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- ...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
- ...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
- ...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} \ap8 KB (1,469 words) - 21:11, 16 September 2022
- ..., 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
- ...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
- <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 \le2 KB (271 words) - 22:06, 12 April 2022
- ...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
- ...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
- 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 number10 KB (1,729 words) - 19:52, 21 October 2023
- ...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
- ...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
- (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
- ...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 odd13 KB (1,957 words) - 12:53, 24 January 2024
- ...{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
- \mathrm{(A)}\ 80-20\pi \qquad \mathrm{(B)}\ 60-10\pi \qquad12 KB (1,781 words) - 12:38, 14 July 2022
- ...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
- 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
- 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
- <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
- ...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 apply6 KB (1,154 words) - 03:30, 11 January 2024
- ...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 expanding13 KB (2,080 words) - 21:20, 11 December 2022
- '''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
- [[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
- ...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
- 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 tr4 KB (729 words) - 01:00, 27 November 2022
- ...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
- ...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
- ...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>. ~Extremelysupercoo3 KB (532 words) - 09:22, 11 July 2023
- {{AIME Problems|year=2004|n=I}} [[2004 AIME I Problems/Problem 1|Solution]]9 KB (1,434 words) - 13:34, 29 December 2021
- ...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
- ...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
- ...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 a7 KB (1,106 words) - 22:05, 7 June 2021
- ...> 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
- 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
- {{AIME Problems|year=2003|n=I}} [[2003 AIME I Problems/Problem 1|Solution]]6 KB (965 words) - 16:36, 8 September 2019
- ...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
- ...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 <mat7 KB (1,127 words) - 09:02, 11 July 2023
- ...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
- 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
- ...can generalize the following relationships for all <i><b>nonnegative</b></i> 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 <cmath17 KB (2,837 words) - 13:34, 4 April 2024
- ...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><p>7 KB (965 words) - 10:42, 12 April 2024
- ...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
- ...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
- ...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 f3 KB (564 words) - 04:47, 4 August 2023
- .../math> for <math>1 \le i \le n</math> and <math>0 \le \theta_{i} < \frac {\pi}{2}</math>. We then have that4 KB (658 words) - 16:58, 10 November 2023
- ...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
- 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 lea2 KB (323 words) - 12:05, 16 July 2019
- ...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
- 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
- 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
- ...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 i6 KB (1,022 words) - 20:23, 17 April 2021
- :<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 by5 KB (874 words) - 22:30, 1 April 2022
- 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 a1 KB (225 words) - 02:20, 16 September 2017
- &= \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
- ...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{\s6 KB (1,010 words) - 19:01, 24 May 2023
- ...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
- ...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
- ...\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
- ...{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 <math9 KB (1,461 words) - 15:09, 18 August 2023
- ...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
- <cmath> f(z) = u(x,y) + i v(x,y). </cmath> + i \frac{\partial v}{\partial x}, \qquad9 KB (1,537 words) - 21:04, 26 July 2017
- ...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> where14 KB (2,102 words) - 22:03, 26 October 2018
