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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

Are you interested in working towards MATHCOUNTS and don’t know where to start? We have you covered! If you have taken Prealgebra, then you are ready for MATHCOUNTS/AMC 8 Basics. Already aiming for State or National MATHCOUNTS and harder AMC 8 problems? Then our MATHCOUNTS/AMC 8 Advanced course is for you.

Summer camps are starting next month at the Virtual Campus in math and language arts that are 2 - to 4 - weeks in duration. Spaces are still available - don’t miss your chance to have an enriching summer experience. There are middle and high school competition math camps as well as Math Beasts camps that review key topics coupled with fun explorations covering areas such as graph theory (Math Beasts Camp 6), cryptography (Math Beasts Camp 7-8), and topology (Math Beasts Camp 8-9)!

Be sure to mark your calendars for the following upcoming events:
[list][*]May 9th, 4:30pm PT/7:30pm ET, Casework 2: Overwhelming Evidence — A Text Adventure, a game where participants will work together to navigate the map, solve puzzles, and win! All are welcome.
[*]May 19th, 4:30pm PT/7:30pm ET, What's Next After Beast Academy?, designed for students finishing Beast Academy and ready for Prealgebra 1.
[*]May 20th, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 1 Math Jam, Problems 1 to 4, join the Canada/USA Mathcamp staff for this exciting Math Jam, where they discuss solutions to Problems 1 to 4 of the 2025 Mathcamp Qualifying Quiz!
[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
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0 replies
jlacosta
May 1, 2025
0 replies
USAJMO problem 2: Side lengths of an acute triangle
BOGTRO   59
N an hour ago by ostriches88
Source: Also USAMO problem 1
Find all integers $n \geq 3$ such that among any $n$ positive real numbers $a_1, a_2, \hdots, a_n$ with $\text{max}(a_1,a_2,\hdots,a_n) \leq n \cdot \text{min}(a_1,a_2,\hdots,a_n)$, there exist three that are the side lengths of an acute triangle.
59 replies
BOGTRO
Apr 24, 2012
ostriches88
an hour ago
high tech FE as J1?!
imagien_bad   60
N 3 hours ago by SimplisticFormulas
Source: USAJMO 2025/1
Let $\mathbb Z$ be the set of integers, and let $f\colon \mathbb Z \to \mathbb Z$ be a function. Prove that there are infinitely many integers $c$ such that the function $g\colon \mathbb Z \to \mathbb Z$ defined by $g(x) = f(x) + cx$ is not bijective.
Note: A function $g\colon \mathbb Z \to \mathbb Z$ is bijective if for every integer $b$, there exists exactly one integer $a$ such that $g(a) = b$.
60 replies
imagien_bad
Mar 20, 2025
SimplisticFormulas
3 hours ago
k Can I make the IMO team next year?
aopslover08   26
N 5 hours ago by steve4916
Hi everyone,

I am a current 11th grader living in Orange, Texas. I recently started doing competition math and I think I am pretty good at it. Recently I did a mock AMC8 and achieved a score of 21/25, which falls in the top 1% DHR. I also talked to my math teacher and she says I am an above average student.

Given my natural talent and the fact that I am willing to work ~3.5 hours a week studying competition math, do you think I will be able to make IMO next year? I am aware of the difficulty of this task but my mom says that I can achieve whatever I put my mind to, as long as I work hard.

Here is my plan for the next few months:

month 1-2: finish studying pre-algebra and learn geometry
month 3-4: learn pre-calculus
month 5-6: start doing IMO shortlist problems
month 7+: keep doing ISL/IMO problems.

Is this a feasible task? I am a girl btw.
26 replies
aopslover08
Yesterday at 7:46 PM
steve4916
5 hours ago
Isosceles everywhere
reallyasian   28
N 6 hours ago by MATHS_ENTUSIAST
Source: 2020 AIME I #1
In $\triangle ABC$ with $AB=AC$, point $D$ lies strictly between $A$ and $C$ on side $\overline{AC}$, and point $E$ lies strictly between $A$ and $B$ on side $\overline{AB}$ such that $AE=ED=DB=BC$. The degree measure of $\angle ABC$ is $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.
28 replies
reallyasian
Mar 12, 2020
MATHS_ENTUSIAST
6 hours ago
No more topics!
Inversely Similiar Triangles
EulerMacaroni   111
N Feb 26, 2025 by bjump
Source: JMO/5 2016
Let $\triangle ABC$ be an acute triangle, with $O$ as its circumcenter. Point $H$ is the foot of the perpendicular from $A$ to line $\overleftrightarrow{BC}$, and points $P$ and $Q$ are the feet of the perpendiculars from $H$ to the lines $\overleftrightarrow{AB}$ and $\overleftrightarrow{AC}$, respectively.

