Happy Memorial Day! Please note that AoPS Online is closed May 24-26th.

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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.

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]
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0 replies
jlacosta
May 1, 2025
0 replies
k i Suggestion Form
jwelsh   0
May 6, 2021
Hello!

Given the number of suggestions we’ve been receiving, we’re transitioning to a suggestion form. If you have a suggestion for the AoPS website, please submit the Google Form:
Suggestion Form

To keep all new suggestions together, any new suggestion threads posted will be deleted.

Please remember that if you find a bug outside of FTW! (after refreshing to make sure it’s not a glitch), make sure you’re following the How to write a bug report instructions and using the proper format to report the bug.

Please check the FTW! thread for bugs and post any new ones in the For the Win! and Other Games Support Forum.
0 replies
jwelsh
May 6, 2021
0 replies
k i Read me first / How to write a bug report
slester   3
N May 4, 2019 by LauraZed
Greetings, AoPS users!

If you're reading this post, that means you've come across some kind of bug, error, or misbehavior, which nobody likes! To help us developers solve the problem as quickly as possible, we need enough information to understand what happened. Following these guidelines will help us squash those bugs more effectively.

Before submitting a bug report, please confirm the issue exists in other browsers or other computers if you have access to them.

For a list of many common questions and issues, please see our user created FAQ, Community FAQ, or For the Win! FAQ.

What is a bug?
A bug is a misbehavior that is reproducible. If a refresh makes it go away 100% of the time, then it isn't a bug, but rather a glitch. That's when your browser has some strange file cached, or for some reason doesn't render the page like it should. Please don't report glitches, since we generally cannot fix them. A glitch that happens more than a few times, though, could be an intermittent bug.

If something is wrong in the wiki, you can change it! The AoPS Wiki is user-editable, and it may be defaced from time to time. You can revert these changes yourself, but if you notice a particular user defacing the wiki, please let an admin know.

The subject
The subject line should explain as clearly as possible what went wrong.

Bad: Forum doesn't work
Good: Switching between threads quickly shows blank page.

The report
Use this format to report bugs. Be as specific as possible. If you don't know the answer exactly, give us as much information as you know. Attaching a screenshot is helpful if you can take one.

Summary of the problem:
Page URL:
Steps to reproduce:
1.
2.
3.
...
Expected behavior:
Frequency:
Operating system(s):
Browser(s), including version:
Additional information:


If your computer or tablet is school issued, please indicate this under Additional information.

Example
Summary of the problem: When I click back and forth between two threads in the site support section, the content of the threads no longer show up. (See attached screenshot.)
Page URL: http://artofproblemsolving.com/community/c10_site_support
Steps to reproduce:
1. Go to the Site Support forum.
2. Click on any thread.
3. Click quickly on a different thread.
Expected behavior: To see the second thread.
Frequency: Every time
Operating system: Mac OS X
Browser: Chrome and Firefox
Additional information: Only happens in the Site Support forum. My tablet is school issued, but I have the problem at both school and home.

How to take a screenshot
Mac OS X: If you type ⌘+Shift+4, you'll get a "crosshairs" that lets you take a custom screenshot size. Just click and drag to select the area you want to take a picture of. If you type ⌘+Shift+4+space, you can take a screenshot of a specific window. All screenshots will show up on your desktop.

Windows: Hit the Windows logo key+PrtScn, and a screenshot of your entire screen. Alternatively, you can hit Alt+PrtScn to take a screenshot of the currently selected window. All screenshots are saved to the Pictures → Screenshots folder.

Advanced
If you're a bit more comfortable with how browsers work, you can also show us what happens in the JavaScript console.

In Chrome, type CTRL+Shift+J (Windows, Linux) or ⌘+Option+J (Mac).
In Firefox, type CTRL+Shift+K (Windows, Linux) or ⌘+Option+K (Mac).
In Internet Explorer, it's the F12 key.
In Safari, first enable the Develop menu: Preferences → Advanced, click "Show Develop menu in menu bar." Then either go to Develop → Show Error console or type Option+⌘+C.

