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Number theory/Advanced
...e can study the set of points where both coordinates lie in some subfield (like the reals or the rationals). One also needs to add a limit point, called th ...ve, they can be added in a way that satisfies the normal laws of addition, like associativity, commutativity and the existence of an identity and inverses.

5 KB (849 words) - 16:14, 18 May 2021
Ellipse
...an ellipse with semimajor and semiminor axes <math>a,b</math> is <math>ab\pi</math>. ...re relatively prime positive integers. Find <math>m+n</math>. ([[2001 AIME I Problems/Problem 5|Source]])

5 KB (892 words) - 21:52, 1 May 2021
Pi
...mber]], as proved by Lindemann in 1882) denoted by the Greek letter <math>\pi </math>. ...^2}=\frac{\pi^2}{6}</math>. Some common [[fraction]]al approximations for pi are <math>\frac{22}{7} \approx 3.14285</math> and <math>\frac{355}{113} \ap

8 KB (1,469 words) - 21:11, 16 September 2022
Logarithm
..., we have <math>\ln (-1)=i\pi</math>. Additionally, <math>\ln (-n)=\ln n+i\pi</math> for positive real <math>n</math>.

4 KB (680 words) - 12:54, 16 October 2023
Roots of unity
...as a positive real number). This leaves us with <math>e^{ni\theta} = e^{2\pi ik}</math>. ...math> ni\theta = 2\pi ik</math>. Solving this gives <math> \theta=\frac{2\pi k}n </math>. Additionally, we note that for each of <math> k=0,1,2,\ldots,

3 KB (558 words) - 21:36, 11 December 2011
Complex analysis
<cmath> f^{(n)}(z_0)=\frac{n!}{2\pi i} \oint_\Gamma \frac{f(z)\; dz}{(z-z_0)^{n+1}}. </cmath> Let <math>f</math> be an [[entire]] function (i.e. holomorphic on the whole complex plane). If <math>\lvert f(z)\rvert \le

2 KB (271 words) - 22:06, 12 April 2022
Riemann Hypothesis
...th>. Then an equivalent statement of the Riemann hypothesis is that <math>\pi(x)=\mathrm{Li}(x)+O(x^{1/2}\ln(x))</math>. ...tion]] to <math>\Re(s)>\frac{1}{2}</math>. Let <math>M(n)=\sum_{i=1}^n \mu(i)</math> be the [[Mertens function]]. It is easy to show that if <math>M(n)\

2 KB (425 words) - 12:01, 20 October 2016
Algebraic number
...ic is called a [[transcendental number]], such as <math>e</math> or <math>\pi</math>.

1,006 bytes (151 words) - 21:56, 22 April 2022
Prime Number Theorem
specifically, it states that the functions <math>\pi(x)</math> and <math>x/\log x</math> are [[asymptotically equivalent]], where <math>\pi(x)</math> is the number

10 KB (1,729 words) - 19:52, 21 October 2023
LaTeX:Symbols
...texify] is an online application which allows you to draw the symbol you'd like and shows you the <math>\text{\LaTeX}</math> code for it! |<math>\pi</math>||\pi||<math>\varpi</math>||\varpi||<math>\rho</math>||\rho||<math>\varrho</math>

16 KB (2,324 words) - 16:50, 19 February 2024
2021 JMC 10
...can <math>|\pi - |e - | e - \pi|||</math> be expressed in terms of <math>\pi</math> and <math>e?</math> ...}2\pi-2e\qquad\textbf{(C) }2e\qquad\textbf{(D) }2\pi \qquad\textbf{(E) }2\pi +e</math>

12 KB (1,784 words) - 16:49, 1 April 2021
2004 AMC 12A Problems
(i) <math>f(1) = 1</math>, and ...rt3}{2} \qquad \text {(D) } \frac {10\pi}{3} - \sqrt3 \qquad \text {(E) }4\pi - 2\sqrt3</math>

13 KB (1,953 words) - 00:31, 26 January 2023
2001 AMC 12 Problems
...that is, <math>A > B > C</math>, <math>D > E > F</math>, and <math>G > H > I > J</math>. Furthermore, ...</math> are consecutive even digits; <math>G</math>, <math>H</math>, <math>I</math>, and <math>J</math> are consecutive odd

13 KB (1,957 words) - 12:53, 24 January 2024
2004 AMC 12B Problems
...{C}) 75+100\pi \quad (\mathrm {D}) 100+100\pi \quad (\mathrm {E}) 100+125\pi</math> ...hrm{(C) \ } \pi c^2 \qquad \mathrm{(D) \ } \pi d^2 \qquad \mathrm{(E) \ } \pi e^2 </math>

