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  • pair M=midpoint(A--Ep); pair C=intersectionpoints(Circle(M,2.5),Circle(A,3))[1];
    2 KB (296 words) - 19:15, 25 December 2020
  • ..._{}</math> is not divisible by the square of any prime number. Find <math>m+n+d.</math> ...], so <math>\angle OEB = \angle OED = 60^{\circ}</math>. Thus <math>\angle BEC = 60^{\circ}</math>, and by the [[Law of Cosines]],
    3 KB (443 words) - 18:31, 4 July 2013
  • ...th>\angle A = 37^\circ</math>, <math>\angle D = 53^\circ</math>, and <math>M</math> and <math>N</math> be the [[midpoint]]s of <math>\overline{BC}</math pair[] M = intersectionpoints(N--E,B--C);
    8 KB (1,206 words) - 15:43, 17 June 2021
  • ...is there at least 1 positive integer <math>n</math> such that <math>mn \le m + n</math>? ...er of <math>\triangle AED</math> is twice the perimeter of <math>\triangle BEC</math>. Find <math>AB</math>.
    11 KB (1,733 words) - 10:04, 12 October 2021
  • ...er of <math>\triangle AED</math> is twice the perimeter of <math>\triangle BEC</math>. Find <math>AB</math>. ...er of <math>\triangle AED</math> is twice the perimeter of <math>\triangle BEC</math>.
    2 KB (317 words) - 14:24, 13 March 2021
  • ...nd <math>m</math> minutes, with <math>0 < m < 60</math>, what is <math>h + m</math>? ...llowing is equal to <math>12^{mn}</math> for every pair of integers <math>(m,n)</math>?
    13 KB (2,105 words) - 12:13, 12 August 2020
  • ...llowing is equal to <math>12^{mn}</math> for every pair of integers <math>(m,n)</math>? \mathrm{(B)}\ P^nQ^m
    14 KB (2,130 words) - 10:32, 7 November 2021
  • ...e from point <math>E</math> of <math>\triangle AEB</math>, <math>\triangle BEC</math>, <math>\triangle CED</math>, <math>\triangle DEA</math>. ...EWX+m\angle EYX+m\angle EYZ+m\angle EWZ=</math><math>360^\circ-m\angle CED-m\angle AEB=180^\circ</math>.
    2 KB (439 words) - 06:54, 19 July 2016
  • ...rtitions the trapezoid into rectangle <math>ADEB</math> and triangle <math>BEC</math>. We will use the triangle to solve for <math>x \times y</math> using
    5 KB (746 words) - 23:56, 31 May 2021
  • <math>\angle ABE</math> is external to <math>\triangle BEC</math> at <math>\angle B</math>. Therefore it is equal to the sum: <math>\a The two angles sum to <math>102^\circ</math>, thus <math>m\angle ECB < 51^\circ</math>
    721 bytes (117 words) - 18:46, 17 July 2011
  • ...ath> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>. ...and <math>\omega</math> are swapped. Thus points <math>F</math> and <math>M</math> map to each other, and are isogonal. It follows that <math>AF</math>
    10 KB (1,676 words) - 21:39, 4 August 2021
  • ...ngles <math>\triangle ABD</math> and <math>\triangle BCE</math>. Let <math>M</math> be the midpoint of <math>\overline{AE}</math>, and <math>N</math> be ...0, 0), B = (16, 0), C = (20, 0), D = (8, 8*sqrt(3)), EE = (18, 2*sqrt(3)), M = (9, sqrt(3)), NN = (14, 4*sqrt(3));
    4 KB (686 words) - 22:51, 31 August 2021
  • <cmath>\frac{1}{2}(AE)(AD)(\sin(m\angle EAD)).</cmath> But, note that <math>\sin(m\angle EAD)=\sin(m\angle CAD)=\frac{O}{H}=\frac{3}{5}</math>. Thus, we see that
    4 KB (630 words) - 16:50, 29 October 2021
  • Drawing it out, we see <math>\angle BDC</math> and <math>\angle BEC</math> are right angles, as they are inscribed in a semicircle. Using the f <cmath>\angle BEC+\angle EBC+\angle ECB=180^{\circ}\implies90^{\circ}+70^{\circ}+\angle ECB=1
    4 KB (641 words) - 01:57, 20 September 2021
  • ...math>EP=EQ=2 \sqrt{5}</math>. Now by the Law of Cosines in <math>\triangle BEC</math>, we have <math>\cos{\left(\angle BEC\right)}=\frac{EB^2+EC^2-BC^2}{2 \cdot EB \cdot EC}=\frac{5}{\sqrt{30}}</mat
    5 KB (871 words) - 00:09, 7 November 2021
  • pair A,B,C,DD,EE,FF, M; B = (0,0); C = (3,0); M = (1.45,0);
    24 KB (3,745 words) - 18:04, 6 October 2021
  • We could use the famous m-n rule in trigonometry in <math>\triangle ABC</math> with Point <math>E</ma <cmath>[AEB] + [BEC] = \frac{1}{2}(x)(5)\sin(\angle DEC) + \frac{1}{2}(x)(15)\sin(180-\angle DE
    11 KB (1,658 words) - 21:53, 6 November 2021
  • ...ath> and <math>n</math> are relatively prime positive integers. Find <math>m+n.</math> ...ta.</math> It follows that <math>\angle CED=\theta</math> and <math>\angle BEC=\angle DEA=180^\circ-\theta.</math>
    12 KB (1,929 words) - 07:01, 4 October 2021
  • ...> where <math>m</math> and <math>n</math> are relatively prime, find <math>m+n.</math> .../math> We claim that <math>EF</math> is the angle bisector of <math>\angle BEC</math>.
    7 KB (744 words) - 20:18, 11 July 2021
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