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Little explorations with HP calculators (no Prime)
03-30-2017, 04:14 PM
Post: #77
RE: Little explorations with the HP calculators
(03-30-2017 12:26 PM)Dieter Wrote:  
(03-29-2017 06:15 PM)Han Wrote:  So \(\angle BED\) , \(\angle BDE \), \( \angle CDE \) and \( \angle AED \) should be the only angles whose measure you cannot determine directly from the properties mentioned. What if you set up an appropriate system of equations involving these angles?

I have tried this as well. Let <BDE = u, <BED = v, <CDE = x and <AED = y and it is possible to set up several linear equations, for instance...

u + v = 162
x + y = 111
u + x = 132
v + y = 141

...but I cannot get four independent ones, so there is no solution for a linear equation system. What do you get?


I had not actually attempted this problem when I posed that quest. My question about considering a system of linear equations was merely that -- a question (i.e. was it tried, and does it work) -- and was not meant as an indication that I had solved it through linear systems.

You are correct, though, that the resulting system is not sufficient for solving the problem (the system does have a solution; however it appears it has an infinite number of solutions assuming we are not restricted to whole degrees for the angles).

I did eventually make some progress on this problem. I think I have a solution but I wonder if it is correct since it is not the cleanest. It is an elementary solution, but nowhere near what Paul Erdos would consider as being from "The Book." But what I ended up doing making use of the law of sines and cosines. Here's an outline of my solution.

Let F be the point along CD such that AF is perpendicular to CD. G be the intersection of the blue lines. Observe that AC and AD are equal in length. Call this length L. Use basic trigonometry determine CF (in terms of L), which then gives you CD (since CD is twice the length of CF). The law of sines applied to triangle CAE will give you the length of CE in terms of L. Then use the law of cosines in triangle CDE to determine ED (again in terms of L). Lastly, apply the law of sines to triangle CGE to determine EG in terms of L. Now you can apply the law of sines to EDG to obtain the angle GDE (which is the same as angle CDE).

I got approximately 76.32 degrees for the angle through a bit of scribbling (very likely I made a arithmetic mistake, I would imagine) and a basic scientific calculator (HP 32S).

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RE: Little explorations with the HP calculators - Han - 03-30-2017 04:14 PM

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