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k a March Highlights and 2025 AoPS Online Class Information
jlacosta   0
Mar 2, 2025
March is the month for State MATHCOUNTS competitions! Kudos to everyone who participated in their local chapter competitions and best of luck to all going to State! Join us on March 11th for a Math Jam devoted to our favorite Chapter competition problems! Are you interested in training for MATHCOUNTS? Be sure to check out our AMC 8/MATHCOUNTS Basics and Advanced courses.

Are you ready to level up with Olympiad training? Registration is open with early bird pricing available for our WOOT programs: MathWOOT (Levels 1 and 2), CodeWOOT, PhysicsWOOT, and ChemWOOT. What is WOOT? WOOT stands for Worldwide Online Olympiad Training and is a 7-month high school math Olympiad preparation and testing program that brings together many of the best students from around the world to learn Olympiad problem solving skills. Classes begin in September!

Do you have plans this summer? There are so many options to fit your schedule and goals whether attending a summer camp or taking online classes, it can be a great break from the routine of the school year. Check out our summer courses at AoPS Online, or if you want a math or language arts class that doesn’t have homework, but is an enriching summer experience, our AoPS Virtual Campus summer camps may be just the ticket! We are expanding our locations for our AoPS Academies across the country with 15 locations so far and new campuses opening in Saratoga CA, Johns Creek GA, and the Upper West Side NY. Check out this page for summer camp information.

Be sure to mark your calendars for the following events:
[list][*]March 5th (Wednesday), 4:30pm PT/7:30pm ET, HCSSiM Math Jam 2025. Amber Verser, Assistant Director of the Hampshire College Summer Studies in Mathematics, will host an information session about HCSSiM, a summer program for high school students.
[*]March 6th (Thursday), 4:00pm PT/7:00pm ET, Free Webinar on Math Competitions from elementary through high school. Join us for an enlightening session that demystifies the world of math competitions and helps you make informed decisions about your contest journey.
[*]March 11th (Tuesday), 4:30pm PT/7:30pm ET, 2025 MATHCOUNTS Chapter Discussion MATH JAM. AoPS instructors will discuss some of their favorite problems from the MATHCOUNTS Chapter Competition. All are welcome!
[*]March 13th (Thursday), 4:00pm PT/7:00pm ET, Free Webinar about Summer Camps at the Virtual Campus. Transform your summer into an unforgettable learning adventure! From elementary through high school, we offer dynamic summer camps featuring topics in mathematics, language arts, and competition preparation - all designed to fit your schedule and ignite your passion for learning.[/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.

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0 replies
jlacosta
Mar 2, 2025
0 replies
k i Peer-to-Peer Programs Forum
jwelsh   157
N Dec 11, 2023 by cw357
Many of our AoPS Community members share their knowledge with their peers in a variety of ways, ranging from creating mock contests to creating real contests to writing handouts to hosting sessions as part of our partnership with schoolhouse.world.

To facilitate students in these efforts, we have created a new Peer-to-Peer Programs forum. With the creation of this forum, we are starting a new process for those of you who want to advertise your efforts. These advertisements and ensuing discussions have been cluttering up some of the forums that were meant for other purposes, so we’re gathering these topics in one place. This also allows students to find new peer-to-peer learning opportunities without having to poke around all the other forums.

To announce your program, or to invite others to work with you on it, here’s what to do:

1) Post a new topic in the Peer-to-Peer Programs forum. This will be the discussion thread for your program.

2) Post a single brief post in this thread that links the discussion thread of your program in the Peer-to-Peer Programs forum.

Please note that we’ll move or delete any future advertisement posts that are outside the Peer-to-Peer Programs forum, as well as any posts in this topic that are not brief announcements of new opportunities. In particular, this topic should not be used to discuss specific programs; those discussions should occur in topics in the Peer-to-Peer Programs forum.

Your post in this thread should have what you're sharing (class, session, tutoring, handout, math or coding game/other program) and a link to the thread in the Peer-to-Peer Programs forum, which should have more information (like where to find what you're sharing).
157 replies
jwelsh
Mar 15, 2021
cw357
Dec 11, 2023
k i C&P posting recs by mods
v_Enhance   0
Jun 12, 2020
The purpose of this post is to lay out a few suggestions about what kind of posts work well for the C&P forum. Except in a few cases these are mostly meant to be "suggestions based on historical trends" rather than firm hard rules; we may eventually replace this with an actual list of firm rules but that requires admin approval :) That said, if you post something in the "discouraged" category, you should not be totally surprised if it gets locked; they are discouraged exactly because past experience shows they tend to go badly.
-----------------------------
1. Program discussion: Allowed
If you have questions about specific camps or programs (e.g. which classes are good at X camp?), these questions fit well here. Many camps/programs have specific sub-forums too but we understand a lot of them are not active.
-----------------------------
2. Results discussion: Allowed
You can make threads about e.g. how you did on contests (including AMC), though on AMC day when there is a lot of discussion. Moderators and administrators may do a lot of thread-merging / forum-wrangling to keep things in one place.
-----------------------------
3. Reposting solutions or questions to past AMC/AIME/USAMO problems: Allowed
This forum contains a post for nearly every problem from AMC8, AMC10, AMC12, AIME, USAJMO, USAMO (and these links give you an index of all these posts). It is always permitted to post a full solution to any problem in its own thread (linked above), regardless of how old the problem is, and even if this solution is similar to one that has already been posted. We encourage this type of posting because it is helpful for the user to explain their solution in full to an audience, and for future users who want to see multiple approaches to a problem or even just the frequency distribution of common approaches. We do ask for some explanation; if you just post "the answer is (B); ez" then you are not adding anything useful.

You are also encouraged to post questions about a specific problem in the specific thread for that problem, or about previous user's solutions. It's almost always better to use the existing thread than to start a new one, to keep all the discussion in one place easily searchable for future visitors.
-----------------------------
4. Advice posts: Allowed, but read below first
You can use this forum to ask for advice about how to prepare for math competitions in general. But you should be aware that this question has been asked many many times. Before making a post, you are encouraged to look at the following:
[list]
[*] Stop looking for the right training: A generic post about advice that keeps getting stickied :)
[*] There is an enormous list of links on the Wiki of books / problems / etc for all levels.
[/list]
When you do post, we really encourage you to be as specific as possible in your question. Tell us about your background, what you've tried already, etc.