- .../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 <math8 KB (1,355 words) - 14:54, 21 August 2020
- ...<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 poin11 KB (1,779 words) - 14:57, 7 May 2012
- 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</math2 KB (460 words) - 13:35, 9 June 2011
- ...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 p3 KB (523 words) - 20:24, 17 August 2023
- ...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
- 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
- <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
- ...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
- ...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
- ...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 qua4 KB (684 words) - 07:28, 3 October 2021
- 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
- ...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
- 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
- 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
- ...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>, so1 KB (228 words) - 18:46, 11 March 2021
- ...<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
- <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
- 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)/(r7 KB (1,274 words) - 15:11, 31 August 2017
- | <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
- 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
- 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
- for(int i = 0; i < 4; ++i){ if(i < 2){11 KB (1,713 words) - 22:47, 13 July 2023
- real eta=pi/2; for(int i = 0; i < 4; ++i)4 KB (641 words) - 21:24, 21 April 2014
- for(int i = 0; i < nrows; ++i) for(int j = 0; j <= i; ++j)5 KB (725 words) - 16:07, 23 April 2014
- 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
- ...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
- <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
- <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
- ...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
- ...<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
- 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\le3 KB (596 words) - 16:19, 28 July 2015
- 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
- ...<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
- ...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>| \alph14 KB (2,199 words) - 13:43, 28 August 2020
- {{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
- 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</ma7 KB (1,167 words) - 21:33, 12 August 2020
- ...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
- ...\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
- ...whose angle at the longer base <math>\overline{AD}</math> is <math>\dfrac{\pi}{3}</math>. The [[diagonal]]s have length <math>10\sqrt {21}</math>, and po pair A=(0,0), D=(4,0), B= A+2 expi(1/3*pi), C= D+2expi(2/3*pi), E=(-4/3,0), F=(3,0);4 KB (629 words) - 22:38, 28 November 2023
- ...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 <cmath>x'=r\cos(\pi/4+\theta)+10 = \frac{\sqrt{2}(x - y)}{2} + 10</cmath>5 KB (725 words) - 22:37, 28 January 2024
- ...ath> and height <math>h</math> has [[volume]] <math>V = \frac{1}{3} \cdot \pi r^2 \cdot h</math>. This is a special case of the general formula for the ...[lateral area]] is <math>\pi rs</math>, and the area of the base is <math>\pi r^2</math>).7 KB (1,128 words) - 20:12, 27 September 2022
- ...\right\rbrace</math>. Then the area of <math>S</math> has the form <math>a\pi + \sqrt{b}</math>, where <math>a</math> and <math>b</math> are positive int ...>S</math> (where [[cis]] denotes <math>\text{cis}\, \theta = \cos \theta + i \sin \theta</math>). Since <math>R</math> is symmetric every <math>60^{\cir6 KB (894 words) - 18:56, 25 December 2022
- In triangle <math>ABC</math>, let <math>I</math> be the incenter and <math>I_a</math> the excenter opposite <math>A</ pair I=incenter(A,B,C);3 KB (437 words) - 15:47, 27 April 2008
- int n = 17; real r = 1; real rad = pi/2; return (r*expi(rad-2*pi*k/n));8 KB (1,318 words) - 12:37, 20 April 2022
- ...or <math>A_3</math> the rotation sends <math>P</math> to <math>e^{\frac{4i\pi}{3}}(P-a)+a</math>. Thus the result of all three rotations sends <math>P</ <math>e^{\frac{4i\pi}{3}}(e^{\frac{4i\pi}{3}}(e^{\frac{4i\pi}{3}}P-1)+1-a)+a</math>2 KB (345 words) - 00:04, 30 January 2021
- for(int i=0;i<=5;i=i+1) draw(circle(7/2+d*i,3/2));71 KB (11,749 words) - 01:31, 2 November 2023
- for ( int i = 1; i <= 7; ++i ) draw((i,0)--(i,6));13 KB (1,821 words) - 22:18, 5 December 2023
- ...by the complex number <math>x</math>. Note that <math>\angle PAB = \frac{\pi}{4}</math> and that <math>PA = \frac{\sqrt{2}}{2}AB</math>, and similarly f p &=& \frac{\sqrt{2}}{2}(b-a)e^{i \frac{\pi}{4}}+a \\2 KB (410 words) - 14:01, 4 March 2023
- (i) The weakest player chooses the first two contestants. pair P=(7,0), Q=(14,0), R=P+7expi(pi/3), M=(10,1.2);3 KB (427 words) - 18:55, 3 July 2013
