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For the problem of 2-dimensional , let O be the centre of the circle . Take a point A on the circumference , and let AD be the diameter with length 2 unit . If we take 2 more points B and C randomly on the circumference to form a triangle , the chance that B and C lie on the same side of AD is 1/2 , while the chance that B and C lie on the opposite sides of AD is also 1/2 . We shall consider the 1st case firstly . Let x be the length of AB , and let y be the length of AC . ( For simplicity let x ≤ y ) Then both x and y varies from 0 to 2 units .
We know that the area of Δ ABC = Δ ABO + Δ OBC - Δ ACO . But how to find the areas of the last 3 Δ s ?
Hi Spicca ,
I really cannot solve this problem . Or we may consider a similar but simpler one of 2- dimensional , perhaps we can get some hint .
From a circle with radius 1 unit , 3 points are picked randomly on its circumference . Find the expected area of the triangle so formed .
Hi thehay95 ,
I think your thread was taken from my thread with the same
name posted 2 years ago .
This problem had already been solved with the help of
bob bundy , you can search the thread and find the solution .
Still thankful to bob bundy !
50 minutes .
Hi kubes ,
It seems your rule can also be applied to numbers with bases other than ten .
For a 2-digited no. ab with base x , which value = a * x + b .
The equation a * x + b = a * b + ( a + b ) ⇒ a * x = a * ( b + 1 )
⇒ x = b + 1
⇒ b = x - 1 .
For examples :
(1) Base 2 :
11 = 3 while 1*1 + 1+1 = 3 also .
(2) Base 8 :
77 = 7 * 8 + 7 = 「63」 while 7 * 7 + 7 + 7 = 「63」 also .
(3) Base 16 :
3 「15 」 = 「 3 * 16 + 15 」= 「63」 while 「 3 * 15 + 3 + 15 」= 「63」also .
(4) Base 100 :
「90 」「99」= 「9099 」 while 「90 * 99 + 90 + 99 」= 「9099 」 also .
Multiplication table of base 「16」( II ) .
( This post may replace # 8 . )
( 2 ) 2 * 1 = 2 , 2 * 2 = 4 , 2 * 3 = 6 , 2 * 4 = 8 , 2 * 5 = 「10」,
2 * 6 = 「12」, 2 * 7 = 「14」, 2 * 8 = 10 , 2 * 9 = 12 , 2 * 「10」= 14 ,
2 * 「11」= 16 , 2 * 「12」= 18 , 2 *「13」= 1「10」, 2 * 「14」 = 1「12」, 2 * 「15」= 1「14」
( 3 ) 3 * 1 = 3 3 * 2 = 6 , 3 * 3 = 9 , 3 * 4 = 「12」, 3 * 5 = 「15」,
3 * 6 = 12 , 3 * 7 = 15 , 3 * 8 = 18 , 3 * 9 = 1「11」, 3 * 「10」= 1「14」,
3 * 「11」= 21 , 3 *「12」= 24 , 3 *「13」= 27 , 3 * 「14」 = 2「10」, 3 * 「15」=2「13」
( 4 ) 4 * 1 = 4 , 4 * 2 = 8 , 4 * 3 = 「12」, 4 * 4 = 10 , 4 * 5 = 14 ,
