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Geometry Problem. Post your solution in the comments box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the figure below to view the complete problem 1045.
Solution by Pradyumna Agashe:Draw perps. BE & CF to AC & BD resply. and call the intersection of the latter two as G.Since CB=CD, CF bisects /_BCD and consequently /_BGE=90° - α and /_GBE = α. By the alternate segment theorem, AB²=BG*BD ---(1) and ///ly. BE²=EG*EC=EG(EG+GC)=EG²+EG*GCSince BCFE is concyclis, EG*GC=BG*GF and thus BE²=EG²+BG*GF= BG² - BE²+EG*GF or 2BE² =BG(BG+GF)=BG*BD/2 or 2*BE²=AB²/2 by (1). In other words, in rt. triangle ABE, BE=AB/2 which makes /_BAE = x = 30°.My thanks to Pradyumna for providing this excellent solution.
Very nice the above solution, but we can polish it little bit: keep the same notations and additionally call M the intersection of BE and CF. See that F is midpoint of BD and E - the one of BM. Since EMFG is cyclic, we have BG*BF=BE*BM ( 1 ). Also, since AB is tangent to the circle (ADG) we get AB*AB=BG*BD=2BG*BF ( 2 ); with (1) we have AB*AB=2BE*BM=4BE*BE, or AB=2BE, getting x=30.Best regards,S. Fulger
I have a different approach. For easy writing let's make alpha = a and 2 alpha = uTriang ACD: sin[90-(a+x}]/CD = sin[90-(2a-x)]/AC which reduces to CD/AC = cos(a+x)/cos2a-x)Triang ABC: sin(x)/BC = sin[180-(a+x)]/AC which reduces to BC/AC = sin(x)/sin(a+x) Note that BC = CDEquating both expressions: sin(a+x)cos(a+x) = sin(x)cos(2a-x) or...... sin(u+2x) = 2sin(x)cos(u-x)after developing and simplifying we get sin(u)cos²(x)= 3sin(u)sin²(x)sin(u) cancels out meaning that x is independent of alpha and finally tan²(x) = 1/3 then x - 30 deg.