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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.

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[*]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
Divisibility..
Sadigly   1
N 8 minutes ago by Mathzeus1024
Source: Azerbaijan Junior MO 2025 P2
Find all $4$ consecutive even numbers, such that the square of their product is divisible by the sum of their squares.
1 reply
Sadigly
5 hours ago
Mathzeus1024
8 minutes ago
official solution of IGO
ABCD1728   7
N 31 minutes ago by ABCD1728
Source: IGO official website
Where can I get the official solution of IGO for 2023 and 2024, there are some inhttps://imogeometry.blogspot.com/p/iranian-geometry-olympiad.html, but where can I find them on the official website, thanks :)
7 replies
ABCD1728
May 4, 2025
ABCD1728
31 minutes ago
Combo geo with circles
a_507_bc   10
N 34 minutes ago by EthanWYX2009
Source: 239 MO 2024 S8
There are $2n$ points on the plane. No three of them lie on the same straight line and no four lie on the same circle. Prove that it is possible to split these points into $n$ pairs and cover each pair of points with a circle containing no other points.
10 replies
a_507_bc
May 22, 2024
EthanWYX2009
34 minutes ago
Vietnam TST #5
IMOStarter   2
N 40 minutes ago by cursed_tangent1434
Source: Vietnam TST 2022 P5
A fractional number $x$ is called pretty if it has finite expression in base$-b$ numeral system, $b$ is a positive integer in $[2;2022]$. Prove that there exists finite positive integers $n\geq 4$ that with every $m$ in $(\frac{2n}{3}; n)$ then there is at least one pretty number between $\frac{m}{n-m}$ and $\frac{n-m}{m}$
2 replies
IMOStarter
Apr 27, 2022
cursed_tangent1434
40 minutes ago
Find max
tranlenhanhbnd   0
an hour ago
Let $x,y,z>0$ and $x^2+y^2+z^2=1$. Find max
$D=\dfrac{x}{\sqrt{2 y z+1}}+\dfrac{y}{\sqrt{2 z x+1}}+\dfrac{z}{\sqrt{2 x y+1}}$.
0 replies
tranlenhanhbnd
an hour ago
0 replies
triangle ABC, # BCDE, BE//AM, BE=AM/2, midpoint (Greece Junior 2014)
parmenides51   8
N 2 hours ago by AylyGayypow009
Let $ABC$ be a triangle and let $M$ be the midpoint $BC$. On the exterior of the triangle, consider the parallelogram $BCDE$ such that $BE//AM$ and $BE=AM/2$ . Prove that line $EM$ passes through the midpoint of segment $AD$.
8 replies
parmenides51
Jul 14, 2019
AylyGayypow009
2 hours ago
a,b,c irrational, f(x)=ax^2+bx+c : [-1,1] to [-1,1] surjective
tom-nowy   1
N 3 hours ago by alexheinis
Consider a quadratic function $f(x) = ax^2 + bx + c$, where the coefficients $a, b,$ and $c$ are all irrational numbers.
Is it possible for this function to have a maximum value of $1$ and a minimum value of $-1$ over the interval $[-1, 1]$?
1 reply
tom-nowy
Yesterday at 11:03 PM
alexheinis
3 hours ago
Inequalities
sqing   0
4 hours ago
Let $ 0\leq x,y,z\leq 2. $ Prove that
$$-48\leq (x-yz)( 3y-zx)(z-xy)\leq 9$$$$-144\leq (3x-yz)(y-zx)(3z-xy)\leq\frac{81}{64}$$$$-144\leq (3x-yz)(2y-zx)(3z-xy)\leq\frac{81}{16}$$
0 replies
sqing
4 hours ago
0 replies
How inflated are current aime/amc problems
derekli   2
N Today at 6:36 AM by Mathgloggers
So I've been working on a math grinding tool in Stellar Learning (https://stellarlearning.app/competitive) and I was wondering how to make an algorithm that can calculate the difficulty of a problem. Specifically I want to know how difficult past AIMEs and AMC 10s and other contests are, compared to our current contests. I'm planning to make a problem ELO system similar to mathdash or something like that. Any help would be appreciated! Again if you would like to support me you may consider joining our developer team! :D
2 replies
derekli
Today at 1:30 AM
Mathgloggers
Today at 6:36 AM
2023 Official Mock NAIME #15 f(f(f(x))) = f(f(x))
parmenides51   1
N Today at 4:53 AM by jasperE3
How many non-bijective functions $f$ exist that satisfy $f(f(f(x))) = f(f(x))$ for all real $x$ and the domain of f is strictly within the set of $\{1,2,3,5,6,7,9\}$, the range being $\{1,2,4,6,7,8,9\}$?

