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Little explorations with HP calculators (no Prime)
04-07-2017, 06:00 PM (This post was last modified: 04-07-2017 06:06 PM by Han.)
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RE: Little explorations with the HP calculators
(04-07-2017 05:41 PM)Gerson W. Barbosa Wrote:  Yes, we can! In fact that was my first approach. I've found cos  = -1/4, cos B = 7/8 and cos Č = 11/16, the sides of the inner triangle being 2sqrt(5/8)x, sqrt(3/2)x and sqrt(47/8)x. Incidentally the area of the inner triangle = 7.
AB and AC must not be collinear. That's the only restriction I am aware of.

My point was that we cannot assume they are equal based on the problem statement. A solution that involves the lengths of the segments must include some explanation why a change in the lengths of the segments does not alter the final outcome. I am referring to the comment about how the "shape" of the triangle does not matter (I edited a previous comment to add some discussion on how volume is invariant under transformations whereas length is not.)

Additional comments:

Your solution suggests that the inner triangle is isosceles -- this is not true in the general case.

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RE: Little explorations with the HP calculators - Han - 04-07-2017 06:00 PM

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