Geometric puzzle

 Posts: 29
 Joined: Wed Feb 13, 2013 10:32 pm
Geometric puzzle
Find t = CB/OA
Not too hard?
Not too hard?
Re: Geometric puzzle
Since we just want ratios, let OC=1, AB=1/2.
Define b = OB.
b^{2}  (b1/2)^{2} = 1/2
b  1/4 = 1/2
b = 3/4
t = (1b) / (b  1/2) = 1
Define b = OB.
b^{2}  (b1/2)^{2} = 1/2
b  1/4 = 1/2
b = 3/4
t = (1b) / (b  1/2) = 1

 Posts: 29
 Joined: Wed Feb 13, 2013 10:32 pm
Re: Geometric puzzle
Right!
But there is something strange.
The black area at the periphery is exactly equal to 7 times the black circle`s area inside???
Any clue why?
But there is something strange.
The black area at the periphery is exactly equal to 7 times the black circle`s area inside???
Any clue why?
Re: Geometric puzzle
JeanFrancois wrote:Right!
But there is something strange.
The black area at the periphery is exactly equal to 7 times the black circle`s area inside???
Any clue why?
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_{Jaap's Puzzle Page}

 Posts: 29
 Joined: Wed Feb 13, 2013 10:32 pm
Re: Geometric puzzle
Thank you for all the explanations.
Now here comes my feeling.
Any proof always start by some feeling.
Sometimes it is not easy to translate the feeling to proof particurlarly when you do not master all the tools.
There is a link with factoring big semiprimes.
The area of an annulus is equal to
Pi*(R^2r^2)
We know that any odd semiprime is equal to n=y^2x^2.
Refering to my picture above :
y= OB
x= OA
We can compute OC as equal to sqrt(2n)
We have an annulus we know its area
We have the value of OC known.
We know that :
OA is an integer
OB is an integer
We know that the black area is equal to the yellow one.
I think that there is only one solution to the problem.
What we need to know is one of the ratios :
 t=CB/OA
 s=black Area outside / black inner circle
I feel that we can find the solution and if we find it then the factorization of any semiprime number will be over (I`m not dreaming ).
I`m waiting for your comments before suggesting something more strange.
Thank you.
Please try to read carefully what I`m saying.
Now here comes my feeling.
Any proof always start by some feeling.
Sometimes it is not easy to translate the feeling to proof particurlarly when you do not master all the tools.
There is a link with factoring big semiprimes.
The area of an annulus is equal to
Pi*(R^2r^2)
We know that any odd semiprime is equal to n=y^2x^2.
Refering to my picture above :
y= OB
x= OA
We can compute OC as equal to sqrt(2n)
We have an annulus we know its area
We have the value of OC known.
We know that :
OA is an integer
OB is an integer
We know that the black area is equal to the yellow one.
I think that there is only one solution to the problem.
What we need to know is one of the ratios :
 t=CB/OA
 s=black Area outside / black inner circle
I feel that we can find the solution and if we find it then the factorization of any semiprime number will be over (I`m not dreaming ).
I`m waiting for your comments before suggesting something more strange.
Thank you.
Please try to read carefully what I`m saying.
Re: Geometric puzzle
I think you're right that there is a link to factoring large odd numbers (not just semiprimes). However, the link doesn't make the factoring any easier.
If you have n=29*31=899, you're going to be looking for y=30,x=1. (y=450,x=449 also works but doesn't help you factor.)
How would you go through the process of finding the factors?
If you have n=29*31=899, you're going to be looking for y=30,x=1. (y=450,x=449 also works but doesn't help you factor.)
How would you go through the process of finding the factors?

 Posts: 29
 Joined: Wed Feb 13, 2013 10:32 pm
Re: Geometric puzzle
There is a "big" difference between whay I`m suggesting and Fermat Method factorization.
Here is my suggestion before going far.
Imagine 2 scenarios :
1. Try to imagine that the black inner circle going to zero. We will have only 2 segments instead of three (OA,AB,BC).
2. Try to imagine the all yellow area being an annulus at the periphery. We will have only 2 segments instead of 3.
Take the values of segments (yellow part). Compute either their arithmetic (or geometric) mean. The value obtained will near the real value we are looking for. That value is the first factor p of n (n=p*q).
I tried for n=91=7*13
The geometric mean is equal to 6.139
The real value is equal to 6.4907 which is almost equal to 7 (p=7 and q=13).
You could try to compute all those values for biggest number (10 digits for example)
I`m waiting for your comments before going far.
Here is my suggestion before going far.
Imagine 2 scenarios :
1. Try to imagine that the black inner circle going to zero. We will have only 2 segments instead of three (OA,AB,BC).
2. Try to imagine the all yellow area being an annulus at the periphery. We will have only 2 segments instead of 3.
Take the values of segments (yellow part). Compute either their arithmetic (or geometric) mean. The value obtained will near the real value we are looking for. That value is the first factor p of n (n=p*q).
I tried for n=91=7*13
The geometric mean is equal to 6.139
The real value is equal to 6.4907 which is almost equal to 7 (p=7 and q=13).
You could try to compute all those values for biggest number (10 digits for example)
I`m waiting for your comments before going far.
Re: Geometric puzzle
I don't understand where those decimal numbers came from.JeanFrancois wrote:I tried for n=91=7*13
The geometric mean is equal to 6.139
The real value is equal to 6.4907 which is almost equal to 7 (p=7 and q=13).
Given n=pq, you can compute the geometric mean sqrt(n) without knowing p and q. In this case, sqrt(91)=9.5394.

 Posts: 29
 Joined: Wed Feb 13, 2013 10:32 pm
Re: Geometric puzzle
After calculations I reached the fact that we can writethundre wrote:I don't understand where those decimal numbers came from.JeanFrancois wrote:I tried for n=91=7*13
The geometric mean is equal to 6.139
The real value is equal to 6.4907 which is almost equal to 7 (p=7 and q=13).
Given n=pq, you can compute the geometric mean sqrt(n) without knowing p and q. In this case, sqrt(91)=9.5394.
n=p*q (p and q odd prime)
as n=f(t).
I mean n will depend only on one single factor t.
Ps : For now I will let the problem aside. I have to solve some problems (health and family) in real life.
Thank you both (Thundre and Jaap) for all your help.