CPhill

Very nice, EP......!!!!!

Remove the ending 5 from 365  = 36

Double this and add1 =   36 * 2  + 1  =  72 + 1  = 73

And that's  how many times can 5 go into 365.....!!!!

Let  one side of the rectangle = x     ....and the other side  =  1.25x

Then the  perimeter  =  2 [ 1x + 1.25x]  = 2 [2.25x] = 4.5x

So the side of the square  =  [4.5 x ] / 4  

And the area of the rectangle  =  x * 1.25x  = 1.25x^2    (1)

And the area of the square  = ( [4.5 x ] / 4)^2  = [ 20.25/16] x^2 = [81/64]x^2    (2)

So the ratio  of  (1) to (2)   =   1.25x^2 / [ (81/64)x^2 ] =  1.25 / (81/64)  = [5/4] / [ 81/64]  =

[5/4] * [ 64/81]  =     [64/4] * [ 5/81]  =  [16*5] / 81  =  80/81    as EP  found!!!

sin θ   = O / H   ......so......

sin 43  = 7.31 / H      rearrange as

H =  [ 7.31 / sin 43 ]  ≈ 10.7185

y^2-16y+64 = -9+64

(y - 8)^2  =  55       sq rt  both sides

 y - 8   =  ± √55     add 8 to both sides

 y  =  ± √55   + 8

P(1 detects) = C(3,1)(.75)(.25)^2  =  9/64

P(2 detects) = C(3,2)(.75)^2(.25)  =  27/64

P(3 detects) = C(3,3)(.75)^3  =  27/64

P(at least one detects)  =    [ 9 + 27 + 27]  / 64   =  63 / 64

<OXW  = < XWO  = 90                             <ZYP  = < OYX  = 90

<PWZ = <XWO  =90                                <ZYP  = <YZP  =90

So < OXW = < PWZ                                 So < OYX  = < YZP

And <XOW  = < ZPW  = 90                      And <XOY = <ZPY

So  ΔXOW  ~ Δ WPZ                                So  ΔXOY  ~  ΔPYZ

XO / OW  = WP / PZ                                XO /OY = PY /PZ

XO * PZ  = WP * OW                                 XO* PZ  = OY* PY

So

WP * OW   = OY* PY

OW / PY  = OY/ WP

OW / PY = [ OP + PY] / [OP + OW]

OP*OW + OW^2  = OP*PY + PY^2

OP[OW − PY] +  OW^2  −  PY^2  = 0

OP[OW − PY]  +  [OW − PY] [[OW + PY]  = 0

[OW − PY] [ OP + (OW + PY) ]  = 0          

And.......since OP , PY and OW  are  all > 0  .....the second factor can never  = 0

So  [OW − PY]  must = 0   →    OW  = PY

Using ...... sin (A - B)  = sinAcosB - sinBcosA ......and we have.....

(cos(pi/6))(sin(pi/5))-(sin(pi/6))(cos(pi/5)) =

sin( pi/5 - pi/6)  =

sin (pi / 30)

I assume that you want the total distance between the time it is dropped and just before it bounces the ninth time

The sum is given by

30  +  30 [1 − .5^8] / [1 − .5 ]  =  89.76m → 89.8 m

Just as EP found....!!!!

Using the triangle inequality, we have that

x+ 2x+ 1 > x + 4

2x > 3

x > 3/2

And

2x + 1 + x + 4 > x

2x > − 5

x > −5/2

And

x + x + 4  > 2x + 1

2x + 4 > 2x + 1   

True for all x

x > 3/2   is the most  restrictive interval

So....as long as x >3/2   a triangle is possible