Live Discussion!
Game Jam: Silent Auction! is going on now!

Stay ahead of learning milestones! Enroll in a class over the summer!

G
Topic
First Poster
Last Poster
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.

Are you interested in working towards MATHCOUNTS and don’t know where to start? We have you covered! If you have taken Prealgebra, then you are ready for MATHCOUNTS/AMC 8 Basics. Already aiming for State or National MATHCOUNTS and harder AMC 8 problems? Then our MATHCOUNTS/AMC 8 Advanced course is for you.

Summer camps are starting next month at the Virtual Campus in math and language arts that are 2 - to 4 - weeks in duration. Spaces are still available - don’t miss your chance to have an enriching summer experience. There are middle and high school competition math camps as well as Math Beasts camps that review key topics coupled with fun explorations covering areas such as graph theory (Math Beasts Camp 6), cryptography (Math Beasts Camp 7-8), and topology (Math Beasts Camp 8-9)!

Be sure to mark your calendars for the following upcoming events:
[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.
[*]May 19th, 4:30pm PT/7:30pm ET, What's Next After Beast Academy?, designed for students finishing Beast Academy and ready for Prealgebra 1.
[*]May 20th, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 1 Math Jam, Problems 1 to 4, join the Canada/USA Mathcamp staff for this exciting Math Jam, where they discuss solutions to Problems 1 to 4 of the 2025 Mathcamp Qualifying Quiz!
[*]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]
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.

Introductory: Grades 5-10

Prealgebra 1 Self-Paced

Prealgebra 1
Tuesday, May 13 - Aug 26
Thursday, May 29 - Sep 11
Sunday, Jun 15 - Oct 12
Monday, Jun 30 - Oct 20
Wednesday, Jul 16 - Oct 29

Prealgebra 2 Self-Paced

Prealgebra 2
Wednesday, May 7 - Aug 20
Monday, Jun 2 - Sep 22
Sunday, Jun 29 - Oct 26
Friday, Jul 25 - Nov 21

Introduction to Algebra A Self-Paced

Introduction to Algebra A
Sunday, May 11 - Sep 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Wednesday, May 14 - Aug 27
Friday, May 30 - Sep 26
Monday, Jun 2 - Sep 22
Sunday, Jun 15 - Oct 12
Thursday, Jun 26 - Oct 9
Tuesday, Jul 15 - Oct 28

Introduction to Counting & Probability Self-Paced

Introduction to Counting & Probability
Thursday, May 15 - Jul 31
Sunday, Jun 1 - Aug 24
Thursday, Jun 12 - Aug 28
Wednesday, Jul 9 - Sep 24
Sunday, Jul 27 - Oct 19

Introduction to Number Theory
Friday, May 9 - Aug 1
Wednesday, May 21 - Aug 6
Monday, Jun 9 - Aug 25
Sunday, Jun 15 - Sep 14
Tuesday, Jul 15 - Sep 30

Introduction to Algebra B Self-Paced

Introduction to Algebra B
Tuesday, May 6 - Aug 19
Wednesday, Jun 4 - Sep 17
Sunday, Jun 22 - Oct 19
Friday, Jul 18 - Nov 14

Introduction to Geometry
Sunday, May 11 - Nov 9
Tuesday, May 20 - Oct 28
Monday, Jun 16 - Dec 8
Friday, Jun 20 - Jan 9
Sunday, Jun 29 - Jan 11
Monday, Jul 14 - Jan 19

Paradoxes and Infinity
Mon, Tue, Wed, & Thurs, Jul 14 - Jul 16 (meets every day of the week!)

Intermediate: Grades 8-12

Intermediate Algebra
Sunday, Jun 1 - Nov 23
Tuesday, Jun 10 - Nov 18
Wednesday, Jun 25 - Dec 10
Sunday, Jul 13 - Jan 18
Thursday, Jul 24 - Jan 22

Intermediate Counting & Probability
Wednesday, May 21 - Sep 17
Sunday, Jun 22 - Nov 2

Intermediate Number Theory
Sunday, Jun 1 - Aug 24
Wednesday, Jun 18 - Sep 3

Precalculus
Friday, May 16 - Oct 24
Sunday, Jun 1 - Nov 9
Monday, Jun 30 - Dec 8

Advanced: Grades 9-12

Olympiad Geometry
Tuesday, Jun 10 - Aug 26

Calculus
Tuesday, May 27 - Nov 11
Wednesday, Jun 25 - Dec 17

Group Theory
Thursday, Jun 12 - Sep 11

Contest Preparation: Grades 6-12

MATHCOUNTS/AMC 8 Basics
Friday, May 23 - Aug 15
Monday, Jun 2 - Aug 18
Thursday, Jun 12 - Aug 28
Sunday, Jun 22 - Sep 21
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)

MATHCOUNTS/AMC 8 Advanced
Sunday, May 11 - Aug 10
Tuesday, May 27 - Aug 12
Wednesday, Jun 11 - Aug 27
Sunday, Jun 22 - Sep 21
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)

AMC 10 Problem Series
Friday, May 9 - Aug 1
Sunday, Jun 1 - Aug 24
Thursday, Jun 12 - Aug 28
Tuesday, Jun 17 - Sep 2
Sunday, Jun 22 - Sep 21 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Monday, Jun 23 - Sep 15
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)

AMC 10 Final Fives
Sunday, May 11 - Jun 8
Tuesday, May 27 - Jun 17
Monday, Jun 30 - Jul 21

AMC 12 Problem Series
Tuesday, May 27 - Aug 12
Thursday, Jun 12 - Aug 28
Sunday, Jun 22 - Sep 21
Wednesday, Aug 6 - Oct 22

AMC 12 Final Fives
Sunday, May 18 - Jun 15

AIME Problem Series A
Thursday, May 22 - Jul 31

AIME Problem Series B
Sunday, Jun 22 - Sep 21

F=ma Problem Series
Wednesday, Jun 11 - Aug 27

WOOT Programs
Visit the pages linked for full schedule details for each of these programs!


