC 12A Problems/Problem 19
  ...ath>L_2</math> and the x-axis, so <math>m=\tan{2\theta}=\frac{2\tan\theta}{1-\tan^2{\theta}}=\frac{120}{119}</math>. We also know that <math>L_1</math>
  2 KB (253 words) - 22:52, 29 December 2021
• 2006 AMC 12A Problems/Problem 20
  ...\frac{1}{729}\qquad \textbf{(C) } \frac{2}{243}\qquad \textbf{(D) } \frac{1}{81} \qquad \textbf{(E) } \frac{5}{243}</math> ==Solution 1==
  5 KB (908 words) - 19:23, 22 September 2022
• 2006 AMC 12A Problems/Problem 21
  <math>S_1=\{(x,y)|\log_{10}(1+x^2+y^2)\le 1+\log_{10}(x+y)\}</math> <math>\log_{10}(1+x^2+y^2)\le 1+\log_{10}(x+y)</math>
  2 KB (280 words) - 17:35, 17 September 2023
• 2006 AMC 12A Problems/Problem 22
  ...of hexagon are visible from a randomly chosen point on the circle is <math>1/2</math>. What is <math>r</math>? ...th>1 / 2</math>, or if the total arc degree measures add up to <math>\frac{1}{2} \cdot 360^{\circ} = 180^{\circ}</math>. Each arc must equal <math>\frac
  2 KB (343 words) - 15:39, 14 June 2023
• 2006 AMC 12A Problems/Problem 23
  <math>\left(\frac{a_1+a_2}{2},\frac{a_2+a_3}{2},\ldots ,\frac{a_{n-1}+a_n}{2}\right)</math> ...h>, and let <math>S=(1,x,x^2,\ldots ,x^{100})</math>. If <math>A^{100}(S)=(1/2^{50})</math>, then what is <math>x</math>?
  3 KB (466 words) - 22:40, 29 September 2023
• 2006 AMC 12A Problems/Problem 24
  ...thrm{(B) \ } 671,676\qquad \mathrm{(C) \ } 1,007,514\qquad \mathrm{(D) \ } 1,008,016\qquad\mathrm{(E) \ } 2,015,028</math> == Solution 1==
  8 KB (1,332 words) - 17:37, 17 September 2023
• 2006 AMC 12A Problems/Problem 25
  How many non-[[empty set | empty]] [[subset]]s <math>S</math> of <math>\{1,2,3,\ldots ,15\}</math> have the following two properties? <math>(1)</math> No two consecutive [[integer]]s belong to <math>S</math>.
  8 KB (1,405 words) - 11:52, 27 September 2022
• 2005 AMC 12A Problems/Problem 20
  For each <math>x</math> in <math>[0,1]</math>, define \begin{array}{clr} f(x) & = 2x, & \text { if } 0 \leq x \leq \frac {1}{2}; \\
  3 KB (437 words) - 23:49, 28 September 2022
• 2005 AMC 12A Problems/Problem 25
  ...<math>y</math>, and <math>z</math> are each chosen from the set <math>\{0,1,2\}</math>. How many [[equilateral]] [[triangle]]s all have their [[vertice === Solution 1 (non-rigorous) ===
  4 KB (498 words) - 00:46, 4 August 2023
• 2005 AMC 12B Problems/Problem 1
  ...C 12B Problems|2005 AMC 12B #1]] and [[2005 AMC 10B Problems|2005 AMC 10B #1]]}} .../math> dollars. They sell all the candy bars at the price of two for <math>1</math> dollar. What was their profit, in dollars?
  1 KB (179 words) - 13:53, 14 December 2021
• 2005 AMC 12B Problems/Problem 2
  ===Solution 1=== {{AMC10 box|year=2005|ab=B|num-b=1|num-a=3}}
  1 KB (145 words) - 13:56, 14 December 2021
• 2005 AMC 12B Problems/Problem 4
  \textbf{(A) }\ 1 \qquad
  1 KB (197 words) - 14:16, 14 December 2021
• 2005 AMC 12B Problems/Problem 5
  ...ile has a pattern consisting of four white quarter circles of radius <math>1/2</math> foot centered at each corner of the tile. The remaining portion o filldraw(Arc((1,0),.5,90,180)--(1,0)--cycle,white,black);
  2 KB (223 words) - 14:30, 15 December 2021
• 2005 AMC 12B Problems/Problem 6
  === Solution 1 === ...th>CD=8</math>, <math>HD=\sqrt{8^2-48}=\sqrt{16}=4</math>, and <math>BD=HD-1</math>, so <math>BD=\boxed{\textbf{(A) }3}</math>.
  2 KB (299 words) - 15:29, 5 July 2022
• 2005 AMC 12B Problems/Problem 7
  == Solution 1== ...h diagonals of <math>6</math> and <math>8</math>. The area is <math>\dfrac{1}{2}\times 6\times8</math>, or <math>\boxed{\mathrm{(D)}\ 24}</math>
  2 KB (357 words) - 20:15, 27 December 2020
• 2005 AMC 12B Problems/Problem 8
  \mathrm{(B)}\ 1 \qquad ...h>a</math>, the only values that satisfy this are <math>0</math> and <math>1</math>, so the answer is <math>\boxed{\mathrm{(C)}\ 2}</math>
  829 bytes (135 words) - 18:23, 9 September 2020
• 2005 AMC 12B Problems/Problem 9
  <math>\textbf{(A) }\ 0 \qquad \textbf{(B) }\ 1 \qquad \textbf{(C) }\ 2 \qquad \textbf{(D) }\ 4 \qquad \textbf{(E) }\ 5 </m ...ence between the mean and median, therefore, is <math>\boxed{\textbf{(B)}\ 1}</math>.
  2 KB (280 words) - 15:35, 16 December 2021
• 2005 AMC 12B Problems/Problem 10
  ...uce a result of <math>250</math>. It just so happens that <math>2005\equiv 1\ (\text{mod}\ 3)</math>, which leads us to the answer of <math>\boxed{\text
  1 KB (204 words) - 14:37, 15 December 2021
• 2005 AMC 12B Problems/Problem 11
  ...}}} \qquad \textbf{(C) }\ {{{\frac{3}{7}}}} \qquad \textbf{(D) }\ {{{\frac{1}{2}}}} \qquad \textbf{(E) }\ {{{\frac{2}{3}}}}</math> == Solution 1==
  4 KB (607 words) - 21:01, 20 May 2023
• 2005 AMC 12B Problems/Problem 12
  <math>\textbf{(A) }\ {{{1}}} \qquad \textbf{(B) }\ {{{2}}} \qquad \textbf{(C) }\ {{{4}}} \qquad \text ===Solution 1===
  2 KB (317 words) - 12:27, 16 December 2021
• 2005 AMC 12B Problems/Problem 13
  ...\log128}{\log4}=\log_{4}{128}=\log_{4}{2^7}=7\cdot\log_{4}{2}=7\cdot\dfrac{1}{2}=\boxed{\mathrm{(D)}\,\dfrac{7}{2}}</math>
  1 KB (203 words) - 19:57, 24 December 2020
• 2005 AMC 12B Problems/Problem 14
  ...ength. Therefore, <math>R+6 = R\sqrt{2} \Rightarrow R = \dfrac{6}{\sqrt{2}-1} = 6+6\sqrt{2} \Rightarrow \boxed{\mathrm{E}}</math>.
  2 KB (278 words) - 21:12, 24 December 2020
• 2005 AMC 12B Problems/Problem 15
  \mathrm{(A)}\ 1 \qquad == Solution 1==
  2 KB (411 words) - 21:02, 21 December 2020
• 2005 AMC 12B Problems/Problem 16
  Eight spheres of radius 1, one per octant, are each tangent to the coordinate planes. What is the rad \mathrm {(C)}\ 1+\sqrt{2}\qquad
  2 KB (364 words) - 04:54, 16 January 2023
• 2005 AMC 12B Problems/Problem 17
  \mathrm{(B)}\ 1 \qquad ...005,0,2005,0)</math> satisfies the equation, so the answer is <math>\boxed{1} \Rightarrow \mathrm{(B)}</math>.
