■ OMK 2009 ■

mathlim

SOALAN 6

Segiempat ABCD dengan sisi AB = 25, BC = 39, CD = 52 dan DA = 60 diletakkan dalam suatu bulatan supaya A, B, C, D berada pada bulatan. Cari diameter bulatan tersebut.


The quadrilateral ABCD with sides AB = 25, BC = 39, CD = 52 and DA = 60 is placed in a circle such that A, B, C, D lie on the circle. Find the diameter of the circle.


ABCD 为一圆内接四边形，已知 AB = 25, BC = 39, CD = 52 及 DA = 60。求此圆的直径。

mathlim

原帖由 yw46 于 18-7-2009 09:43 AM 发表
没有啊...
那个link从recom拿来的...
那边那个人只放最后三题...

原来你放了link。
我登陆不到。
可不可以copy过来。
有没有解答呢？

yw46

Denote the product f(n) = (log23)(log34)(log45)...(logn-1n).
Find the value of f(4) + f(8) + f(16) + ... + f(210).

好像不太对的？？？
是log [ (n-1)(n) ] ???

~HeBe~_@

他的意思是。。。。
f(n) = (log_2  3)(log_3  4)(log_4  5)...(log_(n-1)  n).

~HeBe~_@

f(n) = (log_2 3)(log_3 4)(log_4 5)...[log_(n-1) n]
       = [(lg 3)/(lg 2)][(lg 4)/(lg 3)][(lg 5)/(lg 4)]...[(lg n)/(lg (n-1))]
       = (lg n)/(lg 2)
       = log_ 2 n

   f(4) + f(8) + f(16) + ... + f(2^10)
= (log_2 4) + (log_2 8) + (log_2 16) + ... + (log_2 2^10)
= (log_2 2^2) + (log_2 2^3) + (log_2 2^4) + ... + (log_2 2^10)
= 2 + 3 + 4 + ... + 10
= 54

yw46

小小一个的file上传了半天才成功

mathlim

1. b1 = 1, b2 = a1, ... ..., b100 = a99
S'k = b1 + b2 + ... ... + bk
S'1 = 1, S'k = 1 + Sk-1, k = 2, 3, ... ..., 100
S' = (S'1 + S'2 + ... ... + S'100)/100
     = (100 + S1 + S2 + ... ... +  S99)/100
     = 1 + (S1 + S2 + ... ... +  S99)/100
     = 1 + (S1 + S2 + ... ... +  S99)/99×99/100
     = 1 + 1000×99/100
     = 1 + 990
     = 991

mathlim

2.
a, b, -(a+b)

ax² + bx - (a + b) = 0
(ax + a + b)(x - 1) = 0
x = -(a + b)/a, 1

ax² - (a + b)x + b = 0
(ax - b)(x - 1) = 0
x = b/a, 1

bx² + ax - (a + b) = 0
(bx + a + b)(x - 1) = 0
x = -(a + b)/b, 1

bx² - (a + b)x + a = 0
(bx - a)(x - 1) = 0
x = a/b, 1

- (a + b)x² + ax + b = 0
[-(a + b)x - b](x - 1) = 0
x = -b/(a + b), 1

- (a + b)x² + bx + a = 0
[-(a + b)x - a](x - 1) = 0
x = -a/(a + b), 1

yw46

我的方法是
B^2 - 4AC must be a perfect square bigger  than 0
let a^2 = b^2 + c^2
the 3 unknown will be a^2 , -(b^2) , -(c^2)
这边skip了很多步
so,
(a^2)^2 - 4(b^2)(c^2)
= ( b^2 + c^2 )^2 - 4(b^2)(c^2)
= ( b^2 - c^2 )^2

[ -(b^2) ]^2 - 4(a^2)(-c^2)
= (b^2)^2 + 4(a^2)(c^2)
= ( a^2 - c^2 )^2 + 4(a^2)(c^2)
= ( a^2 + c^2 )^2

[ -(c^2) ]^2 - 4(a^2)(-b^2)
= (c^2)^2 + 4(a^2)(b^2)
= ( a^2 - b^2 )^2 + 4(a^2)(b^2)
= ( a^2 + b^2 )^2



刚才算了一下...
原来...
a , b , -(a+b)
a^2 + 4(b)(a+b) = (2b+a)^2
b^2 + 4(a)(a+b) = (2a+b)^2
(a+b)^2 - 4ab = (a-b)^2

[ 本帖最后由 yw46 于 27-7-2009 08:49 PM 编辑 ]

mathlim

原帖由 yw46 于 27-7-2009 08:42 PM 发表
我的方法是
B^2 - 4AC must be a perfect square bigger  than 0
let a^2 = b^2 + c^2
the 3 unknown will be a^2 , -(b^2) , -(c^2)
...

其实我们的是一样的。

x = -b²,
y = -c²,
b² + c² = - x - y = -(x + y)

你的是 b² + c², -b², -c²。
我的是  -(x + y), x, y

dunwan2tellu

那你们如何确定得到的 roots 一定 disctinct ?

i.e  ax² + bx - (a + b) = 0
Ax² + Bx +C = 0

First player : A = a
2nd Player : C = a
First player : B = -2a

final root = a (repeat root)

mathlim

a ≠ 0, b ≠ 0, a ≠ ±b