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k a March Highlights and 2025 AoPS Online Class Information
jlacosta   0
Mar 2, 2025
March is the month for State MATHCOUNTS competitions! Kudos to everyone who participated in their local chapter competitions and best of luck to all going to State! Join us on March 11th for a Math Jam devoted to our favorite Chapter competition problems! Are you interested in training for MATHCOUNTS? Be sure to check out our AMC 8/MATHCOUNTS Basics and Advanced courses.

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0 replies
jlacosta
Mar 2, 2025
0 replies
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2025 BAMO Problem D/2
BR1F1SZ   4
N 32 minutes ago by cosinesine
Let $\mathcal{S}$ be a finite, nonempty set of points in the plane such that, for every point $A$ in $\mathcal{S}$, there exist points $B,C$ in $\mathcal{S}$ (distinct from $A$) such that $\angle BAC = 125^\circ$. What is the smallest possible number of points in $\mathcal{S}$?
4 replies
BR1F1SZ
5 hours ago
cosinesine
32 minutes ago
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prove that a_50 + b_50 > 20
kamatadu   7
N an hour ago by Marcus_Zhang
Source: Canada Training Camp
The sequences $a_n$ and $b_n$ are such that, for every positive integer $n$,
\[ a_n > 0,\qquad\ b_n>0,\qquad\ a_{n+1}=a_n+\dfrac{1}{b_n},\qquad\ b_{n+1} = b_n+\dfrac{1}{a_n}. \]Prove that $a_{50} + b_{50} > 20$.
7 replies
kamatadu
Dec 30, 2023
Marcus_Zhang
an hour ago
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Reflecting reflections
straight   3
N an hour ago by Ianis
Given $\triangle ABC$ and centroid $G$. $A'$ is the reflection of $A$ over $G$, similarily define $B'$ and $C'$.
1) Prove that if we reflect $A$ over $B'$, it is the same as reflecting $A'$ over $C$

Now, do same for reflecting $C$ over $B'$ and $C'$ over $A$, $B$ over $C'$ and $B'$ over $A$.
2) Is the big triangle similar to $\triangle ABC$?
3 replies
straight
Sep 10, 2024
Ianis
an hour ago
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An isosceles triangle is given, prove a right angle
geometry6   77
N 2 hours ago by bjump
Source: ISL 2020 G1
Let $ABC$ be an isosceles triangle with $BC=CA$, and let $D$ be a point inside side $AB$ such that $AD< DB$. Let $P$ and $Q$ be two points inside sides $BC$ and $CA$, respectively, such that $\angle DPB = \angle DQA = 90^{\circ}$. Let the perpendicular bisector of $PQ$ meet line segment $CQ$ at $E$, and let the circumcircles of triangles $ABC$ and $CPQ$ meet again at point $F$, different from $C$.
Suppose that $P$, $E$, $F$ are collinear. Prove that $\angle ACB = 90^{\circ}$.
