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Let $x,y,z\in R$, and $$\begin{cases} \sqrt{x-3}+\sqrt{y-3}=12\\ \sqrt{y-12}+\sqrt{z-12}=12\\ \sqrt{z-27}+\sqrt{x-27}=12 \end{cases}$$ Find the $x,y,z$.

My try: I want use The geometry to solve it. ( Norbert have solved it) and I think this problem have algebra methods.Thank you

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we only find this $x,y,z$, then we have $$OP^2=ON^2+NP^2-ON\cdot NP=39$$ and $$\dfrac{OP}{\sin{A}}=x\Longrightarrow \sqrt{x}=2\sqrt{13}$$ and use the same methods we easy to find $y,z$

I think of seeing algebra methods? maybe use if $x>y$,then use $$\sqrt{x-3}+\sqrt{y-3}=\sqrt{y-12}+\sqrt{z-12}<\sqrt{x-12}+\sqrt{z-12}?$$ then I can't,

Thank you

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  • $\begingroup$ Thank you,before I hope see the algebraic methods.Thank you $\endgroup$ – user94270 Nov 28 '13 at 14:36
  • $\begingroup$ ok, I hope someone can solve this algebraically. (+1) for geometric idea $\endgroup$ – Norbert Nov 28 '13 at 14:38
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Here is an algebraic solution.

We have the system of equations $$ \sqrt{x-3}+\sqrt{y-3}=12\tag{1} $$ $$ \sqrt{y-12}+\sqrt{z-12}=12\tag{2} $$ $$ \sqrt{z-27}+\sqrt{x-27}=12\tag{3} $$ From $(2)$ and $(3)$ we get that $$ x=27+(12-\sqrt{z-27})^2\qquad y=12+(12-\sqrt{z-12})^2 $$ After substitution in $(1)$ we get $$ \sqrt{24+(12-\sqrt{z-27})^2}=12-\sqrt{9+(12-\sqrt{z-12})^2} $$ After squaring and some simplifications we get $$ \sqrt{z-27}=\sqrt{9+(12-\sqrt{z-12})^2}-6+\sqrt{z-12} $$ Now its time for substitution $t=\sqrt{z-12}-6$: $$ \sqrt{(6+t)^2-15}=\sqrt{(6-t)^2+9}+t $$ Now it is not hard (after squaring twice during derivation) to get $$ t^4-148t^3+384t-192=0\tag{4} $$ This equation has a good solution $t=2$. After division of $(4)$ by $t-2$ it is remains to solve $$ t^3+2t^2-144t+96=0 $$ We can't solve it exactly, but in fact we shouldn't do it. From $(2)$ it is clear that $0\leq\sqrt{z-12}\leq 12$, so $-6\leq t\leq 6$. Consider $f(t)=t^3+2t^2-144t+96$ on the interval $[-6,6]$. Note that $f'(t)=3t^2+4t-144=3(t+2/3)^2-436/3$ is a quadratic function, so $$ \min_{t\in[-6,6]}f'(t)=\min(f'(-6),f'(6),f'(-2/3))=\min(-60,-12,-436/3)<0 $$ In other words $f(t)$ is strictly decreasing on $[-6,6]$, hence there is at most one root $t_0\in[-6,6]$. Since $f(0)=96>0$ and $f(1)=-45<0$, then $0<t_0<1$. This implies that $$ z=12+(6+t)^2<61\\ x=27+(12-\sqrt{z-27})^2>27+(12-\sqrt{z-12})^2>27+5^2=52\\ y=12+(12-\sqrt{z-12})^2>12+5^2=37 $$ Therefore $\sqrt{x-3}+\sqrt{y-3}>\sqrt{49}+\sqrt{34}>7+\sqrt{25}=12$. This contradicts $(2)$, hence $t_0$ is superfluous soltion of $(4)$.

