All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Shapes > Geometry
Two Circles And Maximum Triangle (Posted on 2008-01-11) Difficulty: 3 of 5
The circle x2 + y2 = 4 intersects the x axis respectively at the points E and F. A different circle with variable radius with its center located at F cuts the former circle at point G located above x axis and intersects the line segment EF at the point H.

Determine the maximum area of the triangle FHG.

See The Solution Submitted by K Sengupta    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Solution | Comment 1 of 6

First circle,
   x^2 + y^2 = 2^2
Second circle,
   (x - 2)^2 + y^2 = r^2, where r < 4.
Solving these for point G(x,y),
        8 - r^2
   x = ---------
           4
        r*sqrt(16 - r^2)
   y = ------------------
               4
Area of triangle FGH,

   A = (1/2)(base)(altitude) = (1/2)(r)(y)
        r^2*sqrt(16 - r^2) 
     = --------------------
                8
Setting dA/dr = 0 and solving for r^2 gives,
   r^2 = 32/3
Plugging this into the area formula gives.
   A = 16*sqrt(3)/9 ~= 3.0792
This agrees with Geometer's Sketchpad.
 

  Posted by Bractals on 2008-01-11 12:33:02
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (6)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information