Trurl

Is there a higher mathematical method to solve similar triangles.

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Ok, you are going to laugh at this one. But I am seriously asking this question. I need to think outside the box if I want to break a one-way-function. In this case outside the triangle.

 

My question is: Why can’t I combine the Law of Sines and the property of similar triangles to solve 2 different similar triangles of different sizes?

 

And secondly why can’t I say the proportions in one triangle are not proportional to the same proportion in a similar triangle? I mean if all sides and angles are known in one and the other similar triangle all the angles are known, there has to be a method to solve the segments of the unknown, second similar triangle. Is there a method I do not know of?

 

For example, you know the lengths and angles of triangle one; and you know the angles of a smaller, similar triangle angle two.

 

I know, if you new one side of angle 2 you could solve with similar triangles no problem. If it were that easy the problem wouldn’t be a one-way-function.

 

I just wanted to get advice on the possibility and if anyone sees this as a worthy problem.

 

To be specific I do not know the lengths of the segments on angle 2, but I do know the equations that make up the lengths. Does anyone believe it would be possible to solve angle 2 with only equations as the known for angle 2?

 

I have put much thought into this. Because being able to break a one-way-function is the result. I know that there are infinite similar angles. And yes, I know trigonometry. This has just been bugging me. Why can’t I do this? Or it must be possible? But are there any higher geometry or higher math’s that solve such a problem? I’m sure someone has encountered this problem before.

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What does it mean to "solve 2 different similar triangles"?

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Ok, here is my question simplified:

 

Given: two similar triangles ABC and DEF,

               Are segment AC divided by segment AB proportionate to segment DF divided by segment DE?

 

If in triangle ABC, all lengths are known, and all angles are known;

And in triangle DEF all angles are known;

Using the similarity of AC/AB = DF/DE

DF and DE are known, but AC and AB are known only by equations, can the equation between AC and AB be simplified by relation to the known DF/DE?

 

 

That is my question simplified. I did not include example because the problem relies on AC/AB = DF/DE.

 

Other questions I have just realized that the Law of Sines relates to similar triangles. But what I was not taught in school is:

               d = Sin (a) / A = Sin (b) / B = Sin (c) / C

We were not taught that d equals the proportion and is the circumscribed circle of the triangle.

But this can be discussed later. I just need to know if what I described above is possible. To me it makes sense and is simple. But I can’t remember ever studying similar triangles this way and I don’t know if a text book would list it under its laws. But I think it isn’t there just like we didn’t learn about “d” because the application is limited. I could be completely wrong. But a side divided by the other in one triangle should equal a side divided by another in a second, similar triangle.

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First question: the ratios of the sides are the same.

Second question:  AC and AB known only by equations - what do you have in mind?

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I could make this long, but I will get to the point. I realize most don’t like my Prime equations. I design these triangles as a graphic representation of my problem.

 

Do you agree that triangle yxs is similar to triangle AFC?

 

Do you believe that triangle AFC has no importance other than the fact that it has PNP (also called N) equals the product of 2 unknow Prime numbers?

 

Do you believe that since N is known and that AFC can be solved for and that triangle yxs angles are the same because of the definition of similar triangles?

 

Most importantly, do you believe that y and x are equal to in proportion to N? (This is the most important step, I am still working on.)

 

If y is the larger Prime factor and x is the smallest, we will use similar triangles y/x = AF/FC.

 

All angles and sides of AFC are known. The proportions of y/x can be written with the following equations:

N = 85

y = (N^2/x) + x^2) / N

x = x

N = y * x, but we only know N

 

y/x = AF/FC

 

AF/FC is known as solved by the triangle and is a real number proportion.

 

So (N^2/x) + x^2) / N * (1/x) = proportion AF/FC

 

Calculate it out I am still working on this.

 

I got x = Sqrt(N/ proportion AF/FC)

 

Again I am still verifying everything is mathematically proper. But I post here to get feedback.

SFN20181108SimilarTriangles.gif

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I know no one believes in my problem. But I need someone to follow along with the solution (or disprove my ideas). It is something a problem solver must work through themselves.

 

I know FC does not equal CE. And if we can’t solve for FE or find FC. I will not break the rules of geometry by using my side-angle solution.

 

But I propose a math exercise to go through. Instead of believing my solution, prove the exercise wrong. (Probably not that hard to do, but I think it is worth trying.)

 

FE = CD / cos[angle ECD + angle FCE] = CE / cos[FCE]

This may be solvable with FE the unknown, but we need to solve for CD and CE to for this to work. Angle FCE is also unknown.

