## BdMO National 2021 Secondary Problem 11

Pro_GRMR
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BdMO National 2021 Secondary Problem 11
$$ABCD$$ এমন একটা বর্গ যেন $$A=(0, 0)$$ এবং $$D=(1, 1)$$। $$P\left(\frac{2}{7},\frac{1}{4}\right)$$ বর্গটার ভেতরে একটা বিন্দু। একটা পিঁপড়া $$P$$ বিন্দু থেকে হাঁটা শুরু করে বর্গটার তিনটা বাহু স্পর্শ করার পর আবার $$P$$ বিন্দুতে ফিরে আসে। পিঁপড়াটা দ্বারা সর্বনিম্ন সম্ভাব্য অতিক্রান্ত দূরত্বকে $$\frac{\sqrt{a}}{b}$$ আকারে লেখা যায় যেখানে $$a$$ আর $$b$$ পূর্ণসংখ্যা এবং $$a$$, $$1$$-এর চেয়ে বড় কোনো পূর্ণবর্গ সংখ্যা দিয়ে বিভাজ্য না। $$(a+b)$$-এর মান কত?

Let $ABCD$ be a square such that $A=(0,0)$ and $B=(1,1)$. $P(\frac{2}{7},\frac{1}{4})$ is a point inside the square. An ant starts walking from $P$, touches $3$ sides of the square and comes back to the point $P$. The least possible distance traveled by the ant can be expressed as $\frac{\sqrt{a}}{b}$, where $a$ and $b$ are integers and $a$ not divisible by any square number other than $1$. What is the value of $(a+b)$?
"When you change the way you look at things, the things you look at change." - Max Planck

Anindya Biswas
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Joined:Fri Oct 02, 2020 8:51 pm
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### Solution :

• Lemma 0 : Remember the optics chapter in physics? Let $l$ be a line and $P$ and $Q$ are points on the same side of that line. Let $M$ be a variable point on line $l$. The length $PM+MQ$ is minimum if and only if $\measuredangle (PM,l)=\measuredangle (l,MQ)$.
Proof :
Corollary : Such point $M$ on $l$ for which $PM+MQ$ is minimum, exists uniquely. That means, no other point on $l$ has this property.
• Lemma 1 : Let $ABCD$ be a rectangle. Let $P,Q,R,S$ be points on the sides $AB,BC,CD,DA$ respectively. Here $P$ is a fixed point and $Q,R,S$ are variable points. If the perimeter of $PQRS$ is minimum, then $PQ+QR+RS+SP=2AC$.
Proof :
• Solution to the original problem :
Given, $A\equiv(0,0)$, $D\equiv(1,1)$ and $P\equiv(\frac{2}{7},\frac{1}{4})$
Here, $A$ and $D$ lies on the line $y=x$.
Since $\dfrac{1}{4}<\dfrac{2}{7}$, the square must be on the side of the line $y=x$ where $(1,0)$ is located.
Let's consider the spiral similarity centered at $A(0,0)$ with ratio $\frac{1}{\sqrt{2}}$ and angle $\frac{\pi}{4}$.
After this transformation, the coordinates becomes,
$A\equiv(0,0), B\equiv(1,0), C\equiv(1,1), D\equiv(0,1), P\equiv(\frac{1}{56},\frac{15}{56})$.
Let $E,F,G,H$ be points on the sides $AB,BC,CD,DA$ respectively such that $EG||AD,HF||AB$.
The ant's path must be inscribed in one of these rectangles : $AEGD, EBCG, ABFH, CDHF$.
However, it is easy to check that the rectangle $AEGD$ has the shortest diagonal.
Let $l$ be the length of the ant's path. We can write,
$\begin{split} \dfrac{1}{\sqrt2}l&=2\sqrt{1+\left(\dfrac{1}{56}\right)^2}\\ \Longrightarrow l&=\dfrac{\sqrt{6274}}{28} \end{split}$
So, the answer should be $\boxed{6274+28=6302}$.
Diagram is not drawn to scale
BdMO 2021 National Secondary P11.png (51.23KiB)Viewed 1213 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

F Nishat
Posts:7
Joined:Fri Apr 16, 2021 4:21 pm

### Re: BdMO National 2021 Secondary Problem 11

A different approach.