77 replies
geometry6
Jul 20, 2021
bjump
2 hours ago
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Inequalities
sqing   30
N Today at 1:51 PM by JetFire008
Let $ a,b>0 $ and $ \frac{1}{a}+\frac{1}{b}=1. $ Prove that
$$(a^2-a+1)(b^2-b+1) \geq 9$$$$ (a^2-a+b+1)(b^2-b+a+1) \geq 25$$Let $ a,b>0 $ and $ \frac{1}{a}+\frac{1}{b}=\frac{2}{3}. $ Prove that
$$(a+8)(a^2-a+b+2)(b^2-b+5)\geq1331$$$$(a+10)(a^2-a+b+4)(b^2-b+7)\geq2197$$
30 replies
sqing
Mar 10, 2025
JetFire008
Today at 1:51 PM
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Inequalities
sqing   12
N Today at 12:59 PM by sqing
Let $ a,b,c\geq 0 $ and $a+b+c=1$. Prove that$$a^3b+b^3c+c^3a+\frac{473}{256}abc\le\frac{27}{256}$$Equality holds when $ a=b=c=\frac{1}{3} $ or $ a=0,b=\frac{3}{4},c=\frac{1}{4} $ or $ a=\frac{1}{4} ,b=0,c=\frac{3}{4} $
or $ a=\frac{3}{4} ,b=\frac{1}{4},c=0. $
12 replies
sqing
Mar 22, 2025
sqing
Today at 12:59 PM
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a+b+c=3 ine
jokehim   5
N Today at 9:14 AM by LearnMath_105
Problem. Given non-negative real numbers $a,b,c$ satisfying $a+b+c=3.$ Prove that $$\color{black}{\frac{a\left(b+c\right)}{bc+3}+\frac{b\left(c+a\right)}{ca+3}+\frac{c\left(a+b\right)}{ab+3}\le \frac{3}{2}.}$$Proposed by Phan Ngoc Chau
5 replies
jokehim
Mar 18, 2025
LearnMath_105
Today at 9:14 AM
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Inequalities
sqing   4
N Today at 4:18 AM by sqing
Let $ a,b> 0$ and $ a+b=1 . $ Prove that
$$ \frac{1}{a}+\frac{1}{b}\geq \frac{4+\frac{k}{4096}}{1+ ka^7b^7}$$Where $\frac{8192}{3}\geq k>0 .$
$$ \frac{1}{a}+\frac{1}{b}\geq \frac{\frac{14}{3}}{1+ \frac{8192}{3}a^7b^7}$$
4 replies
sqing
Mar 22, 2025
sqing
Today at 4:18 AM
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Inequalities
lgx57   1
N Yesterday at 2:52 PM by sqing
Let $a,b,c>0$,$\frac{a^2+b^2+c^2}{ab+bc+ca}=2$, find the minimum of

$$\frac{a^3+b^3+c^3}{abc}$$
1 reply
lgx57
Yesterday at 2:25 PM
sqing
Yesterday at 2:52 PM
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A hard inequality
JK1603JK   2
N Yesterday at 2:25 AM by sqing
Let a,b,c\ge 0: a+b+c=3. Prove \frac{1}{abc}+\frac{12}{a^2b+b^2c+c^2a}\ge 5.
2 replies
JK1603JK
Yesterday at 1:40 AM
sqing
Yesterday at 2:25 AM
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Inequalities from SXTX
sqing   11
N Mar 22, 2025 by byron-aj-tom
T702. Let $ a,b,c>0 $ and $ a+2b+3c=\sqrt{13}. $ Prove that $$ \sqrt{a^2+1} +2\sqrt{b^2+1} +3\sqrt{c^2+1} \geq 7$$S
T703. Let $ a,b $ be real numbers such that $ a+b\neq 0. $. Find the minimum of $ a^2+b^2+(\frac{1-ab}{a+b} )^2.$
T704. Let $ a,b,c>0 $ and $ a+b+c=3. $ Prove that $$ \frac{a^2+7}{(c+a)(a+b)} + \frac{b^2+7}{(a+b)(b+c)} +\frac{c^2+7}{(b+c)(c+a)} \geq 6$$S
11 replies
sqing
Feb 18, 2025
byron-aj-tom
Mar 22, 2025
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a^{2000}+b^{2000}=a^{1998}+b^{1998} (Greece Junior 1999 p1)
parmenides51   2
N Mar 22, 2025 by ali123456
Show that if $a,b$ are positive real numbers such that $a^{2000}+b^{2000}=a^{1998}+b^{1998}$ then $a^2+b^2 \le 2$.
