A²+b² > 5c²

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A²+b² > 5c²

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Source: Germany Bundeswettbewerb Mathematik 2006, Day 1, Problem 3

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7579 posts

ZetaX

#1 h Apr 4, 2006, 11:20 AM • 3 Y

Y by Adventure10, Mango247, and 1 other user

Let $a,b,c$ be the sidelengths of a triangle such that $a^2+b^2 > 5c^2$ holds.
Prove that $c$ is the shortest side of the triangle.

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5049 posts

Rust

#2 h Apr 4, 2006, 11:58 AM • 3 Y

Y by Adventure10, Mango247, and 1 other user

Let $b\le c$ , then $a>2c>c+b$ contadition.

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Arne

#3 h Apr 7, 2006, 3:27 PM • 1 Y

Y by Adventure10

Rust wrote:

Let $b\le c$ , then $a > 2c$

Why?

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Shinta

#4 h Apr 7, 2006, 3:38 PM • 2 Y

Y by Adventure10 and 1 other user

Arne wrote:

Rust wrote:

Let $b\le c$ , then $a > 2c$

Why?

Because if $b\le c$ and $a\le 2c$ then $a^2+b^2\le 5c^2$ . Right?

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3660 posts

Arne

#5 h Apr 7, 2006, 3:41 PM • 1 Y

Y by Adventure10

Of course, that was stupid

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lasha

#6 h Sep 2, 2007, 8:56 AM • 2 Y

Y by Adventure10 and 1 other user

Wlog, $b\leq c$ . Then, $c^{2}+a^{2}\geq a^{2}+b^{2}> 5c^{2}$ , or $a > 2c$ . $c\geq b$ follows, that $a > 2c\geq 2b$ . So, $2a = a+a > 2b+2c = 2(b+c)$ , or $a>b+c$ -condradiction. So, $c< b$ and $c < a$ , Q.E.D.

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#7 h Sep 2, 2007, 9:47 AM • 2 Y

Y by Adventure10, Mango247

ZetaX wrote:

Let $a,b,c$ be the sidelengths of a triangle such that $a^{2}+b^{2}> 5c^{2}$ holds.
Prove that $c$ is the shortest side of the triangle.

It is old problem

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