CPhill

See the following

Let circle A have a radius of 9   and let the small circle have a radius , r

Triangle ACD is right with

AC =the hypotenuse = 9 + r

AD is a leg = 9

DC is another leg = 9 -r

So we have this relationship

9^2 + ( 9 - r)^2  = (9 + r)^2

81 + 81 -18r + r^2  =  81 + 18r + r^2     simplify

81 =  36r

r  =  81/36  =  9/4  = the radius of the small circle

Let the circle have a radius of 4

Let A = (0,4)

Let B =  ( 2sqrt 3 , -2 )

AB = sqrt [ (2sqrt 3)^2 + (4--2)^2] = sqrt 48 = 4sqrt (3)  side length of triangle

MC = = (1/2) side length = 2sqrt 3

MB = side length * sqrt 3 =  2 sqrt 3 * sqrt 3 =  6

MB /MC  =  6 / [ 2sqrt 3]  =  3 /sqrt 3 =  sqrt 3

Each term   will  be -1 < term value  < 0

So the floor for each will be -1

Sum = 1000 (-1)  = -1000

r =  .016 / .4  =  .04

Sum  =     .016 / [ 1 -.04] =  .016 / .96  =  1/ 60

a / ( 1-r)  = 3

a^2 / (1 - r^2) = 10  

 [ a / (1-r) ] * [ a/ (1+ r)  ]  =10

[ 3] * [ a / ( 1 + r) ] =10

a/ [ (1+ r)] =10/3

a = (10/3)(1 + r)

So

(10/3)(1 + r) / (1-r)  = 3

(10/3) (1+r) = 4 (1 - r)

10/3 + (10/3)r = 4 -4r

(10/3 + 4)r = 4 -10/3

r =  [ 4 - 10/3 ]/ [ 4 + 10/3] =  [ 2] / [ 22]  =  1 / 11

(n+ 2)/(n ) * (n + 3) / (n + 1) * (n + 4) / ( n + 2)  = 4          simplify as

 (n + 3) *(n + 4)  =  4 (n)(n + 1)

n^2 +7n +12  =  4n^2 + 4n

3n^2 - 3n - 12 = 0

n^2 - n - 4  = 0

n^2 - n +1/4  =   4

(n - 1/2)^2  =  +/- 17/4

n =  [1 + sqrt (17)] /2                     or     n = [ 1 -sqrt (17) ] /2

No integer values for numerator/denominator  are possible

a^3 + 3ab^2  = 675

(675 - a^3) / 3 = ab^2

Subbing this into the second equation we have 

3a^3 - (675 - a^3)/3 = 615      multiply through by 3

9a^3 -675 -+ a^3 = 1845

10a^3  = 2520

a^3 = 2520/10

a = (252)^(1/3) ≈ 6.316

Using the first equation

252 + 3(6.316)*b^2  = 675

18.948b^2  = 423

b = sqrt [ 423 / 18.948 ] ≈ 4.725  or -4.725

a - b =  6.316 - 4.725  ≈ 1.591

or 

a - b = 6.316 - (-4.725) ≈ 11.041

arc length = (1/4) 2pi * 3  =

(3/2) pi ≈ 4.71

(1/6) pi * 21^2 

441 pi / 6 ≈ 230.9 cm^2

3 , 3 , 2 is the only possible triangle

Triangle inequality

Unknown side < 58 + 61

Unknown side < 119

The longest possible length for the remaining side = 118

Area of hexagon =   6 * 1/2 * (side)^2  *sqrt (3) / 2

So

9 = 6 * (1/2) *(side)^2 * sqrt (3) /2

9 = (3/2) sqrt (3) * (side)^2

9 / [ (3/2) sqrt (3)] = side^2

6/sqrt 3 = side^2

2sqrt 3 = side ^2

sqrt [ 2sqrt 3] = side ≈ 1.8612

Call the amount he had at first, A

After he spent 20% on books he had .8A left

After spending (5/8)  of the remainder he had  .8A - (5/8).8A =  .3A  left  

After getting $90 he had  .3A + 90

Amount of money at the end / Amt at first  = 3/2

So

(.3A + 90) / A  = 3/2            cross-multiply

2 (.3A +90) = 3A

.6A + 180 = 3A

180 = 3A -.6A

180 = 2.4A

A =180 / 2.4  =  $75  = what he had at first

BNCD is a triangle...h = the height of the trapezoid

[ BNCD ] = (1/2) h ( CD)

12 = (1/2) h (CD)

24 = h * CD

CD = 24/h

ABMN is a trapezoid

Height of ABMN = (1/2) height of the trapezoid = (1/2)h

[ ABMN] = (1/2) [ (1/2) ( h) (  AB + MN) ] 

8  =  (1/2) [ (1/2)(h) ( AB + (AB + CD) / 2)]

8 = (1/4)h * [ (3/2)AB + (1/2)CD ]

32 = h * ( (3/2) AB + (1/2)CD)

32 = (1/2)h * [ 3AB + CD ]

64 = h * [3AB + CD]

64 = h * [ 3AB + 24/h]

64 = h * 3AB  + 24

40 = h * 3AB

40/3 = h * 3AB

40 / [ 3h ] = AB

CD / AB  =  [ 24/h ] [ 40 / (3h) ] =  72 / 40  =  9 / 5