Little explorations with HP calculators (no Prime)

04062017, 12:47 AM
(This post was last modified: 04072017 08:22 PM by SlideRule.)
Post: #121




RE: Little explorations with the HP calculators
... I share one solution ...
Here's my solution to post #67, page 4, Triangle Problem. ∠EDC = 180°  (∠D + 48°) ≊ 81° where ∠D = acos[(ED² + BD²  BE²)/(2·ED·BD)] ≊ 51° SSS equation BE = sin(84°)/sin(18°)  sin(63°)/sin(39°) ≊ 1.802517077 ED = BE² + BD²  (2·BE·BD·cos18°) ≊ 0.7167358991 SAS equation BD = sin(78°)/sin(18°)  AD ≊ 2.165352128 AD = 1 BEST! SlideRule ps: hopefully, NO typos! 

04062017, 10:43 PM
(This post was last modified: 04072017 06:36 AM by pier4r.)
Post: #122




RE: Little explorations with the HP calculators
The following should be a bit easier than the last one, still I'm not finished. I do not have the 50g at the moment that is busy computing the tournament simulations.
So, if I am not mistaken, I realized that the angle ABC is bisected in E, that CB and AB are equal, that I can apply the Heron's formula, but then it was too late and I started to do rubbish. I am not sure if I can check the problem again until Monday though. Also, the little math handbook that I use to refresh known relationships is not helping, not even the Heron's formula is there. Wikis are great, Contribute :) 

04062017, 11:24 PM
Post: #123




RE: Little explorations with the HP calculators  
04062017, 11:37 PM
Post: #124




RE: Little explorations with the HP calculators
(04062017 11:24 PM)SlideRule Wrote: Please clarify the following: ??= AE, from the illustration. What is AE equal to? CE = AE, look at the image, both are labeled with length 'x'. Bob Prosperi 

04072017, 02:29 AM
(This post was last modified: 04072017 05:15 AM by Gerson W. Barbosa.)
Post: #125




RE: Little explorations with the HP calculators
Area = 7
....... Spoiler ........ ........ The shape of the outer triangle is irrelevant. At first I had made the sides of the outer triangle = 2x, 3x and 4x. Then I used the law of cosines on each side of outer triangles (sides proportional to 3, 4 and 5) in order to determine the angles. By applying the law of the cosines again on these angles and the sides of the inner triangle opposed to them, these sides could be expressed in terms of x. The Heron's formula was then used to compute the areas of both triangles. Finally, the ratio between these two areas is used to find the answer to the problem. But, since the shape of the triangle is irrelevant, why not choosing a righttriangle (now I can do it :), one with angles 90, 60 and 30 degrees and sides 2x, 4x and 2x*sqrt(3)? This makes the solution even more simple: there's no need to use the Heron's formula to calculate the area of the outer triangle; also, the sides of the inner triangle are more easily obtained. Done by hand, except the evaluation and simplification of the area of the inner triangle using the Heron's formula (HP 50g CAS). 

04072017, 09:46 AM
Post: #126




RE: Little explorations with the HP calculators
(04072017 02:29 AM)Gerson W. Barbosa Wrote: Area = 7 While I appreciate the contribution, I do believe that the problem can be solved without trigonometry. Wikis are great, Contribute :) 

04072017, 10:30 AM
Post: #127




RE: Little explorations with the HP calculators
(03312017 07:35 AM)pier4r Wrote: ... The written solution has a typo, but it works http://i.imgur.com/vIdr9Y7.jpg Wikis are great, Contribute :) 

04072017, 12:57 PM
Post: #128




RE: Little explorations with the HP calculators
(04072017 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 (04072017 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 

04072017, 02:37 PM
Post: #129




RE: Little explorations with the HP calculators  
04072017, 02:38 PM
Post: #130




RE: Little explorations with the HP calculators
(04072017 02:29 AM)Gerson W. Barbosa Wrote: Area = 7 I also got 7 for a solution but my solution required merely the insertion of two parallel lines Graph 3D  QPI  SolveSys 

04072017, 02:52 PM
(This post was last modified: 04072017 02:53 PM by Gerson W. Barbosa.)
Post: #131




RE: Little explorations with the HP calculators
(04072017 12:57 PM)SlideRule Wrote:(04072017 02:29 AM)Gerson W. Barbosa Wrote: ...The shape of the outer triangle is irrelevant ... Exactly! As long as the base and the height are kept constant, the area won't change. (04072017 12:57 PM)SlideRule Wrote:(04072017 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? Although apparently misleading, that might have excluded the righttriangle possibility, which is perhaps the most adequate way to solve the problem (at least using trigonometry). An interesting variation of the problem might be the case CF = BF. The solution is also an integer number. Gerson. 

