Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP Posting-Version: version B 2.10 5/3/83 based; site hou2f.UUCP Path: utzoo!watmath!clyde!cbosgd!ihnp4!houxm!hou2f!ven From: ven@hou2f.UUCP (V.VENKATESWARAN) Newsgroups: net.puzzle Subject: Re: Geometry problem Message-ID: <587@hou2f.UUCP> Date: Mon, 18-Nov-85 09:34:31 EST Article-I.D.: hou2f.587 Posted: Mon Nov 18 09:34:31 1985 Date-Received: Tue, 19-Nov-85 04:16:50 EST References: <382@faron.UUCP> Organization: AT&T Bell Labs, Holmdel NJ Lines: 60 Let angle A = 2a , angle B = 2b By the sine rule on AEB get AB = AE sin(180-2b-a)/sin(2b) By the sine rule on ADB get BD = AB sin(2a)/sin(180-2a-b) BD = AE . sin2a . sin(2b+a) ------- --------- sin2b sin(2a+b) We will show that the only solution for sin2b sin(2a+b) = 1 is a = b. ---------------- sin2a sin(2b+a) wlog let a < 45 , a <= b and let b = a+d let f(d) = sin(2a+2d) sin(3a+d) ---------------------- sin2a sin(3a+2d) f(0) = 1 . if (2a+2d) <= 90 (2a+2d) > 2a ; else, 180-2a-2d > 2a hence sin(2a+2d) --------- > 1 sin2a if 3a+d > 90 then sin(3a+d) ---------- also > 1 sin(3a+2d) So examine f(d) for d in 0 to D = 90-3a df/dd = [sin(3a+2d)[2cos(2a+2d)sin(3a+d) + sin(2a+2d)cos(3a+d)] - 2cos(3a+2d)sin(2a+2d)sin(3a+d) ] / C where C = sin(2a)sin(3a+2d)sin(3a+2d) > 0 for d in [0,D] case1 2a+2d > 90 ----- Consider 2sin(3a+2d)cos(2a+2d)sin(3a+d) and -2sin(2a+2d)cos(3a+2d)sin(3a+d) -cos(3a+2d) > 0 and > cos(2a+2d) sin(2a+2d) > sin(3a+2d) The term sin(3a+2d)sin(2a+2d)cos(3a+d) >= 0 for d in [0,D] So C.df/dd > 0 case 2 2a+2d =< 90 ------ C.df/dd = sin(2a+2d)[sin(3a+2d)cos(3a+d)-cos(3a+2d)sin(3a+d)] +sin(3a+d)[sin(3a+2d)cos(2a+2d)-cos(3a+2d)sin(2a+2d)] +sin(3a+2d)sin(3a+d)cos(2a+2d) = sin(2a+2d)sin(d) + sin(3a+d)sin(a) + sin(3a+2d)sin(3a+d)cos(2a+2d) all quantities are >= 0 and sin(3a+d)sin(a) > 0 So for d in [0,D] df/dd > 0 . By the mean value thm. f(d) cannot equal 1 in (0,D]. So there is only one soln. (Case 1 verifies that if angle B is obtuse, AE can never equal BD.)