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Topic: A condition for sides of triangle (Read 1119 times) 

Aryabhatta
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A condition for sides of triangle
« on: May 16^{th}, 2007, 12:12am » 
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A surprising result, which appeared as a problem in a recent IMO. If n >= 3 is a positive integer and t_{1}, t_{2}, ..., t_{n} are positive reals such that [ (t_{1} + t_{2} + ... + t_{n}) (1/t_{1} + 1/t_{2} + ... + 1/t_{n}) ] = n^{2} Then any (t_{i}, t_{j}, t_{k}) form the sides of some triangle. (i,j,k are distinct) ([x] = integral part of x.)

« Last Edit: May 16^{th}, 2007, 12:13am by Aryabhatta » 
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Aryabhatta
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Re: A condition for sides of triangle
« Reply #1 on: May 16^{th}, 2007, 10:10am » 
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In fact we can show the following too I think: a) if n >= 25, then the t_{i,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 t_{i} satisfy the conditions of the problem, then 1 + e > t_{i}/t_{j} for all i,j. i.e if {t_{1}, t_{2}, ..., t_{n},...} is a sequence of t_{i}s which satisfy the conditions of the above problem for every n > M (pick any M) then they must all be equal.

« Last Edit: May 16^{th}, 2007, 10:24am by Aryabhatta » 
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Aryabhatta
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Re: A condition for sides of triangle
« Reply #2 on: May 22^{nd}, 2007, 1:32pm » 
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Maybe this should be moved to Putnam the section...


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Eigenray
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Re: A condition for sides of triangle
« Reply #3 on: May 24^{th}, 2007, 7:50am » 
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Replace n by n+2, and suppose the numbers are a < t_{1} < t_{2} < ... < t_{n} < b We may scale so that t_{i} = 1. Then by AMGM, we have t_{i} n, 1/t_{i} n Thus (a+b+n)(1/a+1/b+n) < (n+2)^{2} + 1, (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 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 a(1+r) + 1/a (1+1/r) < c Since a is real, this requires that the discriminant c^{2}  4(1+r)(1+1/r) > 0, and solving for r we have r < c^{2}/8  1 + {(c^2/8  1)^{2}  1} = 1+1/n+1/(8n^{2}) + (4n+1)/(8n^{2}){8n+1} = 1+{2/n}+O(1/n). In particular, if n 17 (or 19, in the original problem), then the largest ratio r=b/a < 1.409 < 2, so the smallest ratio is > 1/2. 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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Eigenray
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Re: A condition for sides of triangle
« Reply #4 on: May 24^{th}, 2007, 8:04am » 
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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+ ^{'} t_{i})(1/a+1/b+1/c+ ^{'} 1/t_{i}) = n + a/c+c/a+b/c+c/b + ^{'} (t_{i}/t_{j}+t_{j}/t_{i}), where the final ^{'} is over the C(n,2)2 pairs other than {a,c} and {b,c}. Each term in the sum is 2, so we have a/c+c/a+b/c+c/b + n+2(n(n1)/22) ( t_{i})( 1/t_{i}) < n^{2}+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 AMHM, u+v + 2[(1/u+1/v)/2] u+v + 4/(u+v), so we have s + 4/s < 5, and therefore s>1, as desired.

« Last Edit: May 24^{th}, 2007, 8:05am by Eigenray » 
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Aryabhatta
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Re: A condition for sides of triangle
« Reply #5 on: May 24^{th}, 2007, 1:41pm » 
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Well done Eigenray! Here is the proof for the IMO problem I had in mind: We are only dealing with positive reals. Lemma: if (a_{1} + ...+ a_{n})(1/a_{1} + ...+ 1/a_{n}) >= S^{2} then for any real a > 0 (a_{1} + ...+ a_{n}+ a)(1/a_{1} + ...+ 1/a_{n}+ 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' >= S^{2} we are done. Thus if there are some 3 numbers say t_{1}, t_{2}, t_{3} for which we have that (t_{1} + t_{2} + t_{3})(1/t_{1} + 1/t_{2} + 1/t_{3}) >= 10 = 3^{2} + 1 then we have by induction and using above lemma that: (t_{1} + t_{2} + .. + t_{n})(1/t_{1} + 1/t_{2} + ... + 1/t_{n}) >= n^{2} + 1 Now if t_{1}, t_{2}, t_{3} do not form the sides of a triangle, we can show that (t_{1} + t_{2} + t_{3})(1/t_{1} + 1/t_{2} + 1/t_{3}) >= 10 (Will update this post with the proof of the last fact later)


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Eigenray
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Re: A condition for sides of triangle
« Reply #6 on: May 24^{th}, 2007, 2:13pm » 
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I like that your proof gives a slightly stronger result. What argument do you have for the acute triangle when n 25? I have something simpler (bounding a and b near 1 independently) but it only works for n 36.


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Aryabhatta
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Re: A condition for sides of triangle
« Reply #7 on: May 24^{th}, 2007, 7:21pm » 
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I got n >=25 using the following: say x,y,z (among the t_{i}s) 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 (t_{1} + ... + t_{n})(1/t_{1} + ... + 1/t_{n}) >= (n+c)^{2} Now if c is such that for n >=25, (n+c)^{2} >= n^{2}+1 then we are done. For 3 numbers we can take them as 1, x and y where x >=1 and y^{2} >= x^{2} + 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(x^{2}+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(x^{2}+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 + (32sqrt(2))/sqrt(2) Thus if (3+c)^{2} = 9 + (32sqrt(2))/sqrt(2) then (n+c)^{2} >= n^{2} + 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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Aryabhatta
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Re: A condition for sides of triangle
« Reply #8 on: May 25^{th}, 2007, 9:43am » 
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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 t_{i} such that there is at least one right triangle. Here is an attempt to "prove" (i) In the lemma described above, if (a_{1} + ...+ a_{n})(1/a_{1} + ...+ 1/a_{n}) = S^{2} then for any real a > 0 (a_{1} + ...+ a_{n}+ a)(1/a_{1} + ...+ 1/a_{n}+ 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 t_{i} do not form an acute triangle where 1 <= x and y >= sqrt(x^{2} + 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(x^{2}+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 > (1c^{2})/2c > 12 Now (n+c)^^{2} >= n^{2} + 1 iff n >= (1c^{2})/2c > 12 I.e if n >= 13 then the triangles have to be acute. for part (ii) if n <= 12, We start out with a_{1} = 1, a_{2} = 1,a_{3} = sqrt(2) we keep adding new numbers a_{i} for which the equality in the lemma holds (a_{i} = sqrt(K/K')) For these numbers we have the n^{2}+1 > (n+c)^{2} = (1 + 1 + sqrt(2) + a_{4} + .. + a_{n})(1+ 1+ 1/sqrt(2) + ... + 1/a_{n}) and we have a right triangle by using 1,1 and sqrt(2)

« Last Edit: May 25^{th}, 2007, 9:50am by Aryabhatta » 
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