2 replies
parmenides51
Mar 17, 2020
ali123456
Mar 22, 2025
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An inequality
jokehim   3
N Mar 22, 2025 by Indpsolver
Let $a,b,c \in \mathbb{R}: a+b+c=3$ then prove $$\color{black}{\frac{a^2}{a^{2}-2a+3}+\frac{b^2}{b^{2}-2b+3}+\frac{c^2}{c^{2}-2c+3}\ge \frac{3}{2}.}$$
3 replies
jokehim
Mar 21, 2025
Indpsolver
Mar 22, 2025
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Inequalities
sqing   0
Mar 21, 2025
Let $ a,b,c\geq 0 $ and $ a+b+c\geq 2+abc . $ Prove that
$$a^2+b^2+c^2- \frac{2}{5}abc-\frac{1}{2}a^2b^2c^2\geq \frac{48}{25}$$$$a^2+b^2+c^2- \frac{3}{5}abc-\frac{1}{2}a^2b^2c^2\geq \frac{91}{50}$$$$a^2+b^2+c^2- \frac{4}{5}abc-\frac{1}{2}a^2b^2c^2\geq \frac{42}{25}$$$$a^2+b^2+c^2- \frac{8}{5}abc-\frac{1}{2}a^2b^2c^2\geq \frac{3(7\sqrt{21}-27)}{25}$$$$a^2+b^2+c^2- \frac{9}{5}abc-\frac{1}{2}a^2b^2c^2\geq \frac{8}{261+41 \sqrt{41}}$$
0 replies
sqing
Mar 21, 2025
0 replies
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J
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Prove hard inequality
tellmegoodbye   2
N Feb 10, 2013 by sqing
Give $A,B,C$ are angle of one triangle. Prove
\[\frac{\cos B\cos C}{\cos A}+\frac{\cos A\cos C}{\cos B}+\frac{\cos A\cos B}{\cos C}\ge 2\left( {{\cos }^{2}}A+{{\cos }^{2}}B+{{\cos }^{2}}C \right)\]
2 replies
tellmegoodbye
Jan 9, 2013
sqing
Feb 10, 2013
J
Prove hard inequality
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Reveal topic tags
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tellmegoodbye
70 posts
tellmegoodbye
#1 h  Jan 9, 2013, 11:23 AM • 2 Y
Y by Adventure10, Mango247
Give $A,B,C$ are angle of one triangle. Prove
\[\frac{\cos B\cos C}{\cos A}+\frac{\cos A\cos C}{\cos B}+\frac{\cos A\cos B}{\cos C}\ge 2\left( {{\cos }^{2}}A+{{\cos }^{2}}B+{{\cos }^{2}}C \right)\]
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boywholived
632 posts
boywholived
#2 h  Jan 9, 2013, 12:54 PM • 2 Y
Y by Adventure10, Mango247
tellmegoodbye wrote:
Give $A,B,C$ are angle of one triangle. Prove
\[\frac{\cos B\cos C}{\cos A}+\frac{\cos A\cos C}{\cos B}+\frac{\cos A\cos B}{\cos C}\ge 2\left( {{\cos }^{2}}A+{{\cos }^{2}}B+{{\cos }^{2}}C \right)\]
If the triangle is obtuse angled the inequality is not valid.
you can verify it by putting one among the three cosine -ve and the rest two +ve.
You get some -ve >positive which is a contradiction.
I think there should be some additional constraint, just like the triangle is acute angled.
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sqing
41217 posts
sqing
#3 h  Feb 10, 2013, 3:45 AM • 2 Y
Y by Adventure10, Mango247
In an acute triangle $ABC$ , Prove that \[\frac{cosA}{cosBcosC}+\frac{cosB}{cosCcosA}+\frac{cosC}{cosAcosB}\ge4(\frac{cosBcosC}{cosA}+\frac{cosCcosA}{cosB}+\frac{cosAcosB}{cosC})\]\[\ge8(cos^2A+cos^2B+cos^2C).\]
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&p=2782742
In an acute $\triangle ABC$ ,have\[\frac{cosBcosC}{cosA}+\frac{cosCcosA}{cosB}+\frac{cosAcosB}{cosC}\ge \frac{3}{2}\].
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=525540&p=2975648&hilit=Triangle+inequality#p2975648
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&p=3028455
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