AoPS Academy
+1 w
[/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.
by 
grid of squares coloured in the same way as a standard checkerboard. Find the total number of ways to place 
counters on white squares so that each square contains at most one counter and no two counters are in diagonally adjacent white squares.
be a positive integer . Prove that there exists infinitely many pairs of positive integers 
such that 
and :
matrix whose entries come from the set 
is called a silver matrix if, for each 
, the 
-th row and the -th column together contain all elements of 
. Show that:
;
.
the cells of a 
table can be filled with the numbers 
such that the following conditions are satisfied: appears exactly once.
rectangle, one of the numbers is the arithmetic mean of the other two.
is located in the middle cell of the table.
types of candy, numerated type 1, type 2, ... type n. Initially she takes 
candy pieces and places them in a row on a table. Then, she chooses one of the following operations (if available) and executes it:
She eats a candy of type 
, and in its position in the row she places one candy type 
followed by one candy type 
(we consider type 
to be type 1, and type 0 to be type ).
She chooses two consecutive candies which are the same type, and eats them. for which Celeste can leave the table empty for any value of and any configuration of candies on the table.
be a positive integer. Given is a subset 
of 
with 
elements. Prove that there exist three elements 
from such that 
.
be a convex hexagon with 
and 
.
which is equidistant from sides 
and 
.
and 
are the centers of mass of 
and 
, show that 
.
be real numbers satisfying
be a positive integer. On the table, we have 
ornaments in different colours, not necessarily of each colour. Prove that we can hang the ornaments on Christmas trees in such a way that there are exactly ornaments on each tree and the ornaments on every tree are of at most 
different colours.
on each of six problems, and their total score is the sum of the six scores (out of 
). Given two students and 
, we write 
if there are at least five problems on which scored strictly higher than .
such that the following statement is true: for every integer 
, given students 
, \dots, 
satisfying 
, the total score of is always at most points more than the total score of .
be an acute triangle with circumcenter 
such that 
, let 
be the intersection of the external bisector of 
with 
, and let be a point in the interior of such that 
is similar to 
. Show that 
.
as the unit circle, and let 
be the affixes of 
. Let lowercase letters denote the complex affixes for the other points. Suppose without loss of generality the midpoint 
of arc 
has affix 
. Since 
, we find 
. From 
, 
. We compute ![\[p-a^2=-\frac{(a^2-b^2)(a^2-c^2)}{2a^2-b^2-c^2}\]](http://latex.artofproblemsolving.com/8/4/1/841c4ac142862584c5ace5c16c6c44e68c192a5c.png)
![\[q-b^2=\frac{c(a^2-b^2)(c-b)}{a^2-bc}\]](http://latex.artofproblemsolving.com/e/7/6/e76a16ed204bd8548c6e2d87cb7f90d21e9f1f7b.png)
![\[q-p=\frac{(a^2-b^2)(a^2-c^2)(b^2+c^2-bc-a^2)}{(2a^2-b^2-c^2)(a^2-bc)}\]](http://latex.artofproblemsolving.com/5/4/3/54370851091362ca7fdb1819ce2a1ae10105ff18.png)
. From the computations above,![\[\frac{q-p}{p-a^2} \cdot \frac{o-q}{q-b^2}=\frac{(b^2+c^2-bc-a^2)(a^2b^2+a^2c^2-a^2bc-b^2c^2)}{c(a^2-bc)(c-b)(a^2-b^2)} \]](http://latex.artofproblemsolving.com/d/3/2/d3208451e99519703bef9010c8ffede49d89524a.png)
and 
be the midpoints of 
and its minor arc respectively .
are concyclic . Note also that 
is 
-symmedian of 
.
. It is therefore sufficient to prove that 
~ 
. Consider 
, on the other hand , 
. Extend and it meets the circumcircle again at 
then 
~ 
and 
~ 
, 
so 
and 
and we are done .
be the medial triangle of and let 
be the circle of diameter 
passing through 
Let 
be the second intersections of 
with 
, we obtain 
Meanwhile, we also obtain 
Therefore, it is well-known (Characterization 2) that lies on the -symmedian in 
Then since 
lies on the -median in 
, it follows that 
and 
are isogonal WRT 
Therefore, 
Finally, note that 
is the midpoint of arc 
on because 
is the external bisector of 
, Pascal's Theorem on cyclic hexagon 
yields 
Therefore, 
Thus, 
are collinear, and it follows that 
Meanwhile, as is the circle of diameter , we have 
Therefore, 
are concyclic. Hence, 
Thus, ![\[\measuredangle APQ + \measuredangle M_AQO = \measuredangle AXK + \measuredangle KXO = \measuredangle AXO = 90^{\circ}. \; \square\]](http://latex.artofproblemsolving.com/6/9/e/69eb5f6ad7f211fb774dc020a4e348d33d551733.png)
be the circumcircle of triangle 
the tangants from 
intersect eachother at point let the line 
intersect at point 
. At first proof that is unic and it is the midpoint of 
. Let 
be the midponts of arcs 
. Lines 
intersect each other at point . and is the midpoint of segment 
. Triangles 
are similar and points 
are the midpoints of 
so angles 
are equal . We know that
is perpendicular to so 
so we find that 
and denote the image of as 
for all points . The given condition about implies that the circumcircle of 
is tangent to 
at and the circumcircle of 
is tangent to 
at . These circles will invert to lines parallel to and going through 
and 
respectively making 
a parallelogram. Since was on , 
is on the circumcircle of 
and is in fact diametrically opposite the midpoint of minor arc 
(from angle chasing). Finally extend 
to meet at 
. If we let 
denote the circumcenter of (this is not the image of ), then it suffices to show by simple properties of inversion that 
Reflect 
over 
to get 
. Then it suffices to show 
To do this, observe that is also the reflection of across which yields 
making the center of spiral similarty mapping to 
. Hence by the spiral similarity lemma, if denotes the intersection of 
and 
then we have 
and 
are both cyclic. Thus 
as desired.
