Trurl 9 Posted September 4, 2016 Author Share Posted September 4, 2016 Thanks for the link. The following is why I started researching this idea. It may seem silly but does anyone get anything from this? I'm not sure if anyone got anything from the previous posts but this is going to sound brilliant or a bunch of malarkey. Let me start by saying that this idea is just a theory and nothing is proven. This is just an example of how I approach a math problem. It is very much intuitive and visual. The idea: PNP is given and the unit circle will be used to find x and y. (These variables correspond to my previous equations.) The area encompassed by the angles on the unit circle will be used to form equations that find the values of x and y. Imagine a vector solved by the method of parallelogram addition. That is there is a triangle with 2 known sides x and y. Through vector addition and an angle of y/x radians between the sides the area can be found. This area is theoretically equivalent to N in value. The problem is N is the only value we know. To be useful we have to know y/x are at least some properties and proportions. We cannot use vector addition directly because N is only given. But it may be useful to use the unit circle to find which x and y will equal N. So a circle with radius x and a y arc length from an angle of y/x radians, has an area encompassed by the unit circle equal to N. This encompassed area is of angle y/x. If y/x is larger than 2Pi*x the angle encompasses one or more of the circle. That is the idea. I don’t know if it works, but it is how I go about math problems. This would relate to my theory of a logarithmic spiral that would show a pattern in the placement of Prime numbers. I have had this idea for quite some time. As seen in these URL’s: http://www.constructorscorner.net/ideas_and_gadgets/math/math_hunch/hunch_00001/hunches_section0005/trig_parabola.html http://www.constructorscorner.net/ideas_and_gadgets/math/math_hunch/hunch_00001/hunches_section0005/trig_parabola_verified.html http://www.constructorscorner.net/ideas_and_gadgets/math/math_hunch/hunch_00001/hunches_section0008/PrimeRevolutions.html If it is true this is just one step. There must be a way to solve the unknowns x and y in equation form. Remember PNP/y equals a distance of 1 radian. And remember this is only a preliminary idea I want input on. I am not clamming it works. I just think it is interesting enough to consider. Trurl [bJS1] Link to post Share on other sites

Trurl 9 Posted November 9, 2016 Author Share Posted November 9, 2016 I am going to keep this short, because it is just an idea I was working on and I have school work to do and not silly math ideas.But imagine a circle where N is the product of 2 Prime numbers. The radius of the circle is N, so in 1 radian a triangle is formed with a segment opposite the 1 radian angle is N. So an equilateral triangle is formed with all sides equaling N.But we want to know the angle of a triangle with the side opposite the obtuse angle is also of length N, but we have lengths "x" and "y" that are unknown sides. Could we take the N-equilateral triangle and subtract x from one of the end sides? The theory is that the y side would complete the triangle giving us the obtuse's angle in form of an equation with variables in N and x.Can anyone disprove that this will not geometrically solve an unknown triangle of one side N, which is known, and put into equation form x and y? I will add more to this when I have time. I can't insert the picture so it is difficult to visualize what I am describing. I know it sounds stupid, but there is some thought here. Let me know what you think. Trurl Link to post Share on other sites

imatfaal 2481 Posted November 9, 2016 Share Posted November 9, 2016 For starters an equilateral triangle has 60 degree corners - that is 2pi/6 radians not 1 radian. (fyi - it is the curved edge of the segment/sector which measures the same as the radius when the angle is one radian; the straight line forming your equilateral triangle is slightly less than that) Also not sure what other triangle you mean - perhaps a diagram might be useful Link to post Share on other sites

