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Title: A condition for sides of triangle Post by Aryabhatta on May 16th, 2007, 12:12am A surprising result, which appeared as a problem in a recent IMO. If n >= 3 is a positive integer and t1, t2, ..., tn are positive reals such that [ (t1 + t2 + ... + tn) (1/t1 + 1/t2 + ... + 1/tn) ] = n2 Then any (ti, tj, tk) form the sides of some triangle. (i,j,k are distinct) ([x] = integral part of x.) |
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Title: Re: A condition for sides of triangle Post by Aryabhatta on May 16th, 2007, 10:10am In fact we can show the following too I think: a) if n >= 25, then the ti,j,k triplets not only form the sides of a triangle, they form the sides of an acute angled triangle. b) Given an e > 0, there is some N (dependent on e) such that if n >= N and the ti satisfy the conditions of the problem, then 1 + e > ti/tj for all i,j. i.e if {t1, t2, ..., tn,...} is a sequence of tis which satisfy the conditions of the above problem for every n > M (pick any M) then they must all be equal. |
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Title: Re: A condition for sides of triangle Post by Aryabhatta on May 22nd, 2007, 1:32pm Maybe this should be moved to Putnam the section... |
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Title: Re: A condition for sides of triangle Post by Eigenray on May 24th, 2007, 7:50am Replace n by n+2, and suppose the numbers are We may scale so that http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/prod.gif ti = 1. Then by AM-GM, we have Thus (a/b+b/a) + n(a+1/a+b+1/b) < 4n+3, a+1/a + b+1/b < 4 + 1/n, since a/b+b/a http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/ge.gif 2. For n large, a+1/a and b+1/b must be close to 2, which means a and b must be close to 1, so their ratio is close to 1. This shows (b). For (a), suppose a+1/a + b+1/b < c = 4 + 1/n. Writing b=ra, we have Since a is real, this requires that the discriminant and solving for r we have = 1+1/n+1/(8n2) + (4n+1)/(8n2)http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif{8n+1} = 1+http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif{2/n}+O(1/n). In particular, if n http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/ge.gif 17 (or 19, in the original problem), then the largest ratio r=b/a < 1.409 < http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif2, so the smallest ratio is > 1/http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif2. So for any three of our numbers a,b,c, we have (a/c)2 + (b/c)2 > 1, which means that all triangles are acute. Note that the single largest ratio is obtained when all but the smallest and largest numbers are equal, so this doesn't give the best bound for showing all triangles are acute, which involves the sum of the squares of two different ratios. So it may be true for some smaller n as well. |
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Title: Re: A condition for sides of triangle Post by Eigenray on May 24th, 2007, 8:04am Oh, and the original problem (which is very nice): Isolate three of the numbers a,b,c. We need to show a+b>c, or that a/c + b/c > 1. So write (a+b+c+ http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gif' ti)(1/a+1/b+1/c+ http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gif' 1/ti) = n + a/c+c/a+b/c+c/b + http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gif' (ti/tj+tj/ti), where the final http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gif' is over the C(n,2)-2 pairs other than {a,c} and {b,c}. Each term in the sum is http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/ge.gif 2, so we have a/c+c/a+b/c+c/b + n+2(n(n-1)/2-2) http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/le.gif (http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gif ti)(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gif 1/ti) < n2+1. Letting u=a/c, v=b/c, we have u+1/u + v+1/v < 5, and we want to show s=u+v>1. By AM-HM, u+v + 2[(1/u+1/v)/2] http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/ge.gif u+v + 4/(u+v), so we have s + 4/s < 5, and therefore s>1, as desired. |
