The time is now - Spring classes are filling up!

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k a AMC 10/12 A&B Coming up Soon!
jlacosta   0
Nov 1, 2024
There is still time to train for the November 6th and November 12th AMC 10A/12A and AMC 10B/12B, respectively! Enroll in our weekend seminars to be held on November 2nd and 3rd (listed below) and you will learn problem strategies, test taking techniques, and be able to take a full practice test! Note that the “B” seminars will have different material from the “A” seminars which were held in October.

[list][*]Special AMC 10 Problem Seminar B
[*]Special AMC 12 Problem Seminar B[/list]
For those who want to take a free practice test before the AMC 10/12 competitions, you can simulate a real competition experience by following this link. As you assess your performance on these exams, be sure to gather data!

[list][*]Which problems did you get right?
[list][*]Was the topic a strength (e.g. number theory, geometry, counting/probability, algebra)?
[*]How did you prepare?
[*]What was your confidence level?[/list]
[*]Which problems did you get wrong?
[list][list][*]Did you make an arithmetic error?
[*]Did you misread the problem?
[*]Did you have the foundational knowledge for the problem?
[*]Which topics require more fluency through practice (e.g. number theory, geometry, counting/probability, algebra)?
[*]Did you run out of time?[/list][/list]
Once you have analyzed the results with the above questions, you will have a plan of attack for future contests! BEST OF LUCK to all competitors at this year’s AMC 10 and AMC 12!

Did you know that the day after both the AMC 10A/12A and AMC 10B/12B you can join a free math jam where our AoPS team will go over the most interesting problems? Find the schedule below under “Mark your calendars”.

Mark your calendars for these upcoming free math jams!
[list][*]November 20th: Amherst College Info Session, 7:30 pm ET: Matt McGann, Dean of Admission and Financial Aid at Amherst College, and Nathan Pflueger, math professor at Amherst College, will host an info session exploring both Amherst College specifically and liberal arts colleges generally. Topics include opportunities in math, the admission process, and financial aid for both US and international students.
[*]November 7th: 2024 AMC 10/12 A Discussion, Thursday, 7:30 pm ET:
[*]AoPS instructors will discuss problems from the AMC 10/12 A, administered November 6. We will discuss some of the most interesting problems from each test!
[*]November 13th: 2024 AMC 10/12 B Discussion, Wednesday, 7:30 pm ET:
[*]AoPS instructors will discuss problems from the AMC 10/12 B, administered November 12. We will discuss some of the most interesting problems from each test![/list]
AoPS Spring classes are open for enrollment. Get a jump on the New Year and enroll in our math, contest prep, coding, and science classes today! Need help finding the right plan for your goals? Check out our recommendations page!

Don’t forget: Highlight your AoPS Education on LinkedIn!
Many of you are beginning to build your education and achievements history on LinkedIn. Now, you can showcase your courses from Art of Problem Solving (AoPS) directly on your LinkedIn profile!

Whether you've taken our classes at AoPS Online or AoPS Academies or reached the top echelons of our competition training with our Worldwide Online Olympiad Training (WOOT) program, you can now add your AoPS experience to the education section on your LinkedIn profile.

Don't miss this opportunity to stand out and connect with fellow problem-solvers in the professional world and be sure to follow us at: https://www.linkedin.com/school/art-of-problem-solving/mycompany/ Check out our job postings, too, if you are interested in either full-time, part-time, or internship opportunities!

Our full course list for upcoming classes is below:
All classes run 7:30pm-8:45pm ET/4:30pm - 5:45pm PT unless otherwise noted.

