## 2 Our Approach  
Since we consider only three fixed points, we can assume without loss of generality that they are given in the two-dimensional Euclidean space. The three points are denoted by \(L\), \(C\) and \(R\), furthermore let \(d(C,L)=1\), \(d(C,R)=d_{1}\), and \(d(L,R)=d_{2}\) (see Fig. 2). Without loss of generality we assume that \(1\leq d_{1}\leq d_{2}\leq d_{1}+1\), since the actual lengths of the triangle's sides are irrelevant and only their ratios matter. The 2-server problem on \(L\), \(C\) and \(R\) is denoted by \(\Delta(1,d_{1},d_{2})\), where the two servers are on \(L\) and \(R\) initially and the input is given as a sequence \(\sigma\) of points \(\in\{L,C,R\}\). \(\Delta(1,d_{1},d_{2})\) is also used to denote the triangle itself. The cost of an online algorithm \(\mathcal{A}\) for the input sequence \(\sigma\) is denoted by \(ALG_{\mathcal{A}}(\sigma)\) and the cost of the offline algorithm by \(OPT(\sigma)\). Suppose that for some constant \(\alpha\geq 0\), \(E[ALG_{\mathcal{A}}(\sigma)]\leq r\cdot OPT(\sigma)+\alpha\), holds for any input sequence \(\sigma\). Then we say that the competitive ratio of \(\mathcal{A}\) is at most \(r\). We first consider the case that the three points are on a line and both \(d_{1}\) and \(d_{2}\) are integers. In this case, we can design a general online algorithm as follows. The proof is given in the next section. **Lemma 1**: _Let \(n\) be a positive integer. Then there exists an online algorithm for \(\Delta(1,n,n+1)\) whose competitive ratio is at most \(C_{n}=\frac{(1+\frac{1}{n})^{n}-\frac{1}{n+1}}{(1+\frac{1}{n})^{n}-1}\)._  
Note that if triangles \(\Delta_{1}\) and \(\Delta_{2}\) are different, then \"good\" algorithms for \(\Delta_{1}\) and \(\Delta_{2}\) are also different. However, the next lemma says that if \(\Delta_{1}\) and \(\Delta_{2}\) do not differ too much, then one can use an algorithm for \(\Delta_{1}\) as an algorithm for \(\Delta_{2}\) with small sacrifice on the competitive ratio. **Lemma 2**: _Suppose that there are two triangles \(\Delta_{1}=\Delta(1,a_{1},b_{1})\) and \(\Delta_{2}=\Delta(1,a_{2},b_{2})\) such that \(a_{1}\geq a_{2}\) and \(b_{1}\geq b_{2}\) and that the competitive ratio of algorithm \(\mathcal{A}\) for \(\Delta_{1}\) is at most \(r\). Let \(\alpha=\max(\frac{a_{1}}{a_{2}},\frac{b_{1}}{b_{2}})\). Then the competitive ratio of \(\mathcal{A}\) for \(\Delta_{2}\) is at most \(r\cdot\alpha\)._  
_Proof._ For \(\alpha\) let \(\Delta_{\alpha}=\Delta(1/\alpha,a_{1}/\alpha,b_{1}/\alpha)\). Fix an arbitrary input sequence \(\sigma\) and let the optimal offline cost against \(\sigma\) be \(OPT_{1}\), \(OPT_{2}\) and \(OPT_{\alpha}\) for \(\Delta_{1}\), \(\Delta_{2}\) and \(\Delta_{\alpha}\), respectively. Since \(\Delta_{\alpha}\) is similar to \(\Delta_{1}\) and the length of each side is \(1/\alpha\), \(OPT_{\alpha}\) is obviously \((1/\alpha)OPT_{1}\). Since every side of \(\Delta_{2}\) is at least as long as the corresponding side of \(\Delta_{\alpha}\), \(OPT_{2}\geq OPT_{\alpha}=(1/\alpha)OPT_{1}\). Let the expected cost of \(\mathcal{A}\) against \(\sigma\) for \(\Delta_{1}\) and \(\Delta_{2}\) be \(ALG_{1}\) and \(ALG_{2}\), respectively. Note that \(\mathcal{A}\) moves the servers exactly in the same (randomized) way for \(\Delta_{1}\) and \(\Delta_{2}\). Since each side of \(\Delta_{2}\) is at most as long as the corresponding side of \(\Delta_{1}\), \(ALG_{2}\leq ALG_{1}\). We have \(\frac{ALG_{2}}{OPT_{2}}\leq\frac{ALG_{1}}{(1/\alpha)OPT_{1}}=\max(\frac{a_{1} }{a_{2}},\frac{b_{1}}{b_{2}})\cdot\frac{ALG_{1}}{OPT_{1}}\). \(\Box\)  
Thus we can \"approximate\" all triangles, whose \(\alpha\)-value is at most within some constant, by a finite set \(S\) of triangles. More precisely we introduce the notion of an approximation set as follows: Suppose that a target competitive ratio, i.e. a competitive ratio one wishes to achieve, is \(r_{0}\).