阿甲拿到一分Pre-Cal level 的數學競賽試題﹐真的不容易喔﹐20題﹐都出得很活﹐阿甲是都做出來了﹐可是我們學校去比賽的孩子﹐第一名只做對了八題﹐這其中﹐大概還多半是猜的(是選擇題)。其中有三題﹐阿甲選對了答案﹐因為其他答案更不合理﹐可是阿甲卻沒法說服自己那答案怎麼來的﹐所以把它們列在這裡﹐請教高手﹐請各位聰明兄弟姐妹幫阿甲想想吧。
(1) In right triangle ABC a square is drawn upon hypotenuse AC in the plane of Triangle ABC and not overlapping the triangle. Let D be the center of the square. Find the meansure in degree of angle DBC. ===> The answer is 45
A.J.'s note: I know points A,B,C, and the center of the square are on the same circle, and to prove it is 45 degree, do I need to prove the segment BD is the angle bisector to angle B?
結論：Due to blog friend RY's help, I found we should only use the arc and angle's relationship. Just to see how those angles intercept those arcs, remember AC is the diameter of the circle passes through A,B,C,D, and arc ABC and ADC are both semi cicle, and it is 180 degrees, arc AD = arc CD = 90 degrees.
(2) An infinite geometric series sums to 4. The sum of the cubes of the terms of that series sums to 192. What does the sum of the squares of the original series sum to? ===> The answer is 48
A.J.'s note: The sum of an infinite geometric series is a/(1-r) (a is the first term and r is the ratio and r< 1),
so a/(1-r) = 4 and a^3/(1-r^3) = 192. we want a^2/(1-r^2), how to get it from the given two equations?
結論: Actually, there is no short cut here, we just solve a and r from the given 2 equations using substitution method that a=4(1-r). Once we solve a and r, plug in to find what we want.
(3) A subset of the set of intergers {1,2,3,....,99,100} has the property that none of its members is 3 times another. What is the maximum number of members such a subset can have? ===> The answer is 76.
A.J.'s note: I solved this one by listing out all those numbers, I think there got to be some better way than my stupid way, do you have any idea?
阿甲 於 2009/04/21 06:41 AM
~原文中的這些影片已遭移除，所以改爲腦筋健美操。2056註~
2056今天也要來談談數學.....怎麽樣，是不是嚇了大夥兒一跳？別懷疑，不過是從youtube上看到的。大概很多人已經看過了吧。
阿甲老師談過許多數學講座，請看看以下不同的人的乘法。
2056對其中老外的乘法最好奇，也最霧沙沙，數學名師，也許還有各位高手可以講出許多道理來。
老外的算法
聰明人的算法
香港人的算法
天才的算法
2/27/2011
遊魂出題： 請幫忙證明: (1/2) *(3/4) * (5/6) * ... * (999999/1000000) < 1/1000.
遊魂 在 新浪部落 於 2011/02/27 03:47 PM 回應
hint... 解答：
Given: A(n) = 1/2 * 3/4 * 5/6 * ... * 2n-3 / 2n-2 * 2n-1 / 2n (n= 1, 2, .....)
Let: B(n) = 2/3 * 4/5 * 6/7 * ... *2n-2 / 2n-1 * 1 (n= 2, 3,......)
Compare each term in A(n) and B(n): 1/2 < 2/3, 3/4 < 4/5, 5/6 < 6/7, ..., 2n-3 / 2n-2 < 2n-2 / 2n-1, 2n-1 / 2n < 1
Therefore: A < B ---------(1)
Multiply A and B cancels most terms: A * B = 1 / 2n -------- (2)
Multiply both sides of (1) with A, which is greater than zero:
A * A < A * B
A * A < 1 / 2n
A < 1 / sqrt(2n)
QED
(In this case 2n = 1000000)
hint... 在 新浪部落 於 2011/02/28 08:51 AM 回應