AoPS Academy
be a triangle with incenter 
, and let 
be a point on side 
. Points 
and 
are chosen on lines 
and 
respectively such that 
is a parallelogram. Points 
and 
are chosen on side such that 
and 
are the angle bisectors of angles 
and 
respectively. Let 
be the circle tangent to segment 
, the extension of 
past , and the extension of 
past . Prove that is tangent to the circumcircle of triangle .
such that ![\[f(xy-1)+f(x)f(y)=2xy-1,\]](http://latex.artofproblemsolving.com/8/8/8/888ca39f2b7f8cec6d6426bee28d40eade40a66e.png)
for any real numbers 
and 
, consider the 
fractions 
The product of these fractions equals 
, but if you reciprocate (i.e. turn upside down) some of the fractions, the product will change. Can you make the product equal 1? Find all values of for which this is possible and prove that you have found them all.
be a function satisfying
for all integers 
and 
. Show that there exist positive integers 
and such that 
for all integers .
of the triangle 
which doesn't coincide with 
or 
intersectes the altitudes from 
and 
at 
and 
respectively. Let 
be the reflection of in and 
be the reflection of in 
Prove that the circles 
and 
are tangent.
be positive real numbers such that 
Prove that we have![\[(a+b)(b+c)(c+a)\geq 8.\]](http://latex.artofproblemsolving.com/2/a/6/2a63f19de634fce70ff4ebc02ad13d442a20c378.png)
such that for any positive integer 
there exists a positive integer such that the last digits of 
are equal to 
such that for any positive real numbers 
, which are the sides of a triangle, the following inequality holds:![\[
\frac{xy}{x^2 + y^2 + xz} + \frac{yz}{y^2 + z^2 + yx} + \frac{zx}{z^2 + x^2 + zy} \geq C.
\]](http://latex.artofproblemsolving.com/9/1/c/91c8a4af2bd2f3107c8921d2e06f8a09dedd0164.png)
![\[
f : \mathbb{R}^+ \to \mathbb{R}
\]](http://latex.artofproblemsolving.com/e/7/b/e7bfc5b236fb6f00ef328f850b1b14632cbf8416.png)
satisfying the condition that for all 
,![\[
f(x + y^2) \geq f(x) + y.
\]](http://latex.artofproblemsolving.com/3/a/a/3aa20083835682dbd81be692eba65cab19e923e5.png)
be defined as follows:
, and for all positive integers 
,![\[
a_n = a_{\left\lfloor \frac{n}{3} \right\rfloor} + a_{\left\lfloor \frac{n}{2} \right\rfloor}.
\]](http://latex.artofproblemsolving.com/8/0/0/8005effb2b72a59ce55f49a5204169ec8ffa3b60.png)
Find the sum of all values 
for which there exists a unique positive integer such that 
.
the center of 
and let 
the midpoint of minor arc in 
, let 
hit 
at 
respectivily, let 
, let 
the B-excenter of 
, let 
, and denote 
the inverse of 
in 
inversion.
Which is true hence the claim is done. (It is easy to see 
is an Isosceles Trapezoid btw)
are cyclic
hence by angle chase
Same symetric process gives 
cyclic, thus claim finished.
and 
are colinear triples. inversion we get that 
and 
are colinear, note that 
follows by symetry on the perpendicular bisector of since 
is cyclic and 
is also known. Since 
both lie in 
we have by double reim's that 
and 
are cyclic and inverting back gives the result.
is center of homothety of triangles 
and 
, thus 
are colinear, now since 
we get that 
is cyclic, plus since are colinear we can safely say that 
are tangent at as desired, thus we are done 
.
with incenter 
excenter opposite 
as 
we let 
be points on 
respectively with 
Then let 
be the point on 
with 
collinear.
and 
gives 
collinear, similarly for 
lies on the reflection of 
over line 
Thus the homothety at taking to 
takes to 
so the two circles are tangent. and rename points.
the center of and let the midpoint of minor arc in , let hit at respectivily, let , let the B-excenter of , let , and denote the inverse of in inversion.Which is true hence the claim is done. (It is easy to see is an Isosceles Trapezoid btw) are cyclic hence by angle chaseSame symetric process gives cyclic, thus claim finished. and are colinear triples. inversion we get that and are colinear, note that follows by symetry on the perpendicular bisector of since is cyclic and is also known. Since both lie in we have by double reim's that and are cyclic and inverting back gives the result. is center of homothety of triangles and , thus are colinear, now since we get that is cyclic, plus since are colinear we can safely say that are tangent at as desired, thus we are done .
such that: 
.
