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- ...(x)</math> and <math>\cos(x)</math> are periodic with least period <math>2\pi</math>. What is the least period of the function <math>\cos(\sin(x))</math> ...i}{2}\qquad\textbf{(B)}\ \pi\qquad\textbf{(C)}\ 2\pi \qquad\textbf{(D)}\ 4\pi \qquad\textbf{(E)}</math> It's not periodic.15 KB (2,343 words) - 18:26, 25 December 2020
- ...f{(D)}\ 2\sqrt{2}+\sqrt{6} \qquad \textbf{(E)}\ (1+\sqrt{3}) + (1+\sqrt{3})i</math> ...12} = 64</math>, it is easy to see <math>\pm\sqrt{2}</math> and <math>\pm {i} \sqrt{2}</math> as roots. Graphing these in the complex plane, we have fou3 KB (449 words) - 01:54, 11 February 2019
- Q = (10*cos(pi/3), 10*sin(pi/3)); R = (10*cos(5*pi/6), 10*sin(5*pi/6));9 KB (1,539 words) - 15:47, 17 February 2024
- ...th>z_1=18+83i,~z_2=18+39i,</math> and <math>z_3=78+99i,</math> where <math>i=\sqrt{-1}.</math> Let <math>z</math> be the unique complex number with the ...f thinking of complex numbers as purely a real plus a constant times <math>i</math>, let’s graph them and hope that the geometric visualization adds i13 KB (2,252 words) - 15:46, 6 January 2024
- ...lie on the hypotenuse <math>\frac{x}{5} + \frac{y}{2\sqrt{3}} = 1</math>, i.e. <math>a,b</math> must satisfy ...+bi)\cdot\cos(-\frac{\pi}{3})=(-\frac{a+\sqrt{3}b}{2}+\frac{\sqrt{3}a+b}{2}i)</math>. We know that the slope of <math>AC</math> is <math>-\frac{2\sqrt{322 KB (3,622 words) - 17:11, 6 January 2024
- ...distance between any two points labeled <math>i</math> is at least <math>c^i</math>. ...For <math>c\le \sqrt[4]{2},</math> we can make a "checkerboard" labeling, i.e. label <math>(x, y)</math> with <math>1</math> if <math>x+y</math> is eve8 KB (1,495 words) - 12:19, 17 July 2023
- | 86 || I-Can-Do-Math || 6 || 3059.612 || 509.935 | 76 || math-pi || 79 || 45682.709 || 578.262187 KB (10,824 words) - 18:27, 3 February 2022
- for (int i = 0; i < 3; ++i) { pair A = (j,i);14 KB (2,073 words) - 15:15, 21 October 2021
- ...f{(C)}\ 3 \pi \sqrt7 \qquad\textbf{(D)}\ 6\pi \sqrt3 \qquad\textbf{(E)}\ 6\pi \sqrt7</math> ...textbf {(D) } 3\sqrt{3} - \pi \qquad \textbf {(E) } \frac{9\sqrt{3}}{2} - \pi </math>16 KB (2,417 words) - 01:03, 28 April 2022
- ...th>. With <math>\beta</math> being a real number such that <math>0< \beta<\pi/8</math> and <math>x\neq0</math>, the value of <math>\beta</math> is: (a) <math>\frac{\pi}{9}</math>8 KB (1,278 words) - 09:46, 11 January 2018
- ...k=1}^{15}</math> Img<math>\left(\right.</math>cis<math>\left.^{2k-1}\frac{\pi}{36}\right)</math> (a) <math>\frac{2+\sqrt3}{4\sin\frac{\pi}{36}}</math>7 KB (1,127 words) - 18:23, 11 January 2018
- ...ath>z^2=4+4\sqrt{15}i</math> and <math>z^2=2+2\sqrt 3i,</math> where <math>i=\sqrt{-1},</math> form the vertices of a parallelogram in the complex plane <li><math>z^2=4+4\sqrt{15}i</math><p>10 KB (1,662 words) - 12:45, 13 September 2021
- ...st subset of values of <math>y</math> within the closed interval <math>[0,\pi]</math> for which ...+\sin(y)</cmath>for every <math>x</math> between <math>0</math> and <math>\pi</math>, inclusive?15 KB (2,380 words) - 18:52, 7 April 2022
- \textbf{(A) }25\pi \qquad \textbf{(B) }50\pi \qquad14 KB (2,118 words) - 15:36, 28 October 2021
- ...eta</math> is the argument of <math>z</math> such that <math>0\leq\theta<2\pi.</math> ...{4},\frac{\pi}{2},\frac{3\pi}{4},\pi,\frac{5\pi}{4},\frac{3\pi}{2},\frac{7\pi}{4}.</math></li><p>11 KB (1,708 words) - 12:01, 18 March 2023
- ...<math>P_iP_{i+1}</math> is tangent to <math>\omega_i</math> for each <math>i=1,2,3</math>, where <math>P_4 = P_1</math>. See the figure below. The are Let <math>O_i</math> be the center of circle <math>\omega_i</math> for <math>i=1,2,3</math>, and let <math>K</math> be the intersection of lines <math>O_113 KB (2,080 words) - 19:09, 21 October 2023
- ...er of, and is tangent to, circle <math>II</math>. The area of circle <math>I</math> is <math>4</math> square inches. \textbf{(C) }8\sqrt{\pi}\qquad2 KB (306 words) - 18:57, 17 May 2018
- ...heta = \sin (\pi - \theta)</math> and <math>\sin \theta = \sin (\theta + 2\pi)</math>. We can use these facts to create two types of solutions: <cmath>\sin \theta = \sin ((2m + 1)\pi - \theta)</cmath>7 KB (1,211 words) - 00:23, 20 January 2024