  1 KB (159 words) - 21:18, 21 December 2020
• 2005 AMC 12B Problems/Problem 18
  xaxis(-1,15,Ticks(f, 2.0)); yaxis(-1,15,Ticks(f, 2.0));
  2 KB (262 words) - 21:20, 21 December 2020
• 2005 AMC 12B Problems/Problem 19
  <cmath>=(11(a+b))(9(a-b))</cmath> ...e for <math>a-b</math> is <math>1</math>. Now we have <math>(a+b)+(a-b)=11+1</math>, <math>2a=12</math>, <math>a=6</math>, then <math>b=5</math>, <math>
  2 KB (283 words) - 20:02, 24 December 2020
• 2005 AMC 12B Problems/Problem 20
  ...it leads to us getting <math>34</math> with <math>(-3, -2, 2, 6)</math> in 1 set for a total of 9 and the other numbers giving a total of 25. Case 1: <math>a,b,c,d</math> doesn't have 20.
  3 KB (463 words) - 19:28, 6 November 2022
• 2005 AMC 12B Problems/Problem 21
  <math>\mathrm{(A)}\ {{{0}}} \qquad \mathrm{(B)}\ {{{1}}} \qquad \mathrm{(C)}\ {{{2}}} \qquad \mathrm{(D)}\ {{{3}}} \qquad \mathrm ...k+2)d(m) = 80</math>. These numbers are in the ratio 3:4, so <math>\frac{k+1}{k+2} = \frac{3}{4} \implies k = 2 \Rightarrow \boxed{\mathrm{C}}</math>.
  888 bytes (140 words) - 20:04, 24 December 2020
• 2005 AMC 12B Problems/Problem 22
  A sequence of complex numbers <math>z_{0}, z_{1}, z_{2}, ...</math> is defined by the rule <cmath>z_{n+1} = \frac {iz_{n}}{\overline {z_{n}}},</cmath>
  4 KB (660 words) - 17:40, 24 January 2021
• 2005 AMC 12B Problems/Problem 23
  <cmath>\log_{10}(x+y) = z \text{ and } \log_{10}(x^{2}+y^{2}) = z+1.</cmath> == Solution 1 ==
  5 KB (786 words) - 11:36, 19 May 2024
• 2005 AMC 12B Problems/Problem 24
  == Solution 1 (Complex numbers)== ...r that is at a <math>60^{\circ}</math> angle with the complex number <math>1 + 2i</math>. Then, we can find the slope of the line between this new compl
  4 KB (761 words) - 09:10, 1 August 2023
• 2005 AMC 12B Problems/Problem 25
  === Solution 1 === *Case 1: Ant from point <math>F</math> moved to point <math>C</math>
  10 KB (1,840 words) - 21:35, 7 September 2023
• Power of a Point Theorem/Introductory Problem 2
  Applying the Power of a Point Theorem gives <math> 6\cdot x = 4\cdot 1 </math>, so <math> x = \frac 23 </math>
  289 bytes (45 words) - 13:14, 16 July 2017
• Root (polynomial)
  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 \ ...roots as <math>P(x)</math>, given by <cmath>Q(x) = x^n + \frac{c_{n-1}x^{n-1}}{c_n} + \dots + \frac{c_1x}{c_n} + \frac{c_0}{c_n} = \sum_{j=0}^{n} \frac{
  8 KB (1,427 words) - 21:37, 13 March 2022
• Vertical line test
  ...ny vertical line intersects it in, at most, one point, while <math>x^2+y^2=1</math> is not a function (try the line <math>x=0</math>).
  659 bytes (114 words) - 10:41, 27 April 2024
• 2006 AMC 10A Problems
  ==Problem 1== [[2006 AMC 10A Problems/Problem 1|Solution]]
  13 KB (2,028 words) - 16:32, 22 March 2022
• 1995 AHSME Problems/Problem 6
  path p=origin--(0,1)--(1,1)--(1,2)--(2,2)--(2,3); draw(shift(1,0)*p, dashed);
  1 KB (168 words) - 00:49, 14 October 2013
• 2006 AMC 10A Problems/Problem 6
  <math> \textbf{(A) } \frac17\qquad \textbf{(B) } \frac27\qquad \textbf{(C) } 1\qquad \textbf{(D) } 7\qquad \textbf{(E) } 14 </math>
  590 bytes (84 words) - 14:28, 31 May 2023
• 2006 AMC 10A Problems/Problem 8
  == Solution 1 == <math>0=1+-12+c</math>
  2 KB (348 words) - 23:10, 16 December 2021
• 2006 AMC 10A Problems/Problem 13
  ...s to be fair, the amount paid, <math>5</math> dollars, must be <math>\frac{1}{12}</math> the amount of the prize money, so the answer is
  1 KB (207 words) - 09:39, 25 July 2023
• 2006 AMC 12A Problems/Problem 12
  ...ath>20</math> cm. The outside diameter of each of the outer rings is <math>1</math> cm less than that of the ring above it. The bottom ring has an outsi D(CR((0,-47),1.5));
  2 KB (292 words) - 11:56, 17 December 2021
• 2006 AMC 10A Problems/Problem 15
  ...math> laps run by both, or <math>\lfloor 2\cdot 23.8\rfloor = 23 \cdot 2 + 1 =\boxed{\textbf{(D) } 47}</math> meeting points. ...th>(30 \ \text{min})\left(250\frac{\text{m}}{\text{min}}\right)\left(\frac{1}{100\pi}\frac{\text{laps}}{\text{m}}\right)= \frac{75}{\pi}\,\text{laps}\ap
  3 KB (532 words) - 17:49, 13 August 2023
• 2006 AMC 10A Problems/Problem 16
  A circle of radius <math>1</math> is [[tangent]] to a circle of radius <math>2</math>. The sides of <m D(CR((2*t,5),1));
  5 KB (732 words) - 23:19, 19 September 2023
• 2006 AMC 10A Problems/Problem 17
  A=(0,2); B=(1,2); C=(2,2); D=(3,2); H=(0,0); G=(1,0); F=(2,0); E=(3,0);
  6 KB (1,066 words) - 00:21, 2 February 2023
• 2006 AMC 10A Problems/Problem 18
  So, there are <math>6 - 1 = 5</math> choices for the position of the letters.
  2 KB (254 words) - 14:39, 5 April 2024
• 2006 AMC 10A Problems/Problem 19
  <math>\textbf{(A) } 0\qquad\textbf{(B) } 1\qquad\textbf{(C) } 59\qquad\textbf{(D) } 89\qquad\textbf{(E) } 178\qquad</m .../math> degrees. The minimum possible value for the smallest angle is <math>1</math> and the highest possible is <math>59</math> (since the numbers are d
  2 KB (259 words) - 03:10, 22 June 2023
• 2006 AMC 10A Problems/Problem 20
  Six distinct positive integers are randomly chosen between <math>1</math> and <math>2006</math>, inclusive. What is the probability that some ...textbf{(C) } \frac{2}{3}\qquad\textbf{(D) } \frac{4}{5}\qquad\textbf{(E) } 1\qquad</math>
  1 KB (187 words) - 08:21, 17 March 2023
• 2006 AMC 10A Problems/Problem 21
  == Solution 1 (Complementary Counting) == Case <math>1</math>: There is ONLY one <math>2</math> or <math>3</math>. If the <math>2<
  3 KB (525 words) - 20:25, 30 April 2024
• 2006 AMC 10A Problems/Problem 24
  ...{(C) } \frac{1}{4}\qquad\textbf{(D) } \frac{1}{3}\qquad\textbf{(E) } \frac{1}{2}\qquad</math> real r = 1/2;
  2 KB (292 words) - 10:19, 19 December 2021
• Set
  ...set can have multiple indistinct elements, such as the following: <math>\{1,4,5,3,24,4,4,5,6,2\}</math> Such an entity is actually called a multiset. ...th>239</math>, for example. (The standard notation for this set is <math>\{1,3,5,239\}</math>. Note that the order in which the terms are listed is comp
  11 KB (2,021 words) - 00:00, 17 July 2011
• Newman's Tauberian Theorem
  Let <math>\Gamma_-^1</math> be the restriction of the contour to the set Let <math>\Gamma_- = \Gamma_-^1 + \Gamma_-^2</math>, as shown in the diagram
  6 KB (1,034 words) - 07:55, 12 August 2019
• 2006 USAMO Problems
  == Day 1 == === Problem 1 ===
  3 KB (520 words) - 09:24, 14 May 2021
• 1991 AJHSME Problems
  ==Problem 1== <math>1,000,000,000,000-777,777,777,777=</math>
  17 KB (2,246 words) - 13:37, 19 February 2020
• 1990 AJHSME Problems/Problem 1
  draw((0,0)--(10,0)); draw((-1.5,1.5)--(-1.5,2.5)); draw((-1,2)--(-2,2)); draw((1,1)--(3,1)--(3,3)--(1,3)--cycle); draw((1,4)--(3,4)--(3,6)--(1,6)--cycle);
  1 KB (191 words) - 17:12, 29 October 2016
• 2022 USAJMO Problems/Problem 1
  ...ath>a_n-g_n</math> is divisible by <math>m</math> for all integers <math>n>1</math>; ...mid d</math> and <math>m|a+(n-1)d-gr^{n-1}</math> for all integers <math>n>1</math>.
