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Mar 2, 2025
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0 replies
jlacosta
Mar 2, 2025
0 replies
Inequality => square
Rushil   12
N 7 minutes ago by ohiorizzler1434
Source: INMO 1998 Problem 4
Suppose $ABCD$ is a cyclic quadrilateral inscribed in a circle of radius one unit. If $AB \cdot BC \cdot CD \cdot DA \geq 4$, prove that $ABCD$ is a square.
12 replies
+1 w
Rushil
Oct 7, 2005
ohiorizzler1434
7 minutes ago
p + q + r + s = 9 and p^2 + q^2 + r^2 + s^2 = 21
who   28
N 24 minutes ago by asdf334
Source: IMO Shortlist 2005 problem A3
Four real numbers $ p$, $ q$, $ r$, $ s$ satisfy $ p+q+r+s = 9$ and $ p^{2}+q^{2}+r^{2}+s^{2}= 21$. Prove that there exists a permutation $ \left(a,b,c,d\right)$ of $ \left(p,q,r,s\right)$ such that $ ab-cd \geq 2$.
28 replies
who
Jul 8, 2006
asdf334
24 minutes ago
H not needed
dchenmathcounts   44
N an hour ago by Ilikeminecraft
Source: USEMO 2019/1
Let $ABCD$ be a cyclic quadrilateral. A circle centered at $O$ passes through $B$ and $D$ and meets lines $BA$ and $BC$ again at points $E$ and $F$ (distinct from $A,B,C$). Let $H$ denote the orthocenter of triangle $DEF.$ Prove that if lines $AC,$ $DO,$ $EF$ are concurrent, then triangle $ABC$ and $EHF$ are similar.

Robin Son
44 replies
dchenmathcounts
May 23, 2020
Ilikeminecraft
an hour ago
IZHO 2017 Functional equations
user01   51
N an hour ago by lksb
Source: IZHO 2017 Day 1 Problem 2
Find all functions $f:R \rightarrow R$ such that $$(x+y^2)f(yf(x))=xyf(y^2+f(x))$$, where $x,y \in \mathbb{R}$
51 replies
user01
Jan 14, 2017
lksb
an hour ago
No more topics!
IMO ShortList 2003, number theory problem 8
orl   10
N Jan 5, 2024 by IAmTheHazard
Source: IMO ShortList 2003, number theory problem 8
Let $p$ be a prime number and let $A$ be a set of positive integers that satisfies the following conditions:

(i) the set of prime divisors of the elements in $A$ consists of $p-1$ elements;

(ii) for any nonempty subset of $A$, the product of its elements is not a perfect $p$-th power.

What is the largest possible number of elements in $A$ ?
10 replies
orl
Oct 4, 2004
IAmTheHazard
Jan 5, 2024
IMO ShortList 2003, number theory problem 8
G H J
Source: IMO ShortList 2003, number theory problem 8
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orl
3647 posts
#1 • 2 Y
Y by Adventure10, Mango247
Let $p$ be a prime number and let $A$ be a set of positive integers that satisfies the following conditions:

(i) the set of prime divisors of the elements in $A$ consists of $p-1$ elements;

(ii) for any nonempty subset of $A$, the product of its elements is not a perfect $p$-th power.

What is the largest possible number of elements in $A$ ?
Attachments:
This post has been edited 1 time. Last edited by djmathman, May 27, 2018, 3:55 PM
Reason: adjusted wording according to https://anhngq.files.wordpress.com/2010/07/imo-2003-shortlist.pdf
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orl
3647 posts
#2 • 2 Y
Y by Adventure10, Mango247
Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: introduction for the IMO ShortList/LongList project and regardingsolutions :)
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Zhero
2043 posts
#3 • 4 Y
Y by huricane, mijail, Adventure10, Mango247
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v_Enhance
6857 posts
#4 • 8 Y
Y by SK_pi3145, Mathematicsislovely, mijail, PIartist, guptaamitu1, HamstPan38825, Ru83n05, starchan
Solution from Twitch Solves ISL:

Let $D = p-1$. Then the question (thinking in terms of the exponents) can be phrased as follows:

What's the largest multiset of vectors in ${\mathbb F}_p^{D}$ such that no nonempty subset has zero sum?

