Search results

  • pair M=midpoint(A--Ep); pair C=intersectionpoints(Circle(M,2.5),Circle(A,3))[1];
    2 KB (286 words) - 10:16, 19 December 2021
  • ..._{}</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 (484 words) - 13:11, 14 January 2023
  • ...and <math>n</math> is not divisible by the square of any prime. Find <math>m+n</math>. ...}{2}ab\sin C</math> to find this). Now note that <math>\angle FEA = \angle BEC</math> because they are vertical angles, <math>\angle FAE = \angle ECB</mat
    6 KB (974 words) - 13:01, 29 September 2023
  • the point <math>M</math> (midpoint of <math>BD)</math> lies on <math>\ell \implies</math> The point <math>M</math> on the smaller base <math>AC</math> is such that <math>EM = MF.</mat
    50 KB (8,763 words) - 16:41, 26 March 2024
  • ...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 (1,338 words) - 23:15, 28 November 2023
  • ...s <math>AB = BC = CD</math>, <math>m\angle ABC = 70^\circ</math> and <math>m\angle BCD = 170^\circ</math>. What is the degree measure of <math>\angle BA ...AB}</math> and let their intersection be <math>E</math>. Since angle <math>BEC</math> plus angle <math>CYB</math> equals <math>180</math> degrees, quadril
    12 KB (1,944 words) - 17:15, 20 January 2024
  • ...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) - 11:04, 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 (378 words) - 21:38, 19 July 2023
  • ...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) - 13:13, 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) - 11:32, 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 (446 words) - 08:09, 10 April 2023
  • ...the trapezoid into ▭ <math>ADEB</math> and <math>\triangle</math> <math>BEC</math>. We will use the triangle to solve for <math>xy</math> using the Pyt
    5 KB (772 words) - 17:58, 3 March 2024
  • ...lane with <math>A, B, C ,</math> and <math>D</math> such that <math>\angle BEC = 78^{\circ}</math> . Given that <math>\angle EBC > \angle ECB</math>, <mat <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
    1 KB (196 words) - 01:54, 21 December 2021
  • ...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>
    13 KB (2,298 words) - 12:56, 10 September 2023
  • ...and <math>n</math> is not divisible by the square of any prime. Find <math>m+n</math>. ...h>. After simplification, <math>2m^2=17-6k^2</math>. Therefore, <math>k=1, m=\frac{\sqrt{22}}{2}</math>. Finally, note that (using [] for area) <math>[C
    13 KB (2,116 words) - 23:24, 21 March 2024
  • ...h that <math>\angle{FEC}=60^{\circ}</math>. Then <cmath>\angle{BEF}=\angle{BEC}-\angle{FEC}=15^{\circ}</cmath> Since <math>\angle{BEF}=\angle{EBF}</math>, ...s 15</cmath> We know <math>\sin 30 = \frac{1}{2}</math>. In triangle <math>BEC</math>, <math>\sin 15 = \frac{\sqrt{x^2-1}}{x}</math> and <math>\cos 15 = \
    12 KB (1,821 words) - 18:16, 29 October 2023
  • <math>\triangle ADC</math> is congruent to <math>\triangle BEC</math>, so their areas are equal. ...itten as the sum of the figures that make it up, so <math>[ABC] = [ADC] + [BEC] + [CDE]</math>.
    9 KB (1,513 words) - 19:38, 12 November 2023
  • ...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));
    5 KB (854 words) - 16:57, 23 November 2022
  • <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 (656 words) - 01:20, 4 December 2023
  • 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 (654 words) - 11:15, 4 November 2022

View (previous 20 | next 20) (20 | 50 | 100 | 250 | 500)