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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.

Are you ready to level up with Olympiad training? Registration is open with early bird pricing available for our WOOT programs: MathWOOT (Levels 1 and 2), CodeWOOT, PhysicsWOOT, and ChemWOOT. What is WOOT? WOOT stands for Worldwide Online Olympiad Training and is a 7-month high school math Olympiad preparation and testing program that brings together many of the best students from around the world to learn Olympiad problem solving skills. Classes begin in September!

Do you have plans this summer? There are so many options to fit your schedule and goals whether attending a summer camp or taking online classes, it can be a great break from the routine of the school year. Check out our summer courses at AoPS Online, or if you want a math or language arts class that doesn’t have homework, but is an enriching summer experience, our AoPS Virtual Campus summer camps may be just the ticket! We are expanding our locations for our AoPS Academies across the country with 15 locations so far and new campuses opening in Saratoga CA, Johns Creek GA, and the Upper West Side NY. Check out this page for summer camp information.

Be sure to mark your calendars for the following events:
[list][*]March 5th (Wednesday), 4:30pm PT/7:30pm ET, HCSSiM Math Jam 2025. Amber Verser, Assistant Director of the Hampshire College Summer Studies in Mathematics, will host an information session about HCSSiM, a summer program for high school students.
[*]March 6th (Thursday), 4:00pm PT/7:00pm ET, Free Webinar on Math Competitions from elementary through high school. Join us for an enlightening session that demystifies the world of math competitions and helps you make informed decisions about your contest journey.
[*]March 11th (Tuesday), 4:30pm PT/7:30pm ET, 2025 MATHCOUNTS Chapter Discussion MATH JAM. AoPS instructors will discuss some of their favorite problems from the MATHCOUNTS Chapter Competition. All are welcome!
[*]March 13th (Thursday), 4:00pm PT/7:00pm ET, Free Webinar about Summer Camps at the Virtual Campus. Transform your summer into an unforgettable learning adventure! From elementary through high school, we offer dynamic summer camps featuring topics in mathematics, language arts, and competition preparation - all designed to fit your schedule and ignite your passion for learning.[/list]
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0 replies
jlacosta
Mar 2, 2025
0 replies
Inequality
jokehim   1
N 4 minutes ago by no_room_for_error
Problem. Given non-negative real numbers $a,b,c$ satisfying $a+b+c=3.$ Prove that $$\sqrt{a^2+2abc+b^2}+\sqrt{b^2+2abc+c^2}+\sqrt{c^2+2abc+a^2}\le 6.$$Proposed by Phan Ngoc Chau
1 reply
jokehim
Today at 7:34 AM
no_room_for_error
4 minutes ago
Properties of a line
Cathong0324   2
N 21 minutes ago by XAN4
If we connect a triangle's incenter and circumcenter, what are the properties of this line?
2 replies
Cathong0324
Mar 5, 2024
XAN4
21 minutes ago
rhombus, line from bisector x circumcircle points (Puerto Rico TST 2024.7)
Equinox8   1
N 34 minutes ago by pooh123
Let $I$ be the incenter of $\triangle{ABC}$. Angle bisectors $BI$ and $CI$ intersect the circumcircle of $\triangle{ABC}$ at points $S$ and $T$, respectively ($S \neq B$ and $T \neq C$). Line $ST$ intersects sides $AB$ and $AC$ at points $K$ and $L$, respectively. Prove that quadrilateral $AKIL$ is a rhombus.

