CPhill

It's been a joy  - and sometimes a real challenge  - to answer your questions, Tertre.....You have also provided some very good  solutions to many of the questions others have posted on the forum, as well.

So......Thanks for the kudos...and.....

See you on your return.....!!!!

The area  of the tirangle  = (1/2)side^2 * sin (60°)  = (1/2) 3^2 * √3/2 = (9/4)√3  units^2

We can  construct a circle at A  with a radius of 1......the area  bounded by the sides of the triangle and the circle will be  =

pi * (1)^2 * (60/360)  =   pi/6  units ^2

So....the probability that "X"  lies within one unit of "A"   is

[pi / 6]  / [ (9/4)√3]  =

pi /  6   *  4 / [9√3  ]  =

4 pi / [ 54 √3 ] =

 2 pi / [ 27√3]  ≈   .1344 ≈   13.44%

Here's a picture :

The total number of members is given by

(n - 2) ( n + 8)    (1)

In the second formation, we have that   the  number of members, excluding the drummers is n (2n - 3)      (2)

So...since there are at least 4  drummers, the difference  between (1) and (2)  is  ≥ 4

So we have that

(n - 2) ( n + 8) - [ n (2n - 3)  ] ≥ 4

n^2 + 6n - 16   - [2n^2 - 3n] - 4  ≥ 0     simplify

-n^2 + 9n -20  ≥ 0        multiply  through by  -1   and reverse the inequality sign

n^2  - 9n + 20 ≤  0     (1)      let's just set this to  0  and solve

n^2 - 9n + 20   =  0      factor this and we have that

(n   - 5)  ( n - 4)  = 0      set  each factor to 0 and solve for n and we have that

n  = 5     or   n  = 4

The  solution to  the inequality  (1)  is     4 ≤ n ≤ 5

Since  we assume  n to be an integer, the sum of all possible  n  = 4 + 5   =  9

4.In the diagram below, we have \(angleABC= angleACB=angleDEC=angleCDE\), BC=8, and DB=2. Find AB.

Since angle DEC  = angle BEC,    and angle DEC  = angle ABC = angle ECB.....then angles  BEC  and ECB  are equal....then in triangle BEC,   BC  = BE 

Then

BC  = BD  + DE

8  = 2 + DE

6  = DE

And angle ABC  = angle ACB...so...AC  = AB

Note that since angles ACB, ACB, DEC and CDE  are all equal, we have the following similar triangles :

ΔABC  ~  Δ BEC   ~  ΔCED          and  because  angles DEC and CDE are equal, then   CD  = EC

So

AC/BC  = BC /EC   

AC / 8  = 8 / EC

AC * EC   = 8*8

AC * EC  = 64

AC  = 64/EC    (1)

And

AC/BC  = CD/ED

AC/8  = CD / 6

AC/8 = EC/ 6

AC = (8/6)EC

AC = (4/3)EC     (2)

Equating (1)  and (2), we have that

64/EC  = (4/3)EC

(64)(3/4)  = EC^2

48  = EC^2

√48  = EC

4√3  = EC

And  AC  = (4/3)EC   = (4/3) * 4√3   =     16/√3  =   AB

y = 3x^2  - 5x  - 3

Let   A  = (u, 3u^2 - 5u  -3)

Let B =   (v, 3v^2 - 5v - 3)

Using the midpoint, rule, we have that

[ u + v ] / 2  = 0         and      [ 3u^2 - 5u  - 3 + 3v^2 - 5v  - 3 ] / 2  = 0    (2)

u + v  = 0 

v = -u    (1)

Sub (1)  inot (2)  and we have that

[3u^2 - 5u - 3 + 3 (-u)2 - 5(-u) - 3 ] /  2   =  0            multiply  through by 2  and simplify

6u^2  - 6   = 0      divide through by 6

u^2  - 1   = 0     factor

(u + 1) ( u -1)  = 0

Set both factors to 0  and solve for  u  ⇒    u = -1    or  u  = 1

If  u   = -1    then the associated  y  coordinate is  3(1)^2 - 5(-1) - 3   = 5

So A  = (-1, 5)

And the x  coordinate  of  B  is  -u =  1  and  the y coordinate of  B is 3(1)^2 - 5(1) - 3 = -5

So, in this case B  = (1, -5)

The  slope of this line through A and B is    [ -5 - 5]  / [ 1 - -1] =  -10/2  = -5

And the equation of this line is

y  = -5 (x -1) - 5

y = -5x 

So....the  distance between  A  and B   is  √ [ -5 - 5)^2  + ( -1 -1)^2 ]  = √[ 10^2 + 2^2] = √104  

So....the square of the length of AB is just 104

Note that if u  = 1, we have the same results...A and  B are  just "switched"

See the graph, here :  https://www.desmos.com/calculator/e2uvdwwhl3

3.

Since DE  = 2EC....then

EC + 2EC  = DC

3EC  = 20

EC  = 20/3

And DE  = 2EC  = 2(20/3)  = 40/3

Note that triangle  AFB  is similar to  triangle EFD

So

AB/ED  = FB / FD

So

20/(40/3)  = FB / FD

60/40 =  FB / FD

6/4  = FB /FD

3/2 = FB / FD

Then BF  consists of 5 equal parts.....FB is 3  of them and FD  is 2 of  them

And in triangle BDC, FG  is drawn parallel to DC....and we have the following relationship :

BF / BD  = FG / DC

3 / 5   = FG / 20      multiply both sides by 20

20 * 3  / 5    = FG

60 / 5  = FG

12  = FG

2. Find the length of the shortest altitude of a triangle with sides of lengths 10, 24, and 26.

The area  of the triangle will be  (1/2)(10) (24)  = 120  units^2

So....if we let  10  be one altitude....the shortest altitude  will be  drawn  from the intersection of the legs  to the hypotenuse

And since the areas  are equal....we have that

(1/2) (length of the shortest altitude) (26)  = 120    

(13) (length of shortest  altitude)  = 120      divide both sides  by 13

length of the shortest altitude  = 120 / 13   units  .....just as CluelesssPersonn  found  !!!!

The original D  is  ( 4,1)

When reflected across the x axis, 'D'   is  ( 4, -1)

D' and  D" will lie on a line that  will be  perpendicular to  y  = x + 1

So....the slope of this line is  - 1

And the equation of this line   is

y  = - ( x - 4) - 1

y = -x + 4 - 1

y = -x + 3

The  intersection  of these two lines  will be the midpoint  of   D'  and D"

So...setting the y's equal, we have

-x + 3  = x + 1       add  x to both sides,  subtract 1 from each side

2   = 2x     divide both sides by 2

1  = x

And using either line to  find the y coordinate of this intersection point, we have that

y = x + 1

y = 1 + 1

y  = 2

So...the midpoint  of  D' and D"  is  ( 1, 2 )

So...using the midpoint formula with the points ( 4, -1) and (1,2), we can  find  D'  as

(x + 4)/2  =   1                                ( y + -1) / 2  = 2

Multiply  both sides  by 2

x + 4  = 2                                           y - 1  = 4

subtract 4 from both sides                 add 1 to each side

x = -2                                                 y = 5

So...  D"    =  (-2, 5)

Here's a graph that shows this :

https://www.desmos.com/calculator/rbi6tjxihp

We need to  know what the coordinates  of the original  "A" are  ....  

The protractor is considered to be  "cheating"  by Euclidians....!!!!!