Friday, January 6, 2012

Problem 715: Triangle, Altitudes, Parallel, Circumcircle, Angle, Measurement

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the problem 715 details.

Online Geometry Triangle, Altitudes, Parallel, Circumcircle, Angle, Measurement.

5 comments:

  1. Angle B'A'C' = Pi - 2A = 62 deg
    My guess is x = 62 deg.
    Yet to prove.
    Circle A'B'C'(9-point circle)passes through H

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  2. 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=59
    Now 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

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  3. Proof of Problem 719 completed.
    In the usual notation,
    A’D = A’C’ = R sin 2B
    A’E = R sin 2C
    A’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’E
    Follows
    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/2
    Hence H is the midpoint of BC.

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  4. 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

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  5. 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 - C
    By Sine Rule
    A'C'/sin(B-C) = A'F/sin 2C
    Circle A'B'C' is the nine-point circle and its radius is known to be R/2
    chord 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

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