Cyclic permutation of roots

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Cyclic permutation of roots

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Source: Centroamerican Olympiad 2016, problem 3

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388 posts

Jutaro

#1 h Jun 19, 2016, 9:32 PM • 4 Y

Y by Davi-8191, Rounak_iitr, Adventure10, Mango247

The polynomial $Q(x)=x^3-21x+35$ has three different real roots. Find real numbers $a$ and $b$ such that the polynomial $x^2+ax+b$ cyclically permutes the roots of $Q$ , that is, if $r$ , $s$ and $t$ are the roots of $Q$ (in some order) then $P(r)=s$ , $P(s)=t$ and $P(t)=r$ .

This post has been edited 1 time. Last edited by Jutaro, Jun 20, 2016, 8:05 PM
Reason: Typo

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#2 h Jun 19, 2016, 9:58 PM • 2 Y

Y by Adventure10, Mango247

Suppose the three roots of $Q$ are $r,s,t$ . We have that $r+s+t=0$ , $rs+st+rt=-21$ , and $rst=-35$ by Vieta's. If the roots of $Q$ cyclically permute in the polynomials $P(x)=x^3+ax+b$ , then we have that $r^3+ar+b=s$ $s^3+as+b=t$ $t^3+at+b=r$ Adding up our three equations and noting that $r+s+t=0$ from above, we get that $-r^3-s^3-t^3=3b$ By Newton's Identities we have $P_1=0$ , $P_2=42$ and $P_3=-105$ . This implies that $3b=105$ . Dividing both sides by $3$ gives us $\boxed{b=35}$ . Looking back at our original polynomial, note that we can rewrite it as $x^3+35=21x$ , meaning that $r^3+35=21r$ , $s^3+35=21s$ and $t^3+35=21t$ since $r,s,t$ are roots. Thus we have that our cyclic equations turn into $t(21+a)=r$ $r(21+a)=s$ $s(21+a)=t$ Substituting the first into the second and the new second into the third, we get $t(21+a)^3=t$ Clearly $t\neq 0$ since $0$ is not a root of $x^3-21x+35$ . Dividing by $t$ on both sides gives us $(21+a)^3=1$ . Obviously the only real root of this is $-20$ (you can find others with roots of unity) so we have $\boxed{a=-20}$ .

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#3 h Jun 20, 2016, 12:37 AM • 2 Y

Y by Adventure10, Mango247

The problem actually says $P$ is a quadratic polynomial, not cubic. In this case we may easily check that $b = -14$ through a similar method to above, and then we may easily obtain that $a = 2$ through some other manipulations, thus $P(x) = x^2 + 2x - 14$ . In order to verify that it works, we check that $Q(x)$ divides $Q(P(x))$ , and hence $P(r), P(s)$ , and $P(t)$ are also roots of $x$ . After checking that none of the roots of the polynomial $P(x) - x$ are roots of $Q$ , it follows that $P(r) \neq r, P(s) \neq s$ and $P(t) \neq t$ . Finally, we may verify that $P(r), P(s), P(t)$ are all distinct. These three facts together imply that $P$ indeed permutes the roots of $Q$ .

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Jutaro

#4 h Jun 20, 2016, 8:06 PM • 2 Y

Y by Adventure10, Mango247

I apologize for the typo; indeed $P$ is supposed to be a quadratic polynomial.

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#5 h Jun 22, 2016, 2:31 AM • 1 Y

Y by Adventure10

From Vieta relations we get:
{1} $r+s+t=0$
{2} $rs+st+tr=-21$
{3} $rst=-35$

1) $0=(r+s+t)^2=r^2+s^2+t^2+2(rs+st+tr)$
{4} $\Rightarrow r^2+s^2+t^2=42$

2) $r^3+s^3+t^3-3rst=(r+s+t)(r^2+s^2+t^2-rs-st-tr)=0$
{5} $\Rightarrow r^3+s^3+t^3=3rst=-105$

We want that:
$r^2+ar+b=s\longrightarrow r^3+ar^2+b=sr$
$s^2+as+b=t\longrightarrow s^3+as^2+b=st$
$t^2+at+b=r\longrightarrow t^3+at^2+b=rt$

Adding the left-column equations we get:
$\{4\}+a\{1\}+3b=\{1\} \longrightarrow 42=-3b \Rightarrow b=-14$

Adding the right-column we get:
$\{5\}+a\{4\}+3b=\{2\} \longrightarrow 42a=84 \Rightarrow a=-2$

$Q.E.D$
(The checking is done as @juckter did it)

This post has been edited 1 time. Last edited by Packito, Jun 22, 2016, 2:35 AM

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Ilove_mathematics

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Ilove_mathematics

#6 h Nov 23, 2018, 2:27 AM • 2 Y

Y by Adventure10, Mango247

Only use $Vieta$ relations... It's easy to a problem $3)$ ...

