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k a March Highlights and 2025 AoPS Online Class Information
jlacosta   0
Mar 2, 2025
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0 replies
jlacosta
Mar 2, 2025
0 replies
Inspired by BaCaPhe
sqing   2
N 4 minutes ago by lbh_qys
Source: Own
Let $ a,b,c \ge 0 $ and $ ab + bc + ca \ge 4 + abc. $ Prove that
$$  a^2  +  b^2  +  c^2-abc  \ge 4$$$$  a^2  +  b^2  +  c^2  \ge 8$$$$a^2  +  b^2  +  c^2+a^2b^2c^2\ge 8$$$$  a^2  +  b^2  +  c^2+ab+bc+ca  \ge 12$$
2 replies
1 viewing
sqing
4 hours ago
lbh_qys
4 minutes ago
Twin Prime Diophantine
awesomeming327.   18
N 5 minutes ago by EthanWYX2009
Source: CMO 2025
Determine all positive integers $a$, $b$, $c$, $p$, where $p$ and $p+2$ are odd primes and
\[2^ap^b=(p+2)^c-1.\]
18 replies
awesomeming327.
Mar 7, 2025
EthanWYX2009
5 minutes ago
Sharygin 2025 CR P12
Gengar_in_Galar   7
N 9 minutes ago by maths_enthusiast_0001
Source: Sharygin 2025
Circles $\omega_{1}$ and $\omega_{2}$ are given. Let $M$ be the midpoint of the segment joining their centers, $X$, $Y$ be arbitrary points on $\omega_{1}$, $\omega_{2}$ respectively such that $MX=MY$. Find the locus of the midpoints of segments $XY$.
Proposed by: L Shatunov
7 replies
+1 w
Gengar_in_Galar
Mar 10, 2025
maths_enthusiast_0001
9 minutes ago
Problem 1
randomusername   71
N 20 minutes ago by MihaiT
Source: IMO 2015, Problem 1
We say that a finite set $\mathcal{S}$ of points in the plane is balanced if, for any two different points $A$ and $B$ in $\mathcal{S}$, there is a point $C$ in $\mathcal{S}$ such that $AC=BC$. We say that $\mathcal{S}$ is centre-free if for any three different points $A$, $B$ and $C$ in $\mathcal{S}$, there is no points $P$ in $\mathcal{S}$ such that $PA=PB=PC$.

(a) Show that for all integers $n\ge 3$, there exists a balanced set consisting of $n$ points.

(b) Determine all integers $n\ge 3$ for which there exists a balanced centre-free set consisting of $n$ points.

Proposed by Netherlands
71 replies
randomusername
Jul 10, 2015
MihaiT
20 minutes ago
No more topics!
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:
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
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.
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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
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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.
1665 posts
#21
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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
7584 posts
#22
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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
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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
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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
137 posts
#26
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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
1732 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
8853 posts
#28
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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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