[/quote]
,
,
,
,
are integers such that 
.
and 
of 
such that for each integer 
with 
,![\[ n\mid a_i - b_i - c_i
\]](http://latex.artofproblemsolving.com/1/a/e/1ae8d54f068b646ac7a1e43d75e7f26510cfdc95.png)
be an odd prime and 
be integers so that the integers 
are divisible by 
Prove that 
be real numbers such that 
Prove that 
Where 
be a triangle with circumcircle 
.
and 
respectively denote the midpoints of the arcs 
and 
that do not contain the third vertex. Let 
denote the midpoint of arc 
(the arc 
including 
)
be the incenter of 
be the circle that is tangent to 
be the circle that is tangent to 
,
friends, 
people are seated around a circular table in a fixed order. Ali places 
is the circumcenter of 
.
lie on 
,
is tangent to circumcircle of 
.
is tangent to circumcircle of triangle 
.
be the sequence defined as follows: 
,
,
is the smallest integer greater than 
such that in 
there are no arithmetic progressions of length 
.
with 
,
be an integer. In a configuration of an 
board, each of the 
cells contains an arrow, either pointing up, down, left, or right. Given a starting configuration, Turbo the snail starts in one of the cells of the board and travels from cell to cell. In each move, Turbo moves one square unit in the direction indicated by the arrow in her cell (possibly leaving the board). After each move, the arrows in all of the cells rotate 
counterclockwise. We call a cell good if, starting from that cell, Turbo visits each cell of the board exactly once, without leaving the board, and returns to her initial cell at the end. Determine, in terms of 
positive integers 
such that for every index 
,
,
is a multiple of 
.
Y by Davi-8191, A-Thought-Of-God, Adventure10, lian_the_noob12, Mango247, Rounak_iitr, bjump, cubres
Given trapezoid 
with parallel sides 
and 
, assume that there exist points 
on line 
outside segment , and 
inside segment 
such that 
. Denote by 
the point of intersection of and 
, and by 
the point of intersection of and . Let 
be the midpoint of segment , assume it does not lie on line . Prove that belongs to the circumcircle of 
if and only if belongs to the circumcircle of 
.
Proposed by Charles Leytem, Luxembourg
with parallel sides 
and 
, assume that there exist points 
on line 
outside segment , and 
inside segment 
such that 
. Denote by 
the point of intersection of and 
, and by 
the point of intersection of and . Let 
be the midpoint of segment , assume it does not lie on line . Prove that belongs to the circumcircle of 
if and only if belongs to the circumcircle of 
.Proposed by Charles Leytem, Luxembourg
This post has been edited 1 time. Last edited by April, Jul 10, 2009, 8:30 AM
and 
.
,
is cyclic, so 
.
since 
,
is cyclic. Thus, 
.
is cyclic if and only if 
and 
is cyclic if and only if 
.
,
,
,
.![\[ (IJ)(JK)=(FJ)(JE)\Leftrightarrow (a+b)(c)=(b)(c+b+c)\Leftrightarrow ac=b(b+c)\]](http://latex.artofproblemsolving.com/5/d/c/5dc862097522279e390ab106833905b7edca1e9e.png)
![\[ (IJ)(IK)=(IF)(IE)\Leftrightarrow (a+b)(a+b+c)=(a)(a+b+c+b+c)\Leftrightarrow b(b+c)=ac\]](http://latex.artofproblemsolving.com/b/6/7/b6763094598273e350f29788a1c969d7670ff41f.png)
if and only if 
belongs to the circumcircle of 
.
,
is cyclic, so 
.
.
cyclic 
.
.
is cyclic. in deed 
![\[IJ.IK=JF.JE \Leftrightarrow IE.IF=IJ.IK\]](http://latex.artofproblemsolving.com/6/8/d/68d856c23c4f3d25197f25abcae8ce62152563a9.png)
so the equivalence is proved .
is cyclic.
the inversion centered at J and of ratio 
.
and 
and 
,
with 
)
must be a circle so 
is cyclic.
and 
,
is cyclic.![\[ABKI \ cyclic \Leftrightarrow K=K' \Leftrightarrow JDKC\ cyclic \]](http://latex.artofproblemsolving.com/0/9/6/096241c375fa14ba6a8671dacef81eaf1920e617.png)
and 
(As 
is concyclic
(
)
.
is cyclic or 
is cyclic,or 
is cyclic iff 
Also 
is cyclic iff 
From 
as desired.
is a cyclic quadrilateral. Consider the following
and lines 
and 
.