- <math>W_S = \sum_{i=1}^{|D|} \tbinom{6}{x_i}\tbinom{6}{y_i}</math> if and only if there exists ...>S</math> is divisible by 3. Therefore, by the fact that <math>W_S = \sum_{i=1}^{|D|} \tbinom{6}{x_i}\tbinom{6}{y_i}</math>, we have that;26 KB (4,044 words) - 13:58, 24 January 2024
- label("$I$", (2+r) * dir(162), dir(162)); ...ounter-Clockwise}</math>, and <math>\text{Switching}</math>. Let an <math>I</math> signal going clockwise (because it has to be in the ''inner'' circle11 KB (1,934 words) - 12:18, 29 March 2024
- .../math>, <math>yz = 60</math>, and <math>zx = -96 + 24i</math>, where <math>i</math> <math>=</math> <math>\sqrt{-1}</math>. Then there are real numbers < ...you could do the same thing with <math>xy</math> but <math>zx</math> looks like it's easier due to it being smaller. Anyway you get <math>x=20+12i</math>.11 KB (2,077 words) - 20:15, 12 January 2024
- ...enewable energy resources. "How much more does it really cost for a family like ours to switch entirely to renewable energy?" ...Oh," she thinks, "my wormhole allotment was <math>\textit{two}</math>, and I used it up already!"7 KB (1,092 words) - 19:05, 17 December 2021
- for (int i=0; i<7; ++i) { draw((i,0)--(i,7),gray);14 KB (2,191 words) - 03:19, 2 April 2024
- \text{(I) }\frac{97}{98}\qquad \text{(N) }\pi\qquad2 KB (290 words) - 18:27, 30 November 2018
- for(int i=0;i <= 4;i=i+1) draw(shift((4*i,0)) * P);13 KB (2,024 words) - 16:07, 22 April 2024
- ...rac{2\pi k}{n}\right)=\cos\left(\frac{2\pi k}{n}\right)+i\sin\left(\frac{2\pi k}{n}\right)</cmath></center> <center><cmath>(e^{2\pi i k/n})^n=e^{n(2\pi i k/n)}=e^{2\pi i k}=1.</cmath></center>8 KB (1,438 words) - 14:50, 23 June 2022
- ...19</math>, and <math>f\left(\tfrac{1+\sqrt{3}i}{2}\right)=2015+2019\sqrt{3}i</math>. Find the remainder when <math>f(1)</math> is divided by <math>1000< ...math>\frac{1+\sqrt{3}i}{2} = \omega</math> where <math>\omega = e^{\frac{i\pi}{3}}</math> is a primitive 6th root of unity. Then we have4 KB (706 words) - 22:18, 28 December 2023
- D= \frac{1}{8} \pi r^2 - [A_1 A_2 O]=\frac{1}{8} \pi \left(4+2\sqrt{2}\right)- \left(\sqrt{2}+1\right) ...i r^2}{7} - D= \frac{1}{7} \pi \left(4+2\sqrt{2}\right)-\left(\frac{1}{8} \pi \left(4+2\sqrt{2}\right)- \left(\sqrt{2}+1\right)\right)7 KB (1,051 words) - 20:45, 27 January 2024
- ...ide acute triangle <math>ABC</math> such that <math>\angle ADB=\angle ACB+\pi/2</math> and <math>AC\cdot BD=AD\cdot BC</math>. ...>s_0, s_1, \ldots</math> as follows: at step <math>s_i</math>, if <math>L_{i-1}</math> is lit, switch <math>L_i</math> from on to off or vice versa, oth2 KB (452 words) - 17:59, 24 March 2019
- ...f{(C)}\ 3 \pi \sqrt7 \qquad\textbf{(D)}\ 6\pi \sqrt3 \qquad\textbf{(E)}\ 6\pi \sqrt7</math> ...textbf {(D) } 3\sqrt{3} - \pi \qquad \textbf {(E) } \frac{9\sqrt{3}}{2} - \pi </math>16 KB (2,497 words) - 21:14, 12 November 2023
- 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 po <cmath>x=\cos(\theta) + i\sin(\theta)</cmath>1 KB (220 words) - 15:24, 6 July 2021
- Let <cmath>z=\frac{1+i}{\sqrt{2}}.</cmath>What is <cmath>\left(z^{1^2}+z^{2^2}+z^{3^2}+\dots+z^{{1 ...e^{i\pi}+...+e^{2i\pi}\right)\left(e^{\frac{i7\pi}{4}}+e^{i7\pi}+...+e^{2i\pi}\right)</cmath> which can easily be computed as <math>\boxed{36}</math>.5 KB (836 words) - 12:36, 3 December 2023
- surface s=surface(f,(0,0),(pi,2pi),70,Spline); label("$I$",(12,4),SE);7 KB (1,000 words) - 15:03, 23 October 2021
- Like in Solution 1, we determine the coordinates of the three vertices of the tr Like in the other solutions, solve the systems of equations to see that the tria7 KB (1,079 words) - 22:24, 10 November 2023
- <cmath>\frac{a}{b}\cdot\pi-\sqrt{c}+d,</cmath> The area of <math>A</math> is <math>\frac{1}{2} \pi \cdot 2^2 = 2\pi</math>.6 KB (984 words) - 23:52, 11 November 2023
- for (int i = 0; i < 3; ++i) { pair A = (j,i);16 KB (2,477 words) - 15:41, 9 September 2023