MathWOOT Level 1
MathWOOT Level 2
ChemWOOT
CodeWOOT
PhysicsWOOT

Programming

Introduction to Programming with Python
Thursday, May 22 - Aug 7
Sunday, Jun 15 - Sep 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Tuesday, Jun 17 - Sep 2
Monday, Jun 30 - Sep 22

Intermediate Programming with Python
Sunday, Jun 1 - Aug 24
Monday, Jun 30 - Sep 22

USACO Bronze Problem Series
Tuesday, May 13 - Jul 29
Sunday, Jun 22 - Sep 1

Physics

Introduction to Physics
Wednesday, May 21 - Aug 6
Sunday, Jun 15 - Sep 14
Monday, Jun 23 - Sep 15

Physics 1: Mechanics
Thursday, May 22 - Oct 30
Monday, Jun 23 - Dec 15

Relativity
Mon, Tue, Wed & Thurs, Jun 23 - Jun 26 (meets every day of the week!)
0 replies
jlacosta
May 1, 2025
0 replies
(x + y)(y + z)(z + x)/{xyz} if (x+ y}/z=(y + z)/x=(z + x)/y
parmenides51   8
N 5 minutes ago by MathIQ.
Source: 2023 Austrian Mathematical Olympiad, Junior Regional Competition , Problem 1
Let $x, y, z$ be nonzero real numbers with $$\frac{x + y}{z}=\frac{y + z}{x}=\frac{z + x}{y}.$$Determine all possible values of $$\frac{(x + y)(y + z)(z + x)}{xyz}.$$
(Walther Janous)
8 replies
parmenides51
Mar 26, 2024
MathIQ.
5 minutes ago
a! + b! = 2^{c!}
parmenides51   7
N 9 minutes ago by MathIQ.
Source: 2023 Austrian Mathematical Olympiad, Junior Regional Competition , Problem 4
Determine all triples $(a, b, c)$ of positive integers such that
$$a! + b! = 2^{c!}.$$
(Walther Janous)
7 replies
parmenides51
Mar 26, 2024
MathIQ.
9 minutes ago
Geometric inequality with angles
Amir Hossein   7
N 27 minutes ago by MathIQ.
Let $p, q$, and $r$ be the angles of a triangle, and let $a = \sin2p, b = \sin2q$, and $c = \sin2r$. If $s = \frac{(a + b + c)}2$, show that
\[s(s - a)(s - b)(s -c) \geq 0.\]
When does equality hold?
7 replies
Amir Hossein
Sep 1, 2010
MathIQ.
27 minutes ago
IMO 2014 Problem 3
v_Enhance   103
N 34 minutes ago by Mysteriouxxx
Source: 0
Convex quadrilateral $ABCD$ has $\angle ABC = \angle CDA = 90^{\circ}$. Point $H$ is the foot of the perpendicular from $A$ to $BD$. Points $S$ and $T$ lie on sides $AB$ and $AD$, respectively, such that $H$ lies inside triangle $SCT$ and \[
\angle CHS - \angle CSB = 90^{\circ}, \quad \angle THC - \angle DTC = 90^{\circ}. \] Prove that line $BD$ is tangent to the circumcircle of triangle $TSH$.
103 replies
v_Enhance
Jul 8, 2014
Mysteriouxxx
34 minutes ago
Inequalities
sqing   16
N Today at 4:01 PM by DAVROS
Let $ a,b,c\geq 0 ,a+b+c\leq 3. $ Prove that
$$a^2+b^2+c^2+ab +2ca+2bc +  abc \leq \frac{251}{27}$$$$ a^2+b^2+c^2+ab+2ca+2bc  + \frac{2}{5}abc  \leq \frac{4861}{540}$$$$ a^2+b^2+c^2+ab+2ca+2bc  + \frac{7}{20}abc  \leq \frac{2381411}{26460}$$
16 replies
sqing
Yesterday at 12:47 PM
DAVROS
Today at 4:01 PM
Inequalities
sqing   0
Today at 2:26 PM
Let $ a,b,c\geq 0 $ and $ab+bc+ca =1.$ Prove that
$$(a^2+b^2+c^2)(a+b+c-2)\ge 8abc(1-a-b-c) $$$$(a^2+b^2+c^2)(a+b+c-\frac{5}{2})\ge 2abc(1-a-b-c) $$
0 replies
sqing
Today at 2:26 PM
0 replies
A suspcious assumption
NamelyOrange   1
N Today at 1:55 PM by NamelyOrange
Let $a,b,c,d$ be positive integers. Maximize $\max(a,b,c,d)$ if $a+b+c+d=a^2-b^2+c^2-d^2=2012$.
1 reply
NamelyOrange
Today at 1:53 PM
NamelyOrange
Today at 1:55 PM
Maximum value of function (with two variables)
Saucepan_man02   1
N Today at 1:39 PM by Saucepan_man02
If $f(\theta) = \min(|2x-7|+|x-4|+|x-2 -\sin \theta|)$, where $x, \theta \in \mathbb R$, then maximum value of $f(\theta)$.
1 reply
Saucepan_man02
Today at 1:25 PM
Saucepan_man02
Today at 1:39 PM
It is given that $M=1+\frac12+\frac13+\frac14+\cdots+\frac{1}{23}=\frac{n}{23!},
Vulch   3
N Today at 11:58 AM by mohabstudent1
It is given that $M=1+\frac12+\frac13+\frac14+\cdots+\frac{1}{23}=\frac{n}{23!},$ where $n$ is a natural number.What is the remainder when $n$ is divided by $13?$
3 replies
Vulch
Apr 9, 2025
mohabstudent1
Today at 11:58 AM
Vieta's Relations
P162008   8
N Today at 11:53 AM by mohabstudent1
If $\alpha,\beta$ and $\gamma$ are the roots of the cubic equation $x^3 - x^2 + 2x - 3 = 0.$
Evaluate $\sum_{cyc} \frac{\alpha^3 - 3}{\alpha^2 - 2}$
Is there any alternate approach except just bash
8 replies
P162008
Yesterday at 10:11 PM
mohabstudent1
Today at 11:53 AM
[PMO22 Areas I.5] Double log equation
aops-g5-gethsemanea2   1
N Today at 10:16 AM by aops-g5-gethsemanea2
Suppose a real number $x>1$ satisfies $$\log_{\sqrt[3]3}(\log_3 x)+\log_3(\log_{27}x)+\log_{27}(\log_{\sqrt[3]3}x)=1.$$Compute $\log_3(\log_3 x)$.

