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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
old problem from an open contest
Darealzolt   1
N 3 hours ago by alexheinis
Given that $a, b \in \mathbb{R}$ satisfy
\[
a + \frac{1}{a + 2015} = b - 4030 + \frac{1}{b - 2015}
\]and $|a - b| > 5000$. Determine the value of
\[
\frac{ab}{2015} - a + b.
\]
1 reply
Darealzolt
Yesterday at 1:41 AM
alexheinis
3 hours ago
A Collection of Good Problems from my end
SomeonecoolLovesMaths   12
N 3 hours ago by SomeonecoolLovesMaths
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
12 replies
SomeonecoolLovesMaths
May 4, 2025
SomeonecoolLovesMaths
3 hours ago
Geometry Proof
strongstephen   2
N 4 hours ago by greenturtle3141
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.

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)
2 replies
strongstephen
Today at 4:54 AM
greenturtle3141
4 hours ago
BABBAGE'S THEOREM EXTENSION
Mathgloggers   4
N 4 hours ago by Mathgloggers
A few days ago I came across. this interesting result is someone interested in proving this.

$\boxed{\sum_{k=1}^{p-1} \frac{1}{k} \equiv \sum_{k=p+1}^{2p-1} \frac{1}{k} \equiv \sum_{k=2p+1}^{3p-1}\frac{1}{k} \equiv.....\sum_{k=p(p-1)+1}^{p^2-1}\frac{1}{k} \equiv 0(mod p^2)}$
4 replies
Mathgloggers
Apr 29, 2025
Mathgloggers
4 hours ago
geometry
JetFire008   1
N Today at 4:23 AM by ohiorizzler1434
Given four concyclic points. For each subset of three points take the incenter. Show that the four incentres form a rectangle.
1 reply
JetFire008
Yesterday at 4:14 PM
ohiorizzler1434
Today at 4:23 AM
A pentagon inscribed in a circle of radius √2
tom-nowy   2
N Today at 4:20 AM by ohiorizzler1434
Can a pentagon with all rational side lengths be inscribed in a circle of radius $\sqrt{2}$ ?
2 replies
tom-nowy
Today at 2:37 AM
ohiorizzler1434
Today at 4:20 AM
Inequalities
sqing   8
N Today at 3:12 AM by sqing
Let $ a,b,c>0 $ and $ a+b\leq 16abc. $ Prove that
$$ a+b+kc^3\geq\sqrt[4]{\frac{4k} {27}}$$$$ a+b+kc^4\geq\frac{5} {8}\sqrt[5]{\frac{k} {2}}$$Where $ k>0. $
$$ a+b+3c^3\geq\sqrt{\frac{2} {3}}$$$$ a+b+2c^4\geq \frac{5} {8}$$
8 replies
sqing
Sunday at 12:46 PM
sqing
Today at 3:12 AM
trapezoid
Darealzolt   0
Today at 2:03 AM
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 \(m + n\).
0 replies
Darealzolt
Today at 2:03 AM
0 replies
Inequalities
sqing   2
N Today at 1:47 AM by sqing
Let $ a,b,c\geq 0 ,   2a +ab + 12a bc \geq 8. $ Prove that
$$  a+  (b+c)(a+1)+\frac{4}{5}  bc \geq 4$$$$  a+  (b+c)(a+0.9996)+ 0.77  bc \geq 4$$
2 replies
sqing
May 4, 2025
sqing
Today at 1:47 AM
anyone who can help me this 2 problems?
auroracliang   2
N Yesterday at 11:51 PM by ReticulatedPython
1. Let r be the radius of the largest circle which is tangent to the parabola y=x^2 at x=0 and which lies entirely on or inside (that is, above) the parabola, find r.

2. Counting number n has the following property,: if we take any 50 different numbers from 1,2,3, ... n, there always are two numbers with the difference of 7. what is the largest value among all value of n?


thanks a lot
2 replies
auroracliang
Nov 3, 2024
ReticulatedPython
Yesterday at 11:51 PM
What conic section is this? Is this even a conic section?
invincibleee   2
N Yesterday at 11:48 PM by ReticulatedPython
IMAGE