It'll look something like this:
IMAGE
3 replies
slester
Apr 9, 2015
LauraZed
May 4, 2019
k i Community Safety
dcouchman   0
Jan 18, 2018
If you find content on the AoPS Community that makes you concerned for a user's health or safety, please alert AoPS Administrators using the report button (Z) or by emailing sheriff@aops.com . You should provide a description of the content and a link in your message. If it's an emergency, call 911 or whatever the local emergency services are in your country.

Please also use those steps to alert us if bullying behavior is being directed at you or another user. Content that is "unlawful, harmful, threatening, abusive, harassing, tortuous, defamatory, vulgar, obscene, libelous, invasive of another's privacy, hateful, or racially, ethnically or otherwise objectionable" (AoPS Terms of Service 5.d) or that otherwise bullies people is not tolerated on AoPS, and accounts that post such content may be terminated or suspended.
0 replies
dcouchman
Jan 18, 2018
0 replies
Inspired by RMO 2006
sqing   6
N a few seconds ago by sqing
Source: Own
Let $ a,b >0  . $ Prove that
$$  \frac {a^{2}+1}{b+k}+\frac { b^{2}+1}{ka+1}+\frac {2}{a+kb}  \geq \frac {6}{k+1}  $$Where $k\geq 0.03 $
$$  \frac {a^{2}+1}{b+1}+\frac { b^{2}+1}{a+1}+\frac {2}{a+b}  \geq 3  $$
6 replies
sqing
Saturday at 3:24 PM
sqing
a few seconds ago
Simple triangle geometry [a fixed point]
darij grinberg   50
N a few seconds ago by ezpotd
Source: German TST 2004, IMO ShortList 2003, geometry problem 2
Three distinct points $A$, $B$, and $C$ are fixed on a line in this order. Let $\Gamma$ be a circle passing through $A$ and $C$ whose center does not lie on the line $AC$. Denote by $P$ the intersection of the tangents to $\Gamma$ at $A$ and $C$. Suppose $\Gamma$ meets the segment $PB$ at $Q$. Prove that the intersection of the bisector of $\angle AQC$ and the line $AC$ does not depend on the choice of $\Gamma$.
50 replies
darij grinberg
May 18, 2004
ezpotd
a few seconds ago
IMO 2009, Problem 2
orl   143
N 5 minutes ago by ezpotd
Source: IMO 2009, Problem 2
Let $ ABC$ be a triangle with circumcentre $ O$. The points $ P$ and $ Q$ are interior points of the sides $ CA$ and $ AB$ respectively. Let $ K,L$ and $ M$ be the midpoints of the segments $ BP,CQ$ and $ PQ$. respectively, and let $ \Gamma$ be the circle passing through $ K,L$ and $ M$. Suppose that the line $ PQ$ is tangent to the circle $ \Gamma$. Prove that $ OP = OQ.$

Proposed by Sergei Berlov, Russia
143 replies
orl
Jul 15, 2009
ezpotd
5 minutes ago
Interesting inequality
sqing   2
N 19 minutes ago by aidan0626
Source: Own
Let $ (a+b)^2+(a-b)^2=1. $ Prove that
$$0\geq (a+b-1)(a-b+1)\geq -\frac{3}{2}-\sqrt 2$$$$ -\frac{9}{2}+2\sqrt 2\geq (a+b-2)(a-b+2)\geq -\frac{9}{2}-2\sqrt 2$$
2 replies
+2 w
sqing
an hour ago
aidan0626
19 minutes ago
Request for the access of private marathons
Vulch   8
N 4 hours ago by roye
To all AoPS users and admin,
Sometimes I came across the marathon(i .e number theory marathon, functional equation marathon etc) which allow access only after submitting log in request.There is no other way to access the question related to that marathons.It would be glad to open all private marathons publicly.Thank you!
8 replies
Vulch
Yesterday at 6:06 AM
roye
4 hours ago
plz friend admins
JohannIsBach   2
N 5 hours ago by JohannIsBach
hi just so u no im not trying 2 postfarm i just wanted 2 c how many aops admins i could get 2 friend me. if u dont like this then feel free 2 remove it
aops admins so far:
asuth_asuth
gmass
rrusczyk
sbarrack
thats all right now!
plz friend
again, if u dont like this then plz remove if u dont want it here
2 replies
JohannIsBach
5 hours ago
JohannIsBach
5 hours ago
Who is Halp! ? (resolvedd)
A7456321   8
N 5 hours ago by JohannIsBach
Is Halp! a bot? This user has been posting questions in nearly all of my AoPS classes when the user isn't a part of the class, and this user has 150k posts.
8 replies
A7456321
Saturday at 9:22 PM
JohannIsBach
5 hours ago
How to create a poll?
whwlqkd   32
N Yesterday at 6:54 PM by Yiyj
How to create a poll in aops?
32 replies
whwlqkd
Saturday at 12:24 PM
Yiyj
Yesterday at 6:54 PM
banned myself from own blog
Spacepandamath13   8
N Yesterday at 3:55 PM by sultanine
I got curious and decided to see if I can ban myself from my own blog.
can site admins give it back? it says site admins are the administrators of this blog

