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#1 10. 05. 2012 18:50

stuart clark
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no. of triangle

How many triangles are there with integer sides and perimeter $2007$ 

How many of these are equilateral

How any are isosceles

How many are scalene, i.e., neither of the first two kinds?

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#2 10. 06. 2012 14:47 — Editoval vanok (10. 06. 2012 16:05)

vanok
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Re: no. of triangle

Hi ↑ stuart clark:,

Remark: we shall count the triangles of sides has, b, c that only once, about is their permutation of sides.
Thus let us put, without losing the generality, that  a suitable  triangle has the sides  $a, b, c $and that$ a \geq b \geq c$

We know, thanks to the statement that $a+b+c=2007$, where $a, b, c$ are positive, not zeros numbers with furthermore $b+c \geq a \geq b-c$

Let us notice first of $a\in \{669; 1003 \} $


Let us describe now, how form methodically such triangles: we form for one gave all the possible triangles, and we make this for everything has possible.

FORMATIONS OF ALL THE TRIANGLES FOR ONE LOOKED FIX $a$:
We begin with the triangle,

$b =a; c=2007-2a$
then
$ b=a-1; c=2008-2a $
$b=a-2; c=2009-2a$
until
$ b=a-k; c=2007+k-2a$  (with k so big as possible).

So, we are on to have all the triangles of this type.
Let us show now 2 interesting properties.

PROPRETY 1:
By taking into account the condition $b \geq c$, we have
$a - k \geq 2007+ k- 2a$
$3a \geq 2007+2k$
$ k \leq \frac {3a-2007}2$

Let  $K_a$  possible maximum of k.
$K_a = [\frac {3a-2007}2]$

In other words:

For $ a$ odd $K_a = \frac {3a-2007}2$
and for $a$ even $K_a = \frac {3a-2006}2$

So for  a looked we can form $T_a = K_a+ 1$ triangles

PROPRETY 2
$T_{a+2}-T_a=3$
(Thus $T_a; T_{a+2};... $ and $K_a; K_{a+2};... $ forms  the arithmetic  sequences)

Observation:
$T_{669}=1$
$T_{670} = 2$
$T_{671}=4$

$T_{1002}= 500$
$T_{1003}=502$
Conlusion
The sum of number of all suitable triangles is
$\sum_{a=669}^{1003} T_a$
But this is a sum of two aritmetical sequences
For a even
$T_{670}+ T_{672}... + T_{1002}= 2+ 5+... + 500= \frac {167 \cdot (2+502)}2= 42084$
For a odd
$T_{669}+T_{671}+...+ T_ {1003} =1+4+ ...+502= \frac{168 \cdot (1+502)}2= 42252 $                                     
Thus we have $84336$ of all suitable triangles.
Of which $1$ is equilateral
$334$ are isosceles (As shows it the methode used $1003- 669= 334$)
And all the others  $84098$ triangles are scalene.


Srdecne Vanok
The respect, the politeness are essential qualities...and also the willingness.
Do not judge the other one.
Ak odpovedam na nejaku otazku. MOJ PRINCIP NIE JE DAT ODPOVED ALE UKAZAT AKO SA K ODPOVEDI DOSTAT

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#3 10. 06. 2012 18:38

stuart clark
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Re: no. of triangle

↑ vanok:

Thanks vanok

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