  4 KB (792 words) - 00:29, 13 April 2024
• 2006 AIME II Problems/Problem 2
  {{AIME box|year=2006|n=II|num-b=1|num-a=3}}
  1 KB (164 words) - 14:58, 14 April 2020
• 2005 AIME II Problems/Problem 1
  ...given by the [[binomial coefficient]] <math>{n \choose 6} = \frac{n\cdot(n-1)\cdot(n-2)\cdot(n-3)\cdot(n-4)\cdot(n-5)}{6\cdot5\cdot4\cdot3\cdot2\cdot1}< ...ose three cards from <math>n</math> is <math>{n\choose 3} = \frac{n\cdot(n-1)\cdot(n-2)}{3\cdot2\cdot1}</math>.
  1 KB (239 words) - 11:54, 31 July 2023
• 2005 AIME II Problems/Problem 2
  === Solution 1 === *Person 1: <math>\frac{9 \cdot 6 \cdot 3}{9 \cdot 8 \cdot 7} = \frac{9}{28}</math>
  4 KB (628 words) - 11:28, 14 April 2024
• 2005 AIME II Problems/Problem 3
  == Solution 1 == ...ing this equation by <math>\frac{a}{1-r}</math>, we get <math>10 = \frac a{1 + r}</math>. Then <math>a = 2005 - 2005r</math> and <math>a = 10 + 10r</ma
  3 KB (581 words) - 07:54, 4 November 2022
• 2005 AIME II Problems/Problem 4
  == Solution 1 == ...o our previous sum to get an answer of <math>121 + 64 + 276 - 8 - 8 - 11 + 1 = \boxed{435}</math>.
  3 KB (377 words) - 18:36, 1 January 2024
• 2005 AIME II Problems/Problem 13
  ...egers <math>\frac{10}{a}</math> has to be an integer, so <math>a=1,2,5,10,-1,-2,-5,-10</math>. Thus the product of the roots is equal to one of the foll
  4 KB (642 words) - 14:55, 12 August 2019
• 2005 AIME II Problems/Problem 5
  == Solution 1 == ...<math> 44-2+1=43 </math> possibilities for the square case and <math> 12-2+1=11 </math> possibilities for the cube case. Thus, the answer is <math> 43+1
  3 KB (547 words) - 19:15, 4 April 2024
• 2005 AIME II Problems/Problem 6
  ...<math> (n+1) </math> becomes the bottom card of the new stack, card number 1 is on top of this card, and so on, until piles <math> A </math> and <math> == Solution 1 ==
  2 KB (384 words) - 00:31, 26 July 2018
• 2005 AIME II Problems
  == Problem 1 == [[2005 AIME II Problems/Problem 1|Solution]]
  7 KB (1,119 words) - 21:12, 28 February 2020
• 2005 AIME II Problems/Problem 7
  ...}+1)(\sqrt[4]{5}+1)(\sqrt[8]{5}+1)(\sqrt[16]{5}+1)}. </math> Find <math>(x+1)^{48}</math>. == Solution 1==
  2 KB (279 words) - 12:33, 27 October 2019
• 2005 AIME II Problems/Problem 8
  Since <math>\overline{AT}=\overline{TB}=\frac{1}{2}x</math>, ...h>(\overline{PO_3}-\overline{O_3T})(\overline{QO_3}+\overline{O_3T})=\frac{1}{4}x^2</cmath>
  4 KB (693 words) - 13:03, 28 December 2021
• 2005 AIME II Problems/Problem 9
  === Solution 1 === ...here as well, so exactly those values of <math>n</math> congruent to <math>1 \pmod 4</math> work. There are <math>\boxed{250}</math> of them in the giv
  6 KB (1,154 words) - 03:30, 11 January 2024
• 2005 AIME II Problems/Problem 12
  === Solution 1 (trigonometry) === ...",E/2+G/2,(0,1));label("\(y\)",G/2+F/2,(0,1)); label("\(450\)",(O+G)/2,(-1,1));
  13 KB (2,080 words) - 21:20, 11 December 2022
• 2005 AIME II Problems/Problem 10
  draw(box((-1,-1,-1),(1,1,1))); === Solution 1 ===
  3 KB (436 words) - 03:10, 23 September 2020
• 2005 AIME II Problems/Problem 11
  ...ath> a_{k+1} = a_{k-1} - \frac 3{a_k} </math> for <math> k = 1,2,\ldots, m-1. </math> Find <math>m. </math> ==Solution 1==
  3 KB (499 words) - 18:52, 21 November 2022
• Hyperbola
  ...y opening hyperbola, while <math>\frac{(y-k)^2}{b^2}-\frac{(x-h)^2}{a^2} = 1</math> is the standard form of a vertically opening one. Also, the graph of
  666 bytes (112 words) - 10:06, 30 January 2019
• Line
  ...fined as the shortest distance between two points. It is defined to be in 1 direction only, i.e. infinitely thin but also infinitely long. In the [[Car
  674 bytes (106 words) - 18:40, 9 May 2024
• Square root
  .... For example, <math>\left(x^{\frac12}\right)^2 = x^{\frac12 \cdot 2} = x^1 = x</math>, which is exactly what we would have expected. This notion can
  2 KB (275 words) - 20:58, 8 January 2024
• Euler's identity
  ...matics: [[e]], [[imaginary unit | i]], [[pi]], [[zero (constant)| 0]], and 1. ==Proof 1==
  3 KB (452 words) - 23:17, 4 January 2021
• 2006 AMC 12A Problems/Problem 4
  == Solution 1 == 1+2&9&6&3\\
  2 KB (257 words) - 11:20, 2 January 2022
• 2005 AIME I Problems
  == Problem 1 == [[2005 AIME I Problems/Problem 1|Solution]]
  6 KB (983 words) - 05:06, 20 February 2019
• 2005 AIME I Problems/Problem 2
  ...ath> k</math>. For example, <math> S_3 </math> is the [[sequence]] <math> 1,4,7,10,\ldots. </math> For how many values of <math> k </math> does <math> ...501,4)</math>, <math>(668,3)</math>, <math>(1002,2)</math> and <math>(2004,1)</math>, and each of these gives a possible value of <math>k</math>. Thus t
  2 KB (303 words) - 01:31, 5 December 2022
• 2005 AIME I Problems/Problem 3
  In the first case, the three proper divisors of <math>n</math> are <math>1</math>, <math>p</math> and <math>q</math>. Thus, we need to pick two prime In the second case, the three proper divisors of <math>n</math> are 1, <math>p</math> and <math>p^2</math>. Thus we need to pick a prime number
  2 KB (249 words) - 09:37, 23 January 2024
• 2005 AIME I Problems/Problem 4
  === Solution 1 === ...<math>x</math> is maximized for the first case. Thus, <math>x = \frac{69 + 1}{2} = 35</math>, and <math>r = \frac{7 \pm 35}{2} = 21, -14</math>. The lat
  8 KB (1,248 words) - 11:43, 16 August 2022
• 2005 AIME I Problems/Problem 5
  == Solution 1 == ...are <math>8</math> possible places to change from <math>0</math> to <math>1</math> and there is the possibility that there no change occurs). There are
  5 KB (830 words) - 01:51, 1 March 2023
• 2005 AIME I Problems/Problem 6
  == Solution 1 == ...that [[equation]] is nearly equal to <math>(x - 1)^4</math>. Thus, we add 1 to each side in order to complete the fourth power and get
  4 KB (686 words) - 01:55, 5 December 2022
• 2005 AIME I Problems/Problem 7
  === Solution 1 === ...5</math>. <math>PQ=DE</math>, and therefore <math>AB=AP+PQ+BQ=5+\sqrt{141}-1+5=9+\sqrt{141} \rightarrow (p,q)=(9,141) \rightarrow \boxed{150}</math>.
