[ALEXIS_DH]

so my cousin comes to me with this problem.

given point A (-3,8) and B (2,2). point M lays on the X axis (m,0). where is M for the sum of AM plus MB to be shortest.

i´ve solved making AMB a function of n (n=cos of AM, and 5-n=cos of MB ), taking the derivative of that, making 0 the derivative (for minumum lenght).
i´ve got the answer to be M=(1,0). thru numerical aproximation i´ve validated this answer.

now the tricky part. my cousin is in high school pre-cal, so the problem has to be solved geometrically. no calculus allowed yet.

price? my gratitude and the pride of being the RM hero of the hour...

EDIT for clarity

 

[Bicyclist]

Huh? Do you mean line ABM? In which case the answer would be the X value for the line AM when y=0.

 

[ALEXIS_DH]

Huh? Do you mean line ABM? In which case the answer would be the X value for the line AM when y=0.

nope.
AMB. like this.
the point M is on Y=0 unknown X, somewhere between x=-3 and x=2 wherever AMB ends up the shortest


...............................!
.................A............!
...............................!
...............................!
...............................!....B
_____________M_!_______________

 

[Bicyclist]

How is that a line?

 

[ALEXIS_DH]

How is that a line?

oh well, my bad, not a exactly a line.
but the sum of lines AM and MB.

 

[amateur]

Uh, find the slope of the hypotenuse then solve using y=mx+b

 

[ALEXIS_DH]

Uh, find the slope of the hypotenuse then solve using y=mx+b

triangle AMB is not rectangular.

 

[Bicyclist]

Hrm, tricky w/ geometry.

I would write an equation that describes the length of one of the segments as a function of the other.

Basically I'd approach it like one of those problems where, for instance, you have 30 feet of fence and you want to make the biggest yard possible or whatever.

But I'm too lazy to actually do it.

 

[Bicyclist]

Uh, find the slope of the hypotenuse then solve using y=mx+b

Well you don't know where M is so you don't have a definite triangle do you? Although you maybe could nail it down and use the law of cosines, along with the distance formula. But even then you would be guessing as to where M would minimize the distance.

 

[amateur]

triangle AMB is not rectangular.

It can be if you make it. vertical line from A to the x-axis. It isn't triangle ABM, ABM would be the hypotenuse of the triangle you draw.

 

[Bicyclist]

No, that wouldn't make the shortest distance. That would make ABM a line. Segments AM+MB have to have a minimum value, not the whole line of AM.

 

[amateur]

Well you don't know where M is so you don't have a definite triangle do you? Although you maybe could nail it down and use the law of cosines, along with the distance formula. But even then you would be guessing as to where M would minimize the distance.

Use similar triangles.

 

[amateur]

Sonofa...I really should've read the problem closer.

 

[Bicyclist]

OK, I set up the equation root(64 + (3-m)^2) + root(4 + (2-m)^2) = AM+MB.

 

[Bicyclist]

my cousin is in high school pre-cal, so the problem has to be solved geometrically. no calculus allowed yet.

Hey, wait a minute, I'm in precalc and we use integrals and derivatives.

 

[COmtbiker12]

Seems like the simplest way would be to guess and check with x values between -3 and 2 and figure out the distance of AM and BM then figure out which one adds up to the least. Or at least that's what seems to be a fairly easy way to do it, maybe... I don't feel like applying myself too well this weekend.

And like Bicyclist said, I'm in precalc as well and we definitely use derivatives and what not.

 

[ALEXIS_DH]

OK, I set up the equation root(64 + (3-m)^2) + root(4 + (2-m)^2) = AM+MB.

yup, thats similar to what i did.
i made an equation AM+MB=(x^2+4)^.5+((5-x)^2+64)^.5
where x is the distance to the left from (2.0).

derivative of that, and solving for dx=0, gives x=1.
that is the correct answer giving the least possible distance AM+MB=5*5^.5.

but i had to dx.
no dx allowed, according to my cousin. they havent even seen limits yet.

 

[johnbryanpeters]

Why don't you let your esteemed cousin do their own work?

 

[ALEXIS_DH]

Why don't you let your esteemed cousin do their own work?

he is long gone. am at my office right now.

is just i´ve been hurt in my nerd pride by not being able to solve a high school math problem..... and i wanted to know how are you supposed to do it.

 

[johnbryanpeters]

I think you plot AM, ((Ay)*2 + (m - Ax)*2)*(-2), as a function of m, then similarly plot MB, ((By)*2 + (Bx - m)*2)*(-2), and I bet they intersect at m = 1.

Please forgive any errors in the above.

 

[-dustin]

Hey, wait a minute, I'm in precalc and we use integrals and derivatives.

damn. i didn't touch that **** until first semester calc. maybe that's why it took 2 semesters for it to make sense.

 

[ito]

triangle AMB is not rectangular.

Brilliant deduction!

Seeing how it has three sides, how could it be?

The Ito

 

[johnbryanpeters]

Ummm... he's saying it's not a right triangle, I think...

 

[Toshi]

I think you plot AM, ((Ay)*2 + (m - Ax)*2)*(-2), as a function of m, then similarly plot MB, ((By)*2 + (Bx - m)*2)*(-2), and I bet they intersect at m = 1.

Please forgive any errors in the above.

if by "*2" you mean "^2" and by "*(-2)" you mean "^(1/2)" then you've described the lengths of segments AM and BM, sure. but if you plot their intersection you'll find the point where both segments are of equal length. this is not the solution.

of course, i might be just totally misunderstanding your notation, and indeed the concept you're trying to convey, fwiw.

