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 9th (Webinar), 4:00pm PT/7:00pm ET, Learn about Video-based Summer Camps at the Virtual Campus
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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
Concurrent lines
MathChallenger101   3
N 9 minutes ago by pigeon123
Let $A B C D$ be an inscribed quadrilateral. Circles of diameters $A B$ and $C D$ intersect at points $X_1$ and $Y_1$, and circles of diameters $B C$ and $A D$ intersect at points $X_2$ and $Y_2$. The circles of diameters $A C$ and $B D$ intersect in two points $X_3$ and $Y_3$. Prove that the lines $X_1 Y_1, X_2 Y_2$ and $X_3 Y_3$ are concurrent.
3 replies
MathChallenger101
Feb 8, 2025
pigeon123
9 minutes ago
inequality where ab+bc+ca=1/3
StefanS   13
N 14 minutes ago by Namisgood
Source: Macedonia National Olympiad 2009 - Problem 4
Let $a,b,c$ be positive real numbers for which $ab+bc+ca=\frac{1}{3}$. Prove the inequality
\[ \frac{a}{a^2-bc+1}+\frac{b}{b^2-ca+1}+\frac{c}{c^2-ab+1}\ge\frac{1}{a+b+c}\]
13 replies
StefanS
Apr 13, 2012
Namisgood
14 minutes ago
Distant and Difference
USJL   1
N 14 minutes ago by EeEeRUT
Source: 2025 Taiwan TST Round 3 Independent Study 2-C
There are $N$ points on the plane with diameter $D$.
Show that there exist two distinct points $X,Y$ and two not necessarily distinct points $A,B$ not equal to $X$ or $Y$ satisfying that
\[|AX-XY|+|BY-XY|\leq \frac{2D}{N-2}.\]
Proposed by usjl
1 reply
1 viewing
USJL
2 hours ago
EeEeRUT
14 minutes ago
ISI 2016 #1
zizou10   50
N 18 minutes ago by ZeroAlephZeta
Source: ISI BSTAT 2016 #1
In a sports tournament of $n$ players, each pair of players plays against each other exactly one match and there are no draws.Show that the players can be arranged in an order $P_1,P_2, .... , P_n$ such that $P_i$ defeats $P_{i+1}$ for all $1 \le i \le n-1$.
50 replies
zizou10
May 8, 2016
ZeroAlephZeta
18 minutes ago
No more topics!
ARO 2011 10-4
sartt   8
N May 31, 2024 by v_Enhance
Perimeter of triangle $ABC$ is $4$. Point $X$ is marked at ray $AB$ and point $Y$ is marked at ray $AC$ such that $AX=AY=1$. Line segments $BC$ and $XY$ intersectat point $M$. Prove that perimeter of one of triangles $ABM$ or $ACM$ is $2$.

(V. Shmarov).
8 replies
sartt
Apr 28, 2011
v_Enhance
May 31, 2024
ARO 2011 10-4
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sartt
68 posts
#1 • 4 Y
Y by Adventure10, Mango247, NO_SQUARES, Mogmog8
Perimeter of triangle $ABC$ is $4$. Point $X$ is marked at ray $AB$ and point $Y$ is marked at ray $AC$ such that $AX=AY=1$. Line segments $BC$ and $XY$ intersectat point $M$. Prove that perimeter of one of triangles $ABM$ or $ACM$ is $2$.

(V. Shmarov).
This post has been edited 1 time. Last edited by v_Enhance, May 31, 2024, 2:48 PM
Reason: LaTeX
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SnowEverywhere
801 posts
#2 • 5 Y
Y by El_Ectric, thczarif, khina, rayfish, Adventure10
Let $\omega$ denote the $A$-excircle of $\triangle{ABC}$. Let $\omega$ be tangent to $BC$, $AC$ and $AB$ at $D$, $E$ and $F$, respectively. By power of a point, $AE=AF$ and $AE+AF=AC+CD+AB+BE=4$ which implies that $AE=AF=2$. Hence $X$ and $Y$ are the midpoints of $AE$ and $AF$, respectively and $XY$ is the radical axis betwen $\omega$ and the point $A$. This implies that $AM=MD$. Since $M \in (BC)$, we may assume without the loss of generality that $M \in (BD]$. This implies that the perimeter of $\triangle{AMB}$ is $AM+MB+AB=AM+(BD-MD)+AB=AB+BD=AF=2$.
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MBGO
315 posts
#3 • 2 Y
Y by Adventure10, Mango247
sartt wrote:
Perimeter of triangle ABC is 4. Point X is marked at ray AB and point Y is marked at ray AC such that AX=AY=1. BC intersects XY at point M. Prove that perimeter of one of triangles ABM or ACM is 2.
(V. Shmarov).