Answer confirmation
1 reply
aops-g5-gethsemanea2
Today at 10:16 AM
aops-g5-gethsemanea2
Today at 10:16 AM
why $sqrt{(x^2 -1)^{2}}$ should be equal to $1-x^2?$
Vulch   5
N Today at 9:38 AM by vanstraelen
In this link, would anyone explain me why $\sqrt{(x^2 -1)^{2}}$ should be equal to $1-x^2?$
5 replies
Vulch
Today at 5:57 AM
vanstraelen
Today at 9:38 AM
Inequalities
sqing   7
N Today at 9:03 AM by sqing
Let $ a,b,c>0. $ Prove that$$a^2+b^2+c^2+abc-k(a+b+c)\geq 3k+2-2(k+1)\sqrt{k+1}$$Where $7\geq k \in N^+.$
$$a^2+b^2+c^2+abc-3(a+b+c)\geq-5$$
7 replies
sqing
May 20, 2025
sqing
Today at 9:03 AM
Inequalities
sqing   2
N Today at 6:42 AM by sqing
Let $ a,b,c> 0 , a^3+b^3+c^3+abc =4.$ Prove that
$$ (a+b)(c+1) \leq 4$$Let $ a,b> 0 ,  a^3+b^3+ab =3.$ Prove that
$$ (a+b) (a+1) (b+1) \leq 8$$
2 replies
sqing
Today at 5:33 AM
sqing
Today at 6:42 AM
Frameable polygons
anantmudgal09   27
N Apr 13, 2025 by Ilikeminecraft
Source: INMO 2020 P5
Infinitely many equidistant parallel lines are drawn in the plane. A positive integer $n \geqslant 3$ is called frameable if it is possible to draw a regular polygon with $n$ sides all whose vertices lie on these lines, and no line contains more than one vertex of the polygon.

(a) Show that $3, 4, 6$ are frameable.
(b) Show that any integer $n \geqslant 7$ is not frameable.
(c) Determine whether $5$ is frameable.

Proposed by Muralidharan
27 replies
anantmudgal09
Jan 19, 2020
Ilikeminecraft
Apr 13, 2025
Frameable polygons
G H J
Source: INMO 2020 P5
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
anantmudgal09
1980 posts
#1 • 17 Y
Y by amar_04, Fermat_Theorem, GammaBetaAlpha, Ankoganit, hellomath010118, Vietjung, MathInfinite, CaptainLevi16, Kayak, NJOY, mathleticguyyy, ThE-dArK-lOrD, biomathematics, donfluffles1412, Pluto04, Adventure10, cubres
Infinitely many equidistant parallel lines are drawn in the plane. A positive integer $n \geqslant 3$ is called frameable if it is possible to draw a regular polygon with $n$ sides all whose vertices lie on these lines, and no line contains more than one vertex of the polygon.

(a) Show that $3, 4, 6$ are frameable.
(b) Show that any integer $n \geqslant 7$ is not frameable.
(c) Determine whether $5$ is frameable.

Proposed by Muralidharan
This post has been edited 1 time. Last edited by anantmudgal09, Jan 19, 2020, 10:37 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
anantmudgal09
1980 posts
#2 • 20 Y
Y by Kayak, Supercali, amar_04, GammaBetaAlpha, Ankoganit, Wizard_32, hellomath010118, AlastorMoody, NJOY, mathleticguyyy, MathPassionForever, CaptainLevi16, donfluffles1412, Atpar, Aryan-23, Pluto04, Pluto1708, Adventure10, Mango247, math_comb01
Solution 1. (Galois)

The conclusion follows in two steps: fix $n \ge 3$ an integer.

Claim: $n$ is frameable $\implies \cos \frac{2\pi}{n} \in \mathbb{Q}$.

Proof. Suppose the $n$-gon is $A_0A_1A_2\dots A_{n-1}$ with side length $a>0$ and we have a family of parallel lines with distance between consecutive lines as $1$. Let $\ell$ be the line which passes through $A_0$ and $j, k, m$ be the orthogonal distances from $A_1, A_2, A_{n-1}$ to $\ell$. Let $\alpha=\measuredangle (\ell, A_0A_1)$ and $t=a^2$.

Observe that $$\cos \left(\left(\frac{2\pi}{n}+\alpha\right)-\alpha\right)=\cos \left(\frac{2\pi}{n}+\alpha\right)\cos \alpha-\sin \left(\frac{2\pi}{n}+\alpha\right)\sin \alpha=\frac{\sqrt{(t-k)(t-j)}}{t}-\frac{jk}{t}$$which simplifies to $t\cdot \left(\cos^2 \frac{2\pi}{n}-1\right)=-\left(j^2+k^2-2jk\cos \frac{2\pi}{n}\right)$. Similarly, working with $\measuredangle (A_0A_1, A_0A_{n-1})$ one gets $$t\cdot \left(\cos^2 \frac{2\pi}{n}-1\right)=-\left(m^2+j^2+2mj\cos \frac{2\pi}{n}\right).$$Equating, we get the desired result. $\blacksquare$

Claim: $\cos \frac{2\pi}{n} \in \mathbb{Q} \implies n \in \{3, 4, 6\}$.

Proof. Let $z=e^{\frac{2\pi i}{n}}$ and $x=\cos \frac{2\pi}{n}$. Note that $2x=z+z^{-1}$ so $[\mathbb{Q}(z):\mathbb{Q}(x)]=2$ and $[\mathbb{Q}(x):\mathbb{Q}]=1$ since $x \in \mathbb{Q}$. We know from algebraic NT that the minimal polynomial of $z$ is the $n$th cyclotomic polynomial, with degree $\phi(n)$. So $$\phi(n)=[\mathbb{Q}(z):\mathbb{Q}]=[\mathbb{Q}(z):\mathbb{Q}(x)] \cdot [\mathbb{Q}(x):\mathbb{Q}]=2$$forces $\phi(n)=2$ so $n \in \{3, 4, 6\}$. $\blacksquare$

Constructions for these are easy to do. $\blacksquare$


Sketch of Solution 2. (Ingenious descent, official solution)

Again, constructions are easy to do so we won't bother. We prove that $n \geqslant 7$ is not frameable.

Suppose the $n$-gon is $A_1A_2\dots A_n$ with side length $a>0$. Pick a point $O$ away from these, on some line. Construct points $B_1, \dots, B_n$ such that $OA_iA_{i+1}B_i$ is a parallelogram, for all $i$, indices modulo $n$. Note that $B_1B_2\dots B_n$ is a regular $n$-gon, with vertices lying on these lines. Note that $\frac{B_1B_2}{A_1A_2}=\frac{B_1B_2}{OB_2}=\left(2\sin \frac{\pi}{n}\right)=k<1$ as $n>6$. Iterate this procedure $m$ times to get a regular $n$-gon with side length $ak^m$ whose vertices lie on these lines. Since the distance between any two lines exceeds some fixed $\varepsilon>0$, by choosing $m$ sufficiently large, we get $ak^m<\varepsilon$, a contradiction!

