※ 引述《hiei81 (最愛)》之銘言：
※ 引述《hiei81 (最愛)》之銘言：
※ 引述《chaogold (GREECEimoimcominNN5)》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
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