Is this geometry problem solvable? EDIT: Answer is no

[pennylane]

I've never asked for help with "homework", but I'm stuck on this issue I've come across.

It's a triangle problem, summarized in this picture

A triangle is made up of 3 sides of length (a), (c), and (c + b). In the picture, x and y are just vertices.

If a and b are known, can c be found? I thought so because it seems like only one triangle can be made that has those dimensions. I've tried using the law of cosines liberally, but I haven't been able to simplify it. I'm not sure if I'm getting enough independent equations out of it.

I've tried using MATLAB to symbolically solve it, but it hasn't helped.

 

[Spoooon]

If a and b are known, can't you find the length of the line drawn from "X" to "b" and then use that to find c? I don't remember much by way of trig and stuff, so I don't know if that's actually possible.

 

[pennylane]

I don't think so. You normally need 3 pieces of information to solve a triangle right? Any combination of 3 sides and lengths.

You only have two pieces with a and b for the smaller triangle.

 

[smack Down]

I don't think in general there is a solution, for some special values of a,b there may be 1 or zero solutions but for most case C can be any length. depending on what is required prove it false for the special case of b = 0.

 

[homercles337]

Yea, you can. If you find the angle between a & b and you have the vertices x & y, you can find the intersection between two lines (the intersection will be at the upper left). Actually, forget the angle between a & b. b is a line, find the angle at a & c, such that the end segment of b to intersection is equal to the length of a to intersection.

 

[b0mbrman]

This is where I'm at now...probably doesn't help much

http://pics.bbzzdd.com/users/b0mbrman/b0triangle.PNG

Can someone more clever than me grab the torch?

 

[DrPizza]

No. It's not solvable. At least not for a specific value of c. You could solve for a range of possible values (I'll get to that later.)

Here's how to construct triangles with different values of C and the same values for a and b:

For simplicity, let's say that a=7 and b=5.
Pick a value for c, sayyyy 12.

Can you construct a triangle with sides a=7, c=12, and the third side = 17 (equal in length to the sum of b + c)?

Okay, how about for a=7 and b=5, then can c = 13?
That'd give you a triangle with the dimensions 7,13,18
Again, no problem making such a triangle.

Of course, you can specify an inequality for a solution, finding the range of values for c in which the triangle would satisfy the triangle inequality theorem. i.e. if a = 20 and b = 15, then c couldn't equal 1. That would lead to a triangle with sides of 20, 16, and 1.

 

[sao123]

about the only thing you can solve would be a lower bound... which would give you the least possible value for C... but there is not an easy way to solve for a Max.

c > (a-b) / 2

 

[Fenixgoon]

law of cosines?

make a right triangle, use whatever you want to find the angles. then use law of cosines.

 

[DrPizza]

Incidentally, if you write the triangle inequality for each of the three cases in terms of a,b, and c, you can figure out restrictions on a and b (i.e. there is no solution for some values of a and b). Of course, the case where the side length = c is the longest side doesn't make sense; c cannot be greater than a + b + c.
And, if side b is greater than a, then b+c is the longest side of the larger triangle. This leads to a + c > b + c; which reduces to a > b which is a contradiction. Therefore, b cannot be greater than a.

If a = b, then the three sides are b, b+c, and c; (or a, a+c, c) In this case, no triangle is formed; the 3 points are collinear.

Thus, a must be greater than b for any solution to exist.

In order to form a triangle (satisfy the triangle inequality), 2c must be greater than the difference between a and b. i.e. if a is 9 and b is 5, then any value of c which is less than or equal to 2 will not be adequate to form a triangle. Thus, c > (a-b)/2

Is there an upper limit to c? Obviously, as c gets larger, the side b+c becomes the largest side of the triangle. Triangle inequality in this case: a + c > b + c, which is simply a > b which we already know. Thus, there is no upper limit to c, and the lower limit is (a-b)/2

edit: i.e. let a and b, for the sake of argument, be equal to 7 and 5 respectively. Then, I can let c = 1 zillion. The sides of the triangle are: 7, 1 zillion, and (1 zillion + 5). The sum of the two shortest sides is obviously larger than the longest side (by 2). No matter how large a value of c I choose, after a certain point (which makes b+c the longest side), then the sum of the two shortest sides of the triangle will always exceed the longest side of the triangle by the difference between a and b.

 

[mjrpes3]

Originally posted by: sao123but there is not an easy way to solve for a Max.

max has no limit, no?

EDIT: nvm, just saw DrPizza's post.