I honestly don't know where I come up with stuff like this
8 replies
Spacepandamath13
May 24, 2025
sultanine
Yesterday at 3:55 PM
resolved!
JohannIsBach   6
N May 24, 2025 by bpan2021
hi srry if this is in the rong place i didnt no where 2 put it i was wondering how u find a user? i tried using the search but they dont have any posts? dont no wat 2 do...
6 replies
JohannIsBach
May 24, 2025
bpan2021
May 24, 2025
k Introducing myself at AoPS, and what's your magic wand?
asuth_asuth   1193
N May 23, 2025 by Penguin117
Hi!

I'm Andrew Sutherland. I'm the new Chief Product Officer at AoPS. As you may have read, Richard is retiring and Ben Kornell and I are working together to lead the company now. I'm leading all the software and digital stuff at AoPS. I just wanted to say hello and introduce myself! I'm really excited to be part of the special community that is AoPS.

Previously, I founded Quizlet as a 15-year-old high school student. I did Course 6 at MIT and then left to lead Quizlet full-time for a total of 14 years. I took a few years off and now I'm doing AoPS! I wrote more about all that on my blog: https://asuth.com/im-joining-aops

I have a question for all of you. If you could wave a magic wand, and change anything about AoPS, what would it be? All suggestions welcome! Thank you.
1193 replies
asuth_asuth
Mar 30, 2025
Penguin117
May 23, 2025
k how 2 play reaper?
JohannIsBach   3
N May 22, 2025 by JohannIsBach
hi srry if this is in the rong place i dont no where else 2 put it how do u play reaper? and is htere a link 2 the game? just wondering
3 replies
JohannIsBach
May 22, 2025
JohannIsBach
May 22, 2025
k Alcumus reset
Happycat2   1
N May 22, 2025 by bpan2021
Hi! Because you can't reset your alcumus when your in classes can you email sheriff and let them reset it? My class already ended but I think that you have to wait until it completely disappears but I'm signing up for another class so I can't reset it. So can you email sheriff?
1 reply
Happycat2
May 22, 2025
bpan2021
May 22, 2025
All Topics Marked
alpha31415   14
N May 22, 2025 by alpha31415
Excuse me!
for I am a frashman in aops.I wonder how I could get back to mark all my read.All topics in the forum became as if they had been marked.Would someone please help me deal with it???
14 replies
alpha31415
May 16, 2025
alpha31415
May 22, 2025
IMO ShortList 1999, geometry problem 3
orl   13
N May 6, 2024 by asdf334
Source: IMO ShortList 1999, geometry problem 3
A set $ S$ of points from the space will be called completely symmetric if it has at least three elements and fulfills the condition that for every two distinct points $ A$ and $ B$ from $ S$, the perpendicular bisector plane of the segment $ AB$ is a plane of symmetry for $ S$. Prove that if a completely symmetric set is finite, then it consists of the vertices of either a regular polygon, or a regular tetrahedron or a regular octahedron.
13 replies
orl
Nov 13, 2004
asdf334
May 6, 2024
IMO ShortList 1999, geometry problem 3
G H J
Source: IMO ShortList 1999, geometry problem 3
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orl
3647 posts
#1 • 3 Y
Y by Adventure10, megarnie, Mango247
A set $ S$ of points from the space will be called completely symmetric if it has at least three elements and fulfills the condition that for every two distinct points $ A$ and $ B$ from $ S$, the perpendicular bisector plane of the segment $ AB$ is a plane of symmetry for $ S$. Prove that if a completely symmetric set is finite, then it consists of the vertices of either a regular polygon, or a regular tetrahedron or a regular octahedron.
Attachments:
This post has been edited 1 time. Last edited by orl, Nov 14, 2004, 10:18 PM
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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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mathmanman
1444 posts
#3 • 3 Y
Y by Adventure10, Adventure10, Mango247
http://www.kalva.demon.co.uk/imo/isoln/isoln991.html

