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- pair O = (0,0), A = r*expi(pi/3); D(CR(O,r));5 KB (763 words) - 16:20, 28 September 2019
- ...c</math> are legs of right triangle <math>abc</math> with <math>\beta = 90^o</math> and <math>c=1</math>8 KB (1,401 words) - 21:41, 20 January 2024
- triple O=(0,0,0); triple O=(0,0,0);7 KB (1,086 words) - 08:16, 29 July 2023
- ...)",S,W); label("\(15\)",B/2+P/2,N);label("\(20\)",B/2+Q/2,E);label("\(O\)",O,SW); </asy></center> ...h>\triangle APS \cong \triangle CRQ</math>). Quickly we realize that <math>O</math> is also the center of the rectangle.8 KB (1,270 words) - 23:36, 27 August 2023
- ...the center of the dodecagon, which we denote <math>A, M,</math> and <math>O</math> respectively. Notice that <math>OM=1</math>, and that <math>\triangl4 KB (740 words) - 17:46, 24 May 2024
- ...ot \beta = \frac{(o+h)(a+h)}{oa} = \frac{oa +oh +ha +h^2}{oa} = 1+ \frac{h(o+a+h)}{oa} = 1+ \alpha + \beta</math>. From the information provided in the10 KB (1,590 words) - 14:04, 20 January 2023
- ..., <math>BB'</math>, and <math>CC'</math> are concurrent at the point <math>O^{}_{}</math>, and that <math>\frac{AO^{}_{}}{OA'}+\frac{BO}{OB'}+\frac{CO}{ ...bove solutions, find <math>\sum_{cyc} \frac{y+z}{x}=92</math> (where <math>O=(x:y:z)</math> in barycentric coordinates). Now letting <math>y=z=1</math>4 KB (667 words) - 01:26, 16 August 2023
- A circle of radius <math>2</math> is centered at <math>O</math>. Square <math>OABC</math> has side length <math>1</math>. Sides <mat pair O=origin, A=(1,0), C=(0,1), B=(1,1), D=(1, sqrt(3)), E=(sqrt(3), 1), point=B;5 KB (873 words) - 15:39, 29 May 2023
- .... Points <math>A</math> and <math>B</math> on the circle with center <math>O</math> and points <math>C</math> and <math>D</math> on the circle with cent pair X=(-6,0), O=origin, P=(6,0), B=tangent(X, O, 2, 1), A=tangent(X, O, 2, 2), C=tangent(X, P, 4, 1), D=tangent(X, P, 4, 2);4 KB (558 words) - 14:38, 6 April 2024
- ...hen by symmetry, the other rectangle is also centered at the origin, <math>O</math>.3 KB (601 words) - 09:25, 19 November 2023
- ...math>P</math> and <math>Q</math> be the points of tangency of circle <math>O</math> to <math>AC</math> and <math>BD</math> respectively. pair A,B,C,D,P,Q,O,X;8 KB (1,231 words) - 20:06, 26 November 2023
- ...Thus, <math>M=\left(\frac{a+b}{2}, 24\right)</math>. The vector from <math>O</math> to <math>M</math> is <math>\left[\frac{a+b}{2}, 24\right]</math>. Me5 KB (788 words) - 13:53, 8 July 2023
- Call the center of the larger circle <math>O</math>. Extend the diameter <math>\overline{PQ}</math> to the other side of2 KB (272 words) - 03:53, 23 January 2023
- Let the center of the circle be <math>O</math>, and the two chords be <math>\overline{AB}, \overline{CD}</math> and ...pi/6), D=E+48*expi(7*pi/6), A=E+30*expi(5*pi/6), C=E+30*expi(pi/6), F=foot(O,B,A);3 KB (484 words) - 13:11, 14 January 2023
- triple A, B, C, D, O, P; O = (0,0,sqrt(2*sqrt(2)));8 KB (1,172 words) - 21:57, 22 September 2022
- In [[parallelogram]] <math>ABCD</math>, let <math>O</math> be the intersection of [[diagonal]]s <math>\overline{AC}</math> and 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);5 KB (710 words) - 21:04, 14 September 2020
- pathpen = black; pair O = (3.5,3.5); D(O); D(arc(O,1,280,350),EndArrow(4));4 KB (551 words) - 11:44, 26 June 2020
- triple O=(0,0,0), P=(0,0,unit+unit/(r-1)); dot(P); draw(O--P); draw(O--(unit,0,0)--(unit,0,unit)--(0,0,unit)); draw(O--(0,unit,0)--(0,unit,unit)--(0,0,unit));2 KB (257 words) - 17:50, 4 January 2016
- ...math>AE=s</math> and thus circumradius <math>\frac{s}{2}</math>. Let <math>O</math> be its circumcenter. By Inscribed Angles, <math>\angle{BOD'}=2\angle4 KB (609 words) - 22:49, 17 July 2023
- Consider the rhombus <math>OABC</math> on the complex plane such that <math>O</math> is the origin, <math>A</math> represents <math>\text{cis } n^\circ</10 KB (1,514 words) - 14:35, 29 March 2024
