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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.

Summer camps are starting next month at the Virtual Campus in math and language arts that are 2 - to 4 - weeks in duration. Spaces are still available - don’t miss your chance to have an enriching summer experience. 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)!

Be sure to mark your calendars for the following upcoming events:
[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.
[*]May 19th, 4:30pm PT/7:30pm ET, What's Next After Beast Academy?, designed for students finishing Beast Academy and ready for Prealgebra 1.
[*]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
Italian WinterCamps test07 Problem4
mattilgale   89
N 18 minutes ago by cj13609517288
Source: ISL 2006, G3, VAIMO 2007/5
Let $ ABCDE$ be a convex pentagon such that
\[ \angle BAC = \angle CAD = \angle DAE\qquad \text{and}\qquad \angle ABC = \angle ACD = \angle ADE.
\]The diagonals $BD$ and $CE$ meet at $P$. Prove that the line $AP$ bisects the side $CD$.

Proposed by Zuming Feng, USA
89 replies
mattilgale
Jan 29, 2007
cj13609517288
18 minutes ago
Simple triangle geometry [a fixed point]
darij grinberg   49
N 30 minutes ago by cj13609517288
Source: German TST 2004, IMO ShortList 2003, geometry problem 2
Three distinct points $A$, $B$, and $C$ are fixed on a line in this order. Let $\Gamma$ be a circle passing through $A$ and $C$ whose center does not lie on the line $AC$. Denote by $P$ the intersection of the tangents to $\Gamma$ at $A$ and $C$. Suppose $\Gamma$ meets the segment $PB$ at $Q$. Prove that the intersection of the bisector of $\angle AQC$ and the line $AC$ does not depend on the choice of $\Gamma$.
49 replies
darij grinberg
May 18, 2004
cj13609517288
30 minutes ago
Kosovo MO 2010 Problem 5
Com10atorics   19
N 39 minutes ago by CM1910
Source: Kosovo MO 2010 Problem 5
Let $x,y$ be positive real numbers such that $x+y=1$. Prove that
$\left(1+\frac {1}{x}\right)\left(1+\frac {1}{y}\right)\geq 9$.
19 replies
Com10atorics
Jun 7, 2021
CM1910
39 minutes ago
Hard combi
EeEApO   1
N an hour ago by EeEApO
In a quiz competition, there are a total of $100 $questions, each with $4$ answer choices. A participant who answers all questions correctly will receive a gift. To ensure that at least one member of my family answers all questions correctly, how many family members need to take the quiz?

Now, suppose my spouse and I move into a new home. Every year, we have twins. Starting at the age of $16$, each of our twin children also begins to have twins every year. If this pattern continues, how many years will it take for my family to grow large enough to have the required number of members to guarantee winning the quiz gift?
1 reply
EeEApO
2 hours ago
EeEApO
an hour ago
four point lie on circle
Kizaruno   0
5 hours ago
Let triangle ABC be inscribed in a circle with center O. A line d intersects sides AB and AC at points E and D, respectively. Let M, N, and P be the midpoints of segments BD, CE, and DE, respectively. Let Q be the foot of the perpendicular from O to line DE. Prove that the points M, N, P, and Q lie on a circle.
0 replies
+1 w
Kizaruno
5 hours ago
0 replies
Inequalities
sqing   0
5 hours ago
Let $ a,b>0, a^2+ab+b^2 \geq 6  $. Prove that
$$a^4+ab+b^4\geq 10$$Let $ a,b>0, a^2+ab+b^2 \leq \sqrt{10}  $. Prove that
$$a^4+ab+b^4  \leq 10$$Let $ a,b>0,  a^2+ab+b^2 \geq \frac{15}{2}  $. Prove that
$$ a^4-ab+b^4\geq 10$$Let $ a,b>0,  a^2+ab+b^2 \leq \sqrt{10}  $. Prove that
$$-\frac{1}{8}\leq  a^4-ab+b^4\leq 10$$
0 replies
sqing
5 hours ago
0 replies
Range if \omega for No Inscribed Right Triangle y = \sin(\omega x)
ThisIsJoe   0
Today at 2:02 PM
For a positive number \omega , determine the range of \omega for which the curve y = \sin(\omega x) has no inscribed right triangle.
Could someone help me figure out how to approach this?
0 replies
ThisIsJoe
Today at 2:02 PM
0 replies
Interesting question from Al-Khwarezmi olympiad 2024 P3, day1
Adventure1000   1
N Today at 1:10 PM by pooh123
Find all $x, y, z \in \left (0, \frac{1}{2}\right )$ such that
$$
\begin{cases}
(3 x^{2}+y^{2}) \sqrt{1-4 z^{2}} \geq z; \\
(3 y^{2}+z^{2}) \sqrt{1-4 x^{2}} \geq x; \\
(3 z^{2}+x^{2}) \sqrt{1-4 y^{2}} \geq y.
\end{cases}
$$Proposed by Ngo Van Trang, Vietnam
1 reply
Adventure1000
Yesterday at 4:10 PM
pooh123
Today at 1:10 PM
one nice!
MihaiT   3
N Today at 12:53 PM by Pin123
Find positiv integer numbers $(a,b) $ s.t. $\frac{a}{b-2}  $ and $\frac{3b-6}{a-3}$ be positiv integer numbers.
3 replies
MihaiT
Jan 14, 2025
Pin123
Today at 12:53 PM
Acute Angle Altitudes... say that ten times fast
Math-lover1   1
N Today at 12:29 PM by pooh123
In acute triangle $ABC$, points $D$ and $E$ are the feet of the angle bisector and altitude from $A$, respectively. Suppose that $AC-AB=36$ and $DC-DB=24$. Compute $EC-EB$.
1 reply
Math-lover1
Yesterday at 11:30 PM
pooh123
Today at 12:29 PM
Find a and b such that a^2 = (a-b)^3 + b and a and b are coprimes
picysm   2
N Today at 8:28 AM by picysm
it is given that a and b are coprime to each other and a, b belong to N*
2 replies
picysm
Apr 25, 2025
picysm
Today at 8:28 AM
Algebra problem
Deomad123   1
N Today at 8:28 AM by lbh_qys
Let $n$ be a positive integer.Prove that there is a polynomial $P$ with integer coefficients so that $a+b+c=0$,then$$a^{2n+1}+b^{2n+1}+c^{2n+1}=abc[P(a,b)+P(b,c)+P(a,c)]$$.
1 reply
Deomad123
May 3, 2025
lbh_qys
Today at 8:28 AM
Palindrome
Darealzolt   1
N Today at 8:01 AM by ehz2701
Find the number of six-digit palindromic numbers that are divisible by \( 37 \).
1 reply
Darealzolt
Today at 4:13 AM
ehz2701
Today at 8:01 AM
Geometry Proof
strongstephen   17
N Today at 3:59 AM by ohiorizzler1434
Proof that choosing four distinct points at random has an equal probability of getting a convex quadrilateral vs a concave one.
not cohesive proof alert!