.
and 
Hence, both conditions are equivalent. 
implies the four points 
are concyclic. Also 
,
exists too. By PoP, we get 
.
and 
.
lies in the radical axis of 
and 
.
,
,
are concyclic, as desired.
are concyclic. Since 
are concylic, 
.
.
.
,
,
be the midpoint of the side 
.
and 
.
.
be the second point of intersection between 
,
.
and 
art 
,
.
.
be points on a line in that order. Let 
be the midpoint of 
.
.
.
.
are inverses wrt 
,
is cyclic.
is cyclic 
,
,
Thus,
And so we are done. 
and 
.
as desired.
be a circle through 
and center at 
are tangent to 
.
Hence, 
.
Let perpendicular at 
Hence, 
is cyclic. Now, here is a much more natural restatement of the problem:
lie on a circle. Let 
.
be the line through 
.
,
cyclic. We want to show 
.
,
swapping 
gives us that we want to show 
.
.
,
hence 
is cyclic. Now we are ready to PoP:![\[I \in (ABK) \iff JI \cdot JK = JA \cdot JB = JF \cdot JE\]](http://latex.artofproblemsolving.com/e/a/f/eaf4731da2530cc1283036da23e1f14e9d054269.png)
![\[K \in (CDJ) \iff IF \cdot IE = ID \cdot IC = IK \cdot IJ\]](http://latex.artofproblemsolving.com/2/e/1/2e1125e70730d2ccfe7633e98e0e68e364b3b490.png)
and in fact both of these are the same condition as 
and we are done. 
to itself. Note that 
. By butterfly theorem, 
is the reflection of 
if and only if 
.
if and only if it is the inverse of 
are cyclic from the given angle conditions.
QED
so 
so 
is cyclic.
by Power of a Point, and that 
iff 
is cyclic by Power of a Point and it's converse. Thus 
is equivalent to 
by Power of a Point, and 
by Power of a Point and it's converse, so 
.
is equivalent to 
.
and 
respectively. Thus 
and 
,
be equal to 
and 
,
.
is equivalent to 
However, 
which reduces to the same condition, so the claim is proved. 
.
and line 
)
(the PoP condition for 
to be cyclic needs this to be 
)
These are clearly equivalent.
is cyclic too!
,
Again if 
,
Since above conditions match,we are done 
By PoP, 
and by Reim 
cyclic.
Note that ![\[JF\cdot IK-IF^2=FJ\cdot FJ-IF\cdot IJ=FJ(JE-IK)-IJ^2=JF\cdot JE+IK\cdot IF-JI\cdot JK\]](http://latex.artofproblemsolving.com/0/5/4/05466e2b186f63230b9150402d5c3d88d5ac4ab7.png)
which when 
is equal to 
which means that 
![\[IF^2-IK\cdot FJ=IF\cdot FK-IK\cdot IJ\]](http://latex.artofproblemsolving.com/d/5/6/d569c20bed7f4679114045351d6c8f745a9b7c2f.png)
which when 
is equal to 
which means 
Thus, we are done.
Which is equivalent to DECF to be cyclic .
must also be cyclic. From PoP,
if and only if![\[ JI \cdot JK = JA \cdot JB = JE \cdot JF \]](http://latex.artofproblemsolving.com/1/5/0/150fef320c0cd8f26a2ee5a7277d11170ed45122.png)
which is equivalent to 
(from a lemma regarding harmonic bundles).![\[ IE \cdot IF = IC \cdot ID = IK \cdot IJ \]](http://latex.artofproblemsolving.com/f/8/8/f88bc5a8b66b74e1dfd2acd924317dee232c6154.png)
But 
from PoP, so the condition is true if and only if
,
,
concyclic.
cyclic.