 

N/sin[120] = CE/sin[60]

 

To me I drew this triangle to represent the one-way function. I believe it is solvable. This is why I have reposed it. Proving this triangle solvable means that semi-Primes are no longer able to be used in cryptography. It sounds simple enough, but this is no ordinary triangle.

 

I know when I start talking about Prime number solutions or one-way-functions or solving impossible triangles, it draws a red flag. If it were so easy someone would have solved it already. But if you do not believe we can reach a solution, you are probably correct. It may seem like I do not know math because I am always looking for work-arounds. I know the problem is unsolved and I know the probability of finding a solution equals my probability of solving this triangle. But I believe my approach is different. Solved or unsolved semi-Primes, it is still worth the mathematical exercise.

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3 hours ago, Trurl said:

I know when I start talking about Prime number solutions or one-way-functions or solving impossible triangles, it draws a red flag. If it were so easy someone would have solved it already. But if you do not believe we can reach a solution, you are probably correct. It may seem like I do not know math because I am always looking for work-arounds. I know the problem is unsolved and I know the probability of finding a solution equals my probability of solving this triangle. But I believe my approach is different. Solved or unsolved semi-Primes, it is still worth the mathematical exercise.

 

Perhaps for the benefit of us casual spectators, I wonder if you can just outline as simply as you can, the connections among your geometry problem; prime numbers; and one-way functions.

By the way I am a little fuzzy on one-way functions. My understanding is that a one-way function (in computer science) is an invertible mapping between two sets such that one direction is much easier to compute than the other. 

The classic example (again this is according to my limited knowledge of the subject) is that 3 x 5 = 15 is an easy problem. We learned the algorithm in grade school (before Common Core utterly destroyed the teaching of arithmetic fundamentals in the US). And it is a very efficient algorithm, as computer scientists define the term. Probably polynomial I'd guess. (You have a row for each digit, and each row is linear in the digits of the other number. So it's n^2 or something. Just off the top of my head.)

On the other hand, 15 = 3 x 5 is a very difficult problem! The only known way to factor an integer is by trial division. That algorithm grows exponentially with the size of the input. In the CS biz, polynomial = good, exponential = bad. 

If what I said were true, this would be an example of a one-way function. Multiplication is easy and factoring is hard, and they are each other's inverse via the Fundamental theorem of arithmetic. That's the one that says Mrs. Screechy will rap you over the knuckles with her wooden ruler if you don't learn your times tables.

However, here is the beautiful kicker as I understand it. We don't know for certain that factoring is hard! We haven't found a polynomial factoring algorithm, but we haven't proven one doesn't exist! [There's a quantum algorithm that factors in poly time. It's called Shor's algorithmThis is one of the most amazing things I've learned in the past few years].

And more generally, we do not know if any one-way functions exist!

That's everything I've come across about this subject in my readings. Anything I've got wrong please correct me. 

So now, given all this, what does your diagram do?

 

 

 

Edited by wtf

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Like others here, I am a bit mystified understanding what you are trying to do.

In terms of pure plane trigonometry, (no higher maths needed) can you answer the following question.

Are you aware that sometimes the sine rule in plane triangles is ambiguous that is there can be two different solutions to certain triangles when using the sine rule ?

That is why it used to be taught (I fully sympathise with wtf's comment on the 'modern' syllabus) in schools to use the cosuine rule which is never ambiguous.

 

Does this help any ?

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Hypothesis,

 

Given a triangle with known N, the similar triangle formed at any given triangle with the largest side equal to N, will have a similar triangle with sides equal to p and q. This triangle will have the largest side similar to N.

 

This is probably absolutely wrong, but I base it on the fact that the 2 sides multiplied together equal N. I know I am probably breaking a rule of sine and cosine by addition of trigonometry, but my logic is this:

If p and q are the products of N then a triangle the contains N as the largest side, the triangle will have a similar triangle that has sides of the products. After all, a similar triangle is just proportions and since p and q are proportional to the original triangle, multiplication does solve the similar triangle.

 

These similar triangles solve the one-way function of N = p *q. The reason it isn’t easy to visualize is because to find the answer because we did not have the equation to find the relationship between p and q.

 

It is ok if I am wrong. This is after all an impossible one-way-function triangle. But if you read through it, you may understand what I was attempting to do.

 

I will respond to each individual post later. I just wanted to clarify my idea, if possible.

 

Yes, I know there are infinitely many triangles. I’m counting on it so that my sides equal to p and q exist. Again, they always exist. I am just using properties of triangles to simplify an equation that is used to find p and q knowing only N.

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