Alternate solution :
Let us consider the convex hull $S'$ of the points $A_1, A_2, \ldots, A_n$ of $S$.
Since the convex hull of a convex polygon is a convex polygon, we have that $S'$ satisfies the conditions of the problem. Then, $S'$ is the set of the vertices of regular polygon.
Let's show that : $S = S'$. This means that $S'$ does not contain points inside $S$. Indeed, let us consider an inside point $A \in S$, $S'$ would not be invariant with regard to $s_{AA_i}$, where $AA_i = \min_{1 \le j \le n} AA_j$.
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epitomy01
240 posts
#4 • 2 Y
Y by Adventure10, Mango247
I think the problem statement here is different from the real problem.
In the real problem, the points were in a plane, and not in space, so there was nothing about a tetrahedron or an octahedron.
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Moonmathpi496
413 posts
#5 • 2 Y
Y by Adventure10, Mango247
epitomy01 wrote:
I think the problem statement here is different from the real problem.
In the real problem, the points were in a plane, and not in space, so there was nothing about a tetrahedron or an octahedron.
You're right, but the problem appeared as IMO 01 was adopted the main problem...i.e. space to plane.
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mavropnevma
15142 posts
#6 • 3 Y
Y by Adventure10, Mango247, TensorGuy666
Proof for the planar case, as asked in the competition. Denote by $\Omega$ the mass center of $S$; clearly $\Omega$ lies on each axis of symmetry. Since for $A,B \in S$ the perpendicular bisector of $AB$ is an axis of symmetry for $S$, it follows $\Omega A = \Omega B$, hence $S$ is made of concyclic points, lying on a circle of centre $\Omega$.

Consider any three consecutive (on the circle) points $A,B,C \in S$; since the perpendicular bisector of $AC$ is an axis of symmetry for $S$, it follows $B$ lies on it, hence $BA = BC$. Therefore $S$ is a regular polygon, and clearly all regular polygons fulfill the requirements.
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sunny2000
234 posts
#7 • 2 Y
Y by Adventure10, Mango247
orl wrote:
A set $ S$ of points from the space will be called completely symmetric if it has at least three elements and fulfills the condition that for every two distinct points $ A$ and $ B$ from $ S$, the perpendicular bisector plane of the segment $ AB$ is a plane of symmetry for $ S$. Prove that if a completely symmetric set is finite, then it consists of the vertices of either a regular polygon, or a regular tetrahedron or a regular octahedron.

Solution
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vsathiam
201 posts
#8 • 2 Y
Y by Adventure10, Mango247
I guess for the 3-d case it suffices to show that 1) convex hull of S = S is circumscribable in a sphere, and 2) every circle formed by taking 3 points of S is congruent. [this, combined with the 2-D case should be sufficient...]
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goodbear
1108 posts
#9 • 3 Y
Y by chystudent1-_-, Mango247, Mango247
Cute problem!

First, the global part. Let $O$ be the centroid of $S$. Since reflection across the perpendicular bisector of $AB$ fixes $S$, it must also fix $O$, so $AO=BO$ and $O$ is the circumcenter of $S$.