Given that $$AH^2=2\cdot AO^2,$$prove that the points $O,P,$ and $Q$ are collinear.
111 replies
EulerMacaroni
Apr 20, 2016
bjump
Feb 26, 2025
Inversely Similiar Triangles
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Source: JMO/5 2016
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EulerMacaroni
851 posts
#1 • 7 Y
Y by Davi-8191, megarnie, Quidditch, HWenslawski, mathlearner2357, Adventure10, Mango247
Let $\triangle ABC$ be an acute triangle, with $O$ as its circumcenter. Point $H$ is the foot of the perpendicular from $A$ to line $\overleftrightarrow{BC}$, and points $P$ and $Q$ are the feet of the perpendiculars from $H$ to the lines $\overleftrightarrow{AB}$ and $\overleftrightarrow{AC}$, respectively.

Given that $$AH^2=2\cdot AO^2,$$prove that the points $O,P,$ and $Q$ are collinear.
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djmathman
7938 posts
#2 • 9 Y
Y by Generic_Username, sualehasif996, claserken, Ultroid999OCPN, Jc426, megarnie, Danielzh, Adventure10, Mango247
Pretty easy for a JMO5 in my opinion (and certainly easier than JMO1); it took me about 20 minutes to solve during DiffEq today.

Solution
This post has been edited 2 times. Last edited by djmathman, Apr 20, 2016, 9:35 PM
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EulerMacaroni
851 posts
#3 • 7 Y
Y by anantmudgal09, amar_04, myh2910, megarnie, two_steps, Adventure10, Mango247
Anyone else thought this was much easier than JMO/1?

It is well-known that $AH\cdot 2AO=AB\cdot AC$ (just use similar triangles or $\sqrt{bc}$ inversion). Then by Power of a Point,
$$AP\cdot AB=AH^2=AQ\cdot AC$$Consider the transformation $X\mapsto \Psi(X)$ which dilates $X$ from $A$ by a factor of $\dfrac{AB}{AQ}=\dfrac{AC}{AP}$ and reflects about the $A$-angle bisector. Then $\Psi(O)$ clearly lies on $AH$, and its distance from $A$ is $$AO\cdot\frac{AB}{AQ}=AO\cdot\frac{AB}{\frac{AH^2}{AC}}=AO\cdot\frac{AB\cdot AC}{AH^2}=\frac{AO\cdot AH\cdot 2AO}{AH^2}=\frac{2AO^2}{AH}=AH$$so $\Psi(O)=H$, hence we conclude that $O,P,Q$ are collinear, as desired.
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WL0410
408 posts
#4 • 2 Y
Y by megarnie, Adventure10
Length bashing with trig to show PO+OQ=PQ!
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AMN300
563 posts
#5 • 1 Y
Y by Adventure10
coordinate bashed gg
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Wave-Particle
3690 posts
#6 • 1 Y
Y by Adventure10
Complex bash also worked on this.
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Benq
3396 posts
#7 • 1 Y
Y by Adventure10
Barycentric bash also worked on this.
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mathwizard888
1635 posts
#8 • 1 Y
Y by Adventure10
Coordinate bashed with $A=(0,a)$, $B=(b,0)$, $C=(c,0)$, $H=(0,0)$. It didn't turn out that bad, though I did get some messy expressions.
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zephyrcrush78
389 posts
#9 • 2 Y
Y by Adventure10, Mango247
I bashed the area of triangle $\triangle PQO$, which worked out pretty decently I guess.
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mathmaster2012
636 posts
#10 • 2 Y
Y by Adventure10, Mango247
Trig should show that O'A =R, where O' is intersection of PQ and reflection of AH over angle bisector. Is this sufficient?
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room456
146 posts
#11 • 2 Y
Y by Adventure10, Mango247
definitely easier than jmo 1

first wlog R=1, then bc = 2sqrt2, then law of sines and law of cosines on AOP and AOQ to get that angles AOP and AOQ are right.
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mathwizard888
1635 posts
#12 • 2 Y
Y by Adventure10, Mango247
Since the reflection of AH over the angle bisector is AO, this is sufficient, though you should add what I just said to the solution.

oops sniped
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bestwillcui1
2735 posts
#13 • 1 Y
Y by Adventure10
mathwizard888 wrote:
Coordinate bashed with $A=(0,a)$, $B=(b,0)$, $C=(c,0)$, $H=(0,0)$. It didn't turn out that bad, though I did get some messy expressions.

yes. mine was 6 pages , two exceptions when b and c are negatives or o and p are vertical
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FlakeLCR
1791 posts
#14 • 3 Y
Y by seyyed_khandan, Adventure10, Mango247
Did anyone else prove that $P$ and $Q$ were on the diameter opposite of $A$ (immediately implying the solution)?
This post has been edited 1 time. Last edited by FlakeLCR, Apr 20, 2016, 9:43 PM
Reason: asdf
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mathwizard888
1635 posts
#15 • 1 Y
Y by Adventure10
bestwillcui1 wrote:
mathwizard888 wrote:
Coordinate bashed with $A=(0,a)$, $B=(b,0)$, $C=(c,0)$, $H=(0,0)$. It didn't turn out that bad, though I did get some messy expressions.

yes. mine was 6 pages , two exceptions when b and c are negatives or o and p are vertical

Wait dealing with OP vertical is easy, and you know exactly one of b and c is negative because ABC is acute.
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