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Express the length of the hypothenuse as a function of the perimeter.

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ialink
#1
Aug1-11, 02:34 AM
P: 24
1. The problem statement, all variables and given/known data
The altitude perpendicular to the hypothenuse of a right triagle is 12 cm. Express the length of the hypothenuse as a function of the perimeter.

I found a very elegant solution for the problem and even mathematia agrees on that asuming the equations i start with are right. The problem as i will show is checking the solution. When i check it i get an imaginary solution.


2. Relevant equations


After drawing this image I used three equations for start:
  1. p = a + b + h
  2. a^2 + b^2 = h^2
  3. a/12 = h/b or a = 12h/b

p stands for perimeter. a,b,h correspond to the image

3. The attempt at a solution
From eq 1 i derive (p-h)^2 = (a+b)^2 from which the right part can be written as a^2 + b^2 + 2ab

a^2 + b^2 = h^2 (eq 2) and a = 12h/b (eq 3). This results in (p-h)^2 = h^2 + 24h

expanding this results in p^2 + h^2 - 2ph = h^2 + 24h
which equals p^2 = 2ph +24h
which can be written as h(p) = p^2/(2p+24)

Seems very elegant but i trie to check this solution as following
Assume p = 20 h(p) = 25/4 so p-h = a+b = 20-25/4 = 55/4
ab = 12h = 75

So we have a+b=55/4 and ab=75. 2 variables and two equations. Piece of cake so it seems but the resulting equation is b^2 -(55/4)b+75 = 0 wich has only imaginary solutions

I don't know the correct answer, only my work. It's an even problem from review chapter 1 from calculus stewart 5th edition. What am I doing wrong? I spend hours trying to find my mistake.....

I'd very much appreciate help on this one
grt
Ivar
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Hurkyl
#2
Aug1-11, 02:48 AM
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Quote Quote by ialink View Post
Assume p = 20
...
b^2 -(55/4)b+75 = 0 wich has only imaginary solutions
... and since b is a positive real number, our initial assumption must be false. The perimeter cannot be 20.

Standard proof by contradiction -- make an assumption, derive a result that is false, conclude the initial assumption was false.


Assuming all of your work is correct, of course. But I claim once you accept the idea that your initial assumption could be invalid, there's an easy argument that p cannot be 20. (In fact, the easy argument I have in mind shows that p must be bigger than 36)
ialink
#3
Aug1-11, 03:24 AM
P: 24
... and since b is a positive real number, our initial assumption must be false. The perimeter cannot be 20.

Standard proof by contradiction -- make an assumption, derive a result that is false, conclude the initial assumption was false.

Assuming all of your work is correct, of course. But I claim once you accept the idea that your initial assumption could be invalid, there's an easy argument that p cannot be 20. (In fact, the easy argument I have in mind shows that p must be bigger than 36)
Thank you very much. Though you made a little mistake. The minimum is p=57.94. I assume you mean that the minimum is the perimeter from the triangle with angles 90/45/45. I my opinion this results in a exact length of 24 + 24*2^0.5 = 57.94. Indeed if i take 57.95 as perimeter i get Real numbers for answer and indeed the sides are 16.81 and 17.13 acknowledging my my initial assumptions. Taking lower then 57.94 indeed gives imaginary numbers.

Thanks for your help

Hurkyl
#4
Aug1-11, 03:36 AM
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Express the length of the hypothenuse as a function of the perimeter.

Quote Quote by ialink View Post
Though you made a little mistake. The minimum is p=57.94.
57.94 is bigger than 36.

I knew that the minimum was somewhat larger than 36 (in fact, I knew it was larger than 40.97), but I didn't go further than that. The easy argument I had in mind is to make observations a>12, b>12, and h>a.

I thought about suggesting finding the minimum as an exercise -- and I agree with your expectation that the symmetric case is probably where the minimum occurs (especially given the numerical evidence). I bet you can prove it to be the minimum by repeating the argument in your first post, but instead finding b as a function of p... then showing that b is imaginary whenever p is less than your conjectured minimum.
ialink
#5
Aug1-11, 05:01 AM
P: 24
I hope you forgive me my lazyness. Expressing b as a function of p is quitte difficult. Though i was insterested and i used mathematica. It gave me the function b(p). In it is a root. The root of a division to be exact which offcourse can't be less then zero. Solving this gives the proof that p > 24*(1+2^0.5) if we look at the positive domain.

I guess this was what you were meaning, correct?

Hurkyl
#6
Aug1-11, 09:19 AM
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I hope you forgive me my lazyness. Expressing b as a function of p is quitte difficult.
Aww, that's unfortunate. It looks like it does boil down to a quadratic equation as I thought, but I had assumed it wouldn't be particularly difficult to work out.

