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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 fou
    3 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 i
    13 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{3
    22 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 eve
    8 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 || 578.262
    187 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 \qquad
    14 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_1
    13 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}\qquad
    2 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'' circle
    11 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\qquad
    2 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 have
    4 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, oth
    2 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 tria
    7 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>, so
    6 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 number
    8 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 cartesian
    7 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 circl
    2 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:Mathan
    11 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> s
    7 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 s
    5 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</ma
    2 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} 2
    11 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> does
    6 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 we
    2 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</math
    15 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</mat
    6 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 and
    3 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 a
    13 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)</m
    3 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-7
    824 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 a
    13 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}</math
    6 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 thin
    16 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

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