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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.

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[*]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 Adding contests to the Contest Collections
dcouchman   1
N Apr 5, 2023 by v_Enhance
Want to help AoPS remain a valuable Olympiad resource? Help us add contests to AoPS's Contest Collections.

Find instructions and a list of contests to add here: https://artofproblemsolving.com/community/c40244h1064480_contests_to_add
1 reply
dcouchman
Sep 9, 2019
v_Enhance
Apr 5, 2023
k i Zero tolerance
ZetaX   49
N May 4, 2019 by NoDealsHere
Source: Use your common sense! (enough is enough)
Some users don't want to learn, some other simply ignore advises.
But please follow the following guideline:


To make it short: ALWAYS USE YOUR COMMON SENSE IF POSTING!
If you don't have common sense, don't post.


More specifically:

For new threads:


a) Good, meaningful title:
The title has to say what the problem is about in best way possible.
If that title occured already, it's definitely bad. And contest names aren't good either.
That's in fact a requirement for being able to search old problems.

Examples:
Bad titles:
- "Hard"/"Medium"/"Easy" (if you find it so cool how hard/easy it is, tell it in the post and use a title that tells us the problem)
- "Number Theory" (hey guy, guess why this forum's named that way¿ and is it the only such problem on earth¿)
- "Fibonacci" (there are millions of Fibonacci problems out there, all posted and named the same...)
- "Chinese TST 2003" (does this say anything about the problem¿)
Good titles:
- "On divisors of a³+2b³+4c³-6abc"
- "Number of solutions to x²+y²=6z²"
- "Fibonacci numbers are never squares"


b) Use search function:
Before posting a "new" problem spend at least two, better five, minutes to look if this problem was posted before. If it was, don't repost it. If you have anything important to say on topic, post it in one of the older threads.
If the thread is locked cause of this, use search function.

Update (by Amir Hossein). The best way to search for two keywords in AoPS is to input
[code]+"first keyword" +"second keyword"[/code]
so that any post containing both strings "first word" and "second form".


c) Good problem statement:
Some recent really bad post was:
[quote]$lim_{n\to 1}^{+\infty}\frac{1}{n}-lnn$[/quote]
It contains no question and no answer.
If you do this, too, you are on the best way to get your thread deleted. Write everything clearly, define where your variables come from (and define the "natural" numbers if used). Additionally read your post at least twice before submitting. After you sent it, read it again and use the Edit-Button if necessary to correct errors.


For answers to already existing threads:


d) Of any interest and with content:
Don't post things that are more trivial than completely obvious. For example, if the question is to solve $x^{3}+y^{3}=z^{3}$, do not answer with "$x=y=z=0$ is a solution" only. Either you post any kind of proof or at least something unexpected (like "$x=1337, y=481, z=42$ is the smallest solution). Someone that does not see that $x=y=z=0$ is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
Similar, posting "I have solved this problem" but not posting anything else is not welcome; it even looks that you just want to show off what a genius you are.

e) Well written and checked answers:
Like c) for new threads, check your solutions at least twice for mistakes. And after sending, read it again and use the Edit-Button if necessary to correct errors.



To repeat it: ALWAYS USE YOUR COMMON SENSE IF POSTING!


Everything definitely out of range of common sense will be locked or deleted (exept for new users having less than about 42 posts, they are newbies and need/get some time to learn).

