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 triangle with incentre 
and 
. Let lines 
and 
intersect the circumcircle of at 
and 
, respectively. Consider points 
and 
such that 
and 
are parallelograms (with 
, and 
). Let 
be the point of intersection of lines 
and 
. Prove that points 
, and are concyclic.
positive real numbers such that 
. Prove that :
and 
, there cannot exist infinitely many 
such that![\[
\sqrt{n + a} + \sqrt{n + b} \in \mathbb{Z}.
\]](http://latex.artofproblemsolving.com/5/4/d/54d20546de5736770a9929ac86163e3b06553b7d.png)
![\[
x = \sqrt{n + a} + \sqrt{n + b} \in \mathbb{N}.
\]](http://latex.artofproblemsolving.com/6/6/8/668d53723d307556536736fda0e4a19be848834f.png)
![\[
x^2 = (\sqrt{n + a} + \sqrt{n + b})^2 = (n + a) + (n + b) + 2\sqrt{(n + a)(n + b)}.
\]](http://latex.artofproblemsolving.com/1/9/f/19f2627a432354a757d3c9e8ba42e96f83f46726.png)
So:![\[
x^2 = 2n + a + b + 2\sqrt{(n + a)(n + b)}.
\]](http://latex.artofproblemsolving.com/0/7/3/073602693604bbf3cb1f5c1c2e5051e8994a5e78.png)
![\[
\sqrt{(n + a)(n + b)} \in \mathbb{N}.
\]](http://latex.artofproblemsolving.com/1/1/4/11441fab3ba768fee7a0eef4dbf68124d6637aec.png)
![\[
(n + a)(n + b) = k^2.
\]](http://latex.artofproblemsolving.com/0/e/c/0ec5e7bac2aebe9ec507727ed21f7ca028655222.png)
Assume 
. Then we have:![\[
n + a \mid k \quad \text{and} \quad k \mid n + b,
\]](http://latex.artofproblemsolving.com/7/5/f/75fc52b6662df3a4c0cb2f40f49c4b3351c90a4d.png)
or it could also be that 
.![\[
(n + a)k_1 = k \quad \text{and} \quad kk_2 = n + b.
\]](http://latex.artofproblemsolving.com/1/f/5/1f5936241f7bebcf15c68fed5267bbfad2d40c2b.png)
![\[
k_1 k_2 = \frac{n + b}{n + a}.
\]](http://latex.artofproblemsolving.com/d/4/c/d4c1958e3554b58c3def282070f849842cc10912.png)
, we have:![\[
k_1 k_2 = 1 + \frac{b - a}{n + a}.
\]](http://latex.artofproblemsolving.com/3/0/6/306a6a00c44d4e4db1b36c7e588f684f28fdbf44.png)
, 
must be an integer, which is not possible..
be a sequence of positive integers with 
, 
and![\[a_n = a_{n-1}^{a_{n-1}a_{n-2}}-1\]](http://latex.artofproblemsolving.com/1/b/f/1bf1a88979e7d2b33530de979d66a4e81032953d.png)
for all 
. Show that if 
is a prime less than 
for some positive integer 
, then there exists 
such that 
.
and be the incircle and incenter of triangle . Let 
, 
, 
be the tangency points of to 
, 
, 
and let 
be the reflection of about . Assume 
intersects the tangents to at and at points 
and 
. Show that 
.
, find all functions 
such that![\[f(x+y+f(y))=f(x)+f(ay)\]](http://latex.artofproblemsolving.com/7/5/d/75d1ac71fa0c2e53fdfc8e198b109d18e935fe74.png)
holds for all 
.
of the integers with the following property: for any integer 
there is exactly one solution of 
with 
.
be a natural number. The numbers 
are written in a row in some order. For each pair of adjacent numbers, their greatest common divisor (GCD) is calculated and written on a sheet. What is the maximum possible number of distinct values among the 
GCDs obtained? be a finite set of cells in a square grid. On each cell of we place a grasshopper. Each grasshopper can face up, down, left or right. A grasshopper arrangement is Asturian if, when each grasshopper moves one cell forward in the direction in which it faces, each cell of still contains one grasshopper., the number of Asturian arrangements is a perfect square. is the following set:
for all x, y 
0, where is the prime number.
satisfying
for all integers 
.
