CPhill

https://web2.0calc.com/questions/semicircle_5

Let P and Q  lie on a unit circle

The equation of the line through OP  is   y= 4x

The equation of the line through OQ is y = 5x

To find the x coordinate of P

x^2 + (4x)^2   =1

17x^2 =1

x= 1/sqrt 17      y = 4/sqrt17

To find the x coordinate of Q

x^2 + (5x)^2 =1

26x^2 =1

x = 1/sqrt 26     y =5/sqrt 26

To  find  the slope of the line through  PQ

[ 5/sqrt 26 - 4/sqrt 17 ] / [ 1/sqrt 26 - 1/sqrt 17 ]  =

[ 5 sqrt 17 - 4sqrt26 ] / [ sqrt 17 - sqrt 26 ]  =

[4sqrt 26 - 5sqrt 17 ] / [ sqrt 26 -sqrt 17]

[ sqrt 26 + sqrt 17 ] [ 4sqrt 26 - 5sqrt 17] / 9  =

[ 4 *26 + 4sqrt442 -5sqrt 442 - 85] / 9  =

[19 -  sqrt 442 ] / 9

The slope of the line we want  has the negative reciprocal

9 / [ sqrt 442 - 19 ]  =

9 [ sqrt 442 + 19 ] / [ 442 - 361]

 [sqrt 442 + 19]  /  9  ≈ 4.448

Sum of all edges  =  4[ l + w + h ] = 17 →  l + w + h = 17/4     square both sides

l^2 + w^2 + h^2  + 2 ( lw + lh + wh) = 289/16       (1) 

Surface area  =   2 [ lw + lh + wh ] =  15     (2)

Sub (2)  into (1)

l^2 + w^2 +  h^2  +   15 =  289/16

l^2 + w^2 + h^2  =  49/16

The length of the diagonal  = 

sqrt [ l^2 + w^2 + h^2 ]  =  sqrt [ 49/16 ]   = 7 / 4  =   1.75

Let P = (x,y)

PA^2 + PB^2 + PC^2   =

(x-7)^2 + (y + 11)^2  + (x-10)^2 + (y-13)^2 + (x-18)^2 + ( y + 22)^2    .....simplifying, we have

3 x^2 - 70 x + 3 y^2 + 40 y + 1247

3 [ x^2 - 70/3 x + y^2 + 40/3 y ] + 1247       complete the square on  x,y

3[ x^2 - 70/3 x + 4900/36 + y^2 + 40/3y + 1600/36 ] + 1247 - 4900/12 -1600/12

3 [ (x - 70/6]^2 + (y + 40/6)^2 ] +  2116 / 3

Q = (70/6, -40/6 ) =  (35/3 , - 20/3)

k = 2116 / 3

https://web2.0calc.com/questions/help-algebra_31592

Haven't bothered to count them but the path must start at C and  end at D   or vice-versa (beginning and  ending at vertexes with odd valences)

Let these two lines intersect a cirlce with a radius of 1

To find the point on y= x  that lies on the circle

x^2 + x^2  =1

2x^2 =1

x^2= 1/2

x =1/sqrt 2     and y= 1/sqrt 2

To find the point on y= 2x that lies on the circle

x^2 + (2x)^2 = 1

5x^2 =1

x^2 = 1/5

x = 1/sqrt5      y = 2/sqrt 5

The midpoint  between these two points is [  (1/sqrt 2 + 1/sqrt 5) / 2 , (1/sqrt 2 + 2/sqrt 5)/2 ] =

[ (sqrt 2 + sqrt 5)  / (2sqrt 10) ,  (2sqrt 2 + sqrt 5) / (2sqrt 10) ]

The slope of the line drawn from the origin through this point = m  =

[ 2sqrt 2 +sqrt 5 ] / [ sqrt 2 + sqrt 5] =  

[ sqrt 5 - sqrt 2] [ sqrt 5 + 2sqrt 2] / 3  =

[ 5 + sqrt 10 - 4 ] /  3  =

[ 1 + sqrt 10 ] /  3

x^2 + (-4)^2  = 20

x^2 + 16 =20

x^2 = 4

x= 2    or x = -2

A= (-2,4)   B = (2,-4)

AB  =  [ 2 - -2 ]  = 4

Slope = (4 -  - 5)  / ( 5 - - 1)  = 9/6 =  3/2

Eq of line

y = (3/2) (x -5) + 4

2y = 3(x -5) + 8

2y = 3x -7

y = (3/2)x - 7/2

(3/2)x - y = 7/2           mult through by 2/7

(3/7)x - (2/7)y = 1

x / (7/3) - y / (7/2) = 1

r = .036 / .4  =  36 / 400 =  9 /100 =  .09

Sum =   .4 / [ 1 -.09] =  .4 / .91  = 40 /91