let 
be the foot of the altitude from 
to the side 
,
,
be the incenter, 
-
and 
the other intersection points of the circle 
with the lines 
and 
,
.
such that![\[(a^2-b^2)(b^2-c^2) = abc.\]](http://latex.artofproblemsolving.com/7/1/8/7188261c40a45f5e15533100432d36abbf515efd.png)
.
that 
is a prime.
is divided into rectangles with two cells. Two rectangles are called adjacent if they share a segment of length 1 or 2. In each rectangle the amount of adjacent with it rectangles is written.
Y by Davi-8191, anantmudgal09, Math-Ninja, Adventure10, mathematicsy, HWenslawski, megarnie, Mango247
Six points are chosen on the sides of an equilateral triangle 
: 
, 
on 
, 
, 
on 
and 
, 
on 
, such that they are the vertices of a convex hexagon 
with equal side lengths.
Prove that the lines
, 
and 
are concurrent.
Bogdan Enescu, Romania
: 
, 
on 
, 
, 
on 
and 
, 
on 
, such that they are the vertices of a convex hexagon 
with equal side lengths.Prove that the lines
, 
and 
are concurrent.Bogdan Enescu, Romania
This post has been edited 1 time. Last edited by Valentin Vornicu, Oct 3, 2005, 1:59 AM
,
radians around the center of 
brings one triangle to another.
and 
are equilateral and centered at the center of 
Clearly 
,
also lie on these perpendicular bisectors, we have proved that 
are the perpendicular bisectors of the sides of 
;
is isosceles, so its base angle is 
.
.
,
.
and 
intersect at a point K, the exterior angle theorem in triangle 
yields
.
and 
intersect at an angle of 60°, and so do the lines 
.
is the image of the triangle 
under a spiral similitude with rotation angle 60°. [On the exam, I avoided speaking of "angles between lines" without properly defining the orientation of these angles, and thus I rewrote the above argument in a non-transformational way.]
,
,
are all directly similar, and have the angles 
.
(as the hexagon 
.
(again since the hexagon 
,
are congruent, so that 
.
,
.
.
. Anyway, a good first day, although not for me, since I could have done the third one...
and 
are equilateral.
,
and 
.
,
and 
.
.
... respectivly.
we get that : 
...
we easily get from (1) that 
.
.
or 
so 
.
then similarly from (3) 
. Contradiction.
then from (2) 
so 
so 
or 
. Contradiction.
but then from (3) we get 
.
and finally it is obvious that triangles 
.
,
,
![\[\triangle{AOB}=\triangle{DOE}, \triangle{EFP}=\triangle{BCP}, \triangle{CQD}=\triangle{AQF}\]](http://latex.artofproblemsolving.com/8/f/6/8f63a95ff2affd0d7b1e3bf6fe906bc075a439b2.png)
,
,
.
,
.
.
.
,
.
,
.
,
.
..........
****************************************************************************
(*)
Thus,
![$\sqrt [3]{\frac{(a_1-a_2)(a_2-a_3)(a_3-a_1)}{(a_3-a_1)(a_1-a_2)(a_2-a_3)}}=1...iff...l=0$](http://latex.artofproblemsolving.com/5/9/6/5964f2b37bfd776cb7f7f355132de693cc82b435.png)
,
,
From the relations (*) results 
a.s.o. 
meet at 
.
to be the point closer to 
,
to be a point inside the hexagon such that 
is equilateral. We know that 
and 
.
and 
are both rhombus. This gives us 
,
is equilateral.
and 
are equilateral.
.
and 
,![\[ \angle(A_1 C_2, A_2 B_1) = \angle(XC_1, XB_2) = 60^{\circ}. \]](http://latex.artofproblemsolving.com/0/2/3/0231b1be466730261a8e19a06fcfcedb97f39469.png)
Therefore we conclude that 
,
cyclic.
similarly, we could get 
,
all being cyclic.
,
are all congruent. This is enough to forces 
are equilateral.
is a kite, and therefore, 
bisects 
.
bisects 
and 
bisects 
.
concur at the centroid of 
such that 
is a parallelogram. We know that 
.
since they are the same magnitude and are rotations of 
from each other(they form an equilateral triangle).
,
,
should form an equilateral triangle too, implying that 
a parallelogram.
and so forth, we know that 
and 
are equilateral. Consequently, 
,
,
.
as these vectors form hexagon 
as well due to these three vectors having the same magnitude and being rotated 
degrees from each other. Therefore 
which means that these three vectors are also rotated 
is rotated 
.
be the side length of hexagon 
be the points such that 
,
,
are equilateral triangles. There are many observations to be made.
,
,
are equilateral triangles.
and 
,
is a parallelogram and 
.
and 
,
is a parallelogram and 
.
.
circles.
.
.
.
is cyclic. Similarly 
and 
are cyclic. Since 
are radical axises of these 
and 
meet at 
.
,
.
,
,
,
are collinear. Since 
both perpendicular to this line, 
.
to form a closed polygon, namely an equilateral triangle. Thus, by vectors you can translate 
to form a closed polygon. Since the lengths are equal this closed polygon is equilateral.
to form another equilateral triangle, and we see that this one and 
which implies 
meaning that 
form 
angles with each other. Extend 
and 
to intersect at 
and define 
and 
similarly. Then note that 
so we easily obtain that 
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and cyclic variants. In addition, let 
.
is cyclic.![\[\angle AC_1B_1 = 180 - 60 - b = 120 - b \implies \angle AC_1B_2 = 120 - 2b \implies \angle C_2C_1B_2 = 60 + 2b \implies \angle C_1C_2B_2 = 60 - b;\]](http://latex.artofproblemsolving.com/c/8/b/c8b064410ae0e77a3f85818b64ed2e3efd4a6129.png)
this coupled with 
implies 
,
,
;
,
is the circumcenter of triangle 
;
.
,
,
are cyclic; now radical axis finishes (or 
,
so opposite angles are supplementary. Then
This implies that since D,E,F are circumcenters, 
which means that 
is cyclic and other variants, whence radical axis with radical center that point of concurrence finishes. (We're motivated to find angles after that cyclicity, also a common way of concurrence is radical center so we look for circles of four points since they need two points shared with each of the two other circles, from which it is obvious which points to try to prove, especially with a good diagram.)
and 
from them being 120 degrees difference in orientation (idk if rotation describes it accurately, but from here rotation just means difference in orientation). Then this implies that 
is a 120 degree rotation from 
by comparing each addend respectively. Analogously for 
,
we get that the diagonals are perpendicular, namely 
![[asy]
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.
Then 
so we find 
.
.
we find that 
is cyclic.
implying that 
.
,
.
,
is the sum of three unit vectors that are 
is the sum of three unit vectors; these three unit vectors then must form an equilateral triangle. In particular, the lines 
,
,
form an equilateral triangle 
,
.
and cyclic permutations by AAS congruence. Then 
and 
under reflection about 
,
under this reflection. It then follows that this reflection sends 
to 
,
,
but 
since they are equal length and spaced at 
and since they are also equal length they must be at 
This is by writing 
and 
and then translating these four vectors to the origin and angle chasing on the circle centered at the origin they form.
are homothetic, so the concurrency follows.