  4 KB (567 words) - 20:20, 3 March 2020
• 2005 AIME I Problems/Problem 8
  The [[equation]] <math> 2^{333x-2} + 2^{111x+2} = 2^{222x+1} + 1 </math> has three [[real]] [[root]]s. Given that their sum is <math>m/n</ma ...{111x}</math>. Then our equation reads <math>\frac{1}{4}y^3 + 4y = 2y^2 + 1</math> or <math>y^3 - 8y^2 + 16y - 4 = 0</math>. Thus, if this equation ha
  1 KB (161 words) - 19:50, 2 January 2022
• 2005 AIME I Problems/Problem 9
  ...ether contribute a probability of <math>\left(\frac{1}{4}\right)^8 = \frac{1}{2^{16}}</math> ...oduct, <math>\frac{2^6}{3^6} \cdot \frac{5^{12}}{2^{24}3^{12}} \cdot \frac{1}{2^{16}} = \frac{5^{12}}{2^{34}\cdot 3^{18}}</math> and so the answer is <m
  4 KB (600 words) - 21:44, 20 November 2023
• 2005 AIME I Problems/Problem 10
  ...label("B (12,19)",B,(-1,-1));label("C (23,20)",C,(1,-1));label("M",M,(0.2,-1)); label("(17,22)",P,(1,1));
  5 KB (852 words) - 21:23, 4 October 2023
• 2005 AIME I Problems/Problem 11
  ===Solution 1 === ...have <math>r + r\sqrt{2} = 8\sqrt{2}</math> so <math>r = \frac{8\sqrt{2}}{1+\sqrt{2}} = 16 - 8\sqrt{2}</math>. Then the diameter is <math>32 - \sqrt{51
  4 KB (707 words) - 11:11, 16 September 2021
• 2005 AIME I Problems/Problem 12
  ...005 </math> with <math> S(n) </math> [[even integer | even]]. Find <math> |a-b|. </math> == Solution 1==
  4 KB (647 words) - 02:29, 4 May 2021
• 2005 AIME I Problems/Problem 13
  ...particle may only move to <math> (a+1,b), (a,b+1), </math> or <math>(a+1,b+1). </math> === Solution 1 ===
  5 KB (897 words) - 00:21, 29 July 2022
• 2005 AIME I Problems/Problem 14
  === Solution 1 === ...h>E</math> and <math>C</math> to be <math>\frac { - 2 - 0}{2 - 8} = \frac {1}{3}</math>. Because the other two sides are perpendicular, the slope of the
  3 KB (561 words) - 14:11, 18 February 2018
• 2005 AIME I Problems/Problem 15
  == Solution 1== WLOG let E be be between C & D (as in solution 1). Assume <math>AD = 3m</math>. We use power of a point to get that
  5 KB (906 words) - 23:15, 6 January 2024
• 2005 AIME II Problems/Problem 15
  == Solution 1 == Squaring again and canceling yields <math>1 = \frac{x^2}{100} + \frac{(y-12)^2}{75}.</math>
  12 KB (2,000 words) - 13:17, 28 December 2020
• 2005 AIME II Problems/Problem 14
  == Solution 1== ...= \frac{3 \cdot 14^2}{2 \cdot 13^2} = \frac{BC - BE}{BE} = \frac{15}{BE} - 1 \Longrightarrow BE = \frac{13^2 \cdot 15}{463}</math>. The answer is <math>
  13 KB (2,129 words) - 18:56, 1 January 2024
• 2004 AIME I Problems/Problem 15
  ...\text{if }x = 1\\ \frac x{10} & \text{if }x\text{ is divisible by 10}\\ x+1 & \text{otherwise}\end{cases} .... Let <math>d(x)</math> be the smallest <math>n</math> such that <math>x_n=1</math>. (For example, <math>d(100)=3</math> and <math>d(87)=7</math>.) Let
  9 KB (1,491 words) - 01:23, 26 December 2022
• 2004 AIME I Problems/Problem 14
  label("$8$",O1--B1,W);label("$4$",(5*A1+O1)/6,0.8*(1,1)); label("$8$",O1--A1,0.8*(-0.5,2));label("$4\sqrt{5}$",B1/2+A1/2,(0,-1));
  4 KB (729 words) - 01:00, 27 November 2022
• 2004 AIME I Problems/Problem 13
  ...s, 34, </math> with <math> 0 < a_1 \le a_2 \le a_3 \le \cdots \le a_{34} < 1 </math> and <math> r_k>0. </math> Given that <math> a_1 + a_2 + a_3 + a_4 + ...} - x^{19} - x^{17} + 1}{(x - 1)^2} = \frac{(x^{19} - 1)(x^{17} - 1)}{(x - 1)^2} \end{align*}</cmath>
  2 KB (298 words) - 20:02, 4 July 2013
• 2004 AIME I Problems/Problem 12
  ...ordered pair]]s <math> (x, y) </math> such that <math> 0 < x \le 1, 0<y\le 1, </math> and <math> \left[\log_2{\left(\frac 1x\right)}\right] </math> and <math>\left\lfloor\log_2\left(\frac{1}{x}\right)\right\rfloor</math> is even when
  2 KB (303 words) - 22:28, 11 September 2020
• 2004 AIME I Problems/Problem 11
  ===Solution 1=== ...urface area of a cone, we find that <math>A_c=\frac{1}{2}c\cdot \ell=\frac{1}{2}(2\pi x)\left(\frac{5}{3}x\right)=\frac{5}{3}\pi x^2</math>. By subtract
  5 KB (839 words) - 22:12, 16 December 2015
• 2004 AIME I Problems/Problem 10
  A [[circle]] of [[radius]] 1 is randomly placed in a 15-by-36 [[rectangle]] <math> ABCD </math> so that === Solution 1 ===
  5 KB (836 words) - 07:53, 15 October 2023
• 2004 AIME I Problems/Problem 9
  label("\(V_1\)",(1,1.2)); label("\(U_1\)",(3,0.3)); label("\(U_2\)",(1,3)); label("\(V_2\)",(5,3)); MP("7",(D+G)/2,S); MP("6",(D+E)/2,W); MP("9/2"
  4 KB (618 words) - 20:01, 4 July 2013
• 2004 AIME I Problems/Problem 1
  ...00(n + 2) + 10(n + 1) + n = 3210 + 1111n</math>, for <math>n \in \lbrace0, 1, 2, 3, 4, 5, 6\rbrace</math>. Adding these numbers up, we get <math>(0 + 1 + 2 + 3 + 4 + 5 + 6) + 7\cdot28 = \boxed{217}</math>
  2 KB (374 words) - 14:53, 27 December 2019
• 2004 AIME I Problems/Problem 2
  ==Solution 1== ...h>. Therefore, the largest element in <math>A</math> is <math>2 + \frac{m-1}{2}</math>.