4 * 6 = 18 , 4 * 7 = 1「12」, 4 * 8 = 20 , 4 * 9 = 24 , 4 * 「10」= 28 ,
4 *「11」= 2「12」, 4 * 「12」= 30 , 4 * 「13」= 34 , 4 * 「14」 = 38 , 4 * 「15」= 3「12」
( 5 ) 5 * 1 = 5 , 5 * 2 = 「10」, 5 * 3 = 「15」, 5 * 4 = 14 , 5 * 5 = 19 ,
5 * 6 = 1「14」, 5 * 7 = 23 , 5 * 8 = 28 , 5 * 9 = 2「13」, 5 * 「10」= 32 ,
5 * 「11」= 37 , 5 *「12」= 3「12」, 5 * 「13」= 41 , 5 * 「14」= 46 , 5 * 「15」= 4 「11」
( 6 ) 6 * 1 = 6 , 6 * 2 = 「12」, 6 * 3 = 12 , 6 * 4 = 18 , 6 * 5 = 1「14」,
6 * 6 = 24 , 6 * 7 = 2「10」, 6 * 8 = 30 , 6 * 9 = 36 , 6 *「10」= 3「12」,
6 *「11」= 42 , 6 *「12」= 48 , 6 *「13」= 4「14」, 6 * 「14」= 54 , 6 *「15」= 5「10」
( 7 ) 7 * 1 = 7 , 7 * 2 = 「14」, 7 * 3 = 15 , 7 * 4 = 1「12」, 7 * 5 = 23 ,
7 * 6 = 2「10」, 7 * 7 = 31 , 7 * 8 = 38 , 7 * 9 = 3「15」, 7 *「10」= 46 ,
7 *「11」= 4「13」, 7 *「12」= 54 , 7 *「13」= 5「11」, 7 *「14」= 62 , 7 *「15」= 69
( 8 ) 8 * 1 = 8 , 8 * 2 = 10 , 8 * 3 = 18 , 8 * 4 = 20 , 8 * 5 = 28 ,
8 * 6 = 30 , 8 * 7 = 38 , 8 * 8 = 40 , 8 * 9 = 48 , 8 *「10」= 50 ,
8 *「11」= 58 , 8 *「12」= 60 , 8*「13」= 68 , 8*「14」= 70 , 8 *「15」= 78
( 9 ) 9 * 1 = 9 , 9 * 2 = 12 , 9 * 3 = 1「11」, 9 * 4 = 24 , 9 * 5 = 2「13」,
9 * 6 = 36 , 9 * 7 = 3「15」, 9 * 8 = 48 , 9 * 9 = 51 , 9 *「10」= 5「10」,
9 *「11」= 63 , 9 *「12」= 6「12」, 9 *「13」= 75 9 *「14」= 7「14」, 9 *「15」= 87
(「10」)「10」* 1 = 「10」, 「10」* 2 = 14 , 「10」* 3 = 1「14」, 「10」* 4 = 28 , 「10」* 5 = 32 ,
「10」* 6 = 3「12」, 「10」* 7 = 46 , 「10」* 8 = 50 , 「10」* 9 = 5「10」, 「10」*「10」= 64 ,
「10」*「11」= 6「14」, 「10」*「12」= 78 , 「10」*「13」= 82 ,「10」*「14」= 8「12」, 「10」*「15」= 96
(「11」)「11」* 1 = 「11」, 「11」* 2 = 16 , 「11」* 3 = 21 , 「11」* 4 = 2「12」, 「11」* 5 = 37 ,
「11」* 6 = 42 , 「11」* 7 = 4「13」, 「11」* 8 = 58 , 「11」* 9 = 63 , 「11」*「10」= 6「14」,
「11」*「11」= 79 , 「11」*「12」= 84 , 「11」*「13」= 8「15」,「11」*「14」= 9「10」,「11」*「15」= 「10」5
(「12」) 「12」* 1 = 「12」, 「12」* 2 = 18 , 「12」* 3 = 24 , 「12」* 4 = 30 , 「12」* 5 = 3「12」,
「12」* 6 = 48 , 「12」* 7 = 54 , 「12」* 8 = 60 , 「12」* 9 = 6「12」, 「12」*「10」= 78 ,
「12」*「11」= 84 , 「12」*「12」= 90 , 「12」*「13」= 9「12」, 「12」*「14」= 「10」8 , 「12」*「15」= 「11」4
(「13」)「13」* 1 = 「13」, 「13」* 2 = 1「10」, 「13」* 3 = 27 , 「13」* 4 = 34 , 「13」* 5 = 41 ,
「13」* 6 = 4「14」, 「13」* 7 = 5「11」, 「13」* 8 = 68 , 「13」* 9 = 75 , 「13」*「10」= 82 , 「13」*「11」= 8「15」,「13」*「12」= 9「12」,「13」*「13」= 「10」9 ,「13」*「14」= 「11」6 ,「13」*「15」= 「12」3
(「14」)「14」* 1 = 「14」, 「14」* 2 = 1「12」, 「14」* 3 = 2「10」, 「14」* 4 = 38 , 「14」* 5 = 46 ,
