Cross Country course plotted on Cartesian plane (5 points)

[cartemnn]

Hi everyone,
I'm trying to solve this case.
The task is to plot a cross country course on a Cartesian plane.
- The total course distance needs to equal 7000m
- There needs to be 4 points where runners change directions
- The start/finish is (0,0)
- The distance of (0,0) to Point A, and Point D (last point) to (0,0) needs to be equal
- No lines can cross each other (no path can intersect)
- No point can be used multiple times
- Only one point can have the same x or y value as another (can include the start/finish point)
Needs to be solved using trig, Pythagoras, distance formula, or any other geometric strategies - of which I have not found a way to use: can only seem to figure using trial and error which is, you know.
Have a go and get back to me if you get a solution! Be really interesting to see how you guys solve it.

 

[Deleted member 4993]

Hi everyone,
I'm trying to solve this case.
The task is to plot a cross country course on a Cartesian plane.
- The total course distance needs to equal 7000m
- There needs to be 4 points where runners change directions
- The start/finish is (0,0)
- The distance of (0,0) to Point A, and Point D (last point) to (0,0) needs to be equal
- No lines can cross each other (no path can intersect)
- No point can be used multiple times
- Only one point can have the same x or y value as another (can include the start/finish point)
Needs to be solved using trig, Pythagoras, distance formula, or any other geometric strategies - of which I have not found a way to use: can only seem to figure using trial and error which is, you know.
Have a go and get back to me if you get a solution! Be really interesting to see how you guys solve it.

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

https://www.freemathhelp.com/forum/threads/read-before-posting.109846/#post-486520

Please share your work/thoughts about this problem.

 

[cartemnn]

Please show us what you have tried and exactly where you are stuck.

My apologies, here’s my solution:

As aforementioned, I have found a solution to the problem, but I would like to find a way to solve it using geometric strategies such as trigonometry, Pythagoras ++.
The solution I have attached was solved using trial and error, with the point C and B being evaluated using the distance formula.

 

[Dr.Peterson]

What you drew can be figured out using the tools mentioned. With geometry you can find the angles in a regular pentagon; with trigonometry you can find the slope of each line; with the distance formula (and solving an equation) you can find points C and E along the first two such lines; and you can find D as the intersection of two more lines.

Of course, there are many other possible solutions, some of which might require a little less work than this.

 

[cartemnn]

I tried another solution, of which I solved using geometry strategies...
However, as I reached the home straight, I noticed the gradient of the origin to point D, and point D to point C were equal, not meeting the requirement "all runners must change directions at each point", in other words, no gradient of two consecutive lines can be the same.
I have attached the course on a Cartesian plane, and the working out I did (excluding C); please excuse the minor errors I have made in the working, it has not been proof read. Please follow the labels of the first image of the course.


*point C's working has not been digitalised yet.

Could you or some person else advise a course of action?
I could change the coordinates of point C, however, that would mean point B would need to be altered accordingly to ensure the distance between the two points satisfies the distance between point D and point A.
Or, point D could be altered, though, that mean all my calculations are false and that the formula's are ineffective...

 

[cartemnn]

What you drew can be figured out using the tools mentioned. With geometry you can find the angles in a regular pentagon; with trigonometry you can find the slope of each line; with the distance formula (and solving an equation) you can find points C and E along the first two such lines; and you can find D as the intersection of two more lines.

Of course, there are many other possible solutions, some of which might require a little less work than this.

Could you elaborate on how you would solve for the regular pentagon?
*the techniques and calculations required to solve for point B, C and D since A is a given.

 

[Dr.Peterson]

I'd missed the fact that E and C have the same x-coordinate, so it doesn't meet the requirements (though the wording is awkward enough that I might be misinterpreting it). But were you not able to find the slope of AC?

I haven't taken the time to read through your last attempt in detail, but it seems reasonable, except that you should have used 30 rather than 60 degrees for point B, shouldn't you?