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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

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[list][*]May 9th, 4:30pm PT/7:30pm ET, Casework 2: Overwhelming Evidence — A Text Adventure, a game where participants will work together to navigate the map, solve puzzles, and win! All are welcome.
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[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
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0 replies
jlacosta
May 1, 2025
0 replies
Flight between cities
USJL   5
N 10 minutes ago by Photaesthesia
Source: 2025 Taiwan TST Round 1 Mock P5
A country has 2025 cites, with some pairs of cities having bidirectional flight routes between them. For any pair of the cities, the flight route between them must be operated by one of the companies $X, Y$ or $Z$. To avoid unfairly favoring specific company, the regulation ensures that if there have three cities $A, B$ and $C$, with flight routes $A \leftrightarrow B$ and $A \leftrightarrow C$ operated by two different companies, then there must exist flight route $B \leftrightarrow C$ operated by the third company different from $A \leftrightarrow B$ and $A \leftrightarrow C$ .

Let $n_X$, $n_Y$ and $n_Z$ denote the number of flight routes operated by companies $X, Y$ and $Z$, respectively. It is known that, starting from a city, we can arrive any other city through a series of flight routes (not necessary operated by the same company). Find the minimum possible value of $\max(n_X, n_Y , n_Z)$.

Proposed by usjl and YaWNeeT
5 replies
1 viewing
USJL
Mar 8, 2025
Photaesthesia
10 minutes ago
A problem from Le Anh Vinh book.
minhquannguyen   0
16 minutes ago
Source: LE ANH VINH, DINH HUONG BOI DUONG HOC SINH NANG KHIEU TOAN TAP 1 DAI SO
Let $n$ is a positive integer. Determine all functions $f:(1,+\infty)\to\mathbb{R}$ such that
\[f(x^{n+1}+y^{n+1})=x^nf(x)+y^nf(y),\forall x,y>1.\]
0 replies
minhquannguyen
16 minutes ago
0 replies
IMO ShortList 1999, algebra problem 1
orl   42
N an hour ago by ihategeo_1969
Source: IMO ShortList 1999, algebra problem 1
Let $n \geq 2$ be a fixed integer. Find the least constant $C$ such the inequality

\[\sum_{i<j} x_{i}x_{j} \left(x^{2}_{i}+x^{2}_{j} \right) \leq C
\left(\sum_{i}x_{i} \right)^4\]

holds for any $x_{1}, \ldots ,x_{n} \geq 0$ (the sum on the left consists of $\binom{n}{2}$ summands). For this constant $C$, characterize the instances of equality.
42 replies
orl
Nov 13, 2004
ihategeo_1969
an hour ago
q(x) to be the product of all primes less than p(x)
orl   19
N an hour ago by ihategeo_1969
Source: IMO Shortlist 1995, S3
For an integer $x \geq 1$, let $p(x)$ be the least prime that does not divide $x$, and define $q(x)$ to be the product of all primes less than $p(x)$. In particular, $p(1) = 2.$ For $x$ having $p(x) = 2$, define $q(x) = 1$. Consider the sequence $x_0, x_1, x_2, \ldots$ defined by $x_0 = 1$ and \[ x_{n+1} = \frac{x_n p(x_n)}{q(x_n)} \] for $n \geq 0$. Find all $n$ such that $x_n = 1995$.
19 replies
orl
Aug 10, 2008
ihategeo_1969
an hour ago
No more topics!
Soviet Union 2
orl   1
N Sep 5, 2004 by grobber
Source: IMO LongList 1959-1966 Problem 6
Let $m$ be a convex polygon in a plane, $l$ its perimeter and $S$ its area. Let $M\left( R\right) $ be the locus of all points in the space whose distance to $m$ is $\leq R,$ and $V\left(R\right) $ is the volume of the solid $M\left( R\right) .$


a.) Prove that \[V (R) = \frac 43 \pi R^3 +\frac{\pi}{2} lR^2 +2SR.\]

Hereby, we say that the distance of a point $C$ to a figure $m$ is $\leq R$ if there exists a point $D$ of the figure $m$ such that the distance $CD$ is $\leq R.$ (This point $D$ may lie on the boundary of the figure $m$ and inside the figure.)

additional question:

b.) Find the area of the planar $R$-neighborhood of a convex or non-convex polygon $m.$

c.) Find the volume of the $R$-neighborhood of a convex polyhedron, e. g. of a cube or of a tetrahedron.

Note by Darij: I guess that the ''$R$-neighborhood'' of a figure is defined as the locus of all points whose distance to the figure is $\leq R.$
1 reply
orl
Sep 1, 2004
grobber
Sep 5, 2004
Source: IMO LongList 1959-1966 Problem 6
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orl
3647 posts
#1 • 3 Y
Y by Adventure10, Adventure10, Mango247
Let $m$ be a convex polygon in a plane, $l$ its perimeter and $S$ its area. Let $M\left( R\right) $ be the locus of all points in the space whose distance to $m$ is $\leq R,$ and $V\left(R\right) $ is the volume of the solid $M\left( R\right) .$


a.) Prove that \[V (R) = \frac 43 \pi R^3 +\frac{\pi}{2} lR^2 +2SR.\]

Hereby, we say that the distance of a point $C$ to a figure $m$ is $\leq R$ if there exists a point $D$ of the figure $m$ such that the distance $CD$ is $\leq R.$ (This point $D$ may lie on the boundary of the figure $m$ and inside the figure.)

additional question:

b.) Find the area of the planar $R$-neighborhood of a convex or non-convex polygon $m.$

c.) Find the volume of the $R$-neighborhood of a convex polyhedron, e. g. of a cube or of a tetrahedron.

Note by Darij: I guess that the ''$R$-neighborhood'' of a figure is defined as the locus of all points whose distance to the figure is $\leq R.$
This post has been edited 2 times. Last edited by orl, Sep 2, 2004, 12:10 PM
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grobber
7849 posts
#2 • 2 Y
Y by Adventure10, Mango247
God! This is going to be awfully hard to explain in a rigorous manner, but I'll take a whack at it. :)

First of all, let's notice what $M(R)$ "looks like". If we place spheres of radius $R$ centered at each of the vertices of the polygon, then $M(R)$ is precisely the convex hull of the spheres. Alternatively, you may think that we place a sheet of rubber around the whole thing (polygon and spheres), and the set is what we obtain after the sheet has shrinked as much as possible.

a) In order to compute the volume, we need to compute the volume of three parts: the prism having the polygon as the median plane and having a height of $2R$ (this has volume $S\cdot R+S\cdot R=(2\cdot S)R$), the half-cylinders around the edges (this part has volume $\frac 12\sum a_i\cdot \pi R^2=(\frac \pi 2\cdot l)R^2$, where $a_i$ are the sides of the polygon), and whatever remains around the vertices. This last part is made up of sphere slices, which we can slide around on the sides of the polygon so that they form a sphere of radius $R$, so the volume is $(\frac 43\cdot \pi)R^3$.

b) The area is, again the sum of the areas of the three regions we have defined above: the area of the bases of the prism ($2S$), the areas of the half-cylinders, which is $l\cdot \pi R$, and the area of a sphere of radius $R$, which is $4\pi R^2$. The total area is thus $4\pi R^2+(\pi\cdot l)R+2\cdot S$.

c) Something similar to a). I'll be back later to edit this.
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