Actually, the absolute best way to get a helpful response is to take a few examples of problems that you tried to solve but couldn't, and explain what you tried on them / why you couldn't solve them. Here is a great example of a specific question.
-----------------------------
5. Publicity: use P2P forum instead
See https://artofproblemsolving.com/community/c5h2489297_peertopeer_programs_forum.
Some exceptions have been allowed in the past, but these require approval from administrators. (I am not totally sure what the criteria is. I am not an administrator.)
-----------------------------
6. Mock contests: use Mock Contests forum instead
Mock contests should be posted in the dedicated forum instead:
https://artofproblemsolving.com/community/c594864_aops_mock_contests
-----------------------------
7. AMC procedural questions: suggest to contact the AMC HQ instead
If you have a question like "how do I submit a change of venue form for the AIME" or "why is my name not on the qualifiers list even though I have a 300 index", you would be better off calling or emailing the AMC program to ask, they are the ones who can help you :)
-----------------------------
8. Discussion of random math problems: suggest to use MSM/HSM/HSO instead
If you are discussing a specific math problem that isn't from the AMC/AIME/USAMO, it's better to post these in Middle School Math, High School Math, High School Olympiads instead.
-----------------------------
9. Politics: suggest to use Round Table instead
There are important conversations to be had about things like gender diversity in math contests, etc., for sure. However, from experience we think that C&P is historically not a good place to have these conversations, as they go off the rails very quickly. We encourage you to use the Round Table instead, where it is much more clear that all posts need to be serious.
-----------------------------
10. MAA complaints: discouraged
We don't want to pretend that the MAA is perfect or that we agree with everything they do. However, we chose to discourage this sort of behavior because in practice most of the comments we see are not useful and some are frankly offensive.
[list] [*] If you just want to blow off steam, do it on your blog instead.
[*] When you have criticism, it should be reasoned, well-thought and constructive. What we mean by this is, for example, when the AOIME was announced, there was great outrage about potential cheating. Well, do you really think that this is something the organizers didn't think about too? Simply posting that "people will cheat and steal my USAMOO qualification, the MAA are idiots!" is not helpful as it is not bringing any new information to the table.
[*] Even if you do have reasoned, well-thought, constructive criticism, we think it is actually better to email it the MAA instead, rather than post it here. Experience shows that even polite, well-meaning suggestions posted in C&P are often derailed by less mature users who insist on complaining about everything.
[/list]
-----------------------------
11. Memes and joke posts: discouraged
It's fine to make jokes or lighthearted posts every so often. But it should be done with discretion. Ideally, jokes should be done within a longer post that has other content. For example, in my response to one user's question about olympiad combinatorics, I used a silly picture of Sogiita Gunha, but it was done within a context of a much longer post where it was meant to actually make a point.

On the other hand, there are many threads which consist largely of posts whose only content is an attached meme with the word "MAA" in it. When done in excess like this, the jokes reflect poorly on the community, so we explicitly discourage them.
-----------------------------
12. Questions that no one can answer: discouraged
Examples of this: "will MIT ask for AOIME scores?", "what will the AIME 2021 cutoffs be (asked in 2020)", etc. Basically, if you ask a question on this forum, it's better if the question is something that a user can plausibly answer :)
-----------------------------
13. Blind speculation: discouraged
Along these lines, if you do see a question that you don't have an answer to, we discourage "blindly guessing" as it leads to spreading of baseless rumors. For example, if you see some user posting "why are there fewer qualifiers than usual this year?", you should not reply "the MAA must have been worried about online cheating so they took fewer people!!". Was sich überhaupt sagen lässt, lässt sich klar sagen; und wovon man nicht reden kann, darüber muss man schweigen.
-----------------------------
14. Discussion of cheating: strongly discouraged
If you have evidence or reasonable suspicion of cheating, please report this to your Competition Manager or to the AMC HQ; these forums cannot help you.
Otherwise, please avoid public discussion of cheating. That is: no discussion of methods of cheating, no speculation about how cheating affects cutoffs, and so on --- it is not helpful to anyone, and it creates a sour atmosphere. A longer explanation is given in Seriously, please stop discussing how to cheat.
-----------------------------
15. Cutoff jokes: never allowed
Whenever the cutoffs for any major contest are released, it is very obvious when they are official. In the past, this has been achieved by the numbers being posted on the official AMC website (here) or through a post from the AMCDirector account.

You must never post fake cutoffs, even as a joke. You should also refrain from posting cutoffs that you've heard of via email, etc., because it is better to wait for the obvious official announcement. A longer explanation is given in A Treatise on Cutoff Trolling.
-----------------------------
16. Meanness: never allowed
Being mean is worse than being immature and unproductive. If another user does something which you think is inappropriate, use the Report button to bring the post to moderator attention, or if you really must reply, do so in a way that is tactful and constructive rather than inflammatory.
-----------------------------

Finally, we remind you all to sit back and enjoy the problems. :D

-----------------------------
(EDIT 2024-09-13: AoPS has asked to me to add the following item.)

Advertising paid program or service: never allowed

Per the AoPS Terms of Service (rule 5h), general advertisements are not allowed.

While we do allow advertisements of official contests (at the MAA and MATHCOUNTS level) and those run by college students with at least one successful year, any and all advertisements of a paid service or program is not allowed and will be deleted.
0 replies
v_Enhance
Jun 12, 2020
0 replies
k i Stop looking for the "right" training
v_Enhance   50
N Oct 16, 2017 by blawho12
Source: Contest advice
EDIT 2019-02-01: https://blog.evanchen.cc/2019/01/31/math-contest-platitudes-v3/ is the updated version of this.

EDIT 2021-06-09: see also https://web.evanchen.cc/faq-contest.html.

