(I) Determine the maximum area of an ellipse that is inscribed within an integer triangle having sides 13, 14, and 15.
(II) Determine the minimum area of an ellipse that is circumscribed around an integer triangle having sides 13, 14, and 15.
The 13-14-15 triangle consists of two right triangles, each with integer sides, which fact becomes apparent after using Heron's formula to find the area (84) and then finding the length of the perpendicular (12) when the base equals 14.
I'll use this configuration with the left leg equal to 13, and also triangles similar to the 5-12-13 triangle.
Let the horizontal axis of the ellipse equal 2a and the vertical axis equal 2b. Then the altitude cuts the 2a segment in the ratio 5:9 and so (5a/7)/(12-b) = 5/12 which simplifies to a = 7 - 7b/12.
The area of an ellipse equals pi * ab so substitute, take the derivative, and equate to zero to get b = 6, making a = 7/2 and area = pi * 7/2 * 6 = 65.973+ or more than 78 percent of the total area.
This seems big. I have no idea if it's the maximum, or if the above is free of error but I think the decomposition into two right triangles with integer sides is too convenient to be accidental.
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Posted by xdog
on 2022-08-08 10:00:39 |
Part (I), not a solution
