Let $ABCD$ be a square of the plane $P$. Define the minimum and the maximum the value of the function $f: P \to R$ is given by $f (P) =\frac{PA + PB}{PC + PD}$
Problem
Source: 2012 Serbia EGMO TST p4
Tags: geometry, minimum value, maximum value
TuZo
18.09.2018 10:43
Hint: Let be the coordinates of the vertex of the square: $A(-1,-1), B(1,-1), C(1,1), D(-1,1)$, so we get $f(P)=\frac{\sqrt{a+c}+\sqrt{b+c}}{\sqrt{a+d}+\sqrt{b+d}},\text{ where }a={{(x+1)}^{2}},b={{(x-1)}^{2}},c={{(y+1)}^{2}},d={{(y-1)}^{2}}$
WolfusA
18.09.2018 20:03
A useful hint would be exact value of minimum and maximum. Analytic geometry is too nasty.
TuZo
18.09.2018 20:08
It is NOT a hint for the RESULT, it is only a hint to start to solve the problem! Ultimatelly we must see some algebrical calculs with the radicals, nothing analytic geometry in this. If you have a better idea or hint, please write down here!
WolfusA
18.09.2018 22:13
The problem is I have none idea except this inequality $PA+PB\ge AB$.
Kaskade
18.09.2018 22:37
We will use Ptolemy's inequality. By Ptolemy's inequality we have:
$PA\cdot BC+PB\cdot AC\ge PC\cdot AB$
$PA\cdot BD+PB\cdot AD\ge PD\cdot AB$
We now can use the fact that $AB=BC=CD=DA=x$ and $AC=BD=\sqrt{2}\cdot x$ to get $PA+\sqrt{2}\cdot PB\ge PC$ and $PB+\sqrt{2}\cdot PA\ge PD$ and so adding we get $\left(PA+PB\right)\left(1+\sqrt{2}\right)\ge PC+PD$ or $\frac{PA+PB}{PC+PD}\ge\frac{1}{1+\sqrt{2}}=\sqrt{2}-1$
Now we can simply repeat the same argument to get $\sqrt{2}+1\ge\frac{PA+PB}{PC+PD}\ge\sqrt{2}-1$, with the maximum value being attained when $P$ lies on the arc $CD$ of the circumcircle of $ABCD$ and minimum being attained when $P$ lies on the arc $AB$.