Atrisinājums

Izmantosim faktu, ka, ja \(x \in(0; 1)\), tad arī \(x^{2} \in(0; 1)\). Tad \(2-\sqrt{3} \in(0; 1)\), \((2-\sqrt{3})^{2}=7-4 \sqrt{3} \in(0; 1),\ (7-4 \sqrt{3})^{2}=97-56 \sqrt{3} \in(0; 1)\) utt. Šādi iegūstam skaitļus \(h_{1}-a_{1} \sqrt{3}, \quad h_{2}-a_{2} \sqrt{3}\) utt., kur \(a_{n}\) un \(h_{n}\) ir naturāli skaitļi, pie tam \(\left(h_{n}-a_{n} \sqrt{3}\right)^{2}=h_{n}^{2}+3a_{n}^{2}-2h_{n}a_{n} \sqrt{3}=h_{n+1}-a_{n+1} \sqrt{3}\). Tālāk, ja \(h_{n}^{2}-3a_{n}^{2}=1\), tad \(h_{n+1}^{2}-3a_{n+1}^{2}=(h_{n}^{2}+3a_{n}^{2})^{2}-3 \cdot (2h_{n}a_{n})^{2}=(h_{n}^{2}+3a_{n}^{2})^{2}-12h_{n}^{2}a_{n}^{2}=(h_{n}^{2}-3a_{n}^{2})^{2}=1\). Tā kā \(h_{1}^{2}-3a_{1}^{2}=2^{2}-3 \cdot 1^{2}=1\), tad visiem \(nh_{n}^{2}-3a_{n}^{2}=1\).

Vēl ievērosim, ka \(a_{n+1}=2h_{n}a_{n} \geq 2h_{1}a_{n} \geq 4 a_{n}\), tāpēc \(a_{n} \geq 4^{n-1}\). Izvēlēsimies \(n\) tik lielu, ka \(a_{n}>2013^{2}\) un apskatīsim 21.zīm. redzamo vienādsānu trijstūri. Tā perimetram \(P\) ir spēkā nevienādības \(P>4a_{n}\) un \(\sqrt{P}>2 \cdot 2013\). Lielākā atšķirība starp tā malām nav lielāka par

\(\left|\sqrt{a_{n}^{2}+h_{n}^{2}}-2a_{n}\right|=\left|\frac{h_{n}^{2}-3a_{n}^{2}}{\sqrt{a_{n}^{2}+h_{n}^{2}}+2a_{n}}\right|=\frac{1}{\sqrt{a_{n}^{2}+h_{n}^{2}}+2a_{n}}<\)

\(<\frac{1}{\sqrt{a_{n}^{2}+h_{n}^{2}}+a_{n}}=\frac{2}{P}=\frac{1}{0,5 \sqrt{P}} \cdot \frac{1}{\sqrt{P}}<\frac{1}{2013 \cdot \sqrt{P}}\), k.b.j.