For $n=5$, reflect $B_i$ in $O$ to get $B_i'$. This lets us construct a frameable decagon; contradiction! $\blacksquare$

Remark. Easily my favourite problem on the test :)
During test-solve, both me and NVT found the Galois solution, but missed the official one; we feel it is really tricky to come up with. Hope kids liked it as much as I did :D
This post has been edited 3 times. Last edited by anantmudgal09, Nov 14, 2020, 5:14 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
ubermensch
820 posts
#3 • 2 Y
Y by Hexagrammum16, Adventure10
Is it possible to get a bijection between frame-ability and tessellate-ability?
This post has been edited 1 time. Last edited by ubermensch, Jan 19, 2020, 11:10 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
sa2001
281 posts
#4 • 8 Y
Y by fjm30, Supercali, Arhaan, MathInfinite, Aryan-23, alibaba42, Adventure10, Mango247
Hope I'm not missing something.
This can also be done by Niven's theorem, which makes the problem both easy and 'elementary'.
Constructions for $n = 3, 4, 6$ are easy to obtain so we'll omit them.
For $n \ge 5$, label the vertices $1, ..., n$ modulo $n$. Then note that chords $V_2V_n$ and $V_3V_{n-1}$ are parallel and distinct. Their projections on the axis perpendicular to the drawn axes are commensurable (i.e. rational multiples of each other). So their lengths are themselves commensurable. This means $sin(2pi/n)$ and $sin(4pi/n)$ are commensurable. Then $cos(2pi/n)$ is rational. Then Niven's theorem forces $n = 6$, and we're done.
This post has been edited 1 time. Last edited by sa2001, Jan 19, 2020, 12:25 PM
Reason: Grammar
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Wizard_32
1566 posts
#5 • 1 Y
Y by Adventure10
Man I thought that descent could work, just like in the proof to show that the only lattice $n$ gon is a square. But couldn't work out the details in time :(
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Arhaan
829 posts
#6 • 1 Y
Y by Adventure10
sa2001 wrote:
Hope I'm not missing something.
This can also be done by Niven's theorem, which makes the problem both easy and 'elementary'.
Constructions for $n = 3, 4, 6$ are easy to obtain so we'll omit them.
For $n \ge 5$, label the vertices $1, ..., n$ modulo $n$. Then note that chords $V_2V_n$ and $V_3V_{n-1}$ are parallel and distinct. Their projections on the axis perpendicular to the drawn axes are commensurable (i.e. rational multiples of each other). So their lengths are themselves commensurable. This means $sin(2pi/n)$ and $sin(4pi/n)$ are commensurable. Then $cos(2pi/n)$ is rational. Then Niven's theorem forces $n = 6$, and we're done.

I was thinking along those lines in the contest, but I didn't know Niven's theorem so thought it would useless to try to prove when $\sin$ would be rational (also I was caught off guard by the fact that we take $\sin 37^\circ$ to be $\frac{3}{5}$ (that's only an approximation))
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
User526797
130 posts
#7 • 1 Y
Y by Adventure10
anantmudgal09 wrote:
Solution 1. (Galois)

The conclusion follows in two steps: fix $n \ge 3$ an integer.

Claim: $n$ is frameable $\implies \cos \frac{2\pi}{n} \in \mathbb{Q}$.

Proof. Suppose the $n$-gon is $A_0A_1A_2\dots A_{n-1}$ with side length $a>0$ and we have a family of parallel lines with distance between consecutive lines as $1$. Let $\ell$ be the line which passes through $A_0$ and $j, k, m$ be the orthogonal distances from $A_1, A_2, A_{n-1}$ to $\ell$. Let $\alpha=\measuredangle (\ell, A_0A_1)$ and $t=a^2$.

Observe that $$\cos \left(\left(\frac{2\pi}{n}+\alpha\right)-\alpha\right)=\cos \left(\cos \frac{2\pi}{n}+\alpha\right)\cos \alpha-\sin \left(\cos \frac{2\pi}{n}+\alpha\right)\sin \alpha=\frac{\sqrt{(t-k)(t-j)}}{t}-\frac{jk}{t}$$which simplifies to $t\cdot \left(\cos^2 \frac{2\pi}{n}-1\right)=-\left(j^2+k^2-2jk\cos \frac{2\pi}{n}\right)$. Similarly, working with $\measuredangle (A_0A_1, A_0A_{n-1})$ one gets $$t\cdot \left(\cos^2 \frac{2\pi}{n}-1\right)=-\left(m^2+j^2+2mj\cos \frac{2\pi}{n}\right).$$Equating, we get the desired result. $\blacksquare$

Claim: $\cos \frac{2\pi}{n} \in \mathbb{Q} \implies n \in \{3, 4, 6\}$.

Proof. Let $z=e^{\frac{2\pi i}{n}}$ and $x=\cos \frac{2\pi}{n}$. Note that $2x=z+z^{-1}$ so $[\mathbb{Q}(z):\mathbb{Q}(x)]=2$ and $[\mathbb{Q}(x):\mathbb{Q}]=1$ since $x \in \mathbb{Q}$. We know from algebraic NT that the minimal polynomial of $z$ is the $n$th cyclotomic polynomial, with degree $\phi(n)$. So $$\phi(n)=[\mathbb{Q}(z):\mathbb{Q}]=[\mathbb{Q}(z):\mathbb{Q}(x)] \cdot [\mathbb{Q}(x):\mathbb{Q}]=2$$forces $\phi(n)=2$ so $n \in \{3, 4, 6\}$. $\blacksquare$

Constructions for these are easy to do. $\blacksquare$


Sketch of Solution 2. (Ingenius descent, official solution)

Again, constructions are easy to do so we won't bother. We prove that $n \geqslant 7$ is not frameable.

Suppose the $n$-gon is $A_1A_2\dots A_n$ with side length $a>0$. Pick a point $O$ away from these, on some line. Construct points $B_1, \dots, B_n$ such that $OA_iA_{i+1}B_i$ is a parallelogram, for all $i$, indices modulo $n$. Note that $B_1B_2\dots B_n$ is a regular $n$-gon, with vertices lying on these lines. Note that $\frac{B_1B_2}{A_1A_2}=\frac{B_1B_2}{OB_2}=\left(2\sin \frac{\pi}{n}\right)=k<1$ as $n>6$. Iterate this procedure $m$ times to get a regular $n$-gon with side length $ak^m$ whose vertices lie on these lines. Since the distance between any two lines exceeds some fixed $\varepsilon>0$, by choosing $m$ sufficiently large, we get $ak^m<\varepsilon$, a contradiction!