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0 replies
jlacosta
Nov 1, 2024
0 replies
Diofantina quadrática
Ferreirinha   0
5 minutes ago
Source: Own
Find all prime numbers $p$ such that $p\cdot{(2^{p-1}-1)}$ is a perfect square.
0 replies
Ferreirinha
5 minutes ago
0 replies
Functional equation with a parameter
Tintarn   1
N 27 minutes ago by pco
Source: Baltic Way 2024, Problem 1
Let $\alpha$ be a non-zero real number. Find all functions $f: \mathbb{R}\to\mathbb{R}$ such that
\[
xf(x+y)=(x+\alpha y)f(x)+xf(y)
\]for all $x,y\in\mathbb{R}$.
1 reply
Tintarn
an hour ago
pco
27 minutes ago
A bunch of similar triangles all around the circumcircle
Tintarn   0
29 minutes ago
Source: Baltic Way 2024, Problem 12
Let $ABC$ be an acute triangle with circumcircle $\omega$ such that $AB<AC$. Let $M$ be the midpoint of the arc $BC$ of~$\omega$ containing the point~$A$, and let $X\neq M$ be the other point on $\omega$ such that $AX=AM$. Points $E$ and $F$ are chosen on sides $AC$ and $AB$ of the triangle $ABC$ such that $EX=EC$ and $FX=FB$. Prove that $AE=AF$.
0 replies
Tintarn
29 minutes ago
0 replies
A cyclic quadrilateral with perpendicular diagonals and a tangency
Tintarn   0
31 minutes ago
Source: Baltic Way 2024, Problem 11
Let $ABCD$ be a cyclic quadrilateral with circumcentre $O$ and with $AC$ perpendicular to $BD$. Points $X$ and $Y$ lie on the circumcircle of the triangle $BOD$ such that $\angle AXO=\angle CYO=90^{\circ}$. Let $M$ be the midpoint of $AC$. Prove that $BD$ is tangent to the circumcircle of the triangle $MXY$.
0 replies
Tintarn
31 minutes ago
0 replies
No more topics!
A variation of Bondy's theorem
seoneo   4
N Yesterday at 8:29 AM by AshAuktober
Source: Combinatorics by KMS
Let $m$ and $n$ be positive integers such that $m \le \frac{3}{2}n$ and $S = \left\{ 1,2, \ldots , n \right \}$ and
\[ A_1 , A_2 , \ldots , A_m \]be different subsets of $S$. Prove that there is $k \in S$ with the following property; let $B_j = A_j \setminus \left\{ k \right \}$ then in the set
\[ \left\{ B_1, B_2, \ldots, B_m \right \} \]at least $m-1$ elements are different from each other.
4 replies
seoneo
Apr 23, 2018
AshAuktober
Yesterday at 8:29 AM
A variation of Bondy's theorem
G H J
G H BBookmark kLocked kLocked NReply
Source: Combinatorics by KMS
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seoneo
336 posts
#1 • 6 Y
Y by rkm0959, BSWJY, roughlife, Adventure10, Mango247, kiyoras_2001
Let $m$ and $n$ be positive integers such that $m \le \frac{3}{2}n$ and $S = \left\{ 1,2, \ldots , n \right \}$ and
\[ A_1 , A_2 , \ldots , A_m \]be different subsets of $S$. Prove that there is $k \in S$ with the following property; let $B_j = A_j \setminus \left\{ k \right \}$ then in the set
\[ \left\{ B_1, B_2, \ldots, B_m \right \} \]at least $m-1$ elements are different from each other.
This post has been edited 2 times. Last edited by seoneo, Apr 23, 2018, 8:58 AM
Reason: To fix errata.
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MellowMelon
5850 posts
#2 • 7 Y
Y by BSWJY, CZRorz, seoneo, roughlife, enhanced, Adventure10, Mango247
Construct the graph of all subsets of $S$ where two vertices have an edge colored $i$ between them if they are subsets who only differ in that one has $i$ and one does not (i.e. their symmetric difference is $\{i\}$). This is the $n$-hypercube graph. Take the induced subgraph $G$ consisting of the $m$ vertices corresponding to the $m$ subsets $A_1, A_2, \ldots, A_m$. The problem is equivalent to showing that if $m \leq 3n/2$ then some edge color appears at most once in $G$.

We will show the contrapositive instead. Let $H_0$ be a subgraph of the $n$-hypercube graph with exactly two edges of each of the $n$ colors. Choose a cycle in this graph, if any, and let it have length $x_1$. Following this cycle around has us adding or removing one element for each edge. We get the original subset back, so for each color in this cycle, both edges of that color must be in the cycle: one edge added the element, and the other removed it. Remove all edges of this cycle, and let the new graph be $H_1$. No two vertices on the removed cycle can be in the same connected component in $H_1$, since the corresponding subsets differ in an element whose edge color is now gone. Since the cycle had length $x$, $H_1$ has $x_1-1$ more connected components than $H_0$.

From $H_1$, choose a cycle and remove it to get $H_2$. Repeat this until no cycles remain, and let the final graph be $H_r$ (where $r$ may be 0).

For each $H_i$, let its number of connected components be $c_i$, and let its number of edges be $e_i$. We have $c_0 \geq 1$, $e_0 = 2n$, and
\[ c_i = c_{i-1} + x_i - 1, \quad e_i = e_{i-1} - x_i. \]Since $H_r$ has no cycles, it is a forest, and the number of vertices (including isolated ones) is $c_r + e_r$. This is also the number of vertices of $H_0$. Also, since each $x_i$ is at least 4 and there are only $2n$ edges in $H_0$ we could have removed, $r$ is at most $n/2$. We then bound the number of vertices as follows:
\begin{align*}
c_r + e_r &= c_0 + \sum_{i=1}^r (x_i - 1) + e_0 - \sum_{i=1}^r x_i \\
&= c_0 + e_0 - r, \\
&\geq 2n + 1 - n/2 = 3n/2 + 1.
\end{align*}This means $H_0$ had at least $3n/2 + 1$ vertices. Since $H_0$ was an arbitrary graph with two copies of each edge color, it follows that any graph with at most $3n/2$ vertices has some color with one edge, as desired.
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CBMaster
78 posts
#4 • 1 Y
Y by kiyoras_2001
Result of B. Bollobás. See《Combinatorial Problems and Exercises》, Chapter 13, Problem 10 for the solution and more generalized version of the problem.
This post has been edited 1 time. Last edited by CBMaster, Yesterday at 5:58 AM
Reason: .
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nguyenminh21
3 posts
#5
Y by
Result of B. Bollobás. See《Combinatorial Problems and Exercises》, Chapter 13, Problem 10 for the solution and more generalized version of the problem.
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AshAuktober
575 posts
#6
Y by
What does KMS stand for as in the source?
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N Quick Reply
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