Saturday, May 3, 2014

Geometry Problem 1010: Regular Nonagon or Enneagon, Diagonals, Metric Relations

Geometry Problem. Post your solution in the comments box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 1010.

Online Math: Geometry Problem 1010: Regular Nonagon or Enneagon, Diagonals, Metric Relations

Posted by Antonio Gutierrez at 11:38 AM
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3 comments:

  1. Jacob HA (EMK2000)May 3, 2014 at 6:25 PM

    Applying intersecting chords theorem on the circumcircle of the nonagon,
    CJ×DJ = AJ×EJ

    But EJ=AK, DJ=AG, CD=DK, thus
    CJ = CD+DJ = DK+DJ = DK+AG

    CJ×DJ = (DK+AG)×DJ = DK×DJ+AG×DJ = DK×DJ+AG²
    AJ×EJ = AJ×AK

    Hence,
    AJ×AK = DJ×DK+AG²
    AJ×AK − DJ×DK = AG²

    ReplyDelete
  2. Ivan BazarovMay 5, 2014 at 5:26 PM

    <KED=<KDE=1/2*3/9*360=60 so DEK is equilateral. <DAE=1/2*1/9*360=20 and <CDA=1/2*2/9*360=40 so ADJ is isosceles. AGD is also equilateral. Then AJ*AK=EJ*AJ=DJ*CJ=DJ(DJ+CD)
    =DJ^2+DJ*CD=AG^2+DJ*CD=AG^2+DJ*DK

    ReplyDelete
  3. Sumith PeirisSeptember 5, 2015 at 10:50 PM

    We can show that Tr. ADJ is isoceles and Tr. KDE is equilateral and that Tr.s ADK & DEJ are congruent.

    Now since Tr. ACJ & DEJ are similar CJ / EJ = AJ / DJ. So ( DK + DJ) / AK = AJ / DJ. Therefore DJ. DK + DJ ^2. = AK.AJ.

    Hence AJ. AK - DJ. DK = DJ^2 = AD^2 = AG^2.

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete

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