For $n=5$, reflect $B_i$ in $O$ to get $B_i'$. This lets us construct a frameable decagon; contradiction! $\blacksquare$

Remark. Easily my favourite problem on the test :)
During test-solve, both me and NVT found the Galois solution, but missed the official one; we feel it is really tricky to come up with. Hope kids liked it as much as I did :D

I did by official method :wow:
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Hexagrammum16
1774 posts
#8 • 1 Y
Y by Adventure10
ubermensch wrote:
Is it possible to get a bijection between frame-ability and tessellate-ability?

I thought of that, since I recently read about the crystallographic restriction theorem (basically what you're suggesting)

I proved that the pencil of parallel lines must be one of the medians of the triangles whose vertices are part of the polygon. In the contest I could only prove till this but now with some heavy trig bash I think I finished it.
This post has been edited 1 time. Last edited by Hexagrammum16, Jan 19, 2020, 3:30 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
ubermensch
820 posts
#9 • 2 Y
Y by Adventure10, Mango247
Hexagrammum16 wrote:
ubermensch wrote:
Is it possible to get a bijection between frame-ability and tessellate-ability?

I thought of that, since I recently read about the crystallographic restriction theorem (basically what you're suggesting)

I proved that the pencil of parallel lines must be one of the medians of the triangles whose vertices are part of the polygon. In the contest I could only prove till this but now with some heavy trig bash I think I finished it.

Huh... I remember getting a very similar statement, but I couldn't finish it during the exam either... it just reminded me of tesselatablity because of the similar config and of course the values of $n$ which work (I tried to prove that given frame-ability and translating it repeatedly from original polygon along the direction of the perpendicular bisector of each side such that set of parallel lines is the same must imply a way to get a tessellation, but my argument was vague at best)- perhaps you could post your completed solution?
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
mathsworm
765 posts
#10 • 1 Y
Y by Adventure10
I solved only part (a)
I hope the marks distribution is generous enough to give atleast 3 marks for part (a)
For n=3
Consider three consecutive lines. Rotate the one in the one by $30^{\circ}$ clockwise and counterclockwise. Now, we construct the triangle in the obvious way by intersecting the lines.

Assume WLOG that the lines are $x=0,1,-1,2,-2\dots$ in the coordinate plane. Then, the points $(2,1), (-1,2), (-2,-1), (1,-2)$ form the vertices of a square with no two vertices on the same line.

For $n=6$, I used the complex plane, and again, WLOG, let the lines be $Re(z)=0,1,-1,2,-1\dots$. Counter clockwise rotation by $60^{\circ}$ about the origin corresponds to a multiplication by $e=\frac{1}{2}+i\frac{\sqrt{3}}{2}$.
We choose the points $a=4+\sqrt{3}$, $b=ae$, $c=ae^2$. Then, $a,b,c,-a,-b,-c$ form the vertices of a regular hexagon with vertices on different lines.

Hence, we are done.
This post has been edited 2 times. Last edited by mathsworm, Jan 19, 2020, 5:53 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Wizard_32
1566 posts
#11 • 3 Y
Y by mathsworm, amar_04, Adventure10
Just for fun, here's a cool way to construct for $n=4.$

Let the square be $ABCD.$ Consider an arbitrary line $\ell_a$ through $A.$ Draw $\ell_b,\ell_c,\ell_d$ through $B,C,D$ such that they are all parallel to $\ell_a.$ Let $x$ be the distance between $\ell_a,\ell_b,$ which by symmetry is the same as the distance between $\ell_d,\ell_c.$ Let $y$ be the distance between $\ell_b,\ell_d.$

Notice that when $\ell_a \equiv AB, x=0$ while $y>0.$ Further when $\ell_b \equiv BC, y=0$ while $x>0.$ By continuity there would exist a position of $\ell_a$ when $x=y.$ This is the desired configuration.
This post has been edited 1 time. Last edited by Wizard_32, Jan 20, 2020, 3:59 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
biomathematics
2568 posts
#12 • 6 Y
Y by RAMUGAUSS, Ankoganit, amar_04, babu2001, DPS, Adventure10
Easily my favourite problem in INMO 2020. Objectively, (b) is the hardest problem in the paper.

Let $n$ be a framable number, and let $\mathcal{P}$ be an $n$-gon framed in the lines given. Set up a complex coordinate system with the lines being parallel to the $y$-axis and adjacent lines being one unit apart.

Suppose $0$ is the centre of $\mathcal{P}$ (we can do this because the origin will also have its real part rational, and then we can scale and translate), and let $z = a+ib$ be one of its vertices. Then, all its vertices are $z,z\omega, z\omega^2, \cdots, z\omega^{n-1}$, where $\omega = e^{\frac{2i\pi}{n}}$.

We want the real part of $z\omega^k$ to be an integer for all $0 \le k \le n-1$. This implies
$$a \cos \frac{2k\pi}{n} - b \sin \frac{2k\pi}{n} \in \mathbb{Z}$$Put $k=0$ to get $a \in \mathbb{Z}$. Put $k=1$ and $k=n-1$, and add to get
$$a \cos \frac{2\pi}{n}, b \sin \frac{2\pi}{n} \in \mathbb{Q}$$If $a \neq 0$, then we get $\cos \frac{2\pi}{n} \in \mathbb{Q}$. Otherwise $a=0$, so $b \neq 0$ and $b \sin \frac{2i\pi}{n} \in \mathbb{Q}$. Put $k=1,2$ and divide (this is allowed because $\sin \frac{2\pi}{n} \neq 0$) to get $\cos \frac{2\pi}{n} \in \mathbb{Q}$ anyway.

Thus, if $n$ is framable, then $\cos \frac{2\pi}{n} \in \mathbb{Q}$.

Let us now solve the parts given.

(a) This is easy.
(b) and (c) Suppose $n \ge 3$ and $\cos \frac{2\pi}{n}$ is rational. Let $2 \cos \frac{2\pi}{n} = \frac{p}{q}$ for coprime integers $p,q$, with $q > 0$. Then, $\alpha = \cos \frac{2\pi}{n} + i \sin \frac{2\pi}{n}$ has minimal polynomial $qx^2-px+q \in \mathbb{Z}[x]$.