  8 KB (1,437 words) - 21:53, 19 May 2023
• 2004 AIME I Problems/Problem 4
  ...ned by vertex <math>(0,0)</math> form a quarter-[[circle]] with [[radius]] 1. draw(p,CR(A,1));draw(p,CR(B,1));draw(p,CR(C,1));draw(p,CR(D,1));
  3 KB (532 words) - 09:22, 11 July 2023
• 2004 AIME I Problems/Problem 5
  Let <math>q</math> be the number of questions Beta takes on day 1 and <math>a</math> be the number he gets right. Let <math>b</math> be the n <cmath>a + b < 350 - \frac{1}{6}q
  3 KB (436 words) - 18:31, 9 January 2024
• 2004 AIME I Problems/Problem 6
  ...} </math> if <math> i </math> is [[odd integer | odd]] and <math> a_i>a_{i+1} </math> if <math> i </math> is [[even integer | even]]. How many snakelike == Solution 1 ==
  3 KB (562 words) - 18:12, 4 March 2022
• 2004 AIME I Problems/Problem 7
  ...the expansion of the product <math> (1 - x)(1 + 2x)(1 - 3x)\cdots(1 + 14x)(1 - 15x). </math> Find <math> |C|. </math> === Solution 1 ===
  5 KB (833 words) - 19:43, 1 October 2023
• 2004 AIME I Problems/Problem 8
  ...>n</math>-gon in a counterclockwise direction: <math>0, 1, 2, 3, \ldots, n-1.</math> ...med if we choose a vertex number <math>m</math>, where <math>0 \le m \le n-1</math>, and then form the line segments by joining the following pairs of v
  4 KB (620 words) - 21:26, 5 June 2021
• 2004 AIME I Problems
  == Problem 1 == [[2004 AIME I Problems/Problem 1|Solution]]
  9 KB (1,434 words) - 13:34, 29 December 2021
• 2004 AIME II Problems/Problem 15
  ...h and lie on top of the left end, resulting in a <math>256</math> by <math>1</math> strip of quadruple thickness. This process is repeated <math>8</math == Solution 1 ==
  6 KB (899 words) - 20:58, 12 May 2022
• 2004 AIME II Problems/Problem 14
  == Solution 1 == ...0</math> (for example, suppose we counted the solution set <math>(b,c) = (1,9) \Longrightarrow n = 19</math>, but substituting into our original equati
  11 KB (1,857 words) - 21:55, 19 June 2023
• 2004 AIME II Problems/Problem 13
  D(MP("A",A,(1,0))--MP("B",B,N)--MP("C",C,NW)--MP("D",D)--MP("E",E)--cycle); D(D--A--C--E)
  3 KB (486 words) - 22:15, 7 April 2023
• 2004 AIME II Problems/Problem 12
  pair A=(0,0), B=(6,0), D=(1, 24^.5), C=(5,D.y), O = (3,(r^2 + 6*r)^.5); ...dicular]] from <math>D</math> to <math>AB</math>; then <math>AD' = 3 - 2 = 1</math>. By the Pythagorean Theorem, <math>(AD')^2 + (DD')^2 = (AD)^2 \Longr
  3 KB (431 words) - 23:21, 4 July 2013
• 2004 AIME II Problems/Problem 10
  ...> and <math> 2^{40} </math> whose binary expansions have exactly two <math>1</math>'s. If a number is chosen at random from <math> S, </math> the [[prob == Solution 1 (Rigorous)==
  8 KB (1,283 words) - 19:19, 8 May 2024
• 2004 AIME II Problems/Problem 9
  ...} </math> are in geometric progression, and the terms <math> a_{2n}, a_{2n+1}, </math> and <math> a_{2n+2} </math> are in arithmetic progression. Let <m == Solution 1==
  3 KB (538 words) - 21:33, 30 December 2023
• 2004 AIME II Problems/Problem 8
  == Solution 1 == ...ivisors of <math>2004=2^2\cdot 3^1\cdot 167^1</math> is <math>(2+1)(1+1)(1+1)=12</math>.
  2 KB (353 words) - 18:08, 25 November 2023
• 2004 AIME II Problems/Problem 7
  === Solution 1 (Synthetic) === D(MP("E",E,W)--MP("F",F,(1,0))); D(B--G);
  9 KB (1,501 words) - 05:34, 30 October 2023
• 2004 AIME II Problems/Problem 6
  ...d monkeys have at the end of the process are in the [[ratio]] <math> 3: 2: 1, </math>what is the least possible total for the number of bananas? ...h>, and the third monkey got <math>\frac{1}{8}b_1 + \frac{3}{8}b_2 + \frac{1}{12}b_3</math>.
  6 KB (950 words) - 14:18, 15 January 2024
• 2004 AIME II Problems/Problem 5
  ==Solution 1== ...h>w</math>, the entire time be <math>t</math>, and the total work be <math>1</math>.
  4 KB (592 words) - 19:02, 26 September 2020
• 2004 AIME II Problems/Problem 4
  ...to arrange them in an <math>n</math>-digit number, for a total of <math>(2^1 - 2) + (2^2 - 2) + (2^3 -2) + (2^4 - 2) = 22</math> such numbers (or we can ...can form, for a total of <math>(2^0 - 1) + (2^1 - 1) + (2^2 - 1) + (2^3 - 1) = 11</math> such numbers (or we can list them: <math>A0, A00, A0A, AA0, A0
  3 KB (508 words) - 01:16, 19 January 2024
• 2004 AIME II Problems/Problem 3
  ...ed so that three of its faces are visible, exactly <math>231</math> of the 1-cm cubes cannot be seen. Find the smallest possible value of <math> N. </ma ...instance, <math>l - 1 = 1</math> and <math>m - 1 = 11</math> and <math>n - 1 = 3 \cdot 7</math>, among others. However, it should be fairly clear that
  2 KB (377 words) - 11:53, 10 March 2014
• 2004 AIME II Problems/Problem 2
  ...ferent candies will have the form <math>\frac{{10\choose 1}\cdot{10\choose 1}}{{20\choose2}}</math>. It is not difficult to see that these yield the sa {{AIME box|year=2004|num-b=1|num-a=3|n=II}}
  2 KB (330 words) - 13:42, 1 January 2015
• 2004 AIME II Problems/Problem 1
  [[Image:2004 AIME II Problem 1.png]] ...t 2 = 120</math> [[degree]]s, so the area of the [[sector]] is <math>\frac{1}{3}r^2\pi</math>; the rest of the area of the circle is then equal to <math
  2 KB (329 words) - 23:20, 4 July 2013
• 2004 AIME II Problems
  == Problem 1 == [[2004 AIME II Problems/Problem 1|Solution]]
  9 KB (1,410 words) - 05:05, 20 February 2019
• 1983 AIME Problems
  == Problem 1 == Let <math>x</math>, <math>y</math> and <math>z</math> all exceed <math>1</math> and let <math>w</math> be a positive number such that <math>\log_xw=
  7 KB (1,104 words) - 12:53, 6 July 2022
• 2019 AIME I Problems/Problem 2
  ..., 20</math>. Bela then randomly chooses a number <math>B</math> from <math>1, 2, 3,\ldots, 19, 20</math> distinct from <math>J</math>. The value of <mat ==Solution 1==
  5 KB (830 words) - 22:15, 28 December 2023
• 1984 AIME Problems
  == Problem 1 == ...<math>a_3\ldots</math> is an arithmetic progression with common difference 1, and <math>a_1+a_2+a_3+\ldots+a_{98}=137</math>.
  6 KB (933 words) - 01:15, 19 June 2022
• 1986 AIME Problems
  == Problem 1 == [[1986 AIME Problems/Problem 1|Solution]]
  5 KB (847 words) - 15:48, 21 August 2023
• 1987 AIME Problems
  == Problem 1 == [[1987 AIME Problems/Problem 1|Solution]]
  6 KB (869 words) - 15:34, 22 August 2023
• 1988 AIME Problems
  == Problem 1 == ...any order -- the correct five buttons. The sample shown below has <math>\{1, 2, 3, 6, 9\}</math> as its combination. Suppose that these locks are redes
  6 KB (902 words) - 08:57, 19 June 2021
• 1989 AIME Problems
  == Problem 1 == Compute <math>\sqrt{(31)(30)(29)(28)+1}</math>.
