Little explorations with HP calculators (no Prime)

04072017, 07:19 PM
Post: #141




RE: Little explorations with the HP calculators
The claim that "the shape does not matter" is not a trivial statement. It is fairly easy to deduce if we are comparing two triangles who have the same base length and the same height. Then the areas of two such triangles are the same. However, two triangles with different side lengths and having the same area (of 24) can still satisfy the conditions of the problem. Thus a solution using any assumptions such as x=y=z or even just x=z must include some explanation why such a solution extends to the general case when x is not equal to z (or y) while still maintaining an area of 24.
Graph 3D  QPI  SolveSys 

04072017, 07:54 PM
Post: #142




RE: Little explorations with the HP calculators
(04072017 05:12 PM)Han Wrote: Can you explain the deduction process that leads to the conclusion that CB and AB are equal? I suspect that these are not necessarily equal. I suppose I did one assumption too much, after reflecting a bit what I assumed (that E is on the bisector of ABC) can be clearly false. Wikis are great, Contribute :) 

04072017, 07:59 PM
Post: #143




RE: Little explorations with the HP calculators
(04072017 05:40 PM)Dieter Wrote: However, I am not sure if the code box gets a scroll bar if it contains more than exactly 11 lines under all circumstances. It works on chrome on Android, 8 inch screen (portrait or landscape) Wikis are great, Contribute :) 

04072017, 08:52 PM
(This post was last modified: 04072017 08:53 PM by Gerson W. Barbosa.)
Post: #144




RE: Little explorations with the HP calculators
(04072017 06:00 PM)Han Wrote:(04072017 05:41 PM)Gerson W. Barbosa Wrote: Yes, we can! In fact that was my first approach. I've found cos Â = 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. 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. I just trusted the problem statements and chose a convenient shape. I didn't know beforehand the inner triangle thus defined would be isosceles, 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. 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. If I needed a proof, the your solution might have sufficed. Although apparently only geometric solutions matter here, I think all approaches are valid as long they lead to the correct answer. 

04072017, 09:26 PM
Post: #145




RE: Little explorations with the HP calculators
(04072017 08:52 PM)Gerson W. Barbosa Wrote: If I needed a proof, the your solution might have sufficed. Two observations. I do believe (and also th Brilliant community does) that whatever the approach, if it leads to the solution and it is intelligible and not by chance, then it is valid. Just having a "trigonometry free" solution would be interesting (because harder, in general). 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. This if I did not misunderstand you. Wikis are great, Contribute :) 

04072017, 10:00 PM
Post: #146




RE: Little explorations with the HP calculators
(04072017 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.) Graph 3D  QPI  SolveSys 

04072017, 10:04 PM
(This post was last modified: 04072017 10:06 PM by Gerson W. Barbosa.)
Post: #147




RE: Little explorations with the HP calculators
(04072017 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. 

04072017, 10:09 PM
(This post was last modified: 04072017 10:11 PM by Han.)
Post: #148




RE: Little explorations with the HP calculators
(04072017 02:52 PM)Gerson W. Barbosa Wrote:(04072017 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 When you assume that x=y=z, this triangle does not have the same base and height as a triangle for which x, y, and z are not simultaneously equal. The side AC might remain 2x (we may view this as the "base"), but the height of the triangle for which x=y=z will NOT be equal to the height of the triangle for which x, y, and z are not simultaneously equal. Graph 3D  QPI  SolveSys 

04072017, 10:19 PM
(This post was last modified: 04072017 10:20 PM by Han.)
Post: #149




RE: Little explorations with the HP calculators
(04072017 10:04 PM)Gerson W. Barbosa Wrote:(04072017 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. 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 234 triangle). And for the second one, suppose x=1, y=4/3, z=1 (a 244 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. Graph 3D  QPI  SolveSys 

04072017, 10:23 PM
(This post was last modified: 04072017 10:44 PM by Han.)
Post: #150




RE: Little explorations with the HP calculators
As a mathematician I am genuinely interested in an analytical solution and would like to fully understand how your two cases are equivalent to the general case. I hope that my posts are not interpreted as attacks of any sort
Graph 3D  QPI  SolveSys 

04072017, 11:39 PM
(This post was last modified: 04072017 11:40 PM by Gerson W. Barbosa.)
Post: #151




RE: Little explorations with the HP calculators
(04072017 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 234 triangle). And for the second one, suppose x=1, y=4/3, z=1 (a 244 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 234 triangle and a 22*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. 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. 

