Proposed Problem
See also: Complete Problem 284, Collection of Geometry Problems
Level: High School, SAT Prep, College geometry
Saturday, April 25, 2009
Problem 284: Circular Sector 90 degrees, Semicircles, Tangent, Radius
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The two inner circles are given by:
ReplyDelete(x-R/2)^2+y^2=R^2/4
x^2+(y-R+r)^2=r^2
Solve these simultaneously and set the discriminant to zero for the circles to touch each other. This gives us:-R^6+4rR^5-3r^2R^4=0 or -R^2+4rR -3r^2 =0 which can be solved to obtain r = R/3 or r=R of which only the first solution is admissible.
Ajit: ajitathle@gmail.com
We can use Pythagorean Theorem to solve it.
ReplyDeleteSince D, E, C are collinear,OD=R-r, OR=R/2 and DC=r+R/2,
So solving (R-r)^2+(R/2)^2=(r+R/2)^2 will give r=R/3.
See the drawing
ReplyDeleteDefine h=OF
OA=OB=R
D, E and C are aligned
ΔCOD is right in O => DC^2=OC^2+OD^2
OD=r+h
=> (1) : (r+R/2)^2=(R/2)^2+(r+h)^2
=> (1) : (r+R)^2=R^2+4(r+h)^2
OB=OA => R=2r+h => r+h=R-r
(1) : (r+R)^2=R^2+4(R-r)^2
=> r^2+2rR+R^2=R^2+4(R^2-2rR+r^2)
=> 6rR= 3r^2
Therefore 3R=r