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Given an arbitrary disc with radius $200$ on a plane that's been divided into convex $7-$gons each with radius $\leq \frac{1}{2}$, show that it intersects $\geq$ $10^9$ $7-\text {gons}$?

Now, how is this even possible? Since it's a disc, we can consider circles both inside and on the peripheri of the disc. So, considering the area, we get $< 2.10^5$ $7-$gons which intersect and also we have some more polygons that intersect on the sides. How does that make up to $10^9$?

Mathejunior
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    Just to remove confusion, the number is $10^9$ and it shouldn't be mistaken for $10^9~7$. The $7$ is there adjoined with next word "-gon". – Mathejunior Dec 29 '17 at 19:40
  • There are ways to tile the plane with non-congruent convex heptagons, see e.g. here. The point, I think, is that most of the heptagons will have very small areas. However, I don't know how to quantify this to obtain the desired estimate. – Robert Israel Dec 29 '17 at 19:52
  • heptagon is the name of a $7$-gon. –  Dec 29 '17 at 19:52
  • @RobertIsrael Yes, it's not really trivial to quantify this. I can somewhat understand and as they're pretty small, such a vague way of area ratio that I have considered missed out a lot of heptagons as my way doesn't ensure the fact that the heptagons are adjacent to each other. I have considered them as circle and got such a poor bound. – Mathejunior Dec 29 '17 at 19:58
  • You mention disc (not circle) but also intersect instead of (partially) cover. What do you mean for a disc to intersect a heptagon? Did you mean (partially) cover? Or did you mean that the boundary of the disc (the circle) intersects at least one edge of the heptagon? – orlp Dec 29 '17 at 20:25
  • @orlp Honestly, I am not sure of that. I have just copied the question. I have assumed that it means "cover" as just intersection along the circumference won't yield such a great value. Also, had it been the case, the question would have used the word "circle" instead of "disc". That's what I believe. – Mathejunior Dec 29 '17 at 20:30
  • It's easy to prove a lower bound of roughly $160,000$ heptagons based on circle packing. – orlp Dec 29 '17 at 20:37
  • Also do the heptagons have to be convex? – orlp Dec 29 '17 at 20:39
  • @orlp Yes, I have done that circle packing and that comes to be the lower bound but that's extremely weak. And yes, as mentioned, the heptagons are to be convex – Mathejunior Dec 29 '17 at 20:41
  • Where are you copying the question from? – Robert Israel Dec 29 '17 at 21:36
  • @RobertIsrael Some random question paper – Mathejunior Dec 30 '17 at 03:39
  • Anyone with any stronger bound? – Mathejunior Dec 30 '17 at 09:59

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