, and for any positive integer , 
means 
applied times to , and 
means 
applied times to .
centered at , the reflections of across the sides would map to the foot of the perpendicular from to each of the sides of 
. Let the points 
be the feet of the perpendiculars from to sides 
. If we prove that 
is cyclic, then the reflections of across the sides of the quadrilateral would also be cyclic because homothety preserves circles. We have 
since 
is cyclic, 
, and 
since 
is cyclic. So, we have 
. Similarly, we have 
. We have 
and 
, so 
, hence is cyclic.
, , 
and 
to be reflections over 
, 
, 
and 
respectively and , , and to be the intersection of 
, 
, 
and 
with , , and respectively. is cyclic.
= 
and 
= 
.
and 
are cyclic and
is right angled at we get, ![\[\angle ESP = \angle EAP = \angle EAB = 90^{\circ} - \angle ABE = 90^{\circ} - \angle PBE = 90^{\circ} - \angle PQE\]](http://latex.artofproblemsolving.com/e/f/8/ef82eadf0721f7a9270dae779015ac8fadad4d86.png)
.
Hence, quadrilateral is cyclic.
at with a scale factor of 2. So, 
., , and are concyclic and since homothety preserves circles, we can say that , , and are concyclic as well.
across 
be 
respectively, and let 
intersect at 
respectively. Using midlines, we find 
, 
, 
, 
. From similar triangles, we get 
, 
, 
, 
, proving 
with ratio 
. To show 
is cyclic, we need 
, which follows from 
. Since quadrilaterals 
are cyclic, we have 
due to the perpendicular diagonals, thus we are done.
across , , , and be , , , and , respectively. Note 
.
Summing implies
Therefore,
Hence, 
is cyclic, as desired. OoPsOoPs.
be the reflections of across 
, respectively. Let 
be the midpoints of 
, 
, 
, 
, respectively (note that lies on , and so on). Now, take a homothety centered at with scale factor 
. This sends to 
. It remains to show that is cyclic. Fortunately, this is quite easy to see by straight angle chasing. Note that 
, 
, 
, 
are cyclic.
However, 
, so 
. Therefore, is cyclic, so is also cyclic. 
across sides 
as respectively. Now we take a homothety with scale factor 
and as center of homothety. are reflections of E , they get reflected onto sides , let us call these points 
.
is cyclic, this is equivalent to being cyclic.
.since 
.
is cyclic.
are also cyclic.
is cyclic , therefore points are also concyclic.
is any point in any such 
. Prove the result.
be the perpendiculars drawn form to sides 
and respectively. Let 
be the reflection points. Take a homothety 
with center and ratio 2. then,
Let 
denote the midpoints of sides 
. Let 
be the midpoint of 
.Then from the 
, we have that
From 
is cyclic which implies that the reflections are cyclic as well
be the feet of the altitudes from to respectively. is cyclic.
so 
(since points E, P, A, and S are concyclic, as are points P, B, Q, and E). Similarly, 
We want to prove that 
or 
Since 
and 
, we are done as 
. with scale factor 
. It will bring to the quadrilateral in the problem. Since is cyclic, we are done.
be the feet of to 
respectively. It suffices to show that 
is cyclic, as by homothety of scale 
from to 
, the result follows.
are all cylic because of the right angles. Now with some angle chase we get that:
and 
, as desired. 
be a convex quadrilateral such that diagonals 
and 
intersect at right angles, and let 
be their intersection. Prove that the reflections of 
are concyclic.
intersect 
.
is cyclic,and 
,
,
,
,
,
,
because 
,
.
brings the reflections back to the feet of perpendiculars simplifying the solution
,
,
at the points 
,
,
,
,
,
and 
have parallel sides, they are similar. So, we must to show that the quadrilateral 
,
,
we have 
,
,
,
and 
.
