# Geometry Challenge: Drawing Polygons in a Rectangle

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.

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

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

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

$$\color{purple}{\text{Level 4}}$$
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.

$$\color{blue}{ \text{Level 5}}$$
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.

Note by Pranshu Gaba
3 years, 2 months ago

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Level 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: $\frac{n \omega ^2 sin(\frac{2 \pi}{n})}{2(1+ cos(\frac{\pi}{n}))^2}$

- 3 years, 2 months ago

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.

- 3 years, 2 months ago

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 $$l$$

- 3 years, 2 months ago

Your reasoning is correct for a square. In a square, $$a$$ should be $$45^{\circ}$$. Area of square will be $$\frac{w^2}{ 2 \cos^{2}a} = w^2$$, and it will not depend on $$l$$. However, this reasoning does not extend for other $$n$$.

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 $$l$$.

- 3 years, 2 months ago

OOps!correct I think i should sleep .I m sickk

- 3 years, 2 months ago

Nice efforts though!

- 3 years, 1 month ago

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

- 3 years, 2 months ago