Original 2013 post
50 replies
v_Enhance
Feb 15, 2013
blawho12
Oct 16, 2017
<QBC =<PCB if BM = CN, <PMC = <MAB, <QNB = < NAC
parmenides51   1
N 6 minutes ago by dotscom26
Source: 2005 Estonia IMO Training Test p2
On the side BC of triangle $ABC$, the points $M$ and $N$ are taken such that the point $M$ lies between the points $B$ and $N$, and $| BM | = | CN |$. On segments $AN$ and $AM$, points $P$ and $Q$ are taken so that $\angle PMC = \angle  MAB$ and $\angle QNB = \angle NAC$. Prove that $\angle QBC = \angle PCB$.
1 reply
parmenides51
Sep 24, 2020
dotscom26
6 minutes ago
Bosnia and Herzegovina EGMO TST 2017 Problem 2
gobathegreat   2
N 18 minutes ago by anvarbek0813
Source: Bosnia and Herzegovina EGMO Team Selection Test 2017
It is given triangle $ABC$ and points $P$ and $Q$ on sides $AB$ and $AC$, respectively, such that $PQ\mid\mid BC$. Let $X$ and $Y$ be intersection points of lines $BQ$ and $CP$ with circumcircle $k$ of triangle $APQ$, and $D$ and $E$ intersection points of lines $AX$ and $AY$ with side $BC$. If $2\cdot DE=BC$, prove that circle $k$ contains intersection point of angle bisector of $\angle BAC$ with $BC$
2 replies
gobathegreat
Sep 19, 2018
anvarbek0813
18 minutes ago
Another NT FE
nukelauncher   58
N 29 minutes ago by andrewthenerd
Source: ISL 2019 N4
Find all functions $f:\mathbb Z_{>0}\to \mathbb Z_{>0}$ such that $a+f(b)$ divides $a^2+bf(a)$ for all positive integers $a$ and $b$ with $a+b>2019$.
58 replies
nukelauncher
Sep 22, 2020
andrewthenerd
29 minutes ago
Easiest Functional Equation
NCbutAN   7
N 31 minutes ago by InftyByond
Source: Random book
Find all functions $f: \mathbb R \to \mathbb R$ such that $$f(yf(x)+f(xy))=(x+f(x))f(y)$$Follows for all reals $x,y$.
7 replies
NCbutAN
Mar 2, 2025
InftyByond
31 minutes ago
Tennessee Math Tournament (TMT) Online 2025
TennesseeMathTournament   27
N 6 hours ago by sadas123
Hello everyone! We are excited to announce a new competition, the Tennessee Math Tournament, created by the Tennessee Math Coalition! Anyone can participate in the virtual competition for free.

The testing window is from March 22nd to April 5th, 2025. Virtual competitors may participate in the competition at any time during that window.

The virtual competition consists of three rounds: Individual, Bullet, and Team. The Individual Round is 60 minutes long and consists of 30 questions (AMC 10 level). The Bullet Round is 20 minutes long and consists of 80 questions (Mathcounts Chapter level). The Team Round is 30 minutes long and consists of 16 questions (AIME level). Virtual competitors may compete in teams of four, or choose to not participate in the team round.

To register and see more information, click here!

If you have any questions, please email connect@tnmathcoalition.org or reply to this thread!
27 replies
TennesseeMathTournament
Mar 9, 2025
sadas123
6 hours ago
OTIS Mock AIME 2025 airs Dec 19th
v_Enhance   38
N Today at 1:32 AM by smileapple
Source: https://web.evanchen.cc/mockaime.html
Satisfactory. Keep cooking.
IMAGE

Problems are posted at https://web.evanchen.cc/mockaime.html#current now!

Like last year, we're running the OTIS Mock AIME 2025 again, except this time there will actually be both a I and a II because we had enough problems to pull it off. However, the two versions will feel quite different from each other:

[list]
[*] The OTIS Mock AIME I is going to be tough. It will definitely be harder than the actual AIME, by perhaps 2 to 4 problems. But more tangibly, it will also have significant artistic license. Problems will freely assume IMO-style background throughout the test, and intentionally stretch the boundary of what constitutes an “AIME problem”.
[*] The OTIS Mock AIME II is meant to be more practically useful. It will adhere more closely to the difficulty and style of the real AIME. There will inevitably still be some more IMO-flavored problems, but they’ll appear later in the ordering.
[/list]
Like last time, all 30 problems are set by current and past OTIS students.

Details are written out at https://web.evanchen.cc/mockaime.html, but to highlight important info:
[list]
[*] Free, obviously. Anyone can participate.
[*]Both tests will release sometime Dec 19th. You can do either/both.
[*]If you'd like to submit for scoring, you should do so by January 20th at 23:59 Pacific time (same deadline for both). Please hold off on public spoilers before then.
[*]Solutions, statistics, and maybe some high scores will be published shortly after that.
[/list]
Feel free to post questions, hype comments, etc. in this thread.
38 replies
v_Enhance
Dec 6, 2024
smileapple
Today at 1:32 AM
How to get better at AMC 10
Dream9   6
N Today at 1:31 AM by sadas123
I'm nearly in high school now but only average like 75 on AMC 10 sadly. I want to get better so I'm doing like the first 11 questions of previous AMC 10's almost every day because I also did previous years for AMC 8. Is there any specific way to get better scores and understand more difficult problems past AMC 8? I have almost no trouble with AMC 8 problem given enough time (like 23-24 right with enough time).
6 replies
Dream9
Yesterday at 1:17 AM
sadas123
Today at 1:31 AM
An FE. Who woulda thunk it?
nikenissan   112
N Today at 1:04 AM by Marcus_Zhang
Source: 2021 USAJMO Problem 1
Let $\mathbb{N}$ denote the set of positive integers. Find all functions $f : \mathbb{N} \rightarrow \mathbb{N}$ such that for positive integers $a$ and $b,$ \[f(a^2 + b^2) = f(a)f(b) \text{ and } f(a^2) = f(a)^2.\]
112 replies
nikenissan
Apr 15, 2021
Marcus_Zhang
Today at 1:04 AM
AIME score for college apps
Happyllamaalways   56
N Yesterday at 4:45 PM by Countmath1
What good colleges do I have a chance of getting into with an 11 on AIME? (Any chances for Princeton)

Also idk if this has weight but I had the highest AIME score in my school.
56 replies
Happyllamaalways
Mar 13, 2025
Countmath1
Yesterday at 4:45 PM
MIT Beaverworks Summer Institute
PowerOfPi_09   0
Yesterday at 4:30 PM
Hi! I was wondering if anyone here has completed this program, and if so, which track did you choose? Do rising juniors have a chance, or is it mainly rising seniors that they accept? Also, how long did it take you to complete the prerequisites?
Thanks!
0 replies
PowerOfPi_09
Yesterday at 4:30 PM
0 replies
k HOT TAKE: MIT SHOULD NOT RELEASE THEIR DECISIONS ON PI DAY
alcumusftwgrind   8
N Yesterday at 10:13 AM by maxamc
rant lol

Imagine a poor senior waiting for their MIT decisions just to have their hopes CRUSHED on 3/14 and they can't even celebrate pi day...

and even worse, this year's pi day is special because this year is a very special number...

8 replies
alcumusftwgrind
Yesterday at 2:11 AM
maxamc
Yesterday at 10:13 AM
rows are DERANGED and a SOCOURGE to usajmo .
GrantStar   26
N Yesterday at 6:00 AM by joshualiu315
Source: USAJMO 2024/4
Let $n \geq 3$ be an integer. Rowan and Colin play a game on an $n \times n$ grid of squares, where each square is colored either red or blue. Rowan is allowed to permute the rows of the grid and Colin is allowed to permute the columns. A grid coloring is orderly if: [list] [*]no matter how Rowan permutes the rows of the coloring, Colin can then permute the columns to restore the original grid coloring; and [*]no matter how Colin permutes the columns of the coloring, Rowan can then permute the rows to restore the original grid coloring. [/list] In terms of $n$, how many orderly colorings are there?

Proposed by Alec Sun
26 replies
GrantStar
Mar 21, 2024
joshualiu315
Yesterday at 6:00 AM
Geo equals ABsurdly proBEMatic
ihatemath123   73
N Yesterday at 5:38 AM by joshualiu315
Source: 2024 USAMO Problem 5, JMO Problem 6
Point $D$ is selected inside acute $\triangle ABC$ so that $\angle DAC = \angle ACB$ and $\angle BDC = 90^{\circ} + \angle BAC$. Point $E$ is chosen on ray $BD$ so that $AE = EC$. Let $M$ be the midpoint of $BC$.

Show that line $AB$ is tangent to the circumcircle of triangle $BEM$.

Proposed by Anton Trygub
73 replies
ihatemath123
Mar 21, 2024
joshualiu315
Yesterday at 5:38 AM
average FE
KevinYang2.71   74
N Yesterday at 4:55 AM by joshualiu315
Source: USAJMO 2024/5
Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ that satisfy
\[
f(x^2-y)+2yf(x)=f(f(x))+f(y)
\]for all $x,y\in\mathbb{R}$.

Proposed by Carl Schildkraut
74 replies
KevinYang2.71
Mar 21, 2024
joshualiu315
Yesterday at 4:55 AM
IMO ShortList 2002, geometry problem 2
orl   26
N Feb 8, 2025 by HamstPan38825
Source: IMO ShortList 2002, geometry problem 2
Let $ABC$ be a triangle for which there exists an interior point $F$ such that $\angle AFB=\angle BFC=\angle CFA$. Let the lines $BF$ and $CF$ meet the sides $AC$ and $AB$ at $D$ and $E$ respectively. Prove that \[ AB+AC\geq4DE. \]
26 replies
orl
Sep 28, 2004
HamstPan38825
Feb 8, 2025
IMO ShortList 2002, geometry problem 2
G H J
Source: IMO ShortList 2002, geometry problem 2
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orl
3647 posts
#1 • 6 Y
Y by Mathcollege, Adventure10, donotoven, GeckoProd, Mango247, cubres
Let $ABC$ be a triangle for which there exists an interior point $F$ such that $\angle AFB=\angle BFC=\angle CFA$. Let the lines $BF$ and $CF$ meet the sides $AC$ and $AB$ at $D$ and $E$ respectively. Prove that \[ AB+AC\geq4DE. \]
Attachments:
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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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grobber
7849 posts
#3 • 3 Y
Y by Adventure10, Mango247, and 1 other user
Sorry if it's a bit murky. I remember posting this one too, and giving a solution I was proud of (can't remember if I actually posted it). However, I can't find that solution.. [Moderator edit: This was at http://www.mathlinks.ro/Forum/viewtopic.php?t=220 .]

Let $D',E'$ be the images of $D,E$ through the homothety of center $F$ and ratio $4$. We have to show that $D'E'\le AB+AC$, so it would be enough to show $AE'+AD'\le AB+AC$. Again, we notice that it's enough to show $AD'\le AC\ (*)$. Let $X$ be the vertex of the equilateral triangle $CAX$, lying on the opposite side of $CA$ as $B$. Clearly, $AX=AC$, so $(*)$ is equivalent to $FD'\le FX=FA+FC$ (the last equality is well-known, and it follows from Ptolemy's equality applied to the cyclic quadrilateral $AFCX$) or, in other words, $4FD\le FA+FC$. In terms of areas, this means $4S(FAC)\le S(AFCX)\iff 3S(FAC)\le S(XAC)$, and this is clear since for fixed $XAC$, the area $FAC$ reaches its maximum when $FA=FC$, and in this case we have equality in the above inequality.

I think this pretty much ends the proof: we have shown that $4FD\le FX$, which is, as we have shown, equivalent to the initial problem.
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pohoatza
1145 posts
#4 • 10 Y
Y by huricane, alifenix-, Adventure10, Mango247, and 6 other users
I saw this problem these days and I was pretty sure it was an ISL problem.

Lets take the equilateral triangles $ ACP$ and $ ABQ$ on the exterior of the triangle $ ABC$.

We have that $ \angle{APC} + \angle{AFC} = 180$, therefore the points $ A,P,F,C$ are concyclic.

But $ \angle{AFP} = \angle{ACP} = 60 = \angle{AFD}$, so $ D \in (FP)$.
Analoguosly we have that $ E \in (FQ)$.

Now observe that $ \frac {FP}{FD} = 1 + \frac {DP}{FD} = 1 + \frac {[APC]}{[AFC]}\geq 4$, and the equality occurs when $ F$ is the midpoint of $ \widehat{AC}$.

Therefore $ FD \leq \frac {1}{4}FP$, and $ FE \leq \frac {1}{4}FQ$.

So, by taking it metrical, we have that:
$ DE = \sqrt {FD^{2} + FE^{2} + FD \cdot FE}\leq \frac {1}{4}\cdot \sqrt {FP^{2} + FQ^{2} + FP \cdot FQ} = \frac {1}{4}PQ$

But $ PQ \leq AP + AQ = AB + AC$, and thus the problem is solved.
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sayantanchakraborty
505 posts
#6 • 2 Y
Y by ali.agh, Adventure10
This post was also a spam and as I am unable to delete this post,i am writing the proof of $\frac{[APC]}{[AFC]} \ge 3$.

Note that
$(AF-CF)^2 \ge 0 \Rightarrow AF^2+CF^2+AF*CF \ge 3AF*CF \Rightarrow AC^2 \ge 3AF*CF \Rightarrow AP*CP\sin60^{\circ} \ge 3AF*CF\sin120^{\circ} \Rightarrow \frac{[APC]}{[AFC]} \ge 3$.
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AnonymousBunny
339 posts
#7 • 2 Y
Y by Adventure10, Mango247
This is a really nice problem! Thanks to sayantanchakraborty for giving some crucial hints leading to the following solution.

