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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
Number Theory
Foxellar   0
27 minutes ago
It is known that for all positive integers $k$,
\[
1^2 + 2^2 + 3^2 + \ldots + k^2 = \frac{k(k + 1)(2k + 1)}{6}
\]Find the smallest positive integer $k$ such that $1^2 + 2^2 + 3^2 + \ldots + k^2$ is divisible by 200.
0 replies
Foxellar
27 minutes ago
0 replies
Polynomial Minimization
ReticulatedPython   2
N 2 hours ago by lgx57
Consider the polynomial $$p(x)=x^{n+1}-x^{n}$$, where $x, n \in \mathbb{R+}.$

(a) Prove that the minimum value of $p(x)$ always occurs at $x=\frac{n}{n+1}.$
2 replies
ReticulatedPython
Yesterday at 5:07 PM
lgx57
2 hours ago
Compilation of functions problems
Saucepan_man02   0
3 hours ago
Could anyone post some handout/compilation of problems related to functions (difficulty similar to AIME/ARML/HMMT etc)?

Thanks..
0 replies
Saucepan_man02
3 hours ago
0 replies
Nice Number Theory Question (Inspired by AIME)
MathRook7817   5
N 5 hours ago by jasperE3
Let $n$ be a positive integer. Find the smallest value of $n$ such that:

$2^n + 3^n - n$ is divisible by $216$.

5 replies
MathRook7817
Today at 1:57 AM
jasperE3
5 hours ago
Inequalities
sqing   12
N 5 hours ago by sqing
Let $a,b,c> 0$ and $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1.$ Prove that
$$  (1-abc) (1-a)(1-b)(1-c)  \ge 208 $$$$ (1+abc) (1-a)(1-b)(1-c)  \le -224 $$$$(1+a^2b^2c^2) (1-a)(1-b)(1-c)  \le -5840 $$
12 replies
sqing
Jul 12, 2024
sqing
5 hours ago
A pentagon inscribed in a circle of radius √2
tom-nowy   4
N 5 hours ago by jasperE3
Can a pentagon with all rational side lengths be inscribed in a circle of radius $\sqrt{2}$ ?
4 replies
tom-nowy
Yesterday at 2:37 AM
jasperE3
5 hours ago
Geometry Proof
strongstephen   10
N 5 hours ago by martianrunner
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)
10 replies
strongstephen
Yesterday at 4:54 AM
martianrunner
5 hours ago
A Collection of Good Problems from my end
SomeonecoolLovesMaths   24
N Today at 12:46 AM by ReticulatedPython
This is a collection of good problems and my respective attempts to solve them. I would like to encourage everyone to post their solutions to these problems, if any. This will not only help others verify theirs but also perhaps bring forward a different approach to the problem. I will constantly try to update the pool of questions.

The difficulty level of these questions vary from AMC 10 to AIME. (Although the main pool of questions were prepared as a mock test for IOQM over the years)

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5
24 replies
SomeonecoolLovesMaths
May 4, 2025
ReticulatedPython
Today at 12:46 AM
n and n+100 have odd number of divisors (1995 Belarus MO Category D P2)
jasperE3   4
N Yesterday at 9:50 PM by KTYC
Find all positive integers $n$ so that both $n$ and $n + 100$ have odd numbers of divisors.
4 replies
jasperE3
Apr 6, 2021
KTYC
Yesterday at 9:50 PM
Closed form expression of 0.123456789101112....
ReticulatedPython   3
N Yesterday at 8:15 PM by ReticulatedPython
Is there a closed form expression for the decimal number $$0.123456789101112131415161718192021...$$which is defined as all the natural numbers listed in order, side by side, behind a decimal point, without commas? If so, what is it?
3 replies
ReticulatedPython
Yesterday at 8:05 PM
ReticulatedPython
Yesterday at 8:15 PM
primes and perfect squares
Bummer12345   5
N Yesterday at 8:08 PM by Shan3t
If $p$ and $q$ are primes, then can $2^p + 5^q + pq$ be a perfect square?
5 replies
Bummer12345
Monday at 5:08 PM
Shan3t
Yesterday at 8:08 PM
trapezoid
Darealzolt   1
N Yesterday at 7:38 PM by vanstraelen
Let \(ABCD\) be a trapezoid such that \(A, B, C, D\) lie on a circle with center \(O\), and side \(AB\) is parallel to side \(CD\). Diagonals \(AC\) and \(BD\) intersect at point \(M\), and \(\angle AMD = 60^\circ\). It is given that \(MO = 10\). It is also known that the difference in length between \(AB\) and \(CD\) can be expressed in the form \(m\sqrt{n}\), where \(m\) and \(n\) are positive integers and \(n\) is square-free. Compute the value of \(m + n\).
1 reply
Darealzolt
Yesterday at 2:03 AM
vanstraelen
Yesterday at 7:38 PM
Easy one
irregular22104   0
Yesterday at 5:03 PM
Given two positive integers a,b written on the board. We apply the following rule: At each step, we will add all the numbers that are the sum of the two numbers on the board so that the sum does not appear on the board. For example, if the two initial numbers are 2.5, then the numbers on the board after step 1 are 2,5,7; after step 2 are 2,5,7,9,12;...
1) With a = 3; b = 12, prove that the number 2024 cannot appear on the board.
2) With a = 2; b = 3, prove that the number 2024 can appear on the board.
0 replies
irregular22104
Yesterday at 5:03 PM
0 replies
This shouldn't be a problem 15
derekli   2
N Yesterday at 4:09 PM by aarush.rachak11
Hey guys I was practicing AIME and came across this problem which is definitely misplaced. It asks for the surface area of a plane within a cylinder which we can easily find out using a projection that is easy to find. I think this should be placed in problem 10 or below. What do you guys think?
2 replies
derekli
Yesterday at 2:15 PM
aarush.rachak11
Yesterday at 4:09 PM
Geometry Angle Chasing
Sid-darth-vater   6
N Apr 29, 2025 by sunken rock
Is there a way to do this without drawing obscure auxiliary lines? (the auxiliary lines might not be obscure I might just be calling them obscure)

