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Little explorations with HP calculators (no Prime)
04-07-2017, 11:52 PM (This post was last modified: 04-08-2017 12:11 AM by Han.)
Post: #152
RE: Little explorations with the HP calculators
(04-07-2017 11:39 PM)Gerson W. Barbosa Wrote:  
(04-07-2017 10:19 PM)Han Wrote:  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.

There appears to be a confusion here. I never said the areas of the two triangles I used, a 2-3-4 triangle and a 2-2*sqrt(3)-4 triangle are the same. They obviously are not. The relationship between the areas of these triangles and the areas of their respective inner triangles are.


This does clarify some points for me. On the other hand, the bolded statement, however, is not obvious to me. You have certainly demonstrated that the bold statement is true when x=y=z, and y=x*2*sqrt(3)/3 and x=z (the inner area is 7). But I do not see how this statement holds true in general using similar analytic arguments found in your two cases.

As it stands, the relationship between the area of the outer triangle and inner triangle is the same under the two sets of assumption. Why is it also true without either of those assumptions? I think this is what pier4r meant by: "If you just "prove" with a result it is like a proof by example, so, no proof." Does there exist an explanation for the remaining case in a similar style to your solutions?

Quote: When you did your own analysis, you didn't bother to choose a particular shape, you only took care the problem requirements were met. So did I.
Not being a mathematician, I am far from being rigorous though.

I disagree here. You made extra assumptions that reduced the total number of cases without explaining why (or even if) the other case is equivalent to the two you considered. In other words, what about the cases for which your assumptions do not hold?

Further comments:

In mathematics, we sometimes use the phrase "without loss of generality" to indicate that there are other cases, but they are handled similarly with minor but obvious changes. For example, proving that x^2 + y^2 is odd if x and y are of different parity may be reduced to assuming x is odd and y is even. This is allowed because the other case (x even and y odd) would require a proof that looks exactly like the former case, except x and y would be switched. On the other hand, in this geometry problem, the assumptions you made in your two cases are clearly not equivalent to the case where x, y, and z are distinct.

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RE: Little explorations with the HP calculators - Han - 04-07-2017 11:52 PM



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