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.
where 
is an odd integer. Let 
be a function defined on 
taking values in 
such that
for all 
for all 
; b) 
identical-looking coins in a circle. It is known that exactly two of them are fake. Real coins weigh the same, fake ones too, but they are lighter than the real ones. How can you determine in three weighings on a cup scale without weights whether there are fake coins lying nearby or not??
points of general position marked on the coordinate plane (i.e., points among which there are no three lying on the same straight line). Is there a polynomial of two variables 
a) of degree 
; b) of degree such that it equals to zero exactly at these marked points?
be triangle, inscribed in parabola. Tangents in points 
forms triangle 
. Prove that 
. (
is area of triangle 
).
such that there exist integers 
with ![\[x^3_1+x^3_2+\,\ldots\,+x^3_t=2002^{2002}\,?\]](http://latex.artofproblemsolving.com/e/f/a/efaa98e77ac0273284ee3446e4335d8cc4f681cb.png)
and 
doubloons, where and are given mutually prime natural numbers, with 
. A non-negative integer is called lucky if the sum in doubloons can be scored with using no more than 
coins. Find the number of lucky numbers.
with no parallel sides is inscribed in a circle 
. Circles 
are inscribed in triangles 
, respectively. Common external tangents are drawn between 
and 
, and 
, and 
, and and , not containing any sides of quadrilateral . A quadrilateral whose consecutive sides lie on these four lines is inscribed in a circle 
. Prove that the lines joining the centers of and , and , and the centers of and all intersect at one point.
be a quadrilateral inscribed in a circle 
such that 
. Let 
and 
be the midpoints of 
and 
respectively. The line 
meets again at 
. Prove that the tangent at to , the line and the line 
are concurrent.
be a continuous function. A chord is defined as a segment of integer length, parallel to the x-axis, whose endpoints lie on the graph of 
. It is known that the graph of contains exactly 
chords, one of which has length 2025. Find the minimum possible value of .
has been partitioned into disjoint pairs 
(
) so that for all 
, 
equals 
or 
. Prove that the sum ![\[ |a_1-b_1|+|a_2-b_2|+\cdots +|a_{999}-b_{999}| \]](http://latex.artofproblemsolving.com/0/4/4/04422a7d2266b92a6b3f215b698f078cda85091f.png)
ends in the digit 
.
be an acute-angled triangle with circumcircle 
. A circle is internally tangent to at 
and also tangent to 
at 
. Let and 
intersect at 
and 
respectively. Let and be points on line such that 
is the midpoint of 
and 
is the midpoint of 
. Lines 
and 
meet at 
and intersect again at 
and 
respectively. The ray 
meets the circumcircle of triangle 
again at 
.
. be a triangle and 
its circumcentre. A line tangent to the circumcircle of the triangle 
intersects sides at and at . Let 
be the image of under 
. Prove that the circumcircle of the triangle 
is tangent to the circumcircle of triangle .
be a semicircle with diameter and midpoint . Let be a point on different from and .
touches in a point , the segment 
in a point , and additionally the segment . The circle 
touches in a point and additionally the segments 
and .
and 
are perpendicular.
is a cyclic quadrilateral whose diagonals intersect at . The circumcircle of 
meets segment at points and . The circumcircle of 
meets segment at points and 
. Let and be the incenters of 
and 
, respectively. Segments 
and meet at . Prove that the points 
are cyclic.
is a cyclic quadrilateral whose diagonals intersect at . The circumcircle of meets segment at points and . The circumcircle of meets segment at points and . Let and be the incenters of and , respectively. Segments and meet at . Prove that the points are cyclic.
to the bisector of angle 
intersect 
at 
,and define 
analogously.
and 
is cyclic.
and are collinear;
(which can be gotten easily)
and 
are concyclic.Now define new point 
.Again chase some angles to get 
and are concyclic.But since 
are collinear 
.
to the bisector of angle intersect at ,and define analogously. should be distinguished from the one in the Question. is sort of vague.
is perpendicular to bisector of angle 
. Then by some simple angle chasing we are done.
intersects 
at 
respectively, and be the intersection of 
and 
which implies that lies on the perpendicular bisector of 
Meanwhile we have 
which follows that 
is the angle bisector of 
is parallel to 
are collinear, this is because 
Notice that 
Now that bisects 
we get![\[\frac{SQ}{SR}=\frac{PQ}{PR}=\frac{QA}{RB}=\frac{QI}{RJ}.\]](http://latex.artofproblemsolving.com/2/8/c/28c187e921197b9dca57a61269566a97cea00ff3.png)
Hence is parallel to as desired. 
