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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.

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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.
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0 replies
jlacosta
Mar 2, 2025
0 replies
k i Adding contests to the Contest Collections
dcouchman   1
N Apr 5, 2023 by v_Enhance
Want to help AoPS remain a valuable Olympiad resource? Help us add contests to AoPS's Contest Collections.

Find instructions and a list of contests to add here: https://artofproblemsolving.com/community/c40244h1064480_contests_to_add
1 reply
dcouchman
Sep 9, 2019
v_Enhance
Apr 5, 2023
k i Zero tolerance
ZetaX   49
N May 4, 2019 by NoDealsHere
Source: Use your common sense! (enough is enough)
Some users don't want to learn, some other simply ignore advises.
But please follow the following guideline:


To make it short: ALWAYS USE YOUR COMMON SENSE IF POSTING!
If you don't have common sense, don't post.


More specifically:

For new threads:


a) Good, meaningful title:
The title has to say what the problem is about in best way possible.
If that title occured already, it's definitely bad. And contest names aren't good either.
That's in fact a requirement for being able to search old problems.

Examples:
Bad titles:
- "Hard"/"Medium"/"Easy" (if you find it so cool how hard/easy it is, tell it in the post and use a title that tells us the problem)
- "Number Theory" (hey guy, guess why this forum's named that way¿ and is it the only such problem on earth¿)
- "Fibonacci" (there are millions of Fibonacci problems out there, all posted and named the same...)
- "Chinese TST 2003" (does this say anything about the problem¿)
Good titles:
- "On divisors of a³+2b³+4c³-6abc"
- "Number of solutions to x²+y²=6z²"
- "Fibonacci numbers are never squares"


b) Use search function:
Before posting a "new" problem spend at least two, better five, minutes to look if this problem was posted before. If it was, don't repost it. If you have anything important to say on topic, post it in one of the older threads.
If the thread is locked cause of this, use search function.

Update (by Amir Hossein). The best way to search for two keywords in AoPS is to input
[code]+"first keyword" +"second keyword"[/code]
so that any post containing both strings "first word" and "second form".


c) Good problem statement:
Some recent really bad post was:
[quote]$lim_{n\to 1}^{+\infty}\frac{1}{n}-lnn$[/quote]
It contains no question and no answer.
If you do this, too, you are on the best way to get your thread deleted. Write everything clearly, define where your variables come from (and define the "natural" numbers if used). Additionally read your post at least twice before submitting. After you sent it, read it again and use the Edit-Button if necessary to correct errors.


For answers to already existing threads:


d) Of any interest and with content:
Don't post things that are more trivial than completely obvious. For example, if the question is to solve $x^{3}+y^{3}=z^{3}$, do not answer with "$x=y=z=0$ is a solution" only. Either you post any kind of proof or at least something unexpected (like "$x=1337, y=481, z=42$ is the smallest solution). Someone that does not see that $x=y=z=0$ is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
Similar, posting "I have solved this problem" but not posting anything else is not welcome; it even looks that you just want to show off what a genius you are.

e) Well written and checked answers:
Like c) for new threads, check your solutions at least twice for mistakes. And after sending, read it again and use the Edit-Button if necessary to correct errors.



To repeat it: ALWAYS USE YOUR COMMON SENSE IF POSTING!


Everything definitely out of range of common sense will be locked or deleted (exept for new users having less than about 42 posts, they are newbies and need/get some time to learn).

