On 26 Apr., 18:04, The Qurqirish Dragon <qurqiri...@[EMAIL PROTECTED]
> wrote:
> On Apr 26, 4:28 am, perash <mk_...@[EMAIL PROTECTED]
> wrote:
>
> > 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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