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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.

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0 replies
jlacosta
May 1, 2025
0 replies
Six variables
Nguyenhuyen_AG   1
N 21 minutes ago by TNKT
Let $a,\,b,\,c,\,x,\,y,\,z$ be six positive real numbers. Prove that
$$\frac{a}{b+c} \cdot \frac{y+z}{x} + \frac{b}{c+a} \cdot \frac{z+x}{y} + \frac{c}{a+b} \cdot \frac{x+y}{z} \geqslant 2+\sqrt{\frac{8abc}{(a+b)(b+c)(c+a)}}.$$
1 reply
Nguyenhuyen_AG
Today at 5:09 AM
TNKT
21 minutes ago
Anything real in this system must be integer
Assassino9931   3
N 26 minutes ago by Sardor_lil
Source: Al-Khwarizmi International Junior Olympiad 2025 P1
Determine the largest integer $c$ for which the following statement holds: there exists at least one triple $(x,y,z)$ of integers such that
\begin{align*} x^2 + 4(y + z) = y^2 + 4(z + x) = z^2 + 4(x + y) = c \end{align*}and all triples $(x,y,z)$ of real numbers, satisfying the equations, are such that $x,y,z$ are integers.

Marek Maruin, Slovakia
3 replies
Assassino9931
May 9, 2025
Sardor_lil
26 minutes ago
Interesting inequalities
sqing   3
N 35 minutes ago by sqing
Source: Own
Let $ a,b,c\geq 0 , (a+k )(b+c)=k+1.$ Prove that
$$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\geq  \frac{2k-3+2\sqrt{k+1}}{3k-1}$$Where $ k\geq \frac{2}{3}.$
Let $ a,b,c\geq 0 , (a+1)(b+c)=2.$ Prove that
$$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\geq 2\sqrt{2}-1$$Let $ a,b,c\geq 0 , (a+3)(b+c)=4.$ Prove that
$$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\geq  \frac{7}{4}$$Let $ a,b,c\geq 0 , (3a+2)(b+c)= 5.$ Prove that
$$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\geq  \frac{2(2\sqrt{15}-5)}{3}$$
3 replies
sqing
Yesterday at 1:29 PM
sqing
35 minutes ago
Interesting inequalities
sqing   1
N 40 minutes ago by sqing
Source: Own
Let $ a,b>0 $ and $ a+b\leq 1  $ . Prove that
$$\left(\frac{1}{a^3}-1\right)\left(\frac{1}{b^3}-1\right)-k\left(\frac{a}{b}+\frac{b}{a}\right) \geq 49-2k$$Where $24\geq k\in N^+.$
$$\left(\frac{1}{a^3}-1\right)\left(\frac{1}{b^3}-1\right) \geq 49$$$$\left(\frac{1}{a^3}-1\right)\left(\frac{1}{b^3}-1\right)-25\left(\frac{a}{b}+\frac{b}{a}\right) \geq -\frac{13}{12}$$$$\left(\frac{1}{a^3}-1\right)\left(\frac{1}{b^3}-1\right)-26\left(\frac{a}{b}+\frac{b}{a}\right) \geq -\frac{10}{3}$$$$\left(\frac{1}{a^3}-1\right)\left(\frac{1}{b^3}-1\right)-27\left(\frac{a}{b}+\frac{b}{a}\right) \geq -\frac{23}{4}$$
1 reply
sqing
an hour ago
sqing
40 minutes ago
Tetrahedron
4everwise   3
N Yesterday at 10:43 PM by aidan0626
Four balls of radius 1 are mutually tangent, three resting on the floor and the fourth resting on the others. A tetrahedron, each of whose edges have length $s$, is circumscribed around the balls. Then $s$ equals

$\text{(A)} \ 4\sqrt 2 \qquad \text{(B)} \ 4\sqrt 3 \qquad \text{(C)} \ 2\sqrt 6 \qquad \text{(D)} \ 1+2\sqrt 6 \qquad \text{(E)} \ 2+2\sqrt 6$
3 replies
4everwise
Jan 1, 2006
aidan0626
Yesterday at 10:43 PM
Concurrent in a pyramid
vanstraelen   0
Yesterday at 7:13 AM