Since $\angle BFC, \angle CFA, \angle AFB$ are all equal and sum up to $360^{\circ},$ they must each be equal to $120^{\circ}.$ Construct a point $B'$ outside $\triangle ABC$ such that $\triangle ABB'$ is equilateral. Define point $C'$ analogously. Since $\angle AB'B + \angle AFB = 60^{\circ} + 120^{\circ} = 180^{\circ},$ points $A,B',B,P$ are concyclic. Furthermore, since $\angle B'FB = \angle B'AB = 60^{\circ} = 180^{\circ} - \angle BFC,$ points $C,F,D,B'$ are collinear.

I claim that $FB' \geq 4FD.$ This is equivalent to
\begin{align*}
[\triangle AB'C] & \geq 3[\triangle APC] \\
\iff AB' \cdot B'C \cdot  \sin (60^{\circ}) & \geq 3 \cdot AF \cdot CF \cdot \sin (120^{\circ}) \\
\iff AB' \cdot B'C & \geq 3 \cdot AF \cdot CF .\end{align*}
By cosine rule on $\triangle AB'C,$
\begin{align*}
AC^2 & = AB'^2 + B'C^2 - 2 \cdot AB' \cdot B'C \cdot \cos (60^{\circ})  \\
& = AB'^2 + B'C^2 - AB' \cdot B'C \\
& \geq AB' \cdot B'C , \end{align*}
where we have used the trivial inequality $AB'^2 + B'C^2 \geq 2 \cdot AB' \cdot B'C.$ Hence, it suffices to show that
\begin{align*}
AC^2 & \geq 3 \cdot AF \cdot CF \\
\iff AF^2 + CF^2 - 2 \cdot AF \cdot CF \cos (120^{\circ}) & \geq 3 \cdot AF \cdot CF \\ 
\iff AF^2 + CF^2 & \geq 2 \cdot AF \cdot CF,\end{align*}
which is true. Similar arguments show that $FC' \geq 4FE.$

The rest is obvious. Both the dilations centered at $F$ which map to $B$ to $B'$ and $C$ to $C'$ have ratio at least 4, so $B'C' \geq 4DE.$ By triangle inequality, we have that
\[AB'+A'C \geq B'C' \implies AB+AC \geq 4DE. \quad \blacksquare\]

For equality to hold, we need $AF=BF=CF,$ that is, the Fermat point must be the circumcenter of $\triangle ABC.$ This is possible iff $\triangle ABC$ is equilateral, because $ \angle AFB = 2 \angle ACB \implies \angle ACB = 60^{\circ}$ and similarly $\angle ABC= 60^{\circ}.$
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PRO2000
239 posts
#8 • 1 Y
Y by Adventure10
Erect equilateral triangles $AMC$ and $ANC$ outwardly on the sides of $\triangle ABC$. It is well known that $F \in BM$ and $D \in CN$.

$\blacksquare\boxed{\text{Lemma 1}}$ $\frac{1}{FD}=\frac{1}{FA}+\frac{1}{FC}$ and $\frac{1}{FE}=\frac{1}{FA}+\frac{1}{FB}$ .
Proof
Taking $\angle FAC= \alpha$ and $\angle FCA= \beta$ and using $\alpha + \beta =60$ ,$$\frac{FD}{FA}+\frac{FD}{FC}=\frac{sin(\alpha)}{sin(60+\beta)}+\frac{sin(\beta)}{sin(60+\beta)}=1$$and other part is analogously proved.

$\blacksquare\boxed{\text{Lemma 2}}$ $FA+FC \geq 4FD$ and $FA+FB \geq 4FE$
Proof
By lemma 1 , $\frac{FA+FC}{FD}= \left(\frac{1}{FA} + \frac{1}{FC} \right) \cdot ( FA+FC ) \geq 4 \implies FA+FC \geq 4FD$
The other part follows analogously.
Using lemma 2 $$ AB+AC =
 AN+AM 
\geq MN = \sqrt{FN^2+FM^2+FN \cdot FM }
= \sqrt{ (FA+FB)^2+(FA+FC)^2+(FA+FB)\cdot(FA+FC)} 
\geq 4 \cdot \sqrt{FC^2+FD^2+FC \cdot FD}=4DE$$
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mcdonalds106_7
1138 posts
#9 • 2 Y
Y by Adventure10, Mango247
Construct equilateral triangles $ACX$ and $ABY$ outside of $ABC$, so it's well known that $BFDX$ and $CFEY$ are lines. $YAFB$ is cyclic, so consider the tangent at the point $T$, the antipode of $Y$, labeled as line $\ell$. Note that $d(Y,AB):d(Y,\ell)=3:4$, so then $\dfrac{FE}{FY}\le \dfrac 14$ with equality only when $F=T$, and similarly $\dfrac{FD}{FX}\le \dfrac 14$. Let $M$ and $N$ be the points on segments $FY$ and $FX$, respectively, such that $\dfrac{FM}{FY}=\dfrac 14$ and $\dfrac{FN}{FX}=\dfrac 14$. Then since $FE\le FM$ and $FD\le FN$, $DE=\sqrt{FD^2+FE^2+FD\cdot FE}\le \sqrt{FN^2+FM^2+FN\cdot FM}=MN=\dfrac{XY}{4}\le \dfrac{AB+AC}{4}$, as desired.
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bobthesmartypants
4337 posts
#10 • 3 Y
Y by Tsikaloudakis, Adventure10, Mango247
cute

solution
This post has been edited 1 time. Last edited by bobthesmartypants, Apr 4, 2017, 8:08 PM
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Wizard_32
1566 posts
#11 • 2 Y
Y by Adventure10, Mango247
Trigonometry is the best weapon.
orl wrote:
Let $ABC$ be a triangle for which there exists an interior point $F$ such that $\angle AFB=\angle BFC=\angle CFA$. Let the lines $BF$ and $CF$ meet the sides $AC$ and $AB$ at $D$ and $E$ respectively. Prove that \[ AB+AC\geq4DE. \]
Clearly $\angle AFB=\angle BFC=\angle CFA=120^o.$ Now, erect equilateral triangles $ABC', BCA', CAB'$ on the sides, externally. Then $AFBC'. AFCB'$ are cyclic. Hence, $\angle C'FA+\angle AFC=\angle C'BA+\angle AFC=60^o+120^o=180^o,$ and so $C, F, C'$ are collinear. We get two more symmetric results and so $F$ is teh Fermat point given by $AA' \cap BB' \cap CC'.$