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#7 h Mar 30, 2021, 9:36 PM

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By Lagrange's Interpolation Formula, there exists a unique quadratic polynomial whose graph passes through $(r,s)$ , $(s,t)$ , $(t,r)$ . (This already guarantees that the polynomial we will get works). However if this polynomial is $x^2+ax+b$ , in the same way as in the above solutions (fixing typos), we get that the polynomial we want is $x^2+2x-14$ . We are done. $\square$

This post has been edited 2 times. Last edited by Modesti, Mar 30, 2021, 10:13 PM
Reason: Typos

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#8 h Apr 3, 2021, 7:09 AM • 3 Y

Y by Modesti, Mango247, Mango247

There's an issue with your solution, which is that there is no reason for the Lagrange Interpolation polynomial to have leading coefficient $1$ . You can expand this argument to get a pretty nice solution to the problem though.

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#9 h Apr 5, 2021, 1:46 AM

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You're right. Thank you for pointing out that mistake! Here is how I fixed it. (I wonder if there's a nicer way, though).

Let $P_1(x)$ be the quadratic polynomial whose graph passes through $(r,s),(s,t),(t,r)$ . Similarly, let $P_2(x)$ be the quadratic polynomial whose graph passes through $(r,t),(t,s),(s,r)$ . By Lagrange's Interpolation Formula, they are both well-defined and unique. Thus, our problem reduces proving that one among $P_1(x),P_2(x)$ is monic. After that, we find $a,b$ as in the above solutions.

Claim. $P_1(x)\ne P_2(x)$

Proof. Proceed by contradiction. If they are both equal to $P(x)=cx^2+ax+b$ , then $P(x)$ passes through the six points mentioned above. In particular, it passes thorugh $(r,s)$ , $(s,r)$ , from which we get $r+s=-\frac{a+1}{c}$ . By symmetry, this would imply $r+s=r+t=s+t$ , a contradiction.

Now, if $P_1(x)=cx^2+ax+b$ , by doing the same computations and Vieta as in the above solutions, we get $(c,a,b)=\bigg(c,\dfrac{5c-1}{2},-14c\bigg)$ . However, we also have $\sum_{cyc}(cr^2+ar+b)^2=\sum_{cyc}s^2$ . Expanding, using Vieta, and simplifying stuff, this yields $c=\pm1$ . However, if we do the same with $P_2(x)=c'x^2+a'x+b'$ , we get $(c',a',b')=\bigg(c',\dfrac{5c'-1}{2},-14c'\bigg)$ and $c'=\pm1$ .

Finally, by the claim, this implies one of them has leading coefficient 1. We find it is $x^2+2x-14$ , and are done. $\square$

Remark. With this method, we find that the other polynomial that permutates the roots cyclically is $-x^2-3x+14$ .

This post has been edited 5 times. Last edited by Modesti, Apr 5, 2021, 6:26 AM
Reason: Remark

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#10 h Aug 8, 2021, 12:05 AM • 1 Y

Y by Modesti

Quote:

Claim. $P_1(x)\ne P_2(x)$

Is it not enough for showing this to note that $P_1(x)$ and $P_2(x)$ take (by definition) different values for $x = r$ , since the hypothesis states $s \not = t$ ?

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#11 h Aug 8, 2021, 12:11 AM

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baenanabread wrote:

Quote:

Claim. $P_1(x)\ne P_2(x)$

Is it not enough for showing this to note that $P_1(x)$ and $P_2(x)$ take (by definition) different values for $x = r$ , since the hypothesis states $s \not = t$ ?

True, that makes it easier :blush:

:blush:

Thank you!

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lifeismathematics

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lifeismathematics

#12 h Jan 13, 2023, 7:49 PM • 2 Y

Y by dhfurir, CantConcentrate

nice problem

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#13 h Jan 13, 2023, 10:56 PM

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Fun Fact.

The roots are:

$r=2\cos\frac{\pi}{63}+4\cos\frac{2\pi}{63}-2\cos\frac{4\pi}{63}-2\cos\frac{8\pi}{63}+2\cos\frac{10\pi}{63}-2\cos\frac{11\pi}{63}+4\cos\frac{16\pi}{63}-2\cos\frac{\pi}{9}\approx 3.13776$

$s=2\cos\frac{4\pi}{63}+2\cos\frac{11\pi}{63}+2\cos\frac{13\pi}{63}-2\cos\frac{17\pi}{63}-4\cos\frac{19\pi}{63}-2\cos\frac{22\pi}{63}+4\cos\frac{26\pi}{63}+2\cos\frac{4\pi}{9}\approx 2.12108$

$t=-4\cos\frac{5\pi}{63}-2\cos\frac{11\pi}{63}+2\cos\frac{17\pi}{63}-2\cos\frac{20\pi}{63}-4\cos\frac{23\pi}{63}+2\cos\frac{29\pi}{63}-2\cos\frac{31\pi}{63}+2\cos\frac{2\pi}{9}\approx -5.25885$

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