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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    
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Solution Solution | Comment 3 of 6 |
LetE(-2,0), F(2,0) and G:(2cosa,2sina) (any point on given circle)
Radius of variable circle:r²=(2-2cosa)²+4sin²a=8(1-cosa)=>r=4sin(a/2)
Area of FHG=1/2*base*height=1/2*r*2sina=4sin(a/2)sina=8sin²(a/2)cos(a/2)
f(x)=x(1-x²) Max occurs at f'(x)=0 and f''(x)<0
f'(x)=1-x²+x(-2x)=>x=+/- 1/√3
f''(x)=-6x So, max for x=+1/√3
Max Area of FHG=8*1/√3*(2/3)=16/(3√3)=~3.0792 sq. units

  Posted by Praneeth on 2008-01-22 05:38:24
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