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標題: 挑戰100 [列印本頁]
作者: M.N.M.    時間: 07-6-30 21:13
標題: 挑戰100
1.What is the smallest integer x larger than 1 such that x^2 ends in the same three digits as x does ?

2. How many nonempty subsets of {1,2,3...,12} have the property that the sum of  the largest element and the smallest element is 13 ?

3.Let XYZ be a triangle with ∠X=60 ° and ∠Y=45° . A circle with center P passes though points A and B on side XY,C and D on side YZ,

and E and F on side ZX . Suppose AB=CD=EF. Find ∠XPY in degrees.
作者: 傲月光希    時間: 07-6-30 22:14
原文由M.N.M. 於 07-6-30 09:13 PM 發表
2. How many nonempty subsets of {1,2,3...,12} have the property that the sum of  the largest element and the smallest element is 13 ?

Sol)
We consider that the partitions of 13 are (1,12),(2,11),...,(6,7).
Then use each of them to make a subset of {1,2,...,12}.
First,the subsets containing 1 and 12 may contain elements 11,10,...,and 2.
Therefore,the number of the these is C(10,10)+C(10,9)+...+C(10,0)=2^10.

Second,consider the pair (2,11).
Since they play the roles of the largest one and the smallest one,subsets of (2,11) have at most 10 elements.
Thus,the pair has C(8,8)+C(8,7)+...+C(8,0)=2^8 subsets.

Continue in this way.There are 2^0+2^2+...+2^10=((2^2)^6-1)/(2^2-1) subsets...#

(譯)
我們考慮對13的分割有(1,12),(2,11),...,(6,7).
然後用每一個去作一個{1,2,...,12}的子集.
首先，包含1跟12的子集可能包含11,10,...,2這些元素.
因此，這些子集的個數有C(10,10)+C(10,9)+...+C(10,0)=2^10個.

其次，考慮(2,11)這個配對.
因為他們各扮演最大跟最小的數字，所以(2,11)的子集至多有10個元素.
因此，這個配對有C(8,8)+C(8,7)+...+C(8,0)=2^8個子集.

這樣持續下去，總共有2^0+2^2+...+2^10=((2^2)^6-1)/(2^2-1)個子集...#

[ 本文最後由 傲月光希 於 07-6-30 10:52 PM 編輯 ]
作者: himitsuaile    時間: 07-6-30 22:44
2



[ 本文最後由 himitsuaile 於 07-6-30 10:45 PM 編輯 ]
作者: 傲月光希    時間: 07-7-1 00:20
By 超‧激‧萌‧小M編輯
3.Let XYZ be a triangle with ∠X=60 ° and ∠Y=45° . A circle with center P passes though points A and B on side XY,C and D on side YZ,

and E and F on side ZX . Suppose AB=CD=EF. Find ∠XPY in degrees.

See attached picture
(Note:point P is the center of the circle.)
Consider line PG and line PH,which are orthogonal with line ZX and line XY individually.
They must have the same length by the property RHS of triangles.
Next,we see the triangle XGP and the triangle XHP.
PG=PH,PX is the pooled line,and the angleXGP=the angle XHP=90 degrees.
By RHS,they are identical.So angle PXH is 22.5 degrees.

Similarly to argue,we get the angle PYH is 30 degrees.
Thus,the angle XPY is 180-22.5-30=127.5 degrees.

(譯)
如附圖
(注意：P點是圓心)
考慮線段PG跟線段PH，他們分別垂直ZX跟XY.
由RHS三角形性質，他們必定長度相同.
接著，我們先看三角形XGP跟三角形XHP.
PG=PH，PX是共用邊，以及角XGP=角XHP=90度.
由RHS，他們全等，所以角PXH等於22.5度.

用類似的方法去討論，我們知道角PYH是30度.
因此，角XPY等於180-22.5-30=127.5度.

[ 本文最後由 傲月光希 於 07-7-1 08:23 PM 編輯 ]

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