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 an acute-angled triangle in which 
and 
. Let point 
lie on segment 
and point 
lie on segment 
such that 
, 
and 
. Let 
be the circumcenter of triangle 
, 
the orthocenter of triangle , and 
the point of intersection of the lines 
and 
. Prove that , , and are collinear.
be an isosceles triangle with 
, and let 
be its incenter. Incircle touches sides 
at 
, respectively. Foot of altitudes from 
to 
are 
, respectively. Rays 
intersect 
at 
, respectively. Prove that 
touches incircle at 
.
a grid of 9*9, we choose a little square in of this grid three times, we can choose three times the same.
. Find the minimum value of 
+1 w
s such that:
such that 
be the divisors of 
. Find all values of such that 
.
and primes 
that satisfy 
by ![\[\displaystyle f(x) = \left\{ \begin{array}{lr} x+\frac 12 & \text{if}\ \ x < \frac 12\\ x^2 & \text{if}\ \ x \ge \frac 12 \end{array} \right.\]](http://latex.artofproblemsolving.com/c/b/d/cbd9b412893774283a438296cdeb05b80cc878bf.png)
Let 
and 
be two real numbers such that 
. We define the sequences 
and 
by 
, and 
, 
for 
. Show that there exists a positive integer such that ![\[(a_n - a_{n-1})(b_n-b_{n-1})<0.\]](http://latex.artofproblemsolving.com/a/5/c/a5c8844d595bb1186d4804beadd69a3bc1d40ea3.png)
is defined recursively by ![\[a_0=-1,\qquad\sum_{k=0}^n\dfrac{a_{n-k}}{k+1}=0\quad\text{for}\quad n\geq 1.\]](http://latex.artofproblemsolving.com/f/4/b/f4b8f95e49de7274cccd8ee498e0b496f31ddfea.png)
Show that 
for all 
.
be positive integers satisfying the conditions 
and 
Show that there exists a real number 
with the property that all the three numbers 
have their fractional parts lying in the interval ![$ \left(\frac {1}{3}, \frac {2}{3} \right].$](http://latex.artofproblemsolving.com/4/2/b/42ba58ddd6032a3176122f1c9b2015cb6f4ca925.png)
for which each cell of 
table can be filled with one of the letters 
and 
in such a way that:, one third are 
and one third are ; and [/*], one third are and one third are .[/*] table are each labelled 
to in a natural order. Thus each cell corresponds to a pair of positive integer 
with 
. For 
, the table has 
diagonals of two types. A diagonal of first type consists all cells for which 
is a constant, and the diagonal of this second type consists all cells for which 
is constant.
at 
and 
meet at . The circumcircle of 
meets sides and again at 
and 
respectively. Let be the circumcentre of . Show that 
is perpendicular to 
.
is defined as the intersection of the tangents. We can replace the circumcircle of with any circle that passes through and . Indeed, we know that 
, and since 
is cyclic, we have that 
, from which the result follows.
by reflecting 
about its 
angle bisector and taking a homothety at 
. Now since the circumcentre and orthocentre are isogonal conjugates, has to lie on the altitude of , and therefore we are done. 
be the orthocenter of .
is concyclic 
.
, such that goes to . Under this transformation, line 
goes to line because they are isogonal in . Since 
, we have 
.
lies on the circumcircle of 
, we know that
so
we know that
hence we are done 
hence lies on perpendicular bisector of and lies on perpendicular bisector of moreover also lies on these lines
is the orthocentre of hence 
is perpendicular to 
is the 
are concyclic. So 
.
,
since 
are concyclic. Therefore, 
.
(because CFEB is cyclic)
