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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 9th (Webinar), 4:00pm PT/7:00pm ET, Learn about Video-based Summer Camps at the Virtual Campus
[*]April 10th (Math Jam), 4:30pm PT/7:30pm ET, 2025 MathILy and MathILy-Er Math Jam: Multibackwards Numbers
[*]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
quadratic with at least 1 roots
giangtruong13   0
3 minutes ago
Find $m$ to satisfy that the equation $x^2+mx-1=0$ has at least 1 roots $\leq -2$
0 replies
giangtruong13
3 minutes ago
0 replies
Monic Polynomial
IstekOlympiadTeam   21
N 8 minutes ago by zuat.e
Source: Romanian Masters 2017 D1 P2
Determine all positive integers $n$ satisfying the following condition: for every monic polynomial $P$ of degree at most $n$ with integer coefficients, there exists a positive integer $k\le n$ and $k+1$ distinct integers $x_1,x_2,\cdots ,x_{k+1}$ such that \[P(x_1)+P(x_2)+\cdots +P(x_k)=P(x_{k+1})\].

Note. A polynomial is monic if the coefficient of the highest power is one.
21 replies
IstekOlympiadTeam
Feb 25, 2017
zuat.e
8 minutes ago
All numbers occur exactly once
Ankoganit   17
N 26 minutes ago by bin_sherlo
Source: St. Petersburg Olympiad 2015, 2nd Round, Grade 9, also known as Yellowstone Permutation
A sequence of integers is defined as follows: $a_1=1,a_2=2,a_3=3$ and for $n>3$, $$a_n=\textsf{The smallest integer not occurring earlier, which is relatively prime to }a_{n-1}\textsf{ but not relatively prime to }a_{n-2}.$$Prove that every natural number occurs exactly once in this sequence.

M. Ivanov
17 replies
Ankoganit
Sep 24, 2016
bin_sherlo
26 minutes ago
Geometry
MathsII-enjoy   0
an hour ago
Given triangle $ABC$ inscribed in $(O)$, $S$ is the midpoint of arc $BAC$ of $(O)$. The perpendicular bisector $BO$ intersects $BS$ at $I$. $(I;IB)$ intersects $AB$ at $U$ different from $B$. $H$ is the orthocenter of triangle $ABC$. Prove that $UH$ = $US$
0 replies
MathsII-enjoy
an hour ago
0 replies
Inequalities
sqing   25
N 4 hours ago by sqing
Let $ a,b,c,d>0 $ and $(a+c)(b+d)=ac+\frac{3}{2}bd.$ Prove that
$$\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}\geq \frac{20-\sqrt{10}}{3}$$Let $ a,b,c,d>0 $ and $(a+c)(b+d)=ac+\frac{4}{3}bd.$ Prove that
$$\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}\geq \frac{21-\sqrt{6}}{3}$$
25 replies
sqing
Dec 3, 2024
sqing
4 hours ago
Polynomials
CuriousBabu   3
N 4 hours ago by osszhangbanvan
\[ 
\frac{(x+y+z)^5 - x^5 - y^5 - z^5}{(x+y)(y+z)(z+x)} = 0 
\]
Find the number of real solutions.
3 replies
CuriousBabu
Yesterday at 4:09 PM
osszhangbanvan
4 hours ago
Combination
aria123   0
5 hours ago
Prove that three squares of side length $4$ cannot completely cover a square of side length $5$, if the three smaller squares do not overlap in their interiors (i.e., they may touch at edges or corners, but no part of one lies over another).
0 replies
aria123
5 hours ago
0 replies
Geo Mock #6
Bluesoul   3
N Today at 3:16 AM by dudade
Consider triangle $ABC$ with $AB=5, BC=8, AC=7$, denote the incenter of the triangle as $I$. Extend $BI$ to meet the circumcircle of $\triangle{AIC}$ at $Q\neq I$, find the length of $QC$.
3 replies
Bluesoul
Apr 1, 2025
dudade
Today at 3:16 AM
AoPS Volume 2, Problem 262
Shiyul   12
N Today at 2:49 AM by bhavanal
Given that $\color[rgb]{0.35,0.35,0.35}v_1=2$, $\color[rgb]{0.35,0.35,0.35}v_2=4$ and $\color[rgb]{0.35,0.35,0.35} v_{n+1}=3v_n-v_{n-1}$, prove that $\color[rgb]{0.35,0.35,0.35}v_n=2F_{2n-1}$, where the terms $\color[rgb]{0.35,0.35,0.35}F_n$ are the Fibonacci numbers.

Can anyone give me hint on how to solve this (not solve the full problem). I'm not sure how to relate the v series to the Fibonacci sequence.