The points in this are given by
P = (sin2A, sin4A)∀A [0,2π]
Is this a conic section? what is this?
2 replies
invincibleee
Nov 15, 2024
ReticulatedPython
Yesterday at 11:48 PM
Spheres, ellipses, and cones
ReticulatedPython   0
Yesterday at 11:38 PM
A sphere is inscribed inside a cone with base radius $1$ and height $2.$ Another sphere of radius $r$ is internally tangent to the lateral surface of the cone, but does not intersect the larger inscribed sphere. A plane is tangent to both of these spheres, and passes through the inside of the cone. The intersection of the plane and the cone forms an ellipse. Find the maximum area of this ellipse.
0 replies
ReticulatedPython
Yesterday at 11:38 PM
0 replies
Looking for users and developers
derekli   13
N Yesterday at 11:31 PM by DreamineYT
Guys I've been working on a web app that lets you grind high school lvl math. There's AMCs, AIME, BMT, HMMT, SMT etc. Also, it's infinite practice so you can keep grinding without worrying about finding new problems. Please consider helping me out by testing and also consider joining our developer team! :P :blush:

Link: https://stellarlearning.app/competitive
13 replies
derekli
May 4, 2025
DreamineYT
Yesterday at 11:31 PM
trigonometric functions
VivaanKam   12
N Yesterday at 11:06 PM by aok
Hi could someone explain the basic trigonometric functions to me like sin, cos, tan etc.
Thank you!
12 replies
VivaanKam
Apr 29, 2025
aok
Yesterday at 11:06 PM
Real variables inequality
JK1603JK   1
N Apr 3, 2025 by lbh_qys
Let a,b,c\in R then prove that \frac{15}{2}\cdot\frac{a^2+b^2+c^2}{(a+b+c)^2}+\frac{ab}{a^2+b^2}+\frac{bc}{b^2+c^2}+\frac{ca}{c^2+a^2}\ge 4
1 reply
JK1603JK
Apr 3, 2025
lbh_qys
Apr 3, 2025
Real variables inequality
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JK1603JK
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Let a,b,c\in R then prove that \frac{15}{2}\cdot\frac{a^2+b^2+c^2}{(a+b+c)^2}+\frac{ab}{a^2+b^2}+\frac{bc}{b^2+c^2}+\frac{ca}{c^2+a^2}\ge 4
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lbh_qys
558 posts
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JK1603JK wrote:
Let $a,b,c\in \mathbb{R}$ and $a+b+c \neq 0$ then prove that $$\frac{15}{2}\cdot\frac{a^2+b^2+c^2}{(a+b+c)^2}+\frac{ab}{a^2+b^2}+\frac{bc}{b^2+c^2}+\frac{ca}{c^2+a^2}\ge 4$$

WLOG \(a+b+c=5\). Then the original inequality is equivalent to
\[
15\sum \frac{a^2}{5^2}+\sum \frac{2ab}{a^2+b^2}=15\sum \frac{a^2}{\left(\sum a\right)^2}+\sum \frac{(a+b)^2}{a^2+b^2}-3\geq 8,
\]that is,
\[
3\sum \frac{a^2}{5}+\sum \frac{(a+b)^2}{a^2+b^2}\geq 11.
\]
Let
\[
x=2-a,\quad y=2-b,\quad z=2-c,
\]so that
\[
x+y+z=1.
\]Define
\[
q=xy+yz+zx \quad \text{and} \quad r=xyz.
\]Then the inequality is equivalent to proving
\[
3\sum \frac{(2-x)^2}{5}+\sum \frac{(4-x-y)^2}{(2-x)^2+(2-y)^2}\geq 11.
\]
According to the Cauchy--Schwarz inequality,
\[
\sum \frac{(4-x-y)^2}{(2-x)^2+(2-y)^2}\geq \frac{\left(\sum (4-x-y)(6-z)\right)^2}{\sum \left[(2-x)^2+(2-y)^2\right](6-z)^2}.
\]
Since
\[
\sum (2-x)^2=9-2q,
\]\[
\sum (4-x-y)(6-z)=56+2q,
\]\[
\sum \left[(2-x)^2+(2-y)^2\right](6-z)^2=560-80q+2q^2+44r,
\]it suffices to prove
\[
\frac{3(9-2q)}{5}+\frac{(56+2q)^2}{560-80q+2q^2+44r}\geq 11.
\]
Since
\[
3(x+y+z)xyz\leq (xy+yz+zx)^2,
\]we have
\[
r\leq \frac{q^2}{3},
\]and consequently,
\[
\frac{(56+2q)^2}{560-80q+2q^2+44r}\geq \frac{(56+2q)^2}{560-80q+2q^2+44\cdot \frac{q^2}{3}}.
\]Thus, it suffices to prove that
\[
\frac{3(9-2q)}{5}+\frac{(56+2q)^2}{560-80q+2q^2+44\cdot \frac{q^2}{3}}\geq 11.
\]
This simplifies to
\[
(1-3q)q^2\geq 0,
\]which is evidently true. Therefore, the original inequality holds.
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