Printable View
Is there enough information to answer the following question:
Two hikers are in the field out of each other's sight. They each have a cell phone and a compass. They want to meet together someplace along the straight line that connects their current positions. They call each other and agree on a distant landmark visible to each. They report to each other their bearing (Degrees from North) to the landmark. Can they now decide the proper bearing to walk to meetup in the shortest distance?
1) Draw a distant landmark.
2) Draw two rays emanating from the landmark. Make both rays in relatively the same direction, but leave some convenient small angle between them. Extend the rays as far as you dare. I recommend the edge of your drawing surface.
3) Put a hiker on each ray.
4) Grab one hiker and move it almost up to the landmark.
5) Think about the path they would have to walk to meet up. You may wish to sketch this path lightly.
6) Grab that same hiker and drag it back to the edge of your drawing surface.
7) Think about the path they would have to walk to meet up. You may wish to sketch this path lightly.
8) Now what?
Hello, VictorL!
Quote:
Is there enough information to answer the following question?
Two hikers are in the field out of each other's sight.
They each have a cell phone and a compass.
They want to meet together someplace along the straight line that connects their current positions.
They call each other and agree on a distant landmark visible to each.
They report to each other their bearing (degrees from North) to the landmark.
Can they now decide the proper bearing to walk to meet up in the shortest distance?
I say "No" . . . There is insufficient information.
Suppose the landmark is "between" the two hikers, $\displaystyle A$ and $\displaystyle B.$
A diagram might look like this:
Code:L
o
* * :
* * :
: * * :
: * * β :
:α * * :
: * o B
:* *
A o *
* *
* *
* *
* K M *
The landmark is at $\displaystyle L.$
The hikers are at $\displaystyle A$ and $\displaystyle B.$
$\displaystyle A$'s heading to the landmark is $\displaystyle N \alpha^oE$
$\displaystyle B$'s heading to the landmark is $\displaystyle N \beta^o W$
But $\displaystyle A$ can be anywhere on the line $\displaystyle LK.$
And $\displaystyle B$ can be anywhere on the line $\displaystyle LM.$
There is not enough information for their meeting.
Thanks.
I can calculate the angle ALB (=$\displaystyle \alpha + \beta$), so I thought maybe I missed something.
If I knew the distance between A and B? I think that should do it.
No good. Still not unique.
I would suggest having them walk for five minutes, track the distance travelled, and report the angles again.