「14」* 6 = 54 , 「14」* 7 = 62 , 「14」* 8 = 70 , 「14」* 9 = 7「14」, 「14」*「10」= 8「12」, 「14」*「11」= 9「10」,「14」*「12」= 「10」8 ,「14」*「13」= 「11」6 ,「14」*「14」= 「12」4, 「14」*「15」= 「13」2
(「15」)「15」* 1 = 「15」, 「15」* 2 = 1「14」, 「15」* 3 = 2「13」, 「15」* 4 = 3「12」, 「15」* 5 = 4「11」,
「15」* 6 = 5「10」, 「15」* 7 = 69 , 「15」* 8 = 78 , 「15」* 9 = 87 , 「15」*「10」= 96 ,
「15」*「11」= 「10」5 ,「15」*「12」= 「11」4,「15」*「13」= 「12」3 ,「15」*「14」= 「13」2 ,「15」*「15」= 「14」1
Multiplication table of base 「16」
* 1 2 3 4 5 6 7 8 9 「10」 「11」 「12」 「13」 「14」 「15」 *
1 1 2 3 4 5 6 7 8 9 「10」 「11」 「12」 「13」 「14」 「15」 1
2 4 6 8 「10」「12」「14」 10 12 14 16 18 1「10」 1「12」 1「14」 2
3 9 「12」「15」 12 15 18 1「11」 1「14」 21 24 27 2「10」 2「13」 3
4 10 14 18 1「12」20 24 28 2「12」 30 34 38 3「12」 4
5 19 1「14」 23 28 2「13」 32 37 3「12」 41 46 4「11」 5
6 24 2「10」30 36 3「12」 42 48 4「14」 54 5「10」 6
7 31 38 3「15」 46 4「13」 54 5「11」 62 69 7
8 40 48 50 58 60 68 70 78 8
9 51 5「10」 63 6「12」 75 7「14」 87 9
「10」 64 6「14」 78 82 8「12」 96 「10」
「11」 79 84 8「15」 9「10」 「10」5 「11」
「12」 90 9「12」 「10」8 「11」4 「12」
「13」 「10」9 「11」6 「12」3 「13」
「14」 「12」4 「13」2 「14」
「15」 「14」1 「15」
More fractions with corresponding " decimal no. " in base 「16」
2/3 = 0.「10」「10」「10」.... ( by division )
( Or 2/3 = 1 - 1/3 = 0. 「15」「15」「15」....
- 0. 555......
= 0.「10」「10」「10」.... )
3/4 = 3 * 1/4 = 3 * 0.4 = 0. 「12」
2/5 = 2 * 1/5 = 2 * 0.333 ....
= 0.666 ....
3/5 = 1 - 2/5 = 0. 「15」「15」「15」.... - 0.666 ....
= 0. 999 ....
4/5 = 2 * 2/5 = 2* 0.666 ....
= 0. 「12」「12」「12」....
5/6 = 1 - 1/6 = 0. 「15」「15」「15」....
- 0.2「10」「10」「10」....
= 0. 「13」555 ....
2/7 = 2 * 1/7 = 2 * 0.249249....
= 0.4 8 「18」0.4 8 「18」....
= 0.4 9 2 492 ....
3/7 = 3 * 1/7 = 3 * 0.249249....
= 0. 6 「12」「27」.....
= 0. 6 「13」「11」.....
4/7 = 1 - 3/7 = 0. 「15」「15」「15」....
- 0. 6 「13」「11」.....
= 0. 9 2 4 ......
5/7 = 1 - 2/7 = 0. 「15」「15」「15」....
- 0 .492492 .....
= 0. 「11」6 「13」....
6/7 = 1 - 1/7 = 0. 「15」「15」「15」....
- 0.249249....
= 0. 「13」「11」6 ....
3/8 = 3 * 1/8 = 3 * 0.2 = 0.6
5/8 = 1 - 3/8 = 1 - 0.6 = 0. 「10」
7/8 = 1 - 1/8 = 0. 「16」- 0.2 = 0. 「14」
2/9 = 2 * 1/9 = 2 * 0.1「12」7 1「12」7 ....
= 0.2 「24」「14」 2 「24」「14」....
= 0.3 8 「14」3 8 「14」....
4/9 = 2 * 2/9 = 2 * 0.3 8 「14」3 8 「14」....
= 0. 6 「16」「28」 0. 6 「16」「28」....
= 0. 7 0 「28」7 0 「28」....
= 0. 7 1 「12」7 1 「12」....
5/9 = 1 - 4/9 = 0. 8 「14」3 8 「14」3 ....
6/9 = 2/3 = 0.「10」「10」「10」....
7/9 = 1 - 2/9 = 1 - 0.3 8 「14」3 8 「14」....
= 0 .「12」7 1 「12」7 1 ....
8/9 = 1 - 1/9 = 1 - 0.1「12」7 1「12」7 ....
