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 distinct prime numbers. Let ![$\alpha_i=\sqrt{a}, \,\mathrm{A}=\mathbb{Q}[\alpha_1,\alpha_2,\hdots,\alpha_b]$](http://latex.artofproblemsolving.com/0/e/0/0e06919fe05d9f17760890bb2aebb9cfa85d54ce.png)
![$[\mathrm{A}:\mathbb{Q}]=2^b$](http://latex.artofproblemsolving.com/b/e/1/be12fe0e1becd8a9dcddd1eece210f2df8ecd026.png)
![$\mathrm{A}=\mathbb{Q}[\gamma];$](http://latex.artofproblemsolving.com/f/6/3/f639e791853133a444708bf47517457379d347e8.png)
and deduce that the minimum polynomial 
of 
over 
has degree 
factors in ![$\mathbb{Z}_a[X]$](http://latex.artofproblemsolving.com/e/9/9/e994e6e9083e52f44aef280b4346f6994a9d9f02.png)
into a product of polynomials of degree 
either of degree 
. Let 
be a point on side 
, and 
be a point on the extension of 
such that 
. Let 
be the circumcenter of 
, and 
be a point on the side extension of 
satisfying 
. Line BP intersects AC at point Q. Prove that 
be a function from the set of integers to the set of positive integers. Suppose that, for any two integers 
and 
, the difference 
is divisible by 
. Prove that, for all integers and with 
, the number 
is divisible by 
.
be a positive integer, and let 
denote the integers modulo . Determine the number of functions 
satisfying 
for all 
.
such that 
and 
for all reals 
.
such that the following equation holds for every real :![\[ f(f(x) + y) = \lfloor x + f(f(y)) \rfloor. \]](http://latex.artofproblemsolving.com/3/f/b/3fbbb68e9f9c104fe0f701c8ad07e7fbbf583f1e.png)
Note: 
denotes the greatest integer not greater than 
.
such that
holds for all 
.
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.
, the Bank of Cape Town issues coins of denomination 
. Given a finite collection of such coins (of not necessarily different denominations) with total value at most most 
, prove that it is possible to split this collection into 
or fewer groups, such that each group has total value at most 
.
that satisfy the following equation for all real numbers 
:![\[
f\big(f(x - y)\big) = f(x) - f(y).
\]](http://latex.artofproblemsolving.com/c/3/c/c3c423f0ceb2fcb91064c5de218ec88f69c95e6d.png)
such that is unbounded and![\[2f(m)f(n)-f(n-m)-1\]](http://latex.artofproblemsolving.com/0/f/7/0f77b5addbf8f5e603eb1b6923cdac795cd132e9.png)
is a perfect square for all integer 
of four distinct positive integers, we denote the sum 
by 
. Let 
denote the number of pairs 
with 
for which 
divides . Find all sets 
of four distinct positive integers which achieve the largest possible value of .
such that 
divides 
for all positive integers 
and 
with 
.
be the set of positive integers. Find all functions 
such that![\[ m^2 + f(n) \mid mf(m) +n \]](http://latex.artofproblemsolving.com/2/f/4/2f409d1de993f1af8fd839bb8e9f87a57e1b8608.png)
and .
,
is 
.
at 
,
.
lies on the polar of 
,
.
are concurrent, 
,
.
meet the incircle at 
and let ray 
meet line 
.
,
,
and let 
.
,
is a tangent to the incircle. Now 
is harmonic, so 
.
,
be the points of contact of the incircle with the sides 
.
By ceva's and menalaus, we have 
harmonic.
.
,
and 
.
.
.
by applying sine rule
and 
,
we need 
.
is tangent to the incircle and 
so 
where 
and 
.
,
.
is concyclic with diameter 
.
.
is tangent to 
.
= 
= 
= Power of 
lies on 
is a harmonic pencil.
lies on 
.
.
bisects 
internally.
respectively.
and so 
is perpendicular to 
are col-linear and 
so we get that since, 
are perpendicular 
.
in the projective plane and 
and 
and 
,
lies on 
and therefore ![\[ -1 = (X,AP \cap EF;E,F) \overset{A} = (X,D;C,B). \]](http://latex.artofproblemsolving.com/8/8/2/882b6896786d4bcc8b7a66597ad162e20d0868b8.png)
By another well known projective lemma, we get 
.
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[/asy]](http://latex.artofproblemsolving.com/d/5/1/d51a870eff51e10117d36c632d7f2f7df191aaef.png)
and 
to be the second intersection of 
is a harmonic quadrilateral, 
and 
are the tangents to the incircle. From this we have 
,
are collinear. It's well-known that 
;
implies (by another well-known lemma) that 
,
,
.
, hence QED.
,
.
and 
.
,
at 
.
.
collinear 
all lie in a circle. Invert wrt to incircle then 
is fixed. 
swaps to the midpoints of 
respectively. Let it be 
.
and 
. Let 
be 
.
.
is the midpoint of the 
and 
lie on a circle. So 
all lie on circle and we are done.
.
the condition will be satisfied, by the configuration with right angles and bisectors. Let E and F be the the tangency points of the incircle on AC and AB. Now let 
and by the Ceva configuration we have that 
if we wanted to show that 
we only need to prove that 
,
.
,
,
,
.
,
and thus 
,
,
and 
,
.![\[-1=(DZ;FE) \overset{D}{=} (XY;FE) \overset{A}{=} (XD;BC).\]](http://latex.artofproblemsolving.com/4/c/d/4cd7c12276db1d3b72f6f59a930126d8f5750a4f.png)
,
.
.
be the point on the incircle such that 
is tangent to the incircle. Note that 
is the polar of 
which implies 
so by right angles and bisectors 
respectively.
is cyclic.
and 
we get that 
concur, say at G.
.
By radical axis theorem, 
concur at 
is cyclic since 
)
concur, so by Menelaus we get 
Since 
we know 
.
.
.
,
.
,
are known to be concurrent, and hence extending 
at 
.
,
.
:
implying 
,
are tangents to 
,
are collinear. Now the required follows from the right angle-angle bisector-harmonic lemma from the book 
.
be a triangle with circumcenter 
be the midpoints of 
.
is tangent to the circumcircle. Prove 
.
be the reflections of 
.
.
.
share a perpendicular bisector so homothety at 
finishes.
be the intouch points on sides 
,
,
,
is a line, and since![\[(X, D; B, C) = -1\]](http://latex.artofproblemsolving.com/c/3/4/c34fbf809499eb950f0d852773547f2f126c6c3f.png)
and 
and take A in real axis
,
,
,
and by La Hire’s, 
Furthermore, 
is 
but 
which implies that 
are collinear.
By Right angles and bisectors, we are done.