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k a April Highlights and 2025 AoPS Online Class Information
jlacosta   0
Apr 2, 2025
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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0 replies
jlacosta
Apr 2, 2025
0 replies
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max no of different from {1,2,...,2013} so that no 2 have difference equal to 17
parmenides51   6
N 8 minutes ago by mathbetter
Source: JBMO Shortlist 2013 C1
Find the maximum number of different integers that can be selected from the set $ \{1,2,...,2013\}$ so that no two exist that their difference equals to $17$.
6 replies
parmenides51
Apr 24, 2019
mathbetter
8 minutes ago
J
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Brute force in diophantine equation?
primemystic   1
N 13 minutes ago by primemystic
As the title. Are there any method like brute force in solving diophantine equation? Thanks!
1 reply
primemystic
an hour ago
primemystic
13 minutes ago
J
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2 var inequalities
sqing   1
N 18 minutes ago by sqing
Source: Own
Let $ a,b\geq 0 $ and $ a^2+b^2=1 . $ Prove that
$$ \frac{a}{ 1+ab+b }+\frac{b }{ 1+ab+a }+\frac{ab }{1+ ab } \leq 1$$$$ \frac{a}{ 1+ab+b }+\frac{b }{ 1+ab+a }+\frac{ab }{\frac{1}{2}+ ab }\leq \frac{6\sqrt 2 }{7 }- \frac{1 }{14 } $$$$ \frac{a}{ 1+ab+b }+\frac{b }{ 1+ab+a }+\frac{ab }{1+\frac{1}{2} ab }\leq \frac{6(5\sqrt 2-1) }{35 }$$
1 reply
1 viewing
sqing
21 minutes ago
sqing
18 minutes ago
J
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B.Math - geometry
mynamearzo   9
N 19 minutes ago by Apple_maths60
In a triangle $ABC$ , $D$ is a point on $BC$ such that $AD$ is the internal bisector of $\angle A$ . Now Suppose $\angle B$=$2\angle C$ and $CD=AB$ . Prove that $\angle A=72^0$.
9 replies
mynamearzo
Jun 17, 2012
Apple_maths60
19 minutes ago
J
No more topics!
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BG _|_ DF wanted in a square
parmenides51   2
N Aug 24, 2021 by rafaello
Source: 2013 Argentina Cono SUr Training List p13
Let $ABCD$ be a square and $E$ be a point on the side $AB$ . Lines $DE$ , $BC$ meet at $F,$ and lines $CE$ , $AF$ meet at $G$. Prove that the lines $BG$ and $DF$ are perpendicular.
2 replies
parmenides51
Aug 23, 2021
rafaello
Aug 24, 2021
J
BG _|_ DF wanted in a square
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Reveal topic tags
geometry
square
perpendicular
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G H BBookmark kLocked kLocked NReply
Source: 2013 Argentina Cono SUr Training List p13
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parmenides51
30631 posts
parmenides51
#1 h  Aug 23, 2021, 10:02 PM • 2 Y
Y by centslordm, HWenslawski
Let $ABCD$ be a square and $E$ be a point on the side $AB$ . Lines $DE$ , $BC$ meet at $F,$ and lines $CE$ , $AF$ meet at $G$. Prove that the lines $BG$ and $DF$ are perpendicular.
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problemsolver2004
10 posts
problemsolver2004
#2 h  Aug 23, 2021, 11:55 PM • 1 Y
Y by HWenslawski
This one can be solved easily by Cartesian coordinates.
Z K Y
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rafaello
1079 posts
rafaello
#4 h  Aug 24, 2021, 2:17 AM
Y by
Let $S$ be the foot from $G$ to $BC$ and $R$ foot from $G$ to $AD$.
By Ceva's theorem and some similarities,
\begin{align*} \frac{FG}{GA}\cdot \frac{AE}{EB}\cdot \frac{BC}{CF}&=-1\implies \\ \frac{FG}{GA}\cdot \frac{AE}{EB}=\frac{CF}{CB}&=\frac{CF}{CD}=\frac{FB}{EB}\implies\\ FB\cdot GA&=FG\cdot AE. \end{align*}Furthermore, \begin{align*} \frac{FB}{AE}&=\frac{FG}{GA}=\frac{GS}{GR}\implies \\ \frac{GR}{AE}&=\frac{GS}{FB}. \end{align*}Note that $$\frac{BE}{FB}=\frac{AE}{AD}=\frac{AE}{AB}\implies\frac{GR}{RA}= \frac{AB}{FB}=\frac{AE}{BE}.$$Hence, $$\frac{BS}{BE}=\frac{RA}{BE}=\frac{GS}{FB}$$and as $\measuredangle BSG=90^\circ=\measuredangle EBF$, we get that $\triangle GSB\sim\triangle FBE$. Hence, $\measuredangle EFB=\measuredangle BGS=\measuredangle EBG$, we conclude that $GB\perp EF$.
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