Little explorations with HP calculators (no Prime)
04-07-2017, 05:19 PM (This post was last modified: 04-07-2017 05:50 PM by Han.)
Post: #136
 Han Senior Member Posts: 1,865 Joined: Dec 2013
RE: Little explorations with the HP calculators
(04-07-2017 12:57 PM)SlideRule Wrote:
(04-07-2017 02:29 AM)Gerson W. Barbosa Wrote:  ...The shape of the outer triangle is irrelevant ...

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

(04-07-2017 02:29 AM)Gerson W. Barbosa Wrote:  ... the sides of the outer triangle = 2x, 3x and 4x...
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.

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).

The $$n$$-dimensional volume of an object in $$\mathbb{R}^n$$ (Euclidean space) is invariant under unimodular transformations (i.e. transformations of the form $$\mathbf{x} \to \mathbf{A} \cdot \mathbf{x}$$ with $$|\mathbf{A}| =1$$). So volume (in this case, 2-dimensional "volume" is really area) remains the same; but lengths are not preserved. On the other hand, relative positions are preserved. For example, since F is 1/4 the distance of BC from the point B, it will remain 1/4 the distance from B under any unimodular transformation, and B, F and C will remain collinear. The assumption that the outermost sides are 2x, 3x, and 4x, while in my opinion incorrect, do not affect the final result because (in the end) we are looking at ratios of lengths (i.e. relative positions of points on a line segment).