  7 KB (1,045 words) - 20:47, 14 December 2023
• 1990 AIME Problems
  == Problem 1 == [[1990 AIME Problems/Problem 1|Solution]]
  6 KB (870 words) - 10:14, 19 June 2021
• 1991 AIME Problems
  == Problem 1 == [[1991 AIME Problems/Problem 1|Solution]]
  7 KB (1,106 words) - 22:05, 7 June 2021
• 1992 AIME Problems
  == Problem 1 == [[1992 AIME Problems/Problem 1|Solution]]
  8 KB (1,117 words) - 05:32, 11 November 2023
• 1993 AIME Problems
  == Problem 1 == [[1993 AIME Problems/Problem 1|Solution]]
  8 KB (1,275 words) - 06:55, 2 September 2021
• 1994 AIME Problems
  == Problem 1 == [[1994 AIME Problems/Problem 1|Solution]]
  7 KB (1,141 words) - 07:37, 7 September 2018
• 1995 AIME Problems
  == Problem 1 == ...<math>S_{i+2}.</math> The total area enclosed by at least one of <math>S_{1}, S_{2}, S_{3}, S_{4}, S_{5}</math> can be written in the form <math>m/n,</
  6 KB (1,000 words) - 00:25, 27 March 2024
• 1996 AIME Problems
  == Problem 1 == [[1996 AIME Problems/Problem 1|Solution]]
  6 KB (931 words) - 17:49, 21 December 2018
• 1997 AIME Problems
  == Problem 1 == How many of the integers between 1 and 1000, inclusive, can be expressed as the difference of the squares of t
  7 KB (1,098 words) - 17:08, 25 June 2020
• 1998 AIME Problems
  == Problem 1 == [[1998 AIME Problems/Problem 1|Solution]]
  7 KB (1,084 words) - 02:01, 28 November 2023
• 1999 AIME Problems
  == Problem 1 == [[1999 AIME Problems/Problem 1|Solution]]
  7 KB (1,094 words) - 13:39, 16 August 2020
• 2000 AIME I Problems
  == Problem 1 == [[2000 AIME I Problems/Problem 1|Solution]]
  7 KB (1,204 words) - 03:40, 4 January 2023
• 2001 AIME I Problems
  == Problem 1 == [[2001 AIME I Problems/Problem 1|Solution]]
  7 KB (1,212 words) - 22:16, 17 December 2023
• 2002 AIME I Problems
  == Problem 1 == [[2002 AIME I Problems/Problem 1|Solution]]
  8 KB (1,374 words) - 21:09, 27 July 2023
• 2003 AIME I Problems
  == Problem 1 == [[2003 AIME I Problems/Problem 1|Solution]]
  6 KB (965 words) - 16:36, 8 September 2019
• 2000 AIME II Problems
  == Problem 1 == [[2000 AIME II Problems/Problem 1|Solution]]
  6 KB (947 words) - 21:11, 19 February 2019
• 2001 AIME II Problems
  == Problem 1 == [[2001 AIME II Problems/Problem 1|Solution]]
  8 KB (1,282 words) - 21:12, 19 February 2019
• 2002 AIME II Problems
  == Problem 1 == [[2002 AIME II Problems/Problem 1|Solution]]
  7 KB (1,177 words) - 15:42, 11 August 2023
• 2003 AIME II Problems
  == Problem 1 == [[2003 AIME II Problems/Problem 1|Solution]]
  7 KB (1,127 words) - 09:02, 11 July 2023
• 1983 AIME Problems/Problem 1
  Let <math>x</math>, <math>y</math> and <math>z</math> all exceed <math>1</math> and let <math>w</math> be a positive number such that <math>\log_x w == Solution 1 ==
  4 KB (642 words) - 03:14, 17 August 2022
• 1983 AIME Problems/Problem 2
  === Solution 1 === {{AIME box|year=1983|num-b=1|num-a=3}}
  1 KB (184 words) - 20:16, 14 January 2023
• 1983 AIME Problems/Problem 3
  == Solution 1 == ...of this equation are real, since its discriminant is <math>18^2 - 4 \cdot 1 \cdot 20 = 244</math>, which is positive. Thus by [[Vieta's formulas]], the
  3 KB (532 words) - 05:18, 21 July 2022
• 1983 AIME Problems/Problem 4
  ==Solution 1== ...t <math>x = 1</math> and <math>y = 5</math>, such that the answer is <math>1^2 + 5^2 = \boxed{026}</math>.
  11 KB (1,741 words) - 22:40, 23 November 2023
• 1983 AIME Problems/Problem 5
  === Solution 1 === Substituting, <math>w^3-21w+20=0</math>, which factorizes as <math>(w-1)(w+5)(w-4)=0</math> (the [[Rational Root Theorem]] may be used here, along
  4 KB (672 words) - 10:17, 17 March 2023
• 1983 AIME Problems/Problem 6
  === Solution 1 === .../math> and <math>8</math> are greater or less than <math>7</math> by <math>1</math>.
  3 KB (361 words) - 20:20, 14 January 2023
• 1983 AIME Problems/Problem 7
  == Solution 1 == ...three knights are sitting next to each other and subtracting it from <math>1</math>.
  9 KB (1,392 words) - 20:37, 19 January 2024
• 1983 AIME Problems/Problem 8
  === Solution 1 === === Solution 2: Clarification of Solution 1 ===
  2 KB (249 words) - 23:25, 11 May 2024
• 1983 AIME Problems/Problem 9
  == Solution 1 == ...y <math>12</math>. We show this possible with the same methods in Solution 1; thus the answer is <math>\boxed{012}</math>.
  4 KB (722 words) - 20:25, 14 January 2023
• 1983 AIME Problems/Problem 10
  ...hing in common: each is a <math>4</math>-digit number beginning with <math>1</math> that has exactly two identical digits. How many such numbers are the ==Solution 1==
  5 KB (855 words) - 20:26, 14 January 2023
• 1983 AIME Problems/Problem 11
  === Solution 1 === label("A",A,(-1,-1,0));
  5 KB (865 words) - 21:11, 6 February 2023
• 1983 AIME Problems/Problem 12
  ..., but as <math>x</math> and <math>y</math> are different digits, <math>1+0=1 \leq x+y \leq 9+9=18</math>, so the only possible multiple of <math>11</mat == Alternate start to solution 1 ==
  2 KB (412 words) - 18:23, 1 January 2024
• 1983 AIME Problems/Problem 13
  ...le, the alternating sum for <math>\{1, 2, 3, 6,9\}</math> is <math>9-6+3-2+1=5</math> and for <math>\{5\}</math> it is simply <math>5</math>. Find the s === Solution 1 ===
  5 KB (894 words) - 22:02, 5 April 2024
• 1983 AIME Problems/Problem 14
  ...aw(O2--O1); dot(O1); dot(O2); draw(Q--R); label("$Q$",Q,NW); label("$P$",P,1.5*dir(80)); label("$R$",R,NE); label("12",waypoint(O1--O2,0.4),S);</asy> === Solution 1 ===
  13 KB (2,149 words) - 18:44, 5 February 2024
• 1983 AIME Problems/Problem 15
  dotfactor=1; path C1=Circle(O1,1);
  19 KB (3,221 words) - 01:05, 7 February 2023
• 1984 AIME Problems/Problem 1
  ...h>a_3\ldots</math> is an [[arithmetic progression]] with common difference 1, and <math>a_1+a_2+a_3+\ldots+a_{98}=137</math>. == Solution 1 ==
  4 KB (576 words) - 21:03, 23 December 2023
• 1984 AIME Problems/Problem 2
  {{AIME box|year=1984|num-b=1|num-a=3}}
  1 KB (187 words) - 20:05, 29 May 2021
• 1984 AIME Problems/Problem 3
  ...f <math>\triangle ABC</math>, the resulting smaller [[triangle]]s <math>t_{1}</math>, <math>t_{2}</math>, and <math>t_{3}</math> in the figure, have [[a ...(A+(B-A)*3/4--A+(C-A)*3/4); D(B+(C-B)*5/6--B+(A-B)*5/6);D(C+(B-C)*5/12--C+(A-C)*5/12);
