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Law of Cosines
draw(D--(30,4)--(34,4)--(34,0)--D); 1. If the sides of a triangle have lengths 2, 3, and 4, what is the radius of the circle circumscribing the triangle?

6 KB (1,003 words) - 09:11, 7 June 2023
Angle bisector
A=(4,2); D=(3,4);

3 KB (575 words) - 15:27, 19 March 2023
Law of Sines
pair A=(-1,5), B=(-4,-1), C=(4,-1), D, O; *If the sides of a triangle have lengths 2, 3, and 4, what is the radius of the circle circumscribing the triangle?

4 KB (658 words) - 16:19, 28 April 2024
Sequence
...ath>\mathbb{N}</math> corresponds to the sequence <math>X = (x_n) = (0, 1, 4, 9, 16, \ldots)</math>. ...> is called the [[limit]] of <math>(x_n)</math> and is written <math>\lim_{n \to \infty} x_n</math>. The statement that <math>(x_n)</math> converges to

2 KB (413 words) - 21:18, 13 November 2022
Geometric sequence
For example, <math>1, 2, 4, 8</math> is a geometric sequence with common ratio <math>2</math> and <mat ...progression 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

4 KB (644 words) - 12:55, 7 March 2022
Arithmetic sequence
...on difference <math>-8</math>; however, <math>7, 0, 7, 14</math> and <math>4, 12, 36, 108, \ldots</math> are not arithmetic sequences, as the difference ...progression 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

4 KB (736 words) - 02:00, 7 March 2024
Fermat's Last Theorem
...\geq 3</math>, there are no solutions to the equation <math>a^n + b^n = c^n</math>. ...he never published it, though he did publish a proof for the case <math>n=4</math>. It seems unlikely that he would have circulated a proof for the sp

3 KB (453 words) - 11:13, 9 June 2023
2005 USAMO Problems/Problem 4
...ece 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? For <math>0 \le x \le n</math>, it is easy to see that the number of stable tables is <math>(x+1)^2

7 KB (1,276 words) - 20:51, 6 January 2024
Diophantine equation
If <math>n>1</math>, <math>2n, n^2 - 1, n^2 + 1</math> is a Pythagorean triple. ...ny <math>m,n</math>(<math>m>n</math>), we have <math>m^2 - n^2, 2mn, m^2 + n^2</math> is a Pythagorean triple.

9 KB (1,434 words) - 13:10, 20 February 2024
Area
...s 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 ex ...ound using [[trigonometry]] to be of length <math>\frac s2 \cot \frac{180}{n}^{\circ}</math>.

6 KB (1,181 words) - 22:37, 22 January 2023
Floor function
...r+\left\lfloor a+\frac{1}{n}\right\rfloor+\ldots+\left\lfloor a+\frac{n-1}{n}\right\rfloor</cmath> *<math>\lfloor -3.2 \rfloor = -4</math>

3 KB (508 words) - 21:05, 26 February 2024
Pascal's triangle
...he sum of the values on row <math>n</math> of Pascal's Triangle is <math>2^n</math>. ...ved from the combinatorics identity <math>{n \choose k}+{n \choose k+1} = {n+1 \choose k+1}</math>. Thus, any number in the interior of Pascal's Triang

5 KB (838 words) - 17:20, 3 January 2023
Newton's Sums
Consider a polynomial <math>P(x)</math> of degree <math>n</math>, <center><math> P(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0</math></center>

4 KB (690 words) - 13:11, 20 February 2024
1985 IMO Problems/Problem 5
...ath> and <math>BC</math> again at distinct points <math>K</math> and <math>N</math> respectively. Let <math>M</math> be the point of intersection of the {{IMO box|year=1985|num-b=4|num-a=6}}

3 KB (496 words) - 13:35, 18 January 2023
Trigonometry
...=0}^{\infty} (-1)^n \frac{x^{2n}}{(2n)!} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} \cdots</cmath>

8 KB (1,217 words) - 20:15, 7 September 2023
Root-Mean Square-Arithmetic Mean-Geometric Mean-Harmonic mean Inequality
...</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. ...1}{a}}</math> and the harmonic mean's root mean power is -1 as <math>\frac{n}{\frac{1}{x_1}+\cdots+\frac{1}{x_n}}=\sqrt[-1]{\frac{x_1^{-1}+\cdots+x_a^{-

5 KB (912 words) - 20:06, 14 March 2023
Goldbach Conjecture
<math>4 = 2 + 2</math> ...might expect the total number of ways to write a large even integer <math>n</math> as the sum of two odd primes to be roughly

7 KB (1,201 words) - 16:59, 19 February 2024
Riemann Hypothesis
...function, it is easy to see that <math>\zeta(s)=0</math> when <math>s=-2,-4,-6,\ldots</math>. These are called the trivial zeros. This hypothesis is o ...uld hold. The Riemann Hypothesis would also follow if <math>M(n)\le C\sqrt{n}</math> for any constant <math>C</math>.