13 KB (2,049 words) - 13:03, 19 February 2020
2005 AMC 12B Problems
\mathrm{(A)}\ 80-20\pi \qquad \mathrm{(B)}\ 60-10\pi \qquad

12 KB (1,781 words) - 12:38, 14 July 2022
2005 AMC 12B Problems/Problem 22
...z_{n}}</math> is the [[complex conjugate]] of <math>z_{n}</math> and <math>i^{2}=-1</math>. Suppose that <math>|z_{0}|=1</math> and <math>z_{2005}=1</ma Since <math>|z_0|=1</math>, let <math>z_0=e^{i\theta_0}</math>, where <math>\theta_0</math> is an [[argument]] of <math>z_

4 KB (660 words) - 17:40, 24 January 2021
2006 AMC 10A Problems
MP('I', (8,-8), (0,0)); ...qquad\mathrm{(D)} \text{II, by}\ 8\pi\qquad\mathrm{(E)}\ \text{II, by}\ 10\pi</math>

13 KB (2,028 words) - 16:32, 22 March 2022
2006 AMC 10A Problems/Problem 12
MP('I', (8,-8), (0,0)); ...{(D) } II,\,\textrm{ by }\,8\pi\qquad \textbf{(E) } II,\,\textrm{ by }\,10\pi </math>

3 KB (424 words) - 10:14, 17 December 2021
Newman's Tauberian Theorem
<cmath> F(0)-F_T(0) = \frac{1}{2\pi i} \int_\Gamma K(z) (F(z)-F_T(z)) \le\int_{\Gamma_+} \frac{2M}{R^2} \lvert dz \rvert = \frac{2\pi M}{R} . </cmath>

6 KB (1,034 words) - 07:55, 12 August 2019
2005 AIME II Problems/Problem 9
...or equal to <math>1000</math> is <math> (\sin t + i \cos t)^n = \sin nt + i \cos nt </math> true for all real <math> t </math>? ...al number]]s <math>t</math> and all [[integer]]s <math>n</math>. So, we'd like to somehow convert our given expression into a form from which we can apply

6 KB (1,154 words) - 03:30, 11 January 2024
2005 AIME II Problems/Problem 12
...ying using the fact that <math>e^{i \frac{\pi}{4}} = \dfrac{\sqrt{2}}{2} + i \dfrac{\sqrt{2}}{2}</math>, clearing the denominator, and setting the imagi I hope you like expanding

13 KB (2,080 words) - 21:20, 11 December 2022
Euler's identity
'''Euler's Formula''' is <math>e^{i\theta}=\cos \theta+ i\sin\theta</math>. It is named after the 18th-century mathematician [[Leon ...eta) + i\sin(\theta))^n = (e^{i\theta})^n = e^{in\theta} = \cos(n\theta) + i\sin(n\theta)</math>.

3 KB (452 words) - 23:17, 4 January 2021
2005 AIME I Problems/Problem 1
[[Image:2005 AIME I Problem 1.png]] ...ath>. <math>K = 30^2\pi - 6(10^2\pi) = 300\pi</math>, so <math>\lfloor 300\pi \rfloor = \boxed{942}</math>.

1 KB (213 words) - 13:17, 22 July 2017
2005 AIME I Problems/Problem 6
...urth root of 2006. Then the roots of the above equation are <math>x = 1 + i^n r</math> for <math>n = 0, 1, 2, 3</math>. The two non-real members of th Starting like before,

4 KB (686 words) - 01:55, 5 December 2022
2004 AIME I Problems/Problem 14
pair Cxy = 8*expi((3*pi)/2-CE/8); pair L = 8*expi(pi+0.05), R = 8*expi(-0.22); /* left and right cylinder lines, numbers from tr

4 KB (729 words) - 01:00, 27 November 2022
2004 AIME I Problems/Problem 13
...4</math> complex roots of the form <math> z_k = r_k[\cos(2\pi a_k)+i\sin(2\pi a_k)], k=1, 2, 3,\ldots, 34, </math> with <math> 0 < a_1 \le a_2 \le a_3 \l {{AIME box|year=2004|n=I|num-b=12|num-a=14}}

2 KB (298 words) - 20:02, 4 July 2013
2004 AIME I Problems/Problem 11
...the [[Pythagorean Theorem]], we get <math>\ell = 5</math> and <math>A = 24\pi</math>. ...ace area of the original solid, we find that <math>A_f=24\pi - \frac{5}{3}\pi x^2</math>.

