Re: 這個怎麼沒有人po

[hiei81 (最愛)]

※ 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

先吃飯...剩下有空再想...:D

: 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.
: 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.

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推 Dawsen:第一題會被扣１分,因為要在線段上
推 hiei81:-_-!! it just say "on the line BC"...
推 hiei81:然後我證明上會寫「在BC上取一點P使
→ hiei81:MRBP共圓，then NRCP共圓...:)
推 hiei81:更正，是在BC"線段"上..反正題目要玩
→ hiei81:大家一起玩:D
推 darkseer:Problem 3 可以是12n*x, x is not 1,2or5