Kvant Math Problem 2738
Let
Verified: no
Verdicts: UNKNOWN + UNKNOWN
Solve time: 12m06s
Source on kvant.digital
Problem
Distinct real numbers $a_1$, $a_2$, $a_3$, and $b$ are given. It turns out that the equation $$(x-a_1)(x-a_2)(x-a_3)=b$$ has three distinct real roots $c_1$, $c_2$, $c_3$. Find the roots of the equation $$(x+c_1)(x+c_2)(x+c_3)=b.$$
A. Antropov, K. Sukhov
All-Russian Olympiad in Mathematics (XLIX, regional stage)
Exploration
Let
$P(x)=(x-a_1)(x-a_2)(x-a_3).$
The condition says that the equation
$P(x)=b$
has roots $c_1,c_2,c_3$, hence
$P(x)-b=(x-c_1)(x-c_2)(x-c_3).$
Expanding both sides in terms of symmetric polynomials suggests comparing coefficients between the two cubic representations of $P(x)$ and of the factorization through $c_i$. This determines the symmetric sums of $c_i$ in terms of those of $a_i$ and $b$.
The second equation involves
$(x+c_1)(x+c_2)(x+c_3),$
so its coefficients are also symmetric polynomials in $c_i$, hence ultimately expressible through $a_i$ and $b$. The structure suggests that subtraction of $b$ may eliminate the shift and produce a polynomial closely related to $(x+a_1)(x+a_2)(x+a_3)$.
The key point is whether the roles of $a_i$ and $-a_i$ appear symmetrically after this transformation.
Problem Understanding
This is a Type A problem: determine all roots of the equation
$(x+c_1)(x+c_2)(x+c_3)=b,$
given that $c_1,c_2,c_3$ are defined implicitly as the roots of
$(x-a_1)(x-a_2)(x-a_3)=b.$
The structure is symmetric in two stages of “shifted root sets.” The core difficulty is tracking how symmetric polynomials transform when passing from $a_i$ to $c_i$, and then back through the new product $(x+c_1)(x+c_2)(x+c_3)$.
The expected outcome is that the transformation effectively reverses the sign of the original parameters, producing roots $-a_1,-a_2,-a_3$.
Thus the answer is
$\boxed{-a_1,,-a_2,,-a_3}.$
Proof Architecture
The proof proceeds through symmetric polynomial identification.
First, a lemma expresses the symmetric sums of $c_1,c_2,c_3$ in terms of $a_1,a_2,a_3$ and $b$ by comparing coefficients in the identity
$P(x)-b=(x-c_1)(x-c_2)(x-c_3).$
Second, these expressions are substituted into the expansion of
$(x+c_1)(x+c_2)(x+c_3),$
producing a cubic polynomial in $x$ whose coefficients are written in terms of $a_1,a_2,a_3,b$.
Third, a direct comparison shows that subtracting $b$ from this polynomial yields exactly
$(x+a_1)(x+a_2)(x+a_3).$
The hardest point is the correct handling of the constant term when passing from $P(x)$ to $P(x)-b$ and then to the product form in $c_i$.
Solution
Define
$P(x)=(x-a_1)(x-a_2)(x-a_3)=x^3-S_1x^2+S_2x-S_3,$
where
$S_1=a_1+a_2+a_3,\quad S_2=a_1a_2+a_2a_3+a_3a_1,\quad S_3=a_1a_2a_3.$
The given condition that $c_1,c_2,c_3$ are roots of $P(x)=b$ implies
$P(x)-b=(x-c_1)(x-c_2)(x-c_3).$
Expanding the right-hand side gives
$(x-c_1)(x-c_2)(x-c_3)=x^3-T_1x^2+T_2x-T_3,$
where
$T_1=c_1+c_2+c_3,\quad T_2=c_1c_2+c_2c_3+c_3c_1,\quad T_3=c_1c_2c_3.$
Comparing coefficients with
$P(x)-b=x^3-S_1x^2+S_2x-(S_3+b),$
yields
$T_1=S_1,\quad T_2=S_2,\quad T_3=S_3+b.$
Now consider
$(x+c_1)(x+c_2)(x+c_3)=x^3+T_1x^2+T_2x+T_3.$
Substituting the expressions for $T_i$ gives
$(x+c_1)(x+c_2)(x+c_3)=x^3+S_1x^2+S_2x+(S_3+b).$
Subtracting $b$ produces
$(x+c_1)(x+c_2)(x+c_3)-b=x^3+S_1x^2+S_2x+S_3.$
The right-hand side factors as
$x^3+S_1x^2+S_2x+S_3=(x+a_1)(x+a_2)(x+a_3).$
Hence the equation
$(x+c_1)(x+c_2)(x+c_3)=b$
is equivalent to
$(x+a_1)(x+a_2)(x+a_3)=0.$
Therefore the roots are
$x=-a_1,,-a_2,,-a_3.$
$\boxed{-a_1,,-a_2,,-a_3}$
Verification of Key Steps
The critical step is the identification of symmetric sums after shifting by $b$. Starting from
$P(x)-b=(x-c_1)(x-c_2)(x-c_3),$
the constant term on the right is $-c_1c_2c_3$, while on the left it is $-(S_3+b)$. Matching these gives $c_1c_2c_3=S_3+b$, which is the only place where $b$ enters nonlinearly.
A second delicate point is the expansion of $(x+c_1)(x+c_2)(x+c_3)$. The coefficient of $x$ is $T_2=S_2$, which ensures exact agreement with the quadratic symmetric polynomial of the $a_i$ after comparison.
Finally, the factorization
$x^3+S_1x^2+S_2x+S_3=(x+a_1)(x+a_2)(x+a_3)$
follows directly from expanding the right-hand side and matching symmetric coefficients term by term.
Alternative Approaches
A shorter approach uses the observation that the transformation $x\mapsto -x$ interchanges the role of $(x-a_i)$ and $(x+a_i)$ inside cubic symmetric expansions. One can also view the problem as an involution on monic cubic polynomials under the map sending roots to solutions of $P(x)=b$, which preserves symmetric coefficients except for a controlled shift in the constant term. The coefficient-comparison method remains preferable because it isolates the effect of $b$ transparently and avoids implicit functional reasoning.