Stay ahead of learning milestones! Enroll in a class over the summer!

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k a April Highlights and 2025 AoPS Online Class Information
jlacosta   0
Apr 2, 2025
Spring is in full swing and summer is right around the corner, what are your plans? At AoPS Online our schedule has new classes starting now through July, so be sure to keep your skills sharp and be prepared for the Fall school year! Check out the schedule of upcoming classes below.

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[*]April 22nd (Webinar), 4:00pm PT/7:00pm ET, Competitive Programming at AoPS (USACO).[/list]
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0 replies
jlacosta
Apr 2, 2025
0 replies
Hard Inequality Problem
Omerking   1
N 10 minutes ago by lpieleanu
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3$ is given where $a,b,c$ are positive reals. Prove that:
$$\frac{1}{\sqrt{a^3+1}}+\frac{1}{\sqrt{b^3+1}}+\frac{1}{\sqrt{c^3+1}} \le \frac{3}{\sqrt{2}}$$
1 reply
1 viewing
Omerking
Today at 3:51 PM
lpieleanu
10 minutes ago
USAMO 2000 Problem 5
MithsApprentice   22
N 10 minutes ago by Maximilian113
Let $A_1A_2A_3$ be a triangle and let $\omega_1$ be a circle in its plane passing through $A_1$ and $A_2.$ Suppose there exist circles $\omega_2, \omega_3, \dots, \omega_7$ such that for $k = 2, 3, \dots, 7,$ $\omega_k$ is externally tangent to $\omega_{k-1}$ and passes through $A_k$ and $A_{k+1},$ where $A_{n+3} = A_{n}$ for all $n \ge 1$. Prove that $\omega_7 = \omega_1.$
22 replies
MithsApprentice
Oct 1, 2005
Maximilian113
10 minutes ago
f.e with finite number of f(t)=-t
jjkim0336   0
17 minutes ago
Source: own
f:R->R
f(xf(y)+y)=yf(x)+f(f(y)) and there are finite number of t such that f(t)= - t
0 replies
jjkim0336
17 minutes ago
0 replies
Common external tangents of two circles
a1267ab   55
N 18 minutes ago by awesomeming327.
Source: USA Winter TST for IMO 2020, Problem 2, by Merlijn Staps
Two circles $\Gamma_1$ and $\Gamma_2$ have common external tangents $\ell_1$ and $\ell_2$ meeting at $T$. Suppose $\ell_1$ touches $\Gamma_1$ at $A$ and $\ell_2$ touches $\Gamma_2$ at $B$. A circle $\Omega$ through $A$ and $B$ intersects $\Gamma_1$ again at $C$ and $\Gamma_2$ again at $D$, such that quadrilateral $ABCD$ is convex.

Suppose lines $AC$ and $BD$ meet at point $X$, while lines $AD$ and $BC$ meet at point $Y$. Show that $T$, $X$, $Y$ are collinear.

Merlijn Staps
55 replies
a1267ab
Dec 16, 2019
awesomeming327.
18 minutes ago
No more topics!
JBMO Shortlist 2021 G5
Lukaluce   6
N Jun 21, 2024 by Assassino9931
Source: JBMO Shortlist 2021
Let $ABC$ be an acute scalene triangle with circumcircle $\omega$. Let $P$ and $Q$ be interior points of the sides $AB$ and $AC$, respectively, such that $PQ$ is parallel to $BC$. Let $L$ be a point on $\omega$ such that $AL$ is parallel to $BC$. The segments $BQ$ and $CP$ intersect at $S$. The line $LS$ intersects $\omega$ at $K$. Prove that $\angle BKP = \angle CKQ$.

Proposed by Ervin Macić, Bosnia and Herzegovina
6 replies
Lukaluce
Jul 2, 2022
Assassino9931
Jun 21, 2024
JBMO Shortlist 2021 G5
G H J
Source: JBMO Shortlist 2021
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Lukaluce
267 posts
#1 • 2 Y
Y by ImSh95, Rounak_iitr
Let $ABC$ be an acute scalene triangle with circumcircle $\omega$. Let $P$ and $Q$ be interior points of the sides $AB$ and $AC$, respectively, such that $PQ$ is parallel to $BC$. Let $L$ be a point on $\omega$ such that $AL$ is parallel to $BC$. The segments $BQ$ and $CP$ intersect at $S$. The line $LS$ intersects $\omega$ at $K$. Prove that $\angle BKP = \angle CKQ$.

Proposed by Ervin Macić, Bosnia and Herzegovina
This post has been edited 1 time. Last edited by Lukaluce, Jul 2, 2022, 10:08 PM
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StefanSebez
53 posts
#2 • 7 Y
Y by a_507_bc, Vladimir_Djurica, BorivojeGuzic123, GeoKing, ImSh95, Math_.only., Assassino9931
I will prove a more generalized problem
In fact condition $PQ$ parallel to $BC$ is not needed

$\angle CKS=\angle CKL=\angle CAL=\angle ACB=\angle AKB$
Hence $KS$ and $KA$ are isogonal in $\angle BKC$
Since $P$ is intersection of $CS$ and $BA$ and $Q$ is intersection of $BS$ and $CA$, by the isogonal line lemma $KP$ and $KQ$ are isogonal in $\angle BKC$ which is what we needed to prove
Attachments:
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a_507_bc
676 posts
#3 • 1 Y
Y by ImSh95
After proving that $KS$ and $KA$ are isogonal in $\angle BKC$ (*), one might finish with DDIT for the degenerate quadrilateral $BQCP$ as well: note that there is an involution at $K$ interchanging $(KA, KS), (KB, KC), (KP, KQ)$ and because of (*), this involution is a symmetry wrt the angle bisector of $\angle BKC$, so we are done.
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g0USinsane777
48 posts
#4 • 2 Y
Y by ImSh95, Math_.only.
Does anyone have another solution without using isogonality ?
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iammocnhan
403 posts
#5 • 2 Y
Y by ImSh95, Math_.only.
I will use the condition $PQ \parallel BC$
$AS$ intersects $PQ,BC$ at $N,M$; $LS$ intersects $PQ,BC$ at $I,G$.
Easy to see $M$ is the midpoint of $BC$.
Also, $AM,CP,BQ$ are concurrent, therefore $(AS,NM)=-1$
Since $AL \parallel NI \parallel GM$, we have $(LS,IG)=(AS,NM)=-1$
$\Rightarrow (ML,MS,MI,MG)=-1$
Note that $MA=ML$, so $MG$ is the external bisector of $\angle{AML}$
Therefore, $MI$ is the internal bisector of $\angle{AML}$, or $MI$ is the perpendicular bisector of $BC$.
$\Rightarrow \angle{BIM}=\angle{CIM} \Rightarrow \angle{BIP}=\angle{CIQ}$
Moreover, $\angle{BKI}=180^o-\angle{BAL}=180^o-\angle{BPQ}=\angle{BPI}$, hence $BPIK$ is cyclic.
Analogously, $CQIK$ is cyclic, therefore $\angle{BKP}=\angle{BIP}=\angle{CIQ}=\angle{CKQ}$
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trinhquockhanh
522 posts
#6 • 3 Y
Y by GeoKing, ImSh95, Rounak_iitr
https://i.ibb.co/ZKFRsmP/JBMOSL-G6.png
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Reason: .
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Assassino9931
1280 posts
#7
Y by
https://www.matematickitalent.mk/uploads/books/Mam-5P2Z6Uivs9MldDbD4w.pdf

The official solutions use much simpler methods (especially Solutions 2 and 3).
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