Putnam and Beyond Problem 113 Solution

[ehrenfest]

[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?

 

[tiny-tim]

Hi ehrenfest!

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.

 

[ehrenfest]

I got it. When n is large enough

$$\frac{c}{a-b}\left( \frac{c}{a} \right)^{n-1} < 1$$

or

$$(a-b) a^{n-1} > c^n$$

Then we have

$$a^n -b^n = (a-b)(a^{n-1}+ \cdots b^{n-1}) > (a-b)a^{n-1} >c^n$$

which is what we want. Thanks. Please confirm that this is correct.

 

[tiny-tim]

hmm … a bit complicated …

For a simpler proof …

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

 

[ehrenfest]

hmm … a bit complicated …

For a simpler proof …

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

That works too.

EDIT: and wow that is much simpler since you avoid dividing into cases

 

[tiny-tim]

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!

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