We claim the answer is $D \cdot (p-1)$. A construction is given by taking
  • $p-1$ copies of the vector $\left< 1, 0, \dots, 0\right>$;
  • $p-1$ copies of the vector $\left< 0, 1, \dots, 0\right>$;
  • \dots;
  • $p-1$ copies of the vector $\left< 0, 0, \dots, 1 \right>$.
To show it's best possible, suppose we have vectors $v_1$, \dots, $v_N$ with coordinates given as \[ v_k = \left< v_{k,1}, v_{k,2}, \dots, v_{k,D} \right> 	\qquad k = 1, 2, \dots, N.  \]Then, we construct the polynomial \[ 	F(X_1, \dots, X_N) 	= \prod_{i=1}^D 	\left[ 1 - \left( \sum_{k=1}^N X_k v_{k,j} \right)^{p-1} \right] 	- \prod_{i=1}^N (1-X_i) \]If we imagine the $X_i \in \{0,1\}$ as Bernoulli random variables indicating whether $v_k$ is used in a set or not, then $F(X_1, \dots, X_N) \neq 0$ exactly when the $X_i$'s equal to $1$ correspond to a nonempty subset of the vectors which have vanishing sum.
Now assume for contradiction $N < D \cdot (p-1)$. Then the largest degree term is given in the latter product; it is exactly $(-1)^{N+1} X_1 X_2 \dots X_N$. So if we quote Alon's combinatorial nullstellensatz, it now follows that there is a choice of $X_i \in \{0,1\}$ such that $F(X_1, \dots, X_N) \neq 0$ which is a contradiction.
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Math_Is_Fun_101
159 posts
#5 • 2 Y
Y by Derfu37, guptaamitu1
Solution with derfu37 and a hint to use Chevalley's from Al3jandro0000.

We claim the answer is $(p-1)^2$. This is achievable by taking the set
\[ A = \{ q_i^{pj+1}\vert(i,j)\in\{1,\ldots,p-1\}^2 \} \]where $\{q_i\}_{i=1}^{\infty}$ is the sequence of primes. Now we show $|A|>(p-1)^2$ is impossible. Let $n=|A|$. Note that by condition (i) we can represent the elements of $A$ as vectors in $\mathbb F_p^{p-1}$. Let these be
\[ v_i = (e_{i1},\ldots,e_{i(p-1)})\text{ for } 1\le i\le n. \]Now, we define $f_j\colon\mathbb F_p^n\rightarrow\mathbb F_p^n$ as
\[ f_j(x_1,\ldots,x_n) = \sum_{i=1}^n x_i^{p-1}e_{ji}. \]Consider the system of equations $f_j(x_1,\ldots,x_n)=0$ for $1\le j\le p-1$. Note that $(0,\ldots,0)$ is a trivial solution. Suppose for the sake of contradiction that $n>(p-1)^2$. Then, we have
\[ n>(p-1)^2=(p-1)\cdot(p-1)=\sum_{j=1}^{p-1}\deg(f_j). \]Thus, quoting Chevalley's Theorem there must exist a nontrivial solution to the system. However, since $x_i^{p-1}\in{0,1}$, this nontrivial solution specifies a nonempty subset of $A$ for which the product is a perfect $p-$th power, yielding the desired contradiction. $\blacksquare$
This post has been edited 3 times. Last edited by Math_Is_Fun_101, Oct 13, 2021, 7:51 AM
Reason: Mispelled Chevalley's
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Leo.Euler
577 posts
#6
Y by
One can easily rephrase this problem using the prime factorizations of the elements of $A$; this asks for the size of the largest multiset of vectors of $\mathbb{F}_p^{p-1}$ where no nonempty subset has a sum of zero.

The answer is $(p-1)^2$; this can be obtained by taking\[\bigcup_{\textbf{x}\in \mathbb{F}_p^{p-1} \\ |\textbf{x}|=1}\{\textbf{x} \ \text{repeated} \ p-1 \ \text{times}\}.\]Assume for contradiction that there is a solution of vectors $v_1, \ldots, v_N$ with $N>p-1$ that works, where we write\[ v_i = (e_{i1},\ldots,e_{i(p-1)}) \]for each $i$. Now define\[ f_k(x_1,\ldots,x_N) = \sum_{i=1}^N x_i^{p-1}e_{ki} \]for all $1 \le k \le p-1$. Since $(0, \ldots, 0)$ is a common root of all $f_k$, the Chevalley-Warning theorem guarantees a nontrivial root common to all $f_k$. By Fermat's Little Theorem, each $x_i^{p-1}$ is either $0$ or $1$, so there is a nonempty subset of $\{v_1, \ldots, v_N\}$ with a zero sum by using this nontrivial root, a contradiction. Hence $(p-1)^2$ is the size of the largest multiset of vectors of $\mathbb{F}_p^{p-1}$ where no nonempty subset has a sum of zero, and we are done.
This post has been edited 4 times. Last edited by Leo.Euler, Apr 11, 2023, 3:30 PM
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HamstPan38825
8846 posts
#7
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Chevalley-Warning essentially nukes this.

