The polynomial Q(x)=x3−21x+35 has three different real roots. Find real numbers a and b such that the polynomial x2+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.
Problem
Source: Centroamerican Olympiad 2016, problem 3
Tags: algebra, polynomial, roots
20.06.2016 00:58
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)=x3+ax+b, then we have that r3+ar+b=ss3+as+b=tt3+at+b=rAdding up our three equations and noting that r+s+t=0 from above, we get that −r3−s3−t3=3bBy Newton's Identities we have P1=0, P2=42 and P3=−105. This implies that 3b=105. Dividing both sides by 3 gives us b=35. Looking back at our original polynomial, note that we can rewrite it as x3+35=21x, meaning that r3+35=21r, s3+35=21s and t3+35=21t since r,s,t are roots. Thus we have that our cyclic equations turn into t(21+a)=rr(21+a)=ss(21+a)=tSubstituting the first into the second and the new second into the third, we get t(21+a)3=tClearly t≠0 since 0 is not a root of x3−21x+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 a=−20.
20.06.2016 03:37
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)=x2+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)≠r,P(s)≠s and P(t)≠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.
20.06.2016 23:06
I apologize for the typo; indeed P is supposed to be a quadratic polynomial.
22.06.2016 05:31
From Vieta relations we get: {1} r+s+t=0 {2} rs+st+tr=−21 {3} rst=−35 1) 0=(r+s+t)2=r2+s2+t2+2(rs+st+tr) {4} ⇒r2+s2+t2=42 2) r3+s3+t3−3rst=(r+s+t)(r2+s2+t2−rs−st−tr)=0 {5} ⇒r3+s3+t3=3rst=−105 We want that: r2+ar+b=s⟶r3+ar2+b=sr s2+as+b=t⟶s3+as2+b=st t2+at+b=r⟶t3+at2+b=rt Adding the left-column equations we get: {4}+a{1}+3b={1}⟶42=−3b⇒b=−14 Adding the right-column we get: {5}+a{4}+3b={2}⟶42a=84⇒a=−2 Q.E.D (The checking is done as @juckter did it)
23.11.2018 05:27
Only use Vieta relations... It's easy to a problem 3)...
31.03.2021 00:36
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 x2+ax+b, in the same way as in the above solutions (fixing typos), we get that the polynomial we want is x2+2x−14. We are done. ◻
03.04.2021 10:09
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.
05.04.2021 04:46
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 P1(x) be the quadratic polynomial whose graph passes through (r,s),(s,t),(t,r). Similarly, let P2(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 P1(x),P2(x) is monic. After that, we find a,b as in the above solutions. Claim. P1(x)≠P2(x) Proof. Proceed by contradiction. If they are both equal to P(x)=cx2+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=−a+1c. By symmetry, this would imply r+s=r+t=s+t, a contradiction. Now, if P1(x)=cx2+ax+b, by doing the same computations and Vieta as in the above solutions, we get (c,a,b)=(c,5c−12,−14c). However, we also have ∑cyc(cr2+ar+b)2=∑cycs2. Expanding, using Vieta, and simplifying stuff, this yields c=±1. However, if we do the same with P2(x)=c′x2+a′x+b′, we get (c′,a′,b′)=(c′,5c′−12,−14c′) and c′=±1. Finally, by the claim, this implies one of them has leading coefficient 1. We find it is x2+2x−14, and are done. ◻ Remark. With this method, we find that the other polynomial that permutates the roots cyclically is −x2−3x+14.
08.08.2021 03:05
Quote: Claim. P1(x)≠P2(x) Is it not enough for showing this to note that P1(x) and P2(x) take (by definition) different values for x=r, since the hypothesis states s≠t?
08.08.2021 03:11
baenanabread wrote: Quote: Claim. P1(x)≠P2(x) Is it not enough for showing this to note that P1(x) and P2(x) take (by definition) different values for x=r, since the hypothesis states s≠t? True, that makes it easier Thank you!
13.01.2023 22:49
14.01.2023 01:56
Fun Fact. The roots are: r=2cosπ63+4cos2π63−2cos4π63−2cos8π63+2cos10π63−2cos11π63+4cos16π63−2cosπ9≈3.13776 s=2cos4π63+2cos11π63+2cos13π63−2cos17π63−4cos19π63−2cos22π63+4cos26π63+2cos4π9≈2.12108 t=−4cos5π63−2cos11π63+2cos17π63−2cos20π63−4cos23π63+2cos29π63−2cos31π63+2cos2π9≈−5.25885