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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.

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[*]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
angle chasing with 2 midpoints, equal angles given and wanted
parmenides51   5
N 31 minutes ago by breloje17fr
Source: Ukrainian Geometry Olympiad 2017, IX p1, X p1, XI p1
In the triangle $ABC$, ${{A}_{1}}$ and ${{C}_{1}} $ are the midpoints of sides $BC $ and $AB$ respectively. Point $P$ lies inside the triangle. Let $\angle BP {{C}_{1}} = \angle PCA$. Prove that $\angle BP {{A}_{1}} = \angle PAC $.
5 replies
parmenides51
Dec 11, 2018
breloje17fr
31 minutes ago
Problem 4 of Finals
GeorgeRP   2
N an hour ago by Assassino9931
Source: XIII International Festival of Young Mathematicians Sozopol 2024, Theme for 10-12 grade
The diagonals \( AD \), \( BE \), and \( CF \) of a hexagon \( ABCDEF \) inscribed in a circle \( k \) intersect at a point \( P \), and the acute angle between any two of them is \( 60^\circ \). Let \( r_{AB} \) be the radius of the circle tangent to segments \( PA \) and \( PB \) and internally tangent to \( k \); the radii \( r_{BC} \), \( r_{CD} \), \( r_{DE} \), \( r_{EF} \), and \( r_{FA} \) are defined similarly. Prove that
\[
r_{AB}r_{CD} + r_{CD}r_{EF} + r_{EF}r_{AB} = r_{BC}r_{DE} + r_{DE}r_{FA} + r_{FA}r_{BC}.
\]
2 replies
GeorgeRP
Sep 10, 2024
Assassino9931
an hour ago
Interesting functional equation with geometry
User21837561   3
N an hour ago by Double07
Source: BMOSL 2024 G7
For an acute triangle $ABC$, let $O$ be the circumcentre, $H$ be the orthocentre, and $G$ be the centroid.
Let $f:\pi\rightarrow\mathbb R$ satisfy the following condition:
$f(A)+f(B)+f(C)=f(O)+f(G)+f(H)$
Prove that $f$ is constant.
3 replies
User21837561
Today at 8:14 AM
Double07
an hour ago
greatest volume
hzbrl   1
N an hour ago by hzbrl
Source: purple comet
A large sphere with radius 7 contains three smaller balls each with radius 3 . The three balls are each externally tangent to the other two balls and internally tangent to the large sphere. There are four right circular cones that can be inscribed in the large sphere in such a way that the bases of the cones are tangent to all three balls. Of these four cones, the one with the greatest volume has volume $n \pi$. Find $n$.
1 reply
hzbrl
Yesterday at 9:56 AM
hzbrl
an hour ago
Cute matrix equation
RobertRogo   1
N 3 hours ago by loup blanc
Source: "Traian Lalescu" student contest 2025, Section A, Problem 2
Find all matrices $A \in \mathcal{M}_n(\mathbb{Z})$ such that $$2025A^{2025}=A^{2024}+A^{2023}+\ldots+A$$
1 reply
RobertRogo
6 hours ago
loup blanc
3 hours ago
Group Theory
Stephen123980   1
N 4 hours ago by alexheinis
Show that if $G_1,G_2$ are two finite groups with $\gcd(|G_1|,|G_2|)=1,$ then show that $Aut(G_1\times G_2)\cong Aut(G_1)\times Aut(G_2).$
1 reply
Stephen123980
Today at 12:49 PM
alexheinis
4 hours ago
UC Berkeley Integration Bee 2025 Bracket Rounds
Silver08   9
N 4 hours ago by Silver08
Regular Round

