CPhill

Add the equations

p(x + y + z) + 2p  + q(x + y + z) + 2q + r(x + y + z) + 2r  = (x + y + z) + 7 + 5 - 3 + 4 - 8

(p + q + r) ( x + y + z) + 2 (p + q + r) =  (x + y + z) + 5

(-8) (x + y + z) + 2 (-8)  = (x + y + z) + 5

-9  (x + y + z)  =  5 - 2(-8)

-9 (x + y + z)  = 5 + 16

-9 (x + y + z)  =  21

(x + y + z)  =  -21 / 9     =  -7/ 3

1/ ( x + 1) + 3 / (x +7) ≥  1/2

[ (x + 7) +  3(x + 1) ] / [(x + 1)(x+ 7)] - 1/2 ≥  0

[ 4x + 10 ]  / [ ( x + 1 ) (x + 7) ]   - 1/2 ≥  0

[ 2 (4x + 10) - (x + 1) ( x + 7) ] / [ 2 (x + 1) (x + 7) ] ≥  0

[ 8x + 20 - x^2 - 8x - 7 ] / [ 2 ( x + 1) (x + 7) ] ≥  0

[ -x^2 + 13 ] / [ (x + 1) ( x + 7) ]  ≥  0          (1)

We  have  five  possible intervals to  check

(-inf , -7)          check with x = -8   ....this interval makes (1) false

(-7 , -sqrt 13]  check with x =  -5   ....this interval makes (1) true

[-sqrt 13 , -1)  check with x = -2  ......this interval makes (1) false

(-1 , sqrt 13 ]  check with x = 0   ......this interval makes (1) true

[sqrt 13 , inf)  check with x = 5   ......this interval  makes (1) false

Solution comes from  

( -7 , -sqrt 13]  U  ( -1 , sqrt 13]

AC = sqrt (25^2  - 10^2)  = sqrt (525) 

AD = sqrt [ ( CD + 9 )^2 + 10^2 )    =  sqrt [ CD^2 + 18CD + 181 ]

AD  + DC =  AC =  sqrt (525)

sqrt [ CD^2 + 18CD + 181 ] + CD  = sqrt (525)

sqrt [ CD^2 + 18CD + 181 ]  =  sqrt [ 525 ] - CD        square both sides

CD^2 + 18CD + 181  =  525 - 2 CD*sqrt (525)  + CD^2

18CD  + 181  =  525 - 2CD * sqrt (525)

18CD + 2CD * sqrt (525)  = 525 - 181

CD ( 18  + 2sqrt (525))  = 344

CD  =  344 /  [ 18 + 2sqrt (525) ] =  172 / [ 9 + sqrt 525 ]  

AC - CD =  AD

sqrt (525)  -  172 / [ 9 + sqrt 525] =  AD

(sqrt (525) [ 9 + sqrt (525) ]  - 172) / [ 9 + sqrt (525) ] =  AD

[9sqrt (525) + 525 - 172] / [9 + sqrt 525)  = AD

[9 sqrt (525)  + 353 ] / [ 9 + sqrt (525) ] = AD  ≈  17.523

Using the Law of Sines

AD/ sin (ABC) = AB / sin (ADB)

AD/ sin 45°  = AB / sin 105°

24 * sin 105° / sin 45°  = AB    = 12 + 12sqrt (3)

Angle ACD  =angle ADC =  75°

So  AD  = AC = 24

Area of ABC = (1/2) (AC) ( AB) sin 60° =   (1/2) (24) ( 12 + 12sqrt 3) * sqrt (3) / 2  =

12 * 12  ( 1 + sqrt (3)) sqrt (3)  / 2  =

72 ( 1 + sqrt 3) (sqrt 3)  =

72 ( 3 + sqrt 3)

x^2 - y^2 = 5100

(x + y) ( x - y)  = 5100

Divisors of 5100  = 

1 | 2 | 3 | 4 | 5 | 6 | 10 | 12 | 15 | 17 | 20 | 25 | 30 | 34 | 50 | 51 | 60 | 68 | 75 | 85 | 100 | 102 | 150 | 170 | 204 | 255 | 300 | 340 | 425 | 510 | 850 | 1020 | 1275 | 1700 | 2550 | 5100 (36 divisors)

x             y       

1276    1254

 428       422

 260       250

 100         70

   92         58

   76         26

"4" should be the number typed in the  small box outside to  represent those  dogs that got no ribbons

B

7                 7sqrt (3)  =  sqrt (147)

A           D            C

AC =  sqrt  [ 147 - 49 ]  = sqrt [98 ] =  7sqrt (2)

DC = 3.5sqrt (2) = AD  =  sqrt [24.5]

BD  =  sqrt [ 7^2 + 24.5]  = sqrt [ 73.5] 

Using Law of Cosines

BC^2  = BD^2  + DC^2  -  2(BD* DC) cos (BDC)

147  = 73.5 + 24.5 - 2 [ sqrt [ 73.5] * sqrt [ 24.5] ] cos (BDC)

49 / [- 2 sqrt (7203 / 4 )  ] =    cos (BDC)

-49 / sqrt(7203)  = cos (BDC)

sin(BDC) =   sqrt [  1 - (49)^2 / 7203 ]  =  sqrt (6) /  3

tan BDC =  [sqrt (6) / 3 ]  / [ -49 / sqrt (7203) ] =    [sqrt(6)*sqrt (7203) / [ -49 * 3]  =  

(sqrt 2) (sqrt (3) * 49 * sqrt  (3) )  / [ -147]  = 

-sqrt (2) * 147  /  147  = 

   -sqrt (2) 

x + y = (a/2)     square  both sides

x^2 + 2xy + y^2 =  a^2/4

And

xy = -(a + 3)

2xy  = -2(a + 3)

So

x^2  - 2(a + 3) + y^2 = a^2/4

x^2 + y^2 =  a^2/4  - 2(a + 3)  =   (1/4)a^2 - 2a - 6        (1)

Take the derivative  of  (1)  and set    =  0 

(1/2)a  - 2  =  0

(1/2)a =  2

a =  4

The min value  of x^2 + y^2  =  (1/4)(4)^2  - 2(4)  - 6 =    4 - 8 - 6 =   -10

1

2 + 1

3 + 1

4 + 2

5 + 1

I'll let you finish