6. THE BUCKLED RAILROAD TRACK
Railroad designers are always looking for ways to
improve the safety and comfort of train travel.
Suppose technology made it possible to make and
transport straight rails one mile long. If one winter
two pairs of these rails were installed along a two
mile portion of track in the desert, then cach rail
would expand approximately one foot in length in the summer heat. Suppose that
instead of buckling, these rails keep their (straight) shape. In this case, the ends
where they meet could jut upward to form a triangle. How high above the ground
would the ends of the rails be?
Let H be the height (in feet) of the tracks above the ground where the two rails come
together.
a) How large do you think H is? Is it big enough for you to stick your arm
between the ground and the tracks? Is it big enough walk through? Could
you drive a car under the buckled tracks? Make a guess.
b) Draw a picture and calculate H. (Note: a mile is 5280 feet long.)
c) How does your calculated value for H compare with your guess in part (a)?

  5.   y=6; de=7; fg=8

  6.   x=4.2; jl=14.2; mn=13

step-by-step explanation:

when chords cross inside a circle, the product of the segment lengths is a constant.

5. dh×eh = fh×gh

  y = gh = dh×eh/fh = 3×4/2 . . . divide by the coefficient of gh

  y = 6

of course, the total chord length is the sum of the lengths of its segments.

  de = dh +eh = 3+4 = 7

  fg = fh +gh = 2+6 = 8

6. this problem works the same way as the previous one.

  x = lk = mk×nk/jk = 6×7/10 = 4.2

  jl = jk +lk = 10 +4.2 = 14.2

  mn = mk +nk = 6 +7 = 13

comment on chords and secants

you can think of the points h (problem 5) and k (problem 6) as points where the chords meet. a chord is part of a secant line, a line that intersects the circle in two points. one can also consider these points (h or k) to be the points where the respective secant lines meet.

then, the product that is a constant is the product of the distance from the meeting point (h or k) to one circle intersection with the distance from that meeting point to the other circle intersection. for problem 5, the constant is the product hd×he = hf×hg.

it turns out that this rule regarding the products of lengths to points of intersection is also true when the secants intersect outside the circle.

that is, you only need to remember one rule to work both kinds of problems.

i think is g(1)=-1.

hope that i can

here is a big fat