[/quote]
,
by 
,
for every positive integer 
.
leave distinct remainders when divided by 
.
of distinct positive integers such that there exists a positive integer 
for which the number of divisors of 
and of 
are equal.
and 
.
such that ![\[f(2x+y) + f(x+f(2y)) = f(x)f(y) - xy\]](http://latex.artofproblemsolving.com/0/6/2/0623b7c328af6d088cfc5dd288afe3e17a94ff29.png)
for all reals 
and 
.
and 
be two sequences of integers such that 
and 
for all 
.
divides 
for an infinity of integers 
such that 
for all 
are coprime. Consider the sequence 
given by 
.
such that 
is even and 
is an integer. Prove that the numerator in its simplest form of 
is an odd integer greater than 
.
be an integer and let 
be 
.
such that the n subsets 
are also disjoint.
s
.
satisfies 
.
,
.
such that for all irrational numbers 
and 
,
where 
satisfy that 
.![$$\sum_{cyc} \sqrt[4]{\frac{a^3}{b+c}}$$](http://latex.artofproblemsolving.com/e/f/0/ef089c4f2915bbf75b0f60b044bbe9273b301a81.png)
compute the smallest possible value of ![\[\sum_{k=1}^{n}\left\lfloor\frac{a_k}{k}\right\rfloor\]](http://latex.artofproblemsolving.com/5/6/2/562c27c7779285b6439c4087f52933a41f29ee2e.png)
over all permutations 
of 
with incenter 
,
touches sides 
and 
at points 
and 
,
and 
touches 
.
meets 
at 
.
is parallel to 
.
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 
.
,
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.
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 
,
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
,
,
,
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.
