[/quote]
,
,
,
,
for all 
,
and 
in a non-isosceles triangle 
intersect its circumcircle 
at 
and 
,
at 
.
is tangent to 
be the circumcircle of right-angled triangle 
)
at point 
is the midpoint of the minor arc 
,
intersects 
.
.
.
such that 
and 
,
.
,
be the orthocenter of it and 
be any point on the side 
.
are on the segments 
,
and 
are cyclic. The segments 
and 
intersect at 
is a point on 
such that 
is tangent to the circumcircle of triangle 
at 
and 
intersect at 
.
and 
be fixed points on this order on a line. Let 
at the points 
and 
.
and 
be the other points of intersection of 
and 
with 
passes through a fixed point independent of 
and 
.
be a positive integer and let 
be distinct integers in the set 
such that 
for 
.
:![\[
2 \left( 5^x + 6^x - 3^x \right) = 7^x + 9^x.
\]](http://latex.artofproblemsolving.com/f/b/6/fb6d5ebb4f448d8eeb73cce3288fea3a56d7f528.png)
of non-negative real numbers such that the following three conditions are all satisfied:
;
;
of 
[/list]
is a permutation of 
,
.
denote the sum of digits function. Does there exist 
such that![\[
n, s(n), \dots, s^{2024}(n)
\]](http://latex.artofproblemsolving.com/c/3/3/c33803a77912ed4dde5a5c0ae05934496f04707f.png)
are all distinct perfect squares?
having integer coefficients. On each move, Vasya pays him a ruble, and calls an integer 
of his choice, which has not yet been called by him. Petya has to reply with the number of distinct integer solutions of the equation 
.
that satisfies:![\[
\sum_{i=1}^{n} a_{\left\lfloor \frac{n}{i} \right\rfloor} = n^k
\]](http://latex.artofproblemsolving.com/b/b/e/bbe7fe81e366fe390be5d17193d1581cc40dbece.png)
be a positive integer. Prove that for all integers 
,![\[
\frac{c^{a_n} - c^{a_{n-1}}}{n} \in \mathbb{Z}
\]](http://latex.artofproblemsolving.com/0/4/4/04464c1a350e3f9a929ab51534018498680068fc.png)
be real positive numbers such that: 
.
Y by Adventure10
Let 
be a triangle with 
. Let 
be the midpoint of 
. Let 
be the intersection of the angular bisector of 
and the line . Let 
be the point on 
such that 
is perpendicular to . Finally, let 
be the intersection of and 
. Prove that 
divides the line segment 
into two equal parts.
be a triangle with 
. Let 
be the midpoint of 
. Let 
be the intersection of the angular bisector of 
and the line . Let 
be the point on 
such that 
is perpendicular to . Finally, let 
be the intersection of and 
. Prove that 
divides the line segment 
into two equal parts.This post has been edited 1 time. Last edited by parmenides51, Jan 10, 2020, 1:58 PM
,
,
.
,
,
.
.
and thus 
.
to both sides, we have,
,
