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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)!

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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.
[*]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
KJMO 2001 P1
RL_parkgong_0106   1
N 7 minutes ago by JH_K2IMO
Source: KJMO 2001
A right triangle of the following condition is given: the three side lengths are all positive integers and the length of the shortest segment is $141$. For the triangle that has the minimum area while satisfying the condition, find the lengths of the other two sides.
1 reply
RL_parkgong_0106
Jun 29, 2024
JH_K2IMO
7 minutes ago
Cotangential circels
CountingSimplex   5
N 10 minutes ago by rong2020
Let $ABC$ be a triangle with circumcenter $O$ and let the angle bisector of $\angle{BAC}$ intersect $BC$
at $D$. The point $M$ is such that $\angle{MCB}=90^o$ and $\angle{MAD}=90^o$. Lines $BM$ and $OA$ intersect at
the point $P$. Show that the circle centered at $P$ and passing through $A$ is tangent to segment
$BC$.
5 replies
CountingSimplex
Jun 23, 2020
rong2020
10 minutes ago
Inspired by old results
sqing   3
N 17 minutes ago by sqing
Source: Own
Let $a,b,c $ be reals such that $a^2+b^2+c^2=3$ .Prove that
$$(1-a)(k-b)(1-c)+abc\ge -k$$Where $ k\geq 1.$
$$(1-a)(1-b)(1-c)+abc\ge -1$$$$(1-a)(1-b)(1-c)-abc\ge -\frac{1}{2}-\sqrt 2$$
3 replies
sqing
Today at 7:36 AM
sqing
17 minutes ago
Inspired by Philippine 2025
sqing   1
N 22 minutes ago by sqing
Source: Own
Let $ a,b,c,d $ be real numbers . Prove that
$$\frac{(a-1)(b-3)(c-3)(d-1)}{  (a^2+3)(b^2+3)(c^2+3)(d^2+3)} \ge -\frac{7+4\sqrt 3}{144}$$$$\frac{(a-1)(b-2)(c-2)(d-1)}{  (a^2+3)(b^2+3)(c^2+3)(d^2+3)} \ge -\frac{11+4\sqrt 7}{432}$$


1 reply
sqing
37 minutes ago
sqing
22 minutes ago
Combinatorics
P162008   0
an hour ago
$4$ girls and $4$ boys are to be seated in a line. Find the total number of ways such that boys and girls are alternate and a particular boy and girl are never adjacent to each other in any arrangement.
0 replies
P162008
an hour ago
0 replies
Combinatorics
P162008   0
an hour ago
A cricket team comprising of $11$ players named $A,B,C,\cdots,J,K$ is to be sent for batting. If $A$ wants to bat before $J$ and $J$ wants to bat after $G.$ Then find the total number of batting orders if other players could go in any order.
0 replies
P162008
an hour ago
0 replies
[PMO25 Qualifying II.8] A Square Can't Be A Floor
kae_3   2
N 2 hours ago by tapilyoca
Determine the largest perfect square less than $1000$ that cannot be expressed as $\lfloor x\rfloor + \lfloor 2x\rfloor + \lfloor 3x\rfloor + \lfloor 6x\rfloor$ for some positive real number $x$.