= 0 .「14」3 8 「14」3 8 ....
2/ 「10」= 1/5 = 0.333 ....
3/「10」= 5/「10」- 2/ 「10」= 1/2 - 0.333...
= 0.8 - 0.333...
= 0. 7「15」「15」「15」....- 0.333 ....
= 0. 4「12」「12」「12」....
7/「10」= 1 - 3/「10」= 0 .「11」333 ....
9/「10」= 1 - 0.1999 ....
= 0 .「14」666 ....
Fractions converted to " decimal no. " in base 「16」
1/2 = 0.8
1/3 = 0.555 ....
1/4 = 0.4
1/5 = 0.333 ....
1/6 = 0.2「10」「10」「10」....
1/7 = 0.249249....
1/8 = 0.2
1/9 = 0.1「12」7 1「12」7 ....
1/「10 」 = 0.1999 ....
1/ 「11」= 0.1745「13」1745「13」....
1/ 「12」= 0.1555 .....
1/ 「13」= 0.13「11」13「11」....
1/ 「14」= 0.1249249....
1/ 「15」= 0.111 ...
1/ 10 = 0.1
Arithmetic with numbers of base 「16」
Find the values of :
(1) 「10」「11」 + 5「13」
(2) 「10」「11」 - 5「13」
(3) 「10」「11」 * 5「13」
(4) 「10」「11」 / 5「13」
Solutions :
(1) 「10」「11」 + 5「13」
= 「15」「24」
= 「16」8
= 108
(2) 「10」「11」 - 5「13」
= 5 「-2」
= 4 「14」
(3) 「10」「11」 * 5「13」
= 「50」「55」0 + 「130」「143」
= 「50」「185」「143」
= 「50」「193」「15」
= 「62」1 「15」
= 3「14」1 「15」
(4) 「10」「11」 / 5「13」
= 1 + 4 「14」/ 5「13」
Exercise involving digits taken from base ∞
(1) Find the value of 「98 」「76 」「54 」( base「100 」)
in base 「10 」.
Solution : 「98 」「76 」「54 」( base「100 」)
= 98 * 100 ^ 2 + 76 * 100 + 54
= 98 * 10000 + 7600 + 54
= 987654 ( base 「10」)
(2) Find the value of 「15」「15」「15」「14」( base 「16」)
in base 「10 」.
Solution : 「15」「15」「15」「14」( base 「16」)
= 10000 - 2 ( base 「16」)
= 16 ^ 4 - 2 ( base 「10」)
= 65536 - 2
= 65534 ( base 「10」)
(3) Find x if 47 ( base 「x」) + 39 ( base 「x」) = 80 ( base 「x」)
Solution : Expressed in base 10 , the equation becomes
4x + 7 + 3x + 9 = 8x
⇒ 7x + 16 = 8x
⇒ x = 16
(4) Find x if 123 ( base 「x」) = 「51」
Solution : Expressed in base 10 ,
x ^ 2 + 2x + 3 = 51
⇒ x ^ 2 + 2x - 48 = 0
⇒ ( x - 6 ) ( x + 8 ) = 0
⇒ x = 6
or x = - 8 ( rejected )
Thanks ganesh !
( I ) Number and digits
Quantities are represented in numbers , which are formed
by digits . By repetition and different orders of finite digits ,
we can form infinite numbers .
( II ) Digits of numbers with various bases
(1) base 10 ( ten )
The set of digits used in base 10 contains 0 , 1 , 2 ,
3 , 4 , 5 , 6 , 7 , 8 and 9 .
(2) base 2 ( two )
There are only 2 digits in base 2 : 0 and 1 .
(3) base 16 ( sixteen )
For bases > 10 , say base 16 ,the 1st part of the set of
digits will be from 0 to 9 , then for values > 9 , traditionally
we use capital alphabets A , B , C , D , E and F , each
representing a single - digited no. corresponding to
values 10 , 11 , 12 , 13 , 14 and 15 respectively . ( being
2- digited no. of base ten . )
( III ) base ∞ ( infinity )
We shall create a digit system of base ∞ in the
following way : Every digit of base ∞ will consists of
2 parts .
( i ) the value corresponding to a number in base ten ,
which may be single - digited or multiple - digited .
