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 a triangle with circumcircle 
and incentre 
. A line 
intersects the lines 
, 
, and 
at points 
, 
, and 
, respectively, distinct from the points 
, 
, 
, and . The perpendicular bisectors 
, 
, and 
of the segments 
, 
, and 
, respectively determine a triangle 
. Show that the circumcircle of the triangle is tangent to .
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 with integer coefficients. Find all functions ![$\theta : \mathbb{Z}[x] \to \mathbb{Z}[x]$](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 
, and let 
be a point on the extension of 
such that 
and 
are tangent to . The tangent at intersects at 
and the line at 
. Let be the second point of intersection between 
and . Prove that , , are collinear.
be a triangle with incenter , and suppose the incircle is tangent to 
and 
at and . Denote by 
and 
the reflections of and over . Let be the intersection of 
with 
, and let 
be the midpoint of . Prove that 
and 
are perpendicular.
be an acute scalene triangle with circumcenter 
. Let 
be the feet of the altitudes from 
to 
respectively. Let be the midpoint of . Let 
be the circle with diameter . Let 
be the intersection of 
and . Let be the orthocenter of . Let 
be the intersection of and . Let 
be the lines through tangent to 
respectively. Let be the intersection of 
and 
. Let be the intersection of 
and 
. Let 
be the line through parallel to 
and let 
be the reflection of across . Prove that 
is tangent to 
.
be a function from the Euclidean plane to the real numbers such that 
for any acute triangle 
with circumcenter , centroid and orthocenter . Prove that 
is constant.
opposite the vertex be tangent to the side at the point 
. Define the points 
on and 
on analogously, using the excircles opposite and , respectively. Suppose that the circumcentre of triangle 
lies on the circumcircle of triangle . Prove that triangle is right-angled. and , Jaqueline has a ruler that allows tracing the line 
. Jaqueline also has a special object that allows the construction of a circle of diameter . Also, always when two circles (or a circle and a line, or two lines) intersect, she can mark the points of the intersection with a pencil and trace more lines and circles using these dispositives by the points marked. Initially, she has an acute scalene triangle . Show that Jaqueline can construct the incenter of .
be a triangle and 
be the mid-point of 
. Suppose the angle bisector of 
is tangent to the circumcircle of triangle 
at . Prove that 
.
be the angle bisector of 
, Let 
such that, 
, and let 
, then, 
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meet at . Observe 
so 
. From this, we obtain that (1) is thus the midpoint of 
, implying that 
and (2) 
. Hence, 
as desired.
be the angle bisector of , Let such that, , and let , then, 
as mentioned in the question intersect side of 
at .
. This gives 
.
to an arbitrary point beyond the side , we should have 
.
is isosceles, which means 
.
,
are parallel, so 
so 
is an isoscel triangle, so 
,
.
.
