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        riddles >> medium >> prove the rituo= sqrt 2
        
(Message started by: tony123 on Jun 26th, 2007, 5:54pm)
Title: prove the rituo= sqrt 2
Post by tony123 on Jun 26th, 2007, 5:54pm
look and try prove
http://www.mathyards.com/attach/upload2/wh_87858887.PNG
Title: Re: prove the rituo= sqrt 2
Post by tony123 on Jun 27th, 2007, 1:15am
these problem is very hard
Title: Re: prove the rituo= sqrt 2
Post by jollytall on Jun 27th, 2007, 1:37am
IT seems to be really hard to prove. Especially because I got [hide]sqrt(3/2)[/hide].

Edit: It is not that hard, if I do not make a mistake. Then I also get sqrt(2).
Title: Re: prove the rituo= sqrt 2
Post by towr on Jun 27th, 2007, 2:40am

on 06/27/07 at 01:15:15, tony123 wrote:
these problem is very hard

Actually, with the ratio to be proven given (and assuming it's correct), it might almost qualify as easy. Medium would be if you have to find the ratio as well. But even then the method to follow seems quite straightforward.

Given the ratio, the radius of the large (half)circles can be put at sqrt(2), then the radius of the small circle can be put at sqrt(2)-1 (Because heightwise we have the width of one large circle minus that of a small circle and the total is 2)
Now we only have to check the position of one small corner circle. it's center has to be sqrt(2)-1 from the bottom (or top), 2 sqrt(2)-1 from the top (or bottom) midpoint, and 1 from the bottom (or top) midpoint.
Which shouldn't be hard to confirm.

[edit]stupid typos[/edit]
Title: Re: prove the rituo= sqrt 2
Post by tony123 on Jun 27th, 2007, 3:48am
can you help me to prove Sangaku
and  2 Sangaku in these
http://www.mathyards.com/attach/upload2/wh_68869629.PNG

r(yello)/r(blue)=8/2
Title: Re: prove the rituo= sqrt 2
Post by tony123 on Jun 27th, 2007, 3:49am

on 06/27/07 at 03:48:20, tony123 wrote:
can you help me to prove Sangaku
and  2 Sangaku in these
http://www.mathyards.com/attach/upload2/wh_68869629.PNG

r(yello)/r(blue)=8/3
Title: Re: prove the rituo= sqrt 2
Post by pex on Jun 27th, 2007, 3:50am

on 06/27/07 at 02:40:56, towr wrote:
the radius of the small circle can be put at sqrt(1)-1


That's a very small circle...  ;)
Title: Re: prove the rituo= sqrt 2
Post by towr on Jun 27th, 2007, 4:04am

on 06/27/07 at 03:50:13, pex wrote:
That's a very small circle...  ;)
Err.. yes.. Well, it's not like we haven't proved 1=2 ;)
Title: Re: prove the rituo= sqrt 2
Post by towr on Jun 27th, 2007, 4:08am

on 06/27/07 at 03:48:20, tony123 wrote:
can you help me to prove Sangaku
and  2 Sangaku in these
http://www.mathyards.com/attach/upload2/wh_68869629.PNG

r(yello)/r(blue)=8/3
What should be proven here?
Or is it given that the green figure a square?
Title: Re: prove the rituo= sqrt 2
Post by jollytall on Jun 27th, 2007, 4:26am
Still on the first one my calculation:

Let the radius of the large circle be 1, the small a.

The  coordinates of the centre of the one in the right lower corner (measured from the bottom centre) (x,a).
(1-a)2=x2+a2.

The coordinates of the one in the right upper corner (x,y).
x2+y2=(1+a)2

l=y+a

a=(12+/-sqrt(96))/6. a<1, so a=2-sqrt(8/3).
l=sqrt(8/3)
L=2
L/l=sqrt(3/2).

a=(4+/-sqrt(16-8 )/4. a<1, so a=1-sqrt(1/2).
l=sqrt(2)
L=2
L/l = sqrt(2).

Edit: I corrected the typo what Towr mentioned (I calculated like that anyway).
Edit 2: I also corrected the main mistake of l=2-2a and not l=2-a as I did before.
Title: Re: prove the rituo= sqrt 2
Post by towr on Jun 27th, 2007, 5:00am

on 06/27/07 at 04:26:25, jollytall wrote:
The  coordinates of the centre of the one in the right lower corner (measured from the bottom centre) (x,a).
(1-a)2+a2=x2.