 

[Cattlegod]

it could be anything. you need the angle of the C vertex

 

[Itchrelief]

I'm no math whiz, but maybe you need to work with heron's formula and try to do something? I'm assuming that since you never mentioned x and y that we can consider that they are not known. Since we only know the lengths of the sides of the triangle, heron's formula seems to be something that might (?) work.

a page with the formula and some crap i don't understand

you know the perimeter = a + b + 2c = the stuff in the formula (all this crap is wrong, disregard)

one side = c
second = a
third = b + c

plug in. i'm no math whiz but since you're the one doing this and you have matlab cracking, you might be. Like DrPizza said, you might not get one answer but a whole range of them. Again, I'm no good at solving for that, but here's to hoping you are. good luck.

edit: eh, screw that. heron's formula is for area. doesn't help you at all.

 

[Kelemvor]

You weren't given the angle of the AB connection or anything else? If not, you could solve it with variables in the answer but not to get a specific value. You'd need to know an angle of something in order to get the last side.

 

[pennylane]

Yeah, I'm pretty sure the problem isn't solvable analytically. Thanks for all the help guys.

I might try to write some script that tries to find the unknown incrementally, checking to see whether the numbers given actually form a legit triangle or not.

 

[DrPizza]

Originally posted by: fanerman91
Yeah, I'm pretty sure the problem isn't solvable analytically. Thanks for all the help guys.

I might try to write some script that tries to find the unknown incrementally, checking to see whether the numbers given actually form a legit triangle or not.

Did you even read my second post? I spelled it out for you: c> (a-b)/2. And, the conditions on a and b are that a must be greater than b. *sigh*

 

[Alone]

Here.

 

[pennylane]

Edit: I suck at math.

 

[Born2bwire]

Originally posted by: fanerman91

Originally posted by: DrPizza

Originally posted by: fanerman91
Yeah, I'm pretty sure the problem isn't solvable analytically. Thanks for all the help guys.

I might try to write some script that tries to find the unknown incrementally, checking to see whether the numbers given actually form a legit triangle or not.

Did you even read my second post? I spelled it out for you: c> (a-b)/2. And, the conditions on a and b are that a must be greater than b.

Yeah, but that just gives a range right? If I coded it, I could get an actual number, based on those inequalities.

He's telling you there is no unique solution.

 

[BrownTown]

Originally posted by: DrPizza
Did you even read my second post? I spelled it out for you: c> (a-b)/2. And, the conditions on a and b are that a must be greater than b. *sigh*

That was the answer I got after looking at the problem except simply saying ABS(a-b) so you don't have to say that a>b for it to be correct.

EDIT: If you don't want to do math its pretty trivial to just visualize the correct solution. The only mathematically interesting solutions would be at 0 90 and 180 degrees. We can see that and 0 the minimum solution is c= abs(a-b)/2, at 90 it is c= sqrt(a^2+b^2)/2 and at 180 it is c= a+b/2. It can also be see than for any of these solutions you can simply "pull" the point c along the angles 180 45 and 90 respectively and still satisfy the requirements.

 

[pennylane]

Originally posted by: Born2bwire

Originally posted by: fanerman91

Originally posted by: DrPizza

Originally posted by: fanerman91
Yeah, I'm pretty sure the problem isn't solvable analytically. Thanks for all the help guys.

I might try to write some script that tries to find the unknown incrementally, checking to see whether the numbers given actually form a legit triangle or not.

Did you even read my second post? I spelled it out for you: c> (a-b)/2. And, the conditions on a and b are that a must be greater than b.

Yeah, but that just gives a range right? If I coded it, I could get an actual number, based on those inequalities.

He's telling you there is no unique solution.

Oh, I misread the post. Sorry, I'm having a headache right now and got mixed up. I'm not sure what I was thinking.

 

[mjrpes3]

Originally posted by: BrownTown

Originally posted by: DrPizza
Did you even read my second post? I spelled it out for you: c> (a-b)/2. And, the conditions on a and b are that a must be greater than b. *sigh*

That was the answer I got after looking at the problem except simply saying ABS(a-b) so you don't have to say that a>b for it to be correct.

what are you saying? based on that diagram, a has to be greater than b.

 

[mjrpes3]

Originally posted by: fanerman91

Oh, I misread the post. Sorry, I'm having a headache right now and got mixed up. I'm not sure what I was thinking.

if you had actual numbers you could get an actual range. but there is nothing more you can do here that hasn't already been explained by DrPizza.

 

[habib89]

no, you will not be able to find a specific number answer for this because you don't have any numbers for the known values. you can use the law of sines... http://www.ies.co.jp/math/java/trig/seigen/seigen.html

 

[BrownTown]

Originally posted by: mjrpes3
what are you saying? based on that diagram, a has to be greater than b.

no