BdMO National 2021 Secondary Problem 11

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BdMO National 2021 Secondary Problem 11

Unread post by Pro_GRMR » Sun Apr 11, 2021 9:38 pm

\(ABCD\) 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

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Anindya Biswas
Joined:Fri Oct 02, 2020 8:51 pm
Location:Magura, Bangladesh

Solution :

Unread post by Anindya Biswas » Sun Apr 18, 2021 1:19 am

  • 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 :
    Let $Q'$ be the reflection of $Q$ with respect to $l$. Let's assume a variable point $M'\in l$. Let $M$ be the intersection of $PQ$ and $l$.
    $\therefore PM'+M'Q=PM'+M'Q'\geq PQ'=PM+MQ'=PM+MQ$
    So the length $PM'+M'Q$ is minimum if and only if $M'\equiv M$.
    In other words, $P,M',Q'$ must be collinear. But this is true if and only if $\measuredangle PM'A=\measuredangle Q'M'B=\measuredangle BM'Q$.
    Law of reflection.png
    Law of reflection.png (37.99KiB)Viewed 1213 times
    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 :
    Let $A_1P_1BQCR_1D_1$ be the reflection of $APBQCRDS$ with respect to $BC$.
    Let's take similar reflection with respect to line $CD_1$ and line $A_1D_1$ respectively.
    This composition of reflections sends the point $P$ to $P_3$ at the end. Other points are described in the diagram below.
    PQ+QR+RS+SP & = PQ+QR_1+R_1S_1+S_1P_1\\
    & = PQ+QR_1+R_1S_2+S_2P_2\\
    & = PQ+QR_1+R_1S_2+S_2P_3\\
    &\geq PP_3=AA_2=2AC
    The last equality, $PP_3=AA_2$ comes from the fact that $PP_3A_2A$ is a parallelogram.
    And this minimum is achieved when $P,Q,R_1,S_2,P_3$ are collinear.
    The magic rectangle.png
    The magic rectangle.png (178.11KiB)Viewed 1213 times
  • 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,
    \Longrightarrow l&=\dfrac{\sqrt{6274}}{28}
    So, the answer should be $\boxed{6274+28=6302}$.
    BdMO 2021 National Secondary P11.png
    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."
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F Nishat
Joined:Fri Apr 16, 2021 4:21 pm

Re: BdMO National 2021 Secondary Problem 11

Unread post by F Nishat » Mon Apr 19, 2021 2:16 pm

A different approach.
Image*figure not to be scaled*
Let $ABCD$ be the square and $P(\frac{2}{7},\frac{1}{4})$ be the point. We reflect the square and the point $P$ repeatedly as shown in the picture.
The sides of square $ABCD$ are colored pink, blue, green and violet. The segments with same color represent that they are the reflections of a same segment of square $ABCD$.
Let $PE$ be the perpendicular segment from $P$ to $AD$ (not shown in the figure). We see that from $\Delta{PAD}$, $$PE=\frac{1}{28\sqrt{2}}.$$ We now see that we get the shortest path when the ant touches $AD$, $AB$ and $CD$ sides i.e pink, blue and violet segments respectively (Because $P$ is closer to these segments). We see that $PP_1$ touches these $3$ segments, which means $PP_1$ is the shortest path.
Now we draw $\Delta{PP_1P_2}$, which is a right triangle with $PP_2=2\sqrt{2}$ (double of the side length of square $ABCD$) and $P_1P_2=2PE=\frac{1}{14\sqrt{2}}$. So, we have $$PP_1=\sqrt{(2\sqrt{2})^{2}+(\frac{1}{14\sqrt{2}})^2}=\frac{\sqrt{6274}}{28}.$$
Hence, the answer is $6274+28=\boxed{6302}$.

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