Also, $\alpha$ is a root of $x^{n-1}+x^{n-2} + \cdots + 1 \in \mathbb{Z}[x]$ too. This means $qx^2-px+q$ is a factor of $x^{n-1}+x^{n-2} + \cdots + 1$ in $\mathbb{Z}[x]$. This means $q= 1$, so $2 \cos \frac{2\pi}{n}$ is an integer. This forces $\cos \frac{2\pi}{n} = \pm 1, \frac{\pm 1}{2}, 0$, which implies $n=3,4,6$.

Of course, (c) can be solved separately too; $\cos \frac{2\pi}{5} = \frac{\sqrt{5}-1}{4}$ is irrational.
This post has been edited 7 times. Last edited by biomathematics, Jan 20, 2020, 11:40 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Hexagrammum16
1774 posts
#14 • 2 Y
Y by Adventure10, Mango247
Can anyone please send a diagram for n=6?

Thanks @below
This post has been edited 1 time. Last edited by Hexagrammum16, Jan 20, 2020, 3:44 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
biomathematics
2568 posts
#15 • 3 Y
Y by NJOY, Hexagrammum16, Adventure10
Hexagrammum16 wrote:
Can anyone please send a diagram for n=6?

I'll tell you complex coordinates of the $6$ points.

$\sqrt{3}-5i, -\sqrt{3}+5i, 3\sqrt{3}-i, -3\sqrt{3}+i, \sqrt{3}+4i, -\sqrt{3}-4i$. Here, my lines are parallel to the $x$-axis.
This post has been edited 1 time. Last edited by biomathematics, Jan 20, 2020, 11:39 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
zephyr7723
460 posts
#16 • 2 Y
Y by Adventure10, Mango247
Can we directly mention Niven's Theorem in INMO?
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
sa2001
281 posts
#17 • 3 Y
Y by alibaba42, Adventure10, Mango247
Here's the proof for Niven's theorem I'd read when I first found out about the theorem:
First prove by induction that $2cos(nx)$ is an $n$-degree monic integer polynomial in $2cos(x)$ (use $cos((n+2)x) + cos(nx) = 2cos((n+1)x)cosx$). Now, if $q$ is rational, then $2cos(nq*pi)$ is $0$ for some natural number $n$. Thus, $2cos(q*pi)$ is a root of a monic integer polynomial. By rational root theorem, $2cos(q*pi)$ must be an integer if it is rational, which gives us the desired values $cos(q*pi) = -1, -1/2, 0, 1/2, 1$.
Fun fact: We can use the identity $cos(2x) = 2cos^2(x) - 1$ to determine all rational $q$ such that $cos^2(q*pi)$ is rational.
This post has been edited 3 times. Last edited by sa2001, Jan 21, 2020, 5:34 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
BOBTHEGR8
272 posts
#18 • 1 Y
Y by Adventure10
Coordinate Solution-
Let the polygon be $A_1A_2,\cdots,A_n$ . Set up a coordinate system with $A_1\leftrightarrow(0,0)$ and $A_2\leftrightarrow(1,0)$.
We get that $A_3\leftrightarrow(1+$cos$\frac{2\pi}n, $sin$\frac{2\pi}n)$ and $A_4\leftrightarrow(2$cos$^2\frac{2\pi}n+$cos$\frac{2\pi}n, $sin$\frac{2\pi}n(1+2$cos$\frac{2\pi}n)) $
Now let the $n$ be frameable. Let the slope of the parallel lines be $m$. We note that all the $x-intecepts$ of these lines must be rational .
Hence for $A_3$ we get $1+$cos$\frac{2\pi}n + m $sin$\frac{2\pi}n = r$
And for $A_4$ we get $2$cos$^2\frac{2\pi}n+ $cos$\frac{2\pi}n+m$sin$\frac{2\pi}n(1+2$cos$\frac{2\pi}n) =s $ Where $r,s\in\mathbb{Q}$
Eliminating $m$sin$\frac{2\pi}n$ we get that-
$(r-1-$cos$\frac{2\pi}n)(1+2$cos$\frac{2\pi}n)=s-2$cos$^2\frac{2\pi}n- $cos$\frac{2\pi}n$
From this we have $2(r-1)$cos$\frac{2\pi}n=s-r+1$
But as each point lies on only one line the $x-intercept$ of $A_3$ cannot be $1$ (as $n\geq 3$) and hence $r\neq 1$
Hence we get that cos$\frac{2\pi}n$ is rational.
After this the proof is same as in #2 solution 1
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Physicsknight
640 posts
#19 • 1 Y
Y by Adventure10
Answer
$n\in\{3,4,6\} $

Solution
At first, it can be noticed that $3$, $4$ and $6$ are frameable. As we can put the vertices of a triangle, a square and a regular hexagon into the given lines.
$n=3$ is frameable
https://services.artofproblemsolving.com/download.php?id=YXR0YWNobWVudHMvYS9jL2NkZjk5NmNjNTZiNmRmNzcxYTVjNmQzM2JjZDQxOWFjOGVjN2U3LmpwZw==&rn=MjAyMDAxMjdfMTc1MjE3LmpwZw==
$n=4$ is frameable. Let $E $ be midpoint of $A_1(C), A_2(D)$
https://services.artofproblemsolving.com/download.php?id=YXR0YWNobWVudHMvOC8zLzc3ZGZlNzgxMDEwNmI4YzQxOGMxYzkwYmUyY2IzZGMyZjIwMjI2LmpwZw==&rn=MjAyMDAxMjdfMTc1MjI5LmpwZw==
$n=6$ is frameable
//cdn.artofproblemsolving.com/images/6/c/d/6cd0abe6fbd73f59ba60e78ad2fa9a31438bcf61.png
We need to prove that other numbers are not frameable.
Consider $A_1A_2\cdots A_n $ $(n\ge 5) $ is a regular polygon with $n$ sides. Let $\ell_i (i=1,2,\cdots,n)$ be the drawn line passes through $A_i $ and $d_{(i,j)}(1\le i <j\le n)$ be the number of drawn lines between $A_i $ and $A_j $.
Note, $\frac {d (1,2)+1}{d (3,n)+1}=\frac {A_1A_2}{A_3A_n}\implies\frac {A_1A_2}{A_3A_n}\in\mathbb Q$
Let $H_1$, $H_2$ be the points in $\overline {A_1A_2} $ and $\overline {A_nA_3} $ respectively, such that $A_1H_1\perp A_nA_3$ and $A_2H_2\perp A_nA_3$.
$\frac {A_nA_3}{A_1A_2}=1+2\cdot\frac {A_nH_1}{A_nH_1}=1+2\sin\measuredangle A_nA_1H_1$
$\frac {A_nA_3}{A_1A_2}\in\mathbb Q\implies \sin\measuredangle A_nA_1H_1$ must be rational. Now, $\angle A_nA_1H_1=\frac{\pi}{2}-\angle A_1A_nA_3=\frac {\pi}{2}-\frac {2}{n}\cdot \pi=\pi\cdot\frac {n-4}{2n} $
It is easy to see that $0 <\sin\left (\pi\cdot\frac {n-4}{2n}\right)<1(n\ge 5)$
Applying$\text { Niven's theorem}\implies\sin(\pi)\frac {n-4}{2n}=\tfrac 12\implies\frac {n-4}{2n}=\tfrac 16\implies n=6$
If $n\ge 5$, and $n\neq 6$, then $n $ is not frameable. So, $3,4,6$ are all frameable.
This post has been edited 1 time. Last edited by Physicsknight, Feb 4, 2020, 5:53 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Swayx113
2 posts
#20 • 2 Y
Y by Adventure10, Mango247
Making things easier
How about I consider two lines..
A straight perpendicular line which is equal to a units and other slanting lines also equal to k units In other parallels I mean
By this 3 is framable by taking two slant lines and distance between them on the parallel line as k too
Same for 4 as a square of side n.a can be formed where n is a natural number
For 6- 4 slant lines are used which act as transversal and distance between the vertices of parallel line be k
Greater than 6 not possible as those figures will have both slant and perpendiculars which are not equal in length as we know perpendicular is shortest line between two parallel lines


