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.
Best Answer
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 $(1)$.