It's February and we'd love to help you find the right course plan!

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k a February Highlights and 2025 AoPS Online Class Information
jlacosta   0
Feb 2, 2025
We love to share what you can look forward to this month! The AIME I and AIME II competitions are happening on February 6th and 12th, respectively. Join our Math Jams the day after each competition where we will go over all the problems and the useful strategies to solve them!

2025 AIME I Math Jam: Difficulty Level: 8* (Advanced math)
February 7th (Friday), 4:30pm PT/7:30 pm ET

2025 AIME II Math Jam: Difficulty Level: 8* (Advanced math)
February 13th (Thursday), 4:30pm PT/7:30 pm ET

The F=ma exam will be held on February 12th. Check out our F=ma Problem Series course that begins February 19th if you are interested in participating next year! The course will prepare you to take the F=ma exam, the first test in a series of contests that determines the members of the US team for the International Physics Olympiad. You'll learn the classical mechanics needed for the F=ma exam as well as how to solve problems taken from past exams, strategies to succeed, and you’ll take a practice F=ma test of brand-new problems.

Mark your calendars for all our upcoming events:
[list][*]Feb 7, 4:30 pm PT/7:30pm ET, 2025 AIME I Math Jam
[*]Feb 12, 4pm PT/7pm ET, Mastering Language Arts Through Problem-Solving: The AoPS Method
[*]Feb 13, 4:30 pm PT/7:30pm ET, 2025 AIME II Math Jam
[*]Feb 20, 4pm PT/7pm ET, The Virtual Campus Spring Experience[/list]
AoPS Spring classes are open for enrollment. Get a jump on 2025 and enroll in our math, contest prep, coding, and science classes today! Need help finding the right plan for your goals? Check out our recommendations page!

Don’t forget: Highlight your AoPS Education on LinkedIn!
Many of you are beginning to build your education and achievements history on LinkedIn. Now, you can showcase your courses from Art of Problem Solving (AoPS) directly on your LinkedIn profile! Don't miss this opportunity to stand out and connect with fellow problem-solvers in the professional world and be sure to follow us at: https://www.linkedin.com/school/art-of-problem-solving/mycompany/ Check out our job postings, too, if you are interested in either full-time, part-time, or internship opportunities!

Our full course list for upcoming classes is below:
All classes run 7:30pm-8:45pm ET/4:30pm - 5:45pm PT unless otherwise noted.

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0 replies
jlacosta
Feb 2, 2025
0 replies
I need this. Pls help me
Giahuytls2326   12
N 5 minutes ago by Giahuytls2326
Source: Own
Given $\triangle ABC$ , orthocenter $H$, $E$ is the center of Euler's circle of $\triangle ABC$. $X,Y,Z$ is the midpoint of $AH,BH,CH$. From $X,Y,Z$ draw the tangent of $(E)$, cut a line through $E$ perpendicular to sides $BC, CA,AB$ at $M,N,P$ .Prove that $AM,BN,CP$ concurrent
12 replies
Giahuytls2326
Thursday at 4:59 PM
Giahuytls2326
5 minutes ago
Thanks u!
Ruji2018252   4
N 7 minutes ago by sqing
Let $a,b,c>0$ and $a+b+c=abc$
Find minimum (and prove)
\[P=(ab-1)(bc+1)^6(ca-1)\]
4 replies
Ruji2018252
Yesterday at 2:38 PM
sqing
7 minutes ago
Functional Equation
AnhQuang_67   4
N 7 minutes ago by gordian.knot
Find all functions $f:\mathbb{R} \to \mathbb{R}$ satisfying: $$x^2f(x)+f(1-x)=2x-x^4, \forall x \in \mathbb{R} $$
4 replies
AnhQuang_67
Yesterday at 4:02 PM
gordian.knot
7 minutes ago
thanks u!
Ruji2018252   1
N 26 minutes ago by sqing
Let $a,b,c\in \mathbb{R},a,b,c\ne 0$ and $a+b+c=0.$ Find minimum (and prove)
\[C=\dfrac{b^2+c^2-a^2}{b^2+c^2}+\dfrac{c^2+a^2-b^2}{c^2+a^2}+\dfrac{a^2+b^2-c^2}{a^2+b^2}\]
1 reply
1 viewing
Ruji2018252
3 hours ago
sqing
26 minutes ago
No more topics!
Six orthopoles lie on a circle
buratinogigle   2
N Dec 20, 2024 by buratinogigle
Let $ABC$ and $A'B'C'$ be two triangles inscribed circle $(O)$. Prove that orthopoles of $B'C',C'A',A'B'$ with respect to triangle $ABC$ and orthopoles of $BC,CA,AB$ with respect to triangle $A'B'C'$ lie on a circle.
2 replies
buratinogigle
Sep 2, 2012
buratinogigle
Dec 20, 2024
Six orthopoles lie on a circle
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buratinogigle
2300 posts
#1 • 4 Y
Y by doxuanlong15052000, BoleroGirl, Adventure10, Mango247
Let $ABC$ and $A'B'C'$ be two triangles inscribed circle $(O)$. Prove that orthopoles of $B'C',C'A',A'B'$ with respect to triangle $ABC$ and orthopoles of $BC,CA,AB$ with respect to triangle $A'B'C'$ lie on a circle.
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Luis González
4145 posts
#2 • 8 Y
Y by buratinogigle, Omez, langkhach11112, doxuanlong15052000, rmtf1111, MathPassionForever, Adventure10, Mango247
Lemma. Line $\tau$ cuts the circumcircle of $\triangle ABC$ at $P,Q.$ Then Simson lines of $P,Q$ with respect to $\triangle ABC$ meet at the orthopole of $\tau$ with respect to $\triangle ABC.$