The answer is $(p-1)^2$. For construction, pick primes $q_1, q_2, \cdots, q_{p-1}$, and use $q_1, q_1^{p+1}, \cdots, q_1^{1+p(p-2)}$ and similarly for the other $q_i$'s. This obviously satisfies both conditions.

For the proof of maximality, set the $q_i$ to be the same and choose $k = (p-1)^2 + 1$ distinct elements from $A$, which we will label $x_1$ through $ x_k$. For each $j$, let $e_{ij}$ be the exponent of $q_i$ in $x_j$.

Now, for the polynomials $$f_i(X_1, \cdots, X_k) = X_1^{p-1}e_{i1} + X_2^{p-1}e_{i2} + \cdots + X_k^{p-1} e_{ik},$$by Chevalley-Warning there exists a nontrivial solution to the system $$f_1(x_1, x_2, \cdots, x_k) \equiv f_2(x_1, x_2, \cdots, x_k) \equiv \cdots \equiv 0 \pmod p.$$Pick a subset $B$ of all nonzero $x_i$ in this set that satisfy the condition. Then by definition and Fermat's Little Theorem it follows that $\sum_{j \in B} e_{ij} \equiv 0 \pmod p$ for each $i$, thus the product of the elements in $B$ is a perfect $p$th power, contradiction.
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CANBANKAN
1301 posts
#8
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Claim: If $x_j = (x_{j,1},\cdots, x_{j,n}) \in (\mathbb{Z}/p\mathbb{Z})^n$ ($j\in \{1,\cdots,K\}$) such that for all $e_1, \cdots, e_K \in \{0,1\}$ (not all zero), then $\sum e_jx_j \ne 0$. Then $K\le n(p-1)$ (this bound is optimal because I can have $p-1$ copies of $(\underbrace{0, \cdots, 0}_{j}, 1, \underbrace{0, \cdots, 0}_{n-1-j})$ for $j=0,\cdots,n-1$)

Proof: Assume $K\ge n(p-1)+1$. Consider the polynomial

$$P(f_1, \cdots, f_K) = \prod\limits_{j=1}^n (1-(\sum_{k=1}^K f_kx_{k,j})^{p-1})$$
If $\sum e_jx_j \ne 0 \iff$ there exists $t$ such that $\sum_k e_k x_{k,t} \ne 0$. Then $1-(\sum_k e_k x_{k,t})^{p-1}=0$ in $\mathbb{Z}/p\mathbb{Z}$. Hence $P(f_1,\cdots,f_K)=0$ for all $f_1,\cdots,f_k \in \{0,1\}$ that are not all zero, and $P(0,\cdots,0)=1$.

We first flatten the polynomial i.e. replace $f_k^j$ with $f_k$ when $j\ge 2$ (i.e. we take this polynomial and take its remainder mod $f_1^2-f_1, f_2^2-f_2, \cdots, f_K^2-f_K$). Call this polynomial $\tilde P$. Now note $$\tilde P + (1-f_1)(1-f_2)\cdots (1-f_K)$$vanishes in $\{0,1\}^K$, but one can show that $$v\colon [K] \to (\mathbb{Z}/p\mathbb{Z})^{2^K}, v_S = ( \prod\limits_{j\in S} f_j )_{(f_1,\cdots,f_k)\in \{0,1\}^K}$$are linearly independent in $\{0,1\}^K$ (or we induct by seeing this as a linear poly in $f_1$ namely $R(f_2,\cdots,f_K) f_1 + Q(f_2,\cdots,f_K)$ and since this poly is 0 when $f_1\in \{0,1\}$, $R(f_2,\cdots,f_K), Q(f_2,\cdots,f_K)$ satisfy the problem conditions for $K-1$) so $$\tilde P + (1-f_1)(1-f_2)\cdots (1-f_K) \equiv 0$$
This is absurd since $\deg \tilde P < K$
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YaoAOPS
1484 posts
#9
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Yet another C-W nuke solution

Claim: $|A| = (p-1)^2$ is achieveable.
Proof. Let $p_n$ denote the $n$th prime for $1 \le n \le p - 1$. Then $p-1$ occurences of each $p_n$ suffices. $\blacksquare$

