# 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.

• That's an artificial exam question if I've ever seen one. – svavil Sep 29 '18 at 18:56

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 $$(1)$$.

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.

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

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

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

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......