For example I tried rotating triangle MBC 80 degrees around point C (so the BC line segment would now lie on segment AC) but I couldn't get any results. Any help would be appreciated!
6 replies
Sid-darth-vater
Apr 21, 2025
sunken rock
Apr 29, 2025
Geometry Angle Chasing
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Sid-darth-vater
42 posts
#1 • 1 Y
Y by Rounak_iitr
Is there a way to do this without drawing obscure auxiliary lines? (the auxiliary lines might not be obscure I might just be calling them obscure)

For example I tried rotating triangle MBC 80 degrees around point C (so the BC line segment would now lie on segment AC) but I couldn't get any results. Any help would be appreciated!
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vanstraelen
9011 posts
#3 • 2 Y
Y by teomihai, Sid-darth-vater
There is a solution, using the rule of sines.

$\triangle ACM\ :\ \frac{AC}{\sin x}=\frac{CM}{\sin 40^{\circ}}$ and $\triangle BCM\ :\ \frac{BC}{\sin(220^{\circ}-x)}=\frac{CM}{\sin 20^{\circ}}$.
$\sin x=2\sin(220^{\circ}-x)\cos 20^{\circ}$,
$\sin x=\sin(240^{\circ}-x)+\sin(200^{\circ}-x)$,
$\sin x-\sin(240^{\circ}-x)=\sin(200^{\circ}-x)$,
$2\sin(x-120^{\circ})\cos 120^{\circ}=\sin(200^{\circ}-x)$,
$-\sin(x-120^{\circ})=\sin(200^{\circ}-x)$,
$\sin(120^{\circ}-x)=\sin(200^{\circ}-x)$,
$120^{\circ}-x=180^{\circ}-(200^{\circ}-x)$,
$x=70^{\circ}$.
This post has been edited 2 times. Last edited by vanstraelen, Apr 22, 2025, 7:40 PM
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Sid-darth-vater
42 posts
#4
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ah wow, very nice! ty
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sunken rock
4392 posts
#5
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vanstraelen wrote:
There is a solution, using the rule of sines.

$\triangle ACM\ :\ \frac{AC}{\sin x}=\frac{CM}{\sin 40^{\circ}}$ and $\triangle BCM\ :\ \frac{BC}{\sin(220^{\circ}-x)}=\frac{CM}{\sin 20^{\circ}}$.
$\sin x=2\sin(220^{\circ}-x)\cos 20^{\circ}$,
$\sin x=\sin(240^{\circ}-x)+\sin(200^{\circ}-x)$,
$\sin x-\sin(240^{\circ}-x)=\sin(200^{\circ}-x)$,
$2\sin(x-120^{\circ})\cos 120^{\circ}=\sin(200^{\circ}-x)$,
$-\sin(x-120^{\circ})=\sin(200^{\circ}-x)$,
$\sin(120^{\circ}-x)=\sin(200^{\circ}-x)$,
$120^{\circ}-x=180^{\circ}-(200^{\circ}-x)$,
$x=70^{\circ}$.

If $\widehat{AMC}=70^\circ$, with $\widehat{MAC}=70^\circ$ makes $\widehat{ACM}=40^\circ$, but $M$ is supposed to be inside $\widehat{ACB}=20^\circ$! How was that possible?
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mathafou
417 posts
#6
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you misnamed your points ...
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sunken rock
4392 posts
#7
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mathafou wrote:
you misnamed your points ...

Negative, I wrongly took the angle value of $\hat C$! Now I have a nice synthetic proof for it, I shall post it later!
This post has been edited 1 time. Last edited by sunken rock, Apr 29, 2025, 2:19 PM
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sunken rock
4392 posts
#8
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My synthetic solution at https://stanfulger.blogspot.com/2025/04/aops-httpsartofproblemsolvingcomcommuni_29.html

Best regards,
sunken rock
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