5 KB (839 words) - 22:12, 16 December 2015
2004 AIME I Problems/Problem 4
...ed by all of the midpoints is <math>4-4\cdot \left(\frac{\pi}{4}\right)=4-\pi \approx .86</math> to the nearest hundredth. Thus <math>100\cdot k=\boxed{8 ...rea of the square minus the area of the quarter circles, which is <math>4-\pi \approx 0.86</math>, so <math>100k = \boxed{086}</math>. ~Extremelysupercoo

3 KB (532 words) - 09:22, 11 July 2023
2004 AIME I Problems
{{AIME Problems|year=2004|n=I}} [[2004 AIME I Problems/Problem 1|Solution]]

9 KB (1,434 words) - 13:34, 29 December 2021
2004 AIME II Problems
...to the smaller can be expressed in the form <math> \frac{a\pi+b\sqrt{c}}{d\pi-e\sqrt{f}}, </math> where <math> a, b, c, d, e, </math> and <math> f </math {{AIME box|year = 2004|n=II|before=[[2004 AIME I Problems]]|after=[[2005 AIME I Problems]]}}

9 KB (1,410 words) - 05:05, 20 February 2019
1989 AIME Problems
...up heads in exactly <math>3</math> out of <math>5</math> flips. Find <math>i+j^{}_{}</math>. pair A=(0,0),B=(10,0),C=6*expi(pi/3);

7 KB (1,045 words) - 20:47, 14 December 2023
1991 AIME Problems
...th>x^{}_{}</math> satisfy the equation <math>\frac{1}{5}\log_2 x = \sin (5\pi x)</math>? ...The sum of the areas of the twelve disks can be written in the form <math>\pi(a-b\sqrt{c})</math>, where <math>a,b,c^{}_{}</math> are positive integers a

7 KB (1,106 words) - 22:05, 7 June 2021
1995 AIME Problems
...> are the perpendicular bisectors of two adjacent sides of square <math>S_{i+2}.</math> The total area enclosed by at least one of <math>S_{1}, S_{2}, ...math> and the sum of the other two roots is <math>3+4i,</math> where <math>i=\sqrt{-1}.</math> Find <math>b.</math>

6 KB (1,000 words) - 00:25, 27 March 2024
1998 AIME Problems
Given that <math>A_k = \frac {k(k - 1)}2\cos\frac {k(k - 1)\pi}2,</math> find <math>|A_{19} + A_{20} + \cdots + A_{98}|.</math> ...,x_2,x_3,x_4)</math> of positive odd [[integer]]s that satisfy <math>\sum_{i = 1}^4 x_i = 98.</math> Find <math>\frac n{100}.</math>

7 KB (1,084 words) - 02:01, 28 November 2023
2003 AIME I Problems
{{AIME Problems|year=2003|n=I}} [[2003 AIME I Problems/Problem 1|Solution]]

6 KB (965 words) - 16:36, 8 September 2019
2002 AIME II Problems
...math> radians are <math>\frac{m\pi}{n-\pi}</math> and <math>\frac{p\pi}{q+\pi}</math>, where <math>m</math>, <math>n</math>, <math>p</math>, and <math>q< {{AIME box|year = 2002|n=II|before=[[2002 AIME I Problems]]|after=[[2003 AIME I Problems]]}}

7 KB (1,177 words) - 15:42, 11 August 2023
2003 AIME II Problems
...e log. The number of cubic inches in the wedge can be expressed as <math>n\pi</math>, where n is a positive integer. Find <math>n</math>. ...istinct zeros of <math>P(x),</math> and let <math>z_{k}^{2} = a_{k} + b_{k}i</math> for <math>k = 1,2,\ldots,r,</math> where <math>a_{k}</math> and <mat

7 KB (1,127 words) - 09:02, 11 July 2023
1983 AIME Problems/Problem 15
...the minor arc <math>AD</math>, we see that <math>\alpha \in \left(-\frac{\pi}{2}-\alpha_0,\alpha_0\right)</math>. However, since the midpoint of <math>A Let I be the intersection of AD and BC.