We now split the problem into two cases:

Case 1: $S$ is coplanar.
Let $\Omega$ be the circumcircle of $S$. Assume for the sake of contradiction that there are points $A,B,C,D\in S$ in that order around $\Omega$ with
  • $A\ne D$,
  • $A$ and $B$ consecutive,
  • $C$ and $D$ consecutive, and
  • $\overarc{\ensuremath{ADB}}<\overarc{\ensuremath{CAD}}$.
[asy]
draw(unitcircle);
pair A=dir(162),B=dir(234),C=dir(354),D=dir(42);
pair Z=(5/4)*sqrt(A*D), Cp=A*D/C;
draw(-Z--Z,blue+dashed+(1/2)*bp);
draw(C--Cp,blue+(1/2)*bp,Arrow,Margins);
dot("$A$",A,A); dot("$B$",B,B); dot("$C$",C,C); dot("$D$",D,D);
dot("$C'$",Cp,Cp,blue);
[/asy]
Then, reflecting $C$ over the perpendicular bisector of $AD$ gives a point between $A$ and $B$ in $S$, a contradiction.

Therefore, the arcs between consecutive points in $S$ must all be congruent, so $S$ consists of the vertices of a regular polygon, as desired.
$\square$

Case 2: $S$ is not coplanar.
Let $\Omega$ be the circumsphere of $S$. Take the convex polyhedron formed by $S$, i.e. the convex hull of $S$. Let $P\in S$ and let the neighbors of $P$ on the convex hull be $A_1,A_2,\ldots,A_n$ in that order for some $n\ge 3$.

By Case 1, the faces of the convex hull are regular polygons, so $$PA_1=PA_2=\ldots=PA_n.$$Thus, $A_1,A_2,\ldots,A_n$ lie on the circle formed by the intersection of $\Omega$ and the sphere centered at $P$ containing $A_1,A_2,\ldots,A_n$. By Case 1, they then must form a regular polygon. Therefore, $$\angle A_1PA_2=\angle A_2PA_3=\ldots=\angle A_nPA_1,$$so the faces containing these angles have the same number of sides and are congruent regular polygons. It follows that the convex hull of $S$ is a platonic solid.

The cube, dodecahedron, and icosahedron do not work because they are not preserved when reflecting across the perpendicular bisector of a space diagonal. Thus, $S$ consists of the vertices of either a regular tetrahedron or a regular octahedron, as desired.
$\square$
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TheUltimate123
1740 posts
#10 • 1 Y
Y by Mango247
Solved with nukelauncher and Th3Numb3rThr33.

First, we verify the planar case: Let \(G\) be the centroid of the points in \(S\); then \(G\) lies on each perpendicular bisector \(AB\), so \(GA=GB\) for all \(A\), \(B\) in \(S\); i.e.\ the points in \(S\) lie on a circle centered at \(G\). Let the points of \(S\) lie on the circle in the order \(A_1\), \(A_2\), \ldots, \(A_n\). We can verify for each \(i\) by looking at the perpendicular bisector of \(\overline{A_{i-1}A_{i+1}}\) that \(A_iA_{i-1}=A_iA_{i+1}\), so the polygon is regular, as needed.

Then assume the points are not all planar. Let \(G\) be the centroid again, so we can verify analogously that the points in \(S\) lie on a sphere centered at \(G\). Consider the polyhedron formed by the points in \(S\).

Observe that any plane containing at least three points must form a regular polygon. We will verify the following claim:

Claim: If we have a plane containing at least four points, and there is a point \(P\) not on the plane, then there is a point \(P\) on the other side of the plane as \(P\).

Proof. Let the plane contain regular polygon \(A_1A_2\cdots A_k\) with side length \(s\). If there is a point \(P\) ``above'' the plane, then we can select a point \(Q\) on the plane containing \(\triangle PA_1A_2\) yet still ``above'' the plane with \(QA_1=s\).

By \(k\ge4\), we have \(QA_1=A_1A_2<A_1A_3\), so the perpendicular bisector plane of \(\overline{QA_3}\) intersects segment \(A_1A_3\); that is, the reflection of \(A_1\) over the perpendicular bisector plane lies ``below'' the plane, and it must also be in \(S\). \(\blacksquare\)

It follows that every face of the polyhedron is an equilateral triangle, so the polyhedron is either a regular tetrahedron, regular octahedron, or regular icosahedron (by platonic solids). To rule out the icosahedron, take the space diagonal; the two pentagons do not pair up.
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crazyeyemoody907
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#11
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whoa gj raymond and espen
:O
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v_Enhance
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#12 • 1 Y
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Solution from Twitch Solves ISL:

Let $G$ be the centroid of $S$.