- .... Let <math>ABC</math> have circumcenter <math>O</math> and incenter <math>I</math>. Extend <math>AI</math> to meet the circumcircle again at <math>L</m ...cdot IL = PI\cdot QI</math> by Power of a Point. Therefore, <math>2r\rho = PI \cdot QI = (PO + OI)(QO - OI) = (r + d)(r - d)</math>, and we have <math>2r2 KB (308 words) - 06:29, 16 December 2023
- ...h>y\textrm{-coordinate} = \sum_{k = 1}^{\frac{n}{2}}a_k \cdot \sin \frac{2\pi(k-1)}{n}.\textbf{ (1)}</cmath> ...y\textrm{-coordinate} = \sum_{k = \frac{n}{2}+1}^{n}a_k \cdot \sin \frac{2\pi(k-1)}{n}</cmath>4 KB (836 words) - 17:58, 7 December 2022
- Prove that <math>\cos{\frac{\pi}{7}}-\cos{\frac{2\pi}{7}}+\cos{\frac{3\pi}{7}}=\frac{1}{2}</math>. ...{8\pi}{7}} + \cos{\frac{10\pi}{7}} + \cos{\frac{12\pi}{7}} + \cos{\frac{14\pi}{7}} = 0</cmath>5 KB (746 words) - 19:34, 14 October 2023
- == Day I == Prove that <math>\cos{\frac{\pi}{7}}-\cos{\frac{2\pi}{7}}+\cos{\frac{3\pi}{7}}=\frac{1}{2}</math>.2 KB (342 words) - 21:16, 20 August 2020
- int i; for(i=0; i<=7; ++i){draw((i,0)--(i,4),black+0.5);}3 KB (510 words) - 19:01, 3 July 2013
- For what value of <math>n</math> is <math>i + 2i^2 + 3i^3 + \cdots + ni^n = 48 + 49i</math>? Note: here <math>i = \sqrt { - 1}</math>.13 KB (2,105 words) - 13:13, 12 August 2020
- <center><cmath>p(2009 + 9002\pi i) = p(2009) = p(9002) = 0</cmath></center> From the three zeroes, we have <math>p(x) = (x - (2009 + 9002\pi i))(x - 2009)(x - 9002)</math>.2 KB (322 words) - 10:25, 29 July 2020
- ...h>, <math>B = (4,0)</math>, <math>C = (2 \pi + 1, 0)</math>, <math>D = (2 \pi + 1,4)</math>, and <math>E=(0,4)</math>. What is the probability that <math real pi=3.14159265359;5 KB (792 words) - 15:23, 30 November 2021
- ...and when <math>s=1</math>. When <math>s=1</math> and <math>a+bi=\cos\theta+i\sin\theta</math>, ...in(2002\theta) </math> <math> =a-bi= \cos\theta-i\sin\theta= \cos(-\theta)+i\sin(-\theta)</math>4 KB (743 words) - 19:54, 14 March 2024
- For how many values of <math>x</math> in <math>[0,\pi]</math> is <math>\sin^{ - 1}(\sin 6x) = \cos^{ - 1}(\cos x)</math>? int i,j;13 KB (2,030 words) - 03:04, 5 September 2021
- ...frac34i</math> into polar form as <math>\frac{3\sqrt{2}}{4}e^{\frac{\pi}{4}i}</math>. Restated using geometric probabilities, we are trying to find the We multiply <math>z</math> and <math>(\frac{3}{4}+\frac{3}{4}i)</math> to get <cmath>(\frac{3}{4}x-\frac{3}{4}y)+(\frac{3}{4}xi+\frac{3}{42 KB (422 words) - 13:25, 20 January 2020
- ...> be the circle having <math>\overline{CD}</math> as a diameter. Let <math>I</math> be a point outside <math>\triangle ABC</math> such that <math>\overl First, note that <math>AB=37</math>; let the tangents from <math>I</math> to <math>\omega</math> have length <math>x</math>. Then the perimete12 KB (1,970 words) - 22:53, 22 January 2024
- ...in the given square. If <math>r</math> is the ratio of the area of circle I to that of circle II, then <math>r</math> equals: ...the radius of the smaller circle is <math>x</math>, so it's area is <math>\pi x^2</math>.1 KB (191 words) - 22:09, 14 January 2018
- <cmath> \frac{1}{2\pi i} \int\limits_C \frac{f(z)}{z- z_0}dz = f(z_0) .</cmath> as <math>h(t) = r e^{it}+ z_0</math>, for <math>t\in [0,2\pi]</math>. Since <math>\int\limits_{C_r}4 KB (689 words) - 17:19, 18 January 2024
- <cmath> \lvert f'(z_0) \rvert = \biggl\lvert \frac{1}{2\pi i}2 KB (412 words) - 20:30, 16 January 2024
- ...e numbers lie in the interval between <math>\frac{5}{3}</math> and <math>2\pi?</math> label("I",(0.4,3),E);15 KB (2,165 words) - 03:32, 13 April 2024
- We now discovered that <math>b_4=b_2</math>. And as each <math>b_{i+1}</math> is uniquely determined by <math>b_i</math>, the sequence becomes ...t <math>b_k = \sin (kt)</math>, where <math>kt</math> exceeds <math>\frac{\pi}2</math>. Then we'll have <math>b_{k+1} = \sin(kt - t) = \sin ((k-1)t) = b_4 KB (680 words) - 13:49, 23 December 2023
- ...ath> instead of <math>B</math>. <math>\angle AOC_1</math> = <math>\frac {\pi}{7}</math>. Using the [[Law of Cosines]], <math>\overline {AC_1}^2</math> = <math>8 - 8 \cos \frac {\pi}{7}</math>,8 KB (1,266 words) - 20:27, 10 December 2023
- ..._0,\cdots a_n</math> be real numbers in the interval <math>\left(0,\frac {\pi}{2}\right)</math> such that ...tan{\left(a_1 - \frac {\pi}{4}\right)} + \cdots + \tan{\left(a_n - \frac {\pi}{4}\right)}\ge n - 1</cmath>2 KB (322 words) - 13:31, 23 August 2023
- ...primitive <math>n^{th}</math> root of unity <math>\zeta_n = e^{\frac{2\pi i}{n}}</math>. So now the <math>n^{th}</math> roots of unity are <math>1,\zet ...h extension <math>K_i\subseteq K_{i+1}</math> is quadratic (i.e. <math>[K_{i+1}:K_i]=2</math>). We claim that this happens iff <math>\phi(n)</math> (whe5 KB (926 words) - 18:47, 4 March 2022
- ...}\ \frac{0.4}{\pi} \qquad \textbf{(C)}\ 0.4 \qquad \textbf{(D)}\ \frac{4}{\pi} \qquad \textbf{(E)}\ 4</math> Brian: "Mike and I are different species."12 KB (1,817 words) - 15:00, 12 August 2020
- ...\left(\sin(\pi x) \cdot \sin(2 \pi x) \cdot \sin (3 \pi x) \cdots \sin(8 \pi x)\right)</math>. The intersection of the domain of <math>f(x)</math> with ...align*}\sin(\pi x) \cdot \sin(2 \pi x) \cdot \sin (3 \pi x) \cdots \sin(8 \pi x) &= 0\\9 KB (1,434 words) - 17:54, 17 August 2022
