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Little explorations with HP calculators (no Prime)
04-07-2017, 10:19 PM (This post was last modified: 04-07-2017 10:20 PM by Han.)
Post: #149
RE: Little explorations with the HP calculators
(04-07-2017 10:04 PM)Gerson W. Barbosa Wrote:  
(04-07-2017 09:26 PM)pier4r Wrote:  About the proof. I learned the hard way that showing the correct result may be a coincidence, while proving it is something different. If you just "prove" with a result it is like a proof by example, so, no proof.

I have presented two solutions to the problem. Both assume the shape of the figure is irrelevant as long as the required conditions are met, which is quite clear to me. Both give of course the same correct answer to the problem – and this is no coincidence. I didn't need a second solution, I just wanted one that avoided many instances of the law of cosines and two instances of Heron's formula. These are not particularly brilliant or simple solutions I recognize, but if they are not valid solutions then I have no idea wha valid ones should be like.

Please explain how the shape being irrelevant is clear to you. What are the logical deductive steps to conclude that the shape is irrelevant? You wrote in an earlier post that this is true if the base and height are constant. For one triangle, assume that x=y=z. For a second triangle, suppose x, y, and z are not all equal. More specifically, suppose for one triangle we have x=y=z=1 (a 2-3-4 triangle). And for the second one, suppose x=1, y=4/3, z=1 (a 2-4-4 triangle). Can you explain why two such triangles have the same area? I do not see how they would be equal in area. In fact Heron's formula would suggest that they are of different area.

The two solutions you presented are simply two special cases of the problem because you have not presented how the general case may be reduced to either of the two special cases you presented.

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RE: Little explorations with the HP calculators - Han - 04-07-2017 10:19 PM

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