Claim: $(p-1)^2$ is maximal.
Proof. FTSOC assume there exists some $A$ of size $n = p^2 - 2p$. Enumerate $A$ as $a_{1}, a_{2}, \dots, a_n$ and the prime divisors as $p_1, p_2, \dots, p_n$. For each $i$, let $a_i = p_1^{c_{i1}}p_2^{c_{i2}}\dots p_n^{c_{in}}$.
Construct the $p-1$ polynomials $f_1, f_2, \dots f_{p-1}$ in ${\mathbb F}_p$ such that \[ f_i(x_1, x_2, \dots, x_n) = \sum_{j=1}^{n} c_{ij}x_j^{p-1}. \]Note that $\deg f_i = p-1$ and that $f_i$ is $0$ iff the elements of $A$ at its nonzero inputs have a product with $\nu_{p_i}$ divisible by $p$.
Since $n > \deg f_i \cdot (p-1) = (p-1)^2$, and $(0, 0, \dots, 0)$ is one of the common roots of every polynomial, by Chevalley-Warnining there exists another root, contradiction. $\blacksquare$
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Sourorange
107 posts
#11
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Let $D = p-1$. Then the question (thinking in terms of the exponents) can be phrased as follows:

What's the largest multiset of vectors in ${\mathbb F}_p^{D}$ such that no nonempty subset has zero sum?

We claim the answer is $D \cdot (p-1)$. A construction is given by taking
  • $p-1$ copies of the vector $\left< 1, 0, \dots, 0\right>$;
  • $p-1$ copies of the vector $\left< 0, 1, \dots, 0\right>$;
  • $......$;
  • $p-1$ copies of the vector $\left< 0, 0, \dots, 1 \right>$.
To show it's best possible, suppose we have vectors $v_1$, \dots, $v_N$ with coordinates given as \[ v_k = \left< v_{k,1}, v_{k,2}, \dots, v_{k,D} \right> 	\qquad k = 1, 2, \dots, N.  \]Then, we construct the polynomial \[ 	F(X_1, \dots, X_N) 	= \prod_{i=1}^D 	\left[ 1 - \left( \sum_{k=1}^N X_k v_{k,j} \right)^{p-1} \right] 	- \prod_{i=1}^N (1-X_i) \]If we imagine the $X_i \in \{0,1\}$ as Bernoulli random variables indicating whether $v_k$ is used in a set or not, then $F(X_1, \dots, X_N) \neq 0$ exactly when the $X_i$'s equal to $1$ correspond to a nonempty subset of the vectors which have vanishing sum.
Now assume for contradiction $N < D \cdot (p-1)$. Then the largest degree term is given in the latter product; it is exactly $(-1)^{N+1} X_1 X_2 \dots X_N$. So if we quote Alon's combinatorial nullstellensatz, it now follows that there is a choice of $X_i \in \{0,1\}$ such that $F(X_1, \dots, X_N) \neq 0$ which is a contradiction. $\square$
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IAmTheHazard
5000 posts
#13
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The answer is $(p-1)^2$

Interpret each number as a $\mathbb{F}_p^{p-1}$ vector with entries corresponding to their $\nu_q$ modulo $p$: we are asked to find the largest multiset of vectors such that no nonempty subset of them sum to the $0$ vector. For a construction, we may take $p-1$ copies of each of the $p-1$ vectors with one entry equal to $1$ and the rest $0$.

I will now prove that $k:=|A|>(p-1)^2$ is impossible. To that end, suppose we had $k$ vectors $v_1,\ldots,v_k$ and the $j$-th element of $v_i$ was $a_{ij}$. Consider the following polynomial in $\mathbb{F}_p[x_1,\ldots,x_k]$:
$$P(x):=\prod_{i=1}^{k} (1-x_i)-\prod_{j=1}^{p-1} \left(1-\left(\sum_{i=1}^k a_{ij}x_i\right)^{p-1}\right).$$If we restrict $x_i \in \{0,1\}$ for all $i$, then choosing some subset of the vectors and summing them is equivalent to setting some subset of the $x_i$ to $1$ (and the rest $0$). If we choose the empty set, i.e. $x_i=0$ always, then the first and second products are both $1$. Otherwise, the first product clearly vanishes, and by hypothesis, since some entry of their sum is nonzero, Fermat's little theorem guarantees that the second product vanishes as well. Thus $P$ vanishes for any choice of $(x_1,\ldots,x_k) \in \{0,1\}^k$, but since $k>(p-1)^2$ it contains the maximal-degree term $(-1)^kx_1\ldots x_k$, which yields a contradiction by combinatorial nullstellensatz. $\blacksquare$
This post has been edited 1 time. Last edited by IAmTheHazard, Jan 8, 2024, 2:08 AM
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