From now on $t=2$ and we get $$ z=12+(6+t)^2=76\\ x=27+(12-\sqrt{z-27})^2=52\\ y=12+(12-\sqrt{z-12})^2=28 $$

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Actually your geometric idea is wonderful! But I had to correct your shape:enter image description here

$NI || BC,NJ||BC$, so $\angle PIJ = \angle NJA=60$, So $NI=6,NJ=2,IJ=8,NH=2$ therefore $x=(4\sqrt 3)^2+2^2=52$. I am sure you can continue!$y=28,z=76$

EDIT: This solution is for a time that OP said he/she tried the geometric idea but without any success. Then OP edited the post and said he/she succeeded through geometric method.

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  • $\begingroup$ For thoes who can't see how this works clearly, in an equilateral triangle the sum of distances of a point inside the triangle from the sides of the triangle is equal to the length of height of triangle, as $\sqrt 12+ \sqrt 3+\sqrt 27=12 \frac{\sqrt{3}}{2}$ so we can find a point like $N$ inside the triangle such that it satisfies the conditions of the problem $\endgroup$ – hhsaffar Nov 28 '13 at 16:28
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Clearly $OAP=60^\circ$ $ONP=120^\circ$, so from triangle $ONP$ via cosine theorem you can find $OP^2=39$. Since $AONP$ have two right angles it is inscribed in some circle. Clearly $AN$ its diameter. Now apply sine theorem for trianle $ONP$ which is inscribed in the same circle.

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  • $\begingroup$ I think use sinne theorem have $$\dfrac{OP}{\sin{60}}=AN=x$$ it's too easy $\endgroup$ – user94270 Nov 28 '13 at 13:58
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Here's another algebraic solution, similar to Norbert's but taking a somewhat different tack.

For all the square roots to be real, we must have $x,z\ge27$ and $y\ge12$, so let's let

$$\begin{align} x&=27+u^2\\ y&=12+v^2\\ z&=27+w^2\\ \end{align}$$

with $u,v,w\ge0$. Then the equations become

$$\begin{align} \sqrt{u^2+24}+\sqrt{v^2+9}&=12\\ v+\sqrt{w^2+15}&=12\\ u+w&=12\\ \end{align}$$

Thus

$$w=12-u \quad\text{and}\quad v=12-\sqrt{(12-u)^2+15}$$

so the equation to be solved is

$$\sqrt{u^2+24}+\sqrt{\left(12-\sqrt{(12-u)^2+15}\right)^2+9}=12$$

This "simplifies" first to

$$\left(12-\sqrt{u^2-24u+159}\right)^2+9=\left(12-\sqrt{u^2+24}\right)^2$$

which boils down to

$$\sqrt{u^2-24u+159}=\sqrt{u^2+24}-(u-6)$$

This squares and simplifies to

$$2(u-6)\sqrt{u^2+24}=u^2+12u-99$$

and this squares to produce a quartic that factors as

$$3(u-5)(u^3-19u^2+3u+423)=0$$

We thus have the solution $u=5$, which leads to $x=52$, $y=28$ and $z=76$, as Norbert found. The cubic factor looks unpleasant, but we don't have to worry about any of its roots that are less than $0$ or greater than $12$, since those do not satisfy the non-negativity conditions on $u$ and $w$. Now if we go back to the original equation and let

$$f(u)=\sqrt{u^2+24}+\sqrt{\left(12-\sqrt{(12-u)^2+15}\right)^2+9}-12$$

we have

$$f'(u)={u\over\sqrt{u^2+24}}+{12-\sqrt{(12-u)^2+15}\over\sqrt{\left(12-\sqrt{(12-u)^2+15}\right)^2+9}}\cdot{(12-u)\over\sqrt{(12-u)^2+15}}$$

which is clearly positive for $1\le u\le 12$, so we don't have to worry about any roots of the cubic in that range either. Finally, it's easy to see that the cubic decreases from $423$ to $408$ in the interval $0\lt u\lt1$. Thus $u=5$ is the only solution.

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