CPhill

(a + 2) / (a + 1)  =  b / 8

8(a + 2) /(a + 1)  = b

a      b

0    16

1    12

3    10

7      9

-2    0

-3    4

-5    6

-9    7

Let x = AC   and  y = BC      and AB is the hypotenuse

CM = y/2

CN = x/2

Triangles  AMC  and BNC are both right 

AC^2 + ( BC/2) = AM

(AC/2)^2 + BC^2 = BN

So.....by the Pythagorean Theorem we have these two equations

x^2 + (y/2)^2   =1         

(x/2)^2 + y^2  = 1              simplify

x^2 + y^2/4  =1  →   4x^2 + y^2  = 4        (1)

x^2/4 + y^2  =1  →    -x^2 - 4y^2 = -4

Add the last two equations and we  get that

3x^2 - 3y^2  = 0

3x^2 = 3y^2

x^2 = y^2

Sub this  back into (1) for y^2

4x^2 + x^2  = 4

5x^2  = 4

x^2  = 4/5

and

y^2 = 4/5

AC^2 + BC^2  =  AB^2

4/5 + 4/5  = AB^2

AB =   sqrt [ 4/5 + 4/5 ]

AB = sqrt [ 8/5]  =  2sqrt [2/5] = (2/5)sqrt 10

Poltnomial  division

                       x^2

 x^2 - 2x -5   [  x^4 -2x^3 + kx^2  ]

                        x^4 - 2x^3 -5x^2

                       _______________

k - -5  = 0

k + 5  = 0

k = -5

Simplify as

x^2 + 18x + 3     complete the square on x

x^2 + 18x + 81 + 3 - 81

1(x + 9)^2 +(  - 78 )

(r-6)^2 + (r-6)^2  = r^2

2r^2 - 24r + 72  =  r^2

r^2 - 24r + 72 = 0     complete the square on r 

r^2 -24r + 144 = -72 + 144

(r -12)^2 = 72

r - 12  = sqrt [72]

r = 12 + 6sqrt 2

https://web2.0calc.com/questions/geometry_54234

                    B 

A 45             D                      45 C

                   12

Angle B = 90

BD = BD  and is  an altitude

And angle ABD = angle CBD

By AAS, triangle ABD is congruent to triangle CBD

AD = 6 = CD

We have that

AD / DB = BD / CD

6 / BD = BD / 6

BD^2 = 36

BD = 6

Area of ABC  =  (1/2) AC * BD =  (1/2) 12 * 6  =   36

x^2 + (b-3)x + 22  > -4

x^2 + (b-3)x + 26 >  0

Set the discriminant to < 0 

(b - 3)^2  - 4(1)(26) < 0

(b -3)^2 < 104

b -3  <   [ sqrt 104]

b- 3 < ≈ 10.2

b <  ≈13.2  

Greatest  integer value of b = 13

Area = (1/2) (AD) (BC)           Area  = (1/2)(BE)(AC)      Area = (1/2)(CF)(AB)

2A = (12) (BC)                        2A = (20)(AC)                  2A = (CF)(AB)

A/6 = BC                                  A/10 = AC                          2A / CF = AB

By the Triangle Inequality :

AB + BC < AC

2A/CF + A/6 > A/10

2/CF + 1/6 > 1/10

2/CD > 1/10 -1/6

2/CD > -4/60

2/CD > -1/15

-15/2 < CD

BC + AC >  AB

A/6 + A/10 > 2A / CF

CF >  2A / (A/6 + A/10)

CF > 2 / (1/6 + 1/10)

CF > 2/ [ 16/60]

CF > 120/16 

CF > 15/2

AC + AB >  BC

A/10 + 2A/CF > A/6

1/10 + 2/CF > 1/6

2/CF >  1/6 - 1/10

2/CF > 4/60

2/CF >  1/15

CF < 15*2

CF < 30

(a + 2) / ( a + 1)  = b / 8