Proof. Let $A'$ be the orthogonal projection of $A$ on $\tau.$ $S,T,X$ are the orthogonal projections of $Q$ on $CB,BA,AC$ and $M,N,Y$ the orthogonal projections of $P$ on $BC,CA,AB.$ Thus, $A,Q,X,A'$ and $A,P,Y,A'$ are concyclic. Using directed angles $\pmod\pi,$ we have

$\angle AA'X=\frac{\pi}{2}+\angle QA'X=\frac{\pi}{2}+\angle QAC \ , \ \angle AA'N=\angle APN=\frac{\pi}{2}-\angle PAC$

$ \angle NAX'=\angle AA'X-\angle AA'N=\frac{\pi}{2}+\angle QAC- \left (\frac{\pi}{2}-\angle PAC \right)=\angle PAQ$

Let $K \equiv SX \cap MY.$ Since the angle between the Simson lines of $P,Q$ is half the measure of the arc $PQ$ of the circumcircle $(*),$ it follows that $\angle NAX'=\angle NKX$ $\Longrightarrow$ $N,K,A',X$ are concyclic. Consequently

$\angle A'KX=\angle A'NX=\angle APQ=\angle ACQ=\angle XSQ \Longrightarrow KA' \parallel QS$

That is, $KA' \perp BC.$ Similarly, if $B'$ denotes the orthogonal projection of $B$ on $\tau,$ we have $KB' \perp CA$ $\Longrightarrow$ $K$ is the orthopole of $\tau$ with respect to $\triangle ABC.$
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Based on the previous lemma, the pairwise Simson lines of $A',B',C'$ WRT $\triangle ABC$ meet at the orthopoles $A_0,B_0,C_0$ of $B'C',C'A',A'B'$ WRT $\triangle ABC.$ In addition, $\triangle A_0B_0C_0 \sim \triangle A'B'C',$ because of the assertion $(*).$ Let $H,H'$ be the orthocenters of $\triangle ABC,$ $\triangle A'B'C'.$ Then $B_0C_0,$ $C_0A_0,$ $A_0B_0$ pass through the midpoints $X_0,Y_0,Z_0$ of $HA',HB',HC',$ lying on the 9-point circle $(N,\frac{_R}{^2})$ of $\triangle ABC$ $\Longrightarrow$ $\triangle X_0Y_0Z_0 \sim \triangle A'B'C' \sim \triangle A_0B_0C_0.$ If $U$ is midpoint of $\overline{HH'},$ then $UY_0$ is H-midline of $\triangle HH'B'$ $\Longrightarrow$ $UY_0 \parallel B'H' \perp A'C'.$ Similarly, $UZ_0 \perp A'B'$ $\Longrightarrow$ $\angle Y_0UZ_0=\angle B'A'C'=\angle B_0A_0C_0$ $\pmod\pi$ $\Longrightarrow$ $U \in \odot(A_0Y_0Z_0).$ Likewise, $U \in \odot(B_0Z_0X_0)$ $\Longrightarrow$ $U$ is the Miquel point of $\triangle X_0Y_0Z_0$ WRT $\triangle A_0B_0C_0$ $\Longrightarrow$ $U$ is orthocenter of $\triangle X_0Y_0Z_0$ and circumcenter of $\triangle A_0B_0C_0$ (see the topic Own problem and elsewhere) $\Longrightarrow$ $\odot(AY_0Z_0) \cong (N,\frac{_R}{^2}).$

Now, easy angle chase reveals that the angle between a sideline of ABC (A'B'C') and the Simson line of its opposite vertex WRT A'B'C' (ABC) is constant. Let $\lambda$ be the measure of this angle. From $UY_0 \parallel B'H',$ we have then $UA_0=R \cdot \sin \angle (B'H',A_0C_0)=R \cdot \sin (\frac{_{\pi}}{^2}-\lambda)=R \cdot \cos \lambda.$ As a result, orthopoles of $B'C',C'A',A'B'$ WRT $\triangle ABC$ lie on a circle $(U,\varrho)$ centered at the midpoint $U$ of $\overline{HH'}$ with radius $\varrho=R \cdot \cos \lambda.$ By similar reasoning, the orthopoles of $BC,CA,AB$ WRT $\triangle A'B'C'$ lie on the same circle $(U,\varrho).$
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buratinogigle
2300 posts
#3
Y by
Thanks Luis for the interesting solution. I just saw a direct corollary as follows

Corollary. When $A'B'C'$ is the image of $ABC$ under a rotation centered at $O$, let $K$ be the center of the circumcircle of the triangle formed by the Simson lines of $A'$, $B'$, and $C'$ with respect to $\triangle ABC$. Prove that $\angle OKH = 90^\circ$ where $H$ is the orthocenter of $\triangle ABC$.
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