※ 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
于是 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*