Trurl 9 Posted November 18, 2016 Author Share Posted November 18, 2016 Ok, my last post wasn’t very clear. Here is what I was trying to do. N is known. x and y the 2 number that when multiplied together make N. I am trying to simplify the factoring by using a simple vector. This vector might be able to be solved to find the unknown factors x and y. Picture an angle between 2 lines. As the lines increase in distance so the angle stays the same, but the distance between the 2 lines also increases. So I start with a triangle whose segment opposite the (obtuse) angle between the lines equals N. Now I take this same length N and travel along the original 2 lines until the length along the lines approaches a segment between those lines as N. N is the point where the segment between the 2 lines will reach the limit of N. Beyond N the segment is bigger than N. This is where I need your help. Just to see if this is worth pursuing. The other limit (opposite end of the N segment) is 1 Radian with a radius of N. So, on a circle with radius N with 2 sides N and the angle between 1 Radian. I am not sure as you pointed out the triangle of 60 degrees is not 1 radian. But I am referring to the relative angle between side N and side N. So, the tangent of x equals N -x. N is known This is the tricky part! I have not solved this yet. But I am saying I could possibly use the equations I found a pattern in the Prime factors. Which until know is too complex. But Knowing N and the angle opposite N… I am saying there is a possibility of substituting the equations for x with a new simpler equation of x in this geometry. I am not claiming this works. It was just something I was thinking about. Remember this drawing is off. I need to make a clearer drawing. https://1drv.ms/f/s!Ao7PhUWlkaBtgQd7IjIjxkBjv3wz Link to post Share on other sites

imatfaal 2481 Posted November 18, 2016 Share Posted November 18, 2016 1. segment has a definite meaning in maths - it is the area of a circle defined by a line across the circle (chord) and the edge of the circle (arc); it does not appear that you are using this definition. 2. An obtuse angle is between 90 and 180 degrees - ie between pi/2 and pi radians; one radian does not fall in this range. I am going to say what I understand from yours so far A. Start with a triangle ABC - one edge of which between B and C is N long; call this edge a. The angle opposite it is A. - there are infinitely many triangles like this) B. Change the other two sides (whilst keeping length of side a = N) so that the other two sides b and c also have length =N -you now have a triangle with all three sides equalling N - this triangle is unique. Search on "Equilateral Triangle" - it has three identical sides and three identical angles - each side is N long and each angle is 60 degree or pi/3 radians This is where you have lost me - no angle in that triangle is 1 radian. If you take a circle with radius N - and mark off a distance around the circumference of N you will have an angle of 1 radian. This is not the same as you have done Link to post Share on other sites

Trurl 9 Posted November 21, 2016 Author Share Posted November 21, 2016 Yes, my drawing and explanation are off. But don’t disregard the idea yet. This is a one of the graphical representations of my algebraic work posted earlier in this post. Things to remember when looking at this drawing: This is a vector. Both x and y are Prime numbers We are looking for the smallest lengths x and y possible. (They represent Prime numbers after all. There are no angle sides multiples of x and y. Again, these are prime factors. So, there is no 2x or 2y or 3x and 3y. As the lengths of x and y increase, the angle between them decreases. This is significant because the vector addition of x and y is less than N. And for my solution x < y. sin(60 degrees) = 0.866025404sin(1 radian) = 0.841470985 cos(60 degrees) = 0.5cos(1 radian) = 0.540302306 This relates to the same error of my equations. I am working on this problem off and on. I will try and produce a better and corrected drawing. Link to post Share on other sites

Trurl 9 Posted December 10, 2016 Author Share Posted December 10, 2016 https://1drv.ms/i/s!Ao7PhUWlkaBthFy7UgDHUGEIh2Lm Here is my updated drawing. It could still be wrong. I may have some flawed logic. But I am seeing this idea develop, but I cannot explain it. It is abstract. I know if I write it out and keep it as simple as possible, maybe someone else will be able to see it also. That is what I hope. Let me know what you think. This is only the start of the problem. But I need some input that the vector addition with sides x and y (the 2 Prime numbers multiplied together with an angle between them (the angle approaches the limit of Pi radians) have a resultant of N (where N is the product of the 2 Prime numbers x and y). Let me know if this conveys anything. An idea is not as valuable when no one else can understand what you are trying to do. Link to post Share on other sites

imatfaal 2481 Posted December 10, 2016 Share Posted December 10, 2016 Sorry - but just no. If AB=BC=AC then it MUST be an equilateral triangle. And 60 degrees does NOT equal 1 radian - it is almost 5% out. That is not close enough. Link to post Share on other sites