Claim: $FE: FC' \le 1:4.$
Proof: Ptolemy yields $FC'=FA+FB.$ Hence, it suffices to show
$$FA+FB \ge 4FE$$Let $\angle FAB=x.$ Then it suffices to show
$$\frac{FA}{FB}+1 \ge \frac{4FE}{FB} \Leftrightarrow \frac{sin(60^o-x)}{sinx}+1 \ge \frac{4sin(60^o-x)}{sin(60^o+x)}$$$$\Leftrightarrow \left(cosx-\sqrt{3}sinx \right)^2 \ge 0$$which is true. $\square$

Similarly we get $FD:FB' \le 1:4$ and so we get $4DE \le B'C' \le AC'+AB'=AB+AC,$ as desired. $\blacksquare$
This post has been edited 2 times. Last edited by Wizard_32, Oct 30, 2018, 11:37 AM
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mihaig
7339 posts
#12 • 2 Y
Y by Adventure10, Mango247
orl wrote:
Let $ABC$ be a triangle for which there exists an interior point $F$ such that $\angle AFB=\angle BFC=\angle CFA$. Let the lines $BF$ and $CF$ meet the sides $AC$ and $AB$ at $D$ and $E$ respectively. Prove that \[ AB+AC\geq4DE. \]

The problem is a masterpiece.
This post has been edited 1 time. Last edited by mihaig, Aug 8, 2019, 12:10 PM
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alifenix-
1547 posts
#13 • 3 Y
Y by v4913, Adventure10, Mango247
Solution (bash bash bash)
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Spacesam
597 posts
#14
Y by
Construct points $X, Y, Z$ forming equilateral triangles $\triangle BCX$, $\triangle CAY$, $\triangle ABZ$ sticking out from the triangle. Evidently, $F$ is the intersection of the three circumcircles of these equilateral triangles.

Observe additionally that $F \in \overline{AX}$, and in particular $F$ is the concurrence point of $\overline{AX}$, $\overline{BY}$, and $\overline{CZ}$. Note now that $\angle DFE = \angle BFC = 120^\circ$.

Thus, we can calculate \begin{align*}
    DE^2 = DF^2 + FE^2 - 2 \cdot DF \cdot FE \cdot \cos{120^\circ} = DF^2 + FE^2 + DF \cdot FE.
\end{align*}As $F$ varies along $(ABZ)$ with length $AB$ fixed, note that the maximum length of $FE$ occurs when $\overline{FZ} \perp \overline{AB}$, and this is also the case for the minimum length of $FZ$. Thus $\frac{FE}{FZ} \leq \frac{1}{4}$.

As a result, we know \begin{align*}
    DE^2 &= DF^2 + FE^2 + DF \cdot FE \\
    &\leq \frac{1}{16} (FZ^2 + FY^2 + FZ \cdot FY) \\
    &= \frac{1}{16} YZ^2 \\
    &\leq \frac{1}{16} (AZ + AY)^2 \\
    &= \frac{1}{16} (AB + AC)^2,
\end{align*}as desired.
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TheUltimate123
1739 posts
#15 • 3 Y
Y by Eyed, DrYouKnowWho, BorivojeGuzic123
Solved with Alex Zhao, Elliott Liu, Connie Jiang, Groovy (\help), Jeffrey Chen, Nicole Shen, and Raymond Feng.

Externally construct equilateral triangles \(ACY\) and \(ABZ\), so that \(B\), \(F\), \(D\), \(Y\) are collinear and \(C\), \(F\), \(E\), \(Z\) are collinear.

[asy]         size(7cm); defaultpen(fontsize(10pt));         pair A,B,C,Y,Z,F,D,EE;         A=dir(110);         B=dir(220);         C=dir(320);         Y=A+(C-A)*dir(60);         Z=B+(A-B)*dir(60);         F=extension(B,Y,C,Z);         D=extension(B,Y,A,C);         EE=extension(C,Z,A,B);

draw(D--EE);         draw(B--Y,gray);         draw(C--Z,gray);         draw(circumcircle(A,F,C),linewidth(.3));         draw(circumcircle(A,F,B),linewidth(.3));         draw(C--Y--A--Z--B);         draw(A--B--C--cycle,linewidth(.7));

dot("\(A\)",A,dir(105));         dot("\(B\)",B,S);         dot("\(C\)",C,S);         dot("\(F\)",F,dir(265));         dot("\(Y\)",Y,dir(30));         dot("\(Z\)",Z,dir(150));         dot("\(D\)",D,dir(-5));         dot("\(E\)",EE,dir(210));     [/asy]

Observe that \(FY/FD\ge4\) and \(FZ/FE\ge4\). It follows that \begin{align*}     AB+AC=AY+AZ\ge YZ&=\sqrt{FY^2+FZ^2+FY\cdot FZ}\\     &\ge4\sqrt{FD^2+FE^2+FD\cdot FE}=4DE, \end{align*}as needed.
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mihaig
7339 posts
#16
Y by
orl wrote:
Let $ABC$ be a triangle for which there exists an interior point $F$ such that $\angle AFB=\angle BFC=\angle CFA$. Let the lines $BF$ and $CF$ meet the sides $AC$ and $AB$ at $D$ and $E$ respectively. Prove that \[ AB+AC\geq4DE. \]

See also here https://artofproblemsolving.com/community/c6t243f6h2624066_a_refinement_of_imo_shl_2002
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bluelinfish
1445 posts
#17
Y by
First ISL solution in a while. This is the type of problem where if you don't know this property of the Fermat point it's hard to solve (I certainly couldn't do it) and if you know it it's very quick (after I got a hint with the property I solved it within fifteen minutes).

It is well-known that $F$ is the Toricelli/1st Fermat point of $\triangle ABC$. It is a well-known property of $F$ that if $ABG$ and $ACH$ are equilateral triangles erected outward from $AB$ and $AC$, respectively, $C,F,G$ are collinear and $AGBF$ is cyclic (similarly $B,F,H$ are collinear and $AHCF$ is cyclic).

Notice that as $F$ is on minor arc $AB$, the minimum possible value of $\frac{GE}{EF}$ occurs when $F$ is on the midpoint of the arc, as this maximizes $EF$ and minimizes $EG$. In that case, it is easy to show that $\frac{EG}{EF}=3$ and thus $\frac{FG}{FE}=4$, hence it must be true that $\frac{FG}{FE}\ge 4$ and similarly $\frac{FH}{FD}\ge 4$.

