geno3141

The sum of the yellow regions must be the same as the sum of the blue regions for this reason:

The diameter of each of the smaller circles is one-half the diameter of the larger circle.

To make it easy, call the radius of each of the smaller circle "1" so that the radius of the larger circle is "2".

The area of the larger circle is:  Area  =  pi · 2²  =  4·pi.

The area of each of the smaller circles is:  Area  =  pi · 1²  =  1·pi..

So, the total area of all the smaller circles (added together) is:  Area  =  4·pi.

Therefore, the total area of all the smaller circles is the same as the area of the larger circle.

So, the overlap of the smaller circles (the yellow areas) must equal the "underlap"  -- the area of the larger circle that isn't covered by the smaller circles -- the blue areas.

If  a  =  (1, -3)  then  5a  =  (5, -15).

If  b  =  (-1, 2)  then  4b  =  (-4, 8).

4b - 5a  =  (-4, 8) - (5, -15)  =  (-9, 23).

Each side of each tube is a rectangle 4 ft x 11 ft  =  44 ft².

Since both the inside and outside are to be painted, each side  =  88 ft².

Since each tube is hexagonal:  88 ft² x 6  =  528 ft².

Since there are 12 tubes:  528 ft² x 12  =  6336 ft².

Since each can of point covers 40 ft²:  6366 ft² / 40 ft²  =  158.4 cans   --->   159 cans.

If there is a 40% chance that it will rain on Monday, then there is a 60% chance that it won't rain on Monday.

If there is a 30% chance that it will rain on Tuesday, then there is a 70% chance that it won't rain on Tuesday.

If there is a 50% chance that it will rain on Wednesday, then there is a 50% chance that it won't rain on Wednesday.

To rain on exactly two days:

-- rain on Monday (40%) and on Tuesday (30%) but not on Wednesday (50%)  =  0.40 x 0.30 x 0.50  =  0.06

-- rain on Monday (40%) and on Wednesday (50%) but not on Tuesday (70%)  =  0.40 x 0.50 x 0.70  =  0.14

-- rain on Tuesday (30%) and on Wednesday (50%) but not on Monday (60%)  =  0.30 x 0.50 x 0.60  =  0.09

Adding these individual situations:  0.06 + 0.14 + 0.09  =  0.29

If     x  =  t + 3     and     y  =  2t² - 4

then solve the x-equation for t:  x  =  t + 3     --->     t  =  x - 3

and substitute this into the other equation:  y  =  2t² - 4     --->     y  =  2(x - 3)² - 4

                                                                                                      y  =  2(x² - 6x + 9) - 4

                                                                                                      y  =  2x² - 12x + 18 - 4

                                                                                                      y  =  2x² - 12x + 14

The centers of the four smaller circles are the vertices of a square.

The distance from one vertex to its adjacent vertex is 2.

This means that the distance across the square is 2·sqrt(2).

This makes the diameter of the circle 2 + 2·sqrt(2) and its radius 1 + sqrt(2).

You're right -- I messed up -- I thought (3,2) was (2,3) -- my fault ...

A  =  (-2, 0)

B  =  (2, 0)

C  =  (3, 2)

Since there is a right angle at B, AC must be a diameter of the circle.

Therefore, the center of the circle is the midpoint of AC.

The midpoint of AC  =  ( (-2 + 3)/2, (0 + 2)/2 )  =  ( 1/2, 1 )

There is a rather obscure formula for the sum of two inverse tangents:

tan^-1(a) + tan^-1(b)  =  tan^-1(  (a + b / ( 1 - a·b ) )

tan^-1( 1/2 ) + tan^-1( 1/3 )  =  tan^-1(  ( 1/2 + 1/3 ) / ( 1 - ( 1/2 )·( 1/3 ) )

                                        =  tan^-1( (5/6) / (5/6) )

                                        =  tan^-1( 1 )

                                        =  pi/4

Adding this to  tan^-1( 1 )  gives a total of  pi/2  [or 90^o]

alpha     =  tan^-1(1/1)

beta       =  tan^-1(1/2)

gamma  =  tan^-1(1/3)  

The sum will be  pi/2.