Trurl 9 Posted December 20, 2016 Author Share Posted December 20, 2016 Imatfaal, I am not arguing that 60 deg = 1 radian. 1 rad is 180/Pi. I am only looking to find a symbolic value of angle AFC. The entire point of drawing the triangle with side N is to find angle AFC. As for degrees, it would be fine to find angle AFC in degrees. The question is can anyone symbolically find this angle. If you can I will tell you why I want to know its equation for. -1 Link to post Share on other sites

imatfaal 2481 Posted December 21, 2016 Share Posted December 21, 2016 OK - ignoring the elephant in the room that you are contradicting basic geometry... if N = xy which I think you said it did. Then N, x, and y CANNOT form a triangle. There are no primes where xy<x+y; for a triangle the length of the long side must be less than the sum of the other two sides. There is no angle AFC. And it turns out you are contradicting even more basic geometry. Link to post Share on other sites

Trurl 9 Posted January 4, 2017 Author Share Posted January 4, 2017 https://1drv.ms/i/s!Ao7PhUWlkaBthGPYxf06XrWnNzKL Definitely some problems with my geometric representation. Does this look better? Link to post Share on other sites

imatfaal 2481 Posted January 4, 2017 Share Posted January 4, 2017 What have you changed? And can you not upload to the site; I will not download documents onto my windows pc ( I don't mind doing so to my linux box but that is not available at present) Link to post Share on other sites

Trurl 9 Posted January 14, 2017 Author Share Posted January 14, 2017 Ok, sorry I haven’t replied sooner. I have changed that s is now the remainder of N/Pi. That is a 180 the max angle (straight line). I did this because I couldn’t remember the modulus definition. I am saying that if N is much larger than x and y the Prime factors that x would wrap around the half circle (180 degrees) N/Pi times. The remainder would form a triangle where x and y would add to find the remainder of N/Pi as the similar triangle AFC. So triangle AFC is similar to triangle xys. (Yes I know the label isn’t standard.) Why did I do this? There is no guarantee this will make the solution simpler. But I already have proven equations for x and y as related in terms of x and N. I was going to use these equations and place them to the sides of the triangle and use the equations we know about triangles to solve for a pattern. This is a long shot, but I have always pictured the Prime factorization as a logarithmic spiral. That is why I wrapped x at an angle around the circle. I realize this theory has somehow gotten away from me. It is extremely confusing. But I stand behind my underlying theory. As you read my earlier post on this thread, I have found patterns in the Prime factorization. The resulting equations are just impossible to solve to be useful. I thought since I cannot solve the polynomial equation, I would make a graphical representation, but I need to reevaluate this. Perhaps if I map out the entire idea, you guys could assist me in making it makes sense. I am have been busy and it takes time but I think I should explain more, even though this post is an attempt to explain. But in the coming weeks I will work on this. Try not to laugh at me too much. Link to post Share on other sites

Trurl 9 Posted January 16, 2017 Author Share Posted January 16, 2017 There are still problems. s does not equal N/Pi but angle AFC should. That is if the triangles are similar. The only place that the sides of the triangle should be an integer is at x and y. AFC = (remainder(N/Pi)) * Pi (In theory) Link to post Share on other sites

Trurl 9 Posted February 13, 2017 Author Share Posted February 13, 2017 Ok so my graphical representation has many flaws. I’m not finished with it yet. But did you guys look at the last attempt? But back to Post # ___16___ and ____19_____ I know it is no longer the Prime factorization problem if you test for values of x knowing N. I argue that with the test value of x you will know if the equation results in the given number N. The closer to the actual x the closer to the given, N. I know you guys don’t like pseudo-code. I am trying to write a program that will test x values. I know we have been down this road before but this is for an education project I am doing. I would explain the details but it would influence your input. I am going to write a program to efficiently test x. If you graph “x” you know where it approaches N. My arguments are in past posts. But I ask to move away from the Prime Factorization to a trail-and-error computer program. Yes, I believe my equation can eliminate calculations on large Semi-Prime numbers. But this is all you need to know for now. Please try and write pseudo-code if not to prove me wrong but to promote education. Thank you for your participation. Link to post Share on other sites