  4 KB (726 words) - 13:39, 13 August 2023
• 1984 AIME Problems/Problem 4
  == Solution 1 (Two Variables) == \frac{s+68}{n+1}&=56, \\
  2 KB (319 words) - 03:38, 16 January 2023
• 1984 AIME Problems/Problem 5
  == Solution 1 == ...12}{\frac{1}{3 \ln 2} + \frac{2}{2 \ln 2}} = \frac{12 \ln 2}{\frac{1}{3} + 1} = \frac{12 \ln 2}{\frac{4}{3}} = 9 \ln 2</math>. This means that <math>\fr
  6 KB (863 words) - 16:10, 16 May 2024
• 1984 AIME Problems/Problem 6
  == Solution 1 == <cmath>\frac{|-5a + 8 - b|}{\sqrt{a^2+1}}= \frac{|16 - b|}{\sqrt{a^2+1}} \Longleftrightarrow |-5a+8-b| = |16-b|</cmath>We can split this into two
  6 KB (1,022 words) - 19:29, 22 January 2024
• 1984 AIME Problems/Problem 7
  == Solution 1 == ...) = f^2(89) = f^3(94) = \ldots f^{y}(1004)</math>. <math>1004 = 84 + 5(y - 1) \Longrightarrow y = 185</math>. So we now need to reduce <math>f^{185}(100
  4 KB (617 words) - 22:09, 15 May 2024
• 1984 AIME Problems/Problem 8
  The equation <math>z^6+z^3+1=0</math> has complex roots with argument <math>\theta</math> between <math> == Solution 1 ==
  3 KB (430 words) - 19:05, 7 February 2023
• 1984 AIME Problems/Problem 9
  == Solution 1== P=s*markscalefactor*unit(A-B)+B;
  6 KB (947 words) - 20:44, 26 November 2021
• 1984 AIME Problems/Problem 10
  == Solution 1 (Inequalities) == &=30+4(c-1)-(w-4) \\
  7 KB (1,163 words) - 23:53, 28 March 2022
• 1984 AIME Problems/Problem 11
  == Solution 1 == == (Another way to think about Solution 1) ==
  7 KB (1,115 words) - 00:52, 7 September 2023
• 1984 AIME Problems/Problem 12
  == Solution 1 == ...ath>x = 0, \pm 5, \pm 10, \pm 15... \pm 1000</math> so the answer is 400 + 1 = <math>\boxed{401}</math>
  3 KB (588 words) - 14:37, 22 July 2020
• 1984 AIME Problems/Problem 14
  == Solution 1 == ...ites. However, if <math>x = 42</math>, it can also be expressed using case 1, and if <math>x = 40</math>, using case 3. <math>38</math> is the largest e
  8 KB (1,346 words) - 01:16, 9 January 2024
• 1984 AIME Problems/Problem 15
  ...ac{x^2}{8^2-1}+\frac{y^2}{8^2-3^2}+\frac{z^2}{8^2-5^2}+\frac{w^2}{8^2-7^2}=1 </math></div> == Solution 1 ==
  6 KB (1,051 words) - 04:52, 8 May 2024
• 1984 AIME Problems/Problem 13
  Find the value of <math>10\cot(\cot^{-1}3+\cot^{-1}7+\cot^{-1}13+\cot^{-1}21).</math> === Solution 1 ===
  3 KB (473 words) - 12:06, 18 December 2018
• 1995 IMO Problems/Problem 2
  Let <math>a, b, c</math> be positive real numbers such that <math>abc = 1</math>. Prove that <cmath> \frac{1}{a^3(b+c)} + \frac{1}{b^3(c+a)} + \frac{1}{c^3(a+b)} \geq \frac{3}{2}. </cmath>
  6 KB (1,122 words) - 12:23, 6 January 2022
• 1985 AIME Problems/Problem 15
  == Solution 1 (For the visualizers) == ...ath>\sqrt{2}</math> . This pyramid has a volume of 972, because it is also 1/6 of the volume of a cube with side length of 18. Then subtracting 3 congru
  2 KB (245 words) - 22:44, 4 March 2024
• 1985 AIME Problems/Problem 14
  ...got <math>0</math> points, and each of the two players earned <math>\frac{1}{2}</math> point if the game was a tie. After the completion of the tournam == Solution 1==
  5 KB (772 words) - 22:14, 18 June 2020
• 1985 AIME Problems/Problem 13
  ...n</math> be the greatest common divisor of <math>a_n</math> and <math>a_{n+1}</math>. Find the maximum value of <math>d_n</math> as <math>n</math> range == Solution 1==
  4 KB (671 words) - 20:04, 6 March 2024
• 1985 AIME Problems/Problem 12
  ...the vertices of a regular tetrahedron, each of whose edges measures <math>1</math> meter. A bug, starting from vertex <math>A</math>, observes the foll == Solution 1 (Single Variable Recursion) ==
  17 KB (2,837 words) - 13:34, 4 April 2024
• 1985 AIME Problems/Problem 11
  == Solution 1==
  5 KB (932 words) - 17:00, 1 September 2020
• 1985 AIME Problems/Problem 10
  == Solution 1 == ...d our answer will be 100 times the number of integers we can reach between 1 and 10.
  12 KB (1,859 words) - 18:16, 28 March 2022
• 1985 AIME Problems/Problem 9
  == Solution 1== <!-- Images obsoleted Image:1985_AIME-9.png, Image:1985_AIME-9a.png by as label("\(\alpha\)/2",(0.82,-1.25),NE,fontsize(8));
  5 KB (763 words) - 16:20, 28 September 2019
• 1985 AIME Problems/Problem 8
  ...h <math>a_i</math> by an [[integer]] approximation <math>A_i</math>, <math>1\le i \le 7</math>, so that the sum of the <math>A_i</math>'s is also 19 and ...ll of the <math>A_i</math> are 2 or 3, the largest error will be less than 1. So in the best case, we write 19 as a sum of 7 numbers, each of which is
  2 KB (377 words) - 02:17, 16 February 2021
• 1985 AIME Problems/Problem 7
  ...- t^2</math> so we must have <math>s + t^2 = 19</math> and <math>s - t^2 = 1</math>. Then <math>s = 10, t = 3</math> and so <math>d = s^3 = 1000</math>
  1 KB (222 words) - 11:04, 4 November 2022
• 1985 AIME Problems/Problem 6
  == Solution 1== This problem can be done using mass points. Assign B a weight of 1 and realize that many of the triangles have the same altitude. After contin
  5 KB (789 words) - 03:09, 23 January 2023
• 1985 AIME Problems/Problem 5
  ...]s <math>a_1, a_2, a_3, \ldots</math> is chosen so that <math>a_n = a_{n - 1} - a_{n - 2}</math> for each <math>n \ge 3</math>. What is the sum of the f ...{i = 1}^{1488} a_i = (a_1 + a_2 + a_3 + a_4) + \sum_{n = 0}^{247}\sum_{j = 1}^6 a_{6n + j}</cmath>
  2 KB (410 words) - 13:37, 1 May 2022
• 1985 AIME Problems/Problem 4
  ...[[square (geometry) | square]] is constructed inside a square of [[area]] 1 by dividing each side of the unit square into <math>n</math> equal parts, a == Solution 1==
  3 KB (484 words) - 21:40, 2 March 2020
• 1985 AIME Problems/Problem 3
  ...eger]]s which satisfy <math>c=(a + bi)^3 - 107i</math>, where <math>i^2 = -1</math>. ...math> is not divisible by 3, a contradiction. Thus we must have <math>b = 1</math>, <math>3a^2 = 108</math> so <math>a^2 = 36</math> and <math>a = 6</m
  1 KB (205 words) - 18:58, 10 March 2024
• 1986 AIME Problems/Problem 15
  == Solution 1 == ...he slopes of the respective medians; in other words, <math>\tan \theta_1 = 1</math>, and <math>\tan \theta_2 =2</math>.
