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|Three circles (Posted on 2010-09-09)
Given three concentric (having the same center) circles.
Their radii are 1,2 and 3.
Place three points A,B,C - each on a different circle to get a triangle ABC
with a maximal area.
What is this area?
No Solution Yet
Submitted by Ady TZIDON
Rating: 5.0000 (1 votes)
Another approach (spoiler)
| Comment 2 of 3 |
Without loss of generality, let the side BC be parallel to the x axis and at a distance d below it with OB = 2 and OC = 3 so that 0 < d < 2. For the triangle to have maximum area, the point A must then be at the highest point of the unit circle and the three points must have the following coordinates:
A (0, 1) B (-sqrt(4 - d2), -d) C (sqrt(9 - d2), -d)
Just as the line AO is perpendicular to side BC for maximum area, by similar arguments BO is perpendicular to CA and CO is perpendicular to AB, making O the orthocentre of triangle ABC.
Using the gradients of AB and CO: (d + 1)/sqrt(4 - d2) = sqrt(9 - d2)/d
which can be squared to give d2(d + 1)2 = (4 - d2)(9 - d2)
then simplified to d3 + 7d2 - 18 = 0
whose only solution between 0 and 2 is: d = 1.458757..
The area is given by 0.5(d + 1)[sqrt(4 - d2) + sqrt(9 - d2)] and when
d = 1.4587.., the maximum area works out to be 4.904822..
[It does seem surprising that, despite knowing the lengths OA, OB and OC, and knowing that O is the orthocentre, we still need to solve a cubic to find the area. Am I missing something?]
Posted by Harry
on 2010-09-11 20:21:09
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