Geometry/Trig Problem -- Well Constrained but Difficult

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In summary: I am not sure how to use that information.In summary, the problem at hand involves finding the coordinates of a point R on a line L, given the location of another point F and a line segment w with known length connecting F to R. The slope and y-intercept of the line L are also known. The suggested approach is to use a circle with radius w centered at F and the equation of line L to find the intersection point, which will be the coordinates of R. The method involves finding the intersection point between this circle and another circle with radius w and origin at R, which represents all possible points determined by the angle of the connecting segment w.
  • #1
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The image below should explain the problem and the constraints. Basically, I know the location of one point (F) in 2-D space (Cartesian coordinates). A line segment (w) of known length connects this point to another point (R). The coordinates of R are unknown; however, it is known to lie on a line (L), the slope and y-intercept of which are known (tanα and r). My intuition is that this should be solvable and not all that complicated. The slope of segment w is unknown, but there should only be one angle that allows a segment of that length to exactly reach line L. But I cannot figure it out so I am appealing to the community.
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  • #2
steveZ said:
But I cannot figure it out so I am appealing to the community.
Can you show us what you've tried so far? Is this for schoolwork, or a personal project?

Also, are you familiar with how to convert between rectangular and polar coordinates, and vector representations in each set of coordinates?
 
  • #3
steveZ said:
A line segment (w) of known length connects this point to another point (R). The coordinates of R are unknown; however, it is known to lie on a line (L), the slope and y-intercept of which are known (tanα and r). My intuition is that this should be solvable and not all that complicated. The slope of segment w is unknown, but there should only be one angle that allows a segment of that length to exactly reach line L.
The equation of the line is ##y = \tan(\alpha)x + y_0##, where ##y_0## is the known y-intercept. A circle of radius w is ##(x - F_x)^2 + (y - F_y)^2 = w^2## will intersect the line at two points, one of which is the point you seek. One of these line segments will have a slope that is an acute angle, and the other will have an angle that is greater than 90°.
 
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  • #4
Mark44 said:
The equation of the line is ##y = \tan(\alpha)x + y_0##, where ##y_0## is the known y-intercept. A circle of radius w is ##(x - F_x)^2 + (y - F_y)^2 = w^2## will intersect the line at two points, one of which is the point you seek. One of these line segments will have a slope that is an acute angle, and the other will have an angle that is greater than 90°.
Mark, I thought we were supposed to wait for an OP to show some work before we start giving them hints. Have the rules changed?
 
  • #5
No, the rules haven't changed. It looked to me like he had organized the information he had and what was unknown. I was just pointing him in a direction to reach the answer.
 
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  • #6
berkeman said:
Can you show us what you've tried so far? Is this for schoolwork, or a personal project?

Also, are you familiar with how to convert between rectangular and polar coordinates, and vector representations in each set of coordinates?
It's for a personal project. I have tried many things, including setting equations for the two lines equal to each other, substitutions to reduce the number of unknowns, etc. The closest I've come was to get a horrendously complicated quadratic equation with Rx^2 and Rx.
I am not comfortable with polar coordinates or transformations from rectangular to polar. Does this look like the approach that is needed?
 
  • #7
steveZ said:
Does this look like the approach that is needed?
Did you consider Mark's hinted method?
 
  • #8
phinds said:
Mark, I thought we were supposed to wait for an OP to show some work before we start giving them hints. Have the rules changed?
I have filled many sheets of paper with attempts to solve using algebra and trigonometry but did not want to bore you with the many blind alleys I have navigated. This is not a "homework" problem. What is an OP?
 
  • #9
Mark44 said:
The equation of the line is ##y = \tan(\alpha)x + y_0##, where ##y_0## is the known y-intercept. A circle of radius w is ##(x - F_x)^2 + (y - F_y)^2 = w^2## will intersect the line at two points, one of which is the point you seek. One of these line segments will have a slope that is an acute angle, and the other will have an angle that is greater than 90°.
Thank you very much, Mark. This looks promising and I will follow up on it.
 
  • #10
berkeman said:
Did you consider Mark's hinted method?
I absolutely will. I replied to your post before I saw his.
 
  • #11
steveZ said:
What is an OP?
Original post or original poster
 
  • #12
Of course, thanks. I like the idea of thinking of a circle with radius w with origin at point F that includes point R. In fact, the physical meaning of this problem came from realizing that, if you do it backwards and start with known R, there is a circle of radius w comprising possible points (including F) determined by the angle of the connecting segment w. So I should have thought of that.

So now, as I see it, I have two equations for two unknowns. The circle equation has two unknowns, Rx and Ry, but the equation of line L gives me Ry as a function of Rx, so I can substitute. So I have:
w2 = (Rx - Fx)2 + (y0 + tanαRx - Fy)2
I recognize this as an equation I managed to get to yesterday by a longer route. I pursued it then but got a bit lost in the number of terms that result when you square a binomial and a trinomial (10 terms altogether when you multiply it out, including LHS and RHS).

