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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 2 of 6 |

The second circle serves to tell us that the resulting traingle FHG is isoscelese, with one of its equal sides lying coincident with the x-axis and therefore with the diameter of the first circle.

For purposes of finding the area consider the base to be segment HF, and altitude measured from point G to that line.

The altitude intersects HF at some point (x,0), and G has coordinates (x,h) where h is the height of the triangle.

Consider point G' on the circle, with coordinates (x,-h). Chord GG' intersects diameter EF, dividing GG' into two segments of length h and EF into segments of length 2+x and 2-x.

Then h*h = (2-x)(2+x).

So h = sqrt(4-x^2), but we knew that already from G being on the circle.

Since the triangle is isosceles with the base HF being equal to side GF, the length of the base is the same as that of GF: sqrt(h^2 + (2-x)^2).

Substituting for h we get the base equals sqrt(8-4x) = 2 sqrt(2-x).

Thus the area of the triangle is sqrt((4-x^2)(2-x)) = sqrt(8-2x^2-4x+x^3).

Differentiating the above, the numerator, which has to be set equal to zero, is 3x^2-4x-4. Applying the quadratic formula gives solutions as x = 2/3 +/- 4/3.

x=6/3 = 2 results in a zero area "triangle".

x=-2/3 results in

h = sqrt((8/3)(4/3) = 4sqrt(2)/3
base = sqrt(32/3) = 4sqrt(2)/sqrt(3)

Area = 32/(3sqrt(3))/2 = 16sqrt(3)/9 ~= 3.07920143567801


  Posted by Charlie on 2008-01-11 13:22:04
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