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wu :: forums    riddles    medium (Moderators: Icarus, ThudnBlunder, Grimbal, towr, SMQ, Eigenray, william wu)    prove the rituo= sqrt 2 « Previous topic | Next topic » Pages: 1  Reply Notify of replies Send Topic Print    Author  Topic: prove the rituo= sqrt 2  (Read 2836 times) tony123 Junior Member     Posts: 61 prove the rituo= sqrt 2   « on: Jun 26th, 2007, 5:54pm » Quote Modify look and try prove http://www.mathyards.com/attach/upload2/wh_87858887.PNG IP Logged tony123 Junior Member     Posts: 61 Re: prove the rituo= sqrt 2   « Reply #1 on: Jun 27th, 2007, 1:15am » Quote Modify these problem is very hard IP Logged jollytall Senior Riddler     Gender: Posts: 585 Re: prove the rituo= sqrt 2   « Reply #2 on: Jun 27th, 2007, 1:37am » Quote Modify IT seems to be really hard to prove. Especially because I got sqrt(3/2).   Edit: It is not that hard, if I do not make a mistake. Then I also get sqrt(2). « Last Edit: Jun 27th, 2007, 6:26am by jollytall » IP Logged towr wu::riddles Moderator Uberpuzzler Some people are average, some are just mean.     Gender: Posts: 13730 Re: prove the rituo= sqrt 2   « Reply #3 on: Jun 27th, 2007, 2:40am » Quote Modify on Jun 27th, 2007, 1:15am, 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] « Last Edit: Jun 27th, 2007, 4:04am by towr » IP Logged Wikipedia, Google, Mathworld, Integer sequence DB tony123 Junior Member     Posts: 61 Re: prove the rituo= sqrt 2   « Reply #4 on: Jun 27th, 2007, 3:48am » Quote Modify 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 IP Logged tony123 Junior Member     Posts: 61 Re: prove the rituo= sqrt 2   « Reply #5 on: Jun 27th, 2007, 3:49am » Quote Modify on Jun 27th, 2007, 3:48am, 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 IP Logged pex Uberpuzzler     Gender: Posts: 880 Re: prove the rituo= sqrt 2   « Reply #6 on: Jun 27th, 2007, 3:50am » Quote Modify on Jun 27th, 2007, 2:40am, towr wrote: the radius of the small circle can be put at sqrt(1)-1   That's a very small circle...   IP Logged towr wu::riddles Moderator Uberpuzzler Some people are average, some are just mean.     Gender: Posts: 13730 Re: prove the rituo= sqrt 2   « Reply #7 on: Jun 27th, 2007, 4:04am » Quote Modify on Jun 27th, 2007, 3:50am, pex wrote: That's a very small circle...   Err.. yes.. Well, it's not like we haven't proved 1=2 IP Logged Wikipedia, Google, Mathworld, Integer sequence DB towr wu::riddles Moderator Uberpuzzler Some people are average, some are just mean.     Gender: Posts: 13730 Re: prove the rituo= sqrt 2   « Reply #8 on: Jun 27th, 2007, 4:08am » Quote Modify on Jun 27th, 2007, 3:48am, 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? IP Logged Wikipedia, Google, Mathworld, Integer sequence DB jollytall Senior Riddler     Gender: Posts: 585 Re: prove the rituo= sqrt 2   « Reply #9 on: Jun 27th, 2007, 4:26am » Quote Modify 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. « Last Edit: Jun 27th, 2007, 6:27am by jollytall » IP Logged towr wu::riddles Moderator Uberpuzzler Some people are average, some are just mean.     Gender: Posts: 13730 Re: prove the rituo= sqrt 2   « Reply #10 on: Jun 27th, 2007, 5:00am » Quote Modify on Jun 27th, 2007, 4:26am, 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 ? IP Logged Wikipedia, Google, Mathworld, Integer sequence DB jollytall Senior Riddler     Gender: Posts: 585 Re: prove the rituo= sqrt 2   « Reply #11 on: Jun 27th, 2007, 5:44am » Quote Modify That's what I wrote on paper and made a mistake when typed it in IP Logged pex Uberpuzzler     Gender: Posts: 880 Re: prove the rituo= sqrt 2   circles.PNG « Reply #12 on: Jun 27th, 2007, 6:16am » Quote Modify My solution:   hidden: 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. IP Logged jollytall Senior Riddler     Gender: Posts: 585 Re: prove the rituo= sqrt 2   « Reply #13 on: Jun 27th, 2007, 6:18am » Quote Modify For the second I also have 3:8, using the same logic.   And I also corrected the first one, so now having both OK. « Last Edit: Jun 27th, 2007, 6:28am by jollytall » IP Logged Barukh Uberpuzzler     Gender: Posts: 2276 Re: prove the rituo= sqrt 2   « Reply #14 on: Jun 27th, 2007, 6:42am » Quote Modify Nicely done, pex!   BTW, are green and red segments orthogonal? IP Logged towr wu::riddles Moderator Uberpuzzler Some people are average, some are just mean.     Gender: Posts: 13730 Re: prove the rituo= sqrt 2   « Reply #15 on: Jun 27th, 2007, 7:15am » Quote Modify on Jun 27th, 2007, 6:42am, 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. « Last Edit: Jun 27th, 2007, 8:06am by towr » IP Logged Wikipedia, Google, Mathworld, Integer sequence DB pex Uberpuzzler     Gender: Posts: 880 Re: prove the rituo= sqrt 2   « Reply #16 on: Jun 27th, 2007, 7:19am » Quote Modify on Jun 27th, 2007, 6:42am, 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... IP Logged tony123 Junior Member     Posts: 61 Re: prove the rituo= sqrt 2   « Reply #17 on: Jun 28th, 2007, 5:27am » Quote Modify no way IP Logged pex Uberpuzzler     Gender: Posts: 880 Re: prove the rituo= sqrt 2   « Reply #18 on: Jun 28th, 2007, 9:34am » Quote Modify on Jun 28th, 2007, 5:27am, tony123 wrote: no way   IP Logged pex Uberpuzzler     Gender: Posts: 880 Re: prove the rituo= sqrt 2   squareratio.PNG « Reply #19 on: Jun 28th, 2007, 10:56am » Quote Modify 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:   hidden: 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. 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