pixel (pinterface) wrote in mathproofs,

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On Hexagons

I'm sketching out a construction project for my current obsession and ran into some difficulty because I want to be sure of something and suck at proofs (my general math is also out of practice, but I digress). My problem is probably stupidly simple for mathematical genii such as yourselves, so feel free to laugh at my incompetence where appropriate. ;)

three congruent rectangles of length 𝓁 and width 𝓌 forming an equilateral triangle
[large version 693x610 7.5KB]

Given the above diagram (hexagon 𝒽), and that all three rectangles are congruent, what is the smallest regular hexagon ℋ that will enclose the shape? In other words: given a hexagon 𝒽 with parallel but non-equal sides, what is the smallest enclosing regular hexagon ℋ?

I strongly suspect the incircle of ℋ will be the same as the incircle of the three shortest sides (min(𝓁, 𝒸)) of 𝒽, but have no idea how to go about proving it. Has this been shown before? If so, please point me in the right direction. If not, do you have any pointers as to how to prove it?

Update: Now with more math!

Assuming min(𝓁, 𝒸) = 𝓁, and given 𝓇 as the inradius of the triangle formed by the rectangles:
   𝓇′ = 𝓇 + 𝓌
(𝓇′ is inradius of the triangle which would be formed if you extended the outer 𝓁s), can it be shown that 𝓇′ is the inradius of ℋ?

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