so are concyclic.![[asy]
size(7cm); pointpen=black; pathpen=black; defaultpen(fontsize(9pt));
pair A,B,C,D,P,I,J,Q,R,E,F,S;
A=dir(-160); B=dir(-20); C=dir(30); D=dir(110);
P=IP(A--C,B--D);
E=OP(circumcircle(A,P,D),A--B);
F=IP(circumcircle(B,P,C),A--B);
I=incenter(A,D,E); J=incenter(C,F,B);
Q=OP(L(E,I,5,5),circumcircle(A,P,D));
R=OP(L(F,J,5,5),circumcircle(B,P,C));
S=extension(I,E,J,F);
filldraw(anglemark(A,Q,P,3),magenta);
filldraw(anglemark(A,D,P,3),magenta);
filldraw(anglemark(A,C,B,3),magenta);
filldraw(anglemark(P,R,B,3),magenta);
D(unitcircle);
D(circumcircle(A,P,D),blue);
D(circumcircle(B,P,C),blue);
D(A--B--C--D--cycle);
D(A--C); D(B--D); D(D--E); D(C--F);
D(I--J,red+linewidth(1.2)); D(Q--S--R, dashed);
D(Q--R,red+linewidth(1.2)); D(Q--A, dashed);
D(R--B,dashed); D(E--P--F,deepgreen+linewidth(1.1)); D(S--P,dashed);
dot("$A$",A,dir(A));
dot("$B$",B,dir(B));
dot("$C$",C,dir(C));
dot("$D$",D,dir(D));
dot("$P$",P,dir(85)*2);
dot("$E$",E,dir(-90));
dot("$F$",F,dir(-65));
dot("$I$",I,dir(60));
dot("$J$",J,dir(-90));
dot("$Q$",Q,dir(Q));
dot("$R$",R,dir(R));
dot("$S$",S,dir(S));
[/asy]](http://latex.artofproblemsolving.com/b/0/9/b09186dca6c23686ad44501f9b5e5b151c631b99.png)
be the circumcenters of 
and 
. Let 
. Note that 
is the bisector of angle 
. Moreover, 
. Hence 
. Consequently, 
proving 
. Now it is 
proving that are concyclic. 
and 
be the midpoints of shorter arcs and in the circles 
and 
respectively.
is the intersection point of 
and 
.
are collinear.
and 
are similar.
![\[ \angle BJC=90^{\circ}+ \frac{\angle BFC}{2} = 90^{\circ}+ \frac{\angle BPC}{2} =90^{\circ}+ \frac{\angle DPA}{2}=90^{\circ}+ \frac{\angle DEA}{2}= \angle DIA. \]](http://latex.artofproblemsolving.com/c/d/8/cd8de6b2cf8b36ce3357b7c363bd347df42a6924.png)
wrt and 
wrt we get ![\[ \frac{\sin \angle AJB}{\sin \angle CJA} = \frac{\sin \angle AIB}{\sin \angle BID}. \]](http://latex.artofproblemsolving.com/7/5/7/757cf5e106d85c74ef46cb8c8f453317c62c941b.png)
and thus 
is cyclic. Finally note that ![\[ \angle JBA+ \angle AIJ=180^{\circ} \implies \angle EIK = \angle EAK. \]](http://latex.artofproblemsolving.com/3/5/d/35dc1974d57dfd1af84059714dd95de95e7b7bee.png)
We conclude 
is cyclic.
and be the intersections of 
and 
with 
and 
respectively. Let 
. Now, we can get started., and are collinear.![\[2\measuredangle MPD = 2\measuredangle MED = \measuredangle AED = \measuredangle APD = \measuredangle CPB = \measuredangle CFD = 2\measuredangle NFB = 2\measuredangle NPB\]](http://latex.artofproblemsolving.com/6/2/5/625eaa27e2cb276b689fa9335b9e0996d56c9899.png)
This means that 
from which it is quite clear that 
as claimed.
and are parallel.
. This is beacause,![\[\measuredangle APM \sim \measuredangle CPN = \measuredangle CFN = \measuredangle NFB = \measuredangle NPB\]](http://latex.artofproblemsolving.com/1/8/5/185c25fbee05d5771f87db9125c7be3d704cf156.png)
and![\[\measuredangle PMA = \measuredangle PDA = \measuredangle BDA = \measuredangle BCA = \measuredangle BCP = \measuredangle BNP \]](http://latex.artofproblemsolving.com/1/d/a/1da256adabde3233098e85a64cd9af91eabc48f9.png)
and 
. Thus,![\[\frac{MI}{NJ} = \frac{MA}{NB} = \frac{AP}{PB} = \frac{\sin \angle ABP }{\sin \angle PAB}\]](http://latex.artofproblemsolving.com/d/c/5/dc59b40c174546a433835294831296201252c121.png)
Further,![\[\frac{TM}{TN}=\frac{\sin \angle TNM}{\sin \angle TMN}=\frac{\sin \angle FNP}{\sin \angle EMP}=\frac{\sin \angle FBP}{\sin \angle EAP}=\frac{\angle ABP}{\sin \angle PAB}\]](http://latex.artofproblemsolving.com/9/7/c/97c4a1fd6a14ed69cf93fc9c91e1a92c89f363c2.png)
![\[\frac{MI}{MJ}=\frac{TM}{TN}\]](http://latex.artofproblemsolving.com/e/a/3/ea3778d2587633737b58e8d3f9230df1ede9638e.png)
which proves the claim.![\[\measuredangle JIE = \measuredangle NME = \measuredangle PME = \measuredangle PAE = \measuredangle KAE\]](http://latex.artofproblemsolving.com/8/5/f/85fc0a834aeee81940bf10a1b42aced7825e864d.png)
which implies that points 
and are indeed concyclic which was the desired conclusion.
to 
since I'm bad and let 
. Let and denote 
and 
as arc midpoints. We angle chase 
collinear. Then, ![\[\frac{MK}{NK}=\frac{\sin MNK}{\sin NMK}=\frac{\sin ABP}{\sin BAP}=\frac{AP}{BP}\]](http://latex.artofproblemsolving.com/c/f/2/cf25934a48ba5650c366d91c999f8c4447619043.png)
by LoS. But as 
we get that 
, we get 
so 
. To finish, ![\[\measuredangle EIL=\measuredangle EIJ=\measuredangle EMN=\measuredangle EMP=\measuredangle EAP=\measuredangle EAL\]](http://latex.artofproblemsolving.com/d/2/0/d20a20e27c73065421ee75524280abd4d037fe0e.png)
and 
Prove that