 

[Toshi]

and despite staring at this for the better part of an hour i can't come up with an elegant geometric solution.

 

[ALEXIS_DH]

and despite staring at this for the better part of an hour i can't come up with an elegant geometric solution.

tricky isnt it?
i took me a good 15, ti89 enhanced, minutes to come up with an answer...
for the first 10 minutes i was kinda lost, and i thought about cheating and going for the good ol´autocad to find a numerical aproximation.

then he said "no derivatives. geometry only". that made my day.
i was scratching my head hard for like 1 hour, and had it in on my mind a couple hours after he left....

 

[Bicyclist]

I think I win with my above equation. If you plot that on a graphing calc you can determine the minimun value in a predetermined domain and there's your answer.

 

[Bicyclist]

damn. i didn't touch that **** until first semester calc. maybe that's why it took 2 semesters for it to make sense.

Really? I thought that was the point of precalc: to introduce calculus ideas. Otherwise it's just more algebra, trig, and geometry.

 

[Toshi]

I think I win with my above equation. If you plot that on a graphing calc you can determine the minimun value in a predetermined domain and there's your answer.

your solution neither is elegant nor involves geometry beyond the basic definition of segment length in cartesian coordinates, plus it seems to imply an overreliance on technology. be glad i'm not your math teacher ink:

 

[Bicyclist]

Touche, salesman.

Do you have a better solution? And personally I'm not about elegance (to a point) in math, I'm about quickness and simplicity. No reason in using out-dated methods to solve a problem when there's an easier way to think of it.

 

[Toshi]

thanks to a helpful comment from one of my friends working backwards from the solution, who noted that the slopes of AM and BM are "opposite" (in that they're orthogonal -- update: i'm dumb! they're not!) i got started on a train of thought that just led me to the elegant solution i knew was out there:

let theta be the angle opposite side AB, and let AB, AM, BM below represent the length of the segments. the law of cosines thus states:

(AB)^2 = (AM)^2 + (BM)^2 - 2(AM)(BM)cos(theta)

note that when theta = pi/2 the cos(theta) term equals zero, and we get our familiar pythagorean theorem. also note that AM and BM are positive, and that cos(theta) is positive for any theta other than pi/2 mod pi = 0. thus we can see that the whole last term (-2(AM)(BM)cos(theta)) will be negative for all theta not equal to pi/2.

this last observation implies that, for AB constant, AM and BM will be smallest when theta equals pi/2. in other words, setting m such that point M is at the right angle of a right triangle is the solution.

qed.



updated to reflect that this solution is wrong since they're not orthogonal!

 

[Changleen]

According to my housemate, your assumption about AM and BM being orthogonal is incorrect. At least when checking using your solution of (1,0) the gradients of each line AM and BM show that they aren't perpendicular. The gradient of AM at this point is -2 and the gradient of BM is 2, hence not following the gradient of a perpendicular is -1/m thing.

 

[Toshi]

yeah, i just had that pointed out (discussing this problem on my email list). oops! when computing those slopes i flipped a -2 to a -1/2 in my haste to make my proposed solution work.

however, another of my friends came up with a valid, and equally elegant solution. note that the solution isn't mine.

Let C=(2,-2). Clearly len(MB)=len(MC) for any choice of M=(m,0), so we'll minimize len(AM)+len(MC). But this is clearly minimized when A, M, C are colinear (i.e. AMC is a straight line). AC intersects the x-axis at x=1, so choosing m=1 gives the optimal answer.

 

[Changleen]

Both my wife and my housemate, who are both maths grads are beavering away right now... It's kinda cute... :love:

 

[Toshi]

Both my wife and my housemate, who are both maths grads are beavering away right now... It's kinda cute... :love:

i miss this feeling of actually thinking... memorization just doesn't do it for me in the same way.

 

[chicodude]

You guys are ****ing nerds. carryyy on

 

[johnbryanpeters]

if by "*2" you mean "^2" and by "*(-2)" you mean "^(1/2)" then you've described the lengths of segments AM and BM, sure. but if you plot their intersection you'll find the point where both segments are of equal length. this is not the solution.

of course, i might be just totally misunderstanding your notation, and indeed the concept you're trying to convey, fwiw.

Yeah, you got the incorrect notation right. I meant

((Ay)^2 + (m - Ax)^2)^(-2)
and
((By)^2 + (Bx - m)^2)^(-2)

The curves don't intersect at (m, 0), they intersect at (m, <I don't care>), giving the correct value for m. BTW, I have not plotted this, just thought about it.

 

[ALEXIS_DH]

yeah, i just had that pointed out (discussing this problem on my email list). oops! when computing those slopes i flipped a -2 to a -1/2 in my haste to make my proposed solution work.

however, another of my friends came up with a valid, and equally elegant solution. note that the solution isn't mine.

Let C=(2,-2). Clearly len(MB)=len(MC) for any choice of M=(m,0), so we'll minimize len(AM)+len(MC). But this is clearly minimized when A, M, C are colinear (i.e. AMC is a straight line). AC intersects the x-axis at x=1, so choosing m=1 gives the optimal answer.

beautiful. just beautiful.:love:

dx of (x^2+4)^.5+((5-x)^2+64)^.5 is just not the kind of derivative you´d expect pre-cal would teach to solve, definately not this early on the school year.

but i never thought it was going to be so beautifully simple. i was expecting some obscure trig identity or geometric relationship.

 

[Toshi]

ah, elegant proofs. if you really liked that, then you should get this book:

Proofs from The Book

it's a collection of the most elegant proofs that the mathematician Paul Erdos (sorry, missing accents) came across, not necessarily came up with himself, over his career.