CPhill

15. In ∆ ABC, D is a point on AB such that CD bisects ∠ACB, ∠A = 2∠B, AC = 11 and AD = 2

      Find the length of BC.

By Euclid......BC/DB  = 11/2  = 5.5   so  BC  = 5.5DB

And  < A  = 2 <B 

So

sinA/ BC  = sin B /11

sin (2B) /BC  = sin B / 11

[2sinBcosB] / BC  = sinB/ 11

2cosB/BC  = 1/11 

22cosB = BC

cosB  = BC/22 = 5.5DB/22   =  DB/4  

Using the Law  of Cosines....we have

11^2  = (BC)^2 + (DB+2)^2  – 2(BC)(DB+2)cosB

11^2  =  (5.5DB)^2 + (DB+2)^2 – 2(5.5DB)(DB+2)(DB/4)

let x  = DB

121  = 30.25x^2 + (x + 2)^2  – 2.75x^2 * (x + 2)

121 = 30.25x^2 + x^2 + 4x + 4 – 2.75x^3 – 5.5x^2

2.75x^3 - 25.75x^2  – 4x + 117  = 0

275x^3 – 2575x^2 –400x + 11700  =  0

Solving  this for x  produces two  possible positive  solutions   x = 26/11  or x = 9

So  DB  = 26/11   or DB  = 9     →   BC  = 5.5(DB)  = 5.5(26/11) = 13  

Or

BC  = 5.5(9)  = 49.5

But  BC  cannot  = 49.5  because cosB  = BC/22  = 49.5/22  which is impossible

So.......BC  = 13

7.   The smallest  integer  Pythagorean triple is when AD  = 3 , AB  = 4 and DB  = 5

Perimeter of ABD  =  12

Area of ABCD  = AB * AD  =  [ 3 * 4 ]    = 12

Scaling ABCD  up by  a factor of 2  produces   AD  = 6, AB  = 8  and DB  = 10

Perimeter of ABD  =  24

Area  of ABCD  = [ 6 * 8 ]  = 48  = minimum value of S  for integer sides

Let  P/6 be the side length

Area  =  6 *(1/2)(P/6)^2*sin(60)  = 3(P^2/36)√3/ 2   = [ 3√3/72 [ P^2  = [ √3/24 ]P^2

So

P^2 /A  =   P^2 /  [ √3/24 * P^2]  =   1 / [√3/24]  =    24 / √3   =    [ √3 * 24] / [ √3 *√3] =

√3 * 24  / 3  =   8√3 

I'll let you fill in the "tree" part

      cooks pasta                                     doesn't cook  pasta

            1/7                                                            6/7

   2/3                 1/3                                   1/4                  3/4 

ice cream      no ice cream                    ice cream       no ice cream

P [ice cream]  =   (1/7)(2/3)  + (6/7) (1/4)  =     2/21  + 6/28  =

                              2 / 21   +  3 / 14      =     [28 + 63] /  294  =

                              91 / 294   =  13 / 42   ≈ 30.95%

No, thanks....I have plenty.....

See ????

All true, GA......however.....I believe that at least one of your chimps had a brief fling with Occam's Razor.....

The results got mixed reviews :

Mmmmm...he bears a faint resemblance to Sisyphus......except more handsome.....

17 / 45  ≈ 37.8 %

(   (  [ 150 - 125]  /  125 )  * 100  ) %   =

[  ( 25 / 125   ) * 100 ] %  =

  [.20 * 100 ] %  =  20 % increase

Slight correction .....  " 99.99% "   should be   " 99.92% "

9 / 5  = [x + 500] / x      cross-multiply

9x  = 5[x + 500]

9x  = 5x + 2500

4x  = 2500

x  = $ 625  = 2nd place money

x + 500  = $ 1125  = 1st place money

(i)    $ [625 + 1125 ]  =   $ 1750 

(ii)   1125 / 1750  ≈  64.3%