<s>[INST] Find the relationship of the length of triangle's sides. 
Denote the three sides of $\triangle ABC$ to be $a,b,c$. And they satisfy $$a^2+b+|\sqrt{c-1}-2|=10a+2\sqrt{b-4}-22 $$
  Now determine what kind of triangle $\triangle ABC$ is.
  A.Isosceles triangle which its leg and base is not equal.
  B.equilateral triangle
  C.Right triangle
  D.Isosceles Right triangle

The only information I got is from the number in the radical need to be greater than $0$. Then $b\ge4$ and $c\ge 1$. Also $10a+2\sqrt{b-4}-22\ge0 $. But they are all inequalities. What we need is some equalities. It would be great to have some hints.
 [/INST] We have
$$|\sqrt{c-1}-2|=10a+2\sqrt{b-4}-22-a^2-b\tag1$$
from which
$$10a+2\sqrt{b-4}-22-a^2-b\ge 0\tag2$$
follows.
$(2)$ is equivalent to
$$a^2-10a+b-2\sqrt{b-4}+22\le 0,$$
i.e.
$$(a-5)^2-25+(b-4)+4-2\sqrt{b-4}+22\le 0,$$
i.e.
$$(a-5)^2+(\sqrt{b-4}-1)^2\le 0$$
from which
$$a-5=\sqrt{b-4}-1=0$$
i.e.
$$a=b=5$$
follows.
Now you can get $c$ from $(1)$.
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