Proposed Problem

Click the figure below to see the complete problem 307.

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Complete Problem 307

Collection of Geometry Problems

Level: High School, SAT Prep, College geometry

## Sunday, June 21, 2009

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

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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.

ReplyDeleteQED

Connect PC and CO.

ReplyDeleteRevealing 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.

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

ReplyDeleteSumith Peiris

Moratuwa

Sri Lanka