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Little explorations with HP calculators (no Prime)
04-07-2017, 05:41 PM (This post was last modified: 04-07-2017 05:48 PM by Gerson W. Barbosa.)
Post: #138
RE: Little explorations with the HP calculators
(04-07-2017 05:19 PM)Han Wrote:  
(04-07-2017 12:57 PM)SlideRule Wrote:  IF the base and the height remain constant, the AREA of the triangle is also constant, irrespective of an alteration of the "shape" of the original triangle

Then why not give the original dimensions entirely in terms of x? The substitution is plausable; is it implied in the original set of givens?

I approached my attempted solution in a similar manner by asserting the irrelevancy of the SHAPE for the AREA of a triangle but got caught up on the x=y=z substitution.
AB and AC must not be colinear being the only restriction, I think.

BRILL!
SlideRule

I don't think we can assume that x=y=z (which is the implication in assuming the sides are 2x, 3x, 4x).

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.
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RE: Little explorations with the HP calculators - Gerson W. Barbosa - 04-07-2017 05:41 PM



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