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 distinct positive integers, 
. Prove that there exist distinct indices 
and 
such that 
does not divide any of the numbers 
, 
, , 
.
board, 
pieces are placed (vertically or horizontally) so that they occupy entirely 
boxes on the board and do not have a common point.
, let 
be the sum of the digits of . Determines the smallest positive integer 
such that there are infinite positive integers for which you have 
.
, 
, 
be fixed positive integers. There are 
ducks sitting in a ducks ducks picking paper, and ducks picking scissors., , and , the maximum number of moves which could take
be the circumcenter of triangle 
. Line 
intersects the altitude from 
at point 
. Let 
be the midpoints of 
, 
respectively. If 
intersects 
at 
, and the circumcircle of triangle 
meets 
at 
, prove that 
is cyclic. be a triangle with incentre 
and circumcentre . Let 
be the touchpoints of the incircle with , 
, respectively. Prove that 
is the Euler line of 
.
with its incircle 
. Let 
and 
be the points of tangency of with and respectively. On the edges 
and 
, take points 
and 
such that 
is tangent to at 
. Suppose 
is the intersection point of the lines 
and 
. Prove that two lines 
and 
are parallel or coincide.
be a triangle with orthocenter 
and let 
be the midpoint of 
Suppose that and are distinct points on the circle with diameter 
different from 
such that lies on line 
Prove that the orthocenter of 
lies on the circumcircle of 
. Find all functions 
that satisfy
for all integers 
and rational numbers 
.
such that 
( is sum of digit of in base 10)
. We can easily see that 3 angles BCD, CDA, BAD must equal to angle ABC. Therefore, quadrilateral is a rectangle.
such that all four triangles 
, 
, 
and are similar to one-another.
Hence we see
Adding these we get 
.
and 
Adding these we get 
Similarly
a square.
however I have a strong feeling that 
could also be true
. We can easily see that 3 angles BCD, CDA, BAD must equal to angle ABC. Therefore, quadrilateral is a rectangle. can be concave.
https://igo-official.ir/events/
,![\[\square ABCD \text{ is cyclic.}\]](http://latex.artofproblemsolving.com/1/6/5/165e371cea25d9eb70647967ff0afd098495aef0.png)
.![\[\triangle ABO \sim \triangle AOD\]](http://latex.artofproblemsolving.com/5/a/d/5ad40729445de534380a3407948dd22964bac6dc.png)
is a rhombus. On the other hand, it is cyclic too. So, must be a rectangle. similar triangles.
such that all four triangles , , and are similar to one-another. be convex quadrilateral because no concave quadrilateral exists that satisfies the similarity condition. 
is a cyclic quadrilateral. Let diagonals intersect at ,
. 
is a rhombus and cyclic 
is rectangle.
![\[
\text{Find all functions } f : \mathbb{R} \to \mathbb{R} \text{ such that:} \\
f(a^4 + a^2b^2 + b^4) = f\left((a^2 - f(ab) + b^2)(a^2 + f(ab) + b^2)\right)
\]](http://latex.artofproblemsolving.com/2/a/7/2a73a5e58eab4a994fbd83160fb79a0b00152951.png)
are one of such 
