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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.

WOOT early bird pricing is in effect, don’t miss out! If you took MathWOOT Level 2 last year, no worries, it is all new problems this year! Our Worldwide Online Olympiad Training program is for high school level competitors. AoPS designed these courses to help our top students get the deep focus they need to succeed in their specific competition goals. Check out the details at this link for all our WOOT programs in math, computer science, chemistry, and physics.

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[list][*]April 3rd (Webinar), 4pm PT/7:00pm ET, Learning with AoPS: Perspectives from a Parent, Math Camp Instructor, and University Professor
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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
Apr 2, 2025
0 replies
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hard number theory
eric201291   2
N an hour ago by eric201291
Prove:There are no integers x, y, that y^2+9998587980=x^3.
2 replies
eric201291
Wednesday at 2:17 PM
eric201291
an hour ago
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Polynomials
CuriousBabu   5
N 2 hours ago by wh0nix
\[ \frac{(x+y+z)^5 - x^5 - y^5 - z^5}{(x+y)(y+z)(z+x)} = 0 \]
Find the number of real solutions.
5 replies
CuriousBabu
Apr 14, 2025
wh0nix
2 hours ago
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Inequalities
sqing   9
N 2 hours ago by sqing
Let $ a,b $ be reals such that $ a^2+ab+b^2 =3$ . Prove that
$$ \frac{4}{ 3}\geq \frac{1}{ a^2+5 }+ \frac{1}{ b^2+5 }+ab \geq -\frac{11}{4 }$$$$ \frac{13}{ 4}\geq \frac{1}{ a^2+5 }+ \frac{1}{ b^2+5 }+ab \geq -\frac{2}{3 }$$$$ \frac{3}{ 2}\geq \frac{1}{ a^4+3 }+ \frac{1}{ b^4+3 }+ab \geq -\frac{17}{6 }$$$$ \frac{19}{ 6}\geq \frac{1}{ a^4+3 }+ \frac{1}{ b^4+3 }-ab \geq -\frac{1}{2}$$Let $ a,b $ be reals such that $ a^2-ab+b^2 =1 $ . Prove that
$$ \frac{3}{ 2}\geq \frac{1}{ a^2+3 }+ \frac{1}{ b^2+3 }+ab \geq \frac{4}{15 }$$$$ \frac{14}{ 15}\geq \frac{1}{ a^2+3 }+ \frac{1}{ b^2+3 }-ab \geq -\frac{1}{2 }$$$$ \frac{3}{ 2}\geq \frac{1}{ a^4+3 }+ \frac{1}{ b^4+3 }+ab \geq \frac{13}{42 }$$$$ \frac{41}{ 42}\geq \frac{1}{ a^4+3 }+ \frac{1}{ b^4+3 }-ab \geq -\frac{1}{2 }$$
9 replies
+1 w
sqing
Apr 16, 2025
sqing
2 hours ago
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Geometry
AlexCenteno2007   4
N 2 hours ago by sunken rock
Let ABC be an isosceles triangle with AB = AC and M the midpoint of BC. Consider a point E outside the triangle such that BE = BM and CE perpendicular to AB. The point of intersection of the perpendicular bisector of segment EB with the circumcircle of triangle AMB, which is on the same side as A with respect to BE, is point F. Show that angle FME = 90°
4 replies
AlexCenteno2007
Yesterday at 3:49 AM
sunken rock
2 hours ago
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2017 preRMO p25, areas inside a rectangle
parmenides51   1
N Jul 22, 2020 by cowboy_37
Let $ABCD$ be a rectangle and let $E$ and $F$ be points on $CD$ and $BC$ respectively such that area $(ADE) = 16$, area $(CEF) = 9$ and area $(ABF) = 25$. What is the area of triangle $AEF$ ?
1 reply
parmenides51
Aug 20, 2019
cowboy_37
Jul 22, 2020
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2017 preRMO p25, areas inside a rectangle
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Reveal topic tags
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rectangle
areas
area of a triangle
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parmenides51
30629 posts
parmenides51
#1 h  Aug 20, 2019, 1:03 PM • 1 Y
Y by Adventure10
Let $ABCD$ be a rectangle and let $E$ and $F$ be points on $CD$ and $BC$ respectively such that area $(ADE) = 16$, area $(CEF) = 9$ and area $(ABF) = 25$. What is the area of triangle $AEF$ ?
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cowboy_37
13 posts
cowboy_37
#2 h  Jul 22, 2020, 6:04 PM • 1 Y
Y by Mango247
Let us take DE = a , EC = b , BF = c , FC = d
now a(c+d) = 32 , (a+b)c = 50
now let the area of the rectangle = (a+b)(c+d) = m
now mac = 32×50 and 2(16+9+25) - m = ac
From these two eq. we get (m-80)(m-20)=0
So , clearly m = 80
Therefore required answer is 80 - 50 = 30
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