(p.s. I have nothing against using Mathematica, as long as you aren't using it to skip over things you should be learning. )

(p.p.s. the FullSimplify would do a better job if you gave it the second parameter Assumptions -> p>0)
ialink
#7
Aug1-11, 10:15 AM
P: 24
Well what I should be learning is quitte relative since i'm currently not studying. I have a bachelors degree on civil engeneering and 2 years of university (civil engineering). At this point im preparing for the fact that over a year or so i want to study physics. So i grabbed my calculus book from university and i'm doing selfstudy on it now so it's 100% pure interest i'm doing these excersises. Your point still applies but i only use mathematica to check my answers and as you see i check by hand first but this seemed to me to much of a challenge.

Thanks for your help!

grt
Ivar

PS you were right about the assumption but it still gives a quadratic one
Hurkyl
#8
Aug1-11, 11:19 AM
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Oh, since we're playing with Mathematica... I'm sure there's a way to use Reduce to straight out tell you what the range on p is.

I bet if you tried adding inequalities to say that a, b, and h were >= 0, told it to reduce over the reals... and I forget what you can pass in as the other options, but I think one of them would let you ask for what it can tell you in terms of p.
ialink
#9
Aug1-11, 02:20 PM
P: 24
That's something for tomorrow because it's X-files time!!!!

As the famous filosofer said:"I'll be back"

grtz
ivar
EsX_rAptOr
#10
Aug1-11, 03:38 PM
P: 5
Have you ever heard of the 30-60-90 triangle? If I put two of those together, I end up with your triangle:



Where every side is proportional to each other by those constants.

Having this information, finding the length of the hypotenuse as a function of the perimeter is a walk in the park:

[itex]h = \frac{2P}{3+\sqrt{3}}[/itex]

Where h is the hypotenuse and P is the perimeter.

There is no need to make things more complicated than they really are! Ask if you have any questions.
phinds
#11
Aug1-11, 04:11 PM
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Quote Quote by EsX_rAptOr View Post
Have you ever heard of the 30-60-90 triangle? If I put two of those together, I end up with your triangle:



Where every side is proportional to each other by those constants.

Having this information, finding the length of the hypotenuse as a function of the perimeter is a walk in the park:

[itex]h = \frac{2P}{3+\sqrt{3}}[/itex]

Where h is the hypotenuse and P is the perimeter.

There is no need to make things more complicated than they really are! Ask if you have any questions.
Well, that's a very nice solution but is it a solution to the problem presented? I ask because I can't figure out why it is that you are confident that the triangle IS a 30/60/90. I can't see any such constraint.
ialink
#12
Aug2-11, 01:31 AM
P: 24
Well, that's a very nice solution but is it a solution to the problem presented? I ask because I can't figure out why it is that you are confident that the triangle IS a 30/60/90. I can't see any such constraint.
I agree that it may be the solution for this particular case but not for all cases. The formula must cover all possibilities with one angle of 90 and the other two having a sum of 90.

The simpeler the solution the better but this is too simple in my opnion.
phinds
#13
Aug2-11, 02:03 AM
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Quote Quote by ialink View Post
The simpeler the solution the better but this is too simple in my opnion.
Oh, my complaint isn't about the simplicity ... like you, I LOVE simplicity. It's the lack of generality that makes it be not a solution to the question that was asked and elegance isn't worth anything if it doesn't actually solve the problem.
ialink
#14
Aug2-11, 02:15 AM
P: 24
With too simple i meant the same as you. It doesn't cover the question. I think we can agree on that

grtz
Ivar
EsX_rAptOr
#15
Aug2-11, 10:31 AM
P: 5
With what I said above, I simply wanted to point out the important fact that, regardless of their set-up, the triangles will remain proportional to each other by a certain constant which depends on one of the angles.

But since you apparently want an answer to this problem, I will have to post it:



[itex]h = \frac{\sqrt{tan^{2}\alpha + 1}}{\sqrt{tan^{2}\alpha + 1} + tan\alpha + 1}P[/itex]

It is still a simple solution, in my opinion.
ialink
#16
Aug2-11, 03:50 PM
P: 24
Quote Quote by EsX_rAptOr View Post
But since you apparently want an answer to this problem, I will have to post it:

[itex]h = \frac{\sqrt{tan^{2}\alpha + 1}}{\sqrt{tan^{2}\alpha + 1} + tan\alpha + 1}P[/itex]

It is still a simple solution, in my opinion.
Hurkyl already helped me check my answer wich turned out to be right before you started posting so what makes you think i need an answer? I already proofed that h = [itex]\frac{p^2}{2p+24}[/itex]. Maybe first read what the question is before fireing away.

It's not that i don't appreciate help but i'd more appreciated it if you had read the postings before

grtz
Ivar

ps for P = 24+24*2^2 both formulas return 24 (with angle Pi/4 and x = 12, P = 24+24*2^2) but your answer still is wrong cause alpha should not be in the formula. h as a function of p was asked. Fun though that two completely different approaches give the same result
EsX_rAptOr
#17
Aug2-11, 04:25 PM
P: 5
You are right. I apologize.
ialink
#18
Aug3-11, 12:17 PM
P: 24
that's oke


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