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Can we have a triangle with sides $1$, $x$ and $x^2$? And what could $x$ be?

I try to approach this question by making 3 inequalities.
$1+x>x^2$,
$1+x^2>x$,
$x^2+x>1$

and they come with different quadratic inequalities
$x^2-x-1<0$ (solution is $(1+\sqrt 5)/2 > x > (1-\sqrt 5)/2$ ) ;
$x^2-x+1>0$ (solution is $x > (-1+\sqrt 5)/2$ or $x < (-1-\sqrt 5)/2$ )
$x^2+x-1>0$ (no real solution)

Then I start to struggle with the next step.... Thank you

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Struggling with the next step probably has a lot to do with getting the previous step wrong. Nice start, though! –  Jonathan Christensen Dec 4 '12 at 1:12
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5 Answers

Your approach is fine. $1+x > x^2$ has the solution set you have indicated, but since $x$ is the side of a triangle, we cannot have negative $x$. Thus $0 < x < \frac{1 + \sqrt{5}}{2}$. The second inequality, $1 + x^2 > x$ is satisfied by all real numbers, so no further restriction here. The solutions set of $x^2 + x > 1$ is $x < \frac{-1-\sqrt{5}}{2}$ or $x > \frac{-1+\sqrt{5}}{2}$, and only positive $x$ matters. Thus, any $x$ such that:

$$ \frac{-1+\sqrt{5}}{2} < x < \frac{1+\sqrt{5}}{2} $$

will give a valid triangle with sides $1$, $x$ and $x^2$. Approximately, $.618 < x < 1.618$. (Coincidentally, the larger value is the golden ratio, and the lower value is its multiplicative inverse!)

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i don't quite understand why The second inequality, 1+x2>x is satisfied by all real numbers. x^2-x+1>0, the discriminant (assuming x^2-x+1=0) is less than zero. THX, your answer is really useful. –  user51658 Dec 4 '12 at 1:51
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@user51658 This is true because since x^2-x+1 is always greater than zero, it can never equal zero. –  Istvan Chung Dec 4 '12 at 2:59
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@user51658: $x^2-x+1=(x-\frac 12)^2+\frac 34$, so is always greater than zero. –  Ross Millikan Dec 4 '12 at 4:25
    
Arafinwe and RossMillikan are entirely correct... and just to reiterate, consider any $x$ value you like, such as $x=2$. Then wouldn't you agree that $(2)^2 - (2) + 1$ is greater than $0$? –  Shaun Ault Dec 5 '12 at 1:55
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Because you know $x$ must satisfy all three inequalities, your conclusion would prove that no triangle exists, so you should be able to answer the question.

Alas, it turns out your conclusion is not true; try solving the inequalities again, and seeking some way to verify whether or not your solutions are correct.

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So you already have the inequalities:

$$1+x>x^2\Longleftrightarrow x^2-x-1<0\Longleftrightarrow \frac{1-\sqrt 5}{2}<x<\frac{1+\sqrt 5}{2}$$

$$1+x^2>x\Longleftrightarrow x^2-x+1>0\Longrightarrow\,\text{always true}$$

$$x^2+x>1\Longleftrightarrow x^2+x-1>0\Longleftrightarrow x<\frac{-1-\sqrt 5}{2}\,\,\vee\,\,x>\frac{-1+\sqrt 5}{2}$$

The common interval to all the above inequalities is

$$\frac{-1+\sqrt 5}{2}<x<\frac{1+\sqrt 5}{2}$$

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What about $x=1$? –  Jonathan Christensen Dec 4 '12 at 1:03
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Yes, I noticed that before you sent your comment. You rush a little too much to downvote, don't you think? –  DonAntonio Dec 4 '12 at 1:11
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In fact have not downvoted your post, though I certainly wouldn't blame someone for downvoting a false answer. I'm glad to see you've corrected it. –  Jonathan Christensen Dec 4 '12 at 1:15
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You've made a good start, but you are wrong about your inequalities: the second one holds for all real $x$, and the solution that you have written for the second one is in fact the solution for the third one.

Any $x$ which satisfies all three inequalities will be valid. Since the second one holds everywhere, you only need to check the first and third ones, which turn out to both hold for $\frac12 (\sqrt 5-1) < x < \frac12 (\sqrt 5+1)$ .

As a side note, those two numbers are $\varphi -1$ and $\varphi$, where $\varphi$ is the golden ratio. This is not a coincidence. Can you explain way?

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I've been scooped by Shaun. Ah well. Edit: Apparently that deserves a downvote for some reason. –  Jonathan Christensen Dec 4 '12 at 1:12
    
Can you explain why? :-) Please. I think I have an inkling, but that's all. –  LarsH Dec 4 '12 at 3:01
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There is a natural separation between $x\gt 1$ and $x\lt 1$. For if $x\gt 1$, then $x^2\gt x\gt 1$, so $x^2$ is th longest isde.

On the other hand, if $0\lt x\lt 1$, then $x^2$ is the baby of the family.

Case $x \gt 1$: If the inequality $x^2 \lt x+1$ holds, we will be OK. The roots of $x^2-x-1=0$ are $(1\pm \sqrt{5}/2$. So in our range, the desired inequality holds for $1\lt x\lt (1+\sqrt{5})/2$.

Case $0\lt x\lt 1$: Now we want $x^2+x\gt 1$. The roots of $x^2+x-1=0$ are $(-1\pm\sqrt{5})/2$. so in our range, the desired inequality holds if $x\gt (-1+\sqrt{5})/2$.

Of course $x=1$ is fine too. Putting the results together, we see that we can form a triangle precisely if $(-1+\sqrt{5})/2 \lt x \lt (1+\sqrt{5})/2$.

Remark: Consider the numbers $1,x, x^2$. Suppose that we can make a triangle with these sides. The sides can be rewritten as $x^2(1/x^2)$, $x^2(1/x)$, and $x^2(1)$. By scaling, we can make a triangle with these sides iff we can make a triangle with sides $(1/x^2,1/x,1)$. Putting $y=1/x$, we see that there is a triangle with sides $1,x,x^2$ iff there is a triangle with sides $1,y,y^2$. Thus there is a natural correspondence between our triangles that have shortest side equal to $1$ with the triangles that have longest side equal to $1$. We could have used this correspondence to get an instant bound on $x$ for the case $x\lt 1$ from the bound for the case $x\gt 1$: just take the reciprocal.

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+1 helpful remark(s). –  LarsH Dec 4 '12 at 14:45
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