
The impossible circle strikes again
Posted:
Jun 21, 2009 11:03 PM



IA #34 reminds me of an old Math B question. Who remembers June 2004, #33? http://www.nysedregents.org/testing/mathre/regentmathb.html
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 halfchord (by Pyth) is radical 48.
Oops.
Thanks to Pete G. for pointing this out.
Jonathan Halabi HS of American Studies the Bronx

