We are given a rectangle with dimensions \(l \times w\), where \(l \) and \(w\) are positive real numbers and \(l \geq w\). In this note, we will try to find the area of the largest regular polygons that can be drawn completely inside the rectangle.

A shape is said to be drawn completely inside the rectangle if no point of the shape is outside the rectangle.

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What is the area of the largest regular quadrilateral that can be drawn completely inside the rectangle?

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What is the area of the largest regular triangle that can be drawn completely inside the rectangle?

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What is the area of the largest regular hexagon that can be drawn completely inside the rectangle?

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Prove (or disprove): The largest possible regular \(n\)-gon that can be drawn completely in the rectangle is drawn. If \(n\) is even, then two opposite sides of the \(n\)-gon will lie on longer sides \(l\) of the rectangle.

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Prove (or disprove) : There exists a general formula to calculate the area of the largest regular \(n\)-gon that can be drawn completely inside the rectangle in terms of \(l, w, \) and \(n\).

If such a formula exists, find it.

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## Comments

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TopNewestLevel 5: I think formula is \[\frac{n\omega ^2}{4} tan(\frac{\pi}{n})\] for even .I m not sure coz i calculated it orally so i could have missed something conceptually too.

For odd: \[\frac{n \omega ^2 sin(\frac{2 \pi}{n})}{2(1+ cos(\frac{\pi}{n}))^2}\]

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Hmm, something is not right. Your formulae are independent of \(l\). However, the final result does depend on \(l\). You may try the level 2 problem first to see the dependence on \(l\).

Could you elaborate on how you obtained these formulae? Thanks.

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Hey can u tell me how even one will depend on \(l\).

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Let diagnol be D. Hence \(Dcosa=\omega\) Area of square =\(1/2 D^2\) =\(\frac{w^2}{2cos^2a}\).Now cos is a decreasing function and max value of \(a\) is 45 degress coz after that figure will come out of rectangle and at 45 degree opp sides will be rectangle's longer sides.

This can be extended to general .Let there be n sides so angle subtented at centre will be \(\frac{2\pi}{n}\) Let the Longest diagnol be D and it also makes an angle a with vertical .Hence \(Dcosa=w\).Now area of one triangle

=\(\frac{(D/2)^2 Sin(2\pi/n)}{2}\)

So n triangles area= \(\frac{nd^2 Sin(\frac{2\pi}{n})}{8}\)= \(\frac{nw^2 Sin(\frac{2\pi}{n})}{8cos^2(a)}\)

similarily max value of total area will occur at \(a=\pi /n\) because cos is a decreasing function and after \(\pi/n\) figure will come out of rectangle and also 2 opp sides will be on longer sides of rectangle. Simplifying it we get the formula i stated below for even. Note i have taken two points on longer opp sides to generalise and then shown for max that two sides of polygon have to be on two longer sides of rectangle.

can u please tell why it will depend on \(l\)

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Consider a hexagon. Following the same reasoning, the longest diagonal will make an angle \(\pi / 6\) with the vertical. The length of one side of the hexagon will be \(\frac{w}{\sqrt{3}}\). The width of the hexagon will be \(\frac{2w}{\sqrt{3}}\). If \(l > \frac{2w}{\sqrt{3}}\), then area does not depend on \(l\). However, if \(l\) is less than \( \frac{2w}{\sqrt{3}}\), then some of the hexagon will go out of the rectangle, then the area does depend on \(l\).

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