Here's a small mathematical puzzle I came up with recently:

Find the radius of the larger incircle, given that the radius of the smaller incircle is $3-\sqrt{3}$. The hexagon is a regular hexagon, with all sides equal enter image description here

P.S: the drawing was hand drawn and the incircles don't exactly touch all sides of the triangle. Sorry for that :P


The answer is


Easiest derivation:

obviously the ratio of areas of the two triangles is $2:1$; the areas are also proportional to incircle radius times triangle circumference. The circumferences are $2+\sqrt 3$ and $3+\sqrt 3$ in units of the hexagon side so $R = 2 (2+\sqrt 3) (3-\sqrt 3)^2 : [(3-\sqrt 3)(3+ \sqrt 3)] = 2$

  • $\begingroup$ clever approach, but how did you get the circumferences? $\endgroup$ – ThomasL Aug 6 '20 at 20:35
  • $\begingroup$ @ThomasL The sides of the triangles in question can all be related to either the height or the side of a regular triangle; the easiest way to make that totally obvious is splitting the regular hexagon into six regular triangles. $\endgroup$ – Paul Panzer Aug 6 '20 at 20:43

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