For 5 I don't know how to explain tbh
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
ftheftics
651 posts
#21 • 2 Y
Y by hellomath010118, Gerninza
At first of all suppose $\Omega$ be the plane in complex plane consists all equidistant parallel lines .
  • Suppose the parallel lines are parallel to the real axis. This is because we always get a transformation $T(ax+by-c)\to e +si$ where $s$ is a real constant .
  • also suppose two consecutive lines are one unit apart to each other .

  • Suppose $P_0,\cdots ,P_{n-1}$ are n ($n>2$)points of n sided regular polygon . Suppose $P_i =z .\zeta ^i$ for all $i \in \{0,1,2,\cdots ,n -1\}$. Where $z$ is a complex number with $z=x+yi$

    Also denote the parallel lines as $l_i$ iff $P_i \in l_i$ .

So we can see that $l_k \equiv \ Im ({P_k})=y \cos \frac{2\pi k}{n} +x \sin \frac{2\pi k}{n}$ .

So, $l_0 \equiv y \cdots (1) $ .

$l_1 \equiv y \cos \frac{2\pi }{n} +x \sin \frac{2\pi }{n} \cdots (2)$

$l_{n-1} \equiv y \cos \frac{2\pi }{n} -x \sin \frac{2\pi }{n} \cdots (3)$.

From $(2),(3)$ get ,

$x\sin \frac{2\pi}{n} ,y \cos \frac{2\pi}{n} $ both are rational number .

So, from $(2),(1)$ get $y$ is also a rational number .

$\implies \cos \frac{2\pi}{n} $ is rational.

Part (a)

We get $\cos \frac{2\pi }{3} =\frac{-1}{2} \in \mathbb{Q}$ and similarly we get $ \cos \frac{2\pi}{n} $ is rational for $n=4,6$ .

So $n=3,4,6$ are Frameable .

Part (c)

We get $\cos \frac{2\pi}{5} = \frac{\sqrt{5}-1}{2} \in  (\mathbb{R}- \mathbb{Q})$.

So ,$n=5$ is not Frameable .

Part (b).

Claim:

we always get a polynomial $P_n$ ($n\ge 1$)with integer cofficient such that $P_n(2\cos t) = 2 \cos nt$. In general $\deg (P_n)=n$ and its a Monic polynomial.

Proof.

It can be done by simple induction .

We can get some examples for $n=1,2$.

  • $P_1(x)=x $ .
  • $P_2(x)=x^2-2 $ and $P_2(2 \cos t) =2 \cos 2t$.

Note that $2\cos (kx) (2\cos x) - 2\cos ((k-1) x)$.

$\implies \boxed{ P_{k+1} (x)= x P_k(x) -P_{k-1}(x)}$.

Here $\deg (P_{k+1} ) =k+1$.
So our claim is proved $\blacksquare$.

Now $P_n(2\cos \frac{2\pi}{n})=2\cos (2\pi)=2$.

If $2\cos \frac{2\pi} {n} =\frac{p}{q}$ is rational then $P_n-2$ has a rational root then ,

$ | p| \mid 2$ and $|q|  \mid 1$ so,

for $|\frac{p}{q}|$ we have only two chice that is $|\cos \frac{2\pi}{n}| =1,\frac{1}{2}$.

So ,for $n=2,0,3,6$ the above is possible. So, for any $n\ge 7$ are not Frameable .
This post has been edited 13 times. Last edited by ftheftics, Mar 10, 2020, 12:38 PM
Reason: Hj
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
djmathman
7938 posts
#22 • 2 Y
Y by Dem0nfang, Mango247
Orient the diagram so that all parallel lines are horizontal. Suppose a regular $n$-gon is framable, and without loss let it have side length $1$. Let $A$, $B$, $C$, and $D$ be consecutive vertices of the polygon (where possibly $A\equiv D$). Then $ABCD$ is an isosceles trapezoid and, in particular, $AD\parallel BC$. But since all the parallel lines are equidistant, the vertical distance between $A$ and $D$ is a rational multiple of the vertical distance between $B$ and $C$; it follows that the length $AD$ is a rational multiple of the length $BC$, i.e. $AD = \lambda\in\mathbb Q$.

But now a bit of right-triangle trig on trapezoid $ABCD$ yields that $\cos(\tfrac{2\pi}n) = \tfrac{\lambda-1}2$, which is rational. Hence $n\in\{3,4,6\}$, and we proceed as in the other solutions.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
gulkr10sept
27 posts
#23
Y by
djmathman wrote:
Orient the diagram so that all parallel lines are horizontal. Suppose a regular $n$-gon is framable, and without loss let it have side length $1$. Let $A$, $B$, $C$, and $D$ be consecutive vertices of the polygon (where possibly $A\equiv D$). Then $ABCD$ is an isosceles trapezoid and, in particular, $AD\parallel BC$. But since all the parallel lines are equidistant, the vertical distance between $A$ and $D$ is a rational multiple of the vertical distance between $B$ and $C$; it follows that the length $AD$ is a rational multiple of the length $BC$, i.e. $AD = \lambda\in\mathbb Q$.

