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Math Forum / Mathematics / Undergraduate Math / April 2008a triangle
Let T be a triangle of area 6 square inches and perimeter 12 inches. Show that the triangle T can be cut into 2006 triangles each of which has perimeter strictly larger than 6 inches. > Let T be a triangle of area 6 square inches and perimeter 12 inches. Show > that the triangle T can be cut into 2006 triangles each of which has perimeter strictly > larger than 6 inches. Hint: a 3-4-5 right - triangle is an example of a T satisfying the given. Hint 2: you can make an infinite number of sub-triangles with perimeter strictly larger than 10 inches, 6 is too easy. Hint 3: remember: the perimeter of a (non-degenerate) triangle is strictly larger than twice the longest side. I don't get it, how is that possible...? 2006 smaller triangles? Are you combing 2 smaller triangles to make another triangle? > I don't get it, how is that possible...? 2006 smaller triangles? Are > you combing 2 smaller triangles to make another triangle? Sure enough, another of those Googler Gooslers from the house of spam. Make them thin. If you don't include the problem statement, eventually, this thread will turn to gibberish. It's important to keep the essential context. Read about why and other netiquette manners expected of posters at: http://oakroadsystems.com/genl/unice.htm > > Let T be a triangle of area 6 square inches and perimeter 12 inches. Show > > that the triangle T can be cut into 2006 triangles each of which has perimeter strictly > > larger than 6 inches. > > Hint: a 3-4-5 right - triangle is an example of a T satisfying the > given. But not the only one! > Hint 2: you can make an infinite number of sub-triangles with > perimeter strictly larger than 10 inches, 6 is too easy. I don't see that -- all sides of T might be less than 5, thus no two points of T have a distance of 5: Consider the triangle with sidelengths a = b = 4.65269812718231... and c = 2.69460374563538... (more precisely, a=b is solution of (6-a)^2 * (a-3) = 3 and c=12-2a). Among very many triangles, at least one must have very small area (here: area at most 6/2006), hence its perimeter is approximately twice its longest side and thus not much longer than about 9.3 > Hint 3: remember: the perimeter of a (non-degenerate) triangle is > strictly larger than twice the longest side. To maximize the minimal perimeter of the subtriangles, assume that AB is the longest side (of length c), let D be a point on BC with BD=eps, divide BD into 2006 parts by points P[0] = B, P[1], ..., P[n] = D. Then the triangles AP[k]P[k+1] have perimeter at least 2*(c-eps), the same is true for and AP[n]C. Thus we replace the limit length 6 inches with any value less than twice the longest side of T. What can we say about the longest side of T? As noted above, it may happen to be <4.653. By Heron, we have 36 = 6(6-a)(6-b)(6-c) or 6 = (6-a)(6-b)(6-c). At least one of the factors on the RHS is <=cubrt(6) = 1.817... hence at least one side is >= 4.18... We can do better: Wlog. a>=b>=c. Then c<=4 and hence 6-c>=2, (6-a)(6-b)<=3, hence at least one of these two factors is <=sqrt(3)=1.732... and a >= 4.267... But that implies c<=(12-4.267)/2 = 3.86..., (6-a)(6-b) <=2.811..., a >=6-sqrt(2.811...) = 4.3232... We could repeat this; in the limit we will find that a >= L and c<= 6-L/2 where L is determined by 6-sqrt(6/(6-L/2)) = L or 12 = (6-L)^2 * (12-L) sqrt(6/(12-2L)) = 6-L, hence L = 6-cubrt(3) = 4.56775... Since the triangle with a = 6-cubrt(3) = 4.56775... b = c = 3+cubrt(3)/2 = 3.7211... is indeed possible, we see that the conditions A=6, p=12, a>=b>=c allow us to always produce arbitrarily many subtriangles such that each has perimeter >M *if and only if* M<12-2cubrt(3) = 9.1155... hagman |
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