# Putnam triangle problem

1. given a triangle with sides A, B and C, where angle a is equal to 2b ,side A is fixed and angle c is obtuse what is the minimum possible perimetre?

2. The cosine law and the sine law are relevant

3.I tried to mkae function of the square of the preimrtre but the computations became intensive and it was inelegant. Also given the obtuseness constraint the angle b(or was it a) had to be equal or lesser than 30 degrees

I'm looking for something as elegant as possible.

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StatusX
Homework Helper
Look at extreme cases, and how the triangle changes as you vary from one to the other. This should at least give you an idea what the answer is, then you can worry about proving it.

how should i go about proving it? Is the solution computation intensive?

StatusX
Homework Helper
If you don't see a clever way, start doing it the computational way and see how many shortcuts you can make. I don't see a clever way yet either, but if I did, it wouldn't help you if I just told you it.

Hurkyl
Staff Emeritus
Gold Member
1. given a triangle with sides A, B and C, where angle a is equal to 2b ,side A is fixed and angle c is obtuse what is the minimum possible perimetre?

2. The cosine law and the sine law are relevant

3.I tried to mkae function of the square of the preimrtre but the computations became intensive and it was inelegant. Also given the obtuseness constraint the angle b(or was it a) had to be equal or lesser than 30 degrees

I'm looking for something as elegant as possible.
The calculation looks straightforward to me -- what is it that you did, and where did it start to look intractible?

Is there any particular reason why you chose to optimize the perimiter squared, rather than the perimeter?

ok this is what i did. First thing I did was , use the sine law to determine side N. I got what you would expect (sin2a/sina)A=B. Now of course B(sina/sin2a) is constant. I will let sin2a/sina equal u

i then used the law of cosines to determine the side that is opposite to the obtuse angle. I put it into the formula for the perimetre(forget the squares thing) and got something annoyingly complicated.

I decided to backdown as i thought the solution should be more elegant(the other ones i solved were done with virtually no calculations just thought).

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Hurkyl
Staff Emeritus
Gold Member
ok this is what i did. First thing I did was , use the sine law to determine side N. I got what you would expect (sin2a/sina)A=B. Now of course B(sina/sin2a) is constant. I will let sin2a/sina equal u
Side N? I suppose you meant side B.

i then used the law of cosines to determine the side that is opposite to the obtuse angle. I put it into the formula for the perimetre(forget the squares thing) and got something annoyingly complicated.
What expression did you get for side C?

(And is that the only way to compute it?)

What other ways are there to compute it(i'm not really interested in this problem but the fact that i don't know the solution is making me upset).

i have an other idea as to how but its probably gonna turn out ugly anyway.

Hurkyl
Staff Emeritus
Gold Member
What other ways are there to compute it(i'm not really interested in this problem but the fact that i don't know the solution is making me upset).
I did ask two questions, ya know. :tongue:

You know all three angles of the triangle, don't you?

i have an other idea as to how but its probably gonna turn out ugly anyway.
You never know until you try. I meant side B. no we don't know their exact values the idea is the find which angles(value of the paramtre a) gives the lowest perimeter. The only difference is that ti'll be fully sinosoidal and the root will be eliminmated whihc helps greatly but still.

Also how the hell did THIS problem make it onto putnam:

Given a cone of height 3 and radius 1 and cube inscribed inside it what is the side length of the cube?

Anyone?
whoever approved it must have been high(sorry that was rude but still).

Hurkyl
Staff Emeritus
Gold Member
I meant side B. no we don't know their exact values the idea is the find which angles(value of the paramtre a) gives the lowest perimeter.
When I said "know", I meant "have expressed them in terms of the variables in which we are trying to express everything"

The only difference is that ti'll be fully sinosoidal and the root will be eliminmated whihc helps greatly but still.
But what?

Also how the [edited for language] did THIS problem make it onto putnam:

Given a cone of height 3 and radius 1 and cube inscribed inside it what is the side length of the cube?

Anyone?
whoever approved it must have been high(sorry that was rude but still).
Well, IIRC, it was merely an A1 problem. :tongue: By the way, when working the problem, you didn't make that one mistake that's very easy to make, did you?

Yes to the first part. what was the mistake? I actually know nothing about putnam, some of the problems are fun to solve every now and then, but some problems are annoying.

Hurkyl
Staff Emeritus
Gold Member
what was the mistake?
To take a cross section parallel to the faces of the square, and mistakenly drawing this picture:

Code:
      /\
/  \
/----\
/|    |\
/ |    | \
/  |    |  \
/---+----+---\
when in reality it looks more like

Code:
      /\
/  \
/    \
/      \
/  +--+  \
/   |  |   \
/----+--+----\
What you really want to do is draw a cross section along a diagonal.

I drew the corssection along the diagonal like you said i should. I may make logical errors, but i would never make such a gross visual misconception.

i really do need to work on thigns that require more computations. I become very intimidated by them and always back down.

Though i definately know how to solve. Its only a matterof computing things.

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