Trurl 9 Posted February 18, 2017 Author Share Posted February 18, 2017 I will start the pseudo code. Notice that an x of 545 is positive and an x test value of 8756 yields a negative. I am arguing that with these test values the difference from PNP shows if the desired x is higher or lower than the test value. There is no reason that calculus won’t give a value when limit (equation) when PNP approaches 0. I know it takes more than that but do you agree the pseudo code will show an indication where x will fall at PNP? Please join this post. If you don’t believe my code, counter it. There should be a pattern in the pseudo-code. x = 545 y = 6737 PNP = 4639* y (((((x^2*PNP^4 + 2*PNP^2*x^5) + x^8)/ PNP^4) - ((1 - x^2/(2*PNP))))*((PNP^2/x^2))) 545 6737 31252943 (566741960869155702888306342808481973/580236226342089968450) 566741960869155702888306342808481973/580236226342089968450 N[566741960869155702888306342808481973/580236226342089968450, 13] 9.767434971132*10^14 Sqrt[9.76743497113228416002242796713336`13.*^14] 3.125289581964*10^7 PNP - 3.125289581963931252118627996719126077423156917`13.\ 301029995663981*^7 47.18036 test second x x = 8756 y = 6737 PNP = 4639* y (((((x^2*PNP^4 + 2*PNP^2*x^5) + x^8)/ PNP^4) - ((1 - x^2/(2*PNP))))*((PNP^2/x^2))) 8756 6737 31252943 73243982077295884748898890760446202375/74884743323939619512464 N[73243982077295884748898890760446202375/74884743323939619512464, 14] 9.7808951231166*10^14 Sqrt[9.780895123116592692523297300191834`14.*^14] 3.1274422653530*10^7 PNP - 3.127442265353046080516318622627538568244967294`14.\ 301029995663981*^7 -21479.653530 Link to post Share on other sites

Trurl 9 Posted April 21, 2017 Author Share Posted April 21, 2017 It is late 20170420…I still stand behind my triangles. You guys are more experienced mathematicians than me. You should have this thing solved already. I know the problem seems erroneous, but good math comes from new ideas. I mean if we always got a clear answer or it was too easy, there would be no point to doing the math. We’d all be English majors. But following is the latest attempt to show the Prime factorization problem may give clues on how to defeat it. It starts with an equilateral triangle, where all sides equal N. Then BC is subtracted by CE, where CE = the remainder of N / Pi. With alternating angles ECD = 30 degrees. FC = CE/cos(30 degrees). s = CE, which also equals the remainder of N / Pi. AEC is similar to syx. s/AC as x/FC as y/AF. Otherwise stated: s is proportional to AC as x is proportional the FC Currently I have not proven all my values. But they are based on a plan and not just random value assignments. Does this intrigue anyone? No, I am not claiming this works yet. I just wanted some feedback. Link to post Share on other sites

imatfaal 2481 Posted April 21, 2017 Share Posted April 21, 2017 Even being generous and ignoring the rubbish (remainder when divided by pi?) then this line is demonstrably wrong FC = CE/cos(30 degrees). This would only be the case if triangle CEF were a right angled triangle. It cannot be a right angled triangle as CE is an irrational number (whole number minus Pi) and BC is a whole number; half a whole number cannot be an irrational This too.. AEC is similar to syx. Those triangles would only be similar if lines CE and CF were coincident. They are not by definition. This ruins your ratios. And even though it would be a huge abuse of power - and would be immediately reversed by the other moderators - I will be tempted to suspend you with extreme prejudice if you EVER EVER say that 1 radian equals 60 degrees again. This is just arrant nonsense. 60 degrees is one sixth of a circle - 1 radian is one (two*pi)th of a circle; surely you understand that this is not the same? Link to post Share on other sites