Claim: All points of $S$ lie on a sphere $\Gamma$ centered at $G$.
Proof. Each perpendicular bisector plane passes through $G$. So if $A,B \in S$ it follows $GA = GB$. $\blacksquare$

Claim: Consider any plane passing through three or more points of $S$. The points of $S$ in the plane form a regular polygon.
Proof. The cross section is a circle because we are intersecting a plane with sphere $\Gamma$. Now if $A$, $B$, $C$ are three adjacent points on this circle, by taking the perpendicular bisector we have $AB=BC$. $\blacksquare$
If the points of $S$ all lie in a plane, we are done. Otherwise, the points of $S$ determine a polyhedron $\Pi$ inscribed in $\Gamma$. All of the faces of $\Pi$ are evidently regular polygons, of the same side length $s$.

Claim: Every face of $\Pi$ is an equilateral triangle.
Proof. Suppose on the contrary some face $A_1 A_2 \dots A_n$ has $n > 3$. Let $B$ be any vertex adjacent to $A_1$ in $\Pi$ other than $A_2$ or $A_n$. Consider the plane determined by $\triangle A_1 A_3 B$. This is supposed to be a regular polygon, but arc $A_1 A_3$ is longer than arc $A_1 B$, and by construction there are no points inside these arcs. This is a contradiction. $\blacksquare$
Hence, $\Pi$ has faces all congruent equilateral triangles. This implies it is a regular polyhedron --- either a regular tetrahedron, regular octahedron, or regular icosahedron. We can check the regular icosahedron fails by taking two antipodal points as our counterexample. This finishes the problem.
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awesomeming327.
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Let $G$ be the centroid. Note that since the entire figure is symmetric across each perpendicular bisector, the centroid must be on every perpendicular bisector. Thus, $S$ is conspheric.

$~$
Let $T$ be a subset of $S$ that lies on a circle. Let $A_1$, $A_2$, $A_3$, $A_4$, $A_5$ be adjacent points appearing in that order, taking cyclics if needed. By the perpendicular bisector of $A_3A_4$, $A_2A_3=A_4A_5$. By the perpendicular bisector of $A_2A_4$, $A_1A_2=A_4A_5$ which means $A_1A_2=A_2A_3$, and therefore the points on $T$ form a regular polygon.

$~$
Clearly any regular polygon works, so suppose there was some other point $P$. Then its neighbors form a regular polygon and therefore all the angles including $P$ of the convex hull are equal. Furthermore all the faces of this solid are regular polygons so we have ourselves a tetrahedron, cube, octahedron, dodecahedron and icosahedron and it is easy to check only the tetra and octa hedrons work.
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asdf334
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First solve the 2D case. If there are at least three points in a plane then consider the convex hull. By choosing a side of the convex hull and taking the "perpendicular bisector plane" we find that every angle of the convex hull is the same. By choosing a segment between points two vertices apart we find that every side of the convex hull has the same length. Hence the convex hull is a regular polygon.
Take a point $A$ on the convex hull and interior point $B$. Reflection preserves the condition that a point is on the convex hull, which is a contradiction as $A$ will be sent to $B$.
Now we solve the 3D case. We have already found that the points in any nontrivial plane form a convex polygon. Consider the convex hull again.
Consider the gravity center (average of vectors) $O$. Since any plane of symmetry passes through the gravity center it follows $O$ is the center of the sphere through all the points. (Oops, could have made this observation earlier.)
We can actually repeat this logic. Take a point $A$ and its neighbors $A_1,\dots,A_n$. Reflection about the perpendicular bisector plane of $A_iA_j$ sends neighbors of $A_i$ to neighbors of $A_j$, thus $A$ is preserved. Hence $AA_i$ is constant, meaning $A_1,\dots,A_n$ lie on a circle. Thus they form a regular polygon, implying that the faces of the convex hull containing $A$ are actually congruent.
This reduces the search space to platonic solids.
  • Tetrahedrons work.
  • Cubes fail on the space diagonal.
  • Octahedrons work.
  • To disprove dodecahedrons and icosahedrons take two points which are a distance of two edges apart.
We are done. $\blacksquare$
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