Sunday, March 14, 2010

Problem 430: Circumscribed and Inscribed Regular Pentagon, Perpendicular, Areas

Proposed Problem
Click the figure below to see the complete problem 430 about Circumscribed and Inscribed Regular Pentagon, Perpendicular, Areas.

Problem 430: Circumscribed and Inscribed Regular Pentagon, Perpendicular, Areas.
See also:
Complete Problem 430

Level: High School, SAT Prep, College geometry

5 comments:

  1. S/7=OAB-OA'B', S1/7=OA''B''.
    OAB:OA'B':OA''B''=AB^2:A'B'^2:A''B''^2
    AB^2-A'B'^2=A''B''2
    Then, S/7=S1/7
    q.e.d.

    ReplyDelete
  2. name h altitude of A'B'C'D'E'
    from similarity of AOE and A'OE'
    a/R = b/h =>

    R = ah/b (1)

    join E' to B' and name G E'B' meet A"B" we get
    right triangle E'A'G with A'E' = b, A'G = c , GE' =a)
    ( B'B"G is isoceles B'B" = B"G = ...A'A" )

    from similarity of A'OM (M midpoint of A'E')
    and E'A'G ( right tr and angle 36°)
    (b/2)/R = c/a
    substitute R from (1)

    h = b²/2c (2)
    =>
    R = ab/2c (3)

    from similarity of A"ON ( N midpoint of A"E") and GA'E' (c/2)/h1 = c/b ( h1 altitude of pentagon C)

    h1 = b/2 (4)

    Sa = (5∙a ∙R)/2

    Sa = 5a²b/4c ( R from 3)

    Sb = ( 5∙b∙h )/2

    Sb = 5b³/4c ( h from 2)
    =>
    Sj = 5a²b/4c - 5b³/4c = 5b(a² - b²)/4c ( a²-b²=c²)

    Sj = 5bc/4 (5)

    Sc = 5∙c∙h1/2

    Sc = 5bc/4 (6) ( h1 from 4)

    comparing (5) and (6) give the result

    ReplyDelete
  3. the area of a regular pentagon with side s is:
    A=s²(5/4)cotan36
    and we know (previous problem) that c²=a²-b²....
    .-.

    ReplyDelete
  4. Masterstream said…

    The sweetest proof seems to be this:-

    Let a,b,c be the side-lengths of pentagons ABCDE, A'B'C'D'E', A''B''C''D''E'' respectively, as in Problem 429(q.v.); and let S2,S3 be the areas of the inscribed regular pentagon A'B'C'D'E' and the circumscribed regular pentagon ABCDE respectively.

    Since all regular pentagons are similar,
    S3:S2:S1 = a^2:b^2:c^2.

    So S:S1 = (S3–S2):S1 = (a^2–b^2):c^2 = 1 ,
    (since a^2–b^2 = c2, by Problem 429 [q.v.])

    i.e. S = S1 . QED

    ReplyDelete