Quarterfinals

Semifinals

3rd Place Match

Finals
9 replies
Silver08
Today at 2:26 AM
Silver08
4 hours ago
Putnam 2012 A1
Kent Merryfield   14
N 5 hours ago by anudeep
Let $d_1,d_2,\dots,d_{12}$ be real numbers in the open interval $(1,12).$ Show that there exist distinct indices $i,j,k$ such that $d_i,d_j,d_k$ are the side lengths of an acute triangle.
14 replies
Kent Merryfield
Dec 3, 2012
anudeep
5 hours ago
Nice-looking function of class C^2
RobertRogo   1
N 5 hours ago by pi_quadrat_sechstel
Source: "Traian Lalescu" student contest 2025, Section A, Problem 1
Find all functions $f \colon \mathbb{R} \to (0, \infty)$ of class $C^2$ for which there exists an $\alpha>1$ such that $$f''(x)f(x) \geq \alpha \left(f'(x)\right)^2, \; \forall x \in \mathbb{R}$$
1 reply
RobertRogo
6 hours ago
pi_quadrat_sechstel
5 hours ago
Another integral limit
RobertRogo   0
5 hours ago
Source: "Traian Lalescu" student contest 2025, Section A, Problem 3
Let $f \colon [0, \infty) \to \mathbb{R}$ be a function differentiable at 0 with $f(0) = 0$. Find
$$\lim_{n \to \infty} \frac{1}{n} \int_{2^n}^{2^{n+1}} f\left(\frac{\ln x}{x}\right) dx$$
0 replies
RobertRogo
5 hours ago
0 replies
Putnam 1983 B6
Kunihiko_Chikaya   1
N 5 hours ago by pi_quadrat_sechstel
Let $ k$ be a positive integer, let $ m=2^k+1$, and let $ r\neq 1$ be a complex root of $ z^m-1=0$. Prove that there exist polynomials $ P(z)$ and $ Q(z)$ with integer coefficients such that $ (P(r))^2+(Q(r))^2=-1$.
1 reply
Kunihiko_Chikaya
Jun 5, 2008
pi_quadrat_sechstel
5 hours ago
Preparing for Putnam level entrance examinations
Cats_on_a_computer   2
N 6 hours ago by anudeep
Non American high schooler in the equivalent of grade 12 here. Where I live, two the best undergraduates program in the country accepts students based on a common entrance exam. The first half of the exam is “screening”, with 4 options being presented per question, each of which one has to assign a True or False. This first half is about the difficulty of an average AIME, or JEE Adv paper, and it is a requirement for any candidate to achieve at least 24/40 on this half for the examiners to even consider grading the second part. The second part consists of long form questions, and I have, no joke, seen them literally rip off, verbatim, Putnam A6s. Some of the problems are generally standard textbook problems in certain undergrad courses but obviously that doesn’t translate it to being doable for high school students. I’ve effectively got to prepare for a slightly nerfed Putnam, if you will, and so I’ve been looking for resources (not just problems) for Putnam level questions. Does anyone have any suggestions?
2 replies
Cats_on_a_computer
Today at 8:32 AM
anudeep
6 hours ago
Putnam 2016 A2
Kent Merryfield   13
N 6 hours ago by anudeep
Given a positive integer $n,$ let $M(n)$ be the largest integer $m$ such that
\[\binom{m}{n-1}>\binom{m-1}{n}.\]Evaluate
\[\lim_{n\to\infty}\frac{M(n)}{n}.\]
13 replies
Kent Merryfield
Dec 4, 2016
anudeep
6 hours ago
Integration erf(x)
FOL   1
N Today at 1:44 PM by Mathzeus1024
Show that
\[
\int_{-\infty}^{x}e^{-t^2/2}dt=\sqrt{2\pi}\left( \frac{1}{2}+\frac{1}{2}\mathrm{erf}\left(\frac{x}{\sqrt{2}}\right)\right)
\]
1 reply
FOL
May 3, 2023
Mathzeus1024
Today at 1:44 PM
Some Identity that I need help
ItzsleepyXD   2
N Apr 10, 2025 by Tkn
Given $\triangle ABC$ with orthocenter , circumcenter and incenter $H,O,I$ , circum-radius $R$ , in-radius $r$.
Prove that $OH^2 = 2 HI^2 - 4r^2 + R^2$ .
2 replies
ItzsleepyXD
Dec 28, 2024
Tkn
Apr 10, 2025
Some Identity that I need help
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ItzsleepyXD
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Given $\triangle ABC$ with orthocenter , circumcenter and incenter $H,O,I$ , circum-radius $R$ , in-radius $r$.