19 KB (3,221 words) - 01:05, 7 February 2023
1984 AIME Problems/Problem 13
Recall that <math>\cot^{-1}\theta = \frac{\pi}{2} - \tan^{-1}\theta</math> and that <math>\arg(a + bi) = \tan^{-1}\frac{b ...{2} - \arg x + \frac{\pi}{2} - \arg y + \frac{\pi}{2} - \arg z) = 10\cot(2\pi - \arg wxyz)</cmath>

3 KB (473 words) - 12:06, 18 December 2018
1985 AIME Problems/Problem 12
...can generalize the following relationships for all nonnegative integers <math>k:</math> Let <math>\omega = e^{i\pi / 2}</math>. We have that if <math>G(x) = (x+x^2+x^3)^7</math>, then <cmath

17 KB (2,837 words) - 13:34, 4 April 2024
1987 AIME Problems/Problem 14
...eta</math> is the argument of <math>N</math> such that <math>0\leq\theta<2\pi.</math> .../math> so <math>\theta=\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}.</math></li>

7 KB (965 words) - 10:42, 12 April 2024
1988 AIME Problems/Problem 14
...eflection of the asymptote <math>x=0</math> by multiplying this by <math>2-i</math>, getting <math>4+3i</math>. Therefore, the asymptotes of <math>C^*</ ...atrix for a reflection about the polar line <math>\theta = \alpha, \alpha+\pi</math> is:

4 KB (700 words) - 17:21, 3 May 2021
1990 AIME Problems/Problem 12
...576}</math> instead-the motivation for this is to make the expression look like the half angle identity, and the fact that <math>\sqrt{576}</math> is an in ...\pi k}{12}}} \\ &= 12 \sqrt{4\sin^2{\frac{\pi k}{12}}} \\ &= 24\sin{\frac{\pi k}{12}}.\end{align*}</cmath> The rest follows as Solution 1.

6 KB (906 words) - 13:23, 5 September 2021
1990 AIME Problems/Problem 10
...math> and <math>48</math> is <math>144</math>, so define <math>n = e^{2\pi i/144}</math>. We can write the numbers of set <math>A</math> as <math>\{n^8, ...right)</math> respectively, where <math>\text{cis}\,\theta = \cos \theta + i \sin \theta</math> and <math>k_1</math> and <math>k_2</math> are integers f

3 KB (564 words) - 04:47, 4 August 2023
1991 AIME Problems/Problem 15
.../math> for <math>1 \le i \le n</math> and <math>0 \le \theta_{i} < \frac {\pi}{2}</math>. We then have that

4 KB (658 words) - 16:58, 10 November 2023
1991 AIME Problems/Problem 11
...The sum of the areas of the twelve disks can be written in the from <math>\pi(a-b\sqrt{c})</math>, where <math>a,b,c^{}_{}</math> are positive integers a for (int i=0; i<12; ++i)

4 KB (740 words) - 19:33, 28 December 2022
1992 AIME Problems/Problem 10
Let <math>z=a+bi \implies \frac{z}{40}=\frac{a}{40}+\frac{b}{40}i</math>. Since <math>0\leq \frac{a}{40},\frac{b}{40}\leq 1</math> we have th ...c{40}{\overline{z}}=\frac{40}{a-bi}=\frac{40a}{a^2+b^2}+\frac{40b}{a^2+b^2}i</math> so we have <math>0\leq a,b \leq \frac{a^2+b^2}{40}</math>, which lea

2 KB (323 words) - 12:05, 16 July 2019
1994 AIME Problems/Problem 14
...a * pi/180),N); path r = C + .4 * expi(beta * pi/180) -- C - 2*expi(beta * pi/180); ...-(1.5,0));D(r);D(anglemark(C,B,D(A)));D(anglemark((1.5,0),C,C+.4*expi(beta*pi/180)));MP("\beta",B,(5,1.2),fontsize(9));MP("\alpha",C,(4,1.2),fontsize(9))

2 KB (303 words) - 00:03, 28 December 2017
1994 AIME Problems/Problem 13
Using DeMoivre, <math>13 - t = \text{cis}\left(\frac {(2k + 1)\pi}{10}\right)</math> where <math>k</math> is an integer between <math>0</math ...pi}{10}\right) \Rightarrow \bar{t} = 13 - \text{cis}\left(-\frac {(2k + 1)\pi}{10}\right)</math>.

