geno3141

I don't know how to post a diagram. Sorry.

To bisect a line segment:

-- open you compass to a little more than half of the length of the line segment.

-- draw an arc from one of the endpoints through the line segment.

-- draw an arc from the other endpoint through the line segment.

-- there should be two points of intersection of the arcs, one on one side of the line segment;

          the other on the other side.

-- connect these two points of intersection with a line

   -- this line is the perpendicular bisector of the original line segment.

   -- where this line crosses the original line segment is the midpoint of that line segment.

Qestion #2:   [Remember that you have to scale this!]

1)  Find the point X on DA such that DX = 2m. 

     Find the point Y on CB such that CY = 2m.

     Connect X and Y to form line segment XY.

2)  Construct a circle with center A and radius 4 m.

The solution is the point where line segment XY crosses the circle.

Question #1:

1)  Draw a circle of radius 20 cm and center C.

2)  Draw line segment AB.

      Construct the perpendicular bisector of AB.

The perpendicular bisector of AB will intersect the circle in two points; these points are the solution.

Let  A =  lower-left-hand vertex of the triangle.

Let  B =  top vertex of the triangle.

Let  P =  center of the lower-left-hand circle.

Let  Q =  center of the top circle.

Let  X =  point of tangency of AB with circle(P).

Let  y =  point of tangency of AB with circle(Q).

[The left diagonal will be A-X-Y-B.]

Since each circle has radius 2:

PQ  =  4

PX  =  2

QY  =  2

XY  =  4

Triangle(APX) is a 30^o - 60⁰ - 90^o triangle with AP the hypotenuse.

Since PX =2,  AP = 4,  and  AX  =  2·sqrt(3).

YB  =  AX  =  2·sqrt(3). 

AB  =  AX + XY + YB  =  2·sqrt(3) + 4 + 2·sqrt(3)  =  4 + 4·sqrt(3)

angle(TPO) = angle(UPO  =  26^o     --->     angle(TPU  =  52^o  

If minor arc(TU)  =  x,  then major arc(TU)  =  360^o - x

angle(TPU)  =  ½·[ major arc(TU) - minor arc(TU) ]

             52^o  =  ½·[ ( 360^o - x ) - ( x )]

             52^o  =  ½·[ 360^o - 2x ]

             52^o  =  180^o - x

                 x  =  128^o

We have:  log₈(2x),  log₄(x),  and  log₂(x)  forming a geometric progression.

Let's write these all in terms of log₂:

log₄(x)  =  log₂(x) / log₂(4)  =  log₂(x) / 2

log₈(2x)  =  log₂(2x) / log₂(8)  =  log₂(2x) / 3  =  [ log₂(2) + log₂(x) ] / 3  =  1/3 + log₂(x) / 3

So, we now have:  1/3 + log₂(x) / 3,   log₂(x) / 2,   log₂(x)

Just to make the writing simpler, let's write  log₂(x)  as  A:

--->     1/3 + A/3,   A/2,   A

To find the common ratio, we can divide the third term by the second term:

--->     (A) / (A/2)  =  2     (multiply both the numerator and denominator by 2)

We now that this ratio is also true when we divide the second term by the first term:

--->     (A/2) / (1/3 + A/3)   =   2

--->          (3A) / (2 + 2A)   =   2                  (multiplying both the numer and denom by 6)

--->                              3A  =  2(2 + 2A)     (cross multiplying)

--->                              3A  =  4 + 4A

--->                               -A  =  4

--->                                A  =  -4

log₂(x)  =  -4   --->   x  =  2^-4  =  1/16

log₄(x)  =  log₄(1/16)  =  -2

log₈(2x)  =  log₈(1/8)  =  -1

The progression is:  -1, -2, and -4

This might be the right way to solve this problem:

The expression "T[6]" means triangle number 6; the number to the right is the length of each side.

T[1]   --->   1

T[2] = T[1]    --->   1

T[3] = T[2]    --->   1

T[4] = T[1] + T[3]   --->   2   (add the length of a side of triangle 1 to the length of the side of triangle 3)

T[5] = T[4]   --->   2

T[6] = T[5] + T[1]   --->   3   (add the length of a side of triangle 5 to the length of the side of triangle 1)

T[7] = T[6] + T[2]   --->   5

T[8] = T[7] + T[3]   --->   6

T[9] = T[8] + T[4]   --->   8

From now on, the pattern is:  T[n]  =  T[n - 1] + T[n - 5]   ...

If you're writing a program and you have a sequence of numbers, try using a loop; usually, FOR loops are the easiest to write.

Or, do you mean that you want user input?  Then, obviously, use INPUT statements.