CPhill

Since AD bisects  BAC

Let  BD  = 3 - x

Let  CD =  x

So  we have that

BD / AB  = CD / AC

(3 - x) / 4  = x / 5

5 (3 - x)  = 4x

15 - 5x = 4x

15 = 9x

x = 15/9  = 5/3 =  CD

[ ADC ] = (1/2) (CD) ( AB)  =  (1/2) ( 5/3) ( 4)  =   10 / 3

9 = (1/2) ED * height of ADE

18  = ED * height of ADE

18 / ED  = height of ADE

6 = (1/2) ED * height of EDC

12 = ED* height of EDC

12/ED  = height of EDC

So height of    ADE / EDC =  ( 18 /ED ) / ( 12 / ED)  =   18 / 12  = 3 / 2

And the height of EDCB =  height of EDC

So  height of  ABC  =  3 + 2 =  5

And the height of ABC to ADE  =  5 / 3

So the base of ABC to ADE = 5 / 3      since  ABC , ADE  are similar triangles

So the area of  ABC  = area of ABC * (scale factor)^2  =   9 (5/3)^2  = 25  

Simplify as

1 ≤ 4 + 2x < 10 - 8x

Splitting the inequalities, the x's that make this true are

1 ≤ 4 + 2x                4 + 2x   < 10 - 8x

-3 ≤ 2x                      10x < 6

-3/2 ≤ x                        x < 6/10

                                    x < 3/5

So the intervals that DO NOT make this true are

(-inf, -3/2) U [3/5 , inf)

Since CX is an angle bisector

AX / AC  =  BX / BC

AX /10  = BX / 10

AX / BX  = 10 / 10

AX / BX  =  1 / 1

So  triangle ACX and BCX have the same  bases...and they are under the sameheight

So  the ratios of their areas =  1/1

The perpendicular bisectors of each side meet at D  = (6 , 4.55)

The radius  of the circle =  sqrt ( 6^2 + 4.55^2)    =  sqrt ( 56.7025)

Triangle ABC is  isosceles with   AB  = AC

Angle ADB = 90

Angle ABD = 15

Angle BAD = 75 

AB  =  10 / sin 75°  =  10 / cos 15°

AE = EC  =  5 / cos 15°

BE^2  = BC^2  + EC^2  -  2 ( BC * EC) cos (15°)

BE^2 =  20^2  +  25 / [cos 15°]^2  - 2 ( 20 * 5/ cos (15°) * cos 15°

BE^2  = 400 + 25 / [ cos 15°]^2  -  200

BE^2  =  200 + 25 / [ cos 15°]^2    { [ cos 15°]^2  =   (2 + sqrt 3)/4

BE^2  = 200 + 25 * 4 / ( 2 + sqrt 3)

BE^2  = 200 + 100 / (2 + sqrt 3)

BE^2   =  100  ( 2  +  1/ (2 + sqrt 3) )

BE^2  = 100  ( 2  + 2 - sqrt 3 )

BE^2  =  100 ( 4 -sqrt 3)

BE  = 10 sqrt ( 4 -  sqrt 3)

AD is an angle bisector

So

AB / BD  = AC / CD         

2 / BD  = 2 / CD

CD / BD  =   2  / 2

CD  / BD  =   1 / 1

Therefore  the base of triangle ACD = the base of triangle ABD

And they are both under the same  height

So....the ratio of their areas =   1 : 1

Since AD is  an angle bisector

BD / AB  = DC / AC

40 / 60  =  DC / AC

2 / 3  =  DC / AC

(2/3)AC = DC

So  using the Pythagorean Theorem

AB^2  + BC^2  =  AC^2

60^2  +  (BD + DC)^2 =  AC^2

60^2  +  ( 40 +  (2/3 * AC) )^2  = AC^2

3600 +  1600 + (160/3)AC + (4/9)AC^2  = AC^2

(5/9)AC^2   - (160/3)AC  - 5200  =  0

Multiply through by  9

5AC^2  -  480AC  - 46800  = 0

AC  =    [ 480  + sqrt [ 480^2  - 4 * 5 * -46800   ]  /  10    =  

[ 480  +  sqrt [1166400]  ]   / 10  = 

[ 480 + 1080 ] / 10   =

1560  / 10   =

156

Simplify as

x^2 + 12x + y^2 - 16y + 8

Take the derivative with respect to x  and set  to 0

2x + 12 = 0

x = -6

Take the derivative with respect to y and  set  to  0

2y - 16  = 0

y = 8

Minimum  is

(-6)^2  + 12 (-6) + (8)^2 - 16 (8)  + 8  =   -92

semi-perimeter, s ,  =    (12 + 14 + 20)   / 2 =    23

radius of incircle  =   sqrt  [   ( 23 - 12) (23 - 14) (23 - 20)  / 23 ]  = sqrt   [ 11 * 9 * 3 /  23 ] = sqrt (297 / 23) =

3 sqrt (33/23)  =   (3/23) sqrt (33 * 23)  =   (3/23)sqrt (759)  

(1,7)   (2 , -3)

Slope  =  ( -3-7) / (2 -1)  =  -10 / 1   =  -10

Equation of line  using  slope and (1,7)

y = -10 ( x -1 )  + 7

y  = -10x + 10 + 7

10x + y  =   17

Side of square =  6

Perimeter =  24

Area of rectangle =  x *  y  =  24

Perimeter of rectangle     2 ( x + y)  =  24

So     x + y   = 12

So        y  =  12 - x

So

x ( 12 -x)  =  24

-x^2 + 12x  =24

x^2 - 12x + 24  =  0     complete the square on x

x^2 -12x + 36  =  -24 + 36

(x -6)^2  =  12         take the positive root

x - 6  =  sqrt (12)

x = 6 + sqrt (12)  =  6 + 2sqrt (3)   = longer side of rectangle

5

Area  of circumcircle  =   pi  (hypotenuse  / 2)^2  =   pi ( 26/2)^2  = 169 pi

semi-perimeter  =  s =   (26 + 24 + 10) / 2   = 30

radius of  incircle  =  sqrt  [  (30 - 26) (30 -24) (30 -10)  / 30 ]  = sqrt [ (4 * 6 * 20 / 30]  = sqrt (16)   = 4

Area of incircle =   pi * 4^2    = 16 pi

x - y  =   (169 - 16)  pi   =  153 pi

4

Angle YBA  + Angle XAB + Angle AIB  =  180

Angle YBA  + Angle XAB  +  134   =  180

Angle YBA + Angle XAB  =  180 -  134  =  46

2 Angle YBA  +  2 Angle XAB  = 2 * 46

 Angle CBA  +  Angle CAB  = 92

Angle C  =  180  - 92  =  88° 

3

Let  BZ  = x

Let   AZ  =12  - x

Because  CZ is an  angle bisector

BZ /  BC  =  AZ / AC

x /  9.6  =  (12  - x)   /  18

18x  =  9.6 ( 12  - x)

18x  = 115.2 - 9.6x

18x  + 9.6x  = 115.2

27.6x   =  115.2

x =  115.2  / 27.6   =  96 / 23 = BZ