3 KB (375 words) - 23:46, 6 August 2021
1996 AIME Problems/Problem 15
pair B=(0,0), A=expi(pi/4), C=IP(A--A + 2*expi(17*pi/12), B--(3,0)), D=A+C, O=IP(A--C,B--D); === Solution 3 === <!-- I would put this solution first, but needs a bit of formatting for clarity -

5 KB (710 words) - 21:04, 14 September 2020
1996 AIME Problems/Problem 11
...maginary]] part, and suppose that <math>\mathrm {P}=r(\cos{\theta^{\circ}}+i\sin{\theta^{\circ}})</math>, where <math>0<r</math> and <math>0\leq \theta ...y <math>z - 1</math> to both sides, noting that then <math>z \neq \cos 0 + i\sin 0</math> since, then we are multiplying by <math>0</math> which makes i

6 KB (1,022 words) - 20:23, 17 April 2021
1997 AIME Problems/Problem 14
:<math>z^{1997}=1=1(\cos 0 + i \sin 0)</math> Define <math>\theta = 2\pi/1997</math>. By [[De Moivre's Theorem]] the roots are given by

5 KB (874 words) - 22:30, 1 April 2022
1998 AIME Problems/Problem 5
Given that <math>A_k = \frac {k(k - 1)}2\cos\frac {k(k - 1)\pi}2,</math> find <math>|A_{19} + A_{20} + \cdots + A_{98}|.</math> ...uates to an integer ([[triangular number]]), and the [[cosine]] of <math>n\pi</math> where <math>n \in \mathbb{Z}</math> is 1 if <math>n</math> is even a

1 KB (225 words) - 02:20, 16 September 2017
1999 AIME Problems/Problem 11
&= \mbox{Im } \frac{2\cos5 + 2i \sin 5}{(1 - \cos 5) - i \sin 5}\\ &= \mbox{Im } \frac{(2 \cos 5 + 2i \sin 5)[(1 - \cos 5) + i \sin 5]}{(1 - \cos 5)^2 + \sin^2 5}\\

4 KB (614 words) - 04:38, 8 December 2023
1999 AIME Problems/Problem 9
...finition of <math>f(z)</math>, <cmath>f(1+i) = (a+bi)(1+i) = (a-b) + (a+b)i,</cmath> this image must be equidistant to <math>(1,1)</math> and <math>(0, ...re positive, <math>z</math> lies in the first quadrant and <math>\theta < \pi/2</math>; hence by right triangle trigonometry <math>\sin \theta = \frac{\s

6 KB (1,010 words) - 19:01, 24 May 2023
2000 AIME I Problems/Problem 8
...the volume of the liquid can be found by <math>\frac{\pi}{3}r^2h - \frac{\pi}{3}(r')^2h'</math>. ...rac{\pi}{3}\left(\frac{3}{4}r\right)^2 \left(\frac{3}{4}h\right) &= \frac{\pi}{3}\left(r^2h - \left(\frac{rh'}{h}\right)^2h'\right)\\

4 KB (677 words) - 16:33, 30 December 2023
2003 AIME I Problems/Problem 5
...5</math> units. Given that the [[volume]] of this set is <math>\frac{m + n\pi}{p}, </math> where <math> m, n, </math> and <math> p </math> are positive [ ...ach of radius <math>1</math>. Together, their volume is <math> \frac{4}{3}\pi </math>.

2 KB (288 words) - 19:58, 4 July 2013
2003 AIME I Problems/Problem 2
...\ldots - 1^{2} \pi</math>, while the total area is given by <math>100^{2} \pi</math>, so the ratio is <cmath>\frac{100^{2}\pi - 99^{2}\pi + 98^{2}\pi - \ldots - 1^{2}\pi}{100^{2}\pi}</cmath>

4 KB (523 words) - 15:49, 8 March 2021
2003 AIME II Problems/Problem 14
...{3}}{2} i\right) = -\frac{b}{2}-\sqrt{3}+\left(\frac{b\sqrt{3}}{2}-1\right)i</math>. We solve for <math>b</math> and <math>f</math> and find that <math

9 KB (1,461 words) - 15:09, 18 August 2023
2000 AIME II Problems/Problem 9
...we have <math>z = \frac{2\cos 3 \pm \sqrt{4\cos^2 3 - 4}}{2} = \cos 3 \pm i\sin 3 = \text{cis}\,3^{\circ}</math>. ...could let <math>z = a + bi</math> and solve to get <math>z=\cos 3^\circ + i\sin 3^\circ</math>.

