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.
has an incenter 
. Let 
be the midpoint of 
. Suppose that 
. 
meets 
again at point 
. Let points 
and 
be such that is the midpoint of 
and is the midpoint of 
. Point 
lies on the plane such that 
is a parallelogram, and suppose the angle bisectors of 
and 
concur on 
.
and 
meet 
at 
and 
. Prove that 
.
+2 w
be a triangle. Let 
and 
be the intersections of the 
and 
angle bisectors with the opposite sides. Let 
. Let 
where 
is the major 
arc midpiont.
and are coconic on a conic 
intersects 
again at , not equal to 
or , then prove 
and 
concur on 
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 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 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: 
. 
.
, done.
and are given on side of convex quadrilateral (with closer than to ). It is known that 
and 
. Prove that 
.
+ 
..... (1)
- 
....(2)
= 180
=
is cyclic, which implies 
. This also implies that 
. However, we note that 
which implies is cyclic. To finish, we note that 
which implies 
. 
is cyclic because 
.
. We then proceed to angle chase. 
because is cyclic and opposite angles in a cyclic quadrilateral add to 
.
by supplementary angles and 
because the angles in a triangle add up to .
, which shows that 
. Finally, because you just subtract equal angles 
and 
. 
and 
, if we add those equalities to the desired conclusion, we then need to prove that is cyclic., we know that is cyclic, so let 
and 
. Let 
. We then have 
, and by sums of angles in triangle we have 
, so is cyclic as desired. 
, 
is cyclic, wherefore
So is cyclic as well, yielding 
or![\[\angle FAC=\angle BAC-\angle BAE-\angle EAF=\angle CDB-\angle CDF-\angle FDE=\angle EDB,\]](http://latex.artofproblemsolving.com/d/0/1/d01bda978ce395ccda1aa19db70f77153ed72658.png)
as desired. 
is cyclic. Because of this, we can set 
and 
. Now, setting 
and 
, we can angle chase to get that 
and 
. Clearly, 
, so is cyclic. Finally, we have 
, which implies that . We are done. 
be the intersection of 
and 
Since 
is cyclic. As such, 
and 
We also have 
so 
As such, 
As such, 
so 
and is cyclic. As such, 
so 
as desired. 
, we know that 
is cyclic.
, we just need to prove that or that is cyclic
, is cyclic, as desired.
, is a cyclic quadrilateral. It follows that 
However, we also have 
. Thus, 
and 
, so it follows that 
or is cyclic. Then, 
, but we have 
, so 
as desired.
if and only if is cyclic.
and 
is cyclic. This means that 
Therefore, 
which means that 
proving that is cyclic, as desired.
. So, quadrilateral is cyclic.
.
. is cyclic. Hence 
.
is a cyclic quadrilateral. Take point on line such that 
. Then by angle chasing,
which implies quadrilateral being cyclic. Thus
as desired.
. Since is cyclic, 
. 
and 
also equal to so 
and 
. 
so is cyclic. Hence so 
, we know that is cyclic. This also means that 
, implying 
.
. If we substitute in angles (
from earlier and 
from the given), we find that 
. Thus, is cyclic, as the opposite angles in the quadrilateral are supplementary.
by the cyclic condition, and subtracting off some equal angles yields 
, as desired.
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)
.
.
.
,
,
is cyclic, so 
.
.
,
Hence, 
,
Q.E.D.
so, 
,
to intersect 
and 
at 
are cyclic.
is cyclic. Now note that 
.
,
.
after simplifying, so quadrilateral 
,
as required. This was a lot easier in hindsight.
.
.![\[ \measuredangle ABC = \measuredangle ABE = \measuredangle BAE + \measuredangle AEB = \measuredangle BAE + \angle AEF \]](http://latex.artofproblemsolving.com/0/e/8/0e8883554573b4877b050977e489a1b68d3e27fd.png)
and that ![\[ \measuredangle ADC = \angle ADF + \angle FDC \]](http://latex.artofproblemsolving.com/b/3/6/b36a40fce2548385508623223ab4710492d843c6.png)
Finally, ![\[ \measuredangle ABC = \measuredangle BAE + \measuredangle AEF = \measuredangle BAE + \measuredangle ADF = \measuredangle FDC + \measuredangle ADF = \measuredangle ADC \]](http://latex.artofproblemsolving.com/6/6/4/664d33dd786dd8481cb5785135a0014d483438fd.png)
![[asy]
pair a, b, c, d, e, f;
a = dir(140); d = dir(220); f = dir(300); e = dir(30); b = 1.3 * e - 0.3 * f; c = intersectionpoints((100*e-99*f)--(100*f-99*e),circumcircle(a,d,b))[1];
draw(a--e--f--d--cycle,orange); draw(d--c--f,blue); draw(e--b--a,blue); draw(circumcircle(a, b, c),blue); draw(circle((0,0),1),orange);
dot("$A$", a, dir(120));
dot("$D$", d, dir(240));
dot("$F$", f, dir(330));
dot("$E$", e, dir(30));
dot("$B$", b, dir(30));
dot("$C$", c, dir(270));
[/asy]](http://latex.artofproblemsolving.com/c/f/a/cfab0fa694eb45c43d633409951a99b4f16775d8.png)
is cyclic. Because 
implies that 
,
,