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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.This entry contributed by Ajit Athle.Click the figure below to see the problem 761 details.
http://img835.imageshack.us/img835/6848/problem761.pngDraw additional lines per attached sketchNote that ∆ABN is equilateral and BG/BM=BQ/BS= 2/3 => QG⊥BQAnd HG=2.GO and HQ=2.QR …. (Property of Euler line)OR=RQ. √3=1/3 . HR. √3So tan(x)= OR/HR= 1/√3 => x=30And triangle ADE is equilateral
Let X and Y be the midpoints of AC, ABLet L, M be the feet of the altitudes from B, C1/2 = cos 60° = AL/AB = AL/c. So AL = c/2 = AY1/2 = cos 60° = AM/AC = AM/b. So AM = b/2 = AXFollows LX = AX - AL = (b - c)/2 ; MY = AM - AY = (b - c)/2 So LX = MY = d say Next ∆ACM is right angled at M and ∠ACM = 30°(i.e.)∠LCH = 30°So 1/2 = sin 30° = HL/CH, HL = CH/2 = OY by a well known propertySimilar to HL = OY, we have HM = OXFor convenience denote OX = x, OY = yNext HL ∥ OX implies LE/XE = HL/OX = y/x, 1 + LX/XE = y/x, LX/XE = (y - x)/x,(y - x).XE = x. LX = x.dSimilarly HM ∥ OY implies DY/DM = OY/HM = y/x, 1 + MY/DM = y/x, MY/DM = (y - x)/x,(y - x).DM = x. MY = x.dThus XE = DMHence finally,AD = AM + MD = b/2 + MD = AX + XE = AE and so ∆ADE is equilateral(minor typos, if any may be excepted)
Since m(A) = 60 => m(COB) = 120m(CHB) = 180-(B-30)-(C-30) = 120So BHOC are concyclicTherefore m(OHC) = m(OBC) = 30 ( Since triangle OBC is isosceles triangle with OBC=OCB=30 and COB =120)Also m(HCA) = 30 (extend CH to meet AB at P and PCA is a 30-60-90 triangle)So triangle EHC is an isosceles triangle with m(EHC) = m(OHC) = 30 = m(HCE)So m(HEC) = 120=> m(HEA) = 60Therefore DAE is equilateral