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TopNewestLet 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

\[\frac{\sqrt{k}}{2}\left(\sqrt{k}+\sqrt{k-8}\right)\ \times\ \frac{\sqrt{k}}{2}\left(\sqrt{k}-\sqrt{k-8}\right)\] – Kishlaya Jaiswal · 2 years, 3 months ago

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– Sandeep Bhardwaj · 2 years, 3 months ago

Very nice explanation.Log in to reply

– Kishlaya Jaiswal · 2 years, 3 months ago

Thanks. \(\ddot \smile \)Log in to reply

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

@Kishlaya Jaiswal – Raghav Vaidyanathan · 2 years, 3 months ago

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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? – Kishlaya Jaiswal · 2 years, 3 months ago

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– Raghav Vaidyanathan · 2 years, 3 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.Log in to reply

(Btw I am also giving JEE tomorrow.) – Kishlaya Jaiswal · 2 years, 3 months ago

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– Raghav Vaidyanathan · 2 years, 3 months ago

Hey! Then good luck to you too!Log in to reply

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) – Kishlaya Jaiswal · 2 years, 3 months ago

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