| T.R | Title | User | Personal
Name | Date | Lines |
|---|
| 479.1 | c'mon, these are too easy. | METOO::YARBROUGH | | Thu May 01 1986 13:04 | 3 |
| In a 30 degree right triangle the opposite side is half the hypotenuse
so the hypotenuse is 24. 4 inches is outside the circle so the radius
is 20. (Yawn.)
|
| 479.2 | Does a picture help? | KEEPER::KOSTAS | Kostas G. Gavrielidis <o.o> | Fri May 02 1986 02:47 | 31 |
| re. .1
well no,
by your results 12^2 + 20^2 should equal 24^2.
But it does not. 144 + 400 <> 556.
Maybe if I gave a picture ....
|\ <---- tangent touches the circle
/|*\ the angle with R should
/ | *\ be 90 degrees.
/ | * \
R / | * \ 12
/ | * \
/ x | * \ <----- a tangent 12 inches
/ | * \
/)30 | * \
/__)_____|____*____\
R 4
^ ^
| |
center of a point outside the circle
the circle 4 inches a way
I hope this helps.
Kostas G.
|
| 479.3 | Maybe I _was_ awake then! | NMGV01::ASKSIMON | Don't upset the 'Deus ex Machina' | Fri May 02 1986 09:32 | 26 |
| I sympathise with .1. As I see it you have a conflict of
information. I did not reply at first because I thought I
must have misunderstood, until I saw the diagram, which leads
to the impossible triangle I had first envisaged which your
diagram confirms.
Because the 12 inch line is a tangent you have 90 degrees
between it and R. This creates a right angled triangle between
the three points (i.e. not including where the R+4 line
meets the edge of the circle). You have a hypotenuse of R+4 and an
angle at the centre of 30 degrees. This provides two possible
solutions but with a different answer. Firstly, you have
tan(30) = 12/R (perpendicular over base)
and secondly
sin(30) = 12/(R+4) (perpendicular over hypotenuse)
The first equation gives R=20.78 and the second R=20. If you want
the other answer, i.e. approx. 20.78, then you should not have
said "4 inches from" but "somewhere outside the circle" the
remaining information would then have been sufficient to arrive at
the answer by using the above tangent equation. I must still
conditionally agree with .1, though. I.e. that it would then
indeed be too easy.
|
| 479.4 | Some history on this ... | KEEPER::KOSTAS | Kostas G. Gavrielidis <o.o> | Mon May 05 1986 13:13 | 16 |
| re. .1
The interest on this problem arises from the fact that it
once (1951) appeared on a paper set for a competition
participated in by a large number of schools near Boston.
The trouble with it is that it is over-specified.
The three numbers: 4, 12, and 30, form an impossible combination
of data.
If one ignores one of these three numbers, there are three different
solutions that can be obtained.
|