is the problem correct? also since WLOG $M$ is on the external of $BC$ nearer to $B$ than $C$ so we get:
$AM+MB+AB=MD+MB+AB=2MB+BD+AB=AF+2MB>AF=2$ so we get (maybe some kinda) contradiction here...
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El_Ectric
3091 posts
#4 • 2 Y
Y by Adventure10, Mango247
MBGO wrote:
is the problem correct? also since WLOG $M$ is on the external of $BC$ nearer to $B$ than $C$ so we get:
$AM+MB+AB=MD+MB+AB=2MB+BD+AB=AF+2MB>AF=2$ so we get (maybe some kinda) contradiction here...
If $D$ is between $B$ and $M$ then $\triangle ACM$ will have perimeter 2, not $\triangle ABM$.
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Stens
49 posts
#5 • 2 Y
Y by Adventure10, Mango247
In the triangle below, $(AB,BC,CA)=(1.5\bar{3},1.3\bar{3},1.1\bar{3})$, and all the conditions in the problem are satisfied. But clearly the perimeters of both $ABM$ and $ACM$ are $>2$. Have I overlooked something or is the problem actually incorrectly stated?
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62861
3564 posts
#6 • 5 Y
Y by Stens, Adventure10, Mango247, NO_SQUARES, v_Enhance
Stens wrote:
In the triangle below, $(AB,BC,CA)=(1.5\bar{3},1.3\bar{3},1.1\bar{3})$, and all the conditions in the problem are satisfied. But clearly the perimeters of both $ABM$ and $ACM$ are $>2$. Have I overlooked something or is the problem actually incorrectly stated?

The problem should say that segments BC and XY intersect at M.
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anantmudgal09
1980 posts
#7 • 2 Y
Y by Adventure10, Mango247
Very elegant! :)
All-Russian MO 2011/10/4 wrote:
Perimeter of triangle $ABC$ is $4$. Point $X$ is marked at ray $AB$ and point $Y$ is marked at ray $AC$ such that $AX=AY=1$. Line segments $BC$ and $XY$ intersect at point $M$. Prove that perimeter of one of triangles $ABM$ or $ACM$ is $2$.

Proposed by V.Shmarov

Point $E$ is the touching point of the $A$-excircle $\omega$ with line $\overline{BC}$. WLOG, point $E$ lie on segment $CM$. It is clear that $AB+BE=2$. We will show that perimeter of $\triangle ABM$ is $2$.

The main idea is to note that line $\overline{XY}$ is the radical axis of circle $\omega$ and "point-size circle" $A,$ as $X,Y$ bisect the tangents drawn from $A$ to $\omega$ (since $AX=AY=1$).

So, $$MA=ME \Longrightarrow AB+BM+MA=AB+(BM+ME)=AB+BE=2. \, \square$$
This post has been edited 2 times. Last edited by v_Enhance, May 31, 2024, 2:48 PM
Reason: missing space
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JAnatolGT_00
559 posts
#8
Y by
Let $Y$ lie on segment $AC.$ Denote by $U,V,Z$ touch-points of $A-\text{excircle }\omega$ with $AB,AC,BC$ respectively.
Note that $X,Y$ are midpoints of $AU,AV$ and hence $XY$ is radical axis of $A,\omega.$ Thus $$|AB|+|BM|+|AM|=|AB|+|BZ|=|AU|=2.$$
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v_Enhance
6876 posts
#9 • 2 Y
Y by szpolska, OronSH
The solution from my textbook, which I apparently never posted (note that as written in post #6, the problem only works if $M$ is on segments $XY$ and $BC$):

Let $I_A$ be the center of the $A$-excircle, tangent to $\overline{BC}$ at $T$, and to the extensions of $\overline{AB}$ and $\overline{AC}$ at $U$ and $V$. We see that $AU = AV = s = 2$. Then $\overline{XY}$ is the radical axis of the $A$-excircle and the circle of radius zero at $A$. Therefore $AM = MT$.

[asy]
size(12cm); pair A = dir(130); pair B = dir(190); pair C = dir(-10); pair I = incenter(A, B, C); pair L = dir(-90); pair I_A = 2*L-I;; pair T = foot(I_A, B, C); pair U = foot(I_A, A, B); pair V = foot(I_A, A, C); pair X = midpoint(A--U); pair Y = midpoint(A--V); pair M = extension(B, C, X, Y); filldraw(A--B--C--cycle, invisible, blue); draw(B--U, blue); draw(C--V, blue); draw(X--Y, blue); draw(A--M, deepgreen); draw(arc(I_A, abs(I_A-T), -10, 190), grey);
dot("$A$", A, dir(A)); dot("$B$", B, dir(150)); dot("$C$", C, dir(30)); dot("$I_A$", I_A, dir(I_A)); dot("$T$", T, dir(90)); dot("$U$", U, dir(135)); dot("$V$", V, dir(45)); dot("$X$", X, dir(X)); dot("$Y$", Y, dir(Y)); dot("$M$", M, dir(270));
/* -----------------------------------------------------------------+ |                 TSQX: by CJ Quines and Evan Chen                  | | https://github.com/vEnhance/dotfiles/blob/main/py-scripts/tsqx.py | +-------------------------------------------------------------------+ !size(12cm); A = dir 130 B 150 = dir 190 C 30 = dir -10 I := incenter A B C L := dir -90 I_A = 2*L-I; T 90 = foot I_A B C U 135 = foot I_A A B V 45 = foot I_A A C X = midpoint A--U Y = midpoint A--V M 270 = extension B C X Y A--B--C--cycle / 0.1 palecyan / blue B--U / blue C--V / blue X--Y / blue A--M / deepgreen !draw(arc(I_A, abs(I_A-T), -10, 190), grey); */
[/asy]
Assume without loss of generality that $T$ lies on segment $MC$, as opposed to segment $MB$. Then \[ AB + BM + MA  = AB + BM + MT = AB + BT = AB + BU = AU = 2 \]as desired.
This post has been edited 1 time. Last edited by v_Enhance, May 31, 2024, 2:49 PM
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