be the set of positive real numbers. Determine all functions 
and 
![\[f(x+f(xy))+y=f(x)f(y)+1\]](http://latex.artofproblemsolving.com/c/6/a/c6a968a17f6fd5fd637c1ee5d855bdff11e5423a.png)
and 
such that![\[
1^n + 2^n + 3^n + \cdots + n^n = x!
\]](http://latex.artofproblemsolving.com/6/4/b/64b5f5919fd3710cc15e96724918eae216c8e4f0.png)
.
meet at point 
be 
.
meet 
respectively. Let 
lie on a circle.
red cards, 
white cards, and 
blue cards. A player plays all the cards one at a time. With each play he accumulates a penalty. If he plays a blue card, then he is charged a penalty which is the number of white cards still in his hand. If he plays a white card, then he is charged a penalty which is twice the number of red cards still in his hand. If he plays a red card, then he is charged a penalty which is three times the number of blue cards still in his hand. Find, as a function of 
and 
the minimal total penalty a player can amass and all the ways in which this minimum can be achieved.
Y by Adventure10, Mango247
Let 
be a triangle with 
. Let 
and 
be external points to the triangle such that 
and 
are isoscele triangles with right angles at and . Let 
, and let 
and 
be the midpoints of 
and 
, respectively.
Prove that, if three of the points
, 
, , are collinear, then all four are collinear.
be a triangle with 
. Let 
and 
be external points to the triangle such that 
and 
are isoscele triangles with right angles at and . Let 
, and let 
and 
be the midpoints of 
and 
, respectively.Prove that, if three of the points
, 
, , are collinear, then all four are collinear.This post has been edited 1 time. Last edited by bluecarneal, Jul 21, 2019, 6:14 PM
Reason: line break
Reason: line break
are collinear then 
and that means 
.
are collinear then using areas we'll obtain 
is on this line.
being collinear and 
being collinear are easy to prove .Hence I will do 
being collinear.
.this can be proved very easily by cos rule to 
and 
.But I will use a different method. Let the foot of perpendicular from 
.
and 
are cyclic..Let the circumcircle of 
intersect 
..Consider the composition of rotation centered at 
and centered at 
.Since the sum of rotation is 
and since it takes 
.it is easy to observe that 
is a square .Now rotate 
around 
should coincide with 
are collinear
.The rest is trivial.
