Olympiad problems

2024 USA IMO Team Selection Test, 3

Let $n>k \geq 1$ be integers and let $p$ be a prime dividing $\tbinom{n}{k}$. Prove that the $k$-element subsets of $\{1,\ldots,n\}$ can be split into $p$ classes of equal size, such that any two subsets with the same sum of elements belong to the same class. [i]Ankan Bhattacharya[/i]

2024 USA IMO Team Selection Test, 1

Tags: USA TST 2024 , USA TST , algebra

Find the smallest constant $C > 1$ such that the following statement holds: for every integer $n \geq 2$ and sequence of non-integer positive real numbers $a_1, a_2, \dots, a_n$ satisfying $$\frac{1}{a_1} + \frac{1}{a_2} + \cdots + \frac{1}{a_n} = 1,$$ it's possible to choose positive integers $b_i$ such that (i) for each $i = 1, 2, \dots, n$, either $b_i = \lfloor a_i \rfloor$ or $b_i = \lfloor a_i \rfloor + 1$, and (ii) we have $$1 < \frac{1}{b_1} + \frac{1}{b_2} + \cdots + \frac{1}{b_n} \leq C.$$ (Here $\lfloor \bullet \rfloor$ denotes the floor function, as usual.) [i]Merlijn Staps[/i]

2024 USA IMO Team Selection Test, 6

Tags: function , algebra , functional equation , USA TST , USA TST 2024

Find all functions $f\colon\mathbb R\to\mathbb R$ such that for all real numbers $x$ and $y$, \[f(xf(y))+f(y)=f(x+y)+f(xy).\] [i]Milan Haiman[/i]

2024 USA IMO Team Selection Test, 5

Tags: USA TST 2024 , USA TST , number theory

Suppose $a_{1} < a_{2}< \cdots < a_{2024}$ is an arithmetic sequence of positive integers, and $b_{1} <b_{2} < \cdots <b_{2024}$ is a geometric sequence of positive integers. Find the maximum possible number of integers that could appear in both sequences, over all possible choices of the two sequences. [i]Ray Li[/i]

2024 USA IMO Team Selection Test, 2

Tags: geometry , incenter , USA TST , USA TST 2024

Let $ABC$ be a triangle with incenter $I$. Let segment $AI$ intersect the incircle of triangle $ABC$ at point $D$. Suppose that line $BD$ is perpendicular to line $AC$. Let $P$ be a point such that $\angle BPA = \angle PAI = 90^\circ$. Point $Q$ lies on segment $BD$ such that the circumcircle of triangle $ABQ$ is tangent to line $BI$. Point $X$ lies on line $PQ$ such that $\angle IAX = \angle XAC$. Prove that $\angle AXP = 45^\circ$. [i]Luke Robitaille[/i]