AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
be a positive integer, and let 
be a subset of 
. An -partition of into 
parts is a representation of n as a sum 
, where the parts 
belong to and are not necessarily distinct. The number of different parts in such a partition is the number of (distinct) elements in the set 
.-partition of into parts is optimal if there is no -partition of into 
parts with 
. Prove that any optimal -partition of contains at most ![$\sqrt[3]{6n}$](http://latex.artofproblemsolving.com/2/0/1/201489b1f95ad200036698a606f712da7bb465b9.png)
different parts.
other knights where friendship is mutual. Prove that King Arthur can place some of his knights around a round table in such a way that every knight is friends with the 
people adjacent to him and between them there are at least 
friendships of knights that are not adjacent to each other.
cities in a country, where 
is an integer. Some pairs of cities are connected by direct (two-way) flights. For two cities and 
we define:
A 
between and as a sequence of distinct cities 
, 
, such that there are direct flights between 
and 
for every 
;
A 
between and as a path between and such that no other path between and has more cities;
A 
between and as a path between and such that no other path between and has fewer cities. and in the country, there exist a long path and a short path between them that have no cities in common (except and ). Let 
be the total number of pairs of cities in the country that are connected by direct flights. In terms of , find all possible values 
, 
.
.
are fixed positive integers.
are fixed positive real numbers.
be the set of all functions 
with the following properties: is defined for all real numbers and takes only real values.
the following equality holds: 
such that
,
.
, the diagonal 
bisects neither the angle 
nor the angle 
. The point 
lies inside and satisfies ![\[\angle PBC=\angle DBA\quad\text{and}\quad \angle PDC=\angle BDA.\]](http://latex.artofproblemsolving.com/c/2/0/c20761f3eadd054958f40259f3d1c05f26279783.png)
Prove that is a cyclic quadrilateral if and only if 
.
be a cyclic quadrilateral in which the lines 
and 
meet at a point . Let 
be the point of the line 
, different from , such that 
. We construct the parallelograms 
and 
. Prove that the points 
lie on the same circle.
defined on the set of all positive integers and taking non-negative integer values, satisfying the three conditions: 
for at least one ; 
for every positive integers 
and 
; there are infinitely many positive integers such that 
for all 
.
be positive real numbers such that 
. Find the minimum value of the following sum :
knowing that the denominators are positive real numbers.
point 
was chosen such that 
and 
. Prove that if perpendicular from to passes from the intersection of diagonals of , then 
.
balls to 
boxes as she wishes. After that, Aslı and Zehra make alternated moves which consists of taking a ball in any wanted box starting with Aslı. One who takes the last ball from any box takes that box to herself. What is the maximum number of boxes can Aslı guarantee to take herself regardless of Zehra's moves?
, 
. The internal angle bisector of 
and the external angle bisector of intersect the ray at points 
and , respectively. Given that 
, prove that 
.
![\[f(xf(y) + x) = xy + f(x)\]](http://latex.artofproblemsolving.com/9/9/f/99f580ebc50846e6bc2c004667559922749a4dfa.png)
be a convex pentagon such that 
and 
Let be the midpoint of 
and let 
be the circumcenter of triangle 
Given that 
prove that 
be a point on 
such that 
Then 
and hence 
are cyclic. Let 
be a point on 
such that 
Since 
and are parallel to each other, we conclude that 
is a parallelogram. Then 
are collinear. Since 
and 
we conclude that 
and 
are perpendicular. Let 
is also a parallelogram and therefore 
Let 
and 
Then since 
we get 
Hence 
and 
and therefore 
We shall show that 
and 
we get
and since 
we get So, we are done.
be the reflections of 
with respect to . Then 
becomes the perpendicular bisector of 
meaning , so that 
lies on 
. Note that is the midpoint of 
and 
so 
is a parallelogram, implying 
. Then 
so that 
are collinear. Let 
be such that 
. Then given the condition 
this implies 
and clearly 
is isosceles. Now by a simple angle chase, 
, i.e. 
is cyclic. Then 
. But 
. But 
hence 
. Thus, 
is cyclic. Then 
. Now 
. So 
.
in the line 
, just draw the perpendicular from to the line and then draw in the circle with the foot of this perpendicular as the centre, with on the circumference. Then 's antipode is 
.
. and be the reflections of and with respect to , respectively. We find that 
, hence is isosceles. Hence 
, so is the angular bisector of 
and 
. is cyclic, because that would yield 
, as we have to prove.
, we get 
and hence 
is cyclic. be the intersection of the lines 
en 
. Then 
, so 
is also a cyclic quadrilateral. And so and all lie on one circle. Furthermore, since 
is the reflection of with respect to , that quadrilateral is also cyclic.
and so is cyclic and we are done., the condition 
becomes equivalent with lying on a certain circle. So in this way you've got rid of all the ugly conditions, and therefore you get a much easier angle chase 
a point on 
such that 
.
and 
a point such that 
.
a quadrilatere, this one is a parallelogram if and only if his diagonals meet each other in her midpoints.
and 
, and 
is the midpoint of 
, we have 
is a parallelogram and is the midpoint of 
. Otherwise, since we must prove that : 
because the triangle 
is iscoscele. And so we are left to prove that points 
are concyclic.
and is the midpoint of 
and then the triangle 
is iscoscele, hence points 
are concyclic. Even since is the common midpoint of the 
we get that 
are parallelograms. Hence : 
. Moreover, since 
and
and 
hence 
which give us : 
.
which means,
i.e : 
.
