Length of Shortest Side In a Triangle

In summary, the conversation discusses proving that if the sum of the squares of two sides of a triangle is greater than five times the square of the third side, then the third side must be the shortest side. The conversation also includes a proof by assuming the third side is the shortest side.
  • #1

anemone

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If $a,\,b$ and $c$ are the sides of a triangle $ABC$, prove that if $a^2+b^2>5c^2$, then $c$ is the length of the shortest side.
 
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  • #2
Suppose we assume:

\(\displaystyle a>c\implies a^2>c^2\)

\(\displaystyle b>c\implies b^2>c^2\)

These two conditions also imply:

\(\displaystyle ab>c^2\)

Adding the three implications, we obtain:

\(\displaystyle a^2+ab+b^2>3c^2\)

The triangle inequality implies:

\(\displaystyle a^2+2ab+b^2>c^2\)

And, we are given:

\(\displaystyle a^2+b^2>5c^2\)

Adding these last two, there results:

\(\displaystyle a^2+ab+b^2>3c^2\)

As we initially found, this is the result of assuming both \(a\) and \(b\) are greater than \(c\). :)
 
  • #3
Good job, MarkFL! And thanks for participating!

Solution of other:
Assume that $c$ is not the shortest side. Therefore we have $a\le c$, or $a^2 \le c^2$.

Adding $c$ to both sides, and square them yields $(a+c)^2\le 4c^2$.

By the triangle inequality we have $b<a+c$ and hence $b^2<(a+c)^2\le 4c^2$.

We then get $a^2+b^2<5c^2$, which reach to a contradiction and therefore, our assumption is wrong and $c$ is the length of the shortest side of the triangle.
 
  • #4
MarkFL said:
Suppose we assume:

\(\displaystyle a>c\implies a^2>c^2\)

\(\displaystyle b>c\implies b^2>c^2\)

These two conditions also imply:

\(\displaystyle ab>c^2\)

Adding the three implications, we obtain:

\(\displaystyle a^2+ab+b^2>3c^2\)

The triangle inequality implies:

\(\displaystyle a^2+2ab+b^2>c^2\)

And, we are given:

\(\displaystyle a^2+b^2>5c^2\)

Adding these last two, there results:

\(\displaystyle a^2+ab+b^2>3c^2\)

As we initially found, this is the result of assuming both \(a\) and \(b\) are greater than \(c\). :)

Hello Mark

You have assumed c being the shortest side and you have taken the given condition as well. So I am not convinced that
the proof is right. You have taken both the condition and assumption and proved it.

If I have missed something kindly let me know.
 
  • #5
kaliprasad said:
Hello Mark

You have assumed c being the shortest side and you have taken the given condition as well. So I am not convinced that
the proof is right. You have taken both the condition and assumption and proved it.

If I have missed something kindly let me know.

I have assumed that \(c\) is the shortest side and shown how it leads to an implication provided both by the given and the triangle inequality. It seems to me this is sufficient. Is it not?
 
  • #6
MarkFL said:
I have assumed that \(c\) is the shortest side and shown how it leads to an implication provided both by the given and the triangle inequality. It seems to me this is sufficient. Is it not?

But does not prove that this is not true if c is not the shortest side which we need to prove
 
  • #7
kaliprasad said:
But does not prove that this is not true if c is not the shortest side which we need to prove

My apologies. You are absolutely correct. Suppose I arranged things in the following manner:

The triangle inequality implies:

\(\displaystyle a^2+2ab+b^2>c^2\)

And, we are given:

\(\displaystyle a^2+b^2>5c^2\)

Adding these, there results:

\(\displaystyle a^2+ab+b^2>3c^2\)

Now, suppose \(a=3b\) and \(c=2b\):

\(\displaystyle 9b^2+3b^2+b^2>12b^2\)

\(\displaystyle 13b^2>12b^2\)

This is true, even though \(c>b\). I'll try to come up with a sound solution. :)
 

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