Even though this is an AIME problem, a proof is mandatory for full credit. Constants must be ignored as we dont want an infinite number of solutions.
1 reply
parmenides51
Dec 4, 2023
jasperE3
Today at 4:53 AM
Inequalities
sqing   3
N Today at 3:29 AM by sqing
Let $ a,b>0 $ and $\frac{a}{a^2+3}+ \frac{b}{b^2+ 3} \geq \frac{1}{2} . $ Prove that
$$a^2+ab+b^2\geq 3$$$$a^2-ab+b^2 \geq 1 $$Let $ a,b>0 $ and $\frac{a}{a^3+3}+ \frac{b}{b^3+ 3}\geq \frac{1}{2} . $ Prove that
$$a^3+ab+b^3 \geq 3$$$$ a^3-ab+b^3\geq 1 $$
3 replies
sqing
May 7, 2025
sqing
Today at 3:29 AM
exist solutions?
teomihai   6
N Today at 12:05 AM by iwastedmyusername
Find how many perfect squares of five different digits there are, with elements from the set ${0,1,4,6,9}$.
6 replies
teomihai
Yesterday at 5:04 PM
iwastedmyusername
Today at 12:05 AM
A pentagon inscribed in a circle of radius √2
tom-nowy   6
N Yesterday at 11:55 PM by anticodon
Can a pentagon with all rational side lengths be inscribed in a circle of radius $\sqrt{2}$ ?
6 replies
tom-nowy
May 6, 2025
anticodon
Yesterday at 11:55 PM
Menelau's theorem
noneofyou34   6
N Yesterday at 11:10 PM by Shan3t
Please can someone help me prove that orthocenter of a triangle exists by using Menelau's Theorem!
6 replies
noneofyou34
Yesterday at 5:52 PM
Shan3t
Yesterday at 11:10 PM
Two parallels
jayme   3
N Apr 28, 2025 by ND_
Source: own?
Dear Mathlinkers,

1. ABCD a square
2. (A) the circle with center at A passing through B
3. P the points of intersection of the segment AC and (A)
4. I the midpoint of AB
5. Q the point of intersection of the segment IC and (A)
6. X the point of intersection of the parallel to AB through Q and BC
7. Y the point of intersection of the segment AX and (A)

Prove : CY is parallel to IP.

Jean-Louis
3 replies
jayme
Apr 28, 2025
ND_
Apr 28, 2025
Two parallels
G H J
G H BBookmark kLocked kLocked NReply
Source: own?
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jayme
9792 posts
#1
Y by
Dear Mathlinkers,

1. ABCD a square
2. (A) the circle with center at A passing through B
3. P the points of intersection of the segment AC and (A)
4. I the midpoint of AB
5. Q the point of intersection of the segment IC and (A)
6. X the point of intersection of the parallel to AB through Q and BC
7. Y the point of intersection of the segment AX and (A)

Prove : CY is parallel to IP.

Jean-Louis
Z K Y
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ND_
52 posts
#2
Y by
Let \( A \) be \( (0,0) \), \( B \) be \( (2,0) \), \( C \) be \( (2,2) \).
Then \( P = (\sqrt{2}, \sqrt{2}) \), \( I = (1,0) \), \( Q = (8/5, 6/5) \), \( X = (2, 6/5) \), \( Y = (\sqrt{50/17}, \sqrt{18/17}) \).

$$
m_{CY} = \frac{2 - \sqrt{18/17}}{2 - \sqrt{50/17}} = 3.406
$$
$$
m_{IP} = 2 + \sqrt{2} = 3.414
$$
CY and IP are not parallel, they intersect at \( (-190.8, -654.8) \).
Z K Y
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jayme
9792 posts
#3
Y by
Dear,

yes, you are right...

Cancel 5? 6 and 7 in the hyothesis and consider

M the foot of the perpendicular to (AB) through P.

Then IP is parallel to CM.

Sorry...

Sincerely and thank you

Jean-Louis
Z K Y
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ND_
52 posts
#4
Y by
\( AM = \frac{AP}{\sqrt{2}} = \frac{AB}{\sqrt{2}} \), \( AI = \frac{AB}{2} \), \( AP = AB \), \( AC = AB \sqrt{2} \)

\(\Rightarrow \frac{AI}{AM} = \frac{\sqrt{2}}{2} = \frac{AP}{AC} \)

\(\Rightarrow IP \) is parallel to \( MC \)
Z K Y
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