  11 KB (1,722 words) - 09:49, 13 September 2023
• 1986 AIME Problems/Problem 14
  ...espectively. (This would give us the guess that the sides are of the ratio 1:2:3, but let's provide the complete solution.) <cmath>\frac {l^2w^2}{l^2 + w^2} = \frac {1}{\frac {1}{l^2} + \frac {1}{w^2}} = 20</cmath>
  2 KB (346 words) - 13:13, 22 July 2020
• 1986 AIME Problems/Problem 13
  ...ath> and multiplication, the answer is <math>{{2+4-1}\choose2} \cdot{{5+4-1}\choose5}=560</math> ~Slight edits in LaTeX by EthanSpoon ...t has four TT sequences, so when we place them in the spaces, we get <math>1+4=5</math> TT sequences. Since the number of TT sequences stays the same ea
  4 KB (772 words) - 21:09, 7 May 2024
• 1986 AIME Problems/Problem 12
  ...elements. To see why, note that at least <math>\dbinom{6}{0} + \dbinom{6}{1} + \dbinom{6}{2} + \dbinom{6}{3} + \dbinom{6}{4}=57</math> of its subsets h ...<math>a\leq 13</math>, or the subsets <math>\{a,14\}</math> and <math>\{a-1,15\}</math> would have the same sum. So now <math>S</math> must contain 13
  2 KB (364 words) - 19:41, 1 September 2020
• Analytic continuation
  ...h as <math>\sum_{n=1}^{\infty}{n} = \frac{-1}{12}</math> and <math>\sum_{n=1}^{\infty}{n^2} = 0</math>. Interestingly, even though these properties seem
  1 KB (180 words) - 20:12, 19 August 2015
• 1986 AIME Problems/Problem 11
  ...>a_0+a_1y+a_2y^2+\cdots +a_{16}y^{16}+a_{17}y^{17}</math>, where <math>y=x+1</math> and the <math>a_i</math>'s are [[constant]]s. Find the value of <mat === Solution 1 ===
  6 KB (872 words) - 16:51, 9 June 2023
• 1986 AIME Problems/Problem 10
  ===Solution 1 === As in Solution 1, <math>3194 + m \equiv 222(a+b+c) \pmod{222}</math>, and so as above we get
  3 KB (565 words) - 16:51, 1 October 2023
• 1986 AIME Problems/Problem 9
  === Solution 1 === pair D=IP(Ea--(Ea+A-C),A--B),F=IP(Da--(Da+C-B),A--C),Fa=IP(E--(E+A-B),A--C);
  11 KB (1,850 words) - 18:07, 11 October 2023
• 1986 AIME Problems/Problem 8
  == Solution 1 == ...ization]] of <math>1000000 = 2^65^6</math>, so there are <math>(6 + 1)(6 + 1) = 49</math> divisors, of which <math>48</math> are proper. The sum of mult
  3 KB (487 words) - 20:52, 16 September 2020
• 1986 AIME Problems/Problem 7
  The increasing [[sequence]] <math>1,3,4,9,10,12,13\cdots</math> consists of all those positive [[integer]]s whi === Solution 1 ===
  5 KB (866 words) - 00:00, 22 December 2022
• 1986 AIME Problems/Problem 6
  ==Solution 1== ...math> cannot be very large, disregard it for now and solve <math>\frac{n(n+1)}{2} = 1986</math>. The positive root for <math>n \approx \sqrt{3972} \appr
  2 KB (336 words) - 14:13, 6 September 2020
• 1986 AIME Problems/Problem 5
  == Solution 1 ==
  2 KB (338 words) - 19:56, 15 October 2023
• 1986 AIME Problems/Problem 4
  ...g all five [[equation]]s gives us <math>6(x_1 + x_2 + x_3 + x_4 + x_5) = 6(1 + 2 + 4 + 8 + 16)</math> so <math>x_1 + x_2 + x_3 + x_4 + x_5 = 31</math>.
  1 KB (212 words) - 16:25, 17 November 2019
• 1986 AIME Problems/Problem 3
  == Solution 1 == <math>\cot x + \cot y = \frac{1}{\tan x} + \frac{1}{\tan y} = \frac{\tan x + \tan y}{\tan x \cdot \tan y} = 30</math>
  3 KB (545 words) - 23:44, 12 October 2023
• 1986 AIME Problems/Problem 2
  == Solution 1 (Algebra: Generalized) == so <math>\sin\theta=\sqrt{1-\cos^2\theta}=\sqrt{\frac{13}{15}}.</math>
  3 KB (460 words) - 00:44, 5 February 2022
• 1987 AIME Problems/Problem 15
  ...and solving for <math>T_5</math>, <math>T_5 = 1 + T_1 - T_3 + T_2 - T_4 = 1 + \frac {T_1}{441} + \frac {T_2}{441}</math>. Therefore, <math>441T_5 = 441 ...[[hypotenuse]] of triangle <math>ABC</math>. Notice that <math>h - \frac {1}{21}h = \sqrt {440}</math>. (The height of <math>ABC</math> decreased by th
  5 KB (838 words) - 18:05, 19 February 2022
• 1987 AIME Problems/Problem 14
  == Solution 1 (Sophie Germain Identity) == ...]\left[(37^2+9)(43^2+9)\right]\left[(49^2+9)(55^2+9)\right]}=\frac{61^2+9}{1^2+9}=\boxed{373}.</cmath>
  7 KB (965 words) - 10:42, 12 April 2024
• 1987 AIME Problems/Problem 13
  ...e below shows how the sequence 1, 9, 8, 7 is transformed into the sequence 1, 8, 7, 9 by one bubble pass. The numbers compared at each step are underli <center><math>\underline{1 \quad 9} \quad 8 \quad 7</math></center>
  3 KB (514 words) - 21:27, 31 December 2023
• 1987 AIME Problems/Problem 12
  ...ger]] and <math>r</math> is a [[positive]] [[real number]] less than <math>1/1000</math>. Find <math>n</math>. == Solution 1 ==
  4 KB (673 words) - 19:48, 28 December 2023
• 1987 AIME Problems/Problem 11
  === Solution 1=== ...rite down one such sum, with <math>m</math> terms and first term <math>n + 1</math>:
  3 KB (418 words) - 18:30, 20 January 2024
• 1987 AIME Problems/Problem 10
  == Solution 1 == ...rds. Since Bob counts <math>75</math> steps, it takes him <cmath>\frac{75}{1}=75</cmath> seconds to traverse the distance of the escalator moving downwa
  7 KB (1,187 words) - 16:21, 27 January 2024
• 1987 AIME Problems/Problem 8
  == Solution 1== Continue as in Solution 1.
  2 KB (393 words) - 16:59, 16 December 2020
• 1987 AIME Problems/Problem 7
  ==Solution 1== ...)</math>: <math>(0, 3, 4), (1, 3, 4), (2, 3, 4), (3, 3, 4), (3, 2, 4), (3, 1, 4)</math> and <math>(3, 0, 4)</math>.
  3 KB (547 words) - 22:54, 4 April 2016
• 1987 AIME Problems/Problem 6
  ===Solution 1=== ...f the trapezoids are the same. Thus both trapezoids have area <math>\frac{1}{2} \cdot \frac{19}{2}(XY + PQ) = \frac{19}{4}(XY + 87)</math>. This numbe
  3 KB (530 words) - 07:46, 1 June 2018
• 1987 AIME Problems/Problem 5
  <cmath>(3x^2 + 1)(y^2 - 10) = 517 - 10</cmath> ...a multiple of three. <math>169</math> doesn't work either, so <math>3x^2 + 1 = 13</math>, and <math>x^2 = 4</math>. This leaves <math>y^2 - 10 = 39</mat
  1 KB (160 words) - 04:44, 21 January 2023
• 1987 AIME Problems/Problem 4
  == Solution 1 == ...and solving or using the [[Shoelace Theorem]], we get <math>2 \cdot \frac{1}{2}(80 - 48)(15) = \boxed{480}</math>.
  2 KB (371 words) - 17:25, 13 February 2024

View (previous 500 | next 500) (20 | 50 | 100 | 250 | 500)