04072017, 11:52 PM
(This post was last modified: 04082017 12:11 AM by Han.)
Post: #152




RE: Little explorations with the HP calculators
(04072017 11:39 PM)Gerson W. Barbosa Wrote:(04072017 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 234 triangle). And for the second one, suppose x=1, y=4/3, z=1 (a 244 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. 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. 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. Graph 3D  QPI  SolveSys 

04082017, 12:40 AM
(This post was last modified: 04082017 12:41 AM by Gerson W. Barbosa.)
Post: #153




RE: Little explorations with the HP calculators
(04072017 11:52 PM)Han Wrote: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. I am convinced all the infinite cases hold. I only chose a couple of cases I thought were easier to solve. I didn't try a generalization because I didn't think it was necessary. I might return to this problem, but not anytime soon. Meanwhile the solutions I have presented so far should be considered wrong, invalid and useless. 

04082017, 01:17 AM
Post: #154




RE: Little explorations with the HP calculators
(04082017 12:40 AM)Gerson W. Barbosa Wrote: I am convinced all the infinite cases hold. They do hold. Quote:I only chose a couple of cases I thought were easier to solve. I didn't try a generalization because I didn't think it was necessary. Selecting a subset of cases is fine provided there some way to reduce the general case to your two specific cases. Or perhaps we could go in the reverse direction. Is there a way to generalize your specific cases to the more general one where no additional assumptions on x, y, and z need to be made? I do not know myself; this is a genuine question. Quote:I might return to this problem, but not anytime soon. Meanwhile the solutions I have presented so far should be considered wrong, invalid and useless. There is nothing mathematically wrong about your two specific solutions with the additional assumptions about x, y, and z. Graph 3D  QPI  SolveSys 

04082017, 02:44 AM
(This post was last modified: 04082017 02:51 PM by Han.)
Post: #155




RE: Little explorations with the HP calculators
An analytic solution (using Gerson's ideas) is in fact possible. Not wanting to give up, I actually started with Gerson's solution, but without assuming anything about x, y, and z and tried to go as far as I could. It turns out that one can set up the formula
\[ (2x)^2 = (3y)^2 + (4z)^2  2(3y)(4z)\cos(B) \] and solve for \( \cos(B) \). Then, use this formula to deduce that \[ \sin(B) = \frac{\sqrt{(24yz)^2(9y^2+16z^24x^2)^2}}{24yz}. \] Continue this process to determine \( \sin(A) \) and \(\sin(C) \) in terms of x, y, and z. After some very tedious algebra (I ended up using Maple to do a good chunk of it to avoid mistakes in algebraic manipulations), one can deduce ratios of the areas of the subtriangles and obtain the analytic equivalence of the geometric solution. Then I then realized that one could simply just use the law of sines, which has a slightly less tedious set of algebraic manipulations because the formulas for the \( \sin(A)\), \( \sin(B)\), and \(\sin(C)\) are not really necessary. Using the law of sines: \[ \frac{2x}{\sin(B)} = \frac{4z}{\sin(A)} \Longrightarrow \sin(A) = \frac{2z}{x} \cdot \sin(B) \Longrightarrow \underbrace{x\cdot \sin(A)}_{\text{height of } \triangle ADE} = 2\cdot \underbrace{z\cdot \sin(B)}_{\text{height of } \triangle BDF} \] The area of \( \triangle ADE \) is \( \frac{1}{2} \cdot y \cdot (x\cdot \sin(A)) = \frac{1}{2} \cdot x\cdot y \cdot \sin(A) \). The area of \( \triangle BDF \) is \[ \frac{1}{2} \cdot (2 y) \cdot (z\cdot \sin(B)) = \frac{1}{2} \cdot y \cdot (2z)\cdot \sin(B) = \frac{1}{2}\cdot y \cdot (x\cdot \sin(A)) = \underbrace{\frac{1}{2}\cdot x \cdot y \cdot \sin(A)}_{\text{area of } \triangle ADE} .\] Also, \[ \frac{3y}{\sin(C)} = \frac{2x}{\sin(B)} = \frac{4z}{\sin(A)} \Longrightarrow \sin(C) = \frac{3y\sin(B)}{2x} = \frac{3y\sin(A)}{4z} \] The area of \( \triangle CEF \) is \[\frac{1}{2} \cdot x \cdot [ (3z)\cdot \sin(180^\circ  C)] = \frac{1}{2} \cdot x \cdot [ (3z)\cdot \sin(C) ] = \frac{1}{2} \cdot x \cdot 3z \cdot \frac{3y}{4z}\cdot \sin(A) = \underbrace{\frac{1}{2}\cdot x \cdot y \cdot \sin(A)}_{\text{area of } \triangle ADE} \cdot \frac{9}{4}.\] So the area of \( \triangle CEF \) is \( \frac{9}{4}\)th the area of \( \triangle ADE \). Lastly, the area of \( \triangle ABC \) gives us the following identity: \[ 24 = \frac{1}{2} \cdot (3y) \cdot [ 2x\cdot \sin(A) ] \] which implies the area of \( \triangle ADE \) is \[ \frac{1}{2}\cdot x\cdot y \cdot \sin(A) = 4 \] It should be fairly straightforward to deduce that the area of \( \triangle DEF \) is \[ 24  (4+4+9) = 7 \] This analytic solution is really the same as the geometric solution I posted earlier ( http://www.hpmuseum.org/forum/thread795...l#pid71463 ) Graph 3D  QPI  SolveSys 