- ...< \theta < \pi</math>. By symmetry, the interval <math>\pi \leq \theta < 2\pi</math> will also give <math>2</math> solutions. The answer is thus <math>2 ...rm an equilateral triangle, their difference in angle must be <math>\frac{\pi}{3}</math>, so6 KB (987 words) - 19:19, 12 November 2022
- Let <math>\omega=-\tfrac{1}{2}+\tfrac{1}{2}i\sqrt3.</math> Let <math>S</math> denote all points in the complex plane of ...c{3}{2}\sqrt3\qquad\textbf{(D) } \frac{1}{2}\pi\sqrt3 \qquad\textbf{(E) } \pi</math>4 KB (729 words) - 21:23, 15 November 2022
- ...st have <cmath>\frac{f(f(z))-f(z)}{f(z)-z}=-\frac{f(f(z))-f(z)}{z-f(z)}\in i\mathbb R .</cmath>However, <math>f(t)-t=t(t-20)</math>, so ...perpendicular in the complex plane if and only if <math>\frac{a-b}{b-c}\in i\mathbb{R}</math>. To prove this, note that when dividing two complex number8 KB (1,534 words) - 22:17, 28 December 2023
- i (the unit imaginary number) π (the famous number pi that turns up in many interesting areas)3 KB (543 words) - 15:24, 13 June 2019
- ===== pi and e ===== ...the limit <math>\frac{1}{1}-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}...=\frac{\pi}{4}</math>. e is the limit <math>\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}...<35 KB (5,882 words) - 18:08, 28 June 2021
- <cmath> \int_{0}^{\pi/2}\frac{\sin^3x}{\sin^3x+\cos^3x}dx </cmath> Let <math>u=\frac{\pi}{2}-x</math> and <math>du=-dx</math>, the integral than turns into:1 KB (196 words) - 18:32, 14 January 2020
- Let <math> c_i </math> denote the <math> i\text{th} </math> composite integer so that <math> \{c_i\}=4, 6, 8, 9\cdots <cmath> \prod_{i=1}^{\infty}\frac{c_i^2}{c_i^2-1} </cmath>3 KB (445 words) - 18:37, 14 January 2020
- pair O, A, B, C, D, F, G, H, I, P, X; I = rotate(-120, H) * ((G + H) / 2);9 KB (1,380 words) - 16:12, 2 January 2024
- ...then it moves <math>2\pi a</math> units, or one revolution, every <math>2\pi</math> seconds (in other words, it revolves 1 radian per 1 second). the x-coordinate of the point relative to the center is: <math>a\cos(-\frac{\pi}{2}-t)=-a\sin t</math>2 KB (357 words) - 02:41, 27 July 2019
- The set of complex numbers is a set of all numbers ever existed, from pi to 1. ...a real part and an imaginary part, usually written in the from a + b<math>i</math>.2 KB (270 words) - 16:28, 13 June 2022
- ...\sqrt{3})^{2019} \qquad \textbf{(C) }(3+\sqrt{2})^2 \qquad \textbf{(D) }(2\pi)^2 \qquad \textbf{(E) }(3+\sqrt{2})(3-\sqrt{2}) \qquad</math> ...math>. The area of Region <math>A</math> can be expressed as <math>\frac{a\pi+b\sqrt{c}}{d}</math>, where <math>a, b, c, d</math> are positive integers,13 KB (2,059 words) - 02:59, 21 January 2021
- 4. The Infinity Numeral, PI, is the second deity of the Almighty Gmaas's heaven.85 KB (13,971 words) - 18:02, 16 May 2024
- Simplify <math>\left(\frac{-1+i\sqrt{3}}{2}\right)^6+\left(\frac{-1-i\sqrt{3}}{2}\right)^6</math> to the form <math>a+bi</math>. ...s6\cdot\frac{4\pi}{3}+\sin6\cdot\frac{4\pi}{3}\right)=\left(\cos8\pi+\sin8\pi\right)=1+0=1.</math> Thus, the total sum is <math>1+1=\mathbf{2}.</math>1 KB (169 words) - 09:23, 24 July 2020
- A rectangle of length <math>\frac{1}{4} \pi</math> and height 4 is bisected by the x-axis and is in the first and fourt For what value of <math>x</math> <math>(0 < x < \frac{\pi}{2})</math> does <math>\tan x + \cot x</math> achieve its minimum?3 KB (413 words) - 13:10, 21 January 2020
- ...ch the sum of its elements, when divided by 7, leaves the remainder <math> i</math>. Prove that <math> N(1) - N(2) + N(3) - N(4) + N(5) - N(6) = 0</math <cmath> \frac{\pi (a^{2}+b^{2}+c^{2})(b+c-a)(c+a-b)(a+b-c)}{(a+b+c)^{3}}.</cmath>3 KB (478 words) - 14:09, 23 June 2021
- ...n <math>\tan(2x)=\cos(\tfrac{x}{2})</math> have on the interval <math>[0,2\pi]?</math> ...<math>x=\frac{\pi}{4}+\frac{k\pi}{2},</math> and zeros at <math>x=\frac{k\pi}{2}</math> for some integer <math>k.</math> <p>4 KB (615 words) - 04:07, 8 July 2022
- ...get <math>-4\sqrt 2 - 16 \sqrt2 i</math> and <math>-3\sqrt 2 - 12 \sqrt 2 i</math>. This line has a slope of <math>4</math>. Now, back to the cartesian7 KB (1,145 words) - 20:27, 5 November 2023