.
so 
which by 
is equivalent to 
,
are cyclic. 
cyclic is equivalent to 
which due to 
be points on a line in this order.
.
.
and 
.
.
,
respectively on the complex plane.
.
The third condition is equivalent to
.
thus both cyclicities imply each other.
,
,
,
,
,
which implies its cyclicity.
we have ABEF cyclic. Since 
we have CDEF cyclic by Reim's. Then, ![\[ I \in (ABK) \iff JB \cdot JA = JI \cdot JK \]](http://latex.artofproblemsolving.com/8/5/a/85a9475be494a5d76aab4e12225aaf7b86080b41.png)
But since we also have 
by PoP, this is equivalent to 
.
intersects 
,
.
.
,
.![\[\angle DAE=\angle CBF,\]](http://latex.artofproblemsolving.com/f/7/b/f7b162c8b3a21bbc58d7da818a1da158e44c3ac4.png)
we must have that 
,
is cyclic. Now, by Power of a Point, we have that![\[K\in (CDJ) \iff IC*ID=IK*IJ,\]](http://latex.artofproblemsolving.com/8/d/7/8d71552b8156ad6c01425a1a3e481555aa5c7142.png)
however, since ![\[IC*ID=IE*IF,\]](http://latex.artofproblemsolving.com/5/5/a/55adc0a75f14f7be31aed4783a8eb7360a153b9c.png)
so our problem becomes proving that 
.
(by 
and 
,![[asy]
draw((0,0)--(12,0));
dot((0,0)); dot((6,0)); dot((8,0)); dot((4,0)); dot((12,0));
label("$E$", (0,0), S); label("$I$", (6,0), S); label("$F$",(8,0),S); label("$J$",(12,0),S); label("$K$", (4,0),S);
label("$\frac{a+1}{2}$", (2,0), N, red); label("$\frac{a-1}{2}$", (5,0), N, red); label("$1$", (7,0), N, red); label("$\frac{a+1}{a-1}$", (10,0), N, red);
[/asy]](http://latex.artofproblemsolving.com/9/9/6/996f681d3e97298ee8861e3b6fb4a66324d28408.png)
![\[IE*JF=a*1=\frac{a-1}{2}*\frac{2a}{a-1}=IJ*IK,\]](http://latex.artofproblemsolving.com/3/6/f/36f1517f2353baff44b8db5b1889bc5c298d8a89.png)
as desired. This means that 
.
.
,
also.![\[I''C*I''D=I''K*I''J.\]](http://latex.artofproblemsolving.com/b/e/e/bee67f50518ddcbf4a0fdce7a1e4064b8cd37e22.png)
Again, by Reim's, note that ![\[I''C*I''D=I''K*I''J=I''E*I''F \implies \frac{I''E}{I''J}=\frac{I''K}{I''F}.\]](http://latex.artofproblemsolving.com/4/a/9/4a9833b944f7b50acd40561c395574dcb024bb8b.png)
Now, knowing this, and that 
and 
and set up a "number line" with directed lengths and label the (directed) lengths as follows:![[asy]
draw((0,0)--(12,0));
dot((0,0)); dot((6,0)); dot((8,0)); dot((4,0)); dot((12,0));
label("$E$", (0,0), S); label("$I$", (6,0), S); label("$F$",(8,0),S); label("$J$",(12,0),S); label("$K$", (4,0),S);
label("$a'+1$", (2,0), N, red); label("$a'$", (5,0), N, red); label("$1$", (7,0), N, red); label("$\frac{a'+1}{a'}$", (10,0), N, red);
[/asy]](http://latex.artofproblemsolving.com/2/6/6/2662c416200890717aad11a8757e178505d5583b.png)
![\[I''E*JF=\frac{a'+1}{a'}*(2a'+1)=1*\frac{(a'+1)(2a'+1)}{a'}=I''F*JE,\]](http://latex.artofproblemsolving.com/3/f/d/3fd29c4e5d05fb96ac446c2c10a7d1f818c9184e.png)
which means that 
,
.
,
.
as desired.