Let $FG=\alpha FE$ and $FH=\beta FD$, where $\alpha, \beta \ge 4$. Since $\angle EFD = 120^{\circ}$, by LoC on $\triangle FED$ we get $$ED=\sqrt{FE^2+FD^2-2FE\cdot FD \cos{120^{\circ}}}=\sqrt{FE^2+FD^2+FE\cdot FD}.$$Using LoC on $\triangle FGH$, we get

\begin{align*} 
HG &= \sqrt{FG^2+FH^2-2FG\cdot FH\cos{120^{\circ}}} \\ &= \sqrt{FG^2+FH^2+FG\cdot FH} \\ &= \sqrt{(\alpha FE)^2+(\beta FD)^2+(\alpha FE)(\beta FD)} \\ &= \sqrt{\alpha^2 FE^2+\beta^2 FD^2+\alpha\beta FE\cdot FD} \\ & \ge \sqrt{16FE^2+16FD^2+16FE\cdot FD} \\ &= 4\sqrt{FE^2+FD^2+FE\cdot FD} \\ &= 4ED.
\end{align*}
By the Triangle Inequality, $AG+AH\ge GH \ge 4ED$, finishing the problem.
Attachments:
This post has been edited 1 time. Last edited by bluelinfish, Oct 16, 2021, 2:11 AM
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L567
1184 posts
#18 • 1 Y
Y by proxima1681
Let $B', C'$ be such that $ACB', ABC'$ are equilateral. We have that $C,F,C'$ and $B,F,B'$ are collinear.

Claim: $EC' \ge 3EF$

Proof: Note that $\frac{FE}{C'E} = \frac{AF}{AC'}  \frac{\sin \angle FAB}{\sin \angle C'AB} = \frac{AF}{c} \frac{2\sin \angle FAB}{\sqrt{3}}$. Let $AF = x$, $BF = y$. Note that $R$, the circumradius of $(AFBC')$, is equal to $\frac{c}{\sqrt{3}}$.

We have $2R = \frac{y}{\sin \angle FAB} \implies \sin \angle FAB = \frac{y}{2R} = \frac{\sqrt{3}y}{2c}$.

So $\frac{FE}{C'E} = \frac{x}{c} \frac{y}{c} = \frac{xy}{c^2}$.

Observe that $c^2 = x^2 + y^2 + xy \ge 3xy \implies \frac{xy}{c^2} \le \frac{1}{3}$.

So we have $\frac{FE}{C'E} \le \frac{1}{3} \implies C'E \ge 3FE$, as claimed. $\square$.

From the claim, we have $DE \le \frac{B'C'}{4} \le \frac{AB' + AC'}{4} = \frac{AB+AC}{4} \implies AB + AC \ge 4DE$, as desired. $\blacksquare$
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Mahdi_Mashayekhi
689 posts
#19
Y by
Note that ∠AFB = ∠BFC = ∠CFA = 120 so making regular triangles with bases AB and AC is a good move.
Let S and K be outside ABC such that ABS and ACK are regular triangles. Note that AFBS and AFCK are cyclic. Let O1,O2 be reflections of F across AB and AC. FE/ES = [AFB]/[ABS] = [AO1B]/[ASB] ≤ 1/3 so FS ≥ 4FE. Same way we can prove FK ≥ 4FD. so SK ≥ 4DE and SK ≤ AS + AK = AB + AC.
we're Done.
This post has been edited 1 time. Last edited by Mahdi_Mashayekhi, Jan 10, 2022, 7:05 AM
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mihaig
7339 posts
#20
Y by
Try the refinement
https://artofproblemsolving.com/community/c6t243f6h2624066_a_refinement_of_imo_shl_2002
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awesomeming327.
1664 posts
#21
Y by
What.

https://media.discordapp.net/attachments/925784397469331477/952399059321245766/Screen_Shot_2022-03-12_at_7.53.43_PM.png?width=864&height=1170

Let $G$ be on $FD$ extended such that $\angle AGC=60^\circ.$ Let $H$ be on $FE$ extended such that $\angle AHB=60^\circ.$ Note that $AFCG$ is cyclic. Also, $\angle AFD=60^\circ$ and $\angle CFD=60^\circ$ so $\angle CAG=\angle ACG=60^\circ.$ Thus, $ACG$ is equilateral. Similarly, $AHB$ is equilateral. Now, $AB+AC\ge HG.$ Since $\angle HFG$ is obtuse, it suffices to show $HF\ge 4EF$ and $GF\ge 4DF$ to prove that $HG\ge 4ED.$

Note that $\triangle HFA\sim \triangle HAE$ by AA so $\frac{HE}{HF}=\left(\frac{HE}{HA}\right)^2\ge \left(\frac{\sqrt{3}}{2}\right)\ge \frac{3}{4},$ which implies the result that $HF\ge 4EF$. Similarly, $GD\ge 4DF.$ Now, WLOG suppose the parallel line through $E$ parallel to $HG$ lies outside of $\triangle EDF.$ Then this line intersects $FG$ at $J.$ $HG\ge EJ\ge ED$ as desired.
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asdf334
7577 posts
#22
Y by
Construct equilateral triangles $\triangle ABX$ and $\triangle ACY$ outside of $\triangle ABC$ and note that $AXBF$ and $AYCF$ are cyclic. It's easy to see that $FX\ge 4FE$ and $FY\ge 4FD$ so by the Law of Cosines we easily obtain $XY\ge 4DE$ so that
\[AB+AC=AX+AY\ge XY\ge 4DE.\]We are done. $\blacksquare$
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anantmudgal09
1979 posts
#23 • 1 Y
Y by Mango247
Really cute :)
orl wrote:
Let $ABC$ be a triangle for which there exists an interior point $F$ such that $\angle AFB=\angle BFC=\angle CFA$. Let the lines $BF$ and $CF$ meet the sides $AC$ and $AB$ at $D$ and $E$ respectively. Prove that \[ AB+AC\geq4DE. \]

Draw points $K, L$ such that $AKB$ and $ALC$ are equilateral triangles. Clearly, $AFCL, AFBK$ are cyclic quads, and $\angle AFL=\angle ACL=180^{\circ}-\angle AFB=60^{\circ}$ implies $B, F, D, L$ are collinear. Similarly, $C, F, E,$ and $K,$ are collinear. Now $AB+AC=AK+AL \ge KL$ so it suffices to show that $DE \leq \tfrac{1}{4} KL$.

