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k a June Highlights and 2025 AoPS Online Class Information
jlacosta   0
Jun 2, 2025
Congratulations to all the mathletes who competed at National MATHCOUNTS! If you missed the exciting Countdown Round, you can watch the video at this link. Are you interested in training for MATHCOUNTS or AMC 10 contests? How would you like to train for these math competitions in half the time? We have accelerated sections which meet twice per week instead of once starting on July 8th (7:30pm ET). These sections fill quickly so enroll today!

[list][*]MATHCOUNTS/AMC 8 Basics
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For those interested in Olympiad level training in math, computer science, physics, and chemistry, be sure to enroll in our WOOT courses before August 19th to take advantage of early bird pricing!

Summer camps are starting this 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 a transformative 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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0 replies
jlacosta
Jun 2, 2025
0 replies
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Difficult Set Problem
regiszhou   0
2 minutes ago
For $n$ different sets $A_1, A_2, \cdots, A_n$, define the function $f(A_1, A_2, \cdots, A_n)$ to be the maximum number $x$ such that for any $x$ different sets in $A$, the union of any $2$ of them (the $x$ different sets) is not in the $x$ sets. AKA, no set in the chosen $x$ different sets can be expressed as the union of two sets in the chosen $x$ different sets.

For example, $f(\{1\}, \{2\}, \{3\})=3$, but $f(\{1\}, \{2\}, \{1, 2\})=2$ (since if you choose all three, $\{1\} \cup \{2\} = \{1, 2\}$).

Now let $g(n)$ for any positive integer $n$ be the minimum number of $f(A_1, A_2, \cdots, A_n)$ for ALL the different sets $(A_1, A_2, \cdots, A_n)$. Proof that $\sqrt{2n} - 1 \le g(n) \le 2 \sqrt{n} + 1$.
0 replies
regiszhou
2 minutes ago
0 replies
J
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Inspired by SunnyEvan
sqing   1
N 12 minutes ago by sqing
Source: Own
Let $ x,y \geq 0 , \frac{x^2}{x^3+y}+\frac{y^2}{x+y^3} \geq 1 .$ Prove that
$$ (x-\frac{1}{2})^2+(y+\frac{1}{2})^2 \leq \frac{5}{2} $$$$ (x-1)^2+(y+1)^2 \leq 5 $$$$ (x-2)^2+(y+2)^2 \leq 13$$$$ (x-\frac{1}{2})^2+(y+1)^2 \leq \frac{17}{4} $$$$ (x-1)^2+(y+2)^2 \leq 10 $$$$ (x-\frac{1}{2})^2+(y+2)^2 \leq \frac{37}{4} $$
1 reply
1 viewing
sqing
Today at 5:18 AM
sqing
12 minutes ago
J
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tangent circles
m4thbl3nd3r   1
N 28 minutes ago by Lil_flip38
Let $O,H,T$ be circumcenter, orthocenter and A-HM point of triangle $ABC$. Let $AH,AT$ intersect $(O)$ at $K,N$, respectively. Let $XYZ$ be the triangle formed by $TH,BC,KN$. Prove that $(XYZ)$ is tangent to $(O).$
1 reply
m4thbl3nd3r
2 hours ago
Lil_flip38
28 minutes ago
J
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inequality
SunnyEvan   2
N 35 minutes ago by sqing
Let $ a,b > 0 ,$ such that : $ a+b \geq \frac{3(a^4+b^4)}{a^2+b^2+1}\sqrt{\frac{\frac{1}{a}+\frac{1}{b}}{a+b}}.$
Prove that: $$ \frac{3(a^2+b^2)+4}{a^4+b^4} \geq 3(a+b)-1 $$
2 replies
SunnyEvan
Today at 6:53 AM
sqing
35 minutes ago
J
No more topics!
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DC + DL <= BC wanted , <C > 60^o, CA=CB, angle bisector, circumcircle related
parmenides51   1
N Aug 11, 2020 by kiyoras_2001
Source: 2018 Belarus TST 2.2
Given the isosceles triangle $ABC$ ($CA = CB$). The bisector of the angle $\angle B$ intersects the side $AC$ at point $L$ and the circumcircle of the triangle $ABC$ at point $D$. It is known that $\angle C > 60^o$ . Prove that $DC + DL \le BC$ .

(I.Voronovich)
1 reply
parmenides51
Jul 17, 2020
kiyoras_2001
Aug 11, 2020
J
DC + DL <= BC wanted , <C > 60^o, CA=CB, angle bisector, circumcircle related
G H J
Reveal topic tags
geometry
circumcircle
angle bisector
geometric inequality
L
G H BBookmark kLocked kLocked NReply
Source: 2018 Belarus TST 2.2
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parmenides51
30653 posts
parmenides51
#1 h  Jul 17, 2020, 9:46 PM
Y by
Given the isosceles triangle $ABC$ ($CA = CB$). The bisector of the angle $\angle B$ intersects the side $AC$ at point $L$ and the circumcircle of the triangle $ABC$ at point $D$. It is known that $\angle C > 60^o$ . Prove that $DC + DL \le BC$ .

(I.Voronovich)
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
kiyoras_2001
678 posts
kiyoras_2001
#2 h  Aug 11, 2020, 9:05 AM • 1 Y
Y by Mango247
Let $\angle A=2\theta$. Then $\theta\in(0^\circ, 30\circ)$ and we need to prove that
$$DC+DL\le BC$$$\iff$
$$DC+\frac{DC^2}{DB}\le BC$$$\iff$
$$\sin\theta+\frac{\sin^2\theta}{\sin 3\theta}\le \sin 2\theta$$$\iff$
$$\sin\theta+\sin 3\theta\le 2\cos\theta\sin 3\theta$$$\iff$
$$\sin 2\theta\le \sin 3\theta$$which is obvious.
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