This is messier than I had hoped, since I want to code this in Matlab, and it is just a small piece of a much larger problem. By rearranging, and separating out the known quantities, I can get this into the quadratic form:
0 = ax2 + bx + c
Then I can use the quadratic formula to get the two possible values. However, the constant c comprises 5 terms, b has 3 and constant a comprises 2 terms, so it is a bit tedious. Is this what you had in mind with the great hint you gave me earlier? Or am I missing a way to simplify the solution? Thanks again.
 
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  • #13
steveZ said:
w2 = (Rx - Fx)2 + (y0 + tanαRx - Fy)2
I recognize this as an equation I managed to get to yesterday by a longer route. I pursued it then but got a bit lost in the number of terms that result when you square a binomial and a trinomial (10 terms altogether when you multiply it out, including LHS and RHS).
But w, Fx, Fy, and ##\alpha## are known, so you could conceivably substitute in the known values for these constants (including ##\tan(\alpha)##). Or not...
steveZ said:
This is messier than I had hoped, since I want to code this in Matlab, and it is just a small piece of a much larger problem. By rearranging, and separating out the known quantities, I can get this into the quadratic form:
0 = ax2 + bx + c
Then I can use the quadratic formula to get the two possible values. However, the constant c comprises 5 terms, b has 3 and constant a comprises 2 terms, so it is a bit tedious. Is this what you had in mind with the great hint you gave me earlier?
Yes, you will end up with a quadratic equation in Rx, and then use the quadratic formula to solve for Rx.
 
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  • #14
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  • #15
Given alpha, you can make a right triangle, which gives you theta. Since you know the hypotenuse length, you should be able to calculate all parameters of the triangle.
 
  • #16
valenumr said:
Given alpha, you can make a right triangle, which gives you theta. Since you know the hypotenuse length, you should be able to calculate all parameters of the triangle.
I wish that were so, but if you look at the diagram again, I think you will see that it is not so easy.
 
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  • #17
  • #18
valenumr said:
Given alpha, you can make a right triangle, which gives you theta. Since you know the hypotenuse length, you should be able to calculate all parameters of the triangle.
Oh silly me... It looks like your problem only specifies one side length and the opposite angle. I'd be interested to see if this is solvable with the other info provided. I got thrown off a bit by the diagram showing the bottom as horizontal.
 
  • #19
steveZ said:
I wish that were so, but if you look at the diagram again, I think you will see that it is not so easy.
What I'm seeing, one could imagine moving the corner of the triangle to the origin, and reducing the problem to solving a triangle with only one known side length and one known angle (which is not solvable). The offset along R seems irrelevant in the grand scheme of things.
 
  • #20
Ok, it makes sense now... You can move the point R pretty much anywhere along the line L and still keep length w constant because theta is unspecified. There is no unique solution to this problem.
 
  • #21
valenumr said:
Ok, it makes sense now... You can move the point R pretty much anywhere along the line L and still keep length w constant because theta is unspecified. There is no unique solution to this problem.
Darnit.. the point f is specified... So it should be solvable.
 
  • #22
The solution is now in focus... Just think of the bottom left of the triangle as the origin and compute the triangle parameters in that coordinate system. For example, you know the base length is Fx - Rx. That gives you three parameters. It still seems messy.
 
  • #23
I can't see what everyone is going on about. The solution is conceptually trivial (@Mark44 gave the whole solution in post #3 and then expounded on it in post #13), it's just algebraically a mess.
 
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  • #24
phinds said:
I can't see what everyone is going on about. The solution is conceptually trivial (@Mark44 gave the whole solution in post #3 and then expounded on it in post #13), it's just algebraically a mess.
You are exactly right. I have done the solution but, because of the messy algebra, it is easy to make mistakes. I have written code to plot my results and also the circle and line whose intersection we are trying to find. You can see in the image below that there is a small error that I am trying to nail down.
1644087668469.png
 
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  • #25
valenumr said:
Given alpha, you can make a right triangle, which gives you theta. Since you know the hypotenuse length, you should be able to calculate all parameters of the triangle.
I don't see how this is a viable approach.
valenumr said:
You can move the point R pretty much anywhere along the line L and still keep length w constant because theta is unspecified.
No, you can't move R along the line. Doing so would affect the length of w.
steveZ said:
I have written code to plot my results and also the circle and line whose intersection we are trying to find. You can see in the image below that there is a small error that I am trying to nail down.
That error looks pretty significant to me, maybe .3 meter or so. I haven't done the calculations, but I would guess that the incorrect results stem from calculation errors.
 
  • #27
Thread closed briefly for Moderation...
 
  • #28
After a brief Mentor discussion, the thread us re-opened...
 
  • #29
Mark44 said:
I don't see how this is a viable approach.

No, you can't move R along the line. Doing so would affect the length of w.

That error looks pretty significant to me, maybe .3 meter or so. I haven't done the calculations, but I would guess that the incorrect results stem from calculation errors.
It was a mistake I made when doing the algebra. A factor of two on one of the terms that went into the quadratic. I caught it on my third try solving for R. Here is the latest result, which seems correct. Thanks again for your help.
1644108060326.png
 

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  • #30
steveZ said:
It was a mistake I made when doing the algebra.
Easy to do with such a complicated equation to solve, but good job on your correction.
 

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