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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
Thursday at 11:16 PM
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)!

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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
Thursday at 11:16 PM
0 replies
JBMO Shortlist 2022 A2
Lukaluce   13
N 12 minutes ago by Rayvhs
Source: JBMO Shortlist 2022
Let $x, y,$ and $z$ be positive real numbers such that $xy + yz + zx = 3$. Prove that
$$\frac{x + 3}{y + z} + \frac{y + 3}{z + x} + \frac{z + 3}{x + y} + 3 \ge 27 \cdot \frac{(\sqrt{x} + \sqrt{y} + \sqrt{z})^2}{(x + y + z)^3}.$$
Proposed by Petar Filipovski, Macedonia
13 replies
Lukaluce
Jun 26, 2023
Rayvhs
12 minutes ago
A very beautiful geo problem
TheMathBob   4
N an hour ago by ravengsd
Source: Polish MO Finals P2 2023
Given an acute triangle $ABC$ with their incenter $I$. Point $X$ lies on $BC$ on the same side as $B$ wrt $AI$. Point $Y$ lies on the shorter arc $AB$ of the circumcircle $ABC$. It is given that $$\angle AIX = \angle XYA = 120^\circ.$$Prove that $YI$ is the angle bisector of $XYA$.
4 replies
TheMathBob
Mar 29, 2023
ravengsd
an hour ago
Inspired by old results
sqing   6
N an hour ago by Jamalll
Source: Own
Let $ a,b>0 , a^2+b^2+ab+a+b=5 . $ Prove that
$$ \frac{ 1 }{a+b+ab+1}+\frac{6}{a^2+b^2+ab+1}\geq \frac{7}{4}$$$$ \frac{ 1 }{a+b+ab+1}+\frac{1}{a^2+b^2+ab+1}\geq \frac{1}{2}$$$$  \frac{41}{a+b+2}+\frac{ab}{a^3+b^3+2} \geq \frac{21}{2}$$
6 replies
sqing
Apr 29, 2025
Jamalll
an hour ago
A Duality Operation on Decreasing Integer Sequences
Ritangshu   0
an hour ago
Let \( S \) be the set of all sequences \( (a_1, a_2, \ldots) \) of non-negative integers such that
(i) \( a_1 \geq a_2 \geq \cdots \); and
(ii) there exists a positive integer \( N \) such that \( a_n = 0 \) for all \( n \geq N \).

Define the dual of the sequence \( (a_1, a_2, \ldots) \in S \) to be the sequence \( (b_1, b_2, \ldots) \), where, for \( m \geq 1 \),
\( b_m \) is the number of \( a_n \)'s which are greater than or equal to \( m \).

(i) Show that the dual of a sequence in \( S \) belongs to \( S \).

(ii) Show that the dual of the dual of a sequence in \( S \) is the original sequence itself.

(iii) Show that the duals of distinct sequences in \( S \) are distinct.
0 replies
Ritangshu
an hour ago
0 replies
Interesting geometry
polarLines   5
N 3 hours ago by Mathworld314
Let $ABC$ be an equilateral triangle of side length $2$. Point $A'$ is chosen on side $BC$ such that the length of $A'B$ is $k<1$. Likewise points $B'$ and $C'$ are chosen on sides $CA$ and $AB$. with $CB'=AC'=k$. Line segments are drawn from points $A',B',C'$ to their corresponding opposite vertices. The intersections of these line segments form a triangle, labeled $PQR$. Prove that $\Delta PQR$ is an equilateral triangle with side length ${4(1-k) \over \sqrt{k^2-2k+4}}$.
5 replies
1 viewing
polarLines
May 20, 2018
Mathworld314
3 hours ago
Showing that certain number is divisible by 13
BBNoDollar   3
N 3 hours ago by Shan3t
Show that 3^(n+2) + 9^(n+1) + 4^(2n+1) + 4^(4n+1) is divisible by 13 for every n natural number.
3 replies
BBNoDollar
5 hours ago
Shan3t
3 hours ago
Sum of digits is 18
Ecrin_eren   4
N 4 hours ago by maxamc
How many 5 digit numbers are there such that sum of its digits is 18
4 replies
Ecrin_eren
Today at 1:10 PM
maxamc
4 hours ago
Inequality
tom-nowy   0
5 hours ago
Let $0<a,b,c,<1$. Show that
$$ \frac{3(a+b+c)}{a+b+c+3abc} > \frac{1}{1+a} + \frac{1}{1+b} + \frac{1}{1+c} .$$
0 replies
tom-nowy
5 hours ago
0 replies
Logarithm of a product
axsolers_24   2
N 5 hours ago by axsolers_24
Let $x_1=97 ,$ $x_2=\frac{2}{x_1} ,$ $x_3=\frac{3}{x_2} ,$$... , $ $x_8=\frac{8}{x_7}$
then
$ \log_{3\sqrt{2}} \left(\prod_{i=1}^8 x_i-60\right)$
2 replies
axsolers_24
Today at 10:42 AM
axsolers_24
5 hours ago
Inequalities
sqing   1
N 5 hours ago by sqing
Let $ a,b>0 , a^2 + 2b^2 =  a + 2b $. Prove that $$\sqrt{\frac{a}{b( a+2)}} + \sqrt{\frac{b}{a(2b+1)}}  \geq \frac {2}{\sqrt{3}} $$Let $ a,b>0 , a^3 + 2b^3 =  a + 2b $. Prove that $$\sqrt[3]{\frac{a}{b( a+2)}} + \sqrt[3]{\frac{b}{a(2b+1)}}  \geq \frac {2}{\sqrt[3]{3}} $$
1 reply
sqing
5 hours ago
sqing
5 hours ago
Coprime sequence
Ecrin_eren   4
N 5 hours ago by Pal702004
"Let N be a natural number. Show that any two numbers from the following sequence are coprime:

2^1 + 1, 2^2 + 1, 2^4+ 1,2^8+1 ..., 2^(2^N )+ 1."
4 replies
Ecrin_eren
Thursday at 8:53 PM
Pal702004
5 hours ago
Hard Inequality
William_Mai   0
5 hours ago
Given $a, b, c \in \mathbb{R}$ such that $a^2 + b^2 + c^2 = 1$.
Find the minimum value of $P = ab + 2bc + 3ca$.

Source: Pham Le Van
0 replies
William_Mai
5 hours ago
0 replies
Find the minimum
Ecrin_eren   5
N 6 hours ago by Jackson0423
The polynomial is given by P(x) = x^4 + ax^3 + bx^2 + cx + d, and its roots are x1, x2, x3, x4. Additionally, it is stated that d ≥ 5.Find the minimum value of the product:

(x1^2 + 1)(x2^2 + 1)(x3^2 + 1)(x4^2 + 1).

5 replies
Ecrin_eren
Thursday at 9:03 PM
Jackson0423
6 hours ago
Inequalities
sqing   8
N 6 hours ago by sqing
Let $a,b,c> 0$ and $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1.$ Prove that
$$  (1-abc) (1-a)(1-b)(1-c)  \ge 208 $$$$ (1+abc) (1-a)(1-b)(1-c)  \le -224 $$$$(1+a^2b^2c^2) (1-a)(1-b)(1-c)  \le -5840 $$
8 replies
sqing
Jul 12, 2024
sqing
6 hours ago
Find an angle
socrates   3
N Apr 6, 2025 by Nari_Tom
Source: Baltic Way 2016, Problem 18
Let $ABCD$ be a parallelogram such that $\angle BAD = 60^{\circ}.$ Let $K$ and $L$ be the midpoints of $BC$ and $CD,$ respectively. Assuming that $ABKL$ is a cyclic quadrilateral, find $\angle ABD.$
3 replies
socrates
Nov 5, 2016
Nari_Tom
Apr 6, 2025
Find an angle
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Source: Baltic Way 2016, Problem 18
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socrates
2105 posts
#1 • 2 Y
Y by Adventure10, Mango247
Let $ABCD$ be a parallelogram such that $\angle BAD = 60^{\circ}.$ Let $K$ and $L$ be the midpoints of $BC$ and $CD,$ respectively. Assuming that $ABKL$ is a cyclic quadrilateral, find $\angle ABD.$
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george_54
1587 posts
#4 • 3 Y
Y by StarLex1, Adventure10, Mango247
socrates wrote:
Let $ABCD$ be a parallelogram such that $\angle BAD = 60^{\circ}.$ Let $K$ and $L$ be the midpoints of $BC$ and $CD,$ respectively. Assuming that $ABKL$ is a cyclic quadrilateral, find $\angle ABD.$

Let $AB, LK$ intersect at $E$, $EB=a,AB=2a$ and $EK=KL=b, BD=2b$. I draw $BH$ perpendicular to $AB$.

It is: $EB \cdot EA = EK \cdot EL \Leftrightarrow 3{a^2} = 2{b^2} \Leftrightarrow b = \frac{{a\sqrt 6 }}{2}$

We also have, $BH = a\sqrt 3  \Leftrightarrow BD = 2b = a\sqrt 3 \sqrt 2  = BH\sqrt 2 $, thus $\omega=45^0$ and finally $\boxed{\angle ABD=75^0}$
Attachments:
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Reason: typo
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K.N
532 posts
#5 • 1 Y
Y by Adventure10
Yes that's exactly my solution!
Easy problem
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Nari_Tom
117 posts
#6
Y by
Let $F$ be the intersection of $AL \cap BD$. Since $\triangle BAF \sim \triangle FLD$, we have $BF=2*FD$. By some angle chase $\angle BAF=\angle FDA$. So by the secant tangent theorem: $BF*BD=BA^2$ $\implies$ $\frac{2*BD^2}{3}=BA^2$, and rest is just some bash.
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