December 28, 2004

 


»º º

December 28, 2004

 


1

1

1



1

2π

A, B, C

ABC

AB = BC = CA = √ 3.











 




√ 3



  





 



 




0

1

√

3



1/2

1/2

1 = 1 2 .


 

  


1/2






√ 3



1/2



= π · (1/2) ² π · 1 ² = 1

4 .

 










0

2






√ 3

2 − √

3 2 = 1 −

√ 3 2 .







AA

B, C

ABC

A

AB

AC

AA

√

3



A

BC

AA

√

3

AB

BC

AC







√ 3

1

3 .



AA

A, A

A = (cos θ ₁ , sin θ ₁ ), A
= (cos θ ₂ , sin θ ₂ ),

0 ≤ θ ₁ , θ ₂ ≤ 2π.

{(θ ₁ , θ ₂ )|0 ≤ θ ₁ , θ ₂ ≤ 2π}.

(1)

0 ≤ θ 2 ≤ θ 1 ≤ 2π

AA

√

3

2π

3 ≤ θ ₁ − θ ₂ ≤ 4π 3 .


 


(2)

0 ≤ θ ₁ ≤ θ ₂ ≤ 2π

AA

√

3

2π

3 ≤ θ ₂ − θ ₁ ≤ 4π 3 .


 





  








√ 3

1 3 .

- 6

0 2π

3 4π

3 2π

2π 3 4π 3

2π

θ ₁ θ ₂







1.1

 








 







 
 



 



1.2

A, B, C

5, 4, 3

S





SA

SB

(1) 0

(2) 3

(3) 4

(4) 5

(5) ∞

4 -

3 5

A

B C

S



1.3

 #

 

 
#

 DZ


 


 
!"
  Æ


 ! 

%

&





 




1.4

30

10

30

$)* 

25

5

'DZ%-

2

3

1

10 − 1 = 9

9

9 × 3

+

2

= 29

1

?

/





'0
1

1, 2, 3, 4, 5, 6

  2

 

 



2 3 4 5 6 7 8 9 10 11 12

36 1 2

36 3

36 4 36 5

36 6

36 5

36 4 36 3

36 2

36 1

36

 
'0
11(
3

 4
5 
   )




 
1(





 
 


26

 

26

5 ² < 26 < 3 ³ .

,
   
 
7