Solve the equation:
$$\sqrt{x^2+y^2}+ \sqrt{x^2 + \left ( y-4 \right )^2}+ \sqrt{\left ( x-2 \right )^2+ y^2}+ \sqrt{\left ( x-2 \right )^2+\left ( y-4 \right )^2}=4 \sqrt{5}$$
Solution
We may consider $\displaystyle \sqrt{\left ( x-a \right )^2 + \left ( y- b \right )^2}$ as distances between two points. To be more precise let $M(x, y)$ and $A(a, b)$. Now, we are working on the following image.
It is quite obvious that ${\rm AB\Gamma \Delta}$ is a rectangular and its center is ${\rm K}=(1, 2)$. Therefore the initial equation is expressed as:
$$\left ( \rm MA \right )+ \left ( \rm M\Delta \right )+ \left ( \rm MB \right )+ \left ( \rm M\Gamma \right )=4\sqrt{5}$$
However we can easily see that:
$$\left ( \rm MA \right )+ \left ( \rm M\Gamma \right )\geq \left ( \rm A\Gamma \right )=2\sqrt{5} \tag{1}$$
and equality holds when ${\rm M}$ is a point of the segment ${\rm A\Gamma}$.
$$\left ( \rm MB \right )+ \left ( \rm M\Delta \right )\geq \left ( \rm B\Delta \right )= 2\sqrt{5} \tag{2}$$
and equality holds when ${\rm M}$ is a point of the segment ${\rm B\Delta}$.
Adding by parts $(1), \; (2)$ we get that:
$$\left ( \rm MA \right )+ \left ( \rm M\Delta \right )+ \left ( \rm MB \right )+ \left ( \rm M\Gamma \right ) \geq 4\sqrt{5}$$
Equality holds when ${\rm M}$ belongs both at the segment ${\rm A\Gamma}$ and ${\rm B\Delta}$. But there is no other option rather than ${\rm M}$ be ${\rm K}$. Hence the equation has a unique solution which is $(x, y)=(1, 2)$ and the exercise is complete.
$$\sqrt{x^2+y^2}+ \sqrt{x^2 + \left ( y-4 \right )^2}+ \sqrt{\left ( x-2 \right )^2+ y^2}+ \sqrt{\left ( x-2 \right )^2+\left ( y-4 \right )^2}=4 \sqrt{5}$$
Solution
We may consider $\displaystyle \sqrt{\left ( x-a \right )^2 + \left ( y- b \right )^2}$ as distances between two points. To be more precise let $M(x, y)$ and $A(a, b)$. Now, we are working on the following image.
$$\left ( \rm MA \right )+ \left ( \rm M\Delta \right )+ \left ( \rm MB \right )+ \left ( \rm M\Gamma \right )=4\sqrt{5}$$
However we can easily see that:
$$\left ( \rm MA \right )+ \left ( \rm M\Gamma \right )\geq \left ( \rm A\Gamma \right )=2\sqrt{5} \tag{1}$$
and equality holds when ${\rm M}$ is a point of the segment ${\rm A\Gamma}$.
$$\left ( \rm MB \right )+ \left ( \rm M\Delta \right )\geq \left ( \rm B\Delta \right )= 2\sqrt{5} \tag{2}$$
and equality holds when ${\rm M}$ is a point of the segment ${\rm B\Delta}$.
Adding by parts $(1), \; (2)$ we get that:
$$\left ( \rm MA \right )+ \left ( \rm M\Delta \right )+ \left ( \rm MB \right )+ \left ( \rm M\Gamma \right ) \geq 4\sqrt{5}$$
Equality holds when ${\rm M}$ belongs both at the segment ${\rm A\Gamma}$ and ${\rm B\Delta}$. But there is no other option rather than ${\rm M}$ be ${\rm K}$. Hence the equation has a unique solution which is $(x, y)=(1, 2)$ and the exercise is complete.
Hi Tolis. Here is my solution to this problem.
ReplyDeleteUsing Minkowski's Inequality for the left hand side we have
$$\begin{aligned}
4\sqrt{5}=\sqrt{x^2+y^2}&+\sqrt{x^2+(y-4)^2}+\sqrt{(x-2)^2+y^2}+\sqrt{(x-2)^2+(y-4)^2}=\\&=\sqrt{x^2+y^2}+\sqrt{x^2+(4-y)^2}+\sqrt{(2-x)^2+y^2}+\sqrt{(2-x)^2+(4-y)^2}\\&\geq\sqrt{[x+x+(2-x)+(2-x)]^2+[y+(4-y)+y+(4-y)]^2}\\&=\sqrt{16+64}\\&=4\sqrt{5}
\end{aligned}$$
which means we achieve equality.
If we check the equality cases, we see that the $"="$ holds when $x=2-x$ and $y=4-y$ or finally when $x=1,y=2$, which are also the values we are seeking.
This completes the proof.
Thank you George. Nicely done with the Minkowski inequality.
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