Thursday, April 23, 2009

Problem 283: Circular Sector 90 degrees, Semicircle, Circle inscribed, Radius

Proposed Problem

 Problem 283: Circular Sector 90 degrees, Semicircle, Circle inscribed, Radius.

See also: Complete Problem 283, Collection of Geometry Problems

Level: High School, SAT Prep, College geometry

5 comments:

  1. With O as origin, the smallest circle with its centre at (r,h) is represented by (x-r)^2+(y-h)^2=r^2 while the circle with centre at C is:
    (x-R/2)^2+y^2=R^2/4. These two equations when solved simultaneously give: x=(2r+R)h^2 + or -2h^2(2rR-h^2)^(1/2)/(4(h^2+r^2)+R^2-4rR).
    For them to be touching each other at one point (and not intersect) we can conclude that h^2 =2rR. Like wise, the two circles given by:(x-r)^2+(y-h)^2=r^2 and x^2+y^2=R^2 are touching at a point if only if -r^2h^4+2(rRh)^2-r^2R^4+4r^4R^2 =0 where h^2 = 2rR gives us, -4r^4*R^2 + 4r^4*R^2 + 4r^3*R^3 - r^2*R^4 = 0 or 4r^3*R^3 = r^2*R^4 or r = R/4
    Ajit: ajitathle@gmail.com

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  2. can someone give a better answer involved with circle geometry please!!! urgent!!!

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  3. i need like a step by step way to complate this + reasoning.. i dont really understand the above comment... thanks :)

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  4. Applying Euclid's Book II, 13th Proposition to Tr. OCD:(R/2+r)^2 =(R/2)^2+ OD^2 - rR. Now substitute OD = (R - r) in this equation to obtain: R = 4r,QED.

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