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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the problem 715 details.
Angle B'A'C' = Pi - 2A = 62 degMy guess is x = 62 deg. Yet to prove.Circle A'B'C'(9-point circle)passes through H
Pravin, you seem to be 100% correct./_BC'C=/_BB'C=90. Hence we conclude that B, B', C' & C lie on a circle whose diameter is BC. Now prove that H is the midpoint of BC and we're done for then HB=HC=HB' and hence /_HB'C=/_HCB'=/_C and /_C'B'A=/_A (for B,C,B' & C' are concyclic) and thus /_C'B'H=/_A=59Now HC'=HB' if H is the centre of our new circle. Then /_HCB' also =59 and thus /_x=62.All we now need to do is to prove that H bisects BC. Problem #714 must come into play in order to prove this.Ajit
Proof of Problem 719 completed.In the usual notation,A’D = A’C’ = R sin 2BA’E = R sin 2CA’F = R sin 2B.sin 2C/sin(B-C) = A’D sin 2C /sin (B – C)F, D, H, E being concyclic,A’F. A’H = A’D. A’EFollows So A’H = A’E sin(B - C) / sin 2C = R sin(B - C)Now BH = A’B + A’H = c cos B + R sin(B - C)= 2R sin C cos B + R sin(B – C)=R(2 sin C cos B + sin (B – C)]= R sin (B + C)= R sin A = a/2 = BC/2Hence H is the midpoint of BC.
Pravin,Please explain how this comes about:A’F = R sin 2B.sin 2C/sin(B-C) = A’D sin 2C /sin (B – C)Ajit
It is known that angles of triangle A'B'C' are 180-2A,180-2B,180-2C resp'ly.Consider triangle A'C'F<A'C'F=180-<A'C'B'=2C<A'FC'=<CFB'=<AB'F-<ACB = B - CBy Sine Rule A'C'/sin(B-C) = A'F/sin 2CCircle A'B'C' is the nine-point circle and its radius is known to be R/2chord A'C' = 2.(R/2).sin <A'B'C'= R sin (180-2B)= R sin 2B etc Awaiting for a shorter proof of Problem 715