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I believe that I have discovered a mathematical formula

by creax4
Tags: discovered, formula, mathematical
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creax4
#1
Feb26-12, 04:39 AM
P: 6
I believe that I have discovered a mathematical formula related to trigonometry. It is used for finding the ratio of areas of triangles (not just similar triangles, but all kinds). I asked my Mathematics professor if any such formula for finding ratios exist. He said no. I also counter-checked in many websites. But again no. I believe that my formula can be applied as much as the similar triangles proportionality theorem or maybe more.
But the problem is I do not know what further action to take regarding this matter. I don't think patenting is possible but how did the other theorems become very famous?
I would like to receive some advice on what i should do.
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micromass
#2
Feb26-12, 08:31 AM
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I'm sorry to say this, but you're not going to become very famous with a trig formula anymore. It's fair to say that all basic trig formula's have been discovered already or can easily be deduced from other formula's.

Why don't you tell us what the formula is and we'll tell you if it's been done already?

You could always try to publish your work, but the chance that that will happen is zero. However, there are some journals for teachers where teachers and math enthousiasts publish their findings, so perhaps you can have some luck there.
Harrisonized
#3
Feb26-12, 09:12 AM
P: 209
If this formula is so great, why don't you reveal what exactly it does? Do you actually care about mathematics or do you just care about your reputation? If it is reputation you care about, this isn't the way to go. I highly doubt you will get your name on this "formula" that was probably discovered long before you had your revelation. (Probably all of trigonometry has already been "discovered".)

Regarding how people get their names on formulas: they do a lot more than just "discover" a formula. It's like how you don't just synthesize a new chemical compound that isn't in the literature and publish a paper on it. You have to do a lot more than that.

And what on earth gave you the idea that you could patent a formula? You don't want anyone to use this formula without your permission for 20 years? Lol.

creax4
#4
Feb26-12, 09:40 AM
P: 6
I believe that I have discovered a mathematical formula

I am not trying to be famous or gain reputation. And by patenting, I meant getting it published. I'm sorry to get the wrong meaning. Anyways, this theorem is used to find the ratio of area of all triangles formed in one big triangle if the dimensions of the three sides of the big whole triangle is given.
creax4
#5
Feb26-12, 09:46 AM
P: 6
And also its not trigonometry. Its geometry related to triangles and areas
espen180
#6
Feb26-12, 10:11 AM
P: 836
Whatever it is, it won't be new. If there had been any use for the formula you're talking about, it would have been widespread in textbooks. You can safely reveal what you are talking about, like micromass said.
Dadface
#7
Feb26-12, 11:02 AM
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Hello creax
the very fact that you are playing around with mathematics in the way that you are and presumably extending it beyond your normal studies seems to reveal that you have may have a spark of creativity.So what if it proves that this time your discovery is not new or if it contains mistakes.As long as you enjoy it and learn from it then all is well.
Pengwuino
#8
Feb26-12, 12:10 PM
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Quote Quote by creax4 View Post
I believe that I have discovered a mathematical formula related to trigonometry. It is used for finding the ratio of areas of triangles (not just similar triangles, but all kinds). I asked my Mathematics professor if any such formula for finding ratios exist. He said no. I also counter-checked in many websites. But again no. I believe that my formula can be applied as much as the similar triangles proportionality theorem or maybe more.
But the problem is I do not know what further action to take regarding this matter. I don't think patenting is possible but how did the other theorems become very famous?
I would like to receive some advice on what i should do.
One thing you'll find is that as you look into more obscure and possibly useless theorems, the chances you won't find it online or that a single professor won't know about it increases. The concept you described seems pretty dull, so don't put much effort into covering up exactly what your theorem does (it won't hit the NY times front page, for sure). Most research leads to dead ends and there is a good chance your theorem is wrong, so don't be afraid to let people see it. You're not announcing some new idea that would be worth billions of dollars on the drug market or something, so don't worry about other people seeing it :P
creax4
#9
Feb26-12, 12:20 PM
P: 6
Thank you Dadface for your encouraging words



Please open the embedded image.
Here, as you can see, in triangle ABC,
AF= 2cm, FE= 2cm, EB= 1cm, BD= 1cm, DC= 3cm and AC= 3cm
Is it possible to find the ratio of areas of triangles BED, EDF, FDC and AFC and also the fraction of area occupied by each of these triangles in the triangle ABC?
FeX32
#10
Feb26-12, 01:39 PM
P: 67
Your image doesn't work. I also agree with the above comments. And I agree with dadface in terms of creativity.
I work in research, and the 'spark' of creativity is very essential to making any breakthroughs. And yes, many times they lead to dead ends, but not always..
Cheers,
FeX32
#11
Feb26-12, 01:41 PM
P: 67
Another term commonly used is "thinking outside the box"
eddotman
#12
Feb26-12, 02:42 PM
P: 31
Quote Quote by creax4 View Post
Thank you Dadface for your encouraging words



