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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[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
An upper bound for Iran TST 1996
Nguyenhuyen_AG   0
7 minutes ago
Let $a, \ b, \ c$ be the side lengths of a triangle. Prove that
\[\frac{ab+bc+ca}{(a+b)^2} + \frac{ab+bc+ca}{(b+c)^2} + \frac{ab+bc+ca}{(c+a)^2} \leqslant \frac{85}{36}.\]
0 replies
Nguyenhuyen_AG
7 minutes ago
0 replies
Sharygin 2025 CR P2
Gengar_in_Galar   5
N 8 minutes ago by NicoN9
Source: Sharygin 2025
Four points on the plane are not concyclic, and any three of them are not collinear. Prove that there exists a point $Z$ such that the reflection of each of these four points about $Z$ lies on the circle passing through three remaining points.
Proposed by:A Kuznetsov
5 replies
Gengar_in_Galar
Mar 10, 2025
NicoN9
8 minutes ago
Kinda lookimg Like AM-GM
Atillaa   1
N 19 minutes ago by Natrium
Show that for all positive real numbers \( a, b, c \), the following inequality always holds:
\[
\frac{ab}{b+1} + \frac{bc}{c+1} + \frac{ca}{a+1} \geq \frac{3abc}{1 + abc}
\]
1 reply
+1 w
Atillaa
an hour ago
Natrium
19 minutes ago
Dividing Pairs
Jackson0423   0
22 minutes ago
Source: Own
Let \( a \) and \( b \) be positive integers.
Suppose that \( a \) is a divisor of \( b^2 + 1 \) and \( b \) is a divisor of \( a^2 + 1 \).
Find all such pairs \( (a, b) \).
0 replies
+1 w
Jackson0423
22 minutes ago
0 replies
No more topics!
Find n if triangles are equal
shobber   4
N Dec 3, 2016 by vsathiam
Source: APMO 2001
Find the greatest integer $n$, such that there are $n+4$ points $A$, $B$, $C$, $D$, $X_1,\dots,~X_n$ in the plane with $AB\ne CD$ that satisfy the following condition: for each $i=1,2,\dots,n$ triangles $ABX_i$ and $CDX_i$ are equal.
4 replies
shobber
Mar 19, 2006
vsathiam
Dec 3, 2016
Find n if triangles are equal
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Source: APMO 2001
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shobber
3498 posts
#1 • 2 Y
Y by Adventure10, Mango247
Find the greatest integer $n$, such that there are $n+4$ points $A$, $B$, $C$, $D$, $X_1,\dots,~X_n$ in the plane with $AB\ne CD$ that satisfy the following condition: for each $i=1,2,\dots,n$ triangles $ABX_i$ and $CDX_i$ are equal.
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jin
383 posts
#2 • 1 Y
Y by Adventure10
n=4,a very long discuss.
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---Fermat---
159 posts
#3 • 2 Y
Y by Adventure10, Mango247
shobber wrote:
Find the greatest integer $n$, such that there are $n+4$ points $A$, $B$, $C$, $D$, $X_1,\dots,~X_n$ in the plane with $AB\ne CD$ that satisfy the following condition: for each $i=1,2,\dots,n$ triangles $ABX_i$ and $CDX_i$ are equal.

See here: http://mks.mff.cuni.cz/kalva/apmo/asoln/asol015.html
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djb86
445 posts
#4 • 3 Y
Y by vsathiam, Adventure10, Mango247
Here is the solution from our yearbook containing all our training material from 2001. This solution is probably copied from the official solution.

One of the sides $AX_i$ or $BX_i$ is equal to $CD$, thus $X_i$ is on one of the circles of radius $CD$ and centre $A$ or $B$. In the same way $X_i$ is on one of the circles of radius $AB$ with centre $C$ or $D$. The intersection of these four circles has no more than 8 points so that $n\le 8$.

Suppose that circle $S_B$ with centre $B$ and radius $CD$ intersects circle $S_C$ with centre $C$ and radius $AB$ in two points $X_1$ and $X_2$ which satisfy the conditions of the problem. Then in triangles $ABX_1$ and $CDX_1$ we have $BX_1=CD$ and $CX_1=AB$. Since these triangles are congruent then $AX_1=DX_1$, therefore $X_1$ and $X_2$ are on the perpendicular bisector of $AD$. On the other hand $X_1X_2$ is perpendicular to segment $BC$. Then $BC\parallel AD$ and $AB$ and $CD$ are the diagonals or non-parallel sides of a trapezoid.

Suppose that $AB<CD$. Then $BX_1=CD>AB=CX_1$. It follows that the distance from $A$ to the perpendicular bisector of $BC$ must be less than the distance from $D$ to this line otherwise we obtain a contradiction to the condition $AB<CD$. Then for any point $X$ in the perpendicular bisector of $BC$ we have $AX<DX$ and it is not impossible to have $AX=CD$, $DX=AB$. Thus if the circle with centre $A$ and radius $CD$ intersects the circle with centre $D$ and radius $AB$, then the points of intersection of $S_B$ with $S_C$ satisfy the condition of congruence, then the points of intersection of $S_B$ with $S_D$ do not. Thus no more than half of the 8 points of intersection of these circles can satisfy the condition of congruence, i.e. $n\le 4$.

If $n=4$ we have the following example where $ABDX_3X_2C$ is a regular hexagon, and $X_1$ and $X_4$ are the intersections of $X_2X_3$ with respectively $AC$ and $BD$.
[asy] pair A; pair B; pair C; pair D; pair X1; pair X2; pair X3; pair X4;

D=(1,0); C=(-1,0); B=(0.5,sqrt(3)/2);
A=B+C; 
X2=-B; X3=-A; X4=X3+D; X1=X2+C;

draw(A--X1--X4--B--A);
draw(X2--C--D--X3);
draw(circumcircle(A,B,C));
label("$A$",A,NW);
label("$B$",B,NE);
label("$C$",C,W);
label("$D$",D,E);
label("$X_1$",X1,S);
label("$X_2$",X2,S);
label("$X_3$",X3,S);
label("$X_4$",X4,S);
[/asy]
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vsathiam
201 posts
#5 • 2 Y
Y by Adventure10, Mango247
What motivated the solver to get that configuration? While it does have nice symmetric properties, the motivation seems a bit obscure besides that bit about AB, CD being a trapezoid but even that contradicts the diagram.
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