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k a April Highlights and 2025 AoPS Online Class Information
jlacosta   0
Wednesday at 3:18 PM
Spring is in full swing and summer is right around the corner, what are your plans? At AoPS Online our schedule has new classes starting now through July, so be sure to keep your skills sharp and be prepared for the Fall school year! Check out the schedule of upcoming classes below.

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[*]April 22nd (Webinar), 4:00pm PT/7:00pm ET, Competitive Programming at AoPS (USACO).[/list]
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0 replies
jlacosta
Wednesday at 3:18 PM
0 replies
Inspired by JK1603JK
sqing   10
N 13 minutes ago by SunnyEvan
Source: Own
Let $ a,b,c\geq 0 $ and $ab+bc+ca=1.$ Prove that$$\frac{abc-2}{abc-1}\ge \frac{4(a^2b+b^2c+c^2a)}{a^3b+b^3c+c^3a+1} $$
10 replies
+1 w
sqing
Today at 3:31 AM
SunnyEvan
13 minutes ago
Can Euclid solve this geo ?
S.Ragnork1729   31
N 33 minutes ago by PeterZeus
Source: INMO 2025 P3
Euclid has a tool called splitter which can only do the following two types of operations :
• Given three non-collinear marked points $X,Y,Z$ it can draw the line which forms the interior angle bisector of $\angle{XYZ}$.
• It can mark the intersection point of two previously drawn non-parallel lines .
Suppose Euclid is only given three non-collinear marked points $A,B,C$ in the plane . Prove that Euclid can use the splitter several times to draw the centre of circle passing through $A,B$ and $C$.

Proposed by Shankhadeep Ghosh
31 replies
S.Ragnork1729
Jan 19, 2025
PeterZeus
33 minutes ago
Answer is Year
solasky   2
N 43 minutes ago by AshAuktober
Source: Japan MO Preliminary 2021/1
For all relatively prime positive integers $m$, $n$ satisfying $m + n = 90$, what is the maximum possible value of $mn$?
2 replies
solasky
Jun 15, 2024
AshAuktober
43 minutes ago
series and factorials?
jenishmalla   8
N an hour ago by Maximilian113
Source: 2025 Nepal ptst p4 of 4
Find all pairs of positive integers \( n \) and \( x \) such that
\[
1^n + 2^n + 3^n + \cdots + n^n = x!
\]
(Petko Lazarov, Bulgaria)
8 replies
jenishmalla
Mar 15, 2025
Maximilian113
an hour ago
New geometry problem
titaniumfalcon   2
N 3 hours ago by fruitmonster97
Post any solutions you have, with explanation or proof if possible, good luck!
2 replies
titaniumfalcon
Yesterday at 10:40 PM
fruitmonster97
3 hours ago
.problem.
Cobedangiu   2
N 3 hours ago by Lankou
Find the integer coefficients after expanding Newton's binomial:
$$(\frac{3}{2}-\frac{2}{3}x^2)^n (n \in Z)$$
2 replies
Cobedangiu
Today at 6:20 AM
Lankou
3 hours ago
Inequalities
sqing   23
N 3 hours ago by sqing
Let $ a,b,c\geq 0 $ and $a+b+c=1$. Prove that
$$a^2+b^2+ ab +21abc\leq\frac{512}{441}$$Equality holds when $a=b=\frac{38}{21},c=\frac{5}{214}.$
$$a^2+b^2+ ab +19abc\leq\frac{10648}{9747}$$Equality holds when $a=b=\frac{22}{57},c=\frac{13}{57}.$
$$a^2+b^2+ ab +22abc\leq\frac{15625}{13068}$$Equality holds when $a=b=\frac{25}{66},c=\frac{8}{33}.$
23 replies
sqing
Mar 26, 2025
sqing
3 hours ago
Inequalities
sqing   3
N 3 hours ago by sqing
Let $ a, b,c\geq 0 $ and $ 2a+3b+ 4c=11.$ Prove that
$$a+ab+abc\leq\frac{49}{6}$$Let $ a, b,c\geq 0 $ and $ 2a+3b+ 4c=10.$ Prove that
$$a+ab+abc\leq\frac{169}{24}$$Let $ a, b,c\geq 0 $ and $ 2a+3b+ 4c=14.$ Prove that
$$a+ab+abc\leq\frac{63+5\sqrt 5}{6}$$Let $ a, b,c\geq 0 $ and $ 2a+3b+ 4c=32.$ Prove that
$$a+ab+abc\leq48+\frac{64\sqrt{2}}{3}$$
3 replies
1 viewing
sqing
Apr 1, 2025
sqing
3 hours ago
Puzzling p&c question
Hunter87   0
6 hours ago
There are 9 numbered tickets (distinct) to be distributed (1-9), and 7 contestants. Each contestant must get atleast one ticket. Two PARTICULAR contestants (say, A and B) can never get tickets with adjacent numbers and any contestant can get more than 1 ticket. All tickets are to be distributed.
In how many ways can this be done?
0 replies
Hunter87
6 hours ago
0 replies
inequalities
son2007vn   0
Today at 7:13 AM
Prove that :
\[
\frac{2a^2}{(b-c)^2} + \frac{2b^2}{(a-c)^2} + \frac{2c^2}{(b-a)^2} \geq 2 + \sum \left| \frac{(a+b-c)(b+c-a)}{(c-a)^2} \right|
\]
0 replies
son2007vn
Today at 7:13 AM
0 replies
Geo Mock #5
Bluesoul   3
N Today at 7:01 AM by rhydon516
Consider triangle $ABC$ with $AB=13, BC=14, AC=15$. Denote the orthocenter of $\triangle{ABC}$ as $H$, the intersection of $(BHC)$ and $AC$ as $P\neq C$. Compute the length of $AP$.
3 replies
Bluesoul
Apr 1, 2025
rhydon516
Today at 7:01 AM
Inequalities
sqing   0
Today at 3:52 AM
Let $ a,b,c> 0 $ and $  \frac{a}{a^2+ab+c}+\frac{b}{b^2+bc+a}+\frac{c}{c^2+ca+b} \geq 1$. Prove that
$$  a+b+c\leq 3    $$
0 replies
sqing
Today at 3:52 AM
0 replies
Geo Mock #7
Bluesoul   1
N Yesterday at 8:20 PM by vanstraelen
Consider $\triangle{ABC}$ with $\angle{A}=90^{\circ}$ and $AB=10$. Let $D$ be a point on $AB$ such that $BD=6$. Suppose that the angle bisector of $\angle{C}$ is tangent to the circle with diameter $BD$ and say it intersects $AB$ at point $E$. Find the length of $BE$.
1 reply
Bluesoul
Apr 1, 2025
vanstraelen
Yesterday at 8:20 PM
Number of solutions
Ecrin_eren   3
N Yesterday at 8:00 PM by rchokler
The given equation is:

x³ + 4y³ + 2y = (2024 + 2y)(xy + 1)

The question asks for the number of integer solutions.

3 replies
Ecrin_eren
Yesterday at 11:27 AM
rchokler
Yesterday at 8:00 PM
Concyclic points
henderson   3
N Jun 8, 2016 by wu2481632
Let $ABC$ be an isosceles triangle such that $AB=AC$ and let $M$ be the midpoint of $BC$. Consider a point $D$ on the segment $BC$. Let $I$ and $J$ be the incenters of $\triangle ABD$ and $\triangle ACD,$ respectively. Prove that $M, I, J$ and $D$ are concyclic.
3 replies
henderson
Jun 8, 2016
wu2481632
Jun 8, 2016
Concyclic points
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henderson
312 posts
#1 • 3 Y
Y by wu2481632, Adventure10, Mango247
Let $ABC$ be an isosceles triangle such that $AB=AC$ and let $M$ be the midpoint of $BC$. Consider a point $D$ on the segment $BC$. Let $I$ and $J$ be the incenters of $\triangle ABD$ and $\triangle ACD,$ respectively. Prove that $M, I, J$ and $D$ are concyclic.
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Math_CYCR
431 posts
#2 • 2 Y
Y by Adventure10, Mango247
I posted this problem yesterday, also I posted a link to the general case. I erased it so that I can post it in my blog.

Here's my proof for the special case:

Assume WLOG $BD > CD$

First notice that $\angle IDJ=90$

Also $\angle BAI = \angle MAJ$ and $\angle MAI= \angle CAJ$

Now applying trig ceva in $\triangle ABM$ and $\triangle ACM$ we get:

$\frac{ \sin \angle BAI}{ \sin \angle IAM} \times \frac{ \sin \angle IMA}{ \sin \angle IMB} \times \frac{ \sin \angle IBM}{ \sin \angle IBA} =1$

$\Longrightarrow \frac{ \sin \angle BAI}{ \sin \angle IAM} = \frac{ \sin \angle IMB}{ \sin \angle IMA}$

$\frac{ \sin \angle CAJ}{ \sin \angle JAM} \times \frac{ \sin \angle AMJ}{ \sin \angle JMC} \times \frac{ \sin \angle MCJ}{ \sin \angle JCA} =1$

$\Longrightarrow \frac{ \sin \angle CAJ}{ \sin \angle JAM} = \frac{ \sin \angle JMC}{ \sin \angle AMJ}$

$\Longrightarrow \frac{ \sin \angle JMC}{ \sin \angle AMJ} = \frac{ \sin \angle IMA}{ \sin \angle IMB}$

But also $\angle JMC+ \angle AMJ=90= \angle IMA + \angle IMB$. By Two Equal Angles Lemma we get $\angle JMC= \angle IMA$ and $\angle AMJ= \angle IMB$

Therefore $\angle IMJ=90= \angle IDJ$

Done!
This post has been edited 1 time. Last edited by Math_CYCR, Sep 25, 2016, 2:32 PM
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ESCHen99
23 posts
#3 • 2 Y
Y by Adventure10, Mango247
Here: http://artofproblemsolving.com/community/q4h1218150p6080289 is a general solution, the Isosceles case leads $Q$ to be the midpoint.
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wu2481632
4233 posts
#4 • 2 Y
Y by henderson, Adventure10
Solution
This post has been edited 1 time. Last edited by wu2481632, Jun 8, 2016, 7:32 PM
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