The above rules will be applied from next monday (5. march of 2007).
Feel free to discuss on this here.
49 replies
1 viewing
ZetaX
Feb 27, 2007
NoDealsHere
May 4, 2019
Concurrency of two lines and a circumcircle
BR1F1SZ   1
N 5 minutes ago by MathLuis
Source: 2025 Francophone MO Juniors P3
Let $\triangle{ABC}$ be a triangle, $\omega$ its circumcircle and $O$ the center of $\omega$. Let $P$ be a point on the segment $BC$. We denote by $Q$ the second intersection point of the circumcircles of triangles $\triangle{AOB}$ and $\triangle{APC}$. Prove that the line $PQ$ and the tangent to $\omega$ at point $A$ intersect on the circumcircle of triangle $\triangle AOB$.
1 reply
BR1F1SZ
an hour ago
MathLuis
5 minutes ago
IMO Shortlist 2009 - Problem A2
April   93
N 10 minutes ago by ezpotd
Let $a$, $b$, $c$ be positive real numbers such that $\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} = a+b+c$. Prove that:
\[\frac{1}{(2a+b+c)^2}+\frac{1}{(a+2b+c)^2}+\frac{1}{(a+b+2c)^2}\leq \frac{3}{16}.\]
Proposed by Juhan Aru, Estonia
93 replies
April
Jul 5, 2010
ezpotd
10 minutes ago
Product of consecutive terms divisible by a prime number
BR1F1SZ   0
27 minutes ago
Source: 2025 Francophone MO Seniors P4
Determine all sequences of strictly positive integers $a_1, a_2, a_3, \ldots$ satisfying the following two conditions:
[list]
[*]There exists an integer $M > 0$ such that, for all indices $n \geqslant 1$, $0 < a_n \leqslant M$.
[*]For any prime number $p$ and for any index $n \geqslant 1$, the number
\[
a_n a_{n+1} \cdots a_{n+p-1} - a_{n+p}
\]is a multiple of $p$.
[/list]


0 replies
BR1F1SZ
27 minutes ago
0 replies
Fixed and variable points
BR1F1SZ   0
30 minutes ago
Source: 2025 Francophone MO Seniors P3
Let $\omega$ be a circle with center $O$. Let $B$ and $C$ be two fixed points on the circle $\omega$ and let $A$ be a variable point on $\omega$. We denote by $X$ the intersection point of lines $OB$ and $AC$, assuming $X \neq O$. Let $\gamma$ be the circumcircle of triangle $\triangle AOX$. Let $Y$ be the second intersection point of $\gamma$ with $\omega$. The tangent to $\gamma$ at $Y$ intersects $\omega$ at $I$. The line $OI$ intersects $\omega$ at $J$. The perpendicular bisector of segment $OY$ intersects line $YI$ at $T$, and line $AJ$ intersects $\gamma$ at $P$. We denote by $Z$ the second intersection point of the circumcircle of triangle $\triangle PYT$ with $\omega$. Prove that, as point $A$ varies, points $Y$ and $Z$ remain fixed.
0 replies
BR1F1SZ
30 minutes ago
0 replies
Use 3d paper
YaoAOPS   7
N 35 minutes ago by EGMO
Source: 2025 CTST p4
Recall that a plane divides $\mathbb{R}^3$ into two regions, two parallel planes divide it into three regions, and two intersecting planes divide space into four regions. Consider the six planes which the faces of the cube $ABCD-A_1B_1C_1D_1$ lie on, and the four planes that the tetrahedron $ACB_1D_1$ lie on. How many regions do these ten planes split the space into?
7 replies
YaoAOPS
Mar 6, 2025
EGMO
35 minutes ago
Cyclic ine
m4thbl3nd3r   2
N an hour ago by m4thbl3nd3r
Let $a,b,c>0$ such that $a^2+b^2+c^2=3$. Prove that $$\sum \frac{a^2}{b}+abc \ge 4$$
2 replies
m4thbl3nd3r
Yesterday at 3:34 PM
m4thbl3nd3r
an hour ago
Sequence inequality
BR1F1SZ   0
an hour ago
Source: 2025 Francophone MO Seniors P1
Let $a_1, a_2, a_3, \ldots$ be a sequence of positive integers satisfying the following property: for all positive integers $k < \ell$, for all distinct integers $m_1, m_2, \ldots, m_k$ and for all distinct integers $n_1, n_2, \ldots, n_\ell$,
\[
a_{m_1} + a_{m_2} + \cdots + a_{m_k} \leqslant a_{n_1} + a_{n_2} + \cdots + a_{n_\ell}.
\]Prove that there exist two integers $N$ and $b$ such that $a_n = b$ for all $n \geqslant N$.
0 replies
+1 w
BR1F1SZ
an hour ago
0 replies
GCD and LCM operations
BR1F1SZ   0
an hour ago
Source: 2025 Francophone MO Juniors P4
Charlotte writes the integers $1,2,3,\ldots,2025$ on the board. Charlotte has two operations available: the GCD operation and the LCM operation.
[list]
[*]The GCD operation consists of choosing two integers $a$ and $b$ written on the board, erasing them, and writing the integer $\operatorname{gcd}(a, b)$.
[*]The LCM operation consists of choosing two integers $a$ and $b$ written on the board, erasing them, and writing the integer $\operatorname{lcm}(a, b)$.
[/list]
An integer $N$ is called a winning number if there exists a sequence of operations such that, at the end, the only integer left on the board is $N$. Find all winning integers among $\{1,2,3,\ldots,2025\}$ and, for each of them, determine the minimum number of GCD operations Charlotte must use.

