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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Looking for summer camps in math and language arts? Be sure to check out the video-based summer camps offered at the Virtual Campus that are 2- to 4-weeks in duration. There are middle and high school competition math camps as well as Math Beasts camps that review key topics coupled with fun explorations covering areas such as graph theory (Math Beasts Camp 6), cryptography (Math Beasts Camp 7-8), and topology (Math Beasts Camp 8-9)!

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[list][*]April 3rd (Webinar), 4pm PT/7:00pm ET, Learning with AoPS: Perspectives from a Parent, Math Camp Instructor, and University Professor
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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
Fixed point in a small configuration
Assassino9931   1
N 4 minutes ago by mathuz
Source: Balkan MO Shortlist 2024 G3
Let $A, B, C, D$ be fixed points on this order on a line. Let $\omega$ be a variable circle through $C$ and $D$ and suppose it meets the perpendicular bisector of $CD$ at the points $X$ and $Y$. Let $Z$ and $T$ be the other points of intersection of $AX$ and $BY$ with $\omega$. Prove that $ZT$ passes through a fixed point independent of $\omega$.
1 reply
1 viewing
Assassino9931
Yesterday at 10:23 PM
mathuz
4 minutes ago
JBMO Shortlist 2021 G4
Lukaluce   1
N 18 minutes ago by s27_SaparbekovUmar
Source: JBMO Shortlist 2021
Let $ABCD$ be a convex quadrilateral with $\angle B = \angle D = 90^{\circ}$. Let $E$ be the point of intersection of $BC$ with $AD$ and let $M$ be the midpoint of $AE$. On the extension of $CD$, beyond the point $D$, we pick a point $Z$ such that $MZ = \frac{AE}{2}$. Let $U$ and $V$ be the projections of $A$ and $E$ respectively on $BZ$. The circumcircle of the triangle $DUV$ meets again $AE$ at the point $L$. If $I$ is the point of intersection of $BZ$ with $AE$, prove that the lines $BL$ and $CI$ intersect on the line $AZ$.
1 reply
Lukaluce
Jul 2, 2022
s27_SaparbekovUmar
18 minutes ago
incircle with center I of triangle ABC touches the side BC
orl   40
N 2 hours ago by Ilikeminecraft
Source: Vietnam TST 2003 for the 44th IMO, problem 2
Given a triangle $ABC$. Let $O$ be the circumcenter of this triangle $ABC$. Let $H$, $K$, $L$ be the feet of the altitudes of triangle $ABC$ from the vertices $A$, $B$, $C$, respectively. Denote by $A_{0}$, $B_{0}$, $C_{0}$ the midpoints of these altitudes $AH$, $BK$, $CL$, respectively. The incircle of triangle $ABC$ has center $I$ and touches the sides $BC$, $CA$, $AB$ at the points $D$, $E$, $F$, respectively. Prove that the four lines $A_{0}D$, $B_{0}E$, $C_{0}F$ and $OI$ are concurrent. (When the point $O$ concides with $I$, we consider the line $OI$ as an arbitrary line passing through $O$.)
40 replies
orl
Jun 26, 2005
Ilikeminecraft
2 hours ago
Geometric inequality with Fermat point
Assassino9931   2
N 2 hours ago by Quantum-Phantom
Source: Balkan MO Shortlist 2024 G2
Let $ABC$ be an acute triangle and let $P$ be an interior point for it such that $\angle APB = \angle BPC = \angle CPA$. Prove that
$$ \frac{PA^2 + PB^2 + PC^2}{2S} + \frac{4}{\sqrt{3}} \leq \frac{1}{\sin \alpha} + \frac{1}{\sin \beta} + \frac{1}{\sin \gamma}. $$When does equality hold?
2 replies
1 viewing
Assassino9931
Yesterday at 10:21 PM
Quantum-Phantom
2 hours ago
No more topics!
tangent from $A$ to the circle meets the line $BC$ at point
Valentin Vornicu   4
N Mar 12, 2025 by ali123456
Source: JBMO 2005, Problem 2
Let $ABC$ be an acute-angled triangle inscribed in a circle $k$. It is given that the tangent from $A$ to the circle meets the line $BC$ at point $P$. Let $M$ be the midpoint of the line segment $AP$ and $R$ be the second intersection point of the circle $k$ with the line $BM$. The line $PR$ meets again the circle $k$ at point $S$ different from $R$.

Prove that the lines $AP$ and $CS$ are parallel.
4 replies
Valentin Vornicu
Oct 29, 2005
ali123456
Mar 12, 2025
tangent from $A$ to the circle meets the line $BC$ at point
G H J
Source: JBMO 2005, Problem 2
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Valentin Vornicu
7301 posts
#1 • 2 Y
Y by Adventure10, mathematicsy
Let $ABC$ be an acute-angled triangle inscribed in a circle $k$. It is given that the tangent from $A$ to the circle meets the line $BC$ at point $P$. Let $M$ be the midpoint of the line segment $AP$ and $R$ be the second intersection point of the circle $k$ with the line $BM$. The line $PR$ meets again the circle $k$ at point $S$ different from $R$.

Prove that the lines $AP$ and $CS$ are parallel.
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pontios
777 posts
#2 • 5 Y
Y by nquocthuy, primesarespecial, Adventure10, Mango247, and 1 other user
http://www.mathlinks.ro/Forum/viewtopic.php?t=58603


The power of $M$ w.r.t. $k$ is $MA^2$.

Draw the circle $w$ passing through $P,B,R$

The point $M$ is on the radical axis of $w$ and $k$, so the power of $M$ w.r.t. the circle $w$ is $MA^2=MP^2$.

This means that the segment $MP$ is tangent to $w$.

So, the chord $PR$ is shown under equal angles from $P$ and $B$ : $\angle RPA = \angle PBR$

But $\angle PBR = \angle CBR = \angle CSR$ since $C,S,R,B$ are concyclic

So we find that $\angle RPA = \angle CSR$, hence $CS \parallel AP$
Attachments:
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Virgil Nicula
7054 posts
#3 • 4 Y
Y by andrejilievski, primesarespecial, Adventure10, and 1 other user
$MP^2=MA^2=MR\cdot MB\Longrightarrow MP^2=MR\cdot MB\Longrightarrow MPR\sim MBP\ (s.a.s.)$

$\Longrightarrow \widehat {PSC}\equiv \widehat {RSC}\equiv \widehat {RBC}\equiv \widehat {MBP}\equiv \widehat {MPR}\equiv \widehat {APS}$

$\Longrightarrow \widehat {PSC}\equiv \widehat {APS}\Longrightarrow AP\parallel CS.$

Remark. I proposed this problem for J.B.M.O. 2005.
This post has been edited 3 times. Last edited by Virgil Nicula, Feb 9, 2006, 8:58 PM
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Sailor
256 posts
#4 • 2 Y
Y by Adventure10 and 1 other user
Consider an inversion of pole $P$ which invariates circle $k$.
Consequently, $SC$ will transform into the circle circumscribed to $\triangle{BRP}$.
Since $MA^2=MR\cdot{MB}=MP^2$ we get that the line $AP$
is tangent to the aforementioned circle and we are done!
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ali123456
52 posts
#5
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