Kvant Math Problem 27
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Problem
Prove that if $$\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}=0,$$, then $$\frac{a}{(b-c)^2}+\frac{b}{(c-a)^2}+\frac{c}{(a-b)^2}=0.$$.
Exploration
Let
$$ S=\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}. $$
The denominators suggest introducing the differences
$$ x=b-c,\qquad y=c-a. $$
Then
$$ a=c-y,\qquad b=c+x,\qquad a-b=-(x+y). $$
Substituting into $S$ gives
$$ S=\frac{c-y}{x}+\frac{c+x}{y}-\frac{c}{x+y}. $$
After bringing the terms to the common denominator $xy(x+y)$, the numerator becomes
$$ c(x^2+xy+y^2)+x^3+x^2y-xy^2-y^3. $$
The cubic part factors as
$$ x^3+x^2y-xy^2-y^3=(x-y)(x+y)^2. $$
Thus the condition $S=0$ is equivalent to
$$ c(x^2+xy+y^2)+(x-y)(x+y)^2=0. $$
Now compute
$$ T=\frac{a}{(b-c)^2}+\frac{b}{(c-a)^2}+\frac{c}{(a-b)^2} =\frac{c-y}{x^2}+\frac{c+x}{y^2}+\frac{c}{(x+y)^2}. $$
A direct reduction to a common denominator shows that the numerator of $T$ factors as
$$ (x^2+xy+y^2)\Bigl(c(x^2+xy+y^2)+(x-y)(x+y)^2\Bigr). $$
The second factor is exactly the numerator obtained from the condition $S=0$. This appears to be the decisive observation.
The step most likely to hide an error is the factorization of the numerator of $T$, because the conclusion depends on recognizing the factor coming from the hypothesis.
Problem Understanding
We are given
$$ \frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}=0, $$
where the denominators are nonzero, and must prove that
$$ \frac{a}{(b-c)^2}+\frac{b}{(c-a)^2}+\frac{c}{(a-b)^2}=0. $$
This is a Type B problem, a pure proof.
The core difficulty is to extract a useful algebraic consequence from the first identity and show that it forces the second expression to vanish. The expressions are not obviously related, so the proof must reveal a common factor hidden in their numerators.
Proof Architecture
Introduce $x=b-c$ and $y=c-a$, so that $a=c-y$, $b=c+x$, and $a-b=-(x+y)$.
Rewrite the hypothesis in terms of $x$, $y$, and $c$; after clearing denominators, obtain the relation
$$ c(x^2+xy+y^2)+(x-y)(x+y)^2=0. $$
Rewrite the desired expression in the same variables and reduce it to a single fraction.
Show that the numerator of this fraction factors as
$$ (x^2+xy+y^2)\Bigl(c(x^2+xy+y^2)+(x-y)(x+y)^2\Bigr). $$
Use the relation obtained from the hypothesis to conclude that this numerator is zero, hence the whole expression is zero.
The most delicate point is the factorization of the numerator of the second expression.
Solution
Set
$$ x=b-c,\qquad y=c-a. $$
Since the given fractions are defined,
$$ x\neq 0,\qquad y\neq 0,\qquad x+y\neq 0. $$
From the definitions,
$$ a=c-y,\qquad b=c+x,\qquad a-b=-(x+y). $$
The hypothesis becomes
$$ \frac{c-y}{x}+\frac{c+x}{y}-\frac{c}{x+y}=0. $$
Multiplying by $xy(x+y)$ yields
$$ (c-y)y(x+y)+(c+x)x(x+y)-cxy=0. $$
Expanding gives
$$ c(x^2+xy+y^2)+x^3+x^2y-xy^2-y^3=0. $$
Since
$$ x^3+x^2y-xy^2-y^3=(x-y)(x+y)^2, $$
we obtain
$$ c(x^2+xy+y^2)+(x-y)(x+y)^2=0. $$
Now consider
$$ T=\frac{a}{(b-c)^2}+\frac{b}{(c-a)^2}+\frac{c}{(a-b)^2}. $$
Substituting the same expressions for $a$, $b$, and $a-b$ gives
$$ T=\frac{c-y}{x^2}+\frac{c+x}{y^2}+\frac{c}{(x+y)^2}. $$
Multiplying by the common denominator $x^2y^2(x+y)^2$, we find that the numerator of $T$ equals
$$ (c-y)y^2(x+y)^2+(c+x)x^2(x+y)^2+c x^2y^2. $$
Expanding and collecting terms,
$$ \begin{aligned} & (c-y)y^2(x+y)^2+(c+x)x^2(x+y)^2+c x^2y^2 \ &=(x^2+xy+y^2)\Bigl(c(x^2+xy+y^2)+(x-y)(x+y)^2\Bigr). \end{aligned} $$
Hence
$$ T= \frac{(x^2+xy+y^2)\Bigl(c(x^2+xy+y^2)+(x-y)(x+y)^2\Bigr)} {x^2y^2(x+y)^2}. $$
The factor in parentheses is zero by the relation derived from the hypothesis. Therefore the numerator of $T$ is zero, and so
$$ T=0. $$
This completes the proof.
∎
Verification of Key Steps
The first delicate step is the reduction of the hypothesis. Starting from
$$ \frac{c-y}{x}+\frac{c+x}{y}-\frac{c}{x+y}=0, $$
multiplication by $xy(x+y)$ gives
$$ (c-y)y(x+y)+(c+x)x(x+y)-cxy. $$
Expanding the $c$ terms produces
$$ c\bigl(y(x+y)+x(x+y)-xy\bigr) =c(x^2+xy+y^2). $$
The remaining terms are
$$ -y^2(x+y)+x^2(x+y) =(x+y)(x^2-y^2) =(x-y)(x+y)^2. $$
This confirms the relation
$$ c(x^2+xy+y^2)+(x-y)(x+y)^2=0. $$
The second delicate step is the factorization of the numerator of $T$. Writing
$$ N=(c-y)y^2(x+y)^2+(c+x)x^2(x+y)^2+c x^2y^2, $$
the coefficient of $c$ is
$$ y^2(x+y)^2+x^2(x+y)^2+x^2y^2 =(x^2+xy+y^2)^2. $$
The remaining part is
$$ -x^2y^2-2xy^3-y^4+x^4+2x^3y+x^2y^2 =(x-y)(x+y)^2(x^2+xy+y^2). $$
Factoring out $x^2+xy+y^2$ yields exactly
$$ N=(x^2+xy+y^2)\Bigl(c(x^2+xy+y^2)+(x-y)(x+y)^2\Bigr). $$
A careless expansion at this stage could miss the common factor and obscure the connection with the hypothesis.
Alternative Approaches
A different approach starts by clearing denominators in the original condition. Multiplying by $(a-b)(b-c)(c-a)$ gives a symmetric cubic identity in $a$, $b$, and $c$. That cubic can be factored into
$$ (a+b-2c)(b+c-2a)(c+a-2b)=0. $$
Thus one of the three linear relations
$$ a+b=2c,\qquad b+c=2a,\qquad c+a=2b $$
must hold. Substituting any of these relations into the second expression immediately shows that it vanishes.
The method used in the main proof is preferable because it keeps both expressions in the same variables and exhibits a direct factorization linking the desired numerator to the hypothesis. No separate classification of solutions is required.