- The area of a circle whose circumference is <math>24\pi</math> is <math>k\pi</math>. What is the value of <math>k</math>? A <i>palindrome</i>, such as <math>83438</math>, is a number that remains the same when its di13 KB (1,902 words) - 11:20, 5 March 2023
- Fibonacci: sum_{i=1}^{2n-1} F_iF_{i+1} = F_{2n}^2 pen colors(int i){ return rgb(i/n,0.4+i/(2n),1-i/n); } /* shading */55 KB (7,986 words) - 17:04, 20 December 2018
- {{AIME Problems|year=2010|n=I}} [[2010 AIME I Problems/Problem 1|Solution]]8 KB (1,243 words) - 21:58, 10 August 2020
- ...x = 15</math>, the [[volume]] of the resulting [[solid]] is <math>\frac {m\pi}{n\sqrt {p}}</math>, where <math>m</math>, <math>n</math>, and <math>p</mat ...6.5,0), C= (9.75,8.25,0), F=(Fxy.x,Fxy.y,0), G=2*F-A, H=(F.x,F.y,abs(F-A)),I=(F.x,F.y,-abs(F-A));4 KB (636 words) - 16:46, 25 November 2023
- label("$I$",(255.242,5.00321),NE/2); <math>Y = \frac {1}{3}\pi(3)^2 = 3\pi</math>10 KB (1,418 words) - 23:05, 20 October 2021
- Notice that removing as many lemmings as possible at each step (i.e., using the greedy algorithm) is optimal. This can be easily proved throu ...{10})</math>, with <math>0 \le b_i \le 2, 3|a_i-b_i</math> for <math>1 \le i \le 10</math>. Given that <math>f</math> can take on <math>K</math> distinc36 KB (6,214 words) - 20:22, 13 July 2023
- ...If we do this for our <math>f</math> equation above, we get <cmath>f(x) = \pi\left(\frac{\sqrt{x}}{x-1}\right).</cmath> We can use <code>\left</code> and ...example is <code>$\sum_{i=0}^n a_i$</code>, giving <math>\textstyle \sum_{i=0}^n a_i.</math> Note the use of superscripts and subscripts to obtain the8 KB (1,356 words) - 22:35, 26 June 2020
- the inequality <math>a_ia_j \le i+j</math> for all distinct indices <math>i, j</math>. <cmath> \prod_{i=1}^{1005}(4i-1) = 3\times 7 \times \ldots \times 4019. </cmath>11 KB (1,889 words) - 13:45, 4 July 2013
- <math>CE</math> meet at <math>I</math>. Determine whether or not it is possible for pair I = extension(B, D, C, E); ldot(I, "$I$", A-I);7 KB (1,230 words) - 19:47, 31 January 2024
- <cmath>f(\theta)=\frac{\theta}{2\pi}\cdot2\pi=\theta</cmath> ...length where the third jump could land to satisfy the problem. To do this, I can apply average function value with our old buddy calculus,6 KB (1,105 words) - 13:39, 9 January 2024
- Let <math>I</math> be the incenter of acute triangle <math>ABC</math> with <math>AB \ne <i><b>Step 1</b></i>5 KB (792 words) - 01:52, 19 November 2023
- I. Bill is the oldest. ...extbf{(C)}\ 28-4\pi \qquad \textbf{(D)}\ 28-2\pi \qquad \textbf{(E)}\ 32-2\pi</math>13 KB (1,860 words) - 19:58, 8 May 2023
- \textbf{(A)}\ 3 - \frac{\pi}{2} \qquad \textbf{(B)}\ \frac{\pi}{2} \qquad13 KB (1,994 words) - 13:52, 3 July 2021
- So <math>z^4 = 1 \implies z = e^{i\frac{n\pi}{2}}</math> <math>z^2 = \pm i \sqrt{4\sqrt{3} + 7} = e^{i\frac{(2n+1)\pi}{2}} \left(\sqrt{3} + 2\right)</math>2 KB (344 words) - 18:12, 22 August 2021
- {{AIME Problems|year=2011|n=I}} [[2011 AIME I Problems/Problem 1|Solution]]10 KB (1,634 words) - 22:21, 28 December 2023
- Suppose <math>x</math> is in the interval <math>[0, \pi/2]</math> and <math>\log_{24\sin x} (24\cos x)=\frac{3}{2}</math>. Find <ma Like Solution 1, we can rewrite the given expression as2 KB (330 words) - 20:47, 10 December 2023
- ...A_6}</math>, and <math>\overline{A_7 A_8}</math>, respectively. For <math>i = 1, 3, 5, 7</math>, ray <math>R_i</math> is constructed from <math>M_i</ma ...s we have that <math>\cos 2 \angle A_3 M_3 B_1=\cos \left(2\theta + \frac{\pi}{2} \right)=-\sin2\theta</math>.8 KB (1,344 words) - 18:39, 9 February 2023
- ...h <math>j</math>, let <math>w_j</math> be one of <math>z_j</math> or <math>i z_j</math>. Then the maximum possible value of the real part of <math>\sum_ ...the real parts of the blue dots is easily seen to be <math>8+16\cos\frac{\pi}{6}=8+8\sqrt{3}</math> and the negative of the sum of the imaginary parts o5 KB (805 words) - 18:46, 27 January 2024
- ...th>\{a_i,b_i\}</math> (<math>1\leq i\leq 999</math>) so that for all <math>i</math>, <math>|a_i-b_i|</math> equals <math>1</math> or <math>6</math>. Pro Let <math>a_0,a_1,\cdots ,a_n</math> be numbers from the interval <math>(0,\pi/2)</math> such that3 KB (486 words) - 06:11, 24 November 2020
- ...f{(C)}\ \pi \qquad\textbf{(D)}\ \frac{4\pi}{3} \qquad\textbf{(E)}\ \frac{5\pi}{3}</math> ...s hold for all numbers <math>x, y,</math> and <math>z</math>? <cmath>\text{I. x @ (y + z) = (x @ y) + (x @ z)}</cmath> <cmath>\text{II. x + (y @ z) = (x13 KB (2,090 words) - 18:05, 7 January 2021
- Solve for <math>x</math> for all answers in the domain <math>[0, 2\pi]</math>. <math>i = sin(x)</math>8 KB (1,351 words) - 20:30, 10 July 2016
- ...d, or 4th root of unity. These are among the set <math>\{\pm1,\pm i,(-1\pm i\sqrt{3})/2\}</math>. Since complex roots of polynomials come in conjugate p Now suppose <math>z=i</math>. Then <math>4=(a-c)i+(b-d)</math> whereupon <math>a=c</math> and <math>b-d=4</math>. But then <m11 KB (1,979 words) - 17:25, 6 September 2021
