Sunday, June 21, 2009

Problem 307: Regular Nonagon, Midpoints, Side, Arc, Angle

Proposed Problem
Click the figure below to see the complete problem 307.

Problem 307: Regular Nonagon, Midpoints, Side, Arc, Angle.
See also:
Complete Problem 307
Collection of Geometry Problems

Level: High School, SAT Prep, College geometry

Posted by Antonio Gutierrez at 5:12 PM
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4 comments:

  1. TopJune 24, 2009 at 9:08 PM

    By the figure geometry, angle (PON)=40. angle(NOC)=20 and angle (NCO)=70 arising from the fact that the internal angle of the nonagon =140 deg. & ON=OC*sin(70). We can, therefore, say by sine rule in triangle OPN: sin(α)/sin(α+40)=(OC/2)/OC*sin(70) or sin(α)/sin(α+40) = 1/(2sin(70)) which can be solved either by MATLAB or by inspection alone as α=30 deg.
    QED

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  2. AnonymousOctober 23, 2010 at 4:56 PM

    Connect PC and CO.
    Revealing that Angle COM=60 Degrees (1.5 Center Angle, Center Angle=40 Degrees), Angle PCO=30 Degrees. But since PC is perpendicular to NO and OC is perpendicular to BC, COPN is cyclic.
    Therefore, Angle PNO=Angle PCO=a=30 Degrees.

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  3. Sumith PeirisSeptember 5, 2015 at 2:43 AM

    Tr. OMC is equilateral of which CP is an altitude. Tr. OBC is isoceles with ON an altitude. So COPN is cyclic and alpha = 30

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  4. Geek37January 4, 2019 at 2:47 AM

    https://www.youtube.com/watch?v=TQi6kQ02R_w

    ReplyDelete

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