Can They Be Equal?

Yes or No?

#Equal

Note by Lew Sterling Jr
6 years, 2 months ago

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Comments

Let the length and breadth of rectangle be xx and yy. Then as per the question we must have

xy=2(x+y)xy=2(x+y)

By AM-GM inequality we have the result

xy(x+y2)2xy \leq \left(\frac{x+y}{2}\right)^2

Thus,

2(x+y)(x+y2)28x+y2(x+y) \leq \left(\frac{x+y}{2}\right)^2 \Rightarrow 8 \leq x+y

So, we have got our first condition. The trivial solution of the above inequality is the case of square where x=y=4x=y=4 because then area = perimeter = 16.

Now we have to check for the non-trivial solutions.

Since, x+y8x+y \geq 8, we can have the sides of the rectangle as x,kxx,k-x for all k8k \geq 8

The perimeter in this case will be equal to 2k2k and the area will be x(kx)x(k-x)

Equating the two, we get

x(kx)=2kx2kx+2k=0x(k-x) = 2k \Rightarrow x^2-kx+2k=0

The solutions of the above quadratic equation are given as

x=k±k(k8)2x = \frac{k \pm \sqrt{k(k-8)}}{2}

The above solution always exists since we have k8k \geq 8

Thus, we find that there can exist infinitely many rectangles with equal perimeter and area. The dimension of such a rectangle will be given by

k2(k+k8) × k2(kk8)\frac{\sqrt{k}}{2}\left(\sqrt{k}+\sqrt{k-8}\right)\ \times\ \frac{\sqrt{k}}{2}\left(\sqrt{k}-\sqrt{k-8}\right)

Kishlaya Jaiswal - 6 years, 2 months ago

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Very nice explanation.

Sandeep Bhardwaj - 6 years, 2 months ago

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Thanks. ¨\ddot \smile

Kishlaya Jaiswal - 6 years, 2 months ago

Given a perimeter, can we always construct a polygon which has area equal to the perimeter?

@Kishlaya Jaiswal

Raghav Vaidyanathan - 6 years, 2 months ago

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For now, I am only able to prove that we cannot always construct a convex regular polygon which has area equal to perimeter.

Suppose we are required to construct a convex regular polygon of area equal to it's perimeter PP. Assume that the polygon is of nn sides (it's obvious that nNn \in N and n3n\geq 3). Then the side length of this polygon will be Pn\frac{P}{n}. Now we simply need to equate the area and perimeter of this polygon to find that

n4(Pn)2cot(πn)=P\frac{n}{4}\left(\frac{P}{n}\right)^2\cot \left(\frac{\pi}{n}\right) = P tan(πn)=P4n\Rightarrow \tan \left(\frac{\pi}{n}\right) = \frac{P}{4n}

Also, we notice that because x<tanxx(0,π2)x < \tan x \quad \forall x\in \left(0,\frac{\pi}{2}\right) so, we get

πn<tan(πn)=P4nP>4π\frac{\pi}{n} < \tan \left(\frac{\pi}{n}\right) = \frac{P}{4n} \Rightarrow P > 4\pi

So basically we need to check if there always exists some positive integer nn such that

4ntan(πn)=P4n\tan\left(\frac{\pi}{n}\right) = P

And by graph we see that there doesn't always exists a positive integer solution nn.

@Raghav Vaidyanathan While I am working on it, I am just curious to know if you have a general solution?

Kishlaya Jaiswal - 6 years, 2 months ago

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I have JEE mains tomorrow man... I'll get back to you on this later. I just asked the question out of nowhere. Don't know solution to it.

Raghav Vaidyanathan - 6 years, 2 months ago

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@Raghav Vaidyanathan Good Luck.

(Btw I am also giving JEE tomorrow.)

Kishlaya Jaiswal - 6 years, 2 months ago

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@Kishlaya Jaiswal Hey! Then good luck to you too!

Raghav Vaidyanathan - 6 years, 2 months ago

While doing this problem, another interesting thing which I found is that there exists only a unique circle which has it's area equal to perimeter (the one which has radius = 22 units)

Kishlaya Jaiswal - 6 years, 2 months ago
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