- <math>\textbf{(A)}\ \frac{\pi m^2}{2}\qquad \textbf{(B)}\ \frac{3\pi m^2}{8}\qquad20 KB (3,108 words) - 14:14, 20 February 2020
- int i; for(int i=0; i<15; i=i+1) {18 KB (2,551 words) - 18:46, 27 February 2024
- ...-\pi\qquad\text{(C)}\ \frac{\pi} 2\qquad\text{(D)}\ \pi\qquad\text{(E)}\ 2\pi </math> int i;15 KB (2,343 words) - 13:39, 19 February 2020
- ...qquad \mathrm{(D) \ } \frac{5}{6}\pi+2 \qquad \mathrm{(E) \ }\frac{5}{3}\pi+2 </math> int i,j;17 KB (2,488 words) - 03:26, 20 March 2024
- ...}}{2} \qquad\text{and}\qquad y=\frac{-1-i\sqrt{3}}{2},</cmath> where <math>i^2 = -1</math>, then which of the following is not correct? ...c{2\pi k}{3}\right) = \cos\left(\frac{2\pi k}{3}\right)-i\sin\left(\frac{2\pi k}{3}\right),\end{align*}</cmath> using the fact that <math>\cos</math> is3 KB (578 words) - 00:47, 20 March 2024
- ...axis. As <math>\theta</math> sweeps from <math>0</math> to <math>\dfrac{\pi}{2}</math>, what is the probability that <math>b\ge 1</math>? (Proposed by I roll 4 dice with 6, 10, 12, and 20 sides each numbered 1-6, 1-10, 1-12, 1-215 KB (2,444 words) - 21:46, 1 January 2012
- ...m statement tells us that < HPI = < KPJ. By transitivity, we must have < H'PI = < KPJ, implying that H', P, and K are collinear by the converse of Vertic1 KB (204 words) - 18:09, 19 September 2012
- for(int i = 1;i<y.length;++i) { pair p = (a,y[i]);15 KB (2,247 words) - 13:44, 19 February 2020
- ...up of <math>9</math> congruent circular arcs each of length <math>\frac{2\pi}{3}</math>, where each of the centers of the corresponding circles is among ...extbf{(C)}\ 3\pi+4 \qquad\textbf{(D)}\ 2\pi+3\sqrt3+2 \qquad\textbf{(E)}\ \pi+6\sqrt3 </math>14 KB (2,197 words) - 13:34, 12 August 2020
- ...xtbf{(C)}\ 12\pi+\sqrt{3}\qquad\textbf{(D)}\ 10\pi+9\qquad\textbf{(E)}\ 13\pi</math> ...)}\ 2\qquad\textbf{(C)}\ \sqrt{5}\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ \pi</math>18 KB (2,350 words) - 18:48, 9 July 2023
- ...= z,</math> and the imaginary part of <math>z</math> is <math>\sin{\frac{m\pi}{n}}</math>, for relatively prime positive integers <math>m</math> and <mat ...>z</math> will be of the form <math>\sin{\frac{2k\pi}{142}} = \sin{\frac{k\pi}{71}},</math> where <math>k \in \{1, 2, \ldots, 70\}.</math> Note that <mat1 KB (233 words) - 17:15, 30 July 2022
- {{AIME Problems|year=2015|n=I}} [[2015 AIME I Problems/Problem 1|Solution]]10 KB (1,615 words) - 21:48, 13 January 2024
- {{AIME Problems|year=2012|n=I}} [[2012 AIME I Problems/Problem 1|Solution]]10 KB (1,617 words) - 14:49, 2 June 2023
- ...th>\sqrt{111}</math>, and <math>\sqrt{11}</math>. That is, let <math>z=e^{i\theta}</math>, and then: ...g <math>D_1</math> and <math>D_2</math> around <math>A</math> by <math>\pm\pi/3</math>:13 KB (2,052 words) - 18:02, 5 February 2024
- ...t then I had <math>3</math> gold coins left over. How many gold coins did I have? ...{3}\qquad\textbf{(D) }4\sqrt{3}-\frac{2\pi}{3}\qquad\textbf{(E) }4+\frac{4\pi}{3}</math>12 KB (1,771 words) - 21:13, 20 January 2024
- [[Mock AIME I 2012 Problems/Problem 1 |Solution]] [[Mock AIME I 2012 Problems/Problem 2| Solution]]7 KB (1,309 words) - 11:13, 8 April 2012
- ...that a finite set <math>\mathcal{S}</math> in the plane is <i> balanced </i> <math>\mathcal{S}</math> is <i>centre-free</i> if for any three points <math>A</math>, <math>B</math>, <math>C</math> in4 KB (773 words) - 08:14, 19 July 2016
- ...n}</math> are all <math>n</math>th roots of unity. If <math>\omega=e^{2\pi i/n}</math>, then the sum of <math>32</math>nd powers of these roots will be8 KB (1,348 words) - 09:44, 25 June 2022
- ...e <math> M </math> minus the volume fo cone <math> N </math> is <math> 140\pi </math>, find the length of <math> \overline{BC} </math>. ...h>, for terms <math> n\ge3 </math>, <math> S_n=\sum_{i=1}^{n-1}i\cdot S_{n-i} </math>. For example, if the first two elements are <math> 2 </math> and <6 KB (910 words) - 17:32, 27 May 2012
- ...assumes she means the square root of himself, or the square root of <math> i </math>. What two answers should he give? ...[[Euler's identity|Euler's Formula]] that <math> \cos\theta+i\sin\theta=e^{i\theta} </math>.1 KB (170 words) - 20:12, 27 May 2012
- ...h exactly <math>k</math> of the quadrilaterals <math>A_{i}A_{i+1}A_{i+2}A_{i+3}</math> have an inscribed circle. (Here <math>A_{n+j} = A_{j}</math>.) ...> for all <math>i</math>, then quadrilateral <math>A_{i+1}A_{i+2}A_{i+3}A_{i+4}</math> is not tangential.5 KB (871 words) - 18:59, 10 May 2023
- pair A,B,C,D,E,F,G,H,I,J,K,L,M,N,O,P,Q,R; I=(7,8);13 KB (1,835 words) - 08:51, 8 March 2024
- ...qrt2}{2} + \frac{\pi}{4} \qquad\textbf{(E)}\ 1 + \frac{\sqrt2}{4} + \frac{\pi}{8} </math> First, we need to see what this looks like. Below is a diagram.4 KB (701 words) - 17:55, 23 July 2021
- {{AIME Problems|year=2013|n=I}} [[2013 AIME I Problems/Problem 1|Solution]]9 KB (1,580 words) - 13:07, 24 February 2024
- For <math>\pi \le \theta < 2\pi</math>, let ...>\sin \theta = -17/19, 1/3.</math> Since we're given <math>\pi\leq\theta<2\pi,</math> <math>\sin\theta</math> is nonpositive. We therefore use the negati10 KB (1,641 words) - 20:03, 3 January 2024
