Olympiad Maths Prep

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Problem 1286

National olympiad, first round
Algebra Difficulty 6.5 Prove it

1. Let Sk=ak(a−b)(a−c)+bk(b−a)(b−c)+ck(c−a)(c−b)S_{k}=\frac{a^{k}}{(a-b)(a-c)}+\frac{b^{k}}{(b-a)(b-c)}+\frac{c^{k}}{(c-a)(c-b)}. Prove that S0=S1=0,S2=1S_{0}=S_{1}=0, S_{2}=1 and S3=a+b+cS_{3}=a+b+c.

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

Solution. Note that it must hold that a≠b≠c≠aa \neq b \neq c \neq a. We have

S0=1(a−b)(a−c)+1(b−a)(b−c)+1(c−a)(c−b)=b−c−a+c+a−b(a−b)(a−c)(b−c)=0S1=a(a−b)(a−c)+b(b−a)(b−c)+c(c−a)(c−b)=ab−ac−ab+bc+ac−bc(a−b)(a−c)(b−c)=0 \begin{aligned} & S_{0}=\frac{1}{(a-b)(a-c)}+\frac{1}{(b-a)(b-c)}+\frac{1}{(c-a)(c-b)}=\frac{b-c-a+c+a-b}{(a-b)(a-c)(b-c)}=0 \\ & S_{1}=\frac{a}{(a-b)(a-c)}+\frac{b}{(b-a)(b-c)}+\frac{c}{(c-a)(c-b)}=\frac{a b-a c-a b+b c+a c-b c}{(a-b)(a-c)(b-c)}=0 \end{aligned}

Then,

S2=a2(a−b)(a−c)+b2(b−a)(b−c)+c2(c−a)(c−b)=a2(b−c)−b2(a−c)+c2(a−b)(a−b)(a−c)(b−c)=ab(a−b)−c(a2−b2)+c2(a−b)(a−b)(a−c)(b−c)=(a−b)(ab−ac−bc+c2)(a−b)(a−c)(b−c)=(a−b)[a(b−c)−c(b−c)](a−b)(a−c)(b−c)=(a−b)(a−c)(b−c)(a−b)(a−c)(b−c)=1. \begin{aligned} S_{2} & =\frac{a^{2}}{(a-b)(a-c)}+\frac{b^{2}}{(b-a)(b-c)}+\frac{c^{2}}{(c-a)(c-b)}=\frac{a^{2}(b-c)-b^{2}(a-c)+c^{2}(a-b)}{(a-b)(a-c)(b-c)} \\ & =\frac{a b(a-b)-c\left(a^{2}-b^{2}\right)+c^{2}(a-b)}{(a-b)(a-c)(b-c)}=\frac{(a-b)\left(a b-a c-b c+c^{2}\right)}{(a-b)(a-c)(b-c)} \\ & =\frac{(a-b)[a(b-c)-c(b-c)]}{(a-b)(a-c)(b-c)}=\frac{(a-b)(a-c)(b-c)}{(a-b)(a-c)(b-c)}=1 . \end{aligned}

Similarly for S3S_{3} we get

S3=a3(a−b)(a−c)+b3(b−a)(b−c)+c3(c−a)(c−b)=a3(b−c)−b3(a−c)+c3(a−b)(a−b)(a−c)(b−c)=ab(a2−b2)−c(a3−b3)+c3(a−b)(a−b)(a−c)(b−c)=(a−b)[(ab(a+b)−c(a2+ab+b2)+c3](a−b)(a−c)(b−c)=(a−b)[(a2(b−c)+ab(b−c)−c(b2−c2)](a−b)(a−c)(b−c)=(a−b)(b−c)[a2+ab−c(b+c)](a−b)(a−c)(b−c)=(a−b)(b−c)[(a−c)(a+c)+b(a−c)](a−b)(a−c)(b−c)=(a−b)(b−c)(a−c)(a+b+c)(a−b)(a−c)(b−c)=a+b+c \begin{aligned} S_{3} & =\frac{a^{3}}{(a-b)(a-c)}+\frac{b^{3}}{(b-a)(b-c)}+\frac{c^{3}}{(c-a)(c-b)}=\frac{a^{3}(b-c)-b^{3}(a-c)+c^{3}(a-b)}{(a-b)(a-c)(b-c)} \\ & =\frac{a b\left(a^{2}-b^{2}\right)-c\left(a^{3}-b^{3}\right)+c^{3}(a-b)}{(a-b)(a-c)(b-c)}=\frac{(a-b)\left[\left(a b(a+b)-c\left(a^{2}+a b+b^{2}\right)+c^{3}\right]\right.}{(a-b)(a-c)(b-c)} \\ & =\frac{(a-b)\left[\left(a^{2}(b-c)+a b(b-c)-c\left(b^{2}-c^{2}\right)\right]\right.}{(a-b)(a-c)(b-c)}=\frac{(a-b)(b-c)\left[a^{2}+a b-c(b+c)\right]}{(a-b)(a-c)(b-c)} \\ & =\frac{(a-b)(b-c)[(a-c)(a+c)+b(a-c)]}{(a-b)(a-c)(b-c)}=\frac{(a-b)(b-c)(a-c)(a+b+c)}{(a-b)(a-c)(b-c)} \\ & =a+b+c \end{aligned}

We will provide another proof of the last equality. For every x∈Rx \in \mathbb{R}, it holds that

(x−a)(x−b)(x−c)=x3−(a+b+c)x2+(ab+bc+ca)x−abc (x-a)(x-b)(x-c)=x^{3}-(a+b+c) x^{2}+(a b+b c+c a) x-a b c

so

a3−(a+b+c)a2+(ab+bc+ca)a−abc=0b3−(a+b+c)b2+(ab+bc+ca)b−abc=0c3−(a+b+c)c2+(ab+bc+ca)c−abc=0 \begin{aligned} & a^{3}-(a+b+c) a^{2}+(a b+b c+c a) a-a b c=0 \\ & b^{3}-(a+b+c) b^{2}+(a b+b c+c a) b-a b c=0 \\ & c^{3}-(a+b+c) c^{2}+(a b+b c+c a) c-a b c=0 \end{aligned}

If we divide the first equation by (a−b)(a−c)(a-b)(a-c), the second by (b−a)(b−c)(b-a)(b-c), the third by (c−a)(c−b)(c-a)(c-b), and then add the resulting equations, we find that

S3−(a+b+c)S2+(ab+bc+ca)S1−abcS0=0 S_{3}-(a+b+c) S_{2}+(a b+b c+c a) S_{1}-a b c S_{0}=0

Finally, if we take into account that S0=S1=0S_{0}=S_{1}=0 and S2=1S_{2}=1, from the last equation it follows that S3=a+b+cS_{3}=a+b+c.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.