Geometry Challenge: Drawing Polygons in a Rectangle

We are given a rectangle with dimensions l×wl \times w, where ll and ww are positive real numbers and lwl \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.

Level 1 \color{gold} {\text{Level 1}}
What is the area of the largest regular quadrilateral that can be drawn completely inside the rectangle?

Level 2 \color{#EC7300} {\text{Level 2}}
What is the area of the largest regular triangle that can be drawn completely inside the rectangle?

Level 3 \color{#D61F06} {\text{Level 3}}
What is the area of the largest regular hexagon that can be drawn completely inside the rectangle?

Level 4 \color{#69047E}{\text{Level 4}}
Prove (or disprove): The largest possible regular nn-gon that can be drawn completely in the rectangle is drawn. If nn is even, then two opposite sides of the nn-gon will lie on longer sides ll of the rectangle.

Level 5 \color{#3D99F6}{ \text{Level 5}}
Prove (or disprove) : There exists a general formula to calculate the area of the largest regular nn-gon that can be drawn completely inside the rectangle in terms of l,w,l, w, and nn.

If such a formula exists, find it.

Note by Pranshu Gaba
4 years, 2 months ago

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1 vote

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Level 5: I think formula is nω24tan(πn)\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: nω2sin(2πn)2(1+cos(πn))2\frac{n \omega ^2 sin(\frac{2 \pi}{n})}{2(1+ cos(\frac{\pi}{n}))^2}

Gautam Sharma - 4 years, 2 months ago

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

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

Pranshu Gaba - 4 years, 2 months ago

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Hey can u tell me how even one will depend on ll.

Gautam Sharma - 4 years, 2 months ago

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Let diagnol be D. Hence Dcosa=ωDcosa=\omega Area of square =1/2D21/2 D^2 =w22cos2a\frac{w^2}{2cos^2a}.Now cos is a decreasing function and max value of aa 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 2πn\frac{2\pi}{n} Let the Longest diagnol be D and it also makes an angle a with vertical .Hence Dcosa=wDcosa=w.Now area of one triangle

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

So n triangles area= nd2Sin(2πn)8\frac{nd^2 Sin(\frac{2\pi}{n})}{8}= nw2Sin(2πn)8cos2(a)\frac{nw^2 Sin(\frac{2\pi}{n})}{8cos^2(a)}

similarily max value of total area will occur at a=π/na=\pi /n because cos is a decreasing function and after π/n\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 ll

Gautam Sharma - 4 years, 2 months ago

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@Gautam Sharma Your reasoning is correct for a square. In a square, aa should be 4545^{\circ}. Area of square will be w22cos2a=w2\frac{w^2}{ 2 \cos^{2}a} = w^2, and it will not depend on ll. However, this reasoning does not extend for other nn.

Consider a hexagon. Following the same reasoning, the longest diagonal will make an angle π/6\pi / 6 with the vertical. The length of one side of the hexagon will be w3\frac{w}{\sqrt{3}}. The width of the hexagon will be 2w3\frac{2w}{\sqrt{3}}. If l>2w3l > \frac{2w}{\sqrt{3}}, then area does not depend on ll. However, if ll is less than 2w3 \frac{2w}{\sqrt{3}}, then some of the hexagon will go out of the rectangle, then the area does depend on ll.

Pranshu Gaba - 4 years, 2 months ago

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@Pranshu Gaba OOps!correct I think i should sleep .I m sickk

Gautam Sharma - 4 years, 2 months ago

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@Gautam Sharma Nice efforts though!

Nihar Mahajan - 4 years, 2 months ago

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