The Wonder Island is inhabited by Hedgehogs. Each Hedgehog consists of three segments of unit length having a common endpoint, with all three angles between them $120^{\circ}$. Given that all Hedgehogs are lying flat on the island and no two of them touch each other, prove that there is a finite number of Hedgehogs on Wonder Island.
Problem
Source: Baltic Way 1994
Tags: geometry, circumcircle, inradius, search, combinatorics unsolved, combinatorics
Severius
21.02.2006 15:34
Cute problem. Suppose that we had an infinite number of hedgehogs. Then for any arbitrarily small real $r$, we could find two hedgehogs such that the distance between their centers was $<r$. This is easy to prove: if you assume a minimal distance $r$ between any two centers, then we can draw non-intersecting circles of radius $r/2$ and centered at each hedgehog center... but then the Wonder Island contains an infinite area, contradiction (...well, I guess it could be the Wonder Island for a reason, but...). So take two hedgehogs whose centers $A, B$ are very close (at distance $r$, say). It is obvious that they touch. For proof: let $a, b$ be the triangles with vertices as the vertices of $A, B$ respectively (that is, the triangles which contain the hedgehogs). Note that $a, b$ are equilateral with circumradius $1$. Since $AB=r$ and the inradius of $a, b$ is $1/2$, $B$ is in the incircle of $a$ and vice-versa. Now break up each of $a, b$ into three pieces: $a_1, a_2, a_3, b_1, b_2, b_3$, where each piece is bounded by two sides of the hedgehog and the corresponding side of the triangle. Since $A$ is in $b$, it's in some $b_i$, and similarly $B$ is in some $a_j$. Now, taking the sides of $b_i, a_j$, we get our intersection.
ZetaX
21.02.2006 19:01
A search for hedgehog gives me...: http://www.mathlinks.ro/Forum/viewtopic.php?t=27872