Search results

  • ...ion of the circumcircles of <math>\triangle AMN</math> and <math>\triangle ADE</math>. Ray <math>AP</math> meets <math>BC</math> at <math>Q</math>. The ra
    8 KB (1,246 words) - 20:58, 10 August 2020
  • ...ion of the circumcircles of <math>\triangle AMN</math> and <math>\triangle ADE</math>. Ray <math>AP</math> meets <math>BC</math> at <math>Q</math>. The ra ...ince <math>Q</math> lies on the radical axis of <math>(AMN)</math>, <math>(ADE)</math> thus <cmath>\frac{BQ}{CQ}=-\frac{f(B)}{f(C)}=-\frac{BN\cdot BA-BE\c
    9 KB (1,536 words) - 17:54, 30 August 2024
  • ...th>\angle FBA = \angle FAB = \angle FAD = \angle FCD = \angle DAE = \angle ADE = \alpha</math>. And WLOG, <math>MF = 1</math>. Hence, <math>CF = 2</math>, ...th> are concyclic. We know that <math>DE || MC</math> and <math>DE = 1 = MC</math>, so we conclude <math>MCDE</math> is parallelogram. So <math>\angle
    3 KB (567 words) - 00:00, 15 November 2024
  • ...</math>. If there exists a point <math>P</math> outside of <math>\triangle ADE</math> such that <math>AP=PD=28</math>, and there exists a point <math>O</m [[G285 2021 MC-IME I Problem 1|Solution]]
    9 KB (1,577 words) - 22:28, 28 June 2021
  • ...ete quadrilateral. In the diagram these are <math>\triangle ABC, \triangle ADE, \triangle CEF, \triangle BDF.</math> <math>AD||GF \implies \angle ADE + \angle DFG = 180^\circ \implies \angle ADE + \angle AME = 180^\circ \implies</math>
    7 KB (1,213 words) - 15:18, 21 October 2024
  • <cmath>\frac {AM}{MC}= \frac {AB}{BD}.</cmath> <cmath>BM = MC \implies \frac {BD}{CD} = \frac{AB}{AC}.</cmath>
    30 KB (5,152 words) - 16:38, 6 August 2024