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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≤(2x+y)2
Thus,
2(x+y)≤(2x+y)2⇒8≤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≥8, we can have the sides of the rectangle as x,k−x for all k≥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⇒x2−kx+2k=0
The solutions of the above quadratic equation are given as
x=2k±k(k−8)
The above solution always exists since we have k≥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
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∈N and n≥3). Then the side length of this polygon will be nP. Now we simply need to equate the area and perimeter of this polygon to find that
4n(nP)2cot(nπ)=P⇒tan(nπ)=4nP
Also, we notice that because x<tanx∀x∈(0,2π) so, we get
nπ<tan(nπ)=4nP⇒P>4π
So basically we need to check if there always exists some positive integer n such that
4ntan(nπ)=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?
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)
Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
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to ensure proper formatting.2 \times 3
2^{34}
a_{i-1}
\frac{2}{3}
\sqrt{2}
\sum_{i=1}^3
\sin \theta
\boxed{123}
Comments
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≤(2x+y)2
Thus,
2(x+y)≤(2x+y)2⇒8≤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≥8, we can have the sides of the rectangle as x,k−x for all k≥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⇒x2−kx+2k=0
The solutions of the above quadratic equation are given as
x=2k±k(k−8)
The above solution always exists since we have k≥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
2k(k+k−8) × 2k(k−k−8)
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Very nice explanation.
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Thanks. ⌣¨
Given a perimeter, can we always construct a polygon which has area equal to the perimeter?
@Kishlaya Jaiswal
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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 P. Assume that the polygon is of n sides (it's obvious that n∈N and n≥3). Then the side length of this polygon will be nP. Now we simply need to equate the area and perimeter of this polygon to find that
4n(nP)2cot(nπ)=P ⇒tan(nπ)=4nP
Also, we notice that because x<tanx∀x∈(0,2π) so, we get
nπ<tan(nπ)=4nP⇒P>4π
So basically we need to check if there always exists some positive integer n such that
4ntan(nπ)=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?
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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.
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(Btw I am also giving JEE tomorrow.)
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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)