Putnam and beyond prob 113

  • Thread starter ehrenfest
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  • #1
ehrenfest
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[SOLVED] putnam and beyond prob 113

Homework Statement


Let a,b,c be side lengths of a triangle with the property that for any positive integer n, the numbers a^n, b^n, and c^n can also be the side lengths of a triangle. Prove that the triangle is necessarily isosceles.


Homework Equations





The Attempt at a Solution


I can do the case where a,b>1 and c<1 using the triangle inequalities a^n+c^n>b^n>a^n-c^n and squeezing but I cannot do the case where a,b,c>1 or a,b,c<1 unfortunately. Can someone give me a hint?
 

Answers and Replies

  • #2
tiny-tim
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Hi ehrenfest! :smile:

Hint: if it's not isoceles, then name the sides so that a > b > c > 1.

You need to find an n such that a^n > b^n + c^n. :smile:
 
  • #3
ehrenfest
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I got it. When n is large enough

[tex]\frac{c}{a-b}\left( \frac{c}{a} \right)^{n-1} < 1[/tex]

or

[tex](a-b) a^{n-1} > c^n[/tex]

Then we have

[tex]a^n -b^n = (a-b)(a^{n-1}+ \cdots b^{n-1}) > (a-b)a^{n-1} >c^n[/tex]

which is what we want. Thanks. Please confirm that this is correct.
 
Last edited:
  • #4
tiny-tim
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hmm … a bit complicated …

For a simpler proof …

Hint: forget c, and just find an n with a^n > 2(b^n). :smile:
 
  • #5
ehrenfest
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hmm … a bit complicated …

For a simpler proof …

Hint: forget c, and just find an n with a^n > 2(b^n). :smile:

That works too.

EDIT: and wow that is much simpler since you avoid dividing into cases
 
Last edited:
  • #6
tiny-tim
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EDIT: and wow that is much simpler since you avoid dividing into cases

Yes … very simple questions often have very simple answers if you look long enough! :rolleyes:

Moral … simplify the problem as much as possible … then solve it! :smile:
 

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