- ...eta</math> is the argument of <math>z</math> such that <math>0\leq\theta<2\pi.</math> <p> ...<math>3\theta=0,2\pi,4\pi,</math> or <math>\theta=0,\frac{2\pi}{3},\frac{4\pi}{3}.</math> </li><p>4 KB (696 words) - 12:38, 13 September 2021
- <cmath>\sin^2{(\pi x)} + \sin^2{(\pi y)} > 1</cmath> <cmath>\sin^{2}{(\pi x)}+\sin^{2}{(\pi y)}>1</cmath>8 KB (1,412 words) - 06:17, 30 December 2023
- ...quad \textbf{(D) } 3\sqrt3 - \pi \qquad \textbf{(E) } \frac{9\sqrt3}{2} - \pi</math> pair G,H,I,J,K;17 KB (2,392 words) - 12:36, 24 December 2023
- ...root, so is <math>\frac{-1+i\sqrt{3}}{2} \cdot r</math>? (Note that <math>i=\sqrt{-1}</math>) ...3}}{2}</math> = <math>\cos(120^\circ) + i \cdot \sin(120^\circ) = e^{ 2\pi i / 3}</math>.4 KB (726 words) - 16:55, 11 September 2023
- ...th> to be equal to <math>e^{i\frac{2\pi}{3}}</math> and <math>e^{-i\frac{2\pi}{3}}</math>, meaning that all three are equally spaced along the unit circl2 KB (306 words) - 17:45, 28 January 2024
- A = (0, tan(3 * pi / 7)); [[File:2020 AIME I 1.png|450px|left]]6 KB (968 words) - 15:01, 24 January 2024
- ...but exterior to the quadrilateral can be written in the form <math>\frac{a\pi-b}{c},</math> where <math>a,b,</math> and <math>c</math> are positive integ for(int i = 0; i < 3; ++i){18 KB (2,662 words) - 02:08, 9 March 2024
- ...e equation <math>(x + i)^{10} = 1</math> on the complex plane, where <math>i = \sqrt -1</math>. <math>2</math> points from <math>K</math> are chosen, su ...\pi}{13}</math>, <math>\cos \frac{5\pi}{13}</math>, and <math>\cos \frac{7\pi}{13}</math>. What is the least possible sum of the coefficients of <math>P(8 KB (1,223 words) - 15:02, 27 November 2022
- ...s, according to one of the 20 moderators. The activity began at 3:14 PM or pi time on the west coast and ended 106 minutes later at 5:00 PT. ...guarantee a point. The moderators could see who sent in the answer first, I would say that luck plays a incredibly small role in this." - [[User:Mathan11 KB (1,044 words) - 17:36, 11 January 2021
- ...the interior of the hexagon such that <math>\angle AGB=\angle DHE=\frac{2\pi}{3}</math>. Prove that <math>AG+GB+GH+DH+HE\ge CF</math>. ...\triangle EFA</math> and <math>\triangle BCD</math>, and draw points <math>I</math> and <math>J</math> such that <math>IA=IB</math>, <math>JD=JE</math>,1 KB (214 words) - 20:19, 5 July 2020
- ...a</math> is <math>1</math>, is achieved at <math>\theta = \frac{\pi}{2}+2k\pi</math> for some integer <math>k</math>. ...that <math>mx = \frac{\pi}{2}+2a\pi</math> and <math>nx = \frac{\pi}{2}+2b\pi</math>, for integers <math>a, b</math>.9 KB (1,523 words) - 09:12, 3 December 2023
- ...3+i \qquad \textbf{(C) }{-}\sqrt2+\sqrt2 i \qquad \textbf{(D) }{-}1+\sqrt3 i\qquad \textbf{(E) }2i</math> ...ft( \frac{\pi}2 \sin x\right)</math> have in the closed interval <math>[0,\pi]</math>?15 KB (2,383 words) - 09:49, 25 June 2023
- ...3+i \qquad \textbf{(C) }{-}\sqrt2+\sqrt2 i \qquad \textbf{(D) }{-}1+\sqrt3 i\qquad \textbf{(E) }2i</math> Thus, the answer is <math>\boxed{\textbf{(B) }{-}\sqrt3+i}</math>.6 KB (872 words) - 17:36, 4 December 2021
- ...mega = \cos\frac{2\pi}{7} + i \cdot \sin\frac{2\pi}{7},</math> where <math>i = \sqrt{-1}.</math> Find the value of the product <cmath>\prod_{k=0}^6 \lef ...[2023 AIME I Problems|2023 AIME I]]|after=[[2024 AIME I Problems|2024 AIME I]]}}8 KB (1,370 words) - 21:34, 28 January 2024
- How many integer values of <math>x</math> satisfy <math>|x| < 3\pi</math>? ...\qquad \textbf{(C) }64\pi \qquad \textbf{(D) }65\pi \qquad \textbf{(E) }84\pi</math>17 KB (2,418 words) - 12:52, 5 November 2023
- ...ve integers not exceeding <math>100</math> that satisfy the equation <math>i \cdot w^r = z^s.</math> w &= e^{i\cdot\frac{\pi}{6}}, \\5 KB (773 words) - 14:37, 23 February 2023
- [[File:AIME-I-2022-11.png|530px|right]] ...erefore, <math>\frac{CI}{MC}=\frac{AI}{AN}</math>. Let the length of <math>PI=l</math>, then <math>\frac{25-l}{20}=\frac{3+l}{6}</math>. Solving we get <16 KB (2,517 words) - 20:22, 31 January 2024
- ...th>r^2+r+1=(r-\omega)(r-{\omega}^2)</math>, where <math>\omega=e^{i\frac{2\pi}{3}}</math>. Thus, [[File:AIME-I-2022-14a.png|400px|right]]16 KB (2,730 words) - 02:56, 4 January 2023