Or (1-a)2=a2+x2 ?
Title: Re: prove the rituo= sqrt 2
Post by jollytall on Jun 27th, 2007, 5:44am
That's what I wrote [hide]on paper and made a mistake when typed it in[/hide]
Title: Re: prove the rituo= sqrt 2
Post by pex on Jun 27th, 2007, 6:16am
My solution:

[hideb]See the attached diagram. The large circles have radius 1, the small circles have radius r. The width of the rectangle is L = 2; its height equals (diameter of a large circle) - (diameter of a small circle) = 2 - 2r. Thus, we are looking for the radius 2/(2 - 2r) = 1/(1 - r).

Consider the green segment. It is a radius of the large circle, hence it has length 1. On the other hand, it is also a radius of the small circle (length r) plus a segment connecting the two indicated points. We have
(x2 + r2)1/2 + r = 1
=> x2 + r2 = (1 - r)2
=> x2 = 1 - 2r
=> x = (1 - 2r)1/2, as x is clearly positive.

Now, consider the red segment. It is a radius of the large circle plus a radius of the small circle, hence it has length 1 + r. Thus, we have
(x2 + (2 - 3r)2)1/2 = 1 + r
=> 1 - 2r + (2 - 3r)2 = (1 + r)2
=> 4 - 16r + 8r2 = 0
=> r = 1 - (1/2)sqrt(2), as r is clearly less than one.

Thus, the desired ratio is 1/(1 - r) = sqrt(2), as desired.[/hideb]
Title: Re: prove the rituo= sqrt 2
Post by jollytall on Jun 27th, 2007, 6:18am
For the second I also have 3:8, using the same logic.

And I also corrected the first one, so now having both OK.
Title: Re: prove the rituo= sqrt 2
Post by Barukh on Jun 27th, 2007, 6:42am
Nicely done, pex!

BTW, are green and red segments orthogonal?
Title: Re: prove the rituo= sqrt 2
Post by towr on Jun 27th, 2007, 7:15am

on 06/27/07 at 06:42:56, Barukh wrote:
BTW, are green and red segments orthogonal?
No; it's easy to see if you've drawn the triangle betwene those two lines and the midline, that the squares of the two short sides don't add to the square of the long side.
Title: Re: prove the rituo= sqrt 2
Post by pex on Jun 27th, 2007, 7:19am

on 06/27/07 at 06:42:56, Barukh wrote:
Nicely done, pex!

BTW, are green and red segments orthogonal?


Thanks. The segments are not orthogonal: if they were, by Thales' Theorem, we would have x2 + (1 - 2r)2 = (1 - r)2. Here, (1 - r)2 = 1/2, but

x2 + (1 - 2r)2 = sqrt(2) - 1 + (sqrt(2) - 1)2 = 2 - sqrt(2) =/= 1/2,

although it comes close...
Title: Re: prove the rituo= sqrt 2
Post by tony123 on Jun 28th, 2007, 5:27am
no way
Title: Re: prove the rituo= sqrt 2
Post by pex on Jun 28th, 2007, 9:34am

on 06/28/07 at 05:27:19, tony123 wrote:
no way

???
Title: Re: prove the rituo= sqrt 2
Post by pex on Jun 28th, 2007, 10:56am
As jollytall pointed out, the same method of solution can be used for both problems. Here is mine for the second problem, similar to the first one:

[hideb]Let the largest semicircle have radius 1, so that the square has side length 2.  The larger of the full circles has radius R, the smaller circles have radius r. The square has height (radius of large semicircle) + (diameter of medium circle) - (diameter of small circle) = 1 + 2R - 2r. As this has to equal 2, we obtain R = 1/2 + r and the required ratio is R/r = 1 + 1/(2r).

Further, see the diagram. Again, we use the green segment to find x:
1 = r + (x2 + r2)1/2
=>  x2 + r2 = 1 - 2r + r2
=>  x = (1 - 2r)1/2, rejecting the negative root.

Again, we have the red segment left to find r:
R + r = (x2 + (3/2 - 2r)2)1/2
=>  (1/2 + 2r)2 = 1 - 2r + (3/2 - 2r)2
=>  1/4 + 2r + 4r2 = 13/4 - 8r + 4r2
=>  r = 3/10, which is the only possibility, as the equation is linear.

Thus, the required ratio is 1 + 1/(2r) = 8/3 - again, as required.

PS: And, again, the colored segments are not orthogonal.[/hideb]


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