[/quote]
,
such that the difference between any two consecutive terms is the same. That is, 
for all 
,
is the difference of the progression.
positive integer such that 
.
is said to be maximal if 
and 
are not Tlaxcalan arithmetic progressions.
elements?
be an odd prime and 
be integers so that the integers 
are divisible by 
such that:
for any positive integer 
,
are all distinct, and
,
?
lies on a line segment 
between 
and 
and circles are drawn having 
and 
as diameters. A common tangent to both circles touches the circle with 
and the circle with 
.
and the common tangent to both circles at 
and 
denote three distinct integers, and let 
denote a polynomial having integer coefficients. Show that it is impossible that 
and 
.
be a polynomial of degree 
with integer coefficients and let 
be a positive integer. Consider the polynomial 
,
integers 
such that 
.
prove that for each 
there exists 
such that either 
or 
is multiple of 
![\[
S = \left\{\, x \in \mathbb{Q} : x \text{ has a finite decimal expansion} \,\right\}.
\]](http://latex.artofproblemsolving.com/1/c/e/1ce81426f5d169a9b4c254ef91a1fe34250504b4.png)
For each 
,![\[
d(x) = \text{the number of digits after the decimal point in the (reduced) decimal form of } x.
\]](http://latex.artofproblemsolving.com/8/1/9/819d07d26274a051a776fcf9f6d7c668ab8d0dab.png)
Find all functions 
such that, whenever both 
and 
,![\[
f(x) + f(y) = f(x+y) + d(x y).
\]](http://latex.artofproblemsolving.com/5/c/a/5ca142ba21a23dc21db6e3c712bc0a76d561647c.png)
be a finite set of points in the plane such that no three of them are on a line. For each convex polygon 
be the number of vertices of 
be the number of points of 
,
,
vertices respectively. Prove that for every real number 
![\[\sum_{P}{x^{a(P)}(1 - x)^{b(P)}} = 1,\]](http://latex.artofproblemsolving.com/3/2/7/3279b2917f21c8f3c2b819e39a32528ad18e4b4b.png)
where the sum is taken over all convex polygons with vertices in 
be a finite point set in the plane and no three points are collinear. A subset 
of 
g
For each round subset let 
be the number of points from 
and 2 elements are always round, its corresponding polygons are the empty set, a point or a segment, respectively (for which all other points that are not vertices of the polygon are exterior). For each round subset ![\[ P_A(x) = x^{|A|}(1 - x)^{r(A)}.
\]](http://latex.artofproblemsolving.com/e/b/c/ebc6f8fa2cc303e1062c5a95d4f65c9d2691d0ca.png)
,
and 
are three real numbers, all different from 
,
,
.
sign for cyclic summation, this inequality could be rewritten as 
.
such that for all 
,![\[xf(x+f(y))=(y-x)f(f(x)).\]](http://latex.artofproblemsolving.com/e/4/a/e4a3bbb8b91d2aa62d699c24df342fa59be71915.png)
satisfying the following property: if 
,
and 
are integers such that 
,
be a continuous function such that 
show that 
is even.
Y by Davi-8191, mathematicsy, centslordm, donotoven, jhu08, rachelshi, megarnie, Adventure10, Rounak_iitr
In an acute scalene triangle 
, points 
lie on sides 
, respectively, such that 
. Altitudes 
meet at orthocenter 
. Points 
and 
lie on segment 
such that 
and 
. Lines 
and 
intersect at point 
. Compute 
.
Proposed by Zuming Feng
, points 
lie on sides 
, respectively, such that 
. Altitudes 
meet at orthocenter 
. Points 
and 
lie on segment 
such that 
and 
. Lines 
and 
intersect at point 
. Compute 
.Proposed by Zuming Feng
and 
,
and hence 
.
and 
which implies that 
.
and 
,
-
.
.
and 
since both lengths are the radius of the incircle of 
the 
.
be the incenter of 
and 
the foot of perpendicular from 
to 
be symmetrical of 
are collinear.
.
