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.
is chosen. Prove that there exists a prime number that divides infinitely many terms of the sequence 
, where![\[
b_k = a^{k^k} \cdot 2^{2^k - k} + 1.
\]](http://latex.artofproblemsolving.com/1/7/5/1751a60482d729a36c71b77ac9c978e724f40da0.png)
+1 w
and all positive integers 
and 
satisfying 
is interesting if 2018 divides 
(the number of positive divisors of ). Determine all positive integers 
such that there exists an infinite arithmetic progression with common difference whose terms are all interesting.
be an odd prime number and a sequence 
satisfy 
and 
Prove that always exists positive integer satisfying for all positive integers , then 
.
is "not congruence"
and 
Prove that
Solution:
and 
then 
and 
intersect at points 
and 
. A line through intersects and at points 
and 
, respectively. Line 
intersects at point 
, and line 
intersects at point 
. If 
is the circumcenter of 
, prove that 
.
be a triangle with incenter 
, and let be a point on side 
. Points 
and 
are chosen on lines 
and 
respectively such that 
is a parallelogram. Points 
and 
are chosen on side such that 
and 
are the angle bisectors of angles 
and 
respectively. Let 
be the circle tangent to segment 
, the extension of 
past , and the extension of 
past . Prove that is tangent to the circumcircle of triangle .
passes through 
be the Miquel point of 
. We have that 
by spiral similarity so as 
, the two triangles are in fact congruent so 
. As 
clearly lie on the same side of 
, ( lies inside 
) we conclude that 
as desired.
![[asy]
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pair M = (-0.7418154036524869,5.291189386209763), P = (-2.5219236580982507,0.48914284080444176), Q = (0.07919666832617143,0.712700419228189);
pair X = (-1.2118272753134591,1.8859515227576646), Y = (-3.6364372221499246,4.089377965597112), Z = (0.8344687186207644,4.067687193302564);
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dot(P,linewidth(4pt) + dotstyle);
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dot(Q,linewidth(4pt) + dotstyle);
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/* end of picture */
[/asy]](http://latex.artofproblemsolving.com/e/5/6/e56d7aa792666714464bbb0b6b7362b20b530a97.png)
is cyclic. Let 
and be the feet of the -angle bisectors in triangles 
and 
, respectively. Observe that![\[
\angle PAX = \tfrac12\angle BAX = \angle ABX
\]](http://latex.artofproblemsolving.com/6/a/6/6a6f21593eae6d527a151b3a4c60954591da741e.png)
by the given condition, so 
. This yields 
, and analogous reasoning with 
yields 
. Therefore 
, and so quadrilateral 
is cyclic.
is cyclic. Suppose 
intersects again at point (not shown in the diagram). Menelaus and the Angle Bisector Theorem imply![\[
\frac{BY}{YC} = \frac{BP}{PX}\cdot\frac{XQ}{QC} = \frac{BA}{AX}\cdot\frac{XA}{AC} = \frac{BA}{AC}.
\]](http://latex.artofproblemsolving.com/c/1/2/c12c2c55b1f366c82e8702ec45f78587459fe6e6.png)
This means that is the foot of the -external angle bisector of , meaning that , , and 
are concurrent. Now![\[
YA\cdot YM = YB\cdot YC = YP\cdot YQ,
\]](http://latex.artofproblemsolving.com/8/9/9/899d5850b913ba0ffacff298132e450e0a0f8eb9.png)
and so quadrilateral 
is cyclic.
is a center of homothety. Let , 
, and 
intersect 
again at 
, 
, and 
respectively. Then a bit of angle chasing yields![\[
\angle RSP = \angle RAP \equiv \angle XAP = \angle ABX,
\]](http://latex.artofproblemsolving.com/b/3/a/b3af3bf26f6696b0b91a591b4f75b4d7ab7f16b8.png)
and so 
. Analogous reasoning yields 
, and furthermore Reim's Theorem tells us 
. Therefore triangles and 
are homothetic, and their center of homothety is . to 
, and so lies on the line connecting the centers of these two circles. But this line is precisely the perpendicular bisector of 
, and so 
as desired.
, this swaps 
, 
and stays, hence swaps 
. Now , which lies on both circles stays, thus lies on and 
.
be the perpendicular bisector of 
and let 
, which means 
is the angle bisector of 
. 
. By symmetry, is equidistant from the perpendicular bisectors of 
, as desired (sure it could be equidistant to the other arc midpoint but config issues don't matter).
be the circle centered at with radius 
. Consider the "force-overlay polarity" given by doing pole-polar duality wrt , then reflecting over the angle bisector of 
. It is not hard to check that under this transformation, the problem is as follows (angle chase; the two angle conditions act independently):
be an acute triangle, and let the external angle bisector of intersect line at . Let 
be the circle centered at 
through ; define 
similarly. Suppose is a common tangent of and . Show that 
lies on . is the exsimilicenter of the two circles (compare radii and distances). 
and 
.
,
.
that satisfies:
and 
.
![$$(\sqrt{x}+1)[2-(x-6)\sqrt{x-3}]=x+8$$](http://latex.artofproblemsolving.com/b/1/d/b1d283f2e7b8e574b6c0ff72236c52a39d3da9e8.png)
Prove that
Prove that
Denote by 
containing 
again at 
so 
so 
.
which promptly gives us that 
.
and 
to 
,
Thus 
bisects 
and thus 
,
,
denote the reflections of 
,
.
,
.
and 
we see that 
is parallel to the bisector of 
.
,
By easy angle chasing we have 
Then we have 
because we know that 
and 
We conclude that 
is cyclic and we are done.