Friday, October 29, 2010

Problem 533: Euclid's Elements Book XIII, Proposition 10

Geometry Problem
Click the figure below to see the complete problem 533 about the sides of the regular pentagon, regular hexagon and regular decagon inscribed in the same circle.

Euclid's Elements Book XIII, Proposition 10, Math Education
See also:
Complete Problem 533

Level: High School, SAT Prep, College geometry

Posted by Antonio Gutierrez at 7:06 AM
Labels: , , , , , ,

7 comments:

  1. Antonio GutierrezOctober 29, 2010 at 7:44 AM

    Volker Poschel, from Germany, sent the following solution/comment:

    A big THANK YOU for the very good site. HERE IS A SOLUTION for Euklids book XIII 10. A very clever inviting proof.
    I think the german words are easy to understand. But its not a "classical" answer - have you one?

    Volker Pöschel
    Germany
    Visit PoeGot's Math-Club at www.poegot.org/www_seite/www_Geometrie_o.htm

    ReplyDelete
  2. Peter TranOctober 29, 2010 at 5:51 PM

    See attached picture in the hyperlink below
    http://yfrog.com/6wproblem533j
    Let H is the intersection of OA to BD and E is the symmetric of A over BD
    Let x=BE=BA=DA and y=BH=HD=1/2BD=1/2AC
    Note that angle DBA=angle EBH=18 , angle EOB=angle OBE= 36 and BE is the angle bisector of angle OBA

    Since BE is the angle bisector so we have x/R=(R-x)/x or R.x=R^2-x^2
    In right triangle OHB we have BH^2=OB^2-OH^2 or y^2=R^2-(R+x)^2/4 or 4.y^2=3.R^2-2.R.x-x^2
    Replace R.x=R^2-x^2 in above expression we have 4.y^2=R^2 +x^2
    And AC^2=BC^2+AB^2 . Triangle ABC is a right triangle

    Peter Tran

    ReplyDelete
  3. Volker PöschelOctober 30, 2010 at 2:16 AM

    Thanx, a good idea.

    ReplyDelete
  4. Volker PöschelNovember 3, 2010 at 2:59 AM

    You can find the solution in german here: http://www.poegot.org/www_seite/Historie/Viereckring.pdf at page 2 und 3

    ReplyDelete
  5. Peter TranNovember 4, 2010 at 11:07 AM

    To Volker Pöschel

    Thank you for clear graphic and additional explanation solution in Germany .

    Peter Tran

    ReplyDelete
  6. sourav mishraMarch 3, 2011 at 12:55 AM

    TO prove the proposition we need to see that in the given figure if we apply Pythagoras theorem then we shall get〖 4R〗^2 〖sin〗^2 18+R^(2 )=4R^2 〖sin〗^2 36.
    Then we need to see whether the values of sin 18 and sin 36 satisfy the above relation.
    For this we need to simplify the expression after which we shall get 〖sin〗^2 36-〖sin〗^2 18=1/4. to see really if it is true then we substitute θ in place of 18 and 36 respectively and then solve the equation〖 sin〗^2 2θ-〖sin〗^2 θ=1/4. By using identity we get sin3θ.sinθ= 1/4. which then simplifies to 3〖sin〗^2 θ-4〖sin〗^4 θ= 1/4 proceeding step by step we get the following:
    12〖sin〗^2 θ-16〖sin〗^4 θ=1. ⇒ 12〖sin〗^2 θ=1+16〖sin〗^4 θ ⇒ 12〖sin〗^2 θ=〖(1+〖4sin〗^2 θ)〗^2- 8〖sin〗^2 θ ⇒20〖sin〗^2 θ=〖(1+〖4sin〗^2 θ)〗^(2 ) ⇒2√5 sinθ= 1+4〖sin〗^2 θ
    ⇒4〖sin〗^2 θ-2√5 sinθ+1=0. solving this equation as quadratic in sin θ we get sin θ = (√5 ±1)/4 now we get the value of θ as 18 degrees (by applying inverse function of sin) and that of 3θ as 54 degrees. So it shows that the expression 〖 4R〗^2 〖sin〗^2 θ+R^(2 )=4R^2 〖sin〗^2 θ is satisfied by the obtained values of θ. Thus the proposition is proved.
    Q. E. D.

    ReplyDelete
  7. Sumith PeirisMay 8, 2019 at 11:52 AM

    Let the centre of the pentagon be X and let the bisector of < AXC meet circle X at Y. Let AY and XC extended meet at Z. Draw CW // to AZ to meet XY at W.

    AY = YC = CZ = CW = XW = l10

    R^2 = l10(l10 + R) from Tr. XYZ .....(1)

    Now l5^2/4 + (R-l10)^2/4 = l10^2
    i.e. l5^2 = (l10^2 + R^2) - (2R^2 - 2l10^2 - 2Rl10^2)

    From (1) the expression in the 2nd bracket of RHS = 0
    So l5^2 = R^2 + l10^2 and hence

    < ABC = 90

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete

Subscribe to: Post Comments (Atom)