, Q and D are corresponding points on similar figures, so 
,
implies that 
and 
, hence 
and 
.
divide 
in the same ratio externally and internally and it follows as a result that 
and line 
and apply Menelaus' theorem to get 
and then a Ceva's theorem applied to the same triangle and Point 
,
be orthocentre of 
.
and so, 
.
passes through midpoint of 
and hence, 
.
we have that 
are harmonic and now projecting through 
ie 
.
,
.
so 
is a harmonic division, i.e. 
.![\[\dfrac{HQ}{HR}=\dfrac{HS}{HR}=\dfrac{HS}{HD}\cdot\dfrac{HD}{HR}=\dfrac{AE}{AB}\cdot\dfrac{AD}{AP}=\dfrac{AE}{AB}\cdot\dfrac{AB}{AE}=1\]](http://latex.artofproblemsolving.com/7/5/0/750799a2f061b32ad414148192338eb521757e60.png)
from 
,
,
.
.
and 
are the incenter and 
-
respectively. This implies that 
is the tangency point of the incircle of 
and that 
is the tangency point of the 
be the antipode of 
are collinear (this is a standard Yufei configuration) so 
.
so the desired ratio is equal to 
.
are parallel.
and 
where 
is intersection of 
.
and 
yield with angle chasing using cyclic quad 
.
and let 
.
.
for some 
.
are collinear, we must have 
Expanding the determinant, we have ![\begin{align*} t[b(s-b)-c(s-c)]+(s-b)[-a(s-b)]+(s-c)[a(s-c)]&=0\\t[b(s-b)-c(s-c)]&=a[(s-b)^2-(s-c)^2]\\ t\cdot \frac12 [b(a-b+c)-c(a+b-c)]&=a(s-b+s-c)(s-b-s+c)\\ t\cdot\frac12 [(b-c)(a-b-c)]&=-a^2(b-c)\\ t&=\frac{a^2}{s-a}\end{align*}](http://latex.artofproblemsolving.com/5/0/1/501f8927f645a1d9bba6e9a1d55707a00c390d07.png)
so 
We can normalize 
.
.
)
:
,
is the radius of the incircle of 
,
and let it meet 
and 
at 
and 
respectively. Then the homothety mapping 
to 
.
be a quadrilateral whose diagonals meet at 
.
and 
and line 
meets 
and 
at 
and 
respectively. Then 
is a harmonic bundle.
and let 
.
.
.
.
,
.
.
so H is the midpoint of QR, whence the ratio is 1, as desired. 
.
,![\[-1 = (A, H; X, D) \stackrel{P \to \overline{QR}}{\longrightarrow} (P_{\infty}, H; Q, R).\]](http://latex.artofproblemsolving.com/9/6/f/96f79721df0f52f7161eb261706ca098f9c91ed7.png)
Hence 
as requested.
and 
.
,
.![\[(AB;FX_C) \overset{E}{=}(AH;LD) \overset{P}{=} (P_{\infty}H;QR)\]](http://latex.artofproblemsolving.com/5/f/8/5f842cdd0af3e6b983a4125bb3e6a33384ca7295.png)
But this means that ![\[\frac{HQ}{HR} = 1\]](http://latex.artofproblemsolving.com/5/c/7/5c74b477d614e7c232c1f27024449d61d94c3fb8.png)
.
be the antipode of 
,
.
,![\[ \frac{HQ}{HR} = \frac{HQ}{HQ'} = \frac{1}{1} = 1.\]](http://latex.artofproblemsolving.com/e/c/a/ecaeb1649de9de84733c050e43d17e15352c4456.png)
.
.
,
We also have 
,
Multiplying the two ratios above, we obtain 
However, we know that 
are cyclic quadrilaterals, so 
,
is the internal angle bisector of 
.
,
is the external bisector of 
which is equivalent to 
,
,
lies on the incircle so 
which gives us the answer of 
and then apply melaneus and since HQ parallel to AP you just apply thales and you are done
is trivially 
,
is the intersection of ray 
and 
is the intersection of ray 
.
,
is the reflection of 
.
,
.
is a parallelogram (by right angles), and so 
and 
is an isosceles trapezoid with 
.
