※ 引述《hiei81 (最愛)》之銘言： ※ 引述《hiei81 (最愛)》之銘言： ※ 引述《chaogold (GREECEimoimcomin￾N￾N5)》之銘言： : Here are the 6 problems of IMO 2004 (july 12-13 in Athen) : Problem 1 : ABC is acute angle triangle with AB<>AC. The circle with diameter BC : intersects the lines AB and AC respectively at M and N. O is the : midpoint of BC. The bisectors of <BAC and <MON intersect at R. : Prove that the circumcircles of thev triangles BMR and CNR have a : common point lying on the line BC. First note RM=RN but ⊿AMR不全等於⊿ANR => A、M、R、N共圓 => ∠AMR+∠ANR=180' => BMR, CNR外接圓共點於BC上某一點 : Problem 2 : Find all polynomials f with real coefficients such that, for all : reals a,b,c such that ab+bc+ca = 0, we have the relation : f(a-b) + f(b-c) + f(c-a) = 2 f(a+b+c) Let a=b=c=0 => 3f(0) = 2f(0) => f(0)=0 Let b=c=0 => f(a)+f(0)+f(-a)=2f(a) => f is an even function => f(x)= Sigma a x^2k 且a = 0 2k 0 Let s=a-b, t=b-c => c-a= -s-t且a+b+c= √(s^2+st+t^2) => f(s)+f(t)+f(s+t) = 2f(√(s^2+st+t^2)) 因偶函數，可假設s>0, t>0 Let g(x)= Sigma a x^k => f(x)=g(x^2) 2k => g(s^2)+g(t^2)+g(s^2+2st+t^2)=2g(s^2+st+t^2) ... (1) Let s=t => 2g(s^2)+g(4s^2) = 2g(3s^2) if deg(g)>2 for sufficient large s，左式>右式 => deg(g)=2 or 1 => g(x)= ax^2+bx代回(1)均成立 Therefore, f(x) = ax^4+bx^2, a,b屬於R : Problem 3 : Define a "hook" to be a figure made up of six unit squares as shown : in the figure below, or any of the figures obtained by rotations and : reflections to this figure. : [ ][ ][ ] : [ ] [ ] : [ ] : Determine all mxn rectangles that can be covered without gaps and : without overlaps with hooks such that no point of a hook covers area : outside the rectangle Guess: 3N*4M , N,M為自然數 : Problem 4 : Let n>=3 be an integer. Let t[1],...,t[n] be positive real numbers : such that : n^2+1>(t[1]+...+t[n])(1/t[1]+...+1/t[n]) : Show that, for all distinct i,j,k, t[i],t[j],t[k] are the side : lengths of a triangle Without loss of generality, assume that t[1]<=t[2]<=...<=t[n] Let S=(t[1]+...+t[n])(1/t[1]+...+1/t[n]) k+1 1 When n=3, let t[2]=kt[1] => S=((k+1)t[1]+t[3])(----- + -----) k+1 t[3] t[1] kt[1] t[3] =(k+1)^2/k+1 + --------- + (k+1) ---- k t[1] t[3] Let t[3]=ut[1]=> u>=k>=1，且由微分知(k+1)u/k + (k+1)/u 在u=√k時有最小值 又k>=1 => k>=√k => u>=√k =>u越大，S越大 當u=1+k時 S=(2+2k)(1+1/k+1/(k+1))=6+2k+2/k >= 6+4 = 3^2+1 => u>=1+k時，S>=n^2+1，但已知S<n^2+1，故u<1+k => t[3]<t[1]+t[2]三者可為三角形之三邊 考慮原題，僅需證明t[n]<t[1]+t[2]即可 用數學歸納法，n=3時，已知原式成立，設n=k時，原式成立，則n=k+1時 設S'表示1至k項，S表示1至k+1項，若S'>=n^2+1 => S=S'+1+ t[k+1](1/t[1]+1/t[2]+...+1/t[k])+(t[1]+t[2]+...+t[k])/t[k+1] t[k+1] t[1] t[k+1] t[k] =S'+1+ (------ + ------) + ... + (------ + ------) t[1] t[k+1] t[k] t[k+1] >=(k^2+1)+1+2+2+2+...+2 = k^2+2k+1 +1 = (k+1)^2+1 --><-- 故S'<n^2+1 => 由歸納假設知t[k]<t[1]+t[2] 若t[k+1]>=t[1]+t[2]，注意由柯西不等式，S'>=k^2 => S=S'+1+ t[k+1](1/t[1]+1/t[2]+...+1/t[k])+(t[1]+t[2]+...+t[k])/t[k+1] >= k^2+1+t[k+1]/t[1]+t[k+1]/t[2]+(t[1]+t[2])/t[k+1] t[k+1] t[3] t[k+1] t[k] + (------ + ------) + ... + (------ + ------) t[3] t[k+1] t[k] t[k+1] >= k^2+1+(t[1]+t[2])/t[1]+ (t[1]+t[2])/t[2]+1 + 2 + 2 + ... +2 =k^2+1+1+1+1+t[2]/t[1]+t[1]/t[2]+2(k-2) >= k^2+2k+2 = (k+1)^2 + 1 得證 : Problem 5 : In a convex quadrilateral ABCD, the diagonal BD bisects neither <ABC : nor <CDA. A point P lies inside ABCD and satisfies <PBC = <DBA and : <PDC = <BDA. : Prove that ABCD are concyclic if and only if AP = CP. 由已知，B-P-C不共線 若<A=<C=90' 且<QBC=<DBA, <RDC=<BDA => BQ平行RD => P點不存在 所以<A≠<C≠90' 不妨設<A > 90' => <A > <C => <ABD+<ADB < <CDB+<DBC => <ABD < <CBD或<ADB < <CDB 不妨設<ABD < <CBD，若<ADB > <CDB => P點不存在，故<ADB < <CBD (=>) 由共圓性質證得<BPD = 2<C，令圓心O，圓上兩點E、F且B-P-E、D-P-F => 弧BAD = 弧ECF(圓內角) => 弦BE=弦DF => BE、DF等弦心距 <--> => O在<DPE之分角線上 => OP 過弧BF的中點M，又<BDA=<PDC => M亦為弧AC之中點 => OP直線過弧AC之中點 => OP直線垂直平分AC => AP=CP : Problem 6 : A positive integer is alternating if every two consecutive digits in : its decimal representation are of different parity. Find all : positive integers n such that n has a multiple which is alternating. For any positive integer N=2^a*5^b*K, K非2或5之倍數 設X為N之倍數且X is alternating (i) b<=4 若b>=5 => 3125|X => X末五位數可被3125整除 但3125*1, 3125*2, ..., 3125*31皆非alternating (需要驗證一下) => b<=4 (ii) 若b≠0則a<=1 => 若a>=2，20|X => N末兩位皆是偶數 --><-- =>a<=1 (iii) 若b=0, K=1, 對任意a, 存在a位數的X, 用數學歸納法證明 ex: 4|36 => 8|216 => 16|1216 => 32|21216 => ... (iv) 若K=1時X存在，則對任意非2或5倍數之K，X亦存在 K=1時存在一T位數的X，則用10^T+1, 10^2T+10^T+1, ...當乘數來構造 結論：N=2^a*5^b*K, 1<=b<=4且0<=a<=1或 N=2^a*K -- 永不平息的海浪一波一波打來人生的 挑戰與試煉 曾經 我是岩石，盡一切力量承受著 於是 海浪在我身上留下一道道的傷痕 曾經 我是沙子 於是我 隨。波。逐。流 現在 我是海洋 看著一道道海浪打去 沈澱下最美麗的結晶 --

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