- <math>|3-\pi|=</math> ...4 \qquad \textbf{(C) }3-\pi \qquad \textbf{(D) }3+\pi \qquad \textbf{(E) }\pi-3 </math>16 KB (2,451 words) - 04:27, 6 September 2021
- ...ian plane, we use complex numbers. Thus A is 1 and B is <math>2 + 2\sqrt{3}i</math>. ...here we require a clockwise rotation, so we multiply by <math>e^{-\frac{i\pi}{3}}</math> to obtain C. Upon averaging the coordinates of A, B, and C, we8 KB (1,268 words) - 14:10, 31 January 2024
- Now, let <math>w = r_w e^{i \theta_w}</math> and likewise for <math>z</math>. Consider circle <math>O</ ...>w = \frac{-2014 \pm \sqrt{2014^2 - 4(2014^2)}}{2} = -1007 \pm 1007\sqrt{3}i</cmath>6 KB (1,045 words) - 13:08, 21 January 2024
- <i><b>Proof</b></i> Given the triangle <math>\triangle ABC, \omega</math> is the incircle, <math>I</math> is the incenter, <math>B' = \omega \cap AC.</math>15 KB (2,549 words) - 08:36, 2 September 2023
- ...t, when he catches a fly, he places it on <math>a_i</math> for which <math>i</math> is the least such number satisfying the following rules: ...4</math>, then he eats all the flies on <math>a_1</math> through <math>a_{i-1}</math> and then puts his newly caught fly on <math>a_i</math>.10 KB (1,710 words) - 23:23, 10 January 2020
- I shall prove a more general statement about the unit distance graph(<math>V= ...claim that for some choice of <math>0\leq\theta<2\pi</math>, <math>v_i+e^{i\theta}w_j</math> will do the job(a suitable rotation).4 KB (749 words) - 14:09, 29 January 2021
- <math>\textbf{(A)}\ \pi + 2 \qquad \textbf{(B)}\ \frac{2 \pi + 1}{2} \qquad17 KB (2,459 words) - 22:40, 10 April 2023
- triple A,B,C,D,E,F,G,H,I,J,K,L,M,N,O,P; I = (1,1,1);15 KB (2,162 words) - 20:05, 8 May 2023
- <math>\textbf{I:}\ 2x^x \qquad\textbf{II:}\ x^{2x} \qquad\textbf{III:}\ (2x)^x \qquad\textb If <math>i^2=-1</math>, then <math>(i-i^{-1})^{-1}=</math>14 KB (2,124 words) - 13:39, 19 February 2020
- ...through four distinct points of the form <math>(i,j,k)</math>, where <math>i</math>, <math>j</math>, and <math>k</math> are positive integers not exceed If <math> \theta</math> is a constant such that <math> 0 < \theta < \pi</math> and <math> x + \dfrac{1}{x} = 2\cos{\theta}</math>, then for each po17 KB (2,633 words) - 15:44, 16 September 2023
- ...increase in circumference of a circle resulting from an increase in <math>\pi</math> units in the diameter. Then <math>P</math> equals: ...pi\quad\text{(C) } \frac{\pi^2}{2}\quad\text{(D) } \pi^2\quad\text{(E) } 2\pi</math>16 KB (2,571 words) - 14:13, 20 February 2020
- <math>\textbf{(I)}\ x+y < a+b\qquad</math> for(int i=0;i<=5;i=i+1)15 KB (2,190 words) - 15:21, 22 December 2020
- for(int i=0;i<=5;i=i+1) path arc=2*dir(60*i)--arc(2*dir(60*i),1,120+60*i,240+60*i)--cycle;3 KB (482 words) - 11:50, 7 September 2021
- ...qquad\textbf{(C)}\ 4\pi-4 \qquad\textbf{(D)}\ 2\pi+4 \qquad\textbf{(E)}\ 4\pi-2 </math> for(int i=0; i<4; i=i+1)2 KB (287 words) - 03:07, 28 May 2021
- for(int i = 0; i < 4; ++i){ if(i < 2){702 bytes (115 words) - 17:32, 20 April 2014
- ...re positive integers such that <math>\gcd(a+i, b+j)>1</math> for all <math>i, j\in\{0, 1, \ldots n\}</math>, then<cmath>\min\{a, b\}>c^n\cdot n^{\frac{n ...math> and in each cell place a prime <math>p</math> dividing <math>\gcd (a+i, b+j)</math>.2 KB (361 words) - 11:55, 25 June 2020
- <math>\textbf{(A) }\frac{25\sqrt{2}}{\pi}\qquad \textbf{(B) }\frac{50\sqrt{2}}{\pi}\qquad21 KB (3,242 words) - 21:27, 30 December 2020
- Let <math>f(x)=(x^2+3x+2)^{\cos(\pi x)}</math>. Find the sum of all positive integers <math>n</math> for which {{AIME box|year=2014|n=II|before=[[2014 AIME I Problems]]|after=[[2015 AIME I Problems]]}}8 KB (1,410 words) - 00:04, 29 December 2021
- ...the smaller disk, so <math>6\alpha = 2\pi</math>, or <math>\alpha = \frac{\pi}{3}</math>. ...A}</math>, and call this point <math>X</math>. Since <math>\alpha = \frac{\pi}{3}</math> and <math>\angle DXE</math> is right, <cmath>DE = 6</cmath> <cma5 KB (854 words) - 20:02, 4 September 2021
- ...we can find from angle chasing that <math>\angle ABF = \angle EDF = \frac{\pi}4</math>. Therefore, <math>\overline{BF}</math> is the angle bisector of <m {{AIME box|year=2014|n=I|num-b=14|after=Last Question}}10 KB (1,643 words) - 22:30, 28 January 2024
- Of the following five statements, I to V, about the binary operation of averaging (arithmetic mean), <math>\text{I. Averaging is associative }</math>18 KB (2,788 words) - 13:55, 20 February 2020
- If <math>3(4x+5\pi)=P</math> then <math>6(8x+10\pi)=</math> <math>I.\quad y=x-2 \qquad II.\quad y=\frac{x^2-4}{x+2}\qquad III.\quad (x+2)y=x^2-16 KB (2,548 words) - 13:40, 19 February 2020
- This equation is <math>r^6e^{6\theta i}=2^6e^{(\pi\pm 2k\pi) i}</math>. Solving in the usual way, <math>r=2</math> and <math>\theta\in\{\p798 bytes (133 words) - 13:17, 4 February 2016
- draw((0,1)--(cos(pi/14),-sin(pi/14))--(-cos(pi/14),-sin(pi/14))--cycle,dot); draw((-cos(pi/14),-sin(pi/14))--(0,-1/cos(3pi/7))--(cos(pi/14),-sin(pi/14)),dot);14 KB (2,099 words) - 01:15, 10 September 2021
