Friday, March 5, 2010

Problem 429: Circumscribed and Inscribed Regular Pentagon, Perpendicular, Measurement

Proposed Problem
Click the figure below to see the complete problem 429 about Circumscribed and Inscribed Regular Pentagon, Perpendicular, Measurement.

Problem 429: Circumscribed and Inscribed Regular Pentagon, Perpendicular, Measurement.
See also:
Complete Problem 429

Level: High School, SAT Prep, College geometry

4 comments:

  1. From figure geometry: (b/2)/sin(36) =(a/2)/tan(36) or b=acos(36). Moreover, (c+btan(18))cos(18)=b which gives us: c =b(1-sin(18))/cos(18)=btan(36) or c = a*cos(36)*tan(36)=a*sin(36)
    Thus: b^2+ c^2 = a^2[(cos(36))^2 +sin(36))^2]=a^2 QED
    PS: Will someone pl. explain why (1-sin(18))/cos(18)= tan(36)?

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  2. in [0;90[ 18 is the only root of
    (1-sinx)-cosx.tan2x=0
    4sin18=sqr(5)-1
    4cos18=sqr(10+2sqr5)
    4tan36=(sqr(5)-1).sqr(10-2sqr5)
    this numbers are related with the golden ratio
    .-.

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  3. rotate A'B'C'D'E' so that A' is on AB
    mE'A'O=mE"A"O => O,A",A',E' are concyclic => mA"OE'=pi/2=mEE'O => A"O//E'E, note that mE'A"O=mOEE' so A"OEE' is a //gram and A"E'=OE
    consider OA"E' we have OE^2=A"E'^2=OE'^2+OA"^2
    as three regular pentagons are similar to each others, it yields a^2=b^2+c^2

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  4. The sweetest proof seems to be this...
    Let x, y, z be the perpendicular distances from O to AE, A'E', A''B'’ respectively.
    Note that x = the radius of circle O, so x=OA', and also that z=b/2.
    Since all regular pentagons are similar,
    x:y:z= a:b:c.
    But x^2 = y^2 + z^2 (Pythagoras),
    so a^2 = b^2 + c^2. QED

    Corollary: In fact, z=b/2 =(b/2c)×c;
    so y=(b/2c)×b=b^2/2c and x=(b/2c)×a = ab/2c.
    So if d=diameter of circle O,
    then d=2(ab/2c)=ab/c. ∴ ab=cd. How elegant!

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