BdMO National 2021 Higher Secondary Problem 2

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Anindya Biswas
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BdMO National 2021 Higher Secondary Problem 2

Unread post by Anindya Biswas » Sun Apr 11, 2021 9:03 pm

\(u\) আর \(v\) হলো দুটো বাস্তব সংখ্যা। নিচের রাশিটার সর্বনিম্ন মানকে যদি \(n\sqrt{n}\) আকারে লেখা যায়, তাহলে \(10n\)-এর মান বের করো। \[\sqrt{u^2+v^2}+\sqrt{(u-1)^2+v^2}+\sqrt{u^2+(v-1)^2}+\sqrt{(u-1)^2+(v-1)^2}\]

Let $u$ and $v$ be real numbers. The minimum value of \[\sqrt{u^2+v^2}+\sqrt{(u-1)^2+v^2}+\sqrt{u^2+(v-1)^2}+\sqrt{(u-1)^2+(v-1)^2}\] can be written as $n\sqrt{n}$. Find the value of $10n$.
"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."
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Pro_GRMR
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Re: BdMO National 2021 Higher Secondary Problem 2

Unread post by Pro_GRMR » Sun Apr 11, 2021 10:12 pm

Anindya Biswas wrote:
Sun Apr 11, 2021 9:03 pm
Let $u$ and $v$ be real numbers. The minimum value of \[\sqrt{u^2+v^2}+\sqrt{(u-1)^2+v^2}+\sqrt{u^2+(v-1)^2}+\sqrt{(u-1)^2+(v-1)^2}\] can be written as $n\sqrt{n}$. Find the value of $10n$.
Can we use any Idea of symmetry here?
"When you change the way you look at things, the things you look at change." - Max Planck

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Anindya Biswas
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Solution :

Unread post by Anindya Biswas » Mon Apr 12, 2021 3:08 pm

$\text{Lemma :}$ \[\sqrt{a_1^2+b_1^2}+\sqrt{a_2^2+b_2^2}\geq\sqrt{\left(a_1+a_2\right)^2+\left(b_1+b_2\right)^2}\] For all real numbers $a_1,b_1,a_2,b_2$.
$\text{Proof :}$
Let's assume $a_1,b_1,a_2,b_2\in\mathbb{R}$
\[
\begin{equation}
\begin{split}
& (a_1b_1+a_2b_2)^2+(a_1b_2-a_2b_1)^2 \geq (a_1b_1+a_2b_2)^2 \\
& \Longrightarrow \left(a_1^2+b_1^2\right)\left(a_2^2+b_2^2\right)\geq (a_1b_1+a_2b_2)^2 \\
& \Longrightarrow \sqrt{(a_1^2+b_1^2)(a_2^2+b_2^2)}\geq \lvert a_1b_1+a_2b_2\rvert \geq a_1b_1+a_2b_2 \\
& \Longrightarrow a_1^2+b_1^2+a_2^2+b_2^2+2\sqrt{(a_1^2+b_1^2)(a_2^2+b_2^2)} \geq a_1^2+2a_1b_1+b_1^2+a_2^2+2a_2b_2+b_2^2 \\
& \Longrightarrow \left(\sqrt{a_1^2+b_1^2}+\sqrt{a_2^2+b_2^2}\right)^2 \geq \left(a_1+a_2\right)^2+\left(b_1+b_2\right)^2 \\
& \Longrightarrow \sqrt{a_1^2+b_1^2}+\sqrt{a_2^2+b_2^2} \geq \sqrt{\left(a_1+a_2\right)^2+\left(b_1+b_2\right)^2} \\ &\text{Q.E.D.}
\end{split}
\end{equation}
\]

According to this lemma,
\[
\begin{equation}
\begin{split}
& \sqrt{u^2+v^2}+\sqrt{(u-1)^2+v^2}+\sqrt{u^2+(v-1)^2}+\sqrt{(u-1)^2+(v-1)^2} \\
& \geq \sqrt{\left(u+\left(1-u\right)+u+\left(1-u\right)\right)^2+\left(v+\left(1-v\right)+v+\left(1-v\right)\right)^2} \\
& = \sqrt{2^2+2^2} \\
& = 2\sqrt{2}
\end{split}
\end{equation}
\]
And this minimum can be achieved when $u=v=\dfrac12$.
So, according to the question, $n=2$,
Therefore, our answer is $\boxed{10n=20}$
Last edited by Anindya Biswas on Mon Apr 12, 2021 3:46 pm, edited 1 time in total.
"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."
John von Neumann

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Anindya Biswas
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Geometric Interpretation :

Unread post by Anindya Biswas » Mon Apr 12, 2021 3:42 pm

Let's assume we start at $(0,0)$
Then we go to $(u,v)$ along a straight line.
Then we go to $(1,1)$ from there along a straight line.
From there, we go to $(1+u,2-v)$.
Then finally, we reach $(2,2)$ from there.
The total length of the path that we took to reach $(2,2)$ from $(0,0)$ is,
\[\sqrt{u^2+v^2}+\sqrt{(1-u)^2+(1-v)^2}+\sqrt{u^2+(1-v)^2}+\sqrt{(1-u)^2+v^2}\]
The minimum distance from the origin $(0,0)$ to $(2,2)$ is,
\[\sqrt{(2-0)^2+(2-0)^2}=2\sqrt{2}\]
Hence, $2\sqrt{2}$ is the minimum value of the above expression.

Here's the visualization :
BdMO 2021 HS 2.png
The blue path we took vs. the minimum path we could take
BdMO 2021 HS 2.png (142.6 KiB) Viewed 409 times
"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."
John von Neumann

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