2024 USAMO

Part of USAMO

Subcontests

(5)

1

beads on a necklace

m

n

be positive integers. A circular necklace contains

mn

beads, each either red or blue. It turned out that no matter how the necklace was cut into

m

n

consecutive beads, each block had a distinct number of red beads. Determine, with proof, all possible values of the ordered pair

(m, n)

.
Proposed by Rishabh Das

1

Indicator function inequality

n > 2

be an integer and let

\ell \in \{1, 2,\dots, n\}

A_1,\dots,A_k

of (not necessarily distinct) subsets of

\{1, 2,\dots, n\}

\ell

|A_i| \ge \ell

1 \le i \le k

. Find, in terms of

n

\ell

, the largest real number

c

such that the inequality

\sum_{i=1}^k\sum_{j=1}^k x_ix_j\frac{|A_i\cap A_j|^2}{|A_i|\cdot|A_j|}\ge c\left(\sum_{i=1}^k x_i\right)^2

holds for all positive integer

k

, all nonnegative real numbers

x_1,x_2,\dots,x_k

\ell

-large collections

A_1,A_2,\dots,A_k

\{1,2,\dots,n\}

.
Proposed by Titu Andreescu and Gabriel Dospinescu

1

Equal area sum in regular n-gon triangulation

m

be a positive integer. A triangulation of a polygon is

m

-balanced if its triangles can be colored with

m

colors in such a way that the sum of the areas of all triangles of the same color is the same for each of the

m

colors. Find all positive integers

n

for which there exists an

m

-balanced triangulation of a regular

n

-gon.
Note: A triangulation of a convex polygon

\mathcal{P}

n \ge 3

sides is any partitioning of

\mathcal{P}

n-2

n-3

\mathcal{P}

that do not intersect in the polygon's interior.
Proposed by Krit Boonsiriseth

1

Dizzying Set Intersections

S_1, S_2, \ldots, S_{100}

be finite sets of integers whose intersection is not empty. For each non-empty

T \subseteq \{S_1, S_2, \ldots, S_{100}\},

the size of the intersection of the sets in

T

is a multiple of the number of sets in

T

. What is the least possible number of elements that are in at least

50

sets?
Proposed by Rishabh Das

1

d_k-eja Vu

Find all integers

n \geq 3

such that the following property holds: if we list the divisors of

n!

in increasing order as

1 = d_1 < d_2 < \dots < d_k = n!

d_2 - d_1 \leq d_3 - d_2 \leq \dots \leq d_k - d_{k-1}.

Proposed by Luke Robitaille.