- ...>, <math>\sqrt{y}</math>, or <math>\sqrt{z}</math>, the system should look like this: Taking the inverse sine (<math>0\leq\theta\frac{\pi}{2}</math>) of each equation yields a simple system:15 KB (2,208 words) - 01:25, 1 February 2024
- ...<math>z'=\frac{i}{z}</math> where <math>z \in \mathcal{T}</math> and <math>i=\sqrt{-1}</math>. If the area enclosed by <math>\mathcal{T}'</math> is <mat ...<math>75\pi+50</math>. The greatest integer less than or equal to <math>75\pi+50</math> is <math>\boxed{285}</math>.887 bytes (154 words) - 17:37, 4 October 2020
- ...} \right)^n + \left( \frac{-1-i\sqrt{3}}{2} \right)^n </math>, where <math>i = \sqrt{-1}</math>. What is <math>f(2022)</math>? for (int i = 0; i <= 5; ++i) {15 KB (2,233 words) - 13:02, 10 November 2023
- for (int i=1; i<7; ++i) draw((i,0)--(i,5), gray+dashed);16 KB (2,450 words) - 00:13, 12 November 2023
- ...25\pi\qquad\textbf{(C) }26\pi\qquad\textbf{(D) }27\pi\qquad\textbf{(E) }28\pi</math> ...ath> is <math>\pi\cdot\left(\frac{AO_1}{2}\right)^2=\boxed{\textbf{(C) }26\pi}.</math>7 KB (1,026 words) - 13:43, 5 May 2024
- \textbf{Narrow Cone} & 3 & h_1 & & \frac13\pi(3)^2h_1=3\pi h_1 & \\ [2ex] \textbf{Wide Cone} & 6 & h_2 & & \hspace{2mm}\frac13\pi(6)^2h_2=12\pi h_2 &9 KB (1,503 words) - 15:09, 1 August 2023
- ...ft( \frac{\pi}2 \sin x\right)</math> have in the closed interval <math>[0,\pi]</math>? ...<math>\frac{\pi}2 \cos x</math> are both <math>\left[-\frac{\pi}2, \frac{\pi}2 \right],</math> which is included in the range of <math>\arcsin,</math> s7 KB (1,110 words) - 20:10, 5 November 2022
- ...e <math>\cos \frac{2\pi}7,\cos \frac{4\pi}7,</math> and <math>\cos \frac{6\pi}7</math>, where angles are in radians. What is <math>abc</math>? ...\right)=\operatorname{Re}\left(z^{-k}\right)</math> and <math>\sin\frac{2k\pi}{7}=\operatorname{Im}\left(z^k\right)=-\operatorname{Im}\left(z^{-k}\right)9 KB (1,484 words) - 02:25, 21 September 2023
- ...ided regular polygon . Suppose <math>P_i =z .\zeta ^i</math> for all <math>i \in \{0,1,2,\cdots ,n -1\}</math>. Where <math>z</math> is a complex number ...that <math>l_k \equiv \ Im ({P_k})=y \cos \frac{2\pi k}{n} +x \sin \frac{2\pi k}{n}</math> .3 KB (607 words) - 03:12, 22 September 2023
- How many integer values of <math>x</math> satisfy <math>|x|<3\pi?</math> How many values of <math>\theta</math> in the interval <math>0<\theta\le 2\pi</math> satisfy<cmath>1-3\sin\theta+5\cos3\theta = 0?</cmath>15 KB (2,302 words) - 12:31, 27 October 2023
- ...of <math>7\pi</math> and the total area covered by the circles is <math>25\pi</math>. What is the value of <math>r</math>? i. The perpendicular bisectors of the sides of <math>\mathcal{P}</math> all s5 KB (776 words) - 09:35, 8 August 2023
- has more than one solution in the interval <math>(0, \pi)</math>. The set of all such <math>a</math> that can be written ...nect points <math>(0, a_i)</math> to <math>(b_i, 0)</math> for <math>1 \le i \le 14</math> and are tangent to the circle, where <math>a_i</math>, <math>15 KB (2,250 words) - 00:32, 9 March 2024
- \zeta(s)=2^s\pi^{s-1}\sin\left(\pi s\over2\right)\Gamma(1-s)\zeta(1-s) B_1(x)\triangleq\{x\}-\frac12=-\sum_{n=1}^\infty{\sin(2\pi nx)\over\pi n}4 KB (682 words) - 03:56, 13 January 2021
- ...a = 1</math>, <math> b = i</math>, <math> c = - 1</math> and <math> d = - i</math>. ...t {3} - 1}{2}(1 + i)</math> and <math> n = \frac {\sqrt {3} - 1}{2}( - 1 + i)</math>.1 KB (199 words) - 16:40, 29 January 2021
- ==Day I== ...ere <math>\angle AOX</math> is measured in radians in the range <math>[0,2\pi)</math>. Prove that we can find a point <math>X</math>, not on <math>OA</ma2 KB (410 words) - 23:45, 29 January 2021
- ...ere <math>\angle AOX</math> is measured in radians in the range <math>[0,2\pi)</math>. Prove that we can find a point <math>X</math>, not on <math>OA</ma ...e the circle centered at <math>O</math> with radius <math>\sum_{i=1}^{n}a_{i}</math>.2 KB (388 words) - 23:49, 29 January 2021
- ...We have our <math>z^{2021}</math> term but we have these unnecessary terms like <math>z^{2020}</math>. We can get rid of these terms by adding <math>-z^{20 z^3 &= e^{i 0} \\8 KB (1,315 words) - 11:43, 24 October 2023
