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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.

Are you interested in working towards MATHCOUNTS and don’t know where to start? We have you covered! If you have taken Prealgebra, then you are ready for MATHCOUNTS/AMC 8 Basics. Already aiming for State or National MATHCOUNTS and harder AMC 8 problems? Then our MATHCOUNTS/AMC 8 Advanced course is for you.

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[list][*]May 9th, 4:30pm PT/7:30pm ET, Casework 2: Overwhelming Evidence — A Text Adventure, a game where participants will work together to navigate the map, solve puzzles, and win! All are welcome.
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[*]May 20th, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 1 Math Jam, Problems 1 to 4, join the Canada/USA Mathcamp staff for this exciting Math Jam, where they discuss solutions to Problems 1 to 4 of the 2025 Mathcamp Qualifying Quiz!
[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
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0 replies
jlacosta
May 1, 2025
0 replies
IMO Shortlist 2009 - Problem G3
April   48
N 5 minutes ago by Markas
Let $ABC$ be a triangle. The incircle of $ABC$ touches the sides $AB$ and $AC$ at the points $Z$ and $Y$, respectively. Let $G$ be the point where the lines $BY$ and $CZ$ meet, and let $R$ and $S$ be points such that the two quadrilaterals $BCYR$ and $BCSZ$ are parallelogram.
Prove that $GR=GS$.

Proposed by Hossein Karke Abadi, Iran
48 replies
April
Jul 5, 2010
Markas
5 minutes ago
Mount Inequality erupts in all directions!
BR1F1SZ   3
N 6 minutes ago by NumberzAndStuff
Source: Austria National MO Part 1 Problem 1
Let $a$, $b$ and $c$ be pairwise distinct nonnegative real numbers. Prove that
\[
(a + b + c) \left( \frac{a}{(b - c)^2} + \frac{b}{(c - a)^2} + \frac{c}{(a - b)^2} \right) > 4.
\](Karl Czakler)
3 replies
BR1F1SZ
May 5, 2025
NumberzAndStuff
6 minutes ago
Show that (DEN) passes through the midpoint of BC
v_Enhance   24
N 6 minutes ago by Markas
Source: Sharygin First Round 2013, Problem 21
Chords $BC$ and $DE$ of circle $\omega$ meet at point $A$. The line through $D$ parallel to $BC$ meets $\omega$ again at $F$, and $FA$ meets $\omega$ again at $T$. Let $M = ET \cap BC$ and let $N$ be the reflection of $A$ over $M$. Show that $(DEN)$ passes through the midpoint of $BC$.
24 replies
v_Enhance
Apr 7, 2013
Markas
6 minutes ago
2020 EGMO P5: P is the incentre of CDE
alifenix-   49
N 7 minutes ago by Markas
Source: 2020 EGMO P5
Consider the triangle $ABC$ with $\angle BCA > 90^{\circ}$. The circumcircle $\Gamma$ of $ABC$ has radius $R$. There is a point $P$ in the interior of the line segment $AB$ such that $PB = PC$ and the length of $PA$ is $R$. The perpendicular bisector of $PB$ intersects $\Gamma$ at the points $D$ and $E$.

Prove $P$ is the incentre of triangle $CDE$.
49 replies
alifenix-
Apr 18, 2020
Markas
7 minutes ago
No more topics!
Path within S which does not meet itself
orl   5
N Apr 6, 2025 by atdaotlohbh
Source: IMO 1982, Day 2, Problem 6
Let $S$ be a square with sides length $100$. Let $L$ be a path within $S$ which does not meet itself and which is composed of line segments $A_0A_1,A_1A_2,A_2A_3,\ldots,A_{n-1}A_n$ with $A_0=A_n$. Suppose that for every point $P$ on the boundary of $S$ there is a point of $L$ at a distance from $P$ no greater than $\frac {1} {2}$. Prove that there are two points $X$ and $Y$ of $L$ such that the distance between $X$ and $Y$ is not greater than $1$ and the length of the part of $L$ which lies between $X$ and $Y$ is not smaller than $198$.
5 replies
orl
Nov 11, 2005
atdaotlohbh
Apr 6, 2025
Path within S which does not meet itself
G H J
Source: IMO 1982, Day 2, Problem 6
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orl
3647 posts
#1 • 1 Y
Y by Adventure10
Let $S$ be a square with sides length $100$. Let $L$ be a path within $S$ which does not meet itself and which is composed of line segments $A_0A_1,A_1A_2,A_2A_3,\ldots,A_{n-1}A_n$ with $A_0=A_n$. Suppose that for every point $P$ on the boundary of $S$ there is a point of $L$ at a distance from $P$ no greater than $\frac {1} {2}$. Prove that there are two points $X$ and $Y$ of $L$ such that the distance between $X$ and $Y$ is not greater than $1$ and the length of the part of $L$ which lies between $X$ and $Y$ is not smaller than $198$.
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SCP
1502 posts
#3 • 2 Y
Y by Adventure10, Mango247
There is sth wrong ith my idea:

The length of $X,Y>4*99-1$ is quite trivial in one direction, but if they ean the shortest distance, we just take points very close to the boundary and with angles close to $180$ degrees except in the corners, what is the mistake in those ideas?
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mavropnevma
15142 posts
#4 • 2 Y
Y by Adventure10, Mango247
See John Scholes' solution at http://mks.mff.cuni.cz/kalva/.
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Phorphyrion
397 posts
#6
Y by
The condition should be $A_0\neq A_n$. That's the statement appearing on the official IMO website.
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Lukariman
406 posts
#7
Y by
The solution is difficult to understand because it involves compact sets. Can anyone explain the properties of compact sets
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atdaotlohbh
186 posts
#8
Y by
The first one to post a solution!
Start walking from $A_0$ to $A_n$ along $L$. Paint in red all points on the boundary of $S$ that are within distance $\frac{1}{2}$ from the current point. Obviously at any point the red points are a finite union of red segments (with possibly length zero). Suppose such points $X$ and $Y$ do not exist.

The key observation: If $AB$ and $CD$ are opposite sides of squares, then we must completely paint $AB$ before we start painting $CD$ (or vice versa).
Proof: Suppose it is false and $AB$ was partly painted when we firstly painted a point on $CD$. Let $X$ be a red point on $AB$ such that on some side of it on $AB$ its close is not painted (More simply, an endpoint of a segment). Say $X$ was painted when passing $M$ on $L$. We must return to paint the neighbourhood of $X$, and as we always paint segments, we will point $X$ once again as well, say when passing through $N$. Notice that from triangle inequality $MN \leq MX+NX \leq \frac{1}{2}+\frac{1}{2}=1$. Also, to get from $M$ to a point which painted $CD$ a bit we must have travelled at least $99$, and to get back to $N$ we need $99$ again. Thus, the distance travelled between $M$ and $N$ is at least 198, and so $M$ and $N$ work as $X$ and $Y$ in the problem statement.

Now we finish by proving that the key observation is impossible. To do that, let the square be $ABCD$ and $AB$ painted before $CD$, $AD$ painted before $BC$. WLOG $B$ was painted before $D$, but $B$ is a part of $BC$, thus before $B$ was painted $AD$ was completely done already, and hence $D$ also, which is the desired contradiction.

Notice we never used the fact that $L$ doesn't intersect itself.
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