Let $A_{1}A_{2}\ldots A_{2n}$ be a convex polygon and let $P$ be a point in its interior such that it doesn't lie on any of the diagonals of the polygon. Prove that there is a side of the polygon such that none of the lines $PA_{1}$, $\ldots$, $PA_{2n}$ intersects it in its interior.
Problem
Source: Romanian I TST 2007
Tags: geometry, geometric transformation, rotation, combinatorics proposed, combinatorics
e.lopes
13.04.2007 21:08
with this lemma, the problem is easy: if P is in interior of $A_{1}A_{2}..A_{n+1}$, we have that the lines $PA_{1}$, $PA_{n+1}$ and $PA_{i}$ (where i = n+2, ... , 2n) will intersect the sides $A_{1}A_{2}$, $A_{2}A_{3}$, ... , $A_{n}A_{n+1}$. so, we have n+1 lines intersecting n sides. so, two lines will intersect the same side! (impossible, if we suppose that the problem isn't correct)
maggot
15.04.2007 13:11
Wow, that's nice
prowler
16.04.2007 04:16
We had a little bit harder problem to solve on our TST: http://www.mathlinks.ro/Forum/viewtopic.php?t=137070 proposed by me
pohoatza
20.04.2007 21:30
another solution i thought of should be by considering the figure formed by $A_{k}A_{k+1}A_{n+k}A_{n+k+1}$, considering just the two areas of the triangles $A_{k}OA_{k+1}$ and $A_{n+k}OA_{n+k+1}$, where $O \in A_{k}A_{n+k}\cap A_{k+1}A_{n+k+1}$. now it is obvious that these $n$ pairs of two triangles formed by the "big" diagonals and with one common vertices cover the all polygon, therefore $P$ must be in one of the $2n$ triangles, let that be $A_{i}O_{i}A_{i+1}$, therefore the side $A_{i+n}A_{i+n+1}$ is the side which isn't cut by the $PA_{k}$ type lines.
sombrhiver
14.06.2007 05:16
The problem with this second solution is that I didn't know how to prove that the pairs of triangles formed with two consecutive 'big' diagonals cover up the entire polygon. It seems it is not quite an obvious thing. I saw the official solution (I think they had two solutions, in fact, but I can't remember the other one), which involved the rotation of a diagonal and, considering P to be on, let's say, the right, by rotating the diagonal 2n times, the point moved to the left with respect to the diagonal. Frankly, I don't see how this goes. Why does that mean that the point necessarily belongs to one "papillote"? Someone please enlighten me. Even if my question may be stupid for you
sombrhiver
11.07.2007 21:00
Can no one comment on this solution to the problem?
pohoatza
11.07.2007 22:41
Oh, sorry because I didn't see your message at first. This "lemma" is quite well-known since 2006, it is a basic lemma used in an official solution to IMO2006/pb6, as well you can see here, it is mentioned in fedja's solution: http://www.mathlinks.ro/viewtopic.php?p=572824#572824.