CPhill

Starting off 

Dave  had   D  game cards

Perry had D + 72

Perry gave (1/3) of his cards to Dave

Dave now has  (1/3)(D + 72) + D  = (4/3)D + 24

Perry now has (2/3) (D + 72)  = (2/3)D + 48

Dave then  gave 3/8 of his cards to Perry

Dave now has (5/8) [ (4/3)D +24 ] =  (5/6)D + 15

Perry now has  (2/3)D + 48 +  (3/8)[(4/3)D + 24] =  (2/3)D + (1/2)D + 48 +9  = (7/6)D + 57

And the difference at the end was that Perry had 114 more cards than Dave

So

  Perry's  - Dave's   =  114           in equation  form we have

[ (7/6)D + 57]  -  [  (5/6)D + 15 ]  =  114    simplify

(2/6)D + 57 -15  = 114

(1/3)D  + 42  = 114

(1/3)D = 114 - 42

(1/3)D =  72     multiply through  by 3

D = 216 =  what Dave started with

D + 72 = 216 + 72 =  288  = what Perry started with

The hypotenuse of the large right triangle at the left =  sqrt (1^2 + (1+2)^2) = sqrt (10)

Looking  at the small right triangle at the bottom left, notice that it is congruent to  the small right triangle at the upper left

Call the longer leg of one of these triangles, x

The sin of the greater acute angle in these triangles = 3 / sqrt (10)

By the Law of Sines, we have

sin (90) /1  =  3/sqrt (10) / x

1 = [3 sqrt (10)] / x

x = 3/sqrt (10)

Call the smaller leg in either one of these triangles, y

The sin of the smaller acute angle in these triangles = 1/sqrt (10)

Again, by the Law of Sines, we have 

sin (90) / 1 =   1/sqrt (10)  / y

1 =[ 1/sqrt (10)] / y

y =  1/sqrt (10)

The length of the  side of the square =

[The hypotenuse of the large right triangle on the left]   - [  x ]  - [  y  ]  =

sqrt (10 )  - 3/sqrt (10)  - 1/sqrt (10)

sqrt (10)  - [ 3 + 1] /sqrt (10)

(10 - 4) /sqrt (10)  =    6/sqrt (10) = side of the shaded square

Shaded area =  [ 6 / sqrt (10) ] ^2 =   36 / 10  =   3.6

Area of large square  =   9

Ratio of areas =  3.6 / 9 =   36/90  = 4 / 10 =   2/ 5

Thanks for the reference , jonathan....I knew there was a method to solve such a thing, but as I remember, it was a little involved

You're correct.....it even LOOKS depressing  (LOL!!!)

Rearrange as

z^3 + 7z + 8 = 0

By inspection , I can see that   z = -1   is one root

Using synthetic division  to  find the remaining  polynomial we have

-1  [  1    0     7     8  ]

              -1    1    -8

   _______________

        1   -1     8    0

The  remaining polynomial is  z^2  - z  +  8

The   other two (non-real) roots are

[ 1 + sqrt ( 1 - 4(1) (8) ] /  2  =   [ 1 - i sqrt ( 31)] / 2   

And    [ 1 + i sqrt (31) ] / 2

If the x coordinate of the  vertex is 3.....then we have that   -b / (2a)  =  3

Which neans that  -b = 6a   ..so....   b = -6a

We know the y coordinate of the  vertex is  2  so  this means that

2 = a(3)^2  - 6a (3) + c    simplifying

2 = 9a - 18a + c

2 = -9a + c   

c = 2 + 9a       (1)

And the parabola contains the point (2,1) is on the parabola...so we have  

1 = a(2)^2 - 6a(2) + c

1 = 4a - 12a + c  

1 =  -8a + c   

c =  1 + 8a       (2)

Equate (1)  and (2)

2 + 9a  = 1 + 8a

2-1 = 8a - 9a

1 = -a

a = -1

And c = 1 + 8(-1)  = -7

And b = -6 (-1)  =  6

(a, b, c) = (-1 , 6, -7 )

Each term in the first series is   generated by 2^n  

Each term in the second series is generated by 4 + 2n

All terms in each series will be even

The smalest positive difference  between any two terms  (one from A and one from B ) =  2

Ex:  "8" from ""A"  and "6" from "B"

Let   B  be the number of blue matbles that we need to  add

So

Total Blue  / Total Number of marbles =   5/8

[5 + B ] / [ 20 + B ]   = 5/8     

5 + B =  (5/8) ( 20 + B]    nultiply both sides by 8/5

8 + (8/5) B  = 20 + B

(8/5)B - B = 20 - 8

(3/5)B = 12

B = 12 * (5/3)  =  60 / 3 =  20

We need to add 20 blue marbles

Proof

[5 + 20 ] / [ 20 + 20]  =  25 / 40  =  5 / 8

Thanks, plaintainmountain....I didn't think of that !!!

Good job, history !!!