Note: A rhombus is a parallelogram with all sides equal.
1 reply
Equinox8
Mar 12, 2025
pooh123
34 minutes ago
Collinear proof
Wiselady   2
N an hour ago by Wiselady
Let ABC be any triangle. Let side BC pass through point C to point D such that CD=AC. Let P be the second intersection point of the circumscribed circle ACD with the circle with BC as the diameter. Let BP and AC meet at point E and let CP and AB meet at point F. Prove that D, E, F lie on the same straight line.
IMAGE
2 replies
Wiselady
Today at 1:33 AM
Wiselady
an hour ago
Happy 3.14
JetFire008   2
N an hour ago by whwlqkd
Happy PI day!! :clap: :trampoline:
2 replies
JetFire008
an hour ago
whwlqkd
an hour ago
Inequalities
sqing   0
an hour ago
Let $a,b,c\ge \frac{1}{2}$ and $\left(\frac{1}{a}+\frac{1}{b}-\frac{1}{c}\right)\left(\frac{1}{a}-\frac{1}{b}+\frac{1}{c}\right)\le 1. $ Prove that
$$a+b+c\geq 2$$Let $a,b,c\ge \frac{1}{2}$ and $ \left(a+\frac{1}{a}+\frac{1}{b}-\frac{1}{c}\right)\left(a+\frac{1}{a}-\frac{1}{b}+\frac{1}{c}\right)\le \frac{9}{2}. $ Prove that
$$a^2+b^2+c^2\geq 1$$Let $a,b\ge \frac{1}{2}$ and $ \left( \frac{1}{a}-\frac{1}{b}+2\right)\left( \frac{1}{b}-\frac{1}{a}+2\right) \le   \frac{20}{9}. $ Prove that
$$ a+b\geq 2$$Let $a,b\ge \frac{1}{2}$ and $a^2+b^2=1. $ Prove that
$$\left(\frac{2}{a}+\frac{1}{b}-1\right)\left(\frac{2}{a}-\frac{1}{b}+1\right)\ge \frac{13}{3}$$
0 replies
sqing
an hour ago
0 replies
IOQM P2 2024
SomeonecoolLovesMaths   10
N an hour ago by BackToSchool
The number of four-digit odd numbers having digits $1,2,3,4$, each occuring exactly once, is:
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SomeonecoolLovesMaths
Sep 8, 2024
BackToSchool
an hour ago
Find a line in inequality gemetry
William_Mai   1
N 3 hours ago by William_Mai
M point is given inside a angle xOy. Draw a line d that passes through M and intersects Ox and Oy at the A và B with the smallest possible of AB.
(Note: line d not passing through O)

IMAGE
1 reply
William_Mai
Mar 9, 2025
William_Mai
3 hours ago
2017 BCSMC Round 2 #1 (BC)^2 - (AB)^2 = 1081 in right ABC
parmenides51   2
N 4 hours ago by BackToSchool
In right triangle $ABC$, $\angle ABC = 90^o$, $AC = 37$, and $(BC)^2 - (AB)^2 = 1081$. Find $AB + BC$.
2 replies
parmenides51
Jan 24, 2024
BackToSchool
4 hours ago
Inequalities
sqing   10
N 4 hours ago by sqing
Let $ a,b $ be reals such that $   a^2+b^2=1. $ Prove that
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sqing
Mar 9, 2025
sqing
4 hours ago
Inequalities from SXTX
sqing   6
N 6 hours ago by sqing
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
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sqing
Feb 18, 2025
sqing
6 hours ago
Difficult Geometry Optimization Problem.
ReticulatedPython   4
N Today at 8:00 AM by bhontu
Consider three concentric circles with radii $x$, $y$, and $z.$ Point $X$ is chosen on the circle with radius $x$, point $Y$ is chosen on the circle with radius $y$, and point $Z$ is chosen on the circle with radius $z.$ Given that $x<y<z$, find the maximum and minimum values of $$XY^2+YZ^2+XZ^2$$in terms of $x$, $y$, and $z.$
4 replies
ReticulatedPython
Yesterday at 3:09 PM
bhontu
Today at 8:00 AM
2017 BCSMC Round 2 #3 yellow pig walks on a number line
parmenides51   3
N Today at 4:20 AM by MathPerson12321
A yellow pig walks on a number line starting at $17$. Each step the pig has probability $\frac{8}{17}$ of moving $1$ unit in the positive direction and probability $\frac{9}{17}$ of moving $1$ unit in the negative direction. Find the expected number of steps until the yellow pig visits $0$.
3 replies
parmenides51
Jan 24, 2024
MathPerson12321
Today at 4:20 AM
number theory
IOQMaspirant   6
N Today at 3:19 AM by Yiyj1
Prove 6|(a + b + c) if and only if 6|(a^3 + b^3 + c^3)
6 replies
IOQMaspirant
Mar 11, 2025
Yiyj1
Today at 3:19 AM
2017 BCSMC Round 2 #1 (BC)^2 - (AB)^2 = 1081 in right ABC
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parmenides51
30627 posts
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In right triangle $ABC$, $\angle ABC = 90^o$, $AC = 37$, and $(BC)^2 - (AB)^2 = 1081$. Find $AB + BC$.
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CubeAlgo15
35 posts
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Add \( 2AB \) to both sides, then we have \( BC^2 + AB^2 = 1081 + 2(AB)^2 \) rearranging gives \( 37^2 - 1081 = 2(AB)^2 \)
So AB = 12 and BC = 35, thus our answer is \( \boxed{47} \).
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BackToSchool
1638 posts
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$$(BC)^2 - (AB)^2 = 1081$$$$(BC)^2 + (AB)^2 = (AC)^2 = 37^2 = 1369$$$$(BC)^2 = \frac {1081+1369}{2} = 1225, BC = 35$$$$(AB)^2 = \frac {- 1081+1369}{2} = 144, AB = 12$$$$AB + BC = \boxed {47}$$
This post has been edited 1 time. Last edited by BackToSchool, 2 hours ago
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