But now a bit of right-triangle trig on trapezoid $ABCD$ yields that $\cos(\tfrac{2\pi}n) = \tfrac{\lambda-1}2$, which is rational. Hence $n\in\{3,4,6\}$, and we proceed as in the other solutions.

Nice but i could not understand the last paragraph
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
themathematicalmantra
145 posts
#24
Y by
what i did was that for Part A i did a periodic tiling for 3,4,6
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
phoenixfire
372 posts
#25
Y by
Consider this plane to be a ruled sheet of paper. Assume that a regular $n$-gon has been placed on the paper. Let $O$ be the center of this polygon. Rotate the plane through an angle $360^{\circ} / n$ centered at the point $O$. Then each vertex of the $n$-gon lies at the intersection point of two grid lines: the old and the new (obtained after the rotation) parallel lines. Now, this is just embedding in the unit lattice, but this is possible only for $n=3,4$ or $6$.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
HamstPan38825
8868 posts
#26
Y by
The key is that the result is equivalent to $\cos \frac{2\pi}n$ being rational. To see this, consider any four consecutive vertices $A_1A_2A_3A_4$ and note that $A_1A_4=rA_2A_3$ for some rational $r$. By dropping the altitudes from $A_2, A_3$ this is equivalent to the conclusion.

On the other hand, we must have $n \in \{3, 4, 6\}$ by Niven's theorem. It is easy to see that regular $n$-gons for these values of $n$ actually work.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
abeot
128 posts
#27 • 1 Y
Y by centslordm
Consider the problem in the complex plane. By arbitrarily rotating and scaling, we may assume that the lines are parallel to the real axis of the complex plane, and that the points are defined as \[ \text{cis } \theta, \text{cis} \left( \theta + \frac{2\pi}{n} \right), \dots, \text{cis} \left( \theta + (n-1) * \frac{2\pi}{n} \right) \]in particular, then we require \[ \frac{\sin \left( \theta + 2\cdot \frac{2\pi}{n} \right) - \sin \left( \theta - 2\cdot \frac{2\pi}{n} \right)}{\sin \left( \theta + \frac{2\pi}{n} \right) - \sin \left( \theta - \frac{2\pi}{n} \right)} \]to be rational; using sum to product, this is equivalent to \[ \frac{\cos \theta \sin\left( 2\cdot \frac{2\pi}{n} \right)}{\cos \theta \sin \left( \frac{2\pi}{n} \right)} = \frac{\sin \left( 2\cdot \frac{2\pi}{n} \right)}{\sin \left( \frac{2\pi}{n} \right)} \]but now, just use the fact $\sin \left( 2\cdot \frac{2\pi}{n} \right) = 2 \sin \left( \frac{2\pi}{n} \right) \cos \left( \frac{2\pi}{n} \right)$. Thus, we have \[ \cos \left( \frac{2\pi}{n} \right) \in \mathbb{Q} \]which, by Niven's theorem, implies that $n = 3, 4, 6$.

Now the construction for $n = 3$ is just $1, w, w^2$ where $w^2 + w + 1 = 0$ with the lines parallel to the real axis through each point (they are obviously equally spaced apart).

For $n = 4$, we consider $0, 2+i, -1+2i, 1+3i$ and the lines parallel to the real axis intersecting the imaginary axis at integer multiples of $i$.

For $n = 6$, consider the angle $\theta < 60^\circ$ satisfying $2\sin \theta = \sin \left( 60^\circ - \theta \right)$; in particular, for this $\theta$, $\sin \theta = \frac{\sqrt{3}}{\sqrt{28}}$. Then consider the points \[ A_1 = 0, A_2 = \text{cis } \theta, A_3 = \text{cis } \theta + \text{cis } \left( 60 + \theta \right), \dots \]notice that by construction, the imaginary component of $A_2 - A_1$ is half that of $A_6 - A_1$. By symmetry, it is also the case that $A_5 - A_4$ has an imaginary component half of $A_6 - A_4$.
I claim that the imaginary component of $A_2A_3$ is also a rational multiple of $A_1A_2$. Indeed, it is just \[ \sin \left( \theta + 60^\circ \right) = \sin \theta \cos 60^\circ + \cos \theta \sin 60^\circ = \frac{6\sqrt{3}}{2\sqrt{28}} \]so then just take the lines parallel to the real axis, intersecting the imaginary axis at multiples of $\frac{\sqrt{3}}{\sqrt{28}}$. $\blacksquare$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Saucepan_man02
1356 posts
#28
Y by
Here we go:

By sufficient dilating and rotating, consider the vertices of the regular $n$-gon to be in unit circle, with the black lines parallel to the imaginary axis. Consider the points to be of the form: $\text{cis}(\theta + \frac{2 \pi k}{n})$.
We must have: $\frac{\text{cis} (\theta + \tfrac{2\pi k}{n}) - \text{cis} (\theta - \tfrac{2 \pi k}{n})}{\text{cis} (\theta + \frac{2\pi}{n}) - \text{cis} (\theta - \frac{2 \pi}{n})} \in \mathbb Q.$ Expanding it and plugging $k=2$ gives us: $\cos(\tfrac{2 \pi}{n}) \in \mathbb Q$.

Due to Niven's theorem, we must have $n=3, 4, 6$ and we are done.
This post has been edited 1 time. Last edited by Saucepan_man02, Jan 15, 2025, 10:50 AM
Reason: EDIT
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Ilikeminecraft
658 posts
#29
Y by
Note that this implies that $\frac{\cos(\theta + \ell \cdot \frac{2\pi}{n}) - \cos(\theta - \ell \cdot \frac{2\pi}{n})}{\cos(\theta + \frac{2\pi}{n}) - \cos(\theta - \frac{2\pi}{n})} \in\mathbb Q.$ Simplifying implies $\frac{\sin(\frac{2\pi}{n}\cdot\ell)}{\sin(\frac{2\pi}{n})}\in\mathbb Q.$ If we take $\ell = 2,$ then we have $\cos(\frac{2\pi}{n}) \in\mathbb Q.$ Nixen implies only $n = 3, 4, 6$ work.
Z K Y
N Quick Reply
G
H
=
a