How large an equilateral triangle can you fit inside a 2-by-2 square?
@skullpatrol @ParthKohli @ganeshie8
Largest area-wise?
Yeah that's what I was thinking of putting my answer as.
Can be done by having a centroid in the middle of that square and then connect the segments to obtain an equaliterial triangle and then all you need is to know the length of the base and height of that Triangle. ThatS all. Cheers
The base would be 2 and so is the height thus the area would be 2*2*0.5 = 2
According to http://mathworld.wolfram.com/TrianglePacking.html the largest equilateral triangle that can be fit into a square with side length \(x\) looks like this: |dw:1480196475184:dw| The Pythagorean theorem tells us that \[\left(y\sqrt2\right)^2=2y^2=x^2+(x-y)^2\implies y^2=2x^2-2x\]and so the area of the triangle would be \[\frac{\sqrt3}{4}\left(y\sqrt2\right)^2=\sqrt3(x^2-x)\]With \(x=2\), the maximum area would be \(2\sqrt3\approx3.464\).
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