2 KB (425 words) - 12:01, 20 October 2016
Quadratic residues
...ulo]] <math>m</math> if there is some integer <math>n</math> so that <math>n^2-a</math> is [[divisibility | divisible]] by <math>m</math>. ...p-1}{2}}</math>, so <math>\left(\frac{-1}{p}\right)=1 \iff p \equiv 1 \mod 4</math>

5 KB (778 words) - 13:10, 29 November 2017
Power of a Point Theorem
Let <math>P</math> be a point, and let <math>S</math> be an <math>n</math>-sphere. Let two arbitrary lines passing through <math>P</math> inter ...th> intersect at <math>R</math>. If <math>AR:BR=1:4</math> and <math>CR:DR=4:9</math>, find the ratio <math>AB:CD</math>.

5 KB (827 words) - 17:30, 21 February 2024
2004 AIME I
* [[2004 AIME I Problems/Problem 4]] {{AIME box|year=2004|n=I|before=[[2003 AIME I]], [[2003 AIME II|II]]|after=[[2004 AIME II]]}}

1 KB (135 words) - 18:15, 19 April 2021
Collatz Problem
...sitive integer]] <math>n</math>, the sequence <math>\{n,f(n),f(f(n)),f(f(f(n))),\ldots\}</math> contains 1. This conjecture is still open. Some people h ==Properties of <math>f(n)</math> ==

1 KB (231 words) - 19:45, 24 February 2020
2004 AIME II
* [[2004 AIME II Problems/Problem 4]] {{AIME box|year=2004|n=II|before=[[2004 AIME I]]|after=[[2005 AIME I]], [[2005 AIME II|II]]}}

1 KB (135 words) - 12:24, 22 March 2011
2005 AIME I
* [[2005 AIME I Problems/Problem 4 | Problem 4]] {{AIME box|year=2005|n=I|before=[[2004 AIME I|2004 AIME I]], [[2004 AIME II|II]]|after=[[2005 AIME

1 KB (154 words) - 12:30, 22 March 2011
2006 AIME I
* [[2006 AIME I Problems/Problem 4]] {{AIME box|year=2006|n=I|before=[[2005 AIME I]], [[2005 AIME II|II]]|after=[[2006 AIME II]]}}

1 KB (135 words) - 12:31, 22 March 2011
2005 AIME II
* [[2005 AIME II Problems/Problem 4]] {{AIME box|year=2005|n=II|before=[[2005 AIME I]]|after=[[2006 AIME I]], [[2006 AIME II|II]]}}

1 KB (135 words) - 12:30, 22 March 2011
Mock AMC
| n/a | n/a

51 KB (6,175 words) - 20:58, 6 December 2023
Bertrand's Postulate
...em. A more widely known version states that there is a prime between <math>n</math> and <math>2n</math>. ...closer look at the [[combinations|binomial coefficient]] <math>\binom{2n}{n}</math>. Assuming that the reader is familiar with that proof, the Bertrand

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
Modular arithmetic/Introduction
draw((0,0), linewidth(4)); <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>

15 KB (2,396 words) - 20:24, 21 February 2024
Perfect square
...so that <math>m^2=n</math>. The first few perfect squares are <math>0, 1, 4, 9, 16, 25, 36</math>. ...th>n</math> 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
Modular arithmetic/Intermediate
...th>n</math>), if the difference <math>{a - b}</math> is divisible by <math>n</math>. ...ntegers modulo <math>n</math>''' (usually known as "the integers mod <math>n</math>," or <math>\mathbb{Z}_n</math> for short). This structure gives us

14 KB (2,317 words) - 19:01, 29 October 2021
2006 AIME I Problems
{{AIME Problems|year=2006|n=I}} == Problem 4 ==

7 KB (1,173 words) - 03:31, 4 January 2023
2006 AIME I Problems/Problem 15
=== Solution 4 === {{AIME box|year=2006|n=I|num-b=14|after=Last Question}}

6 KB (910 words) - 19:31, 24 October 2023
2006 AIME I Problems/Problem 14
...> is not divisible by the square of any prime. Find <math> \lfloor m+\sqrt{n}\rfloor. </math> (The notation <math> \lfloor x\rfloor </math> denotes the triple M=(B+C)/2,S=(4*A+T)/5;