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Title: Re: A condition for sides of triangle Post by Aryabhatta on May 24th, 2007, 1:41pm Well done Eigenray! Here is the proof for the IMO problem I had in mind: We are only dealing with positive reals. Lemma: if (a1 + ...+ an)(1/a1 + ...+ 1/an) >= S2 then for any real a > 0 (a1 + ...+ an+ a)(1/a1 + ...+ 1/an+ 1/a) >= (S+1)2 Proof: Put K = Sigma a_i and K' = Sigma 1/a_i. Now we have (K+a)(K'+1/a) = KK' + 1 + K/a + aK' Now K/a + aK' >= 2sqrt(KK') by AM >= GM. Since KK' >= S2 we are done. Thus if there are some 3 numbers say t1, t2, t3 for which we have that (t1 + t2 + t3)(1/t1 + 1/t2 + 1/t3) >= 10 = 32 + 1 then we have by induction and using above lemma that: (t1 + t2 + .. + tn)(1/t1 + 1/t2 + ... + 1/tn) >= n2 + 1 Now if t1, t2, t3 do not form the sides of a triangle, we can show that (t1 + t2 + t3)(1/t1 + 1/t2 + 1/t3) >= 10 (Will update this post with the proof of the last fact later) |
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Title: Re: A condition for sides of triangle Post by Eigenray on May 24th, 2007, 2:13pm I like that your proof gives a slightly stronger result. What argument do you have for the acute triangle when n http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/ge.gif 25? I have something simpler (bounding a and b near 1 independently) but it only works for n http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/ge.gif 36. |
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Title: Re: A condition for sides of triangle Post by Aryabhatta on May 24th, 2007, 7:21pm I got n >=25 using the following: say x,y,z (among the tis) are not the sides of an acute angled triangle, and say we find a c such that (x+y+z)(1/x+1/y+1/z) >= (3+c)2 then by using the above lemma, we find that (t1 + ... + tn)(1/t1 + ... + 1/tn) >= (n+c)2 Now if c is such that for n >=25, (n+c)2 >= n2+1 then we are done. For 3 numbers we can take them as 1, x and y where x >=1 and y2 >= x2 + 1 We can show that (1+x+y)(1+1/x+1/y) is a increasing function of y (if y is the largest) So we consider the case when y = sqrt(x2+1) (1 + x + y)(1 + 1/x + 1/y) = 3 + (x+1/x) + (x/y + y/x) + (y + 1/y) >= 7 + (y + 1/y) Now y = sqrt(x2+1) >= sqrt(2) We have that y + 1/y is increasing for y >= 1 so y + 1/y >= 1/sqrt(2) + sqrt(2) So we have that (1 + x + y)(1 + 1/x + 1/y) >= 7 + 3/sqrt(2) = 9 + (3-2sqrt(2))/sqrt(2) Thus if (3+c)2 = 9 + (3-2sqrt(2))/sqrt(2) then (n+c)2 >= n2 + 1 for n >= 25. This is a crude (and not very simple) estimate and i think we can certainly do better. |
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Title: Re: A condition for sides of triangle Post by Aryabhatta on May 25th, 2007, 9:43am I just realised, we can actually give a best estimate for the acute angled triangle problem! We can show that: i) If n >= 13, then the triangles must all be acute. ii) if n < 13, we can find ti such that there is at least one right triangle. Here is an attempt to "prove" (i) In the lemma described above, if (a1 + ...+ an)(1/a1 + ...+ 1/an) = S2 then for any real a > 0 (a1 + ...+ an+ a)(1/a1 + ...+ 1/an+ 1/a) >= (S+1)2 In fact equality occurs if a = sqrt(K/K') (for definition of K and K' see proof of Lemma) Now suppose (wlog) 1,x,y among the ti do not form an acute triangle where 1 <= x and y >= sqrt(x2 + 1) Then consider (1+x+y)(1+1/x+1/y) which is increasing in y. So it is enough to consider the case when y = sqrt(x2+1) expanding we get 1 + x + 1/x + y + 1/y + x/y + y/x putting x = tan(z) where z is in [pi/4, pi/2) we can show that this is increasing in z (and hence in x) (i just used an online plotter, didn't try proving it, should be straight forward i suppose) Thus min occurs when x= 1 and y = sqrt(2) i.e (1+x+y)(1+1/x+1/y) >= 5 + 3sqrt(2) = (3+c)2 say. For this c, we see that (I used a calculator) 13 > (1-c2)/2c > 12 Now (n+c)^2 >= n2 + 1 iff n >= (1-c2)/2c > 12 I.e if n >= 13 then the triangles have to be acute. for part (ii) if n <= 12, We start out with a1 = 1, a2 = 1,a3 = sqrt(2) we keep adding new numbers ai for which the equality in the lemma holds (ai = sqrt(K/K')) For these numbers we have the n2+1 > (n+c)2 = (1 + 1 + sqrt(2) + a4 + .. + an)(1+ 1+ 1/sqrt(2) + ... + 1/an) and we have a right triangle by using 1,1 and sqrt(2) |
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