BdMO National Higher Secondary 2020 P3

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Mursalin
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BdMO National Higher Secondary 2020 P3

Unread post by Mursalin » Mon Feb 08, 2021 4:08 pm

\(R\) 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 \(\frac{1}{a}\).
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Mehrab4226
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Re: BdMO National Higher Secondary 2020 P3

Unread post by Mehrab4226 » Tue Feb 09, 2021 7:39 pm

Let, the sides of the rectangle be $a,b$
Given,
$2(a+b) = 1$
Or,$a+b = \frac{1}{2}$
Or,$\frac{a+b}{2} = \frac{1}{4}$

Now, applying AM-GM we get,
$\frac{a+b}{2} \geq \sqrt{ab}$

Or,$\frac{1}{4} \geq \sqrt{ab}$

Or,$\frac{1}{16} \geq ab$

Now the maximum area of the rectangle is $\frac{1}{16}$ units.
So, our required area must be greater than or equal to this.

Now let us take a square(Squares are also considered rectangles) with an area $\frac{1}{16}$ units. Clearly, the sides are $\frac{1}{4}$ and the perimeter is $1$. So our square is a rectangle we are talking about but also with the maximum area. We claim that the circumcircle of the square is the required area.

At first, let us find the area of the circle. Very simple as the diagonal is the diameter of the circumcircle. So, the radius of the circle is $\frac{1}{4\sqrt{2}}$
Thus the area is equal to $\frac{1}{32} \times \pi$
So, $\frac{1}{a} = 32$

Now for why that is the minimum area. Let us assume that the circle is not our required minimum area. Then it must be smaller than the circle. Then let us take the area and make it shaded. So there will be a part inside the circle without the shade. We can easily rotate our square to the unshaded part and find a valid rectangle(the square itself) not in the region. A contradiction! $\square$


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

Naeem588
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Re: BdMO National Higher Secondary 2020 P3

Unread post by Naeem588 » Tue Apr 06, 2021 12:33 pm

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.

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Mehrab4226
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Re: BdMO National Higher Secondary 2020 P3

Unread post by Mehrab4226 » Wed Oct 13, 2021 10:21 am

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

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