Given a pyramid $(T,ABCD)$ where $ABCD$ is a parallelogram.
The intersection of the diagonals of the base is point $S$.
Point $A$ is connected to the midpoint of $[CT]$, point $B$ to the midpoint of $[DT]$,
point $C$ to the midpoint of $[AT]$ and point $D$ to the midpoint of $[BT]$.
a) Prove: the four lines are concurrent in a point $P$.
b) Calulate $\frac{TS}{TP}$.
0 replies
vanstraelen
Yesterday at 7:13 AM
0 replies
Triangle on a tetrahedron
vanstraelen   2
N Friday at 7:51 PM by ReticulatedPython

Given a regular tetrahedron $(A,BCD)$ with edges $l$.
Construct at the apex $A$ three perpendiculars to the three lateral faces.
Take a point on each perpendicular at a distance $l$ from the apex such that these three points lie above the apex.
Calculate the lenghts of the sides of the triangle.
2 replies
vanstraelen
Friday at 2:43 PM
ReticulatedPython
Friday at 7:51 PM
shadow of a cylinder, shadow of a cone
vanstraelen   2
N Friday at 6:33 PM by vanstraelen

a) Given is a right cylinder of height $2R$ and radius $R$.
The sun shines on this solid at an angle of $45^{\circ}$.
What is the area of the shadow that this solid casts on the plane of the botom base?

b) Given is a right cone of height $2R$ and radius $R$.
The sun shines on this solid at an angle of $45^{\circ}$.
What is the area of the shadow that this solid casts on the plane of the base?
2 replies
vanstraelen
Friday at 3:08 PM
vanstraelen
Friday at 6:33 PM
Cube Sphere
vanstraelen   4
N Friday at 2:37 PM by pieMax2713

Given the cube $\left(\begin{array}{ll} EFGH \\ ABCD \end{array}\right)$ with edge $6$ cm.
Find the volume of the sphere passing through $A,B,C,D$ and tangent to the plane $(EFGH)$.
4 replies
vanstraelen
Friday at 1:10 PM
pieMax2713
Friday at 2:37 PM
parallelogram in a tetrahedron
vanstraelen   1
N Friday at 12:19 PM by vanstraelen
Given a tetrahedron $ABCD$ and a plane $\mu$, parallel with the edges $AC$ and $BD$.
$AB \cap \mu=P$.
a) Prove: the intersection of the tetrahedron with the plane is a parallelogram.
b) If $\left|AC\right|=14,\left|BD\right|=7$ and $\frac{\left|PA\right|}{\left|PB\right|}=\frac{3}{4}$,
calculates the lenghts of the sides of this parallelogram.
1 reply
vanstraelen
May 5, 2025
vanstraelen
Friday at 12:19 PM
Regular tetrahedron
vanstraelen   7
N May 6, 2025 by ReticulatedPython
Given the points $O(0,0,0),A(1,0,0),B(\frac{1}{2},\frac{\sqrt{3}}{2},0)$
a) Determine the point $C$, above the xy-plane, such that the pyramid $OABC$ is a regular tetrahedron.
b) Calculate the volume.
c) Calculate the radius of the inscribed sphere and the radius of the circumscribed sphere.
7 replies
vanstraelen
May 4, 2025
ReticulatedPython
May 6, 2025
volume 9f a pentagonal base pyramid circumscribed around a right circular cone
FOL   1
N May 6, 2025 by Mathzeus1024
A pentagonal base pyramid is circumscribed around a right circular cone, whose height is equal to the radius of the base. The total surface area of the pyramid is d times greater than that of the cone. Find the volume of the pyramid if the lateral surface area of the cone is equal to $\pi\sqrt{2}$.
1 reply
FOL
Jul 22, 2023
Mathzeus1024
May 6, 2025
Geometry books
T.Mousavidin   4
N Apr 30, 2025 by compoly2010
Hello, I wanted to ask if anybody knows some good books for geometry that has these topics in:
Desargues's Theorem, Projective geometry, 3D geometry,
4 replies
T.Mousavidin
Apr 29, 2025
compoly2010
Apr 30, 2025
Tetrahedrons and spheres
ReticulatedPython   4
N Apr 28, 2025 by soryn
Let $OABC$ be a tetrahedron such that $\angle{AOB}=\angle{AOC}=\angle{BOC}=90^\circ.$ A sphere of radius $r$ is circumscribed about tetrahedron $OABC.$ Given that $OA=a$, $OB=b$, and $OC=c$, prove that $$r^2+\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \ge \frac{9\sqrt[3]{4}}{4}$$with equality at $a=b=c=\sqrt[3]{2}.$
4 replies
ReticulatedPython
Apr 21, 2025
soryn
Apr 28, 2025
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:
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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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