,
are concyclic.![\[\angle EPA = \tfrac12\pi = \pi - \tfrac12\pi = \pi - \angle ESA\]](http://latex.artofproblemsolving.com/5/8/1/581a91b1f43afd2e670376c4ce6ba283d8d92b95.png)
Thus, 
are concyclic.
.
respectively.
be the circles centered at 
with radius 
.
and 
.
centered at 
)
Radical Axis of 
.
respectively .
.
.
then 
,
,
,
is cyclic and thus 
,
.
,
is cyclic as well, so thus 
,
.
,
,
.
,
is cyclic as well, so thus 
,
.
,
,
be foot of altitudes from 
is cyclic. with scaling with center of 
we are done.
denote the foot of the perpendicular from 
respectively. Let the reflection of 
be 
.
are 
is cyclic. Noting that 
are all cyclic,
Hence, 
,
has isogonal conjugate 
)
are reflections of 
i
.
)
around 
r
c
,
,
,
,
,![\[\angle EIF = \angle EAF = 90^{\circ} - \angle ABE = 90^{\circ} - \angle FBE = 90^{\circ} - \angle FGE\]](http://latex.artofproblemsolving.com/a/f/b/afb2432543a5a44ddaacc9691dfec2ef27260541.png)
![\[\angle EIH = \angle EDH = 90^{\circ} - \angle ECD = 90^{\circ} - \angle ECH = 90^{\circ} - \angle EGH\]](http://latex.artofproblemsolving.com/6/1/f/61fe26475e1cac3fa4cff58f973bef20db36ad4b.png)
Therefore, ![\[\angle FIH = \angle FIE + \angle EID = 90^{\circ}- \angle FGE + 90^{\circ} - \angle EGH = 180^{\circ} - \angle FGH \]](http://latex.artofproblemsolving.com/b/6/8/b68579ae0c28249035e60c08a1653f126f4d4676.png)
so quadrilateral 
![[asy]
import olympiad;
size(250);
pair A = (-3,0), B = (0,3), C = (3,0), D = (0,-2);
pair E = intersectionpoint(A--C, B--D);
pair E_AB = 2*foot(E, A, B) - E;
pair E_BC = 2*foot(E, B, C) - E;
pair E_CD = 2*foot(E, C, D) - E;
pair E_DA = 2*foot(E, D, A) - E;
pair O = circumcenter(E_AB, E_CD, E_DA);
real r = abs(O - E_AB);
pair K = (E + E_AB)/2;
pair L = (E + E_BC)/2;
pair M = (E + E_CD)/2;
pair N = (E + E_DA)/2;
draw(A--B--C--D--cycle, linewidth(1.2) + red);
draw(A--C, dashed + blue);
draw(B--D, dashed + green);
draw(E--E_AB, dotted + orange);
draw(E--E_BC, dotted + purple);
draw(E--E_CD, dotted + cyan);
draw(E--E_DA, dotted + magenta);
draw(circle(O, r), linewidth(1.2) + heavygreen);
dot("$A$", A, SW, red);
dot("$B$", B, NW, blue);
dot("$C$", C, SE, green);
dot("$D$", D, S, orange);
dot("$E$", E, NE, purple);
dot("$F$", E_AB, W, cyan);
dot("$G$", E_BC, NE, magenta);
dot("$H$", E_CD, SE, yellow);
dot("$I$", E_DA, S, brown);
dot("$O$", O, NE, heavygreen);
dot("$K$", K, NW, red);
dot("$L$", L, NE, blue);
dot("$M$", M, SE, green);
dot("$N$", N, SW, orange);
[/asy]](http://latex.artofproblemsolving.com/d/f/e/dfe32174dfcdf803339fdefd8469adae72976474.png)
and
Then from 
we have 
,
and 
are also cyclic. Also, since 
is the midpoint of 
is the midpoint of 
,
and similarly for
Since 
to 
;
is cyclic or
which is true since 
and 
are right triangles, so we are done.