We will show that $FE \leq \tfrac{1}{4}FK$ and $FD \leq \tfrac{1}{4}FL$. It suffices to prove the following two lemmas to finish:

Lemma 1. Point $W$ lies on arc $\widehat{YZ}$ of the circumcircle of equilateral triangle $XYZ$ not containing $X$ and line $XW$ meets $YZ$ at point $T$. Then $WX \geq 4WT$.

Proof: Indeed, it is enough to show $XT \ge 3WT$. Now $XT$ is larger than the $X$-median of $\triangle XYZ$ and $WT$ is smaller than the length it achieves when $W$ is antipodal to $X$. For rigour, this follows as $XW \cdot XT$ is fixed by shooting lemma. When $W$ is antipodal, equality is achieved, proving the lemma.

Lemma 2. In obtuse triangle $XYZ$ with obtuse angle at $X$, points $Y_1, Z_1$ lie on rays $XY, XZ$ such that $XY \geq 4XY_1$ and $XZ \geq 4XZ_1$. Then $YZ \geq 4Y_1Z_1$.

Proof: Scale by a factor of $4$ to assume $XY_1 \leq XY$ and $XZ_1 \le XZ$. Now $Y_1Z_1<Y_1Z$ as $\angle Y_1Z_1Z>\angle Y_1XZ>90^{\circ}$ and $Y_1Z<YZ$ as $\angle YY_1Z>\angle YXZ>90^{\circ}$, so $Y_1Z_1<YZ$ unless $Y_1=Y$ and $Z_1=Z$, proving the claim.

Finally, by combining Lemma 1 and Lemma 2 in triangle $FKL$ for points $D$ and $E$, the conclusion follows.
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pikapika007
294 posts
#24
Y by
r poblem

Construct equilateral triangles $ABX$ and $ACY$ so that both are not in the of $ABC$. Then it is well known that $A$, $F$, $X$ and $B$, $F$, $Y$ are collinear, and moreover $AXBF$, $AYCF$ are cyclic. Now we can obtain $FX\ge 4FE$, $FY\ge 4FD$ and hence by LOC $XY \ge 4DE$. To finish,
\[AB+AC=AX+AY\ge XY\ge 4DE\]as desired. $\square$
This post has been edited 1 time. Last edited by pikapika007, Jul 18, 2023, 4:13 AM
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lian_the_noob12
173 posts
#25
Y by
Point $F$ is $\textbf{First Fermat Point}$ and construction can easily be found from the theorem thonk:/
This post has been edited 3 times. Last edited by lian_the_noob12, Dec 12, 2023, 5:31 PM
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dudade
136 posts
#26
Y by
Note $F$ is the Fermat Point. Thus, let $X$ and $Y$ be points such that $\triangle ABX$ and $\triangle ACY$ are equilateral triangles lying outside $\triangle ABC$.

Claim. $FX \geq 4 \cdot FE$ and $FY \geq 4 \cdot FD$.
Proof. We will prove this with area ratios. Note, $AB^2 = AF^2 + FB^2 + AF \cdot FB$, by Law of Cosines.
\begin{align*}
\dfrac{[AXB]}{[AFB]} = \dfrac{\tfrac{\sqrt{3}}{4} \cdot AB^2}{\tfrac{1}{2} \cdot AF \cdot FB \cdot \sin\left(120^{\circ}\right)} = \dfrac{AF^2 + FB^2 + AF \cdot FB}{AF \cdot FB} = \dfrac{AF}{FB} + \dfrac{FB}{AF} + 1 \geq 3.
\end{align*}Thus, $[AXB] \geq 3 \cdot [AFB]$, thus $XE \geq 3 \cdot EF$ and $FX \geq 4 \cdot FE$. Then, $FY \geq 4 \cdot FD$ follows, as desired. $\blacksquare$

Note that by triangle inequality this clearly implies $XY \geq 4 \cdot DE$. But, $AB + AC = AX + AY \geq XY$, by triangle inequality. Therefore $AB + BC \geq 4 \cdot DE$, as desired.
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EpicBird08
1730 posts
#27
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Clearly $\angle AFB = \angle BFC = \angle CFA = 120^\circ.$ Erect equilateral triangles $\triangle ACP$ and $\triangle ABQ$ outside of $\triangle ABC.$
Let $AF = x$ and $FB = y.$ Observe that $AFBQ$ is cyclic as $\angle AQB + \angle AFB = 60^\circ + 120^\circ = 180^\circ.$ Thus by Ptolemy on $AFBQ,$ we get $FQ = FA + FB = x + y.$ Since $\triangle FAE \sim \triangle FQB$ (by simple angle-chasing), we get $FE \cdot FQ = FA \cdot FB,$ so $FE = \frac{FA \cdot FB}{FA + FB} = \frac{xy}{x+y}.$ Therefore, $$\frac{FE}{FQ} = \frac{xy}{(x+y)^2} \le \frac{xy}{4xy} = \frac{1}{4}$$by AM-GM on the denominator. Similarly, $\frac{FD}{FP} \le \frac{1}{4}.$ Therefore, $$AB + AC = AQ + AP \ge QP \ge 4DE,$$as desired.
This post has been edited 1 time. Last edited by EpicBird08, Nov 29, 2024, 7:14 AM
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HamstPan38825
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By similar triangles and angle bisector theorem, we may compute \[EF = AE \cdot \frac{BF}{AB} = AB \cdot \frac{AF}{AF+BF} \cdot \frac{BF}{AB} = \frac{AF \cdot BF}{AF+BF}.\]Now let $a = AF$, $b = BF$, and $c = CF$, and observe that $EF = \frac{ab}{a+b} \leq \frac{a+b}4$ while $FD \leq \frac{a+c}4$. From here, it is very much feasible to directly expand $(AB+AC)^2 \geq 16 DE^2$ using Law of Cosines, but here is a comparatively nicer finish.

Erect equilateral triangles $BCX$, $ACY$, and $ABZ$ outside triangle $ABC$ such that $F = \overline{AX} \cap \overline{BY} \cap \overline{CZ}$, and note that $EF \leq \frac 14 FZ$, et cetera. So \[DE^2 = EF^2+DF^2 + DE \cdot EF \leq \frac{FZ^2+FY^2 + FZ \cdot FY}{16} = \frac{ZY^2}{16} \leq \frac{(AB+AC)^2}{16}\]as needed.

Remark: For some reason, this felt quite hard for G2.
This post has been edited 1 time. Last edited by HamstPan38825, Feb 8, 2025, 10:49 PM
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