- <math>\textbf{(A) }\frac{4}{\pi}\qquad \textbf{(B) }\frac{\pi}{\sqrt{2}}\qquad15 KB (2,366 words) - 07:52, 26 December 2023
- .../cmath> These roots are <math>w = e^{i \pi /3}</math> and <math>w = e^{2i \pi /3}</math>. ...}{2} + \frac{\sqrt{3}}{2}i + 1 = \frac{1}{2} + \frac{\sqrt{3}}{2}i = e^{i \pi /3}.</cmath>3 KB (472 words) - 20:04, 30 October 2021
- have a solution <math>(x, y)</math> inside Quadrant I if and only if <math>\textbf{(A)}\ 36\pi \qquad16 KB (2,291 words) - 13:45, 19 February 2020
- ...(2*cos(pi/5)-sin(pi/10),cos(pi/10)), C=(1,0), D=(0,0), E1=(-sin(pi/10),cos(pi/10)); ..., G=intersectionpoints(A--C,E1--B)[0], H=intersectionpoints(B--D,A--C)[0], I=intersectionpoints(C--E1,D--B)[0], J=intersectionpoints(E1--C,D--A)[0];7 KB (1,146 words) - 14:09, 13 April 2024
- \textbf{(B)} \ \pi \qquad \textbf{(D)} \ 4 \pi \qquad15 KB (2,309 words) - 23:43, 2 December 2021
- <math>\textbf{(A)} \ \pi \qquad \textbf{(B)} \ \frac{2}{\pi} \qquad17 KB (2,500 words) - 19:05, 11 September 2023
- <math>\textbf{(A) }\frac{\pi}{4}\qquad \textbf{(B) }\frac{3\pi}{4}\qquad17 KB (2,664 words) - 01:34, 19 March 2022
- <math>\textbf{(A) }\frac{1}{\pi^2}\qquad \textbf{(B) }\frac{1}{\pi}\qquad15 KB (2,432 words) - 01:06, 22 February 2024
- \textbf{(D) }\frac{\pi}{16}\qquad <math>\begin{array}{l}\textbf{I. }\text{"If the wild pig on planet beta has a long nose, then the pink elep17 KB (2,835 words) - 14:36, 8 September 2021
- The lengths in inches of the three sides of each of four triangles <math>I, II, III</math>, and <math>IV</math> are as follows: \hbox{I}& 3,\ 4,\ \hbox{and}\ 5\qquad &17 KB (2,732 words) - 13:54, 20 February 2020
- ...th>x</math> and <math>y</math> are acute angles such that <math>x+y=\frac{\pi}{4}</math> and <math>\tan{y}=\frac{1}{6}</math>, find the value of <math>\t <math>\text{(I) }\frac{1}{6}\qquad2 KB (266 words) - 21:30, 4 February 2023
- ...math> is a nonnegative real number, and <math>x_1+x_2+x_3+\cdots+x_{2007}=\pi</math>. Find the value of <math>a+b</math>.427 bytes (72 words) - 12:24, 6 April 2024
- \textbf{(C) }\ 1: \pi \qquad \textbf{(D) }\ 3: \pi \qquad19 KB (2,873 words) - 18:57, 16 August 2023
- The area of a circle inscribed in an equilateral triangle is <math>48\pi</math>. The perimeter of this triangle is: If <math>S = i^n + i^{-n}</math>, where <math>i = \sqrt{-1}</math> and <math>n</math> is an integer, then the total number26 KB (3,950 words) - 21:09, 31 August 2020
- small circle is <math>9\pi</math>. Find the area of the shaded <math>\textit{lune}</math>, the region ...d_1d_2 \ldots d_6}</math> have period exactly six? Do not include patterns like <math>0.323</math> and <math>0.17</math> that have shorter periods.11 KB (1,648 words) - 09:55, 20 December 2021
- where <math>i^2 = -1</math> as usual. your answer as an exact multiple of <math>\pi</math> (and not as a6 KB (890 words) - 22:14, 7 November 2014
- string[] let = {"$A$","$B$","$C$","$D$","$E$","$F$","$G$","$H$","$I$","$J$","$K$","$L$"}; for(int i=0;i<=11;i=i+1)13 KB (1,957 words) - 12:08, 13 January 2024
- for(int i=0; i<3; i+=1) draw((0,i+0.05)--(0,i+0.95));12 KB (1,897 words) - 22:45, 18 March 2024
- ...nce between the points is at least <math>\frac12</math> is <math>\frac{a-b\pi}{c}</math>, where <math>a,b,</math> and <math>c</math> are positive integer <cmath>(\text{cos}(a\pi)+i\text{sin}(b\pi))^4</cmath>13 KB (2,117 words) - 12:33, 24 August 2023
- ...aw(0.9*dir(90-30*i)--dir(90-30*i));label("$"+(string) i+"$",0.78*dir(90-30*i));} ...s <math>20\pi</math>, so that is <math>20\pi</math> traveled on a <math>60\pi</math> arrow path. This is a ratio of 1/3, so the angle it carves is 120 de9 KB (1,380 words) - 09:00, 1 December 2023
- <cmath>(\text{cos}(a\pi)+i\text{sin}(b\pi))^4</cmath> Let <math>\cos(a\pi) = x</math> and <math>\sin(b\pi) = y</math>. Consider the binomial expansion of the expression:5 KB (793 words) - 20:32, 26 May 2022
- ...nce between the points is at least <math>\frac12</math> is <math>\frac{a-b\pi}{c}</math>, where <math>a,b,</math> and <math>c</math> are positive integer ...t{0.25 - x^2} dx = 4\left(\frac{1}{4} - \frac{\pi}{16}\right) = 1 - \frac{\pi}{4}.</cmath>12 KB (1,981 words) - 18:33, 3 September 2023
- ...{4\pi}{3} \qquad\textbf{(D) } \dfrac{7\pi}{5} \qquad\textbf{(E) } \dfrac{3\pi}{2} </math> ...C--A,B--D), I=IP(D--B,E--C), J=IP(C--E,D--A); D(MP("F",F,dir(126))--MP("I",I,dir(270))--MP("G",G,dir(54))--MP("J",J,dir(198))--MP("H",H,dir(342))--cycle14 KB (2,247 words) - 11:38, 26 March 2024
- ...c{\sqrt{2}}{2} \qquad\textbf{(D)}\; 4(\pi-\sqrt{3}) \qquad\textbf{(E)}\; 2\pi-\dfrac{\sqrt{3}}{2}</math> <cmath>f(i,j) = \begin{cases}\text{mod}_5 (j+1) & \text{ if } i = 0 \text{ and } 0 \le j \le 4 \text{,}\\13 KB (2,064 words) - 13:39, 1 October 2022
- Let <math>x=e^{i\pi/6}</math>, a <math>30^\circ</math> counterclockwise rotation centered at th <cmath>1-x=1-\frac{\sqrt{3}}{2}-\frac{1}{2}i</cmath>13 KB (2,117 words) - 11:30, 8 April 2023
