Geometric puzzle

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JeanFrancois
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Geometric puzzle

Post by JeanFrancois » Tue Feb 19, 2013 6:07 pm

Find t = CB/OA
puzzleannul.gif
puzzleannul.gif (16.83 KiB) Viewed 7108 times
Not too hard?

thundre
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Re: Geometric puzzle

Post by thundre » Tue Feb 19, 2013 6:39 pm

Since we just want ratios, let OC=1, AB=1/2.

Define b = OB.

b2 - (b-1/2)2 = 1/2
b - 1/4 = 1/2
b = 3/4

t = (1-b) / (b - 1/2) = 1
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jaap
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Re: Geometric puzzle

Post by jaap » Tue Feb 19, 2013 7:19 pm

Expand
Let the radii be r1 = OA, r2 = OB, r3 = OC.

That the areas are equal gives us the equation:
r12 + (r32-r22) = r22-r12

The lengths being equal gives
r3 = 2(r2-r1)

and we want to find (r3-r2)/r1.

From the first equation, we get
r32 = 2 r22 - 2 r12 = 2(r2-r1)(r2+r1)
= r3(r2+r1) by substituting the second equation.

Therefore r3 = r1+r2, and so (r3-r2)/r1 = 1.

JeanFrancois
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Re: Geometric puzzle

Post by JeanFrancois » Wed Feb 20, 2013 1:58 am

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
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Re: Geometric puzzle

Post by jaap » Wed Feb 20, 2013 10:51 am

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?
Expand
From my previous post, we have:
r3 = 2(r2-r1)
and
r3 = r1+r2

This leads to
2(r2-r1) = r1+r2
r2 = 3 r1

and substituting this in either of the first equations gives
r3 = 4 r1

The area of the outer ring is pi times
r32 - r22 = 16 r12 - 9 r12 = 7 r12
which is 7 times the area of the inner disc.

JeanFrancois
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Re: Geometric puzzle

Post by JeanFrancois » Wed Feb 20, 2013 6:34 pm

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 semi-primes.

The area of an annulus is equal to
Pi*(R^2-r^2)

We know that any odd semi-prime is equal to n=y^2-x^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 semi-prime number will be over (I`m not dreaming :D ).

I`m waiting for your comments before suggesting something more strange.

Thank you.

Please try to read carefully what I`m saying.

thundre
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Re: Geometric puzzle

Post by thundre » Wed Feb 20, 2013 11:18 pm

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?
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jaap
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Re: Geometric puzzle

Post by jaap » Thu Feb 21, 2013 12:49 am


JeanFrancois
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Re: Geometric puzzle

Post by JeanFrancois » Thu Feb 21, 2013 2:17 am

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.

thundre
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Re: Geometric puzzle

Post by thundre » Thu Feb 21, 2013 9:49 am

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).
I don't understand where those decimal numbers came from.

Given n=pq, you can compute the geometric mean sqrt(n) without knowing p and q. In this case, sqrt(91)=9.5394.
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JeanFrancois
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Re: Geometric puzzle

Post by JeanFrancois » Fri Feb 22, 2013 5:20 pm

thundre wrote:
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).
I don't understand where those decimal numbers came from.

Given n=pq, you can compute the geometric mean sqrt(n) without knowing p and q. In this case, sqrt(91)=9.5394.
After calculations I reached the fact that we can write
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.

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