CPhill

For convenience let  Q   = (0, 1) 

Let the center lie at (0,0)

Note that in the equation of the ellipse, "b"    = 1

Then  1/2  the  focal length  =  1/sqrt (3) 

So F1F2  = 2/sqrt (3) =   (2/3)sqrt (3)

We can find "a"   as  sqrt  [ 1^2 + (1/sqrt(3))^2  ]  = sqrt (1 + 1/3)   =  sqrt [ 4/3 ] =  2/sqrt (3)

a^2 =  4/3

A line from Q through R  will have a slope of  tan (-60)  =  -sqrt (3)

So.....the equation of this line is   y  = -sqrt (3)  +  1      (1)

To find the positive x values  of R, sub  (1)  into the equation of the ellipse

x^2 / ( 4/3)  +   [ -sqrt (3)x  + 1]^2  =  1

(3/4)x^2  +  3x^2  -2sqrt (3)x  + 1   = 1

(15/4)x^2  -  2sqrt (3)  x  =  0

x   ( 15/4x -  2sqrt (3)  )  =  0

The second  factor set to 0  will give ys the  x coordinate of R

(15/4)x  - 2sqrt (3)   =  0

x = (4/15)2sqrt (3)

x =  (8/15)sqrt (3) 

And the y coordinate  of  R = -sqrt (3) (8/15)sqrt (3)  + 1  =  -24/15 + 1  =  -9/15   = -3/5

And  QR  = PQ =   sqrt   [ (8/15sqrt (3)^2   +  (1 - -3/5)^2  ]    = 

sqrt[ (8/5)^2  ( 1/3 + 1 ]  =  (8/5)sqrt [ 1 +1/3  ]  =  (8/5) sqrt  ( 4/3)  =  (16/5) sqrt (1/3)  = (16sqrt (3) / 15)  =

(16/15)sqrt (3)

So

PQ   / F1F2  =     (16/15) sqrt (3)    / [ (2/3)sqrt (3)  ]  =    (16/15) (3/2)  =  8/5

The distance. D.  from  the center of any  hexagon to its edge can ne found as

(1/2) side length  /  sin 30°  =   D /sin 60°

3.5 / (1/2) = D / (sqrt (3)  / 2)

7 (sqrt (3) / 2 )  = D

The  side   of   our  triangle is twice this  =   7sqrt (3)  

So....O1O2O3  forms an equilateral  triangle  with a side length of 7sqrt (3)

It's    area   =  (1/2) (7sqrt (3))^2  sqrt (3)  / 2   =    [ 147 sqrt (3)  /  4   ]   units^2

THX, mimi997   !!!

Nice, cryptoaops   !!!!

We  have a   9 - 12  - 15  right triangle

So  x +  y =  15  =  base length of this triangle

The  area of this triangle  =(1/2) (9)(12)  =   (1/2) ( 108)  =  54

The area can also be expressed as

54   =  (1/2) base * z

54  = (1/2) 15 * z

108 = 15 * z

z = 108/15 =  36/5

So

x + y + z  =    15  + 36/5 =     [ 75 + 36 ]  /  5  =   111  / 5

Imgaine that we  have a rectangle with a height of  6

From C, draw a perpendicular to the base of the rectangle

We have  three right triangles 

sqrt (S^2  - 2^2)   is the width of the  rectangle

And  sqrt (S^2  - 6^2)  +  sqrt (S^2 - 4^2)   also represents the width

So

sqrt (S^2  -  4)  = sqrt ( S^2 - 36) + sqrt ( S^2 - 16)        square both sides

S^2 - 4 =  S^2 - 36  + S^2 -16   + 2 sqrt [ (S^2  -36) ( S^2 - 16) ]

48 - S^2  =  2sqrt  [(S^2 - 36) (S^2 -16)  ]        square both sides again

S^4  - 96S^2  + 2304  = 4 S^4 - 208S^2  + 2304

3S^4 - 112S^2  =  0

S^2  ( 3S^2  - 112)   = 0

The second factor will produce a positive result for S

S^2  =   112/3

S =  sqrt  (112 / 3)  =   sqrt  (16 * 7) / sqrt (3)  =  4sqrt (7/3) ≈  6.11   (as jugoslav found  !!!)

Just a simple proportion

2 / 100  =   50   /  N          cross-multiply

2N  = 100 * 50

2N =  5000

N  =5000/2   =  2500 salmon

Another way to see this is to  note that 1/50  of the salmon are tagged

So....there are 50 times as  many as we  find  tagged

So...if we  find  50 tagged....then  50 * 50    = 2500  

Assuming that x is a positive ineger

Let x  be  some "large positive integer "

The ordering of the  numbers  is

2, 3 , 5 , x , 2x

And the median is  5.....just  as   cryptoaops   found  !!!

 Note  that not just  ANY integer will produce this....for example...if  x  = 1   we have

1 2 2 3 5    and the median is  2  

And if x  = 2  we  have    2  2  3  4  5  and the median    = 3

If x =   3  we have   2 3 3  5  6       median  = 3

If x =  4  we have   2 3 4 5  8       median = 4

In fact  x  must be an integer greater than  4 to have a median of 5

If we inscribe  the hexagon in a circle, let  the radius  = R

By AAS  we  have 8 congruent  30-60-90  right triangles

The side opposite the 60° angle = R/2

And the side opposite  the   30°  angle =  R / ( 2sqrt (3) )

So.....the area  of  these  8 right triangles  is

8 *(1/2) (R/2) (R  /(2sqrt (3))  =   R^2/sqrt (3)

And  the  side of the  hexagon  = R....so its area =  3sqrt (3)R^2 /2

So   H  /  S    =        3sqrt(3) R^2 / 2               3 sqrt (3) sqrt (3)             9

                              _____________  =          ______________   =     ____

                               R^2 / sqrt (3)                             2                            2

Exactly what   cryptoaops found  !!!!

Good job, cryptoaops  !!!!

A(1)  =  p(1)   + q^1  =   p +  q  =  4  ⇒      q =   4 - p      (1)

4A(2)  = 4 [ p(2)  + q^2 ] =  8p + 4q^2  =  37         (2)

Sub (1)  into (2)  for  q 

8p + 4 (4 - p)^2    = 37

8p + 4 (p^2   - 8p +16)  =  37

8p + 4p^2  - 32p + 64   = 37

4p^2  - 24p + 27  =  0        factor

(2p - 9)(2p - 3)  =  0

Set each factor to  0  and  solve  for  p   and we  get that  p =  9/2    or  p  =3/2

When  p =  9/2, q  =  4 - (9/2)  = -1/2   and     p  - q =  9/2 -  - 1/2  =  10/2  = 5

When p=  3/2 , q = 4 - (3/2)  = 5/2   and p - q  =  3/2 - 5/2  =  -2/2  =   - 1

So max value  of    p - q   =    5