# What is the maximum area of a square inscribed in an equilateral triangle?

What is the maximum area of a square inscribed in an equilateral triangle?

Please post the approach to solve the above question.

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The square of maximum area occurs when upper corners of square touches the sides of the equilateral triangle and the bottom side of the square is on one side of the triangle. Then you can find the relation between the area of square and the equilateral triangle.

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Why is it true that "upper corners of square touches the sides of the equilateral triangle and the bottom side of the square is on one side of the triangle"? –  Han de Bruijn Oct 30 '13 at 15:25
@HandeBruijn because if you choose any square inside the triangle that does not fit the definition that I gave, an equilateral triangle of smallest area which covers this square will have less area than the first equilateral triangle. In the other words, try to draw an equiateral triangle of smallest area which covers a square. I mean, the ratio of the areas of a max area square and an equilateral triangle which covers it will be constant always. If you agree with this then there is no problem. –  nikamed Oct 30 '13 at 15:41

Building up on newzad's answer, we might consider the side of the equilateral circle to be $s$, and the find the length of the square, $l$, in terms of $s$.

Now, one of the methods using trigonometry is very simple. You can see that $l\cot\angle BAC + l+l\cot\angle ABC = AB$ or $2l\cot 60^\circ + l = s$ or: $$l = \frac{s}{1 + 2\cot 60^\circ} = \frac{\sqrt3 s}{2+\sqrt3}$$

If trigonometry is not allowed then we could achieve the result by seeing the similarity of $\triangle CDE$ and $\triangle CAB$, where $CH$ is the height or $h$, we have: $$\frac{h - l}{l} = \frac{h}{s}$$

Considering that in a equilateral triangle, $h = \frac{\sqrt3 s}{2}$, we again have: $$l=\frac{\sqrt3 s}{2+\sqrt3}$$

I would like to tell you that these approaches work for any triangle, albeit minor changes.

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