04072017, 02:57 PM
Post: #132




RE: Little explorations with the HP calculators
(04072017 02:38 PM)Han Wrote:(04072017 02:29 AM)Gerson W. Barbosa Wrote: Area = 7 Euclides, the one with a moustache in the photo, was my math teacher in high school. I think I was sleeping when he was explaining the Elements. I paid more attention to his trigonometry classes, though :) 

04072017, 04:11 PM
(This post was last modified: 04072017 04:23 PM by Han.)
Post: #133




RE: Little explorations with the HP calculators
It would be nice if there was a feature to hide parts of a post until a user explicitly clicks a spoiler button. Anyway draw a line through E and parallel to AB. Let G be the point of intersection of this line with BC and note G is the midpoint of BC. EGC has area 1/4*24=6. Note that EG has length 1.5y. Hence ADE has area 4. BDF has half the height as ADE (a second line through F parallel to AB helps to see this) but twice the base length. Therefore its area is also 4. Lastly EFG (compared vs BDF) has area 3 using a ratio of their bases.
Graph 3D  QPI  SolveSys 

04072017, 05:01 PM
Post: #134




RE: Little explorations with the HP calculators
(04072017 04:11 PM)Han Wrote: It would be nice if there was a feature to hide parts of a post until a user explicitly clicks a spoiler button. Anyway draw a line through E and parallel to AB. Let G be the point of intersection of this line with BC and note G is the midpoint of BC. EGC has area 1/4*24=6. Note that EG has length 1.5y. Hence ADE has area 4. BDF has half the height as ADE (a second line through F parallel to AB helps to see this) but twice the base length. Therefore its area is also 4. Lastly EFG (compared vs BDF) has area 3 using a ratio of their bases. Very nice! Hidden text (sort of) 

04072017, 05:12 PM
(This post was last modified: 04072017 05:12 PM by Han.)
Post: #135




RE: Little explorations with the HP calculators
(04062017 10:43 PM)pier4r Wrote: The following should be a bit easier than the last one, still I'm not finished. I do not have the 50g at the moment that is busy computing the tournament simulations. 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. Graph 3D  QPI  SolveSys 

04072017, 05:19 PM
(This post was last modified: 04072017 05:50 PM by Han.)
Post: #136




RE: Little explorations with the HP calculators
(04072017 12:57 PM)SlideRule Wrote:(04072017 02:29 AM)Gerson W. Barbosa Wrote: ...The shape of the outer triangle is irrelevant ... I don't think we can assume that x=y=z (which is the implication in assuming the sides are 2x, 3x, 4x). EDIT (Additional comments): 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, 2dimensional "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). 

04072017, 05:40 PM
Post: #137




RE: Little explorations with the HP calculators
(04072017 04:11 PM)Han Wrote: It would be nice if there was a feature to hide parts of a post until a user explicitly clicks a spoiler button. You could use a code box and place the hidden text after 11 or 12 blank lines: Code:
However, I am not sure if the code box gets a scroll bar if it contains more than exactly 11 lines under all circumstances. Dieter 

04072017, 05:41 PM
(This post was last modified: 04072017 05:48 PM by Gerson W. Barbosa.)
Post: #138




RE: Little explorations with the HP calculators
(04072017 05:19 PM)Han 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 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. AB and AC must not be collinear. That's the only restriction I am aware of. 

04072017, 06:00 PM
(This post was last modified: 04072017 06:06 PM by Han.)
Post: #139




RE: Little explorations with the HP calculators
(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. My point was that we cannot assume they are equal based on the problem statement. A solution that involves the lengths of the segments must include some explanation why a change in the lengths of the segments does not alter the final outcome. I am referring to the comment about how the "shape" of the triangle does not matter (I edited a previous comment to add some discussion on how volume is invariant under transformations whereas length is not.) Additional comments: Your solution suggests that the inner triangle is isosceles  this is not true in the general case. Graph 3D  QPI  SolveSys 

04072017, 06:58 PM
(This post was last modified: 04072017 07:44 PM by SlideRule.)
Post: #140




RE: Little explorations with the HP calculators
Yes, the assumption x=y=z appears problematic.
IF x=y=z is true, then the original depiction of the inscribed triangle to the circumscribed triangle is curious (see my depiction). I'm soooooooooo confused! [attachment=4657] BEST! SlideRule 

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