4 KB (675 words) - 13:42, 4 April 2024
Holomorphic function
<cmath> f(z) = u(x,y) + i v(x,y). </cmath> + i \frac{\partial v}{\partial x}, \qquad

9 KB (1,537 words) - 21:04, 26 July 2017
University of South Carolina High School Math Contest/1993 Exam/Problems
...athrm{(D) \ } \frac{\sqrt{3}}2 - \frac{\pi}6 \qquad \mathrm{(E) \ } \frac{\pi}6 </cmath> If <math>(1 + i)^{100}</math> is expanded and written in the form <math>a + bi</math> where

14 KB (2,102 words) - 22:03, 26 October 2018
Mock AIME 1 2006-2007/Problems
.../math>, <math>CA=1</math>, and <math>AB=3</math>. If <math>\angle A=\frac{\pi}{n}</math> where <math>n</math> is an integer, find the remainder when <mat ...et circles <math>A''</math>, <math>B''</math>, <math>C''</math>, and <math>I</math> have radii <math>a</math>, <math>b</math>, <math>c</math>, and <math

8 KB (1,355 words) - 14:54, 21 August 2020
2006 Romanian NMO Problems
...<math>a_i \in \{ -1, 1 \}</math>, such that <center><math>n = \sum_{1\leq i < j \leq k } a_ia_j</math>.</center> ...ngle with the base <math>BC</math>. We know that <math>\angle ABD = \frac{\pi}{2}</math>. Let <math>M</math> be the midpoint of <math>BC</math>. The poin

11 KB (1,779 words) - 14:57, 7 May 2012
2005 Canadian MO Problems/Problem 3
For <math>i = 1, 2, 3</math>, let <math>m_i</math> be the line perpendicular to <math>l ...\left(\frac{1}{2}, \frac{\pi}{3}\right), V_3 = \left(\frac{1}{2}, \frac{2\pi}{3}\right) \in S</math>. It is easy to see that for any point <math>P</math

2 KB (460 words) - 13:35, 9 June 2011
Volume
...volume of a [[sphere]] of [[radius]] <math>r</math> is <math>\frac 43 r^3\pi</math>. ...[[cylinder]] of height <math>h</math> and radius <math>r</math> is <math>\pi r^2h</math>. (Note that this is just a special case of the formula for a p

3 KB (523 words) - 20:24, 17 August 2023
Right angle
...of every right angle is 90 [[degree (geometry) | degrees]] or <math>\frac \pi 2</math> [[radian]]s. When drawing diagrams, we denote right angles with a pair A, B, C, D, I;

673 bytes (91 words) - 23:59, 11 June 2022
1970 IMO Problems/Problem 1
Let <math>I</math> be the incenter of <math>ABC</math>, and let <math>I_{c}</math> be i c & = &q \left[ \cot\left(\frac{\pi - A}{2}\right) + \cot \left(\frac{\pi - B}{2}\right) \right]\\

2 KB (380 words) - 22:12, 19 May 2015
2007 iTest Problems
<cmath>\sum_{i=1}^{100}(i!)^{2}</cmath> ...4\quad\mathrm{(F)}\,5\quad\mathrm{(G)}\,6\quad\mathrm{(H)}\,7\quad\mathrm{(I)}\,2007</math>

33 KB (5,177 words) - 21:05, 4 February 2023
Straight angle
...e of every straight angle is 180 [[degree (geometry) | degrees]] or <math>\pi</math> [[radian]]s. pair A, B, I;

568 bytes (78 words) - 13:50, 12 June 2022
2005 AMC 10A Problems
...D) \ } \frac{2}{3}\pi+\frac{\sqrt{3}}{3}\qquad \mathrm{(E) \ } \frac{2}{3}\pi+\frac{\sqrt{3}}{2} </math> for(int i=0;i<=4;i=i+1)

14 KB (2,026 words) - 11:45, 12 July 2021
1985 IMO Problems/Problem 1
...es of arcs, <math> \angle DTA = \angle DCO = \frac{\angle DCB}{2} = \frac{\pi}{2} - \frac{\angle DAB}{2} </math>. It follows that <math>AT = AD </math>. ...</math> such that <math>AT = AD </math>. Then <math>\angle DTA = \frac{ \pi - \angle DAB}{2} = \angle DCO</math>, so <math>DCOT </math> is a cyclic qua

4 KB (684 words) - 07:28, 3 October 2021
2004 AMC 10A Problems
A grocer stacks oranges in a pyramid-like stack whose rectangular base is 5 oranges by 8 oranges. Each orange above for(int i=0; i<=5; ++i)

15 KB (2,092 words) - 20:32, 15 April 2024
1985 AIME Problems
...rotated about the other leg, the volume of the cone produced is <math>1920\pi \;\textrm{cm}^3</math>. What is the length (in cm) of the hypotenuse of the ...itive integers which satisfy <math>c=(a + bi)^3 - 107i</math>, where <math>i^2 = -1</math>.

