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Weird calculus/geometry question

Hi, I'm a senior in calculus. I know this thread list is supposed to be for university students, but there's no calculus listed in the high school section. My tutor referred me to this site because we couldn't solve this before the session ended. The problem actually sounds like a geometry problem, and maybe that's why I'm having a hard time solving it. Anyways, on to the problem. You'll have to see the diagram to make sense of what I say here. I'm sorry, but I had to manufacture it in a paint program.

There is a rectangle (ABCD) measuring 14 x 8.5. Triangle RST is inside of the rectangle, with R between A and B, S between B and C, and T between A and D. Segments AR and RS each have variable length x. I am trying to find an expression for the length of RT in terms of x (then I am to analyze that function, but I don't anticipate any difficulties there).

So far, I've noticed that RBS and ART are right triangles no matter what x is, so I can figure out that the length of RB is 8.5-x, and use Pythagoreas' Theorem (one of the things I did manage to remember from geometry) to find that BS=x^2 - (8.5-x)^2, and that AT=RT^2 - x^2. Unfortunately, I haven't been able to go anywhere except for in circles with this information. I'm so stuck--little help, please?

Re: Weird calculus/geometry question

Got an update. My friend and I went in and asked the teacher for some help at lunch. She said that the square of the length of RT equals (2x^3)/(2x-8.5). She said that showing why that is is still up to us. This helped me get through the other parts of the problem, but I still can't figure out where that expression comes from. I feel like I'm missing something obvious or easy or something. I'm trying to work backwards from the expression, but I can't find any of the measures I figured-out at first in there no matter how I manipulate it. Maybe I'm still going about this all wrong, but I still don't have any better ideas.

Re: Weird calculus/geometry question

I don't see anything in the problem that tells you where T is - other than it's on segment AD. Did you give us all the information in the problem statement?

- Hollywood

Re: Weird calculus/geometry question

Yes, there's not enough information. T can be anything, you didn't constrain T in any way.

Re: Weird calculus/geometry question

There was no restriction on T. I don't know, but maybe there's a misprint or something. I figure that T must be fixed. If it isn't, then the length of RT would depend on two variables. Fixing T solves the lack of information issue (maybe there are other lacks?). I guess I should have asked the teacher today, but I'm pretty convinced fixing T is necessary to solve this. I'm also assuming that S is not fixed, but this was not given in the problem statement either. I guess since I kind of have the length of AT in terms of x, the length of TD could be given as 14-AT so that, in a sense, TD could be expressed in terms of x. But my measure of AT is given in terms of RT and x, so I think this just puts me in another roundabout. Its not really a big deal. I got the other parts (the "calculus parts") of the problem because the result itself was provided, I was just bugged by being stuck on something that didn't look like it should be so difficult. I appreciate you guys taking a look at it, and if anybody gets any ideas I'll check back in, but I'm pretty much ready to write this one off. Sometimes its just not worth the frustration. I'm sure I'll kick myself for not getting it once I see the solution.

Re: Weird calculus/geometry question

If T is fixed, the problem becomes trivially easy.

You should ask the teacher, problem cannot be solved with given information.

Re: Weird calculus/geometry question

I figured it out, but something about that last reply bothers me. You know, I posted my work, my thoughts, everything I had on this problem. I even explicitly said that I could just be missing something easy or obvious. I was only looking for help with my thoughts, not answers. I went further, got a hint from the teacher, it didn't help, so I even updated the thread and explained that I was still having trouble and just couldn't see the solution path. You, "Super" Member, could have pointed out my flaws or oversights (whatever it was that was holding me back from being productive with the exercise) quite easily and been a great help. Instead, you just paraphrase the only relevant reply I got, then throw "trivially easy" at me. If it wasn't worth your time to at least provide a minor suggestion, then why was it worth your time to be of absolutely no assistance? Pumping your post count? Enjoy implying insults? I don't get it, but it was unnecessary, certainly not helpful, and even somewhat belittling, "RICHARD."

I registered here because I was told it was a great site with extremely helpful people who weren't elitists that looked down on even stupid/obvious/trivial questions, as long as you showed you gave the problem an honest shot (check) and didn't show up looking to be handed answers (check). It appears I was misinformed. This may be a great site overall, but you've ruined it for me. I'm inactivating my membership on principle.

Re: Weird calculus/geometry question

Hi powerofcoffee,

My apologies if my use of the word "trivial" may have confused or offended you. Mathematicians and the like tend to have a bad habit of loosely throwing around the word "trivial," even in published literature and textbooks. Sorry if it offended you.

The truth is, RT cannot be found with the given information, because we don't know what AT is. If AT is fixed, then by Pythagorean theorem,

$\displaystyle RT = \sqrt{AT^2 + x^2}$

If AT is not fixed, then we do not have enough information to find RT in terms of just x, unless we have some other constraint that I don't know about (e.g. is triangle RST isosceles? Right?). This is because T is free to move about segment AD without changing anything else.

Hope this clears it up for you,

richard1234

Re: Weird calculus/geometry question

This problem looks an awful lot like the paper folding problem where the length of the fold is to be minimized ... since the OP left in a huff, I guess we'll never know.

Re: Weird calculus/geometry question

Ah, possibly. If so, then we can find AT and RT (in terms of x).

Re: Weird calculus/geometry question

Let $\displaystyle AT = ST = y$, so that $\displaystyle RT^2 = x^2 + y^2$.

If we "slide" segment AB so that B coincides with S, then by Pythagorean theorem,

$\displaystyle 8.5^2 + (y - BS)^2 = y^2 \Rightarrow 8.5^2 + y^2 - 2y BS + BS^2 = y^2$. Cancel $\displaystyle y^2$, solve for y, you should get

$\displaystyle y = \frac{8.5^2 + BS^2}{2BS} \Rightarrow y^2 = \frac{(8.5^2 + BS^2)^2}{4BS^2}$ (I decided to square y since we'll be plugging it in for RT).

Since $\displaystyle BS^2 = x^2 - (8.5 - x)^2 = 17x - 8.5^2$, substituting we obtain

$\displaystyle y^2 = \frac{(17x)^2}{4(17x - 8.5^2)} = \frac{17x^2}{4x - 17}$

Then $\displaystyle x^2 + y^2 = x^2 + \frac{17x^2}{4x - 17} = \frac{4x^3}{4x - 17} = \frac{2x^3}{2x - 8.5}$.

Ahh, now we have arrived at a solution! skeeter, thanks for pointing out the "paper-folding" scenario. @powerofcoffee, hope this solution clears everything up. Next time, make sure to indicate *all* of the information in the problem -- the fact that we didn't know whether RST was a right angle left both hollywood and me confused.

Re: Weird calculus/geometry question

Reminds me of the old story: a professor mentioned in a lecture that some result was trivial, and one of the students raised his hand and asked, "Is that result really trivial?" After an hour and a half of calculations, filling and erasing the chalkboard many times, the professor answered, "Yes, it is indeed trivial."

I don't know if you're still here, powerofcoffee, but you might look at richard1234's post again. He's rejecting the idea that T is some fixed point on AD because he figures you would have been able to calculate it easily. And given an expression for AT and AR=x, you would have been able find RT easily, and without using other information in the problem. That's also part of what he means by "trivial".

Your information is correct - we don't look down on stupid/obvious/trivial questions (your words, not mine). I figure that if you go through the effort of posting your question here, then it's not trivial for you.

Maybe we could all try to be a little more patient with one another....

- Hollywood