12 replies
Shiyul
Apr 9, 2025
bhavanal
Today at 2:49 AM
Find the angle
pythagorazz   4
N Today at 2:41 AM by dudade
Let $X$ be a point inside equilateral triangle $ABC$ such that $AX=\sqrt{2},BX=3$, and $CX=\sqrt{5}$. Find the measure of $\angle{AXB}$ in degrees.
4 replies
pythagorazz
Yesterday at 9:07 AM
dudade
Today at 2:41 AM
Inequalities
sqing   12
N Today at 2:11 AM by sqing
Let $ a,b,c $ be real numbers so that $ a+2b+3c=2 $ and $ 2ab+6bc+3ca =1. $ Show that
$$-\frac{1}{6} \leq ab-bc+ ca\leq \frac{1}{2}$$$$\frac{5-\sqrt{61}}{9} \leq a-b+c\leq \frac{5+\sqrt{61}}{9} $$
12 replies
sqing
Apr 9, 2025
sqing
Today at 2:11 AM
Probability
Ecrin_eren   4
N Today at 1:20 AM by huajun78
In a board, James randomly writes A , B or C in each cell. What is the probability that, for every row and every column, the number of A 's modulo 3 is equal to the number of B's modulo 3?

4 replies
Ecrin_eren
Apr 3, 2025
huajun78
Today at 1:20 AM
Algebra book recomndaitons
idk12345678   3
N Today at 12:50 AM by idk12345678
Im currently reading EGMO by Evan Chen and i was wondering if there was a similar book for olympiad algebra. I have egmo for geo and aops intermediate c&p for combo, and the intermediate number thoery transcripts for nt, but i couldnt really find anything for alg
3 replies
idk12345678
Yesterday at 10:45 PM
idk12345678
Today at 12:50 AM
Help with Competitive Geometry?
REACHAW   2
N Today at 12:09 AM by alextheadventurer
Hi everyone,
I'm struggling a lot with geometry. I've found algebra, number theory, and even calculus to be relatively intuitive. However, when I took geometry, I found it very challenging. I stumbled my way through the class and can do the basic 'textbook' geometry problems, but still struggle a lot with geometry in competitive math. I find myself consistently skipping the geometry problems during contests (even the easier/first ones).

It's difficult for me to see the solution path. I can do the simpler textbook tasks (eg. find congruent triangles) but not more complex ones (eg. draw these two lines to form similar triangles).

Do you have any advice, resources, or techniques I should try?
2 replies
REACHAW
Yesterday at 11:51 PM
alextheadventurer
Today at 12:09 AM
Paper-folding trisectors
Johann Peter Dirichlet   2
N Jul 19, 2006 by rem
Source: Brazil Math Olympiad - 2000
A rectangular piece of paper has top edge $AD$. A line $L$ from $A$ to the bottom edge makes an angle $x$ with the line $AD$. We want to trisect $x$. We take $B$ and $C$ on the vertical ege through $A$ such that $AB = BC$. We then fold the paper so that $C$ goes to a point $C'$ on the line $L$ and $A$ goes to a point $A'$ on the horizontal line through $B$. The fold takes $B$ to $B'$. Show that $AA'$ and $AB'$ are the required trisectors.
2 replies
Johann Peter Dirichlet
Mar 3, 2006
rem
Jul 19, 2006
Paper-folding trisectors
G H J
Source: Brazil Math Olympiad - 2000
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Johann Peter Dirichlet
376 posts
#1 • 2 Y
Y by Adventure10, Mango247
A rectangular piece of paper has top edge $AD$. A line $L$ from $A$ to the bottom edge makes an angle $x$ with the line $AD$. We want to trisect $x$. We take $B$ and $C$ on the vertical ege through $A$ such that $AB = BC$. We then fold the paper so that $C$ goes to a point $C'$ on the line $L$ and $A$ goes to a point $A'$ on the horizontal line through $B$. The fold takes $B$ to $B'$. Show that $AA'$ and $AB'$ are the required trisectors.
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treegoner
637 posts
#2 • 1 Y
Y by Adventure10
This is a very nice way to trisect an angle in origamic geometry (which in Euclidean geometry, we can't always do). The solution is an application of easy reflection and some computation with ratios. I will post the solution later. Sorry!!! :!:
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rem
1434 posts
#3 • 2 Y
Y by Adventure10, Mango247
Here's the solution. Let $\angle{C'AA'}=\alpha$, then $\angle{BAA'}= 90-x+\alpha$, $BA'$ is horizontal, so $\angle{ABB'}=90^{o}$, hence $\angle{BB'A}=90^{o}-\angle{A'AB}=x-\alpha$
By reflection properties, $AB=A'B', AB'=BA'$, $AA'$ is common so $\triangle{AA'B}$ is congruent to $\triangle{A'AB}$ hence $\angle{B'AA'}=\angle{BA'A}=x-\alpha$, also $\angle{AB'A'}=\angle{ABA'}=90^{o}$
Now in $\triangle{C'AA'}$ $C'B'=B'C'$ and $AB'\perp C'A'$ hence $\angle{C'AB'}=\angle{B'AA'}=x-\alpha$
So $\angle{C'AA'}=2\angle{C'AB'}=2(x-\alpha)$, but $\angle{C'AA'}=\alpha$, so:
$2(x-\alpha)=\alpha$
$\alpha=\frac{2x}{3}$
So $\angle{C'AB'}=\angle{B'AA'}=\frac{x}{3}$, $\angle{A'AD}=x-\frac{2x}{3}=\frac{x}{3}$ hence $\angle{C'AB'}=\angle{B'AA'}=\angle{A'AD}$ so $AA'$ and $AB'$ are valid trisectors. QED
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