Proposed Problem

See also: Complete Problem 285, Collection of Geometry Problems

Level: High School, SAT Prep, College geometry

## Saturday, April 25, 2009

### Problem 285: Circular Sector 90 degrees, Semicircles, Circle, Tangent, Radius

Labels:
circle,
circular sector,
common tangent,
radius,
semicircle,
tangent

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We can use Descarte's Theorem for 4 mutually tangent circles: 2(a^2+b^2+c^2+d^2)=(a+b+c+d)^2

ReplyDeletewhere a,b.c & d are the circle curvatures given by a=-1/R (the curvature is negative for the outermost circle),b=2/R,c=3/R (by problem 284)and d=1/r. On substitution we get, d=2c or r=R/6

Ajit: ajitathle@gmail.com

Simpler:

ReplyDeleteBuild semicircles with centres C and D.

Build a circle with center F' with radius R/6 and touching the first two.

Extent OF' with another R/6 to a point Z

But OF' = CD = 5/6R (OCF'D is a rectangle ; radius of the D semicircle is R/3)

Now OZ is equal to OB, therefore Z is on the O quarter circle (since the centers of the O circle and the F' circles and Z are on the same line)

M.

Even simpler is the following: Let AD = DE = a. (R/2+a)^2=(R-a)^2+(R/2)^2 giving a = R/3. Now, 2R/3 = OD = CF = R/2 + r or r = R/6

ReplyDeleteAjit, this is only valid if you can show tht DFCO is a rectangle, which you have not shown.

ReplyDeleteHow would you show that DFCO is a rectangle?

ReplyDeleteCenters and tangency points are colinear...

ReplyDelete