# 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

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.

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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!

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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Ok, so I know nobody likes my hypothesis. But don’t let it discourage you, I think there is a relevant problem here in this triangle. The only thing I am trying to do is relate a one-way-function, where N is the product of two Prime numbers p and q. The triangle shows my thought of coming up with equations. I posted earlier about finding p knowing only N. Imatfaal agreed there was a pattern but deemed it useless. I argue that only the polynomial was too complex to solve for p, it still showed a relationship between N and p, giving the distance a test p was from N. It would be good for a computer loop. Like the equations or not, they show patterns. Patterns never before used. So, if I say I have a pattern and that pattern is ugly, what should such a pattern look like?

Simply put, the job of this triangle is to simplify my Prime patterns. Does it do that? I don’t know. But I designed it to do so.

In my previous post of “Prime Products One More Time”, look on page 3, posted Nov. 6, 2017. CE is equal to p. N mod p = 0. We don’t know CE; only N. We get a new length N – CE. N – CE has an unknown length FE. FE is the remainder when subtracting q from N-CE. As an added challenge, we do not know the value of FE or q.

CE is sliding along AC’. It increases from the perpendicular CD to CE until it is displaced along a length FE from E. AF equals q.

Now to the part you are not going to like. Using vector edition:

AC = N [Absolute value [ AC – AE – (AC-CE) – CE]] = FE

I know this is mostly likely wrong, but it helped me imagine 2 vectors of p and q added together at an unknown angle with a side opposite that angle equal to N.

Also, I don’t think CE = CF in all cases. I would solve for FC  then solve CE.

__________________________________________________________________-

So why did I give the new hypothesis?

I believe that if you take any triangle that has N as its largest side, the sum of those sides are factors fall on an angle that forms N. Yes, I know the term factors is not correct, because these are decimal values. That is until p and q fall along those segments of the triangle.

Yes, I know this isn’t vector addition. I am not adding p and q at different angles to get N. I am only using a triangle with longest length N and stating a similar triangle with lengths p and q exist on that similar triangle.

So, these are two different approaches; one vector addition; one similar triangles.

This is what my triangle is supposed to solve. Does it do it? The odds are against it. After all, it is a one -way-function. But know that I just didn’t throw this thing together with willy-nilly lines. I know in the development the logic in the formation can not be explained why I chose it. I mean, why did I develop a triangle in this way to solve an impossible function? Well, I know the teacher says: “show your work.” But I cannot show my reasoning; only if something works or doesn’t work.

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On 11/14/2018 at 1:06 AM, wtf said:

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.

Did the previous explanation explain this impossible triangle? I might be able to solve the triangle knowing only N because of the geometric constructions of the angles surrounding N. I don’t know if they still teach using tools like a straight edge and compass. But along with the constructions, I have equations that given p and q in terms of N.

I know it isn’t very believable. I’m not claiming this will give a correct solution. But I do think relating my previous posts equations to a geometric figure will help to simplify it. I can see it, but I don’t think others are interested. That is ok. But if I am going to prove my equations useful, I need them to work and be simplified.

I like your description of the one-way-function. That’s how I’d define it. I believe that one-way-functions exist for us. My concern is RSA and cryptography. For example, if my triangle worked, we’d have to rethink one-way-functions and RSA cryptography. With the Prime factorization problem, many have tried, so we all believe it is impossible to solve with patterns. I would like to see it solved by similar triangles or vectors. I will get laughed at in the process, but I realize the impossibility of the problem. It may not be humanly possible to solve patterns of semi-Primes, but I thought I came up with a good model. I know my equations are too complex to solve for p, but the equation does show a pattern in semi-Primes. I just need to get others to see the potential of the problem.

So if what I am explaining does not make sense or is not explained enough let me know. If something is plain wrong or breaks rules let me know. I am seriously trying as hard as I can to break the Prime factorization problem.

On 11/14/2018 at 6:13 AM, studiot said:

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 ?

No I’m not aware of the ambiguity of the Law of Sines. It has been 20 years since I had a trig class. The rules and Laws are “imbedded” in my mind. By that I mean that I know trig, I just don’t remember how I learnt it. I’m usually good after reviewing a Law or identity.

I thought the ambiguity was the tangent of angles above 180; a difference in the direction of the vector.

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39 minutes ago, Trurl said:
On ‎11‎/‎14‎/‎2018 at 11:13 AM, studiot said:

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 ?

No I’m not aware of the ambiguity of the Law of Sines. It has been 20 years since I had a trig class. The rules and Laws are “imbedded” in my mind. By that I mean that I know trig, I just don’t remember how I learnt it. I’m usually good after reviewing a Law or identity.

I thought the ambiguity was the tangent of angles above 180; a difference in the direction of the vector.

I f you are given two sides and the angle which is not included in the triangle two solutions may be possible by the sine rule.

I'm sorry, I can't do a sketch tonight but I can try to describe it.

Draw a long horizontal line - you don't yet know its length so make it long.

draw a second line to intersect the first one, towards the left hand end, at the given angle to the first and of the given length.

The leaves the third side of the triangle to construct Do this by setting a compass to the second given length and set the centre point of the compasses to the free end of the sloping line.

scribe an arc,  cutting the horizontal line in one two or no places.

The case of two cuts describe two different triangles which satisfy the initial data and arise because in (180-a) = sin a

I will post a sketch when I can.

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Here you go, hope this helps.