[/quote]
,
such that 
is a divisor of 
.
,
and 
are equal, and they intersect at 
.
and 
intersect at point 
lying inside triangle 
,
and 
intersect at point 
lying inside triangle 
.
.
denote the set of positive rational numbers. Find, with proof, all functions 
such that, for all positive rational numbers 
and 
we have ![\[f(x)=f(x+y)+f(x+x^2f(y)).\]](http://latex.artofproblemsolving.com/9/e/a/9ea71d2ee277dc3c689ba2d791b451e29353edb3.png)
be a triangle with circumcenter 
.
be the intersections of the perpendicular bisectors of 
and 
.
and 
and the point 
is given. Let 
be set of all 
segments of real line of type ![$[i, j]$](http://latex.artofproblemsolving.com/a/6/4/a644edd67caa59eaded765c417b6d9822f782ec2.png)
,
and 
are integers, 
.
is said to be valuable if the intersection of any two segments from 
is either empty, or is a segment of nonzero length belonging to 
satisfying 
,![\[ \frac{a^4+5b^4}{a(a+2b)} + \frac{b^4+5c^4}{b(b+2c)} + \frac{c^4+5a^4}{c(c+2a)} \geq 1- ab-bc-ca \]](http://latex.artofproblemsolving.com/f/e/8/fe85cf52b883c214806ab837a2eb25cebb511d7b.png)
such that there exists a unique integer 
such that 
with the following property:![\[ n!\mid a^n + 1
\]](http://latex.artofproblemsolving.com/e/2/3/e23d39aab685888993d355d8c93d4e6475f53487.png)
satisfy that: 
and natural number 
Y by FaThEr-SqUiRrEl
A convex quadrilateral is drawn on a blackboard. Three boys made one claim each: Alexey said, “This quadrilateral can be cut by its diagonal into two acute triangles”. Boris replied: “This quadrilateral can be cut by its diagonal into two right triangles”. And Charlie concluded: “This quadrilateral can be cut by its diagonal into two obtuse triangles”. It turned out that just one of them was wrong. Name the boy who certainly was right, and prove that he was.
(Frenkin B.R.)
(Frenkin B.R.)
This post has been edited 1 time. Last edited by parmenides51, May 20, 2022, 2:52 PM
right triangles. We now have three cases:
(they must be opposite vertices because otherwise a diagonal would split the angle). The other two vertices must have angles 
and 
for some 
is obtuse and 
are preserved when diagonal 
and 
so diagonal 
so 
which are formed by diagonal 
.
obtuse. Contradiction.
possible cases, Charlie being wrong is not possible. Therefore, if exactly 