and let be the feet of the 
bisector on . Then 
is the apollonius circle(let its be ) and 
is the center of . Let it intersect 
at 
. Then 
is the symmedian in . From the given angle condition it is clear that is the midpoint of . Since and 
are orthogonal , inversion around 
swaps 
. Hence 
. 
.
be an acute triangle with circumcenter such that , let be the intersection of the external bisector of with , and let be a point in the interior of such that is similar to . Show that . is the 
of . So, 
and 
. So, it just suffices to show that 
. So we just have to show that 
is tangent to at .
and 
be the bisector of 
where 
. So, is the Circumcenter of the 
Also 
and 
is harmonic. So, Combining MC'laurin and PoP we get that 
inversion and reflect across the internal bisector of . 
is the reflection of over , 
is a parallelogram. is the intersection of the external bisector of and the circumcircle. is the reflection of over . ![\[\angle QPA + \angle OQB = 90^{\circ} \iff \angle AQ'P' + \angle AB'Q' - \angle AO'Q' =90^{\circ} \iff \angle AO'Q' = \angle RQ'P'\]](http://latex.artofproblemsolving.com/9/9/5/995cfd6f3b05af6cac54a35c4bc0efd59cc5ec0c.png)
which equal to 
.
meet 
again at , the foot of the -altitude be , the midpoint of be , and 
. Clearly, is the -Dumpty point.-inversion. Notice that 
is the reflection of over , 
is the reflection of over , 
is the midpoint of arc , and 
is the second intersection between and .
from midlines, 
so is the circumcenter of 
, which implies 
. Furthermore, we have 
which means 
is the perpendicular bisector of 
. and are also symmetric about . Now, since lies on , we have 
as desired. 
is the 
Dumpty point of .
and 
and let the angle bisector intersect at 
we get that is just the intersection of the tangents from and 
to as is the midpoint of Appolonius circle is (I actually didnt know that and had to prove it but I'm lazy now xD)
, it is clear that 
is the 
Symmedian in 
because 
is orthogonal to 
is the 
point so if we let be the intersection of the 
with then is the midpoint of . (We cite this as well-known but it is easy as inversion maps with some point 
satisfying that 
is a parallelogram, which concludes the proof of the claim). be the feet of the interior bisector of and 
be the midpoint of 
. See that 
. But now, as 
this implies that is the center of 
. Thus, 
is tangent to . Now, by LaHire, as 
we also get that 
is tangent to . Thus, we get 
, while at the same time 
, so we are done as 
ends the problem.
is dumpty. After performing inversion and reflecting over the angle bisector of 
, lies on the perpendicular bisectors of 
thus, 
and are tangent to each other where 
is an isosceles trapezoid.![\[\measuredangle OQB+\measuredangle QPA=180-\measuredangle QPO+\measuredangle QPA=90\]](http://latex.artofproblemsolving.com/2/6/1/261943a575cc16f82382e73bfb5c2bd1d5b9057d.png)
As desired.
Let 
be the major arc midpoint of 
. Rename as and as 
(see that this is the -Dumpty point). Let be the reflection of about midpoint of 
.
is tangent to 
.
. Then see that and are swapped; is swapped with reflection of over midpoint of (call it 
). So we just need to prove that 
which is just because 
is a parallelogram. ![\[\angle AD_AX \overset{\sqrt{bc}}= \angle AN_AA'=180 ^{\circ}-\angle XON_A=\angle M_AOX=90 ^{\circ}-\angle OXB\]](http://latex.artofproblemsolving.com/0/6/2/0629cbdf049ed1cd84bcd559ece24d75316d18bd.png)
as required.
are roots of 
,
and 
: Prove that :
and 
and![\[P=a_1^2+a_2^2+...+a_{2024}^2-(a_1a_3+a_2a_4+...+a_{2022}a_{2024})\]](http://latex.artofproblemsolving.com/d/b/b/dbb88435dcbba1f5f43f99cdf1ee42aeea3f24ea.png)
Find maximum 
at 
(call it 
and that 
be the center of ![\[
\angle OQB=\angle QPS=90-\angle APQ
\]](http://latex.artofproblemsolving.com/4/f/c/4fc5ba107b79c72143f734f00256f7787ecd7a8a.png)
as desired.
and 
which is precisely 
where 
which is 
,
=
,
is right angled at 
,
=
,
=