\(R\) হলো এমন সব আয়তের সেট যাদের কেন্দ্র মূলবিন্দুতে এবং পরিসীমা \(1\) (একটা আয়তের কেন্দ্র হলো তার কর্ণদুটোর ছেদবিন্দু)। \(S\) হলো এমন একটা ক্ষেত্র যার ভিতরে \(R\)-এর সবগুলো আয়তই আছে। \(S\)-এর সর্বনিম্ন সম্ভাব্য ক্ষেত্রফলকে \(\pi a\) আকারে লেখা যায় যেখানে \(a\) একটা বাস্তব সংখ্যা। \(\frac{1}{a}\)-এর মান বের করো।
Let \(R\) be the set of all rectangles centered at the origin and with perimeter \(1\) (the center of a rectangle is the intersection point of its two diagonals). Let \(S\) be a region that contains all of the rectangles in \(R\) (a region \(A\) contains another region \(B\) if \(B\) is completely inside of \(A\)). The minimum possible area of \(S\) has the form \(\pi a\), where \(a\) is a real number. Find \(\frac{1}{a}\).
BdMO National Higher Secondary 2020 P3
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- Mehrab4226
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Re: BdMO National Higher Secondary 2020 P3
The Mathematician does not study math because it is useful; he studies it because he delights in it, and he delights in it because it is beautiful.
-Henri Poincaré
-Henri Poincaré
Re: BdMO National Higher Secondary 2020 P3
But here S doesn't contain all rectangles.
Like it doesn't contain the rectangles of base > 0.25
It only contains all possible squares.
Like it doesn't contain the rectangles of base > 0.25
It only contains all possible squares.
- Mehrab4226
- Posts:230
- Joined:Sat Jan 11, 2020 1:38 pm
- Location:Dhaka, Bangladesh
Re: BdMO National Higher Secondary 2020 P3
Ok we can assume that we have a rectangle with length $\frac{1}{2}$ and breadth $0$. The circle that comprises all those rectangles has an area of $\frac{1}{16} \pi$. So, $\frac{1}{a} = 16$. No other rectangle with the given criteria can be outisde this circle.
The Mathematician does not study math because it is useful; he studies it because he delights in it, and he delights in it because it is beautiful.
-Henri Poincaré
-Henri Poincaré