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• 1985 IMO Problems/Problem 5
  ...ath> is cyclic. <math>OM_1 B M_2</math> is also cyclic with diameter <math>BO</math> and thus <math>M</math> must lie on the same circumcircle as <math>B The radius of <math>\omega</math> is <math>MO'' = BO'' = \frac {BN}{2 \sin \gamma}.</math>
  3 KB (496 words) - 13:35, 18 January 2023
• 2006 AIME I Problems/Problem 14
  <cmath>BO=\sqrt{TB^2-TO^2}=3</cmath>
  6 KB (980 words) - 21:45, 31 March 2020
• 2005 AIME II Problems/Problem 12
  Construct <math>BO, AO.</math> Let <math>\angle{FOB} = \alpha.</math> Also let <math>FB = x</m Let <math>FB = x</math>. Then <math>EM = 400+x</math>. Since <math>FB/BO = \frac{x}{450\sqrt{2}}</math>, and <math>MO/EM = FB/OB</math>, we have <cm
  13 KB (2,080 words) - 13:14, 23 July 2024
• 1992 AIME Problems
  ...ac{BO}{OB'}+\frac{CO}{OC'}=92</math>, find <math>\frac{AO}{OA'}\cdot \frac{BO}{OB'}\cdot \frac{CO}{OC'}</math>.
  8 KB (1,117 words) - 05:32, 11 November 2023
• 1983 AIME Problems/Problem 15
  Since <math>BO = 5</math>, <math>BM = 3</math>, and <math>\angle BMO</math> is right, <mat
  20 KB (3,497 words) - 15:37, 27 May 2024
• 1992 AIME Problems/Problem 14
  ...ac{BO}{OB'}+\frac{CO}{OC'}=92</math>, find <math>\frac{AO}{OA'}\cdot \frac{BO}{OB'}\cdot \frac{CO}{OC'}</math>. <cmath>\frac{AO}{OA'}=\frac{K_B+K_C}{K_A}</cmath> <cmath>\frac{BO}{OB'}=\frac{K_A+K_C}{K_B}</cmath> <cmath>\frac{CO}{OC'}=\frac{K_A+K_B}{K_C}
  4 KB (667 words) - 01:26, 16 August 2023
• 1995 AIME Problems/Problem 12
  ...\theta</math> is the angle formed by two [[perpendicular]]s drawn to <math>BO</math>, one on the plane determined by <math>OAB</math> and the other by <m
  8 KB (1,172 words) - 21:57, 22 September 2022
• Mock AIME 2 2006-2007 Problems/Problem 12
  ...onals of <math>ABCD</math>. If <math>AD=4,</math> <math>BC=6</math>, <math>BO=1,</math> and the [[area]] of <math>ABCD</math> is <math>\frac{a\sqrt{b}}{c Also, from the Power of a Point Theorem, <math>DO \cdot BO=AO\cdot CO\Rightarrow CO=\frac{3a}{2}</math>
  2 KB (311 words) - 10:53, 4 April 2012
• Mock AIME 2 2006-2007 Problems
  ...DB]}{[ABC]}=\frac12.</math> If <math>AD=4,</math> <math>BC=6</math>, <math>BO=1,</math> and the area of <math>ABCD</math> is <math>\frac{a\sqrt{b}}{c},</
  5 KB (848 words) - 23:49, 25 February 2017
• Qq808casino
  ...ll the popular games like baccarat, blackjack, roulette, dragon tiger, sic bo, fan tan and more. Besides, there is a selection of providers where you can
  2 KB (278 words) - 12:34, 21 September 2024
• Euler line
  <math>CO' = HO', BO = OH \implies OO'</math> is the midline of <math>\triangle BHC \implies \tr <math>CO' = HO', BO = OH \implies OO'</math> is the midline of <math>\triangle BHC \implies \tr
  59 KB (10,203 words) - 04:47, 30 August 2023
• 1985 IMO Problems/Problem 1
  Since <math>AO=EO=BO</math>, we obtain two squares, <math>AOED</math> and <math>BOEC</math>. <math>AD+BC=AO+BO=AB</math>
  4 KB (684 words) - 07:28, 3 October 2021
• 2007 AIME II Problems/Problem 3
  ...BE</math> is a cyclic quadrilateral. Pythagorean Theorem yields that <math>BO=AO=\frac{13\sqrt{2}}{2}</math>. Now, using Ptolemy's Theorem, we get that <cmath>AO\cdot BE + BO\cdot AE = AB\cdot AO</cmath>
  6 KB (933 words) - 00:05, 8 July 2023
• Circumradius
  .../math>, <math>BC=a</math>, <math>AC=b</math>, <math>BE=h</math>, and <math>BO=R</math>. We know that <math>\angle BAD</math> is a right angle because <ma
  4 KB (712 words) - 19:27, 17 September 2024
• 1989 USAMO Problems/Problem 4
  ...ss through the circumcenter <math>O</math>. Extend <math>AO</math>, <math> BO</math>, <math>CO</math> to <math>D</math>,<math>E</math>,<math>F</math> on Since <math>AO = BO = CO = R</math>, we let <math>\alpha = \angle OAC = \angle OCA</math>, <mat
  2 KB (290 words) - 19:11, 18 July 2016
• Asymptote: Useful functions
  ...angle AOB</math>. The function will draw an arc from line AO or l1 to line BO or l2, counterclockwise, with radius <math>|\textrm{radius}|</math> and lab
  4 KB (645 words) - 05:21, 13 August 2024
• 2000 AMC 12 Problems/Problem 17
  ...heta}=\frac{\overline{AB}}{\overline{BO}}, \cos{\theta}=\frac{1}{\overline{BO}}, \tan{\theta}=\overline{AB}</cmath>
  7 KB (1,072 words) - 00:38, 20 August 2024
• 2003 AMC 12B Problems/Problem 22
  <cmath>\frac{1}{2} AO \cdot BO = 60 = \frac{1}{2} ON \cdot AB = \frac{\sqrt{8^2 + 15^2}}{2} \cdot ON \Long
  4 KB (551 words) - 14:17, 23 June 2022
• 2003 AMC 12B Problems/Problem 21
  ...t <math>\triangle BDO</math> is a <math>30-60-90</math> triangle, as <math>BO = 5</math>, <math>BD = \frac{5 \sqrt{3}}{2}</math>, <math>DO = \frac52</mat
  3 KB (461 words) - 10:23, 20 September 2024
• 2008 AIME I Problems/Problem 14
  From the diagram, we see that <math>BQ = OT + BO \sin\theta = 9 + 9\sin\theta = 9(1 + \sin\theta)</math>, and that <math>QP
  8 KB (1,339 words) - 13:29, 3 September 2024

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