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Find the possible value from the following.

I'm not able to end up on a concrete note, as I'm unable to get the essence of question, still not clear to me.

$x$, $y$, $z$ are distinct reals such that $y=x(4-x)$, $z=y(4-y)$, $x=z(4-z)$. The possible values of $x+y+z$ is:

$$\begin{array}{l} A.\ 4 && C.\ 7 \\ B.\ 6 && D.\ 9 \end{array}$$

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Just a remark: nothing is linear in this question :-) –  Siminore Jun 26 '12 at 10:09
    
What is the context of the question? I find that more than one of the solutions are possible. –  Zander Jun 26 '12 at 12:23
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If we let $f(t)=t(4-t)$, then we're initially looking for a solution to $f(f(f(x)))-x=0$, which is an 8th degree equation. A bit of sketching shows that it has 8 real solutions between 0 and 4. However, two of these (0 and 3) are also solutions to $f(x)=x$, and those don't work because $x$, $y$, and $z$ are specified to be distinct. So we can use polynomial division to divide $f(f(f(x))-x$ by $f(x)-x$ and get a 6th degree polynomial whose 6 roots, in groups of 3, are the possible $(x,y,z)$ triples. –  Henning Makholm Jun 26 '12 at 12:53
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5 Answers

up vote 6 down vote accepted

To get some grip on the problem I considered the functions $f(x):=4x-x^2$ and $$g(x):=f\bigl(f\bigl(f(x)\bigr)\bigr)-x=63 x - 336 x^2 + 672 x^3 - 660 x^4 + 352 x^5 - 104 x^6 + 16 x^7 - x^8\ .$$ Plotting $g$ revealed 8 real zeros of g, two of them being $0$ and $3$ and the other six irrational.

Now obviously $x=y=z=0$ with sum $0$ and $x=y=z=3$ with sum $9$ solve the equations (but not the problem). Furthermore the data show that we can put $$x=0.4679111137620438,\quad y=1.6527036446661105,\quad z=3.8793852415690395$$ with sum $6$ and $$x=0.7530203962825327,\quad y=2.445041867912943,\quad z=3.801937735808046$$ with sum $7$.

Altogether this suggests that the possible sums $x+y+z$ are $0$, $6$, $7$, and $9$.

In order to attack the problem algebraically we introduce the elementary symmetric functions $s_1$, $s_2$, $s_3$ of $x$, $y$, $z$.

Adding the three equations gives $s_1=4s_1-(s_1^2-2 s_2)$ or $$s_2=(s_1^2-3s_1)/2\ ,\qquad(1)$$ and multiplying them gives $$s_3=s_3(4-x)(4-y)(4-z)\ .$$ Since we may assume $s_3=xyz\ne0$ we can conclude $1=64-16s_1+4s_2-s_3$, or $$s_3=63+2s_1^2-22s_1\ .\qquad(2)$$ To get a third equation we multiply the three equations of the form $$y-z=(x-y)(4-x-y)$$ and divide by $(y-z)(z-x)(x-y)\ne0$. Thus we get $$1=(4-(x+y))(4-(y+z))(4-(z+x))\ ,$$ which can be rewritten as $$1=64-32 s_1+4s_1^2 + 4s_2-s_1 s_2+s_3\ .\qquad(3)$$ Eliminating $s_2$ and $s_3$ from $(3)$ by means of $(1)$ and $(2)$ finally gives the equation $${1\over2}(6-s_1)^2(7-s_1)=0$$ which has the solutions found numerically before.

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Nice answer! But this only gives necessary conditions. Two things remain to be proved: that the roots are indeed distinct, and that triples $(x,y,z)$ with sum 6 and 7 exist. The former can be proven by noticing that 3 is prime therefore a 3-cycle that is not a fixed point must have 3 distinct values. This also implies that the roots of $g$ which are not fixpoints (their sum is $13=16-0-3$) can only consist of strict 3-cycles, so that because $13=6+7$ there must be one 3-cycle of sum 6 and one of sum 7. –  Generic Human Jun 26 '12 at 15:48
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Expand $y$ and $z$ to get a polynomial in $x$ by itself. $$ x = y(4-y)(4-y(4-y))=x(4-x)(x-2)^2(4-x(4-x)(x-2)^2) $$ In fact it's easier if you write $a=x-2,b=y-2,c=z-2$ then the original expressions become $b=2-a^2$, etc. and you get $$ a = 2-(2-(2-a^2)^2)^2 $$ This gives a degree 8 polynomial in $a$. Two of the roots are integers you can find by inspection. Factoring them out leaves a degree 6 polynomial which factors as two cubics. Then if the roots of each cubic give a solution for $(a,b,c)$ (they do), then you can read off their sum from the coefficients. As I said in the comments, though, this seems to give two different solutions from the choices.

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Composing the functions, we get $$ \begin{align} 0 &=x^8-16x^7+104x^6-352x^5+660x^4-672x^3+336x^2-63x\\ &=x(x-3)(x^3-7x^2+14x-7)(x^3-6x^2+9x-3)\tag{1} \end{align} $$ The roots $x=0$ and $x=3$ lead to indistinct $x$, $y$, and $z$.

$x^3-7x^2+14x-7$ has 3 real roots in $[0,4]$ whose sum is $7$ (the negative of the coefficient of $x^2$).

$x^3-6x^2+9x-3$ has 3 real roots in $[0,4]$ whose sum is $6$ (the negative of the coefficient of $x^2$).

$x$, $y$, and $z$ all satisfy $(1)$.

$t(4-t)$ rotates the roots of the cubics.

Thus, the possible values of $x+y+z$ are $6$ and $7$.

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Hint: If you graph $y=f(x)$ versus $x$, which is the same function so that $z=f(y)$ and $x=f(z)$, you will see that it is a parabola opening downwards with maximum at $(2,4)$ and $x$-intercepts at $0$ and $4$. Then try some integer values between the intercepts and you should quickly get the answer.

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Is there any non-graphical method to do it? Thanks. –  Bazinga Jun 26 '12 at 10:27
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There are no integer values that lead to solutions. $0$ and $3$ don't give distinct $x$, $y$ and $z$; $1$ leads to $3$; $2$ and $4$ both lead to $0$. –  Henning Makholm Jun 26 '12 at 13:02
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Alternative hint: look for a fixed point.

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But the problem explicitly specifies that the values are distinct. –  Henning Makholm Jun 26 '12 at 12:20
    
@Henning, serves me right for not paying attention. –  Gerry Myerson Jun 27 '12 at 1:43
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