Answer Confirmation
2 replies
kae_3
Feb 21, 2025
tapilyoca
2 hours ago
solve in R
zolfmark   1
N 2 hours ago by Mathzeus1024
x+1/y=9/2 and y+1/z=11/4 and z+1/x=12/5
1 reply
zolfmark
Feb 16, 2016
Mathzeus1024
2 hours ago
not so hard problem (own)
BinariouslyRandom   1
N 3 hours ago by BinariouslyRandom
An open rectangular box is made from exactly $900 m^2$ of cardboard, with 5 rectangular faces instead of 6 and there is no leftover cardboard. What is the largest volume this box can hold given that $100 m^2$ of cardboard is needed to cover the box?
1 reply
BinariouslyRandom
3 hours ago
BinariouslyRandom
3 hours ago
Minimize z.
Entrepreneur   3
N 3 hours ago by no_room_for_error
Minimize $z = 6x + 3y.$ Subject to the constraints:
$$\begin{cases}
4x+y\ge80,\\
x+5y \ge115,\\
3x+2y\le150,\\
x,y\ge0.
\end{cases}$$
3 replies
Entrepreneur
May 21, 2025
no_room_for_error
3 hours ago
Inequalities
sqing   18
N Today at 7:48 AM by sqing
Let $ a,b,c\geq 0 ,a+b+c\leq 3. $ Prove that
$$a^2+b^2+c^2+ab +2ca+2bc +  abc \leq \frac{251}{27}$$$$ a^2+b^2+c^2+ab+2ca+2bc  + \frac{2}{5}abc  \leq \frac{4861}{540}$$$$ a^2+b^2+c^2+ab+2ca+2bc  + \frac{7}{20}abc  \leq \frac{2381411}{26460}$$
18 replies
sqing
May 21, 2025
sqing
Today at 7:48 AM
Inequalities
sqing   18
N Today at 7:20 AM by sqing
Let $ a,b>0   $ . Prove that
$$ \frac{a}{a^2+a +2b+1}+ \frac{b}{b^2+2a +b+1}  \leq  \frac{2}{5} $$$$ \frac{a}{a^2+2a +b+1}+ \frac{b}{b^2+a +2b+1}  \leq  \frac{2}{5} $$
18 replies
sqing
May 13, 2025
sqing
Today at 7:20 AM
helpppppppp me
stupid_boiii   1
N Today at 6:56 AM by vanstraelen
Given triangle ABC. The tangent at ? to the circumcircle(ABC) intersects line BC at point T. Points D,E satisfy AD=BD, AE=CE, and ∠CBD=∠BCE<90 ∘ . Prove that D,E,T are collinear.
1 reply
stupid_boiii
Yesterday at 4:22 AM
vanstraelen
Today at 6:56 AM
Algebra Polynomials
Foxellar   2
N Today at 5:43 AM by Foxellar
The real root of the polynomial \( p(x) = 8x^3 - 3x^2 - 3x - 1 \) can be written in the form
\[
\frac{\sqrt[3]{a} + \sqrt[3]{b} + 1}{c},
\]where \( a, b, \) and \( c \) are positive integers. Find the value of \( a + b + c \).
2 replies
Foxellar
Today at 4:52 AM
Foxellar
Today at 5:43 AM
Volume of right parallelepiped with minimal surface area
WakeUp   2
N Feb 2, 2011 by m.candales
Source: Romanian JBMO TST 2001
Determine a right parallelepiped with minimal area, if its volume is strictly greater than $1000$, and the lengths of it sides are integer numbers.
2 replies
WakeUp
Jan 19, 2011
m.candales
Feb 2, 2011
Volume of right parallelepiped with minimal surface area
G H J
G H BBookmark kLocked kLocked NReply
Source: Romanian JBMO TST 2001
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WakeUp
1347 posts
#1 • 2 Y
Y by Adventure10, Mango247
Determine a right parallelepiped with minimal area, if its volume is strictly greater than $1000$, and the lengths of it sides are integer numbers.
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mszew
1046 posts
#2 • 2 Y
Y by Adventure10, Mango247
Volume: $V=abc\ge 1001$
half Area $A_2=\frac{A}{2}=ab+bc+ca$

WLOG: $a \le b \le c$

$a=b=10$ $c=11$, $V=1100$, $A_2=320$ Then $a \le 10$

If $a\le 3$ then $334 \le bc \le A_2$
then $a \in \{4,5,6,7,8,9,10\}$

Notice that $\frac{1001}{a} \le bc$
and $A_2=a(b+c)+bc$
Checking some cases to find $(a,b,c)=(7,12,12)$
Z K Y
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m.candales
186 posts
#3 • 2 Y
Y by Adventure10, Mango247
The right answer is $(8,9,14)$
Z K Y
N Quick Reply
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