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TYSM!! @Alan! It's correct. Also, thanks for leaving some of the work for me to solve. Really appreciate it.

AD is an angle bisector

So

AB / BD  = AC / CD         

2 / BD  = 2 / CD

CD / BD  =   2  / 2

CD  / BD  =   1 / 1

Therefore  the base of triangle ACD = the base of triangle ABD

And they are both under the same  height

So....the ratio of their areas =   1 : 1

Since AD is  an angle bisector

BD / AB  = DC / AC

40 / 60  =  DC / AC

2 / 3  =  DC / AC

(2/3)AC = DC

So  using the Pythagorean Theorem

AB^2  + BC^2  =  AC^2

60^2  +  (BD + DC)^2 =  AC^2

60^2  +  ( 40 +  (2/3 * AC) )^2  = AC^2

3600 +  1600 + (160/3)AC + (4/9)AC^2  = AC^2

(5/9)AC^2   - (160/3)AC  - 5200  =  0

Multiply through by  9

5AC^2  -  480AC  - 46800  = 0

AC  =    [ 480  + sqrt [ 480^2  - 4 * 5 * -46800   ]  /  10    =  

[ 480  +  sqrt [1166400]  ]   / 10  = 

[ 480 + 1080 ] / 10   =

1560  / 10   =

156

Simplify as

x^2 + 12x + y^2 - 16y + 8

Take the derivative with respect to x  and set  to 0

2x + 12 = 0

x = -6

Take the derivative with respect to y and  set  to  0

2y - 16  = 0

y = 8

Minimum  is

(-6)^2  + 12 (-6) + (8)^2 - 16 (8)  + 8  =   -92

semi-perimeter, s ,  =    (12 + 14 + 20)   / 2 =    23

radius of incircle  =   sqrt  [   ( 23 - 12) (23 - 14) (23 - 20)  / 23 ]  = sqrt   [ 11 * 9 * 3 /  23 ] = sqrt (297 / 23) =

3 sqrt (33/23)  =   (3/23) sqrt (33 * 23)  =   (3/23)sqrt (759)  

(1,7)   (2 , -3)

Slope  =  ( -3-7) / (2 -1)  =  -10 / 1   =  -10

Equation of line  using  slope and (1,7)

y = -10 ( x -1 )  + 7

y  = -10x + 10 + 7

10x + y  =   17

Side of square =  6

Perimeter =  24

Area of rectangle =  x *  y  =  24

Perimeter of rectangle     2 ( x + y)  =  24

So     x + y   = 12

So        y  =  12 - x

So

x ( 12 -x)  =  24

-x^2 + 12x  =24

x^2 - 12x + 24  =  0     complete the square on x

x^2 -12x + 36  =  -24 + 36

(x -6)^2  =  12         take the positive root

x - 6  =  sqrt (12)

x = 6 + sqrt (12)  =  6 + 2sqrt (3)   = longer side of rectangle

5

Area  of circumcircle  =   pi  (hypotenuse  / 2)^2  =   pi ( 26/2)^2  = 169 pi

semi-perimeter  =  s =   (26 + 24 + 10) / 2   = 30

radius of  incircle  =  sqrt  [  (30 - 26) (30 -24) (30 -10)  / 30 ]  = sqrt [ (4 * 6 * 20 / 30]  = sqrt (16)   = 4

Area of incircle =   pi * 4^2    = 16 pi

x - y  =   (169 - 16)  pi   =  153 pi

4

Angle YBA  + Angle XAB + Angle AIB  =  180

Angle YBA  + Angle XAB  +  134   =  180

Angle YBA + Angle XAB  =  180 -  134  =  46

2 Angle YBA  +  2 Angle XAB  = 2 * 46

 Angle CBA  +  Angle CAB  = 92

Angle C  =  180  - 92  =  88° 

3

Let  BZ  = x

Let   AZ  =12  - x

Because  CZ is an  angle bisector

BZ /  BC  =  AZ / AC

x /  9.6  =  (12  - x)   /  18

18x  =  9.6 ( 12  - x)

18x  = 115.2 - 9.6x

18x  + 9.6x  = 115.2

27.6x   =  115.2

x =  115.2  / 27.6   =  96 / 23 = BZ

2

Because BY is an  angle bisector

BC  / CY  =  AB / AY

BC / 8  =  12   / 10

BC  =  8 * 12   /  10      =  9.6

*****

Surface area  =   b^2  + 4 * ( b^2 /3)

432 = (7/3)b^2

b^2 = 432 ( 3/7)  = 1296 / 7

b = sqrt (1296/ 7)  = 36 /sqrt 7

To find the  slant  height, s, of the pyramid

area of face = 1/2 b * s

b^2 / 3 =  (1/2) 36 / sqrt 7 *  s

1296 / (7*3)  = 18 / sqrt 7 * s

s = 1296 / 21 * sqrt 7 / 18 =  24 / sqrt 7

To find the height, h, of the pyramid by the Pythagorean Theorem

h  =  sqrt  ( s^2  - (b/2)^2 )

h =  sqrt [ (24/sqrt 7)^2  - (18/sqrt 7)^2 ]

h =   6

Volume of pyramid = 

(1/3)  base area * height  =

(1/3) (b^2) * h

(1/3) (1296  / 7) * 6  =  2592 / 7 units^3   ≈ 370.29  units^3  

****

  Avg of nth row of Pascal's Triangle =  2^n  / ( n + 1)

So

2^n / (n + 1)   = 2

2^n  = 2(n + 1)

n = 3

                        1                            Avg  = 1

                    1      1                        Avg =  1

                  1     2     1                    Avg  = 4/3

                1   3     3     1                 Avg  =  2

n =  3

-7/4

5005

5115

5225

5335

5445

5555

5665

5775

5885

5995

Simplify as follows: 

\((3x - 27)^3 + (27x - 3)^3 = (3x + 27x - 30)^3\)

\((3x - 27)^3 + (27x - 3)^3 = (30x - 30)^3\)

\((3(x-9))^3 + (3(9x - 1))^3 = (3(10x - 10))^3\)

\((x-9)^3 + (9x-1)^3 = (10x-10)^3\)

\(-270x^3+2730x^2-2730x+270 = 0\)

\(9x^3 - 91x^2 +91x - 9 = 0\)

Notice that x = 1 is a solution to the equation. Now, divide the left-hand side by (x-1). The result will be quadratic, and you can use the quadratic formula to solve for the other 2 x values. 

Notice that there is only 1 path to get to B from A. 

Then there is 1 way to get to both C and D from B. 

Because you have to get to E from either C or D, there are 1 + 1 = 2 ways to get to E. 

Then there are 2 ways to get to F and G because you have to go through E. 

So then there are 2 + 2 = 4 ways to get to H.

And there is only 1 way to get to I from H, so there are a total of \(\color{brown}\boxed{4}\) paths from A to I.

As follows:

The question specifies that \(k \ne 1\) 

x = 1 is certainly a solution, but there are two other real number solutions to the original cubic equation!