[/quote]
,
specifies the greatest common divisor of 
and 
.
be a prime number. Define the set![\[
M_q = \left\{ x \in \mathbb{Z}^* \,\middle|\, \sqrt{x^2 + 2q^{2025} x} \in \mathbb{Q} \right\}.
\]](http://latex.artofproblemsolving.com/9/8/b/98b266a869cf5686cea86be51152ef234486993e.png)
.
be an odd integer. Prove that there exists a prime number 
such that![\[
p \mid 2^{\varphi(n)} - 1 \quad \text{but} \quad p \nmid n.
\]](http://latex.artofproblemsolving.com/3/5/1/3514b64ecd5f7b2586d5bfeae59b7c9504d5b455.png)
be a triangle for which there exists an interior point 
such that 
.
and 
meet the sides 
and 
at 
and 
respectively. Prove that ![\[ AB+AC\geq4DE. \]](http://latex.artofproblemsolving.com/3/a/4/3a49c5f8bd2930adca7bde2601dd045d4bf05ff6.png)
such that ![\[f(xf(y)+f(x+y)+1)=(y+1)f(x+1)\]](http://latex.artofproblemsolving.com/4/9/e/49ec422b640766de653d9d640bbc45065a6ac375.png)
holds for all 
.
and 
be points on a plane such that 
Let 
be a point on that plane such that 
Prove that ![$$AP \in \left[\frac{5-\sqrt{5}}{10}, \frac{-1+\sqrt{5}}{2}\right] \cup \left[\frac{5+\sqrt{5}}{10}, \frac{1+\sqrt{5}}{2}\right].$$](http://latex.artofproblemsolving.com/7/8/e/78e428cbc99ce4e9448f5c4cec86f50567123812.png)
has exactly 
ordered pairs 
as solutions?
such that 
such that
for all nonnegative real numbers 
with 
.
be three non-negative real numbers satisfying 
such that any real numbers 
and 
satisfy
Proposed by EmilXM
board has some of its cells coloured red, blue, or green. Each cell is coloured with at most one colour, and some cells may remain uncoloured. Additionally, there is at least one cell of each colour. Two coloured cells are said to be friends if they have different colours and lie in the same row or in the same column. The following conditions are satisfied:
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 
creates 
and 
so diagonal 
so 
which are formed by diagonal 
.
obtuse. Contradiction.
possible cases, Charlie being wrong is not possible. Therefore, if exactly 