- <center><math>C_m=\mathcal{C}(t;z_0+r_me^{it},0,2\pi)</math></center> <center><math>2\pi f(w)=\oint_{C_2}\frac{f(z)}{z-w}\,dz-\oint_{C_1}\frac{f(z)}{z-w}\,dz</math>8 KB (1,471 words) - 22:02, 12 April 2022
- ...ss than or equal to <math>n</math>. Suppose <math>\pi(a)^{\pi(b)}=\pi(b)^{\pi(a)}=c</math>. For some fixed <math>c</math> what is the maximum possible nu ...ll positive integers <math>n<1000</math> such that <cmath>\sum_{i=1}^{n}a_{i}<4n.</cmath>4 KB (651 words) - 20:18, 6 March 2021
- pair K=(0,0),B=(1,0),A=(-1,0),L=(0,0.5),M=(sqrt(2)/2,.25),I=(2*sqrt(2)/3,1/3),E=(sqrt(2)/3,1/3),P=(0,0.25); draw(K--I,red);3 KB (462 words) - 07:58, 6 September 2021
- ...} \qquad\textbf{(D)} ~\frac{\pi+2\sqrt{3}-3}{4} \qquad\textbf{(E)} ~\frac{\pi+3-2\sqrt{2}}{4}</math> ...f{(C)} ~\frac{(2-\sqrt{2})\pi}{8-2\pi} \qquad\textbf{(D)} ~\frac{\pi}{32-8\pi}\qquad</math>13 KB (2,097 words) - 17:38, 29 April 2021
- ...)} ~\frac{3\sqrt{3}}{2}-\frac{\pi}{3} \qquad\textbf{(E)} ~2\sqrt{3}-\frac{\pi}{3}</math> Let <math>n=(e^{({\sin(\pi)+\cos(\pi)})\pi})^{e^{\frac{i\pi}{2}}}</math>, find the remainder when<cmath>\left \lfloor{\sum_{k=0}^{20} 211 KB (1,691 words) - 18:56, 25 April 2022
- <cmath>P(x)=1+\cos(x)+i\sin(x)-\cos(2x)-i\sin(2x)+\cos(3x)+i\sin(3x)</cmath> ...qrt{-1}</math>. For how many values of <math>x</math> with <math>0\leq x<2\pi</math> does6 KB (1,019 words) - 04:02, 9 October 2023
- <b><i>Claim</b></i> <b><i> Proof</b></i>14 KB (2,254 words) - 18:26, 8 February 2024
- ...}} e^{i \frac{\beta}{2}} = e^{i \frac{\alpha + \beta}{2}} = \frac{3}{5} + i \frac{4}{5}</math>. & = \frac{1}{2i} \left( e^{i \frac{\alpha}{2}} - e^{-i \frac{\alpha}{2}}14 KB (2,217 words) - 00:28, 29 June 2023
- for (int i = 0; i <= 5; ++i) { draw((0,i)--(5,i));15 KB (2,224 words) - 13:10, 20 February 2024
- ...math>8</math> units tall. The volume of the cup can be written as <math>k \pi</math> cubic units. Find <math>k</math>. #David: Chandler told two truths. I am the oldest person in the room.7 KB (1,100 words) - 18:40, 11 July 2021
- <cmath> \theta_c = \pm \frac{\pi}{3} - \theta_a </cmath> From here, we determine <math>\theta_c = \pm \frac{\pi}{3}</math> and <math>c = \frac{1}{2} \pm \frac{\sqrt{3}}{2}</math>. Then we2 KB (377 words) - 14:08, 9 August 2021
- Let <math>\Gamma</math> be a circle with centre <math>I</math>, and <math>ABCD</math> a convex quadrilateral such that each of ...AB - \angle IAC</cmath> <cmath>\angle IZX = \angle IAB</cmath> Since <math>I</math> is the incenter of quadrilateral <math>ABCD</math>, <math>AI</math>7 KB (1,196 words) - 10:30, 18 June 2023
- pair C=(0,0),D=(cos(pi/12),sin(pi/12)),E=rotate(150,D)*C,F=rotate(-30,E)*D,A=rotate(150,F)*E,B=rotate(-30,A)* .../math>, where <math>a</math> and <math>b</math> are real numbers and <math>i = \sqrt{-1}</math>, is the complex number <math>\overline{w} = a - bi</math15 KB (2,452 words) - 19:37, 7 June 2023
- for (int i=1; i<7; ++i) draw((i,0)--(i,5), gray+dashed);14 KB (2,191 words) - 19:57, 12 November 2023
- ...centimeters. If the surface area that is iced can be expressed as <math>m\pi,</math> find <math>m.</math> ..._i\right),</math> where <math>r_i</math> is the remainder when <math>2^i+3^i</math> is divided by 10.4 KB (695 words) - 12:37, 6 June 2022
- ...<math>a = 0</math> are <math>C \cos x</math>; those with <math>a = -\frac{\pi}{2}</math> are <math>C \sin x</math>. The initial condition <math>f'(0) = 0 ...1})</math> can be computed in terms of <math>c_{i-1}</math> and <math>f(c_{i-1})</math> using the given differential equation), until <math>b = c_n</mat6 KB (969 words) - 20:33, 15 December 2023
- I like to make pics with Asymptote like this one: since I can still type my normal LaTeX stuff around it:2 KB (253 words) - 00:28, 19 November 2023
- ...three, we need to make sure the roots are in the form of <math>e^{i\frac{k\pi}{9}}</math>, so we only have to look at <math>D,E</math>. If we look at choice <math>E</math>, <math>x=e^{i\frac{\pm2\pi}{9}}</math> which works perfectly, the answer is just <math>E</math>8 KB (1,325 words) - 02:14, 11 April 2024