, are concyclic and hence we get the desired result which is :
.
with 
.Suppose 
and so 
.That implies 
. Also from 
we've 
. So solving we get 
. Now from the condition 
we get 
and basically so, 
. Now 
.Also as 
so, 
. Now take 
. From the condition 
directly we get 
. Now from the condition directly we've 
. Now just eliminating 
from above two expressions we get 
. Here we're asked to show 
. So we need to show 
.Suppose 
. If 
then 
, which is indeed true, so 
, that's what all we needed to show,so we're done.
be the midpoint of 
.Then 
and 
,so 
is cyclic.Thus 
since 
.Let 
be the midpoint of .Then it is well known that 
so is the circumcenter of 
.Consequently from straightforward angle chasing we get that 
are the internal angle bisectors of 
and 
respectively.Now applying sine rule in 
and 
we get
.Dividing the expressions and using the fact that 
we get 
.Thus 
,as desired.
about to get 
respectively.
implies that 
are concyclic. Now we have by the length conditions, 
are collinear and 
.
are concyclic and since 
by Fact 5 we have 
. This completes the proof. 
about to get respectively. implies that are concyclic. Now we have by the length conditions, are collinear and . are concyclic and since by Fact 5 we have . This completes the proof. 
about to get respectively.
?)
on 
such that 
and 
. Then observe that 
so 
are perpendicular bisectors of 
respectively. Hence bisects and 
respectively.
. From 
, we get 
. Moreover, if 
, then notice that![\begin{align*}
\angle EAQ &= 180^{\circ}-\angle ABC \\
&= 180^{\circ}-\angle CDE \\
&= \angle DCP \\
&= \angle DBP \\[4pt]
\angle AEP &= 180^{\circ}-\angle BCD = \angle BPD
\end{align*}](http://latex.artofproblemsolving.com/a/6/f/a6fd783cb04b2bcde5e38130cb8b0170f57f3e70.png)
So 
which means 
or 
. Now 
and 
or 
are tangents to 
. Let be the midpoint of 
. Then by symmedian lemma, 
. Finally, angle chase
so we are done.
![[asy]
size(8cm);
pair A,B,D,P,E,M,C,O,X,T;
A=dir(215);
B=dir(305);
P=dir(120);
E=0.2*P+0.8*A;
M=midpoint(P--B);
C=2*M-E;
D=dir(100);
T=2*M-D;
O=circumcenter(B,C,D);
X=foot(O,D,M);
T=2*X-D;
filldraw(E--P--D--cycle,gray);
filldraw(C--B--T--cycle,gray);
draw(A--B--C--D--P--cycle,linewidth(1.2));
draw(unitcircle);
draw(B--P);
draw(E--C);
draw(B--D);
draw(D--T);
draw(E--D);
draw(T--C);
draw(T--B);
draw(circumcircle(B,T,C),dashed);
dot("$A$",A,dir(200));
dot("$B$",B,dir(-30));
dot("$C$",C,dir(0));
dot("$D$",D,dir(80));
dot("$E$",E,dir(180));
dot("$P$",P,dir(120));
dot("$M$",M,2*dir(240));
dot("$T$",T,dir(270));
[/asy]](http://latex.artofproblemsolving.com/4/e/7/4e7dc83b81dadcf4c2f9820a9c6965173d297e7d.png)
on 
such that 
.Now by the problem stipulation we have that 
is parallelogram.So lies on 
.Because 
we have 
and so 
.Whence it suffices to show 
or 
is cyclic.
be the reflection of about .By the problem condition we know 
is cyclic.Whence 
.Now a simple claim:
is the reflection of 
about 
and whence 
.Now compute: 
because .Whence 
is cyclic and we are done 
be the reflection of wrt .
, we get that 
are collinear and thus, 
. be the midpoint of , be the midpoint of . Thus, as 
, we have 
cyclic quadrilateral. Hence,
and as 
and hence we conclude that 
is a cyclic quadrilateral. Thus, 
we are done.
across to . Let be a point on such that 
. By the given length condition, 
as well.
, 
are collinear.
. So, since is midpoint of hypotenuse in 
, 
. Since , this means 
be midpoints of and . Since we're given 
, this means 
is cyclic. are midpoints, there is a homothety centered at with ratio that takes 
to 
and so the reflection of across , call it , lies on 
, reflecting stuff across , we have that 
and so is cyclic.
and so we are done. 
and be the reflections of and over 
Notice lies on 
as 
is a parallelogram and lies on as 
Also, the length condition implies 
is cyclic.
and 
are the reflections of 
and 
over 
respectively; hence, is cyclic. Since 
is a parallelogram, 
and be the reflections of and across . We have 
, so . Thus, is a cyclic trapezoid, and 
is also a cyclic trapezoid which is the reflection across .
so collinear. ![\[\angle BDB'=\angle CDE=\angle ABC=180^\circ-\angle BAE=180^\circ-\angle BAB'\]](http://latex.artofproblemsolving.com/3/5/a/35ac324cd0c7acd8909fd11d4712e42639874ad5.png)
so is cyclic. Note that 
so ![\[\angle ADB=\angle AB'B=90^\circ-\frac12 \angle B'AB=\frac12 \angle BDB'=\frac12 \angle CDE\]](http://latex.artofproblemsolving.com/a/6/7/a67e2530bbd8e1eea6bcb4caee79ed5c63f4e915.png)
as desired.
(such that triangle 
is isosceles) and then proving the two claims
collinear, where 
is the midpoint of 
, and
collinear, where is the reflection of over ?
cyclic.) Unfortunately, I wasn't able to make much headway proving either claim, but they intuitively do not feel hard to prove. (Thus I suspect that they're actually wrong -- which might make one of the funniest coincidences in geometry.)
such that:
such that 
.
at 
intersect 
.
,
meet this circle again at 
.
intersect 
at the distinct points 
.
and the perpendicular bisector of 
.
.
i.e. 
.
.
be the midpoint of 
,
is cyclic.
,
,
.
is cyclic.
where 
.