- ...ce on each half. The area of one of these unpainted faces is <math>a\cdot\pi + b\sqrt{c}</math>, where <math>a</math>, <math>b</math>, and <math>c</math [[File:2015 AIME I 15.png|400px|right]]9 KB (1,407 words) - 19:37, 17 February 2024
- Let <math>x = \cos 1^\circ + i \sin 1^\circ</math>. Then from the identity <cmath>\sin 1 = \frac{x - \frac{1}{x}}{2i} = \frac{x^2 - 1}{2 i x},</cmath>9 KB (1,351 words) - 17:26, 16 January 2024
- .../math> are equally spaced on a minor arc of a circle. Points <math>E,F,G,H,I</math> and <math>A</math> are equally spaced on a minor arc of a second cir pair A,B,C,D,E,F,G,H,I,O;5 KB (782 words) - 16:04, 21 July 2023
- ...dth such value will be when <math>k = 100</math> and <math>n = \lfloor 200\pi \rfloor = \lfloor 628.318 \rfloor = \boxed{628}</math>. Thus, we need <math>-1<n<0.</math> So we have <math>-1+2\pi k <n <0+2\pi k.</math>6 KB (1,034 words) - 21:29, 14 January 2024
- for(int i=0;i <= 4;i=i+1) draw(shift((4*i,0)) * P);14 KB (2,180 words) - 22:25, 25 April 2024
- ...{(C)}\ \pi+2\sqrt{2}\qquad\textbf{(D)}\ 2\pi+\sqrt{2}\qquad\textbf{(E)}\ 2\pi+2\sqrt{2}</math> ...;++i){draw(rotate(i*90,(0,0))*arc((1/2,1/2),sqrt(1/2),-45,135));dot(rotate(i*90,(0,0))*(1/2,1/2));}</asy>5 KB (761 words) - 23:26, 7 September 2022
- ...t are integers lie entirely within the region bounded by the line <math>y=\pi x</math>, the line <math>y=-0.1</math> and the line <math>x=5.1?</math> for(int i=0;i<6;++i)for(int j=0;j<=3*i;++j)dot((i,j));3 KB (497 words) - 08:50, 9 March 2024
- ...C)}\ \pi+2\sqrt{2} \qquad\textbf{(D)}\ 2\pi+\sqrt{2} \qquad\textbf{(E)}\ 2\pi+2\sqrt{2}</math> <cmath>A = \frac{\pi}{4} +\frac{1}{2}</cmath>1 KB (188 words) - 22:57, 9 January 2024
- ...t are integers lie entirely within the region bounded by the line <math>y=\pi x</math>, the line <math>y=-0.1</math> and the line <math>x=5.1?</math> ...C)}\ \pi+2\sqrt{2} \qquad\textbf{(D)}\ 2\pi+\sqrt{2} \qquad\textbf{(E)}\ 2\pi+2\sqrt{2}</math>14 KB (2,037 words) - 19:09, 29 July 2023
- real r = (pi/2 - asin(1/5))/2; B = 40*dir(r*180/pi);9 KB (1,526 words) - 02:31, 29 December 2021
- -1. GMAAS looks like this when he is mad: https://cdn.artofproblemsolving.com/images/7/4/b/74b21 ...eorem. You wouldn't waste that much money to solve one problem. Therefore, I proved that you cannot use the Games theorem to solve problems. But using t69 KB (11,805 words) - 20:49, 18 December 2019
- ...TEST = 2006</math>. What is the largest possible value of the sum <math>I + T + E + S + T + 2006</math>? ...}\,-\frac{\sqrt{26}}{26}\quad\mathrm{(H)}\,\frac{\sqrt{2}}{2}\quad\mathrm{(I)}\,\text{none of the above}</math>31 KB (4,811 words) - 00:02, 4 November 2023
- ...c{\pi}{2} - \beta.</math> (One could probably cite this as well-known, but I have proved it here just in case.) ...\frac{AO}{AQ}.</math> We also have <math>\angle PAH = \angle OAQ = \frac{\pi}{2} - \beta,</math> so <math>\triangle PAH\sim\triangle OAQ</math> by SAS s10 KB (1,733 words) - 19:15, 14 June 2020
- ...1,a_2,\cdots,a_{1990}\}=\{1,2,\cdots,1990\}</math> and <math>\theta=\frac{\pi}{995}</math>. ...th>a_1,a_2,\cdots,a_{1990}</math> such that <math>\sum_{i=1}^{1990}a_i^2e^{i\theta}=0</math>.3 KB (522 words) - 13:54, 30 January 2021
- ...p+1}sin^{p}\left ( \frac{\pi}{3} \right )(i)^{p}\left ( cos\left ( \frac{p\pi}{3} \right ) \right )}{3}</math> ...1}sin^{p}\left ( \frac{\pi}{3} \right )(i)^{p+1}\left ( sin\left ( \frac{p\pi}{3} \right ) \right )}{3}</math>3 KB (443 words) - 21:54, 22 November 2023
- ...ondecreasing order. (E.g., if <math>n=4,</math> then the partitions <math>\pi</math> are <math>1+1+1+1,</math> <math>1+1+2,</math> <math>1+3, 2+2,</math> ...artition <math>1+1+2+2+2+5,</math> then <math>A(\pi)=2</math> and <math>B(\pi) = 3</math>).5 KB (975 words) - 14:32, 30 August 2018
- <div class=li><span class=num>(i)</span> <math>a_1 = \frac{1}{2}</math></div> ...less than or equal to <math>r</math>, <math>e.g.,</math> <math>[6] = 6, [\pi] = 3, [-1.5] = -2.</math> Indicate on the <math>(x,y)</math>-plane the set3 KB (499 words) - 12:17, 11 August 2016
- ...these <math>z</math> such that <math>z^{24}=1</math> are <math>e^{\frac{ni\pi}{12}}</math> for integer <math>0\leq n<24</math>. So <math>z^6=e^{\frac{ni\pi}{2}}</math>3 KB (459 words) - 04:19, 2 February 2021
- ...ath> units of line segment <math>\overline{AB}</math> has volume <math>216\pi</math>. What is the length <math>\textit{AB}</math>? ...t{3}-\frac{2\sqrt{3}\pi}{9}\qquad\textbf{(E)}\ \frac{4}{3}-\frac{4\sqrt{3}\pi}{27}</math>15 KB (2,285 words) - 18:02, 28 October 2023
- ...+i),\frac{1}{\sqrt{8}}(-1+i),\frac{1}{\sqrt{8}}(1-i),\frac{1}{\sqrt{8}}(-1-i) \right\}.</cmath> ...e of odd multiples of <math>\pi/4</math>, i.e. <math>\pi/4,3\pi/4,5\pi/4,7\pi/4</math>. When we draw these 6 complex numbers out on the complex plane, we18 KB (2,878 words) - 01:47, 16 December 2023
- ...th> units of line segment <math>\overline{AB}</math> has volume <math>216 \pi</math>. What is the length <math>AB</math>? ...+i),\frac{1}{\sqrt{8}}(-1+i),\frac{1}{\sqrt{8}}(1-i),\frac{1}{\sqrt{8}}(-1-i) \right\}.</cmath> For each <math>j</math>, <math>1\leq j\leq 12</math>, an15 KB (2,418 words) - 16:58, 7 November 2022