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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.

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    $\begingroup$ That's an artificial exam question if I've ever seen one. $\endgroup$ – svavil Sep 29 '18 at 18:56
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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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HINT:

Write the equality as $$a^2-10a+22+b+|\sqrt{c-1}-2|-2\sqrt{b-4}=0$$ and since we know that $a$ is real, $$22+b+|\sqrt{c-1}-2|-2\sqrt{b-4}\le25\\b-2\sqrt{b-4}\le3-|\sqrt{c-1}-2|\le3.$$ But if $f(b)=b-2\sqrt{b-4}$, $f'(b)=1-\dfrac1{\sqrt{b-4}}=0$ for stationary points, resulting in $b=5$ as a minimum, and $f(b)=3$.

Hence that is the only value for $b$, meaning that $c=\cdots\,\,?$


Spoiler:

The triangle is equilateral.

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Note that: $$a^2+b+|\sqrt{c-1}-2|=10a+2\sqrt{b-4}-22 \iff \\ (a-5)^2+|\sqrt{c-1}-2|+(\sqrt{b-4}-1)^2=0 \Rightarrow \\ a=5, b=5, c=5.$$

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$(a-5)^2+(\sqrt{b-4}-1)^2+|\sqrt{|c-1|}-2|=0$

This lead $a=b=c=5$.

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Noodling:

$a^2$ and $10a$ in equations imply that maybe I should attempt to complete the square.

So I get

$a^2 - 10a +25+b +|\sqrt{c-1} - 2|=(a-5)^2 + b+ |\sqrt{c-1} -2|= 2\sqrt{b-4} + 3$

And, well this seems a bit weird but that is an even number in front of the $\sqrt{b-4}$ and we have $\sqrt{b-4}$ and $b$ variables to deal with so we can complete the square again with $v = \sqrt{b-4}$ and $v^2 = b-4$.

(In the back of my mind I'm worrying about the $\sqrt{c-1}$ which is just a single term; I'm not sure at this point what will happen.)

$(a -5)^2 + (b-4) - 2\sqrt{b-4} + 1 +|\sqrt{c-1} - 2|=3-4 + 1$

$(a-5)^2 + (\sqrt{b-4} - 1)^2 +|\sqrt{c-1} - 2| = 0$.

Oh......

We have three things that can't be negative adding up to $0$. So they must each equal $0$.

So $(a-5)^2 = 0$ and $a =5$. And $(\sqrt{b-4}-1)^2=0$ and $b=5$. And $|\sqrt{c-1} -2| = 0$ so $c=5$.

Well.... okay then......

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