- Like the solutions above we can know that <math>|z_1| = |z_2| = \sqrt{10}</math> ...\sqrt{10}}e^{-i\theta} </math>, <math>\frac{1}{z_2}= \frac{1}{\sqrt{10}}e^{i\theta}</math>.7 KB (1,164 words) - 11:20, 1 January 2024
- for(int i = 0; i < 3; ++i){ lVs.push(rho*lVs[i]);13 KB (2,080 words) - 11:27, 25 October 2023
- for (int i = 0; i < 7; ++i) { for (int j = 0; j < i; ++j) {21 KB (3,265 words) - 17:06, 15 November 2023
- ...} \right)^n + \left( \frac{-1-i\sqrt{3}}{2} \right)^n </math>, where <math>i = \sqrt{-1}</math>. What is <math>f(2022)</math>? -1 + i\sqrt{3} &= 2e^{\frac{2\pi i}{3}}, \\5 KB (866 words) - 22:17, 27 October 2023
- ...+i</math> in the complex plane, by an angle of <math>270^{\circ} = \frac{3\pi}{2} \text{ rad}</math> counterclockwise. ...)e^{\frac{3\pi i}{2}} + (3+i) = (-4-3i)(-i) + (3+i) = 4i-3i^2+3+i = 4i-3+3+i = 5i,</math> which corresponds to <math>\boxed{\textbf{(B)}\ (0,5)}</math>2 KB (278 words) - 13:32, 5 April 2023
- Let <math>ABC</math> be a triangle with incenter <math>I</math> and let <math>D</math> be an arbitrary point on the side <math>BC</m (i) no three points in <math>P</math> lie on a line and3 KB (492 words) - 14:07, 24 December 2022
- I. If <math>\gcd(a,14)=1</math> or <math>\gcd(b,15)=1</math> or both, then <m ...xt{I, II, and III}\qquad\textbf{(B)}~\text{I only}\qquad\textbf{(C)}~\text{I and II only}\qquad\textbf{(D)}~\text{III only}\qquad\textbf{(E)}~\text{II a13 KB (2,107 words) - 22:19, 20 April 2024
- int i, j; for(i=0; i<7; i=i+1)3 KB (452 words) - 14:33, 21 January 2024
- AF + AE e^{i \left( \pi - \theta \right)} + EP e^{i \left( \frac{3 \pi}{2} - \theta \right)} - PF i .17 KB (2,612 words) - 14:54, 3 July 2023
- ...mega = \cos\frac{2\pi}{7} + i \cdot \sin\frac{2\pi}{7},</math> where <math>i = \sqrt{-1}.</math> Find the value of the product<cmath>\prod_{k=0}^6 \left ...>z_n = \left(\textrm{cis }\frac{2n\pi}{7}\right)^3 + \textrm{cis }\frac{2n\pi}{7} + 1</math>.9 KB (1,284 words) - 23:37, 31 January 2024
- for (int i = 0; i < 6; ++i) { pair next = current + side * dir(angle * i);13 KB (1,886 words) - 22:08, 10 April 2024
- ...number of circles needed to make the total shaded area at least <math>2023\pi</math>? for(int i = 0; i < 4; ++i) {16 KB (2,411 words) - 00:18, 7 May 2024
- ...})=70T</math>. Also, <math>\text{gcd}(a_i,a_j)=1</math> for all <math>1\le i<j\le2022</math>. Find the number of possible sequences <math>S</math>. ...<math>a+b\cos(\frac{\pi}{c})</math>, where <math>\text{cis}(x) = \cos(x) + i\sin(x)</math>. Find <math>a+b+c</math>.1 KB (217 words) - 21:59, 31 May 2023
- {{AIME Problems|year=2024|n=I}} [[2024 AIME I Problems/Problem 1|Solution]]8 KB (1,307 words) - 20:00, 6 February 2024
- ...educes two complicated equations to one linear and one quadratic equation. I can then easily find a non-zero solution and even get the closed form. ...s <math>\left( 1 - \frac{1}{8^2 \cdot 18 \pi^4} , 1 - \frac{1}{8 \cdot 18 \pi^3} \right) = \left( 1 - 8.9 \cdot 10^{-6}, 1 - 2.2 \cdot 10^{-4} \right)</m3 KB (515 words) - 03:57, 4 February 2024
- ...en the hypotenuse of the triangle and the curve of the semicircle be <math>I</math>. Then <math>OI</math> and <math>OB</math> are the radii of the semic ...area <math>= \frac{75\sqrt{3}}{4} + \frac{25}{2} \pi = \frac{75\sqrt{3}+50\pi}{4}</math>. Our answer <math>= 75 + 3+ 50 + 4 = \boxed{132}</math>.2 KB (276 words) - 20:22, 1 July 2023
- ...Pre-algebra class. He enters every class by posting a meme and announcing "I have arrived!" Sseraj is the human servant of Gmaas and is one of only 5 en - 2024 Multiverse war I erupts and Zzgurkk 3T is destroyed, along with many of Gmaas's bibles.88 KB (14,928 words) - 13:54, 29 April 2024
- 1. The system has the digits 3,1,4,(one),5,9,p,i, and (gmaas). It is in base eight. i-7824 bytes (144 words) - 20:58, 3 November 2023
- ...\right),Rsin\left( \frac{2\pi}{n}i \right) \right\rangle</math> for <math>i=0,1,2,...,(n-1)</math>. <math>A=\left\langle Rcos\left( \frac{2\pi}{n}a \right),Rsin\left( \frac{2\pi}{n}a \right) \right\rangle</math>5 KB (955 words) - 02:21, 21 November 2023