04082017, 03:37 AM
(This post was last modified: 04082017 03:37 AM by Han.)
Post: #156




RE: Little explorations with the HP calculators
Here is a proof (using linear algebra) that reduces the general case to the special case provided by Gerson.
Define the barycentric coordinates \(\mathbf{\lambda} = (\lambda_1, \lambda_2, \dotsm, \lambda_n ) \) of a point \( \mathbf{P}\in \mathbf{R}^m \) relative to a set of points \( \{ \mathbf{Q}_1, \mathbf{Q}_2, \dotsm, \mathbf{Q}_n \} \) as the solution to the equation \[ \mathbf{P} =\sum_{k=1}^n \lambda_k \cdot \mathbf{Q}_k \] (This is actually a system of equations in the coordinates of the points.) We can even require that \( \sum \lambda_k = 1 \) by embedding these points in a hyperplane \( x_{m+1} = 1 \) of \( \mathbb{R}^{m+1} \) though this is not necessary for this problem. Let \(f \) be the linear transformation \( f(\mathbf{x}) = \mathbf{A}\cdot \mathbf{x} \) with \( \mathbf{A} = 1 \) (determinant equal to 1; i.e. \(f\) is a unimodular transformation). Observe that \( f \) preserves volume (since \(  \mathbf{A} = 1 \)), and \( f \) preserves barycentric coordinates because: \[ f(\mathbf{P}) = \mathbf{A}\cdot \mathbf{P} = \mathbf{A} \cdot \sum_{k=1}^n \lambda_k \cdot \mathbf{Q}_k = \sum_{k=1}^n \lambda_k \cdot \mathbf{A}\cdot \mathbf{Q}_k = \sum_{k=1}^n \lambda_k \cdot f(\mathbf{Q}_k) \] Thus, it suffices to show that every nondegenerate triangle ABC with \[ \begin{align*} D & = \frac{1}{3}\cdot A + \frac{2}{3}\cdot B + 0 \cdot C\\ E & = \frac{1}{2}\cdot A + 0\cdot B + \frac{1}{2}\cdot C\\ F & = 0\cdot A + \frac{1}{4}\cdot B + \frac{3}{4}\cdot C\\ \end{align*} \] can be transformed into a 306090 right triangle (with points A'=f(A), B'=f(B), and C'=f(C)) having the same area and D', E', and F' satisfying the three equations analogous to those for D, E, and F above. Let \( \mathbf{u} \) be the vector from A to C, and \( \mathbf{v} \) be the vector from A to B. Similarly let \( \mathbf{u}' \) be the vector from A' to C', and \( \mathbf{v}' \) be the vector from A' to B'. Then the transformation matrix \( \mathbf{A} \) is the solution to \[\mathbf{A}\cdot \begin{bmatrix} \mathbf{u} & \mathbf{v} \end{bmatrix} = \begin{bmatrix} \mathbf{u}' & \mathbf{v}' \end{bmatrix} \] This solution always exists provided that ABC (and likewise A'B'C') is nondegenerate (i.e. vertices are in general position). Graph 3D  QPI  SolveSys 

04082017, 09:45 AM
(This post was last modified: 04082017 09:48 AM by pier4r.)
Post: #157




RE: Little explorations with the HP calculators
(04082017 12:40 AM)Gerson W. Barbosa Wrote: Meanwhile the solutions I have presented so far should be considered wrong, invalid and useless. I am sorry if I came across harshly, it was not my intention. I wanted just to point out that when one solves a case, has a proof for that case, not in general (unless otherwise stated and argued). For me your solution is surely an interesting approach "out of the box" that I could not find before months of thinking, maybe, maybe I would still fail after that period. So, your solution is pretty great. @Han: amazing effort you put there. Thanks for sharing! Wikis are great, Contribute :) 

04082017, 03:01 PM
Post: #158




RE: Little explorations with the HP calculators


04082017, 05:19 PM
(This post was last modified: 04082017 05:20 PM by Han.)
Post: #159




RE: Little explorations with the HP calculators
Great post, Gerson! For anyone who is curious about the tedious algebra using Heron's formula as shown by Gerson, here is a text output from Maple 17:
Code: > restart: Graph 3D  QPI  SolveSys 

04082017, 06:23 PM
(This post was last modified: 04082017 06:33 PM by pier4r.)
Post: #160




RE: Little explorations with the HP calculators
Very nice!
I am still shocked that the problem requires, aside thinking out of the box, such amount of passages. Ok the first solutions on the site are showing: a method of proportion, like Han used, a method with trigonometry, like Gerson but less general, and a very nice method with vector scalar and cross product. Wikis are great, Contribute :) 

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