Let $ ABCDEF$ be a convex hexagon with all the sides of length 1. Prove that one of the radii of the circumcircles of triangles $ ACE$ or $ BDF$ is at least 1.
Problem
Source: Romanian TST 1 2008, Problem 3
Tags: geometry, circumcircle, geometry proposed
mto
02.05.2008 03:32
Lemma: Let MNPQ be a convex quadrilateral where MN=NP=1 and 2.[MQP] =< [MNP] then the radius of circumstance of MPQ is at least 1 Apply the lemma, if all the radius are less than 1, takes the sum on all the angles we come to a contradiction that 2*2*180 > (6-2)*180
Goblin
02.05.2008 04:22
This problem is from Kvant (M951).
deadly_dozen
02.05.2008 12:01
mto wrote: Lemma: Let MNPQ be a convex quadrilateral where MN=NP=1 and 2.[MQP] =< [MNP] then the radius of circumstance of MPQ is at least 1 Apply the lemma, if all the radius are less than 1, takes the sum on all the angles we come to a contradiction that 2*2*180 > (6-2)*180 If your Lemma is not in "iff" form, but only "if" you can't use it like that. If it is iff provide a proof of your lemma.
mto
02.05.2008 12:40
Goblin wrote: This problem is from Kvant (M951). Well, it was quite straightforward to discover the result in the lemma when I compared the angles ABC and AOC where O is the center of circumstance circle of ACE
crazyfehmy
07.06.2008 23:51
Let $ m(B)+m(D)+m(F)\geq360$ . $ O$ be the circumcenter of triangle $ AEC$. If circumradii of $ AEC$ is $ <1$ then $ FE>EO$ , $ DC>OC$ , $ AB>OB$ and $ m(FOE)>m(EFO)$ , $ m(EOD)>m(EDO)$ , $ m(AOB)>m(ABO)$ and $ \frac{m(B)+m(D)+M(F)}{2}<180$ and this is a contraction.