Please open the embedded image.
Here, as you can see, in triangle ABC,
AF= 2cm, FE= 2cm, EB= 1cm, BD= 1cm, DC= 3cm and AC= 3cm
Is it possible to find the ratio of areas of triangles BED, EDF, FDC and AFC and also the fraction of area occupied by each of these triangles in the triangle ABC?
Your image doesn't seem to work, but based on your labelling, I'm assuming that you've created a large triangle, ABC, and inscribed on its edges a smaller triangle, DEF, thus forming four interior triangles. The given data seems to completely specify the lengths of the outer triangle and the locations of D, E and F. The area of the outer triangle and its interior angles are calculable since all its sides are known, and since we know where D, E and F are, we can use basic trig (law of cosines) to find the lengths of the sides of DEF. From there, we know all the lengths of every line segment, and all areas are calculable.
alan2
#13
Feb26-12, 02:57 PM
P: 206
I think that every one of us has discovered something only to find that it was already discovered. Best advice that I ever received from a professor was "If you discover something then you own it regardless of how many people discovered it before you". Keep thinking!
creax4
#14
Feb27-12, 12:53 AM
P: 6
Thank you all.
eddotman, your speculations were correct.

But to find the area of all the triangles, you need to know the base and height or the length of all the three sides of the triangles.

Here, the height is not given at all. Also, besides the big triangle, the length of only one side of the triangles are given. So is it possible to find the ratios and fractions?
If yes, please could you tell me how?
eddotman
#15
Feb27-12, 04:52 AM
P: 31
Quote Quote by creax4 View Post
Thank you all.
eddotman, your speculations were correct.

But to find the area of all the triangles, you need to know the base and height or the length of all the three sides of the triangles.

Here, the height is not given at all. Also, besides the big triangle, the length of only one side of the triangles are given. So is it possible to find the ratios and fractions?
If yes, please could you tell me how?
So, in this problem, we're given pretty much all the lengths of the big triangle (I think only one length, AE, is not given) and one length of the inner triangle (EF). Recall that whenever you inscribe a triangle inside another triangle, you create a bunch of similar triangles. In other words, for the four inner triangles you've created, all except the innermost triangle will share an angle with the big triangle. This means that certain length ratios are equal. For example, AC/AB = AF/AE and BD/DE = BC/AC and CF/DF = AC/AB. Thus, it's quite simple to use these ratios to solve for the lengths of the inner triangle. Once we know all the lengths, then we can easily find all the areas (either by law of cosines or Heron's formula).
creax4
#16
Feb27-12, 09:18 AM
P: 6
Quote Quote by eddotman View Post
So, in this problem, we're given pretty much all the lengths of the big triangle (I think only one length, AE, is not given) and one length of the inner triangle (EF). Recall that whenever you inscribe a triangle inside another triangle, you create a bunch of similar triangles. In other words, for the four inner triangles you've created, all except the innermost triangle will share an angle with the big triangle. This means that certain length ratios are equal. For example, AC/AB = AF/AE and BD/DE = BC/AC and CF/DF = AC/AB. Thus, it's quite simple to use these ratios to solve for the lengths of the inner triangle. Once we know all the lengths, then we can easily find all the areas (either by law of cosines or Heron's formula).


eddotman,
This is the image. I think you got the figure wrong. Please right click the image icon and go to the image url.
micromass
#17
Feb27-12, 09:40 AM
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P: 18,333
Quote Quote by creax4 View Post


eddotman,
This is the image. I think you got the figure wrong. Please right click the image icon and go to the image url.
This is really an elementary exercise in trigonometry.

|AB|=5, |AC|=3, |BC|=4. (note this is a right triangle, which could make it easier, but we won't use this fact for generality).
By the cosine rule we have

[tex]|AC|^2=|AB|^2+|BC|^2-2|AB||BC|\cos(B)[/tex]

Thus

[tex]\cos(B)=0.8[/tex]

and thus

[tex]\sin(B)=\sqrt{1-0.8^2}=0.6[/tex]

[tex]Area(BDE)=\frac{1}{2}|EB||ED|\sin B = 0.3[/tex]

[tex]Area(BDF)=\frac{1}{2}|FB||FD|\sin B = 0.9[/tex]

[tex]Area(BCF)=\frac{1}{2}|FB||BC|\sin B = 3.6[/tex]

[tex]Area(ABC)=\frac{1}{2}|AB||AC|\sin B = 6[/tex]

Now the ratio's are easily calculated.


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