- I. If <math>\gcd(a,14)=1</math> or <math>\gcd(b,15)=1</math> or both, then <m ...xt{I, II, and III}\qquad\textbf{(B)}~\text{I only}\qquad\textbf{(C)}~\text{I and II only}\qquad\textbf{(D)}~\text{III only}\qquad\textbf{(E)}~\text{II a13 KB (1,959 words) - 10:29, 4 April 2024
- ...number of circles needed to make the total shaded area at least <math>2023\pi</math>? ...(2^2\pi-1^2\pi)+(4^2\pi-3^2\pi)+(6^2\pi-5^2\pi)+ \cdots + (n^2\pi-(n-1)^2 \pi)</math> for any even number <math>n</math>.6 KB (967 words) - 07:01, 28 January 2024
- <cmath>|(1+a+a^2-b^2)+ (b+2ab)i|=4</cmath> Thus <math>z=i\sqrt{19}/2</math>, and so the answer is <math>\boxed{\textbf{(B)}~21}</math6 KB (1,055 words) - 21:58, 28 February 2024
- pair F, G, H, I, J; I = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 3*72 + 180);19 KB (2,967 words) - 16:56, 24 February 2024
- <i><b>Solution</b></i> <i><b>Solution</b></i>12 KB (2,104 words) - 14:11, 24 February 2024
- ...c{\arctan \frac{1}{2}}{\pi}\qquad\textbf{(E)}~\frac{2\arcsin \frac{1}{4}}{\pi}</math> ...<math>1</math> unit away from <math>S</math> is <math>\frac{4 \alpha }{ 2 \pi}</math>.4 KB (587 words) - 18:15, 2 January 2024
- &=(75a-117b)+(117a+75b)i+48\left(\dfrac{2+3i}{a+bi}\right) \\ &=(75a-117b)+(116a+75b)i+48\left(\dfrac{(2+3i)(a-bi)}{(a+bi)(a-bi)}\right) \\12 KB (1,957 words) - 09:04, 18 May 2024
- ...ts of <math>P_n(x)</math> will be in the form <math>x=e^{\frac{k}{n+1}2\pi i}</math> for <math>k=1,2,\cdots,n</math> with the only real solution when <m ...c{2}{1005}2\pi i},e^{\frac{3}{1005}2\pi i},\cdots,e^{\frac{1004}{1005}2\pi i}</math> for a total of <math>\textbf{1004}</math> complex roots.3 KB (442 words) - 20:51, 26 November 2023
- ...<math>a+b\cos(\frac{\pi}{c})</math>, where <math>\text{cis}(x) = \cos(x) + i\sin(x)</math>. Find <math>a+b+c</math>.377 bytes (75 words) - 13:13, 14 December 2023
- ...e <math>A</math> with the circumcircle of <math>ABC</math>. The line <math>PI</math> intersects for the second time the circumcircle of <math>ABC</math>1 KB (191 words) - 04:59, 26 March 2024
- ...and let <math>a_n = a_{n-1} - \frac{a_{n-3}}{8}</math>. Find <cmath>\sum_{i=0}^\infty a_i.</cmath> such <math>0 \le x \le \pi</math>640 bytes (94 words) - 21:32, 15 December 2023
- ...of <cmath>y=4 g(f(\sin (2 \pi x))) \quad\text{ and }\quad x=4 g(f(\cos (3 \pi y))).</cmath> ...th> as Function 1 and function <math>x = 4g \left( f \left( \cos \left( 3 \pi y \right) \right) \right)</math> as Function 2.8 KB (1,315 words) - 19:41, 18 February 2024
- ...sin}^n x + \text{cos}^n x.</math> For how many <math>x</math> in <math>[0,\pi]</math> is it true that Find <math>i + 2i^2 +3i^3 + . . . + 2002i^{2002}.</math>10 KB (1,606 words) - 01:46, 31 December 2023
- <math>\sin</math> and <math>\cos</math> are easy to define. I prefer the unit circle definition as it makes these proofs easier to unders Note: I've omitted <math>\theta</math> because it's unnecessary and might clog thin16 KB (2,796 words) - 13:12, 21 January 2024
- for(int i=0; i<360; i+=30) { dot(dir(i), 4+black);8 KB (1,395 words) - 17:26, 9 February 2024
- <i><b>Standard Solution</b></i> <i><b>Short Solution</b></i>32 KB (5,375 words) - 11:43, 5 May 2024
- ...^1{\ln ^3\left( 1-x \right) \ln ^2\left( 1+x \right) \text{d}x}=2\underset{I}{\underbrace{\int_0^1{\frac{\ln ^2\left( \dfrac{2}{1+x} \right) \ln ^3\left ...\left( 1+x \right) ^2}}\text{d}x=\frac{1}{4}\zeta \left( 3 \right) +\frac{\pi ^2}{6}-\frac{1}{3}\ln ^32-\ln ^22-\text{2}\ln 2</cmath>17 KB (2,704 words) - 04:58, 29 January 2024
- ...\frac{\sqrt{3}-2\sin \theta}{2\sqrt{3}-2\tan \theta}=\lim_{\theta\to\frac{\pi}{3}}cos^3{x}=\frac{1}{8}</math>. This means that <math>y=\frac{3\sqrt{3}}{8 pair A=(0.5,0); pair B=(0,sin(pi/3));13 KB (2,178 words) - 11:34, 14 May 2024
- ...- \omega^k)^2 + 1\right) = \prod_{k=0}^{12} \left((1 + i) - \omega^k)((1 - i) - \omega^k\right)</cmath> <cmath>((1 + i)^{13} - 1)(1 - i)^{13} - 1)</cmath>5 KB (771 words) - 10:36, 19 April 2024
- <i><b>Igor Fedorovich Sharygin</b></i> (13/02/1937 - 12/03/2004, Moscow) - Soviet and Russian mathematician and t <i><b>Solution</b></i>29 KB (4,997 words) - 18:06, 16 May 2024