Prove that $OH^2 = 2 HI^2 - 4r^2 + R^2$ .
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ItzsleepyXD
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Tkn
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To show the above equality, we need to prove these two claims:
Claim 1. $OH^2=9R^2-(a^2+b^2+c^2)$
Proof. This one is easy by vector bash with the Euler line. Note that
$$\frac{OG}{OH}=\frac{1}{3}$$where $G$ denotes the centroid of $\triangle{ABC}$. It is easy to see that
$$\overrightarrow{OH}=3\overrightarrow{OG}=3\left(\frac{\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}}{3}\right).$$Comparing modulus of both sides:
\begin{align*}
    |\overrightarrow{OH}|^2&=\overrightarrow{OH}\cdot \overrightarrow{OH}\\
    &=3R^2+2R^2\left(\cos(2\angle{A})+\cos(2\angle{B})+\cos(2\angle{C})\right)\\
    &=3R^2+6R^2-4R^2\left(\sin^2(\angle{A})+\sin^2(\angle{B})+\sin^2(\angle{C})\right)\\
    &=9R^2-(a^2+b^2+c^2)
\end{align*}The last step requires the sine's law on $\triangle{ABC}$, as in the form:
$$\frac{a}{\sin(\angle{A})}=\frac{b}{\sin(\angle{B})}=\frac{c}{\sin(\angle{C})}=2R.$$where $a,b$ and $c$ are side lengths of $BC,CA$ and $AB$ respectively.
Claim 2. $HI^2=2r^2-4R^2\cos(\angle{A})\cos(\angle{B})\cos(\angle{C})$
Proof. Given that $M$ is a midpoint of the segment $\overline{BC}$.
We use the well-known lemma to find, $AH=2OM=2R\cos(\angle{A})$. And,
$$AI=2r\cos\left(\frac{\angle A}{2}\right)=4R\sin\left(\frac{\angle{B}}{2}\right)\sin\left(\frac{\angle{C}}{2}\right).$$The rest is just some bashing of cosine law:
\begin{align*}
    HI^2&=AI^2+AH^2-2AI\cdot AH\cos\left(\frac{\angle{B}-\angle{C}}{2}\right)\\
    &=16R^2\sin^2\left(\frac{\angle{B}}{2}\right)\sin^2\left(\frac{\angle{C}}{2}\right)+4R^2\cos^2(\angle{A})-16R^2\cos(\angle{A})\sin\left(\frac{\angle{B}}{2}\right)\sin\left(\frac{\angle{C}}{2}\right)\cos\left(\frac{\angle{B}-\angle{C}}{2}\right)\\
    &=4R^2\left(\cos^2(\angle{A})+(4-4\cos(\angle{A}))\sin^2\left(\frac{\angle{B}}{2}\right)\sin^2\left(\frac{\angle{C}}{2}\right)-\cos(\angle{A})\sin(\angle{B})\sin(\angle{C})\right)\\
    &=4R^2\left(\cos^2(\angle{A})+8\sin^2\left(\frac{\angle{A}}{2}\right)\sin^2\left(\frac{\angle{B}}{2}\right)\sin^2\left(\frac{\angle{C}}{2}\right)-\cos(\angle{A})\sin(\angle{B})\sin(\angle{C})\right)\\
    &=4R^2\left(8\sin^2\left(\frac{\angle{A}}{2}\right)\sin^2\left(\frac{\angle{B}}{2}\right)\sin^2\left(\frac{\angle{C}}{2}\right)+\cos(\angle{A})\left(\cos(\angle{A})-\sin(\angle{B})\sin(\angle{C})\right)\right)\\
    &=4R^2\left(8\sin^2\left(\frac{\angle{A}}{2}\right)\sin^2\left(\frac{\angle{B}}{2}\right)\sin^2\left(\frac{\angle{C}}{2}\right)-\cos(\angle{A})\cos(\angle{B})\cos(\angle{C})\right)\\
    &=2r^2-4R^2\cos(\angle{A})\cos(\angle{B})\cos(\angle{C})
\end{align*}Which is complete.
Then, it just equivalent of showing:
$$8R^2(1+\cos(\angle{A})\cos(\angle{B})\cos(\angle{C}))=a^2+b^2+c^2$$Note that $a^2=4R^2\sin^2(\angle{A})$, similarly for $b$ and $c$. Simplify and gives
$$2\left(1+\cos(\angle{A})\cos(\angle{B})\cos(\angle{C})\right)=\sin^2(\angle{A})+\sin^2{(\angle{B})}+\sin^2(\angle{C})$$or equivalent to
$$1=2\cos(\angle{A})\cos(\angle{B})\cos(\angle{C})+\cos^2(\angle{A})+\cos^{2}(\angle{B})+\cos^2(\angle{C})$$This is obviously true due to
\begin{align*}
    2\cos(\angle{A})\cos(\angle{B})\cos(\angle{C})&=-\cos^2(\angle{C})+\cos(\angle{A}-\angle{B})\cos(\angle{C})\\
    &=-\cos^2(\angle{C})+\frac{1}{2}\left(\cos(\angle{A}-\angle{B}+\angle{C})+\cos(\angle{A}-\angle{B}-\angle{C})\right)\\
    &=-\cos^2(\angle{C})+\frac{1}{2}(\cos(180^{\circ}-2B)+\cos(180^{\circ}-2A))\\
    &=-\cos^2(\angle{C})-\frac{1}{2}(\cos(2\angle{A})+\cos(2\angle B))\\
    &=-\cos^2(\angle{C})-\cos^2(\angle{A})-\cos^2(\angle{B})+1
\end{align*}Which actually solves the problem.
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