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 an acute-angled triangle with circumcircle 
and circumcentre 
. Points 
and 
lie on such that 
and 
. Let 
meet 
at 
, and 
meet 
at 
.
and 
have an intersection lie on line 
.
be a positive integer. Alex plays on a row of 9 squares as follows. Initially, all squares are empty. In each turn, Alex must perform exactly one of the following moves:
Choose a number of the form 
, with 
a non-negative integer, and place it in an empty square.
Choose two (not necessarily consecutive) squares containing the same number, say . Replace the number in one of the squares with 
and erase the number in the other square.
, while the other squares are empty. Determine, as a function of , the maximum number of turns Alex can make.
be a convex hexagon with 
and 
.
which is equidistant from sides 
and 
.
and 
are the centers of mass of 
and 
, show that 
.
be a point in the plane, distinct from the vertices of 
. Consider 
the reflections of with respect to lines 
and 
, in this order.
are collinear if and only if lies on the circumcircle of . does not lie on the circumcircle of and the centroids of triangles and 
coincide, prove that is equilateral.
given that there exist positive integers 
satisfying
and , we define 
as follows:
where 
are all the positive divisors of . For example,
.
Find all positive integers such that 
.
Find all positive integers such that 
.
be an odd prime number. How many -element subsets 
of 
are there, the sum of whose elements is divisible by ?
be real numbers satisfying
be a fixed square and 
a variable point on segment 
. The square 
is constructed such that 
is on segment 
and 
is on segment 
. Let 
be the intersection point of lines 
and 
. Find the locus of as varies along segment .
be a positive integer. In an 
board, two opposite sides have been joined, forming a cylinder. Determine whether it is possible to place queens on the board such that no two threaten each other when: 
. 
.
denote the number of positive integers less than that are relatively prime to . Prove that there exists a positive integer for which the equation 
has at least 
solutions in .
that can be written in the form
where 
are pairwise coprime positive integers such that 
, 
, and 
.
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.
and![\[ f(x+y)+f(x^2+f(y))=f(f(x))^2+f(x)+f(y)+y,\forall x,y\in\mathbb{R}\]](http://latex.artofproblemsolving.com/c/8/9/c895ae7fdf8d7f284ac9fc94cc077d6edad6cbf0.png)
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 
,
=
,
is right angled at 
,
=
,
=