### Quad in square

Posted:

**Fri Sep 21, 2012 9:38 am**On every side of a square with sides measuring one,choose one point. The four points will form a quadrilateral of perimeter $h$. Prove that $2\sqrt{2}\leq h$.

The Official Online Forum of BdMO

https://matholympiad.org.bd/forum/

Page **1** of **1**

Posted: **Fri Sep 21, 2012 9:38 am**

On every side of a square with sides measuring one,choose one point. The four points will form a quadrilateral of perimeter $h$. Prove that $2\sqrt{2}\leq h$.

Posted: **Sat Sep 22, 2012 1:24 pm**

Posted: **Sun Sep 23, 2012 10:30 am**

suppose, we want to get the quad having the shortest perimeter.

what is we can make two of the sides $0$ that will be the shortest. because, straight line is the shortest distance between two points.

so, let the $4$ vertices of the quad be on the $2$ points diagonally opposite vertices of the square.

$h=2 \sqrt2$

so, $2\sqrt2 \ngtr h$

what is we can make two of the sides $0$ that will be the shortest. because, straight line is the shortest distance between two points.

so, let the $4$ vertices of the quad be on the $2$ points diagonally opposite vertices of the square.

$h=2 \sqrt2$

so, $2\sqrt2 \ngtr h$

Posted: **Fri Sep 28, 2012 12:27 pm**

We will use a simple lemma to prove it.

Lemma: In any right angled triangle with the sides $a,b,c$ where $c$ is a hypotenuse.Then $\sqrt{2}c \geq a+b$.(Prove the lemma yourself by $A.M-G.M$.)

Now applying it in $\Delta EBF$, we get, $\sqrt{2}EF \geq EB+BF$.

Likewise,$\sqrt{2}FG \geq FC+CG$,$\sqrt{2}HG \geq HD+GD$,$\sqrt{2}EH \geq AE+AH$.(look the figure above).Summing these inequalities we get what we wanted.

Lemma: In any right angled triangle with the sides $a,b,c$ where $c$ is a hypotenuse.Then $\sqrt{2}c \geq a+b$.(Prove the lemma yourself by $A.M-G.M$.)

Now applying it in $\Delta EBF$, we get, $\sqrt{2}EF \geq EB+BF$.

Likewise,$\sqrt{2}FG \geq FC+CG$,$\sqrt{2}HG \geq HD+GD$,$\sqrt{2}EH \geq AE+AH$.(look the figure above).Summing these inequalities we get what we wanted.

Posted: **Sun Mar 11, 2018 10:07 pm**

Please specify the diagramsakibtanvir wrote: ↑Fri Sep 28, 2012 12:27 pmWe will use a simple lemma to prove it.

Lemma: In any right angled triangle with the sides $a,b,c$ where $c$ is a hypotenuse.Then $\sqrt{2}c \geq a+b$.(Prove the lemma yourself by $A.M-G.M$.)

Now applying it in $\Delta EBF$, we get, $\sqrt{2}EF \geq EB+BF$.

Likewise,$\sqrt{2}FG \geq FC+CG$,$\sqrt{2}HG \geq HD+GD$,$\sqrt{2}EH \geq AE+AH$.(look the figure above).Summing these inequalities we get what we wanted.