BdMO National 2021 Secondary Problem 11

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Pro_GRMR
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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 $(a+b)$?
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Anindya Biswas
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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 $\measuredangle PM'A=\measuredangle Q'M'B=\measuredangle BM'Q$.
    $\text{Q.E.D.}$
    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.
    \[
    \begin{equation}
    \begin{split}
    PQ+QR+RS+SP & = PQ+QR_1+R_1S_1+S_1P_1\\
    & = PQ+QR_1+R_1S_2+S_2P_2\\
    & = PQ+QR_1+R_1S_2+S_2P_3\\
    &\geq PP_3=AA_2=2AC
    \end{split}
    \end{equation}
    \]
    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.
    \(\text{Q.E.D.}\)
    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,
    \[
    \begin{equation}
    \begin{split}
    \dfrac{1}{\sqrt2}l&=2\sqrt{1+\left(\dfrac{1}{56}\right)^2}\\
    \Longrightarrow l&=\dfrac{\sqrt{6274}}{28}
    \end{split}
    \end{equation}
    \]
    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
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F Nishat
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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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