6 KB (980 words) - 21:45, 31 March 2020
2006 AIME I Problems/Problem 13
...h> S_n=\sum_{k=1}^{2^{n-1}}g(2k). </math> Find the greatest integer <math> n </math> less than 1000 such that <math> S_n </math> is a [[perfect square]] .../math> but not by <math>2^n, \ldots,</math> and <math>2^{n-1}-2^{n-2} = 2^{n-2}</math> elements of <math>S</math> that are divisible by <math>2^1</math>

10 KB (1,702 words) - 00:45, 16 November 2023
2006 AIME I Problems/Problem 10
...ough the tangent point of the other two circles. This clearly will cut the 4 circles into two regions of equal area. Using super advanced linear algebra {{AIME box|year=2006|n=I|num-b=9|num-a=11}}

4 KB (731 words) - 17:59, 4 January 2022
2006 AIME I Problems/Problem 8
pair A=(0,0), B=(4.2,0), C=(5.85,-1.6), D=(4.2,-3.2), EE=(0,-3.2), F=(-1.65,-1.6), G=(0.45,-1.6), H=(3.75,-1.6), I=(2.1, label("$\mathcal{Q}$",(4.2,-1),NW);

5 KB (730 words) - 15:05, 15 January 2024
2006 AIME I Problems/Problem 7
for(int i=0; i<4; i=i+1) { pair A=(1/3,4), B=A+7.5*dir(-17), C=A+7*dir(10);

4 KB (709 words) - 01:50, 10 January 2022
2006 AIME I Problems/Problem 5
<math> b = 4 </math> <math>b=4</math>

3 KB (439 words) - 18:24, 10 March 2015
2006 AIME I Problems/Problem 4
...oduct <math> 1!2!3!4!\cdots99!100!. </math> Find the remainder when <math> N </math> is divided by <math>1000</math>. ...<math>5</math> dividing it, for <math>76</math> extra; every n! for <math>n\geq 50</math> has one more in addition to that, for a total of <math>51</ma

2 KB (278 words) - 08:33, 4 November 2022
2006 AIME I Problems/Problem 3
...math> so we take <math>N_0 = 25</math> and <math>n = 2.</math> Then <cmath>N = 7 \cdot 10^2 + 25 = \boxed{725},</cmath> and indeed, <math>725 = 29 \cdot ...ow that <math>N<1000</math> (because this is an AIME problem). Thus, <math>N</math> has <math>1,</math> <math>2</math> or <math>3</math> digits. Checkin

4 KB (622 words) - 03:53, 10 December 2022
2022 AIME I Problems/Problem 1
Q(20) &= &\hspace{1mm}-800 + 20c + d &= 53, &&(4) ...]</math> from <math>5\cdot[(1)+(2)]</math> to get <cmath>b+d=5\cdot(54+54)-4\cdot(53+53)=\boxed{116}.</cmath>

4 KB (670 words) - 13:03, 13 November 2023
2021 JMC 10
...bf{(A) } 1 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 3 \qquad\textbf{(D) } 4 \qquad\textbf{(E) } 7</math> ==Problem 4==

12 KB (1,784 words) - 16:49, 1 April 2021
2006 AMC 12B Problems
...ne <math>x\spadesuit y = (x + y)(x - y)</math>. What is <math>3\spadesuit(4\spadesuit 5)</math>? == Problem 4 ==

13 KB (2,058 words) - 12:36, 4 July 2023
2006 AMC 12A Problems
== Problem 4 == [[2006 AMC 12A Problems/Problem 4|Solution]]

15 KB (2,223 words) - 13:43, 28 December 2020
2005 AMC 12A Problems
...8 \qquad (\mathrm {B}) \ -4 \qquad (\mathrm {C})\ 2 \qquad (\mathrm {D}) \ 4 \qquad (\mathrm {E})\ 8 == Problem 4 ==

13 KB (1,971 words) - 13:03, 19 February 2020
2004 AMC 12A Problems
== Problem 4 == [[2004 AMC 12A Problems/Problem 4|Solution]]

13 KB (1,953 words) - 00:31, 26 January 2023
2003 AMC 12A Problems
Members of the Rockham Soccer League buy socks and T-shirts. Socks cost $4 per pair and each T-shirt costs $5 more than a pair of socks. Each memb <math> \mathrm{(A) \ } 4.5\qquad \mathrm{(B) \ } 9\qquad \mathrm{(C) \ } 12\qquad \mathrm{(D) \ } 18

13 KB (1,955 words) - 21:06, 19 August 2023
2002 AMC 12A Problems
<math>(2x+3)(x-4)+(2x+3)(x-6)=0 </math> <math> \mathrm{(A) \ } \frac{7}{2}\qquad \mathrm{(B) \ } 4\qquad \mathrm{(C) \ } 5\qquad \mathrm{(D) \ } 7\qquad \mathrm{(E) \ } 13 </

12 KB (1,792 words) - 13:06, 19 February 2020

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