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  • ...o hours it's left stuck to the wall. One morning, at around <math>9</math> o' clock, Tony sticks the spider to the wall in the living room three feet ab
    71 KB (11,749 words) - 01:31, 2 November 2023
  • ...math>AB=BC=CD=DE=EA=2</math> and <math>\angle ABC=\angle CDE=\angle DEA=90^o</math>. The plane of <math>\triangle ABC</math> is parallel to <math>\overl Now, note that <math>\angle CDE=\angle DEA=90^o</math>. This means that there exists some vector <math>DE</math> parallel t
    3 KB (470 words) - 19:46, 17 July 2023
  • label("$O$", (0,0), SW); ...a semicircle of area <math> 2\pi</math>. The circle has its center <math> O</math> on hypotenuse <math> \overline{AB}</math> and is tangent to sides <m
    13 KB (1,821 words) - 22:18, 5 December 2023
  • ...ABC</math> to <math>\triangle A'B'C', T(ABC) = A'B'C'.</math> Denote <math>O</math> the center of <math>T.</math> ...this center is point <math>O,</math> so these circles contain point <math>O</math>. Similarly for another circles.
    28 KB (4,863 words) - 00:29, 16 December 2023
  • pair A, B, C, D, O, P, Q, R, SS; O = (0,0) ;
    2 KB (410 words) - 14:01, 4 March 2023
  • ...90^\circ</math>, either <math>O</math> is on side <math>BC</math> or <math>O</math> and <math>A</math> are on opposite sides of line <math>BC</math>. In
    8 KB (1,470 words) - 22:24, 18 June 2022
  • ...h> ions from the acid (proton donor) react with <math>OH^-</math> or <math>O^{2-}</math> ions from the base (proton acceptor) to form water. The cations
    358 bytes (65 words) - 20:10, 23 January 2017
  • Using the letters <math>A</math>, <math>M</math>, <math>O</math>, <math>S</math>, and <math>U</math>, we can form five-letter "words"
    10 KB (1,539 words) - 13:34, 23 July 2024
  • Using the letters <math>A</math>, <math>M</math>, <math>O</math>, <math>S</math>, and <math>U</math>, we can form five-letter "words" Let <math>A = 1</math>, <math>M = 2</math>, <math>O = 3</math>, <math>S = 4</math>, and <math>U = 5</math>. Then counting backw
    1 KB (178 words) - 04:20, 4 November 2022
  • pair O=(0,0); path inner=Circle(O,r1), outer=Circle(O,r2);
    14 KB (2,137 words) - 15:29, 9 June 2024
  • pair O=(0,0), E=dir(0), NE=dir(60), NW=dir(120); draw(O -- (O+E) -- (O+E+4*NE) -- (O+E+4*NE+2*NW) -- (O-3*E+4*NE+2*NW));
    4 KB (654 words) - 22:49, 5 August 2024
  • ...of a regular tetrahedron <math>ABCD</math>. Prove that angle <math>PAQ<60^o</math>.
    2 KB (273 words) - 18:53, 3 July 2013
  • ...rically opposite A, and let <math>Q</math> be the plane through the center O of the sphere perpendicular to <math>BB'</math> and passing through the mid
    2 KB (360 words) - 23:13, 18 July 2016
  • ...f points A, B, C, and D, respectively, with respect to an arbitrary origin O. Let us also for simplicity define <math>a^2 = a \cdot a = ||a||^2</math>,
    6 KB (1,096 words) - 23:07, 26 August 2017
  • ...XY is our axis of symmetry and it intersects with CD at a point O. Point O is our origin of reference whose coordinates are (0,0). Let our point P be on the axis of symmetry at z distance from the origin O.
    2 KB (410 words) - 15:25, 23 March 2020
  • ...ry triangle <math>ABC</math>. Let <math>ABC</math> have circumcenter <math>O</math> and incenter <math>I</math>. Extend <math>AI</math> to meet the circ
    2 KB (308 words) - 06:29, 16 December 2023
  • ...h>I\cdot M \cdot O=2001</math>. What is the largest possible sum <math>I+M+O</math>?
    14 KB (2,035 words) - 20:19, 14 October 2024
  • ...X=r*expi(pi/3), X1=r*expi(-pi/12), Y=r*expi(4*pi/3), Y1=r*expi(11*pi/12), O=(0,0), P, P1; draw(pica, circ1);draw(pica, B--A--P--Y--X);dot(pica,P^^O);
    5 KB (848 words) - 23:41, 6 July 2020
  • ...ahedron <math>PABC</math> (i.e., <math>\angle APB=\angle BPC=\angle CPA=90^o</math>) is <math>S</math>, determine its maximum volume.
    2 KB (358 words) - 23:15, 18 July 2016
  • ...X=r*expi(pi/3), X1=r*expi(-pi/12), Y=r*expi(4*pi/3), Y1=r*expi(11*pi/12), O=(0,0), P, P1; draw(pica, circ1);draw(pica, B--A--P--Y--X);dot(pica,P^^O);
    3 KB (510 words) - 19:01, 3 July 2013

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