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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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Perfect Square Function
Miku3D   16
N 11 minutes ago by MathLuis
Source: 2021 APMO P5
Determine all Functions $f:\mathbb{Z} \to \mathbb{Z}$ such that $f(f(a)-b)+bf(2a)$ is a perfect square for all integers $a$ and $b$.
16 replies
+1 w
Miku3D
Jun 9, 2021
MathLuis
11 minutes ago
J
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2 var inquality
sqing   4
N 14 minutes ago by sqing
Source: Own
Let $ a,b\geq 0 $ and $ a+ b+2ab=4 . $ Prove that
$$ 3(a+b-2)(2 - ab) \ge (a-1)(b-1)(a-b)$$$$ 9 (a+b-2)(3 - 2ab) \ge 2\sqrt 5(a-1)(b-1)(a-b)$$$$9(a+b-2)(6 - 5ab) \ge2\sqrt {14} (a-1)(b-1)(a-b)$$
4 replies
3 viewing
sqing
2 hours ago
sqing
14 minutes ago
J
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2 var inquality
sqing   2
N 26 minutes ago by lbh_qys
Source: Own
Let $ a,b\geq 0 $ and $ a+ b+ab=3 . $ Prove that
$$ (a^2+b^2+a+b-4)(2 - ab)\ge\sqrt 7(a-1)(b-1)(a-b)$$$$ 3 (a^2+b^2+a+b-4)(5 - 2ab)\ge 20(a-1)(b-1)(a-b)$$$$ 6(a^2+b^2-2)(5 - 2ab)\ge 35(a-1)(b-1)(a-b)$$$$ 2(a^2+b^2-2)(3 - ab)\ge 7(a-1)(b-1)(a-b)$$
2 replies
sqing
an hour ago
lbh_qys
26 minutes ago
J
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Inspired by JK1603JK
sqing   2
N 27 minutes ago by sqing
Source: Own
Let $ a,b,c\geq 0 $ and $ ab+bc+ca=3. $ Prove that
$$ (a+b+c-3)(12-5abc)\ge 2(a-b)(b-c)(c-a)$$$$6(a+b+c-3)(5-2abc)\ge 5(a-b)(b-c)(c-a)$$$$2(a+b+c-3)(9-5abc)\ge 3(a-b)(b-c)(c-a)$$$$3(a+b+c-3)(14-5abc)\ge 7(a-b)(b-c)(c-a)$$
2 replies
sqing
Yesterday at 2:11 PM
sqing
27 minutes ago
J
No more topics!
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Mn=ae+af
Amir.S   5
N Jan 28, 2012 by Luis González
Source: 17-th Iranian Mathematical Olympiad 1999/2000
Circles $ C_1$ and $ C_2$ with centers at $ O_1$ and $ O_2$ respectively meet at points $ A$ and $ B$. The radii $ O_1B$ and $ O_2B$ meet $ C_1$ and $ C_2$ at $ F$ and$ E$. The line through $ B$ parallel to $ EF$ intersects $ C_1$ again at $ M$ and $ C_2$ again at $ N$. Prove that $ MN = AE + AF$.
5 replies
Amir.S
Dec 14, 2005
Luis González
Jan 28, 2012
J
Mn=ae+af
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Reveal topic tags
geometry
rectangle
trapezoid
geometry proposed
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G H BBookmark kLocked kLocked NReply
Source: 17-th Iranian Mathematical Olympiad 1999/2000
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Amir.S
786 posts
Amir.S
#1 h  Dec 14, 2005, 12:42 PM • 2 Y
Y by Adventure10, Mango247
Circles $ C_1$ and $ C_2$ with centers at $ O_1$ and $ O_2$ respectively meet at points $ A$ and $ B$. The radii $ O_1B$ and $ O_2B$ meet $ C_1$ and $ C_2$ at $ F$ and$ E$. The line through $ B$ parallel to $ EF$ intersects $ C_1$ again at $ M$ and $ C_2$ again at $ N$. Prove that $ MN = AE + AF$.
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Andreas
578 posts
Andreas
#2 h  Dec 14, 2005, 3:12 PM • 2 Y
Y by Adventure10, Mango247
$\angle BAE = 90^o = \angle BAF$.

$AB$ is in common between $\Delta BAF$ and $\Delta BAE$, so $A \in FE$.

Now $\angle AEB = \angle EBN$ and $\angle BNE = \angle BAE = 90^o$.

For ASA $\Delta BAE \equiv \Delta BNE$ $\Longrightarrow$ $AE = BN$.

Also $\Delta BAF \equiv \Delta BMF$ $\Longrightarrow$ $AF = BM$.

Adding this $MN = AF + AE$.
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Mashimaru
340 posts
Mashimaru
#3 h  Feb 11, 2009, 5:07 PM • 1 Y
Y by Adventure10
Since $ \widehat{BAF}=\widehat{BAE}=\widehat{ABM}=\widehat{ABN}$ we obtain that $ ABMF$ and $ ABNE$ are rectangles, i.e., $ AF=MB, AE=NB$. Thus: $ MN=MB+BN=AE+AF$ and we are done.
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Mashimaru
340 posts
Mashimaru
#4 h  Feb 11, 2009, 6:15 PM • 2 Y
Y by Adventure10, Mango247
I think the problem has been stated wrong. It should have been $ O_1B \cap (O_2) = \{B,E\}$ and $ O_2B \cap (O_1) = \{B,F\}$
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Amir Hossein
5452 posts
Amir Hossein
#5 h  Jan 28, 2012, 8:11 PM • 2 Y
Y by Adventure10, Mango247
Please someone post a solution to this problem (to the correct version). Thanks.
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Luis González
4147 posts
Luis González
#6 h  Jan 28, 2012, 9:30 PM • 2 Y
Y by Amir Hossein, Adventure10
Isosceles triangles $\triangle BFO_1$ and $\triangle BEO_2$ with apices $O_1,O_2$ are clearly similar $\Longrightarrow$ $\angle FO_1E=\angle EO_2F$ $\Longrightarrow$ $EFO_1O_2$ is cyclic $\Longrightarrow$ $\angle BFE= \angle BO_1O_2.$ But $\angle BO_1O_2=\frac{_1}{^2}\angle BO_1A=\angle BMA,$ thus $\angle BMA=\angle BFE$ $\Longrightarrow$ $AM \parallel BF,$ i.e. $ABFM$ is an isosceles trapezoid with congruent diagonals $AF=BM.$ Similarly, $AE=BN$ and the conclusion follows.
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