The above rules will be applied from next monday (5. march of 2007).
Feel free to discuss on this here.
49 replies
ZetaX
Feb 27, 2007
NoDealsHere
May 4, 2019
Integer equation in 3 variables
Kimchiks926   2
N 4 minutes ago by MuradSafarli
Source: Latvian TST for Baltic Way 2019 Problem 15
Determine all tuples of integers $(a,b,c)$ such that:
$$(a-b)^3(a+b)^2 = c^2 + 2(a-b) + 1$$
2 replies
Kimchiks926
May 29, 2020
MuradSafarli
4 minutes ago
Interesting inequality
sqing   2
N 7 minutes ago by SunnyEvan
Source: Own
Let $ a,b,c\geq 2  . $ Prove that
$$(a^2-1)(b-1)(c^2-1) -\frac{9}{4}abc\geq -9$$$$(a^2-1)(b-1)(c^2-1) -\frac{11}{5}abc\geq -\frac{43}{5}$$$$(a^2-1)(b-1)(c^2-1) -2abc\geq -7$$$$(a-1)(b^2-1)(c-1) -\frac{3}{4}abc\geq -3$$$$(a-1)(b^2-1)(c-1) -\frac{3}{5}abc\geq -\frac{9}{5}$$$$(a-1)(b^2-1)(c-1) -\frac{1}{2}abc\geq -1$$
2 replies
1 viewing
sqing
43 minutes ago
SunnyEvan
7 minutes ago
Interesting inequality
sqing   1
N 10 minutes ago by sqing
Source: Own
Let $ a,b,c\geq 2  . $ Prove that
$$(a^2-1)(b-\frac{3}{2})(c^2-1) - \frac{9}{4}abc\geq -15$$$$(a^2-1)(b-\frac{3}{2})(c^2-1) - 2abc\geq -\frac{73}{6}$$$$(a^2-1)(b-\frac{3}{2})(c^2-1) - abc\geq -\frac{7}{2}$$$$(a^2-2)(b-\frac{3}{2})(c^2-2) - abc\geq -6$$
1 reply
1 viewing
sqing
19 minutes ago
sqing
10 minutes ago
Hard problem involving circumcenter and concurrent lines
GeoMetrix   6
N 11 minutes ago by bin_sherlo
Source: AQGO 2020 Problem 3
Let $\triangle{ABC}$ be a triangle with circumcenter $O$. Let $M,N$ be the midpoints of $\overline{AB}$ and $\overline{AC}$ respectively and let $T$ be the projection of $O$ on $\overline{MN}$. Let $D$ be the projection of $A$ on $\overline{BC}$. Let $\overline{TD}$ intersect $\odot(BOC)$ at points $U$ and $V$. Let $\odot(AUV)$ intersct $\overline{MN}$ at points $X,Y$. Let $\overline{AY}$ intersect $\odot(AMN)$ at $R$ and $\overline{AX}$ intersect $\odot(AMN)$ at $S$. Then show that $\overline{AO},\overline{RS},\overline{MN}$ are concurrent.

Proposed by GeoMetrix
6 replies
GeoMetrix
Jun 20, 2020
bin_sherlo
11 minutes ago
Oi! These lines concur
Rg230403   17
N 18 minutes ago by L13832
Source: LMAO 2021 P5, LMAOSL G3(simplified)
Let $I, O$ and $\Gamma$ respectively be the incentre, circumcentre and circumcircle of triangle $ABC$. Points $A_1, A_2$ are chosen on $\Gamma$, such that $AA_1 = AI = AA_2$, and point $A'$ is the foot of the altitude from $I$ to $A_1A_2$. If $B', C'$ are similarly defined, prove that lines $AA', BB'$ and $CC'$ concurr on $OI$.
Original Version from SL
Proposed by Mahavir Gandhi
17 replies
Rg230403
May 10, 2021
L13832
18 minutes ago
Find the period
Anto0110   1
N 22 minutes ago by Anto0110
Let $a_1, a_2, ..., a_k, ...$ be a sequence that consists of an initial block of $p$ positive distinct integers that then repeat periodically. This means that $\{a_1, a_2, \dots, a_p\}$ are $p$ distinct positive integers and $a_{n+p}=a_n$ for every positive integer $n$. The terms of the sequence are not known and the goal is to find the period $p$. To do this, at each move it possible to reveal the value of a term of the sequence at your choice.
If $p$ is one of the first $k$ prime numbers, find for which values of $k$ there exist a strategy that allows to find $p$ revealing at most $8$ terms of the sequence.
1 reply
Anto0110
Yesterday at 7:37 PM
Anto0110
22 minutes ago
inequality
senku23   2
N an hour ago by User21837561
Let x,y,z in R+ prove that 8(x^3+y^3+z^3)2≥9(x^2+yz)(y^2+zx)(z^2+xy).
2 replies
senku23
2 hours ago
User21837561
an hour ago
Differentiable functional
bakerbakura   2
N an hour ago by Gryphos
Find all differentiable functions $ f;\mathbb{R}\to\mathbb{R}$ such that, for all real numbers $ a,b,t$ with $ 0<t<1$, $ t^2f(a)+(1-t^2)f(b)\geq f(ta+(1-t)b)$
2 replies
bakerbakura
Jan 11, 2010
Gryphos
an hour ago
function equation
cipher703516247   2
N an hour ago by luutrongphuc

