CPhill

I tried a similar procedure  as you, 3Mathketeers  and I got the  same answer.....but notice that a square of  5 will not fit  inside the  semi-circle      [a 4 x 5  rectangle will !! ]

In short, I believe that this is an impossible  problem as set up by the poster.....

AC is the hypotenuse

The circumradius =  (1/2) (hypotenuse)   = (3sqrt (3) / 2)  = (3/2)sqrt 3

The circumference  of O =    2pi * circumradius  =  2pi (3/2)sqrt (3) =   3 sqrt (3)  pi

Let the sides of the triangle =   2

Then AM = MB  =  1

And MN = 1

Let PM = NQ = x

Angle AMP  =120°

By the intersecting  chord theorem

AM * MB =  PM * MQ

AM * MB =  PM ( MN + NQ)

1 *    1 =    x ( 1 + x)

1 =  x + x^2

x^2  + x - 1  =  0  

x^2  + x   = 1               complete the square  on x

x^2 + x + 1/4  =  1 + 1/4

(x + 1/2) =  5/4                 take the  positive root of  both sides

x + 1/2 = sqrt (5) / 2

x =    [ sqrt (5)  - 1 ]  /  2 =   PM =   phi

By the Law of Cosines

AP^2  =   PM^2 +AM^2  - 2 ( AM * PM) * cos (AMP)               {cos AMP = cos 120° = -1/2 }

AP^2  =  phi^2  + 1^2  - 2 (1 * phi) (-1/2)

AP^2 =  phi^2 + 1 + phi          { 1 + phi  = Phi }

AP^2 =  phi^2 + Phi               { phi^2 + Phi =  2 }

AP^2 =  2

AP = sqrt (2) 

AP  / AM  =  sqrt (2) / 1  = sqrt (2)

                  R

                  m

P       8       X     24       Q

                 36 - m

                   S

Intersecting Chords Theorem

Let RX =  m     and SX =  36 - m 

PX * QX =  RX * SX

8  *  24  =   m ( 36 - m)

192  =  36m - m^2

m^2 - 36m + 192  = 0

m^2 - 36m =  -192         complete the square  on m

m^2 - 36m + 324  = -192 + 324

(m - 18)^2  =  132          take the  positive root

m - 18 =  sqrt [ 132]

m =  sqrt [132] + 18 =     2sqrt33 + 18  =  2 [ sqrt 33 + 9 ]  =  smallest positive value of RX   

             P

                 6       

12    B         Y

A 

Secant-Tangent Theorem

Let  AB = x   and PB = 12 - x

We have that

PY^2  = PB ( PA)

6^2 = (12-x) (12)

36  = 144 - 12x

12x =  144 - 36

12x  = 108

x = 108 / 12    =  9   = AB

Let C be the center of circle SAR  and let D be the center of circle  RBT

Draw AC , CD  and DB

We have the irregular pentagon  ACDBP

Since PB is tangent to circle RBT....then angle  DBP =  90

And since PA is tangent  to circle SAR....then  angle  CAP also = 90 

Angle BDT  = 16

So angle CDB = 180 - 16 =   164

Angle  SDA = 33

So angle ADC = 180 - 33 = 147

The sum of the  interior angles  of a  pentagon  =  540

So angle  ABP =   540 - 147 - 164  - 90  - 90  =  49°

x = 51

By Vieta

The sum of the   roots  =  -4 / 3

So

r + s = -4/3          square both sides

r^2  + 2rs + s^2  =  16/9

And the product of the  roots   = -12/ 3 =  -4  = rs

So

2rs = 2 (-4) =  -8

So

r^2 + 2rs + s^2 = 16/9

r^2 - 8 + s^2  = 16/9

r^2 + s^2  =  16/9 + 8  =  88/9

CORRECTED  !!!!....thx to The 3Mathketeers for  catching  my error  !!!!

39^2  =  (40 -1) (40 -1) =  40^2  - 2(40) + 1  =  1600 - 79

He subtracts 79

Very nice, The 3Mathketeers  !!!!!

This one stumped me........your approach and  explanation is excellent  !!!!