Assume that for pairwise distinct real numbers $a,b,c,d$ holds: $$ (a^2+b^2-1)(a+b)=(b^2+c^2-1)(b+c)=(c^2+d^2-1)(c+d).$$Prove that $ a+b+c+d=0.$
Problem
Source: Poland - Second Round 2020 P1
Tags: algebra, Poland
62861
08.02.2020 23:21
The equation $(a^2 + b^2 - 1)(a + b) = (b^2 + c^2 - 1)(b + c)$ is
\begin{align*}
a^3 + a^2 b + a b^2 + b^3 - (a + b) & = c^3 + c^2 b + c b^2 + b^3 - (c + b)\\
\iff (a^3 - c^3) + (a^2 - c^2) b + (a - c) b^2 - (a - c) & = 0\\
\iff (a - c) \big[a^2 + ac + c^2 + (a + c) b + b^2 - 1\big] & = 0\\
\iff a^2 + b^2 + c^2 + ab + bc + ca & = 1
\end{align*}as $a \ne c$. Similarly $d^2 + b^2 + c^2 + db + bc + cd = 1$, so subtracting these two, we get
\begin{align*}
(a^2 - d^2) + (a - d)b + (a - d)c & = 0\\
\iff (a - d) \big[a + d + b + c\big] & = 0\\
\iff a + b + c + d & = 0
\end{align*}as $a \ne d$.
Tintarn
09.02.2020 21:45
Or in one line: \[0=(b-d)(a^2+b^2-1)(a+b)+(d-b-a+c)(b^2+c^2-1)(b+c)+(a-c)(c^2+d^2-1)(c+d)=(a-c)(d-a)(b-d)(a+b+c+d).\]
sqing
21.02.2020 10:19
pggp wrote: Assume that for pairwise distinct real numbers $a,b,c,d$ holds: $$ (a^2+b^2-1)(a+b)=(b^2+c^2-1)(b+c)=(c^2+d^2-1)(c+d).$$Prove that $ a+b+c+d=0.$
Attachments:
