CPhill

Let  ABCDEFGH  be the octagon  and let it be inscribed in a circle

We can  find   the radius of the circle as follows

(2sqrt 2)^2  = 2r^2 - 2r^2* (1/sqrt2)

8 = r^2 (2 - 2/sqrt 2)

8 = r^2 (2 -sqrt (2))

8 / (2 -sqrt 2) = r^2

8 (2 + sqrt 2) / (4 -2)  = r^2

4 (2 + sqrt 2) = r^2

r = 2sqrt [ 2 +sqrt 2] =  sqrt [ 8 + sqrt 32]

Let A=   (0 , sqrt [8  + sqrt 32[ )

Let E = ( 0 . -sqrt [8 +sqrt 32] )

Let C = (sqrt [8 + sqrt 32] , 0]

Let D =  ( sqrt [8 + sqrt32] /sqrt 2 , -sqrt[8-sqrt32] /sqrt 2)  = (sqrt [4 + sqrt 8] , -sqrt [4 +sqrt 8] )

There are only four diagonals of differing lengths

The easiest one  to  calculate   is AE  = 2r =   2sqrt [ 8 +  sqrt32]   = 2sqrt [8 + 4sqrt 2 ]

AD =  sqrt [  ( sqrt [4 + sqrt 8[)^2 +  ( sqrt [8 +sqrt 32) ]  + sqrt [4 +sqrt 8] ^2 ] = sqrt [ 24 + 24sqrt 2]

AB = sqrt (8) 

AC =  sqrt [  ( sqrt [8 + sqrt 32 ] )^2 +  ( sqrt [8 +sqrt 32 ] )^2 ]= sqrt [ 16 + 8sqrt 2 ]

Sum of squares of differing length diagonals

AE^2 + AD^2 + AB^2 + AC^2  = 

4 [8 + 4sqrt 2] + [24 + 24sqrt 2]  + 8  + 16 + 8sqrt 2 = 80 + 48sqrt 2  

\( \log_{8a} 4$ if $a = 1/4\)

log _8(1/4) 4  =

log ₂  4 =               {use the change-of-base theorem }

log 4  / log 2   =

log 2^2 / log 2  =

2log 2  / log 2  =

2

Very nice.....I would have to rely on a graph to solve this.....your approach is WAY more elegant !!!

Thanks, NotThat Smart !!!!

\(\log_2 (\log_3 x) = 2 \log_4 (x) \)

To elaborate a little, we can use the change of base theorem to write

log ( logx / log 3 )            2 log (x)

______________    =    ________                { log 4  = log 2^2  = 2log 2 }

  log 2                             log 4

log ( log x / log 3 )           2log (x)

______________  =     _______

      log 2                        2log 2

log ( logx / log 3 ) =      log (x)

This implies that

logx / log 3  = x

log x = xlog 3

log x  = log 3^x

x = 3^x

Note here that the graphs of both functions never intersect

https://www.desmos.com/calculator/82kerdz2ip

So, like NotThatSmart, I don't find any solutions, either !!!

\(\cfrac{1}{1 + \cfrac{1}{4 + \cfrac{1}{4 + \cfrac{1}{2}}}}.\)

Workng from the bottom-up

4 + 1/2  =  9/2

4 + 1/ (9/2)  =   4 + 2/9  =  (38/9)

1 + 1/(38/9)  =    1 + 9/38  = 47/38

1 / (47/38)  =   

38 / 47

a^3   - 1/a^3 

a^2 - a -1 = 0

a^2 -a + 1/4 = 1  + 1/4

(a - 1/2)^2  = 5/4            take both roots

a -1/2  = sqrt 5/2         or   a -1  = -sqrt 5/2

a = [1 + sqrt 5]/2          a = [ 1 -sqrt 5]/2

a^3  = [ 1 +sqrt 5] /2  *  [1 + sqrt 5]/2  *[1 +sqrt 5] /2

a^3  = [ 1 + 2sqrt 5 + 5 ] / 4   *  [ 1 + sqrt 5 ] /2

a^3 = [ 1 + 2sqrt 5 + 5 + sqrt 5 + 10 + 5sqrt 5] / 8

a^3  = [16 + 8sqrt 5] / 8  =  2 + sqrt 5

1/a^3  =   1/ [2 + sqrt 5]  =  [2-sqrt 5] / [4-5]  = sqrt 5 -2

So in one case  a^3 - 1/a^3  =  [2 + sqrt 5] -[ sqrt 5  - 2]  =  4

In the other case

a^3 = [ 1 -sqrt 5]/2 * [1 -sqrt 5]/2 * [1 -sqrt 5]/2

a^3  = [ 1 - 2sqrt 5 + 5]/4 * [ 1 -sqrt 5]/2

a^3  =  [ 1 -2sqrt 5 + 5 - sqrt 5 +10 -5sqrt 5] /8

a^3  = [ 16 - 8sqrt 5]/8 = 2 -sqrt 5

1/a^3  =  1/[2 - sqrt 5]  =  [2 + sqrt 5] / [4-5] = -2-sqrt 5

So in the other case  a^3 - 1/a^3 =  [2-sqrt 5 ] -  [- 2 - sqrt 5 ] =  4     (same result !!!)

4 / (4 + 9)  * 351    = 108 were cats

9 / (4 + 9) * 351  =   243 were dogs

2A = BE * AC

2A = 20 * AC

2A/20 = AC

2A = AD * BC

2A = 18 * BC

2A/18 = BC

2A  = CF * AB

2A / CF  = AB

AC  + BC   >  AB

2A/20 + 2A/18  >  2A/CF

1/20 + 1/18  > 1/CF

38 / 360 > 1/CF

19 / 180 > 1/CF

CF >  180  / 19  ≈ 9.47

Apparently CF  has no max length.....it's min length   = 10

Can  someone else verify this ???

These two graphs don't  produce  a bounded area

https://www.desmos.com/calculator/awnzsje8th

https://web2.0calc.com/questions/angles_109