## Saturday, May 3, 2014

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

Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 1010.

1. 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

2. <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

3. 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