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−  Let <math> n \geq 2</math> be a positive integer and <math> \lambda</math> a positive real number. Initially there are <math> n</math> fleas on a horizontal line, not all at the same point. We define a move as choosing two fleas at some points <math> A</math> and <math> B</math>, with <math> A</math> to the left of <math> B</math>, and letting the flea from <math> A</math> jump over the flea from <math> B</math> to the point <math> C</math> so that <math> \frac {BC}{AB} = \lambda</math>.
 
   
−  Determine all values of <math> \lambda</math> such that, for any point <math> M</math> on the line and for any initial position of the <math> n</math> fleas, there exists a sequence of moves that will take them all to the position right of <math> M</math>.
 
− 
 
−  ==Solution==
 
−  Write <math>n = N+1</math>; we claim the answer is\[
 
−  \boxed{\left\{\lambda\colon \lambda \geq \frac 1N\right\}}.
 
−  \]Place the points on the number line; let <math>p</math> denote the coordinate of point <math>P</math>. For a configuration <math>x_1\leq x_2\leq \cdots \leq x_{N+1}</math> of points <math>X_1,X_2,\ldots, X_{N+1}</math>, define its cap by <math>x_{N+1}</math> and its score by <math>X_1X_{N+1} + X_2X_{N+1} + \cdots + X_NX_{N+1}</math>. Notice now that we can get all the points past <math>M</math> if and only if we can get the cap to exceed <math>m</math>; only if clearly holds, and for if just notice that if <math>X_{N+1}</math> is past <math>M</math>, we just reflect everything over <math>X_{N+1}</math> and win. So, we just want to find <math>\lambda</math> so that the cap can get arbitrarily large.
 
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−  Lemma 1: <math>\lambda \geq \frac 1N</math> all work.
 
− 
 
−  Proof: First, note that we can get a configuration in which no two fleas are at the same point. To do this, first take the leftmost point and reflect over the rightmost point. Then, we have a point strictly to the right of all other points. Suppose the points are <math>P_1,P_2,\ldots, P_{N+1}</math>, respectively. Then, at each point, reflect the leftmost point over the rightmost one <math>N</math> times, until we have <math>N+1</math> points all at distinct places.
 
− 
 
−  Suppose our current configuration is <math>X_1,X_2,\ldots, X_{N+1}</math>, and let <math>d_1 = X_1X_2</math>, <math>d_2 = X_2X_3, \ldots</math>, and write <math>d = \min(d_i)>0</math>. Reflect <math>X_1</math> over <math>X_{N+1}</math>. Then, the configuration becomes <math>X_2,X_3,\ldots, X_{N+1}, X_1'</math>, and the pairwise distances become <math>d_2,d_3,\ldots, d_N, \lambda(d_1+d_2+\cdots+d_N)</math>, which are again all <math>\geq d</math>. So, at each step, we know that the cap has coordinate increasing by at least <math>d</math> every time, so we can get the cap to be arbitrarily large. <math>\Box</math>
 
− 
 
−  Lemma 2: <math>\lambda < \frac 1N</math> all fail.
 
− 
 
−  Proof: Define the value <math>V</math> of a configuration to be\[
 
−  V = C + \frac{\lambda S}{1N\lambda},
 
−  \]where <math>C</math> is the cap and <math>S</math> is the score, respectively. We claim that the value is nonincreasing. Indeed, suppose we have a configuration <math>X_1,X_2,\ldots, X_{N+1}</math>, where <math>x_1\leq x_2\leq \cdots \leq x_{N+1}</math>. Suppose we reflect <math>X_i</math> over <math>X_j</math> with <math>i<j</math>. Write <math>\alpha = X_iX_j</math>, and <math>\beta =X_jX_{N+1}</math>. If the cap does not change, then clearly the score decreases, since <math>X_i</math> moves closer to <math>X_{N+1}</math> and the other distances do not change. Otherwise, we see that the cap increases by <math>\lambda \alpha  \beta</math>.
 
− 
 
−  Furthermore, since we went from <math>X_1,X_2,\ldots, X_{N+1}</math> to <math>X_1,X_2,\ldots, X_{i1},X_{i+1},\ldots, X_{N+1},Q</math>, where <math>Q</math> is where <math>X_i</math> jumped to, then the score changes from\[
 
−  \sum_{k=1}^N x_{N+1}x_k = Nx_{N+1}  \sum_{k=1}^N x_k
 
−  \]to\[
 
−  \sum_{\substack{k=1\\ k\neq i}}^{N+1}q  x_k = Nq  \sum_{\substack{k=1\\k\neq i}}^{N+1} x_k,
 
−  \]so we compute that <math>\Delta S = N(qx_{N+1})  x_{N+1}+x_i = N(\lambda\alpha\beta)  (\alpha+\beta)</math>. So,\[
 
−  \Delta V = \Delta C + \frac{\lambda \Delta S}{1N\lambda} = (\lambda\alpha\beta) + \frac{\lambda (N(\lambda\alpha\beta)\alpha\beta)}{1N\lambda} \leq \lambda\alpha + \frac{\lambda(N\lambda1)\alpha}{1N\lambda} = 0.
 
−  \]So, <math>V</math> is indeed nonincreasing. So, if <math>V_0</math> is the initial value, since <math>S\geq 0</math> always we have that\[
 
−  C\leq V \leq V_0
 
−  \]implying that <math>m = V_0</math> indeed works, and that we cannot move all the fleas to the right of <math>M</math>. <math>\Box</math>
 
− 
 
−  Thus, with both lemmas proven, we may conclude.
 