※ 引述《FAlin (FA（バルシェ应援）)》之铭言： : 1. Given triangle ABC the point J is the centre of the excircle opposite the : vertex A. This excircle is tangent to the side BC at M, and to the lines AB : and AC at K and L, respectively. The lines LM and BJ meet at F, and the lines : KM and CJ meet at G. Let S be the point of intersection of the lines AF and : BC , and let T be the point of intersection of the lines AG and BC. Prove : that M is the midpoint of ST. : (The excircle of ABC opposite the vertex A is the circle that is tangent to : the line segment BC, to the ray AB beyond B, and to the ray AC beyond C.) : 2. If positive reals a_2, a_3,...,a_n satisfy a_2 * a_3 * ... * a_n = 1 : and n > 2 prove that : (a_2 + 1)^2 * (a_3 + 1)^3 * ... * (a_n + 1)^n > n^n : 3. The "liar's" guessing game is a game played between two players A and B. : The rules of the game depend on two positive integers k and n which are known : to both players. : At the start of the game A chooses integers x and N with 1≦x≦N. Player keeps : x secret, and truthfully tells N to player B. Player B now tries to obtain : information about x by asking player A questions as follows: each question : consists of B specifying an arbitrary set S of positive integers (possibly one : specified in some previuos question), and asking A whether x belongs to S. : Player B may ask as many questions as he wishes. After each question, player A : must immediately answer it with yes or no, but is allowed to lie as many times : as she wants; the only restriction is that, among any k+1 consecutive answers, : at least one answer must be truthful. : After B has asked as many questions as he wants, he must specify a set X of at : most n positive integers. If X belongs to B, then wins; otherwise, he loses. : Prove that: : 1. If n≧2^k, then B can guarantee a win. : 2. For all sufficiently large k , there exists an integer n ≧(1.99)^k such : that B cannot guarantee a win. 第一题 首先证明∠JFL = ∠JZL = ∠A/2 三角形FBM中 ∠FBM = ∠B + ∠FBA = ∠B + 90 - ∠B/2 = 90 + ∠B/2 ∠FMB = ∠LMC = ∠C/2 ∴∠BFM = ∠A/2 ∴JFAL四点共圆 同理 AGJK四点共圆 而∠ALJ = ∠AKJ = 90 ∴AGLJKF六点共圆 接着证明 BS = BA ∠AFJ = ∠AKJ = 90 ∴∠SFJ = 90 (六点共圆条件使用) 而 ∠SBF = ∠JBM = ∠JBK = ∠ABF 所以 三角形SBF全等三角型ABF 於是 SM = SB + BM = SA + BK = AK 同理 MT = CM + CT = CL + CA = AL 显然有 AK = AL 证毕 第二题 看到一堆数相乘 = 1 ， 可令 a_2 = x_3 / x_2 (反之亦可) a_3 = x_4 / x_3 .... a_n = x_2 / x_n 代换，则原式为 (x_2 + x_3)^2 * (x_3 + x_4)^3 * ... * (x_n + x_2)^n ＞ x_2^2 * x_3^3 * ... * x_n^n * n^n 将左式配项 x_k ( x_k + x_k+1 )^k = ( k-1*----- + x_k+1 )^k k-1 ≧ ( k^k / (k-1)^(k-1) )* x_k^k-1 * x_k+1 把k代入2~n然後相乘，则有 (x_2 + x_3)^2 * (x_3 + x_4)^3 * ... * (x_n + x_2)^n ≧ x_2^2 * x_3^3 * ... * x_n^n * n^n 但是等号成立时，x_2 = x_3 , (x_3)/2 = x_4 , (x_4)/3 = x_5 ... (x_n)/n-1 = x_2 1 左右相乘会有 ------ = 1 矛盾 (∵ n > 2) (n-1)! 所以等号不成立 --

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