Question

Buna! Problema a fost data la etapa on-line a concursului Valeriu Alaci, la clasa a 9a.


Cate numere reale x exista astfel ca {x}; [x] si x sa fie in progresie aritmetica?


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Mie mi-a dat o infinitate, insa raspunsul este "doua numere". Dar nu reusesc sa inteleg cum... Eu am ajuns la 0<=x<3

Answer 1

Răspuns:

$x \in \{0, 1.5\}$

Explicație pas cu pas:

$\displaystyle \textrm{Pentru ca 3 termeni sa fie in progresie aritmetica, }\\\textrm{cel din mijloc trebuie sa fie media aritmetica a celorlalti 2.}\\ 1[x] = \frac{\{x\} + x}{2}\Bigg | \cdot 2 \\\\ 2[x] = \{x\} + x\\\\ 2[x] = \{x\} + [x] + \{x\}\\\\ 1[x] = 2\{x\}\\\\ x - \{x\} = 2\{x\} \\\\ x = 3\{x\}\\\\ Deoarece \\ 0 \leq \{x\} < 1\\\\ 0 \leq 3\{x\} < 3\\\\ 0 \leq x < 3$

$\textrm{Daca } x\in [0, 1)\\\implies [x] = 0\implies x = \{x\}\\\\ \{x\} = 3\{x\}\\\\\implies \{x\} = 0 \\\\\implies x = 0\\\\\textrm{Daca } x\in [1, 2) \\\implies [x] = 1 \implies x = [x] + \{x\} = 1 + \{x\}\\\\ x = 3\{x\} \implies 1 + \{x\} = 3\{x\}\\\\ 2\{x\} = 1\Bigg | \div 2\\\\ \{x\} = 0,5 \implies x = 1 + 0,5 = 1,5\\\\\textrm{Daca } x \in [2, 3) \implies [x] = 2\implies x = [x] + \{x\} = 2 + \{x\}\\\\x = 3\{x\}\\\\ 2 + \{x\} = 3\{x\}\\\\ 2\{x\} = 2\Bigg | \div 2\\\\ \{x\} = 1 \nless 1 \implies \textrm{Nu exista solutii in [2, 3)}\\\\ x \in \{0, 1.5\}$