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Topic: The impossible circle strikes again
Replies: 11   Last Post: Jun 23, 2009 5:49 PM

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Jonathan Halabi

Posts: 119
Registered: 8/16/08
The impossible circle strikes again
Posted: Jun 21, 2009 11:03 PM
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IA #34 reminds me of an old Math B question. Who remembers June 2004, #33?

In this case, a chord parallel to the diameter is 3/4 the length of the
diameter. Call the Diameter 16, the radius 8, the chord 12. From the center
of the circle to the endpoint of the chord is a radius, 8. Drop a
perpendicular from the endpoint of the chord to the diameter. It meets the
diameter 6 units from the center. By Pythagoras, the perpendicular is
2sqr(7) units long.

Or, try this. Let the radius be 8, the perpendicular be 4, then the
half-chord (by Pyth) is radical 48.


Thanks to Pete G. for pointing this out.

Jonathan Halabi
HS of American Studies
the Bronx

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