Find all the functions $f: \mathbb R^{+} \to \mathbb R^{+}$such that:
$$f(xy^n +f(y)^{2n}) )=f(x)f(y)^n +(yf(y))^n $$
2 replies
cipher703516247
Feb 14, 2025
luutrongphuc
an hour ago
gcd (a^n+b,b^n+a) is constant
EthanWYX2009   78
N an hour ago by SimplisticFormulas
Source: 2024 IMO P2
Determine all pairs $(a,b)$ of positive integers for which there exist positive integers $g$ and $N$ such that
$$\gcd (a^n+b,b^n+a)=g$$holds for all integers $n\geqslant N.$ (Note that $\gcd(x, y)$ denotes the greatest common divisor of integers $x$ and $y.$)

Proposed by Valentio Iverson, Indonesia
78 replies
EthanWYX2009
Jul 16, 2024
SimplisticFormulas
an hour ago
Maximum Cost of Identifying Two Digits in a Three-Digit Number
Borisaurus   0
an hour ago
Source: 2025 MBL QQ P7 Variant
You find yourself on a flight to Bosnia and Herzegovina. You are seated next to a strange man wearing a hat—similar to the one worn by Indiana Jones. The man spends most of the flight rambling about some pyramids.

However, as you approach your destination, he tells you the following:

"I have thought of a $three-digit$ number. I will allow you to ask me questions. A question consists of you saying a three-digit number, to which I respond 'yes' if at least two of its digits are correct, or 'no' otherwise. (A digit is correct if it matches the corresponding digit in my number.)

You can ask as many questions as you like, but they come at a price. The game ends when you have correctly identified two digits of my number. If
you guess correctly, I will sell you my precious pyramid. You will have to pay a number of Bosnian marks equal to the number of questions you asked.

What is the maximum number of Bosnian marks you will have to pay using the best strategy?"
0 replies
Borisaurus
an hour ago
0 replies
BMN is equilateral iff rectangle ABCD is square
parmenides51   3
N 2 hours ago by Tsikaloudakis
Source: 2004 Romania NMO SL - Shortlist VII-VIII p8 https://artofproblemsolving.com/community/c3950157_
Consider a point $M$ on the diagonal $BD$ of a given rectangle $ABCD$, such that $\angle AMC = \angle  CMD$. The point $N$ is the intersection point between $AM$ and the parallel line to $CM$ that contains $B$. Prove that the triangle $BMN$ is equilateral if and only if $ABCD$ is a square.

Valentin Vornicu
3 replies
parmenides51
Sep 16, 2024
Tsikaloudakis
2 hours ago
D1015 : A strange EF for polynomials
Dattier   1
N 2 hours ago by Dattier
Source: les dattes à Dattier
Find all $P \in \mathbb R[x,y]$ with $P \not\in \mathbb R[x] \cup \mathbb R[y]$ and $\forall g,f$ homeomorphismes of $\mathbb R$, $P(f,g)$ is an homoemorphisme too.
1 reply
Dattier
Mar 16, 2025
Dattier
2 hours ago
Cutting a big square into smaller squares
nAalniaOMliO   4
N 2 hours ago by anudeep
Source: Belarusian National Olympiad 2020
A $20 \times 20$ checkered board is cut into several squares with integer side length. The size of a square is it's side length.
What is the maximum amount of different sizes this squares can have?
4 replies
nAalniaOMliO
Jan 29, 2025
anudeep
2 hours ago
Line Perpendicular to Euler Line
tastymath75025   54
N Mar 17, 2025 by ohiorizzler1434
Source: USA TSTST 2017 Problem 1, by Ray Li
Let $ABC$ be a triangle with circumcircle $\Gamma$, circumcenter $O$, and orthocenter $H$. Assume that $AB\neq AC$ and that $\angle A \neq 90^{\circ}$. Let $M$ and $N$ be the midpoints of sides $AB$ and $AC$, respectively, and let $E$ and $F$ be the feet of the altitudes from $B$ and $C$ in $\triangle ABC$, respectively. Let $P$ be the intersection of line $MN$ with the tangent line to $\Gamma$ at $A$. Let $Q$ be the intersection point, other than $A$, of $\Gamma$ with the circumcircle of $\triangle AEF$. Let $R$ be the intersection of lines $AQ$ and $EF$. Prove that $PR\perp OH$.

Proposed by Ray Li
54 replies
tastymath75025
Jun 29, 2017
ohiorizzler1434
Mar 17, 2025
Line Perpendicular to Euler Line
G H J
Source: USA TSTST 2017 Problem 1, by Ray Li
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ike.chen
1162 posts
#51 • 1 Y
Y by Jalil_Huseynov
Remarks: This problem took me way too long (around three hours). Amusingly enough, I finally found this solution while watching a ballet performance ;).