NOTE: By choosing four distinct points, that means no three points lie on the same line on the Gaussian Plane.
NOTE: The probability of each point getting chosen don’t need to be uniform (as long as it is symmetric about the origin), you just need a way to choose points in the infinite plane (such as a normal distribution)

Start by picking three of the four points. Next, graph the regions where the fourth point would make the quadrilateral convex or concave. In diagram 1 below, you can see the regions where the fourth point would be convex or concave. Of course, there is the centre region (the shaded triangle), but in an infinite plane, the probability the fourth point ends up in the finite region approaches 0.

Next, I want to prove to you the area of convex/concave, or rather, the probability a point ends up in each area, is the same. Referring to the second diagram, you can flip each concave region over the line perpendicular to the angle bisector of which the region is defined. (Just look at it and you'll get what it means.) Now, each concave region has an almost perfect 1:1 probability correspondence to another convex region. The only difference is the finite region (the triangle, shaded). Again, however, the actual significance (probability) of this approaches 0.

If I call each of the convex region's probability P(a), P(c), and P(e) and the concave ones P(b), P(d), P(f), assuming areas a and b are on opposite sides (same with c and d, e and f) you can get:
P(a) = P(b)
P(c) = P(d)
P(e) = P(f)

and P(a) + P(c) + P(e) = P(convex)
and P(b) + P(d) + P(f) = P(concave)

therefore:
P(convex) = P(concave)
17 replies
strongstephen
May 6, 2025
ohiorizzler1434
Today at 3:59 AM
Strategy game based modulo 3
egxa   1
N Apr 18, 2025 by Euler8038
Source: All Russian 2025 9.7
The numbers \( 1, 2, 3, \ldots, 60 \) are written in a row in that exact order. Igor and Ruslan take turns inserting the signs \( +, -, \times \) between them, starting with Igor. Each turn consists of placing one sign. Once all signs are placed, the value of the resulting expression is computed. If the value is divisible by $3$, Igor wins; otherwise, Ruslan wins. Which player has a winning strategy regardless of the opponent’s moves?
1 reply
egxa
Apr 18, 2025
Euler8038
Apr 18, 2025
Strategy game based modulo 3
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G H BBookmark kLocked kLocked NReply
Source: All Russian 2025 9.7
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egxa
210 posts
#1
Y by
The numbers \( 1, 2, 3, \ldots, 60 \) are written in a row in that exact order. Igor and Ruslan take turns inserting the signs \( +, -, \times \) between them, starting with Igor. Each turn consists of placing one sign. Once all signs are placed, the value of the resulting expression is computed. If the value is divisible by $3$, Igor wins; otherwise, Ruslan wins. Which player has a winning strategy regardless of the opponent’s moves?
This post has been edited 1 time. Last edited by egxa, Apr 18, 2025, 5:24 PM
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Euler8038
5 posts
#2
Y by
Igor can just put the multiplication sign between 59 and 60 as his first move and he's guaranteed to win.
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