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

Labels:
90,
circular sector,
radius,
semicircle,
tangent

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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

ReplyDeletedrawingDefine 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