[/quote]
,
and incentre 
intersects the lines 
,
,
at points 
,
,
,
,
,
,
,
of the segments 
,
,
,
.
be a positive integer. Prove that there are infinitely many primes, each of which divides at least one term of the sequence 
,
.
![$\mathbb{Z}[x]$](http://latex.artofproblemsolving.com/4/b/7/4b7d768fffbb7a1a6163af392ed348f2d49d8783.png)
denote the set of single-variable polynomials in ![$\theta : \mathbb{Z}[x] \to \mathbb{Z}[x]$](http://latex.artofproblemsolving.com/0/3/2/032a637d4ae21da2c0ddd8dc127723807f11cc5f.png)
(i.e. functions taking polynomials to polynomials)![$p, q \in \mathbb{Z}[x]$](http://latex.artofproblemsolving.com/7/e/7/7e7eb925258a33e456ef4ff722c1f943dc155d10.png)
,
;![$p \in \mathbb{Z}[x]$](http://latex.artofproblemsolving.com/0/2/6/0267b98c16b400dda1f081877e9b84fe672ce0e7.png)
,
has an integer root if and only if 
does.
of positive real numbers that satisfy the system:![\[
\begin{aligned}
11bc - 36b - 15c &= abc \\
12ca - 10c - 28a &= abc \\
13ab - 21a - 6b &= abc.
\end{aligned}
\]](http://latex.artofproblemsolving.com/1/7/0/170e1cfaee90f5c37428fa75192ac889261a8f83.png)
that verify the following equation for all 
:
is a prime number
is a function for which the expression 
for all 
is always a perfect square. Prove that 
for all 
.
such that 
.
be a quadrilateral inscribed in a circle 
,
be a point on the extension of 
and 
are tangent to 
and the line 
.
and 
and 
and 
the reflections of 
,
be the midpoint of 
and 
are perpendicular.
.
be the feet of the altitudes from 
to 
respectively. Let 
be the circle with diameter 
be the intersection of 
and 
be the intersection of 
be the lines through 
respectively. Let 
and 
.
and 
.
be the line through 
and let 
be the reflection of 
is tangent to 
.
be a function from the Euclidean plane to the real numbers such that 
for any acute triangle 
with circumcenter 
is constant.
.
on 
on 
lies on the circumcircle of triangle 
.
Y by Adventure10, Mango247
In triangle 
, the incircle, with center 
, touches the sides 
at point 
. Line 
meets 
at 
. The tangent line from to the incircle (different from ) intersects 
at 
. If 
and intersect at point 
, prove that 
.
Proposed by Hooman Fattahimoghaddam
, the incircle, with center 
, touches the sides 
at point 
. Line 
meets 
at 
. The tangent line from to the incircle (different from ) intersects 
at 
. If 
and intersect at point 
, prove that 
.Proposed by Hooman Fattahimoghaddam
.
.
we get 
Thus we have 
.
implying that 
.
,
,
,
be the second tangent from 
.
,
,
.
and 
,
this implies that 
is tangent to the incircle. By Brianchon's, 
we have 
colinear and since 
we have 
symetric w.r.t. 
has an incircle and the angle bisectors concur at 
is cyclic .So 
,
so 
![[asy]import graph; size(8cm);
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draw((-1.26,3.94)--(-0.62,-1.14), linewidth(0.8));
draw((-0.62,-1.14)--(6.04,-1.12), linewidth(0.8));
draw((6.04,-1.12)--(-1.26,3.94), linewidth(0.8));
draw(circle((0.8240641549073694,0.502372949317706), 1.6380290344455826), linewidth(0.8));
draw((-1.26,3.94)--(4.500133160918301,-1.1246242247419869), linewidth(0.8));
draw((0.8240641549073694,0.502372949317706)--(-1.26,3.94), linewidth(0.8));
draw((-0.9244071321537639,1.2762316114705021)--(0.8240641549073694,0.502372949317706), linewidth(0.8));
draw((0.8240641549073694,0.502372949317706)--(0.8180665158025027,2.499586771238265), linewidth(0.8));
draw((0.8180665158025027,2.499586771238265)--(-0.9244071321537639,1.2762316114705021), linewidth(0.8));
draw((0.8240641549073694,0.502372949317706)--(0.8289831388371004,-1.1356486992827717), linewidth(0.8));
draw((0.8240641549073694,0.502372949317706)--(4.500133160918301,-1.1246242247419869), linewidth(0.8));
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draw((-0.3845588582412495,2.4959752836285243)--(-0.9244071321537639,1.2762316114705021), linewidth(0.8));
draw((-0.3845588582412495,2.4959752836285243)--(0.8180665158025027,2.499586771238265), linewidth(0.8));
draw(circle((0.21975264833305966,1.4991741164731156), 1.1656778988299026), linewidth(0.8));
dot((-1.26,3.94),linewidth(4pt) + dotstyle);
label("$A$", (-1.3850408997846597,4.078986433915431), NE * labelscalefactor);
dot((-0.62,-1.14),linewidth(4pt) + dotstyle);
label("$B$", (-1.0114335048127107,-1.3179350436719013), NE * labelscalefactor);
dot((6.04,-1.12),linewidth(4pt) + dotstyle);
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dot((0.8289831388371004,-1.1356486992827717),linewidth(4pt) + dotstyle);
label("$D$", (0.8728548204304464,-1.0425819070603026), NE * labelscalefactor);
dot((0.8180665158025027,2.499586771238265),linewidth(4pt) + dotstyle);
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dot((-0.9244071321537639,1.2762316114705021),linewidth(4pt) + dotstyle);
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dot((4.500133160918301,-1.1246242247419869),linewidth(4pt) + dotstyle);
label("$Z$", (4.298247839878729,-1.6060480473876127), NE * labelscalefactor);
dot((-0.3845588582412495,2.4959752836285243),linewidth(4pt) + dotstyle);
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clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
[/asy]](http://latex.artofproblemsolving.com/0/4/7/04765e2cb6c31c0c51f95e265742497491ca212b.png)
and 
.
is cyclic.![\[\angle AIY = \angle EIY = \angle EXY=\frac{1}{2}\angle AXY = \frac{1}{2}(180^{\circ}-2\angle CXD) = \frac{1}{2}(180^{\circ}-(180^{\circ}-2\angle C)) = \frac{1}{2}\cdot 2\angle C=\angle ACZ\implies AIZC\text{ is cyclic}\]](http://latex.artofproblemsolving.com/a/c/3/ac39a62e2e74aee0dbc5c842d63efbc6fa17e3da.png)
Finally,![\[\angle AZI = \angle ACI=\angle ZCI = \angle ZAI\implies IA=IZ\implies \triangle AIF\cong\triangle IDZ\implies AF=DZ\implies AF+BF=BD+DZ\implies AB=BZ\]](http://latex.artofproblemsolving.com/5/c/e/5ce6aa474f67436dc2a67591c016d74ff238bb5c.png)
as desired.