Let $\omega$ denote the Nine-Point Circle of $ABC$ and $N_9$ be the Nine-Point center of $ABC$. It's clear that $$Pow_{\Gamma}(P) = PA^2 = Pow_{(AMN)}(P) = PM \cdot PN = Pow_{\omega}(P)$$and $$Pow_{\Gamma}(R) = RA \cdot RQ = Pow_{(AEF)}(R) = RE \cdot RF = Pow_{\omega}(R)$$so $PR$ is the Radical Axis of $\Gamma$ and $\omega$. Because $H, N_9, O$ are collinear on the Euler Line, the desired result follows immediately from properties of Radical Axes. $\blacksquare$
This post has been edited 1 time. Last edited by ike.chen, Aug 21, 2022, 9:46 AM
Reason: Format
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Tafi_ak
309 posts
#52 • 1 Y
Y by Jalil_Huseynov
We Claim that $PR$ is the radical axis of $(ABC), (EFM)$. For the proof, notice that $P$ is the radical center of $(AMN), (ABC), (EFM)$. Therefore the radical axis of $(ABC), (EFM)$ passes through $P$. Similarly $R$ is the radical center of $(BCF), (ABC), (EFM)$. So the radical axis of $(ABC), (EFM)$ passes through $R$. $\blacksquare$

It is very well known that the center of $(ABC)$ and $(EFM)$ are collinear with $H$. Hence done.
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lazizbek42
548 posts
#53
Y by
$K$ is midpoint $BC$.$D$ feet of the altitudes from $A$
$K,H,Q$ collinear.
$AQEF$,$EFDK$,$AQDK$ cyclic.
by radical axis $AQ,EF,DK$ concurrent.
$R,B,C$ collinear.
Carnot theorem.
$$HP^2-OP^2=HR^2-OR^2$$$$( AP^2 +AH^2 - 2AP*AH*cos(\angle PAH))-(PA^2+OA^2)=RD^2+HD^2-RK^2-OK^2 $$This expression can be clearly proved because it will all be expressions related to $ABC$.
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Mogmog8
1080 posts
#54 • 2 Y
Y by centslordm, megarnie
By Radical Axis on $(AEF),(BCEF),$ and $\Gamma,$ we see $R$ lies on $\overline{BC}.$ Then, the power of $R$ with respect to the Nine-Point circle of $\triangle ABC,$ $(N_9),$ is $$RF\cdot RE=RB\cdot RC=\textrm{pow}_{\Gamma}(R)$$as $\triangle RFB\sim\triangle RCE.$ Also, $\angle PAN=180-\angle CBA=\angle AMP$ so $\triangle AMP\sim\triangle NAP$ and $$\textrm{pow}_{(N_9)}=PM\cdot PN=PA^2=\textrm{pow}_{\Gamma}.$$Hence, $\overline{PR}$ is the radical axis of $(N_9)$ and $\Gamma.$ $\square$
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JAnatolGT_00
559 posts
#55
Y by
$R$ is radical center of $\Gamma, (AFHE), (BCEF), (MFNE)$ and therefore $R=BC\cap EF.$
$P$ is radical center of $\Gamma, (AMN), (MFNE).$ Hence $PR$ is radical axis of $\Gamma, (MFNE).$
But $OH$ passes through centers of these circles, so we are done.
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MathLuis
1451 posts
#56
Y by
Another speedrun les goooo.
By radax on $(AQEF), (BFEC), (ABC)$ we have that $R$ lies on $BC$ now let $D$ the projection from $A$ to $BC$, then its known that $DFMEN$ is cyclic becuase they lie on the nine point circle and since the nine point center lies on $OH$ is enough to show that $PR$ is the radical axis of the nine-point circle and $(ABC)$. Now by PoP
$$RE \cdot RF=RB \cdot RC \implies R \; \text{lying inside the radax of} \; (ABC), (DFMEN)$$Now by angle chasing knowing that $D$ and $A$ are symetric w.r.t. $MN$ we have that
$$\angle RAM=\angle ACB=\angle ANM=\angle DNM \implies PD \; \text{tangent to} \; (DFMEN) \implies PD^2=PN \cdot PM$$Wait, rememebr when i said $D$ and $A$ are symetric w.r.t. $MN$, hence $PD=PA$ meaning that $PA^2=PM \cdot PN$ hence $P$ lies on the radax of $(ABC), (DFMEN)$
Thus, we are done :blush:
This post has been edited 1 time. Last edited by MathLuis, Feb 17, 2022, 5:04 PM
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pi271828
3363 posts
#58
Y by
Let $\omega$ denote the nine-point circle. Notice that since the center of $\omega$ is the centroid $G$, and $O$, $H$, $G$ are collinear, it suffices to prove that $PR$ is the radical axis of $\omega$ and $\Gamma$. Let $\gamma$ denote $(OMAN)$. Note that $\mathrm{Pow}(P, \omega) = PM \cdot PN = \mathrm{Pow}(P, \gamma)$. Now, since $A$ is the antipode of $O$ across $\gamma$, we have that $PA$ is tangent to $\gamma$ and therefore $\mathrm{Pow}(P, \gamma) = \mathrm{Pow}(P, \Gamma)$. It suffices to prove that $R$ lies on the radical axis. Note that $BFEC$ is cyclic. Therefore by radical axis theorem, $AQ, EF, BC$ all concur at point $R$. Because $R$ is the radical center, we are done.
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IAmTheHazard
5000 posts
#59 • 1 Y
Y by centslordm
Basically was told the fact that nuked the problem but I think typing this up will help me remember it:

$P$ lies on the orthic axis of $\triangle ABC$, since $PA^2=PM\cdot PN$. More concretely, $\overline{PR}$ is the radical axis of $(ABC)$ and the 9-point circle, since $RE\cdot RF=RB\cdot RC$ (where $R$ lies on $\overline{BC}$ because of radical center) and $PA^2=PM\cdot PN$, so $\overline{PR}$ is perpendicular to $\overline{ON_9}$ which is just the Euler line. $\blacksquare$
This post has been edited 1 time. Last edited by IAmTheHazard, Jul 3, 2023, 2:12 PM
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IAmTheHazard
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#60 • 1 Y
Y by centslordm
Here's a complex solution (some calculations omitted, but I did do them):

First note that by radical center on $(AEF)$, $(BCEF)$, and $(ABC)$, we find that $R$ is also the intersection of $\overline{AQ}$ and $\overline{BC}$. Since it is well-known that $Q$, the $A$-antipode, and the midpoint of $\overline{BC}$ are collinear, we set up the equation
$$\frac{q+a}{q-\frac{b+c}{2}}=\frac{\frac{1}{q}+\frac{1}{a}}{\frac{1}{q}-\frac{\frac{1}{b}+\frac{1}{c}}{2}}.$$This can be neatly cross-multiplied, and since $q=-a$ is a factor we divide out $q+a$ and obtain a linear equation in $q$, which we can solve to yield
$$q=\frac{2a+b+c}{2+\frac{a}{b}+\frac{a}{c}}.$$Now let $P'=\overline{AA} \cap \overline{BC}$. By complex intersection,
$$p'=\frac{a^2(b+c)-2abc}{a^2-bc},$$and since $P$ is the midpoint of $\overline{AP}$ for homothety reasons we find that
$$p=\frac{a^3+a^2b+a^2c-3abc}{2(a^2-bc)}.$$We then calculate $R$, which is surprisingly neat since a lot of $2+\tfrac{a}{b}+\tfrac{a}{c}$ terms die:
$$r=\frac{a^2b+a^2c+ab^2+ac^2-2abc-b^2c-bc^2}{2(a^2-bc)},$$so
$$p-r=\frac{a^3-ab^2-ac^2-abc+b^2c+bc^2}{2(a^2-bc)}.$$We would like to show that this is a pure imaginary multiple of $a+b+c$, so using the Euclidean algorithm we try to divide out $a+b+c$ from the numerator and find that it actually equals $(a+b+c)(a-b)(a-c)$, so we want to show that
$$\frac{(a-b)(a-c)}{a^2-bc}=-\frac{(\frac{1}{a}-\frac{1}{b})(\frac{1}{a}-\frac{1}{c})}{\frac{1}{a^2}-\frac{1}{bc}},$$but this is obvious after multiplying the numerator and denominator of the RHS by $a^2bc$, so we're done. $\blacksquare$
This post has been edited 2 times. Last edited by IAmTheHazard, Aug 29, 2023, 12:08 AM
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trinhquockhanh
522 posts
#61
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https://i.ibb.co/hZvgFkX/Screenshot-2023-08-22-200814.png
It's easy to prove that $R\in BC,$ also note that $PR$ is the radical axis of $(O)$ and the Euler