7 KB (1,071 words) - 19:24, 23 February 2024
Asymptote: Advanced Configuration
There are two things that Asymptote users commonly would like to do with their figures: make the images output in a different format, and I like to make pics with Asymptote like this one:

12 KB (1,931 words) - 13:53, 26 January 2020
Asymptote: Reference
returns the complex number <math>e^{i\theta}</math>, i.e. (cos(theta),sin(theta)) where theta is measured in radians. star=expi(0)--(scale((3-sqrt(5))/2)*expi(pi/5))--expi(2*pi/5)--

7 KB (1,205 words) - 21:38, 26 March 2024
2007 AIME I Problems/Problem 4
...30} </math>, <math> b(t) = \frac{t \pi}{42} </math>, <math>c(t) = \frac{t \pi}{70}</math>. ...frac{ 2 t \pi}{105} </math>. These are simultaneously multiples of <math>\pi </math> exactly when <math>t </math> is a multiple of <math>105</math>, so

1 KB (228 words) - 18:46, 11 March 2021
2007 AIME II Problems
...<math>a_{i}</math> is [[odd]], and <math>a_{i}>a_{i+1}</math> if <math>a_{i}</math> is [[even]]. How many four-digit parity-monotonic integers are ther ...> divides <math>n_{i}</math> for all <math>i</math> such that <math>1 \leq i \leq 70.</math>

9 KB (1,435 words) - 01:45, 6 December 2021
2007 AMC 12A Problems
<math>\mathrm{(A)}\ 8.64\qquad \mathrm{(B)}\ 12\qquad \mathrm{(C)}\ 5\pi\qquad \mathrm{(D)}\ 17.28\qquad \mathrm{(E)}\ 18</math> ...bx^{2} + cx + d</math> has real [[coefficient]]s, and <math>f(2i) = f(2 + i) = 0.</math> What is <math>a + b + c + d?</math>

11 KB (1,750 words) - 13:35, 15 April 2022
2007 USAMO Problems/Problem 6
pair A=(2,8), B=(0,0), C=(13,0), I=incenter(A,B,C), O=circumcenter(A,B,C), p_a, q_a, X, Y, X1, Y1, D, E, F; real r=abs(I-foot(I,A,B)), R=abs(A-O), a=abs(B-C), b=abs(A-C), c=abs(A-B), x=(((b+c-a)/2)^2)/(r

7 KB (1,274 words) - 15:11, 31 August 2017
LaTeX:Commands
| <math>\textstyle 2^{23}</math>||2^{23}||<math>\textstyle n_{i-1}</math>||n_{i-1} | <math>a^{i+1}_3</math>||a^{i+1}_3||<math>x^{3^2}</math>||x^{3^2}

12 KB (1,898 words) - 15:31, 22 February 2024
2005 PMWC Problems
for (int i=0;i<6;i=i+1){ draw(dir(60*i)..3*dir(60*i)..cycle);

9 KB (1,449 words) - 20:49, 2 October 2020
1998 PMWC Problems
for(int i = 0; i < n; ++i){ for(int j = n-i; j > 0; --j){

11 KB (1,738 words) - 19:25, 10 March 2015
2000 PMWC Problems
for(int i = 0; i < 4; ++i){ if(i < 2){

11 KB (1,713 words) - 22:47, 13 July 2023
2001 PMWC Problems
real eta=pi/2; for(int i = 0; i < 4; ++i)

4 KB (641 words) - 21:24, 21 April 2014
2002 PMWC Problems
for(int i = 0; i < nrows; ++i) for(int j = 0; j <= i; ++j)

5 KB (725 words) - 16:07, 23 April 2014
2003 PMWC Problems
pair M=(-1,0), N=(1,0),a=4/5*expi(pi/10),b=expi(37pi/100); for(int i = -2; i <= 2; ++i)

7 KB (918 words) - 16:15, 22 April 2014
Root (operation)
...t[n]{x}=\sqrt[n]{|x|}\left(\cos\frac{\theta+2\pi k}{n}+i\sin\frac{\theta+2\pi k}{n}\right)</math> , where <math>k=0,1,2,...,n-1</math> and <math>x\in\mat ...h>\sqrt[4]{16}=\sqrt[4]{16}\left(\cos\frac{0+2\pi\cdot0}{4}+i\sin\frac{0+2\pi\cdot0}{4}\right)\implies\boxed{2}</math>

3 KB (532 words) - 16:52, 20 May 2020
2007 AMC 10A Problems
<math>\mathrm{(A)}\ 8.64\qquad \mathrm{(B)}\ 12\qquad \mathrm{(C)}\ 5\pi\qquad \mathrm{(D)}\ 17.28\qquad \mathrm{(E)}\ 18</math> pair A=(1,3), B=(2,3), C=(2,2), D=(3,2), Ep=(3,1), F=(2,1), G=(2,0), H=(1,0), I=(1,1), J=(0,1), K=(0,2), L=(1,2);