( ii ) the above number enclosed in a pair of brackets
"「 " and " 」" . ( 「 and 」 are bracket signs
usually used in writing Chinese . )
Thus the set of digits of base ∞ will be : 「0」, 「 1」,
「2 」, 「 3」, 「 4」, 「5 」, 「 6」, 「7 」, 「8 」, 「9 」,
「10 」, 「11 」...........「99」,「100」, and so on up to 「∞」.
This means the shapes of these digits are mainly
borrowed from the numbers used in base 10 .
( IV ) Combined digit system of base 10 and base ∞
By adopting digits 0 , 1 , ...... 8 , and 9 of base 10 , together
with digits 「10 」, 「11 」, 「12」... up to 「∞」of base ∞,
a more convenient digit system can be formed .
For bases ≦ 10 , the set of digits of base 10 will be enough ,
if single digits with values > 10 are required , then digits
such as 「10 」, 「11 」etc will also be used .
For example , the set of digits of base 16 will be from
0 , 1 , ... to 9 , together with 「10 」, 「11 」, ... up to 「15」.
Comparing with using alphabets A , B , ... to F in the traditional
way , the advantage of using digits taken from base ∞ is that
it will be more intuitive , the exact value of the digits can
be obtained at a glance .
( V ) Borrowing of digits from higher bases
Although the set of digits for a base 「n 」will be from
0 to 「n - 1」, but it is sometimes convenient to borrow digits
from a higher base .
For example , in base 10 , to denote the no. 34 we may also
write 2 「14」, 1 「24」 or 「34」 .
He has passed away some days ago.
Really regretted to learn about this .
Hi chen.aavaz ,
According to the constraints " for any two sandwiches, there is at least one ingredient at both of them but not at the 3rd one. " the ways in your 1st , 2nd , 3rd , 5th rows etc. will not be permitted .
E.g. in the 1st row , ( p,q) and (r,s) have no common ingredient .
Hi chen. aavaz ,
Let the 4 ingredients be denoted by p , q , r and s .
For (ii) , there are 4 C 3 = 4 choice for the sandwich containing 3 ing. , say p , q and r . Then one of the remaining sandwich will contain 1 ing. from either p or q or r , say p ( there are 3 C 1 = 3 ways .) together with s which is left behind . The last sandwich must contain s , together
with either q or r but not p . ( there are 2 C 1 = 2 ways .)
Thus the total possible ways of (ii) = 4 * 3 * 2 = 24 .
I have assumed that the order of the sandwiches will not be considered , otherwise
the total possible ways will be 34 * 6 = 204 .
Hi chen. aavaz ,
For (i) , there are 4 C 2 = 6 ways .
For (ii) , there are 4 C 3 * 3 C 1 * 2 C 1 = 4 * 3 * 2 = 24 ways .
For (iii) , there are 4 C 3 = 4 ways .
Therefore the total no. of possible ways = 6 + 24 + 4 = 34 .
Hi chen. aavaz ,
The no. of ingredients contained in each sandwich should be
either 2 or 3 , but excluding the case of any 3 ingredients in all
the 3 sandwiches .
Thus the possible cases may be divided into 3 types :
(i) ( 3 , 3 , 2 ) i.e. 2 sandwiches both with 3 ingredients while
the remaining sandwich with only 2 ingredients .
(ii) ( 3 , 2 , 2 ) , 1 sandwich with 3 ingredients while the 2
remaining sandwiches both with 2 ingredients .
(iii) ( 2 , 2 , 2 ) , all the 3 sandwiches each with 2 ingredients .
Hi taylorn5683 ,
17) 14
18) 5
19) 20
20) 1
Thanks bobbym and thickhead ,
For Related problem (III) , should there be a formula to calculate
the area of overlapping area of A and B if the angle between
the 3 centers being x degrees where 0 ≤ x < 60 ?
Related problem (III)
2 circles A and B both with radius 1 unit rotate freely and randomly
outside a circle E also with radius 1 unit on its circumference .
Find the expected value of the overlapping area of A and B .
Although problem (III) states that both A and B are movable .
But if one of them , say A is fixed touching E at a certain
point , with only B movable , the answer should be the same .
Thus the problem will become simpler .
A possible way to solve the problem is to consider the angle
between the centers of the 3 circles .
Hi bobbym ,
No , I don't know the answer .
Hi bobbym ,
I mean the circumferences of A and B rotate on the circumference of E . ( Just touch
at 1 point . )
Thanks thickhead ,
What will be the probability if A cannot get through the centre of E but B can ?
For the original problem , what will be the probability if both A and B cannot get
through the centre of E ?