CPhill

Connect  AO,  BO , PO

Tangents drawn from a single point outside a circle are equal

OP = OP

AO = BO

AP = BP

So by SSS.....triangles APO and BPO  are congruent

Then because angle OPA = 30° then angle  APB = 60°

And angle AOB is supplemental to angle APB =  120°

And central angle APB = measure of minor arc AB = 120°

Not as difficult  as it seems if we think about this

sum [ ( sin 5° + sin 10° + sin 15° +  ..... + sin 180° )  + ( sin355° + sin 350° + sin 345° + .... +sin 180° ) ]  =  0

And sin 0° = sin 360°  = 0 

Adding all these, the sum  =  0

Let AC  =  AB

tan (angle BEA) =  AB / EA  =  AB / [1/3) AB)   =   1/ (1/3)  = 3

arctan 3 ≈ 71.6°

Angle  BED ≈  180  - 71.6  ≈  108.4°

P ( 3 show the same number )  =   C(5,3) (1/6)^3 ( 5/6) (4/6)   =  25/ 972 = 200/7776

P( 4 show the same number)  = C(5,4) (1/6)^4 ( 5/6)   = 25/7776

P( 5 show the same number ) =  C(5,5) ( 1/6)^5  = 1/7776

Total probability  = (200 + 25 + 1) / 7776  =   226 /7776 = 113/ 3888

Yep.....good to BE back !!!!

Nice work  guys  !!!!

Set  the discriminant =  2c + 3

We have

5^2 - 4c  = 2c + 3

25 - 3  = 6c

22= 6c

22 / 6  = c  =  11 / 3

We will have exactly real one solution  whenever the discriminant   =  0 

We have the form  ax^2  + bx + c = 0

So....the  discriminant is

b^2  - 4ac

4^2 - 4 (3)(c)  = 0

16 - 12c  =  0 

16  =12c

16/12  = c

4/3  = c

\( sqrt(3 - sqrt(8 - sqrt(x)))\)

The least   value in  the domain  will  be  when x = 0

So we have  

sqrt ( 3 - sqrt ( 8 - sqrt ( 0 ) ) )  =  sqrt ( 3 - sqrt 8 ) =  sqrt ( 3 - 2sqrt 2)

Since we cannot take the square root of a negative, the greatest value in the domain = 64  because

sqrt ( 3 - sqrt ( 8 - sqrt (64)) )  =  sqrt ( 3 - sqrt ( 8  - 8) ) = sqrt ( 3 - 0 ) =  sqrt 3

See the graph here : https://www.desmos.com/calculator/0nd567etzs

Nice job, BB   !!!!