13 KB (2,058 words) - 17:54, 29 March 2024
1995 AHSME Problems
<math>\text{(i)}</math> <math>4,000 \leq N < 6,000;</math> for(int i = 0; i < 5; ++i) { pair P = dir(90-i*72); dot(P); label("$"+string(i+1)+"$",P,1.4*P); }

17 KB (2,387 words) - 22:44, 26 May 2021
1959 AHSME Problems
...i p^2}{27}\qquad\textbf{(D)}\ \frac{\pi p^2}{81}\qquad\textbf{(E)}\ \frac{\pi p^2\sqrt3}{27} </math> ...owing is not true for the equation<math> ix^2-x+2i=0</math>, where <math>i=\sqrt{-1}</math>

22 KB (3,345 words) - 20:12, 15 February 2023
1991 USAMO Problems
...<math>\,S\,</math> of numbers, let <math>\,\sigma(S)\,</math> and <math>\,\pi(S)\,</math> denote the sum and product, respectively, of the elements of <m <cmath> \sum \frac{\sigma(S)}{\pi(S)} = (n^2 + 2n) - \left( 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}

3 KB (512 words) - 19:17, 18 July 2016
1976 IMO Problems/Problem 2
I shall prove by induction that <math>P_n(x)</math> has <math>2^n</math> dist ...l}{2^n+1}\cdot\pi</math>. As we can choose the range <math>0\leq\theta\leq\pi</math> to ensure no duplications, we get that, upon rearranging, <math>0\le

3 KB (596 words) - 16:19, 28 July 2015
2008 AMC 12A Problems/Problem 23
The solutions of the equation <math>z^4+4z^3i-6z^2-4zi-i=0</math> are the vertices of a convex polygon in the complex plane. What is ...^{4}=1+i=2^{\frac{1}{2}}\cos \frac {\pi}{4} + 2^{\frac{1}{2}}i\sin \frac {\pi}{4} </cmath>

1 KB (199 words) - 01:38, 10 November 2019
2008 AMC 12A Problems/Problem 25
...<math>z_n = 2e^{i\pi/6}z_{n-1}</math>, so <math>z_{n-1} = \frac{1}{2}e^{-i\pi/6}z_n</math>. This is the reverse transformation. We have <cmath> z_{99} = \frac{1}{2}e^{-i\pi/6}z_{100}</cmath>

5 KB (745 words) - 10:58, 9 December 2022
2008 AMC 12B Problems
...uad \textbf{(C)}\ \pi \qquad \textbf{(D)}\ 2\pi \qquad \textbf{(E)}\ 10^{2\pi}</math> ...ers and <math>i^2 = - 1</math>. Suppose that <math>f(1)</math> and <math>f(i)</math> are both real. What is the smallest possible value of <math>| \alph

14 KB (2,199 words) - 13:43, 28 August 2020
2008 AIME I Problems
{{AIME Problems|year=2008|n=I}} ...e party is now <math>58\%</math> girls. How many students now at the party like to dance?

9 KB (1,536 words) - 00:46, 26 August 2023
2008 AIME II Problems
Let <math>a = \pi/2008</math>. Find the smallest positive integer <math>n</math> such that ...efine a ''move'' for the particle as a counterclockwise rotation of <math>\pi/4</math> radians about the origin followed by a translation of <math>10</ma

7 KB (1,167 words) - 21:33, 12 August 2020
2008 AIME I Problems/Problem 5
...math>\sqrt {r^{2} + h^{2}}</math>. Thus, the length of the path is <math>2\pi\sqrt {r^{2} + h^{2}}</math>. ...of the path is 17 times the circumference of the base, which is <math>34r\pi</math>. Setting these equal gives <math>\sqrt {r^{2} + h^{2}} = 17r</math>,

1 KB (230 words) - 20:18, 4 July 2013
2008 AIME I Problems/Problem 8
...\arctan\frac {1}{4} + \arctan\frac {1}{5} + \arctan\frac {1}{n} = \frac {\pi}{4}.</cmath> We now have <math>\arctan{\dfrac{23}{24}} + \arctan{\dfrac{1}{n}} = \dfrac{\pi}{4} = \arctan{1}</math>. Thus, <math>\dfrac{\dfrac{23}{24} + \dfrac{1}{n}}{

3 KB (490 words) - 22:36, 28 November 2023

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