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Can They Be Equal?

Yes or No?

Note by Llewellyn Sterling
1 year, 11 months ago

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Let the length and breadth of rectangle be $$x$$ and $$y$$. Then as per the question we must have

$xy=2(x+y)$

By AM-GM inequality we have the result

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

Thus,

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

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

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

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

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

Equating the two, we get

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

The solutions of the above quadratic equation are given as

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

The above solution always exists since we have $$k \geq 8$$

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

$\frac{\sqrt{k}}{2}\left(\sqrt{k}+\sqrt{k-8}\right)\ \times\ \frac{\sqrt{k}}{2}\left(\sqrt{k}-\sqrt{k-8}\right)$ · 1 year, 11 months ago

Very nice explanation. · 1 year, 11 months ago

Thanks. $$\ddot \smile$$ · 1 year, 11 months ago

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

@Kishlaya Jaiswal · 1 year, 11 months ago

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 $$P$$. Assume that the polygon is of $$n$$ sides (it's obvious that $$n \in N$$ and $$n\geq 3$$). Then the side length of this polygon will be $$\frac{P}{n}$$. Now we simply need to equate the area and perimeter of this polygon to find that

$\frac{n}{4}\left(\frac{P}{n}\right)^2\cot \left(\frac{\pi}{n}\right) = P$ $\Rightarrow \tan \left(\frac{\pi}{n}\right) = \frac{P}{4n}$

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

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

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

$4n\tan\left(\frac{\pi}{n}\right) = P$

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

@Raghav Vaidyanathan While I am working on it, I am just curious to know if you have a general solution? · 1 year, 11 months ago

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. · 1 year, 11 months ago

Good Luck.

(Btw I am also giving JEE tomorrow.) · 1 year, 11 months ago

Hey! Then good luck to you too! · 1 year, 11 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 = $$2$$ units) · 1 year, 11 months ago