Note: The number $\operatorname{gcd}(a, b)$ denotes the greatest common divisor of $a$ and $b$, while the number $\operatorname{lcm}(a, b)$ denotes the least common multiple of $a$ and $b$.
0 replies
BR1F1SZ
an hour ago
0 replies
Balanced grids
BR1F1SZ   0
an hour ago
Source: 2025 Francophone MO Juniors/Seniors P2
Let $n \geqslant 2$ be an integer. We consider a square grid of size $2n \times 2n$ divided into $4n^2$ unit squares. The grid is called balanced if:
[list]
[*]Each cell contains a number equal to $-1$, $0$ or $1$.
[*]The absolute value of the sum of the numbers in the grid does not exceed $4n$.
[/list]
Determine, as a function of $n$, the smallest integer $k \geqslant 1$ such that any balanced grid always contains an $n \times n$ square whose absolute sum of the $n^2$ cells is less than or equal to $k$.
0 replies
BR1F1SZ
an hour ago
0 replies
Radiant sets
BR1F1SZ   0
an hour ago
Source: 2025 Francophone MO Juniors P1
A finite set $\mathcal S$ of distinct positive real numbers is called radiant if it satisfies the following property: if $a$ and $b$ are two distinct elements of $\mathcal S$, then $a^2 + b^2$ is also an element of $\mathcal S$.
[list=a]
[*]Does there exist a radiant set with a size greater than or equal to $4$?
[*]Determine all radiant sets of size $2$ or $3$.
[/list]
0 replies
BR1F1SZ
an hour ago
0 replies
Classic Diophantine
Adywastaken   4
N 2 hours ago by mrtheory
Source: NMTC 2024/6
Find all natural number solutions to $3^x-5^y=z^2$.
4 replies
Adywastaken
Yesterday at 3:39 PM
mrtheory
2 hours ago
Where are the Circles?
luminescent   43
N 3 hours ago by Amkan2022
Source: EGMO 2022/1
Let $ABC$ be an acute-angled triangle in which $BC<AB$ and $BC<CA$. Let point $P$ lie on segment $AB$ and point $Q$ lie on segment $AC$ such that $P \neq B$, $Q \neq C$ and $BQ = BC = CP$. Let $T$ be the circumcenter of triangle $APQ$, $H$ the orthocenter of triangle $ABC$, and $S$ the point of intersection of the lines $BQ$ and $CP$. Prove that $T$, $H$, and $S$ are collinear.
43 replies
luminescent
Apr 9, 2022
Amkan2022
3 hours ago
Divisibilty...
Sadigly   0
3 hours ago
Source: Azerbaijan Junior NMO 2025 P2
Find all $4$ consecutive even numbers, such that the square of their product divides the sum of their squares.
0 replies
Sadigly
3 hours ago
0 replies
Quadratic system
juckter   35
N 4 hours ago by shendrew7
Source: Mexico National Olympiad 2011 Problem 3
Let $n$ be a positive integer. Find all real solutions $(a_1, a_2, \dots, a_n)$ to the system:

\[a_1^2 + a_1 - 1 = a_2\]\[ a_2^2 + a_2 - 1 = a_3\]\[\hspace*{3.3em} \vdots \]\[a_{n}^2 + a_n - 1 = a_1\]
35 replies
juckter
Jun 22, 2014
shendrew7
4 hours ago
IMO ShortList 2002, geometry problem 2
orl   27
N Apr 10, 2025 by ZZzzyy
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 \[ AB+AC\geq4DE. \]
27 replies
orl
Sep 28, 2004
ZZzzyy
Apr 10, 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:
This post has been edited 2 times. Last edited by orl, Sep 27, 2005, 4:41 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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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
7361 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 • 4 Y
Y by v4913, Adventure10, Mango247, Alex-131
Solution (bash bash bash)
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Spacesam
596 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
1740 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
7361 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
1449 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
695 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
7361 posts
#20
Y by
Try the refinement
https://artofproblemsolving.com/community/c6t243f6h2624066_a_refinement_of_imo_shl_2002
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awesomeming327.
1717 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
7585 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
1980 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
298 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
139 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
1752 posts
#27
Y by
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
8863 posts
#28
Y by
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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ZZzzyy
2 posts
#29
Y by
Let $C'$ be the point outside of $\triangle ABC$, and $C'AB$ is equilateral, by simple angle chase, we have that $C, F, C'$ colinear, and $F$ lies on $(ABC')$. Define $B'$ similarly, notice by triangle inequality, $$B'C' \leq AB' + AC' = AB + AC$$Claim:
In a equilateral triangle $ABC$, the line passes through $C$ is $l$, let $E = \overline{AB} \cap l$, $F = (ABC) \cap l$, then $\frac {CF}{EF} \geq 4$, equality case holds iff $CF \perp AB$.
Proof:
Let $l'$ be the perpendicular to $AB$ from $C$, and $E' = \overline{AB} \cap l', F' = (ABC) \cap l'$. So $CF' = 2R$ is a diameter, and it is trivial that $F'E' = \frac 1 2 R$, so $\frac {CE'}{E'F'} = 3$ and $\frac {CF'}{E'F'} = 4$. Now we can see that $CE$ is the hypotenuse of rt$\triangle CEE'$, so $CE \geq CE'$, also we have $CEE' \sim CF'F$, thus $\frac {CE} {CE'} = \frac {CF'}{CF} = \frac{CE' + E'F'}{CE + EF} \geq 1$ $\Rightarrow \frac {CF' + CF + E'F'} {CF + CF' + EF} \geq 1$ $\Rightarrow CE' + CE + E'F' \geq CE' + CE + EF$ $\Rightarrow E'F' \geq EF$, so $\frac {CE}{EF} \geq \frac{CE'}{E'F'} = 3$, so $\frac{CE}{EF} + 1 = \frac{CE + EF} {EF} = \frac{CF}{EF} \geq 3 + 1 = 4$ as desired. The equality only holds when $F' = F, E' = E$, so when $CF \perp AB$.

Now back to the problem, by the claim, we have that $\frac{C'F}{EF}, \frac{B'F}{DF} \geq 4$, WLOG let $\frac{C'F}{EF} < \frac{B'F}{DF}$, let the parallel line to $DE$ passing through $C$ intersecting $B'D$ at $B''$, then it is trivial that $B''C \leq B'C'$, equality holds when $B'' = B'$. Now we can finish the problem:
$$ 4DE \leq \frac{C'F}{EF} \cdot DE = B''C' \leq B'C' \leq AB + AC$$as desired.
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