Question (2 Viewers)

[kurt.physics]

A triangle has its lengths in an arithmatic progression, with difference d. The area of the triangle is t. Find the lenghs and angles of the triangle.

 

[tommykins]

I don't think you can do this question without one sort of numeric value.

Lengths would be a-d, a, a+d

Area = a(a^2-d^2) = a^3 - ad^2 = t

You can't really do anything else to it..

 

[Iruka]

Well, if t=6, then the 3,4,5 right angled triangle fits the criteria mentioned. So if t isn't 6, you can just rescale this triangle until the area is 6.

But of course, we can do something similar with a 3,5,7 triangle (or any of an infinite number of different triangles, as long as we pick three numbers in arithmetic progression that obey the triangle inequality.) So I don't think that you can define a unique triangle using just this information.

So, my suggestion is, let the sides be 1-d, 1, and 1+d. Find the angles (using the cosine rule, I guess) in terms of d, and then use t in Heron's formula to find out by how much you must scale the triangle by. You can construct some sort of a general solution this way.

 

[kurt.physics]

I don't think you can do this question without one sort of numeric value.

Lengths would be a-d, a, a+d

Area = a(a^2-d^2) = a^3 - ad^2 = t

You can't really do anything else to it..

Do you have to get a numerical answer? You could use herons formula to get a quadratic for value a. Then you could use the cosine rule for non-right angled triangles to find the angles.

This is apart of the book "Solving Mathematical Problems: A personal perspective" by Terrence Tao. He didnt mension if you had to givw a numerical answer.

Do you believe that you would need a numerical answer to the question if it was say in the International Mathematical Olympiad?

The question wrote is exactly the same as it is in the book.

 

[kurt.physics]

Well, if t=6, then the 3,4,5 right angled triangle fits the criteria mentioned. So if t isn't 6, you can just rescale this triangle until the area is 6.

But of course, we can do something similar with a 3,5,7 triangle (or any of an infinite number of different triangles, as long as we pick three numbers in arithmetic progression that obey the triangle inequality.)

Maybe you are meant to express the angles and side lengths in terms of t and d?

The question is linked to this post

 

[Iruka]

I think I was editing my last post while you were posting, Kurt.

Like I said, find the angles in terms of d, and then use a scale factor (shall we call it x?) and the fact that the area of the triangle is t to find out what the scale factor should be.

Interesting question.

 

[kurt.physics]

I think I was editing my last post while you were posting, Kurt.

Like I said, find the angles in terms of d, and then use a scale factor (shall we call it x?) and the fact that the area of the triangle is t to find out what the scale factor should be.

Interesting question.

Im not quite sure what your talking about.

 

[Iruka]

Suppose that the triangle has sides x-d, x, x+d. We take x as our unit. Then, in our measuring system, the length of the sides are 1-d*, 1, 1+d*, where d* = d/x. That is, we are introducing a new parameter, d*, to make the algebra easier by combining two of our unknowns into one parameter. (Applied mathematicians do this sometimes when they have large systems of DEs to solve.)

We find the cosines of the angles of the triangle using the cosine rule in terms of d*. Then we use the other piece of information given (about t being the area of the triangle) to find d* in terms of t. We can then find the cosines of the angles (and hence the angles themselves) purely in terms of t, and since we know that x = d/d*, we can also find the lengths of the sides in terms of d and t.

Hope that makes more sense.

 

[hollyy.]

lol at your sig kurt.physics; my maths teacher had the dropping subjects application forms ready on peoples desks after a rather nasty exam

 

[tommykins]

Do you have to get a numerical answer? You could use herons formula to get a quadratic for value a. Then you could use the cosine rule for non-right angled triangles to find the angles.

This is apart of the book "Solving Mathematical Problems: A personal perspective" by Terrence Tao. He didnt mension if you had to givw a numerical answer.

Do you believe that you would need a numerical answer to the question if it was say in the International Mathematical Olympiad?

The question wrote is exactly the same as it is in the book.

Wow, I gave my opinion/what I could do in my head, you don't have to be so patronising.

 

[kurt.physics]

Wow, I gave my opinion/what I could do in my head, you don't have to be so patronising.

Oh no, no. I didnt mean to come across in such a manor. It was a serious question. Do you reckon that if you or me would have been given that question in the IMO, would they give full points for having an algebraic answer?

Sorry for comming across in that way, it wasnt my intentions.

 

[tommykins]

Oh no, no. I didnt mean to come across in such a manor. It was a serious question. Do you reckon that if you or me would have been given that question in the IMO, would they give full points for having an algebraic answer?

Sorry for comming across in that way, it wasnt my intentions.

Of course not, but I'm not IMO level, s o I wouldn't have the mathematical thinking they have.

