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Little explorations with HP calculators (no Prime)
04-07-2017, 10:00 PM
Post: #146
RE: Little explorations with the HP calculators
(04-07-2017 08:52 PM)Gerson W. Barbosa Wrote:  The problem makes no requirement about the shape of the triangle . It only states that AE = CE, DB = 2*AD and CF = 3*BF. Provided these conditions are met, the area relationship between both triangles is preserved, no mater the shape. That's what the problem implies.

The problem statement may imply that, but unless it is a trivial consequence (and I do not think that it is) this implication must be proved. Notice that the problem also seems to suggest that x, y, and z may each be any positive value as there are no restrictions on x, y, and z in the statement itself. However, we would be wrong to conclude that x, y, and z may each be any positive real number because some combinations of x, y, and z result in invalid triangles as the sum of two sides may actually be shorter than the third. This is why it is important to always prove any statement that seems to be implied (as opposed to being explicitly stated) by the problem.

Quote:I just trusted the problem statements and chose a convenient shape. I didn't know beforehand the inner triangle thus defined would be isosceles


Is is only isosceles if you assume that the lengths x and z are equal and that y = x*2*sqrt(3)/3. Nothing in the problem statement suggests that x and z are equal. The fact that they used two separate variable names suggests that x and z are generally not equal. The isosceles property is not pertinent to the solution; my point was simply that the assumption (x=z, y=x*2*sqrt(3)/2) gives more properties to the (inner) triangle that would not necessarily be a property of the general case.

Quote:but after calculating the lengths of the three sides this fact became evident and I took advantage of it to avoid the use of the more complicated Heron's formula.

Then the solution is a special case solution (namely when x=z and y=x*2*sqrt(3)/3), which is fine. However it is not a complete solution to the problem, which asks for the general case when x may or may not be equal to z (and similarly for the relationship between x and y). Unless there is some trivial explanation that reduces the general case to the case where x=z and y=x*2*sqrt(3)/3, such a reduction must be proved.

Quote:The invariance of the areas relationship regardless the shape needs indeed a proof, but that is not up to the solver, as the problem itself suggests it is true.

I do not see how the problem statement indicates that the invariance is guaranteed. Yes, it is a consequence of the relative lengths of the sides, but one that is not obvious (to me) and needs proof.

Quote:Although apparently only geometric solutions matter here, I think all approaches are valid as long they lead to the correct answer.

"All solutions matter" indeed (this may be a reference that only Americans will likely get, unless you follow American politics and social issues). I am equally interested in analytical solutions just as much as geometric ones. In the case of an analytical solution using trigonometry, it appears that we have only solved a special case. (Setting y=k1*x and z=k2*x may give us a general solution; I have not tried this.)

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RE: Little explorations with the HP calculators - Han - 04-07-2017 10:00 PM



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