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  • 2. Note the commonality with [[1969 IMO Problems/Problem 5]]. In fact, of [[1969 IMO Problems/Problem 5]].
    6 KB (1,054 words) - 17:09, 11 December 2024
  • ...E<BF </math> and <math> E </math> between <math> A </math> and <math> F, m\angle EOF =45^\circ, </math> and <math> EF=400. </math> Given that <math> BF=p+q\ ...math>). Then <math>\tan \angle EOG = \frac{x}{450}</math>, and <math>\tan \angle FOG = \frac{y}{450}</math>.
    13 KB (2,080 words) - 12:14, 23 July 2024
  • ...must add up to <math>180^{\circ}</math>. By the [[Law of Cosines]], <math>\angle APB=\cos^{-1}\left(\frac{{-11}}{24}\right)</math>. Also, angles <math>QPA</ ...). In other words, since <math>\angle TP'N = 90^\circ</math>, then <math> \angle TPN = 90^\circ</math>.
    14 KB (2,351 words) - 20:06, 8 December 2024
  • ...s, do not intersect. Suppose that <math>AB=36, AC=72,\,</math> and <math>\angle B=90^\circ.\,</math> Then the area of the set of all fold points of <math> ...they are not on the paper. It follows that <math>\angle APB, \angle BPC, \angle CPA > 90^{\circ}</math>; the [[locus]] of each of the respective conditions
    4 KB (717 words) - 21:20, 3 June 2021
  • ...<math>AB=13, BC=14,</math> and <math>CA=15,</math> and the [[tangent]] of angle <math>PAB</math> is <math>m/n,</math> where <math>m_{}</math> and <math>n_{ ...h>, where <math>m</math> and <math>n</math> are sides on either side of an angle, <math>\beta</math>. So,
    7 KB (1,184 words) - 12:25, 22 December 2022
  • ==Solution 2 Length-chasing (Angle-chasing but for side lengths) == ...re <math>a_4</math> in the previous question to a. We do some "side length chasing" and get <math>4a - 4 = 2a + 5</math>. Solving, we get <math>a = 4.5</math>
    3 KB (485 words) - 23:31, 18 January 2024
  • ...</math> and <math>\angle BCD=90^{\circ}-\frac{\theta}{2}</math>, so <math>\angle DBC=\theta+30^{\circ}</math>. We are now able to use the Law of Sines on <m Cross-multiplying and using double-angle formulae gives that this statement is equivalent to
    6 KB (1,080 words) - 18:28, 21 September 2014
  • ...ex]] of the angle, and the rays themselves are called the [[side]]s of the angle. ...If there is no ambiguity, this notation can be shortened to simply <math>\angle B</math>.
    4 KB (597 words) - 17:39, 9 May 2024
  • <math>\angle GHC = \angle AHB</math> (Vertical angles are equal). <math>\angle F = \angle B</math> (Both are 90 degrees).
    9 KB (1,446 words) - 11:43, 18 September 2024
  • ...= 2</math>. Point <math>E</math> lies on side <math>CA</math> such that [[angle]] <math>DEF = 60^{\circ}</math>. The area of triangle <math>DEF</math> is ...angle CED = \angle AFE</math>. Similarly, we find that <math>\angle EDC = \angle AEF</math>. Thus, <math>\triangle AEF \sim \triangle CDE</math>. Setting up
    4 KB (673 words) - 21:14, 6 August 2022
  • ...chasing, we find out that <math>\angle AEB=75^{\circ}</math>, thus <math>\angle ABE=15^{\circ}</math>. <math>\frac{AE}{EB}=\sin{15^{\circ}}=\frac{\sqrt{6}- [[Category:Introductory Geometry Problems]]
    4 KB (710 words) - 12:59, 1 November 2024
  • {{duplicate|[[2000 AMC 12 Problems|2000 AMC 12 #21]] and [[2000 AMC 10 Problems|2000 AMC 10 #19]]}} WLOG, let a side of the square be <math>1</math>. Simple angle chasing shows that the two right triangles are [[similar triangles|similar]]. Thus
    5 KB (804 words) - 00:22, 13 May 2024
  • ...meet at <math>P</math>, and bisectors of <math>\angle B</math> and <math>\angle C</math> meet at <math>Q</math>. What is the area of hexagon <math>ABQCDP</ ...ac{\angle D+\angle A}{2}=180^{\circ}-\angle APD</math>, so <math>\angle P=\angle Q=90^{\circ}</math>.
    13 KB (2,244 words) - 21:31, 13 October 2024
  • ...isosceles trapezoid]] with <math>\overline{AD}||\overline{BC}</math> whose angle at the longer base <math>\overline{AD}</math> is <math>\dfrac{\pi}{3}</math ...have to be moved even farther outwards from <math>A</math> to maintain the angle of <math>\frac{\pi}{3}</math> and <math>\omega</math> could not touch it, a
    4 KB (648 words) - 16:07, 9 December 2024
  • ...erality, assume <math>AB >AC</math>. It is sufficient to prove that <math>\angle OFA = 90^{\circ}</math>, as this would immediately prove that <math>A,P,O,F ...{DA} \iff \frac{DA}{AE} = \frac{DF}{FE} \iff AF \text{ bisects exterior } \angle EFD
    20 KB (3,565 words) - 10:54, 1 May 2024
  • ...eals <math>\angle NBC = \angle NAC = \alpha</math> and <math>\angle MBC = \angle MAC = 90^\circ - \alpha</math>. Additionally <math>NB = \frac{4}{5}</math> By the Angle Bisector Formula,
    11 KB (1,862 words) - 20:23, 23 May 2024
  • ...and suppose that <math>\triangle CFE</math> is equilateral. What is <math>\angle ACB</math>? Let <math>\angle BAE = \angle ACD = x</math>.
    4 KB (697 words) - 19:33, 19 August 2024
  • ...MB</math> and <math>\angle CMB</math> respectively and <math>\angle AMB + \angle CMB = 180^\circ</math>. By properties of [[tangent (geometry)|tangents]] to ...it's bit easier to use use the right triangle of <math>O_1MO_2</math> than chasing the area ratio. The inradius of <math>\triangle{ABC}</math> can be calculat
    14 KB (2,210 words) - 19:19, 29 September 2024
  • ...th> on <math>\overline{AB}</math> is chosen such that <math>\angle APC = 2\angle ACP</math> and <math>CP = 1</math>. The ratio <math>\frac{AP}{BP}</math> ca ...ath>. It now follows that <math>\angle{DOA} = 2\angle ACP = \angle{APC} = \angle{DPB}</math>. Hence <math>ODP</math> is isosceles and <math>OD = DP = 2</mat
    10 KB (1,507 words) - 23:31, 18 November 2023
  • Let <math>ABC</math> be a triangle with <math>\angle A = 90^{\circ}</math>. Points <math>D</math> ...sides <math>AC</math> and <math>AB</math>, respectively, such that <math>\angle
    7 KB (1,230 words) - 18:47, 31 January 2024

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