I'm normally a slow starter with questions that are hard, but I get it out due time if I think of a good idea (as my algebraic skills are really quick imo).

Did you get the answer out?

 

[kurt.physics]

Of course not, but I'm not IMO level, s o I wouldn't have the mathematical thinking they have.

I'm normally a slow starter with questions that are hard, but I get it out due time if I think of a good idea (as my algebraic skills are really quick imo).

Did you get the answer out?

I used IMO because the question is really hard and i would expect it to be in the IMO, and also the IMO is a good example for marks and stuff. But just lets generalise it to say the HSC or what ever.

 

[mangonaley]

Kurt, I'm quite sure Tommykins is right.

The reason why, at any level in fact, you can't be expected to give a numerical solution, is because there isn't one to be given. By looking at the algebraic expressions that tommykins derived, you can see that there would be a million possible numerical values that would fit those equations.

However, be careful because if there is anything at all else give to you in that question, then it is likely that there will be a (or perhaps several) numerical solutions.

p.s. are you sure this is an ext II question? which topic does it fall under?

 

[kurt.physics]

p.s. are you sure this is an ext II question? which topic does it fall under?

No, im not sure. I put it in this forum as the most intelegent students will be frequenting the ext 2 section say compared to the general section.

 

[roadrage75]

i'd be thinking about what we know first....

we know that a>2d. why? because the sum of two sides of a triangle must be greater than the third side.

so using the cos rule, letting e be angle opposite a....

you get cose = (((a+d)^2 + (a-d)^2 - a^2))/(2(a-d)(a+d))
= (a^2 + 2d^2)/((2a^2 -2d^2)


using the fact that a>2d, we get cose<1
and now letting a = nd, when n is very large, we see as n --> infinity, cose --> 1/2

hence 0<e<pi/3


again using cos rule, letting f be the angle opposite a+d

you get cosf = ((a^2 + (a-d)^2 - (a+d)^2)/((2a)(a-d))
= (a^2 - 4ad)/(2a^2 - 2ad)
= (a - 4d)/(2a - 2d), a doesn't =0

at a = 2d, cosf = -1, hence f = pi

as letting a = nd, as a gets very large, cosf --> 1/2

hence pi/3 < f < pi


again using cos rule, getting g be angle opposite a-d you get cosg = ((a^2 + (a+d)^2 - (a-d)^2)/((2a)(a+d))
= (a^2 + 4ad)/(2a^2 + 2ad)
= (a + 4d)/(2a + 2d), a doesn't =0

since a>2d, we can see that 0<cosg<pi/3 (using same methdos as previously

not that this tells us much.... but it tells us that the largest angle, (opposite a+d) is the only angle which is greater than pi/3.

 

[3unitz]

we have sides: l, l+d, l-d
let @ be the angle between l+d and l-d

t = 1/2(l-d)(l+d)sin@
=> sin@ = 2t / (l-d)(l+d) ---------(1)

and from cos rule:
cos@ = [(l+d)^2 + (l-d)^2 - l] / 2(l+d)(l-d) ---------(2)

(1)^2 + (2)^2 = 1:

[2t / (l-d)(l+d)]^2 + {[(l+d)^2 + (l-d)^2 - l^2] / 2(l+d)(l-d)}^2 = 1

[4t / 2(l-d)(l+d)]^2 + [(2l^2 + 2d^2 - l^2) / 2(l+d)(l-d)]^2 = 1

(4t)^2 + (l^2 + 2d^2)^2 = [2(l+d)(l-d)]^2

16t^2 + l^4 + 4l^2d^2 + 4d^4 = 4 (l^2 - d^2)^2

16t^2 + l^4 + 4l^2d^2 + 4d^4 = 4 (l^4 - 2l^2d^2 + d^4)

16t^2 + l^4 + 4l^2d^2 + 4d^4 = 4l^4 - 8l^2d^2 + 4d^4

16t^2 - 3l^4 + 12l^2d^2 = 0

3l^4 - 12l^2d^2 - 16t^2 = 0

let l^2 = u

3u^2 - (12d^2)u - 16t^2 = 0

u = {(12d^2) + sqrt[(12d^2)^2 + 4(3)(16t^2)]} / 6

l = sqrt {{(12d^2) + sqrt[(12d^2)^2 + 4(3)(16t^2)]} / 6}

hence lengths of sides are (in terms of d and t):

sqrt {{(12d^2) + sqrt[(12d^2)^2 + 4(3)(16t^2)]} / 6}
sqrt {{(12d^2) + sqrt[(12d^2)^2 + 4(3)(16t^2)]} / 6} + d
sqrt {{(12d^2) + sqrt[(12d^2)^2 + 4(3)(16t^2)]} / 6} - d

from this you can get the angles by cosine rule

is this the correct form for the answer? =/