Trurl 9 Posted April 23, 2017 Author Share Posted April 23, 2017 Even being generous and ignoring the rubbish (remainder when divided by pi?) then this line is demonstrably wrong We are already in agreement that is wrong. I just didn’t update the drawing. I wanted to show I was using the same drawing. It takes longer to update the drawing. Those triangles would only be similar if lines CE and CF were coincident. They are not by definition. This ruins your ratios. You are right. I meant to say AFC is similar to syx. This would only be the case if triangle CEF were a right angled triangle. It cannot be a right angled triangle as CE is an irrational number (whole number minus Pi) and BC is a whole number; half a whole number cannot be an irrational FC = CE/cos(30 degrees) should be: CE + (CE/cos(30 degrees)) = FC, where CE = s = [remainder of N/PI) Proof: http://www.constructorscorner.net/ideas_and_gadgets/math/scos_ssin/scosinep1.htm Thanks again for bearing with me as I work through this problem. I will update later with a clearer drawing and definitions. I may seem that I’m an idiot throwing math together. But there is a design around this problem. This geometry might prove to be impossible to solve. Also it is confusing. Be assured that I am not intentionally trying to irritate you with bad math. I believe these corrections will let you better see how I approached the problem. Link to post Share on other sites

imatfaal 2481 Posted April 23, 2017 Share Posted April 23, 2017 Similar triangles looks ok from first glance Not going to bother looking through that dreadful proof page - just grok the distances. CF is NOT 1.866 times longer than CE. You cannot uniquely (or even partially) determine triangles with just one angle. All you know is that angle ECF is 30 degrees - until you can bring some other constraints into your calculations you cannot determine a ratio of side lengths. Draw a DECENT diagram - with measured angles and lengths and this would be obvious. Better still spend time at Khan Academy and learn basic maths and geometry before trying to solve one of old problems in subject Link to post Share on other sites

Trurl 9 Posted May 8, 2017 Author Share Posted May 8, 2017 Ok so it looks like my triangle theory has failed. But it doesn’t matter. If an idea does not work it is just a dud. Here is why I designed the triangle the way I did. S = r * theta. In radians of course, but for the problem I converted it to degrees since my triangle is equilateral. I was looking for a triangle whose sides are proportional to the vector that with resultant of N, where N is the product of 2 Primes. I theorized that Pi radians or 180 degrees would divide into N and leave the remained of N/Pi (converted to degrees). And from this remainder N – the remainder; and CE = remained of N/Pi = s (on the unknown triangle) would solve the proportions of x and y, where x * y = N. We agreed that triangle syx was similar to triangle ACE. But I believe that triangle ECD is similar to triangle EBA. Of course, I have not proved it yet, but if true the triangle diagram would be useful. I began to think about drawing an accurate diagram. I am in the process I just wanted to run the idea by the message board. CE is much smaller part of N than my drawing shows. I am not claiming the triangles are similar, but I am going to do the work to find out. If this message is hard to understand, give it some leeway. I have not had a class in trig in 20 years. Also, it is difficult to explain why I choose to draw the diagram as I did. Over the next few weeks I will post the end. Right or wrong. Most likely wrong because of the difficulty of the problem. Even if this idea is a dud, I stand behind my previously posted equations. There is a pattern shown by the equations. It is just unfortunate that it is complex. But my next step will be to simplify the equations. My patterns come out of very simple patterns in multiplication. I want to post the patterns on my website to show how simple they are. I know that N = x *y is supposed to be a one-way function because there are 2 unknowns. However, I don’t believe in one-way functions. Yes, I know I’ve wasted a lot of time on an impossible problem, but it was geometry that gave me a lead. The problem is that no one believes your problem until you can prove it. So, in a few weeks, I will conclude my work on this ever-confusing drawing and post an improved diagram. Link to post Share on other sites

imatfaal 2481 Posted May 9, 2017 Share Posted May 9, 2017 Ok so it looks like my triangle theory has failed. But it doesn’t matter. If an idea does not work it is just a dud. Here is why I designed the triangle the way I did. S = r * theta. In radians of course, but for the problem I converted it to degrees since my triangle is equilateral. I was looking for a triangle whose sides are proportional to the vector that with resultant of N, where N is the product of 2 Primes. I theorized that Pi radians or 180 degrees would divide into N and leave the remained of N/Pi (converted to degrees). And from this remainder N – the remainder; and CE = remained of N/Pi = s (on the unknown triangle) would solve the proportions of x and y, where x * y = N. We agreed that triangle syx was similar to triangle ACE. But I believe that triangle ECD is similar to triangle EBA. Of course, I have not proved it yet, but if true the triangle diagram would be useful. I began to think about drawing an accurate diagram. I am in the process I just wanted to run the idea by the message board. CE is much smaller part of N than my drawing shows. I am not claiming the triangles are similar, but I am going to do the work to find out. If this message is hard to understand, give it some leeway. I have not had a class in trig in 20 years. Also, it is difficult to explain why I choose to draw the diagram as I did. Over the next few weeks I will post the end. Right or wrong. Most likely wrong because of the difficulty of the problem. Even if this idea is a dud, I stand behind my previously posted equations. There is a pattern shown by the equations. It is just unfortunate that it is complex. But my next step will be to simplify the equations. My patterns come out of very simple patterns in multiplication. I want to post the patterns on my website to show how simple they are. I know that N = x *y is supposed to be a one-way function because there are 2 unknowns. However, I don’t believe in one-way functions. Yes, I know I’ve wasted a lot of time on an impossible problem, but it was geometry that gave me a lead. The problem is that no one believes your problem until you can prove it. So, in a few weeks, I will conclude my work on this ever-confusing drawing and post an improved diagram. I was looking for a triangle whose sides are proportional to the vector that with resultant of N, where N is the product of 2 Primes. This is initial problem - take any two numbers greater than two and multiply them together and the product will ALWAYS be greater than the sum. BUT for a triangle the length of two shorter sides must SUM to greater than the third. This is an internal and irresolvable contradiction. The long side is the product of the two shorter and thus will be greater than the sum of the two shorter sides BUT AT THE SAME TIME for it to be a properly formed triangle it must also be less than the sum of the two shorter sides. If you get a contradiction like this you know you must restart with new propositions. I theorized that Pi radians or 180 degrees would divide into N and leave the remained of N/Pi (converted to degrees). And from this remainder N – the remainder; and CE = remained of N/Pi = s (on the unknown triangle) would solve the proportions of x and y, where x * y = N. The problem with introducing pi like this is that pi is an irrational number (it is actually transcendental). An integer multiplied or divided by an irrational will lead to another irrational. I think - but have not checked - that your construction will have sides all of which will be irrational lengths. All primes are integers - no primes are irrational We agreed that triangle syx was similar to triangle ACE. But I believe that triangle ECD is similar to triangle EBA. Of course, I have not proved it yet, but if true the triangle diagram would be useful. Triangle ECD would be similar to EBA for only one version (ie one possible prime if the idea worked) and that would be when angle ABE equalled angle EDC with both being 60 degrees or 2pi/6. Angles AEB and CED are equal. But that is the limit of similarity in the general case So, in a few weeks, I will conclude my work on this ever-confusing drawing and post an improved diagram. First things first. Explain how you will construct triangle - you have three sides explain the relationship between them and your primes. And let's check you can actually construct a triangle within those parameters. Remember if the angles are all 60 degrees then you must have equal side lengths - non-negotiable; and vice versa. Also remember three side lengths uniquely determine a triangle - so that is a good starting point. Link to post Share on other sites

Trurl 9 Posted May 11, 2017 Author Share Posted May 11, 2017 I haven’t confirmed the Prime number part, but the triangles are drawn correctly. The only change is that triangle EDC is a right triangle and for triangle ABE to be similar another 4.789 chord must be drawn at a 30 degree angle from CD. So this new angle is similar. Similar to triangle ABE 4.789 comes from the fact that CE equals [the remainder of (N / Pi ) ] * 85 All of the equilateral sides of the main triangle = N = 85 I have a drawing but it is from AutoCAD 14 and an AutoCad 14 dxf file. I am working to convert. Also there are 2 more similar triangles. Triangle BED is similar to triangle AEC The problem is I can’t find out if triangle BED is also similar to triangle CEF We know s = CE and triangle syx is similar to triangle AFC Anyways I am now confused from looking at hundreds of angles. I need help to find if I have enough information to solve triangle AFC. Please help! My drawing is ready. I just have to fight to get it into readable format. This is more difficult than I thought: 32 and 64 bit; AutoCad 14 ; dxf; dwg; So now you will have to believe me that the values work. It really isn't that impressive. Especially unimpressive if I can still not solve for triangle syx as I originally intended to do. https://1drv.ms/f/s!Ao7PhUWlkaBtgQd7IjIjxkBjv3wz Here is a link if you can open AutoCad 14 drawings. Link to post Share on other sites

imatfaal 2481 Posted May 11, 2017 Share Posted May 11, 2017 I haven’t confirmed the Prime number part, but the triangles are drawn correctly. The only change is that triangle EDC is a right triangle and for triangle ABE to be similar another 4.789 chord must be drawn at a 30 degree angle from CD. So this new angle is similar. Similar to triangle ABE 4.789 comes from the fact that CE equals [the remainder of (N / Pi ) ] * 85 All of the equilateral sides of the main triangle = N = 85 I have a drawing but it is from AutoCAD 14 and an AutoCad 14 dxf file. I am working to convert. Also there are 2 more similar triangles. Triangle BED is similar to triangle AEC The problem is I can’t find out if triangle BED is also similar to triangle CEF We know s = CE and triangle syx is similar to triangle AFC Anyways I am now confused from looking at hundreds of angles. I need help to find if I have enough information to solve triangle AFC. Please help! My drawing is ready. I just have to fight to get it into readable format. This is more difficult than I thought: 32 and 64 bit; AutoCad 14 ; dxf; dwg; So now you will have to believe me that the values work. It really isn't that impressive. Especially unimpressive if I can still not solve for triangle syx as I originally intended to do. https://1drv.ms/f/s!Ao7PhUWlkaBtgQd7IjIjxkBjv3wz Here is a link if you can open AutoCad 14 drawings. The only change is that triangle EDC is a right triangle and for triangle ABE to be similar another 4.789 chord must be drawn at a 30 degree angle from CD. So this new angle is similar. Similar to triangle ABE If EDC is right triangle with Angle EDC (per your autocad) as 90degs then triangle ABE can never be similar. Angle ABE is 60 degrees, Angle BAE is less than 60, and Angle AEB is the same as Angle CED (and cannot be 90 degrees) . To summarise EDC is a right triangle and ABE cannot have right angle. Therefore not similar Triangle BED is similar to triangle AEC It is not. Angle AEC is same as Angle BED. Angles EAC and DBE are less than 60 degrees. Angle ACE is 60 degrees thus for triangles to be similar then EDB must be 60 degrees and it is clearly not. The problem is I can’t find out if triangle BED is also similar to triangle CEF Think not. Cannot be bothered to prove Link to post Share on other sites

Trurl 9 Posted May 15, 2017 Author Share Posted May 15, 2017 Think not. Obviously, you didn’t look at the drawing I posted. This is a science and engineering forum. There must be some viewer who has access to Autocad. I know you think, I’m stupid claiming to work with Prime numbers. And you don’t think I have a math background. I enjoy you reading my problem and telling me when something just doesn’t work. However, you should have viewed the drawing before dismissing my comments. If I am wrong your judgment is correct. However, there is no wrong. There are wrong techniques, but failures just mean that I try other approaches. In this problem, I am not asking you to find Prime numbers, I am asking the group to find techniques to find triangles with limited given. If you want to reverse a one-way function, you will have to use new ideas, because the old ones don’t work either. I understand why you think my problem is crap. But don’t think of it as supposed to solve Prime numbers. Think of it as a geometry problem where limited information is known about the angle. Yes, I could be wrong, but I believe I am right-on about the similar angles. The question is does it help me solve the unknown values of the triangle I need. As you said before a drawing will show everything. I don’t have any programs to draw triangles other than AutoCad 14. The AutoCad 14 files I shared in my last post should open in the current version. I am 20 years behind when it comes to CAD. They are just too expensive to buy. I have the student edition of Solid Works, but I have-to draw as I learn. So, if you can view my drawing. It may be awhile before I have a drawing in universal format. Link to post Share on other sites

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