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Here is the picture of what I meant (the small circles are all the same radius, I'm just really bad at paint)enter image description here

So I realised that you can connect the circle centers to make a hexagon with all sides equal to $2r$ where $r$ is the radius of the small circle, but in order to prove that the radius of the big circle is 3 times larger than $r$, i have to prove that the hexagon is regular, but I don't know how to prove that the angles of the hexagon are all equal.

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  • $\begingroup$ Show that the sides are equal. Also the $7$th circle can fit at the center. $\endgroup$ – farruhota Feb 13 '18 at 13:45
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    $\begingroup$ Sorry you said the small circles all touch each other.. I think that's impossible. Do you mean each circle touches the bigger circle and touches its two neighbors? If that's the case, It is obvious the hexagon is regular by symmetry the rotational symmetry. Also if that's the case you should put a better figure, its really confusing $\endgroup$ – David Jaramillo Feb 13 '18 at 13:45
  • $\begingroup$ @DavidJaramillo Yes sorry, exactly as you said, each circle touches the bigger circle and touches its two neighbors should be the text of the problem. $\endgroup$ – Ayy Lmao Feb 13 '18 at 13:48
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By symmetry, the triangle formed by the centers of three tangent circles is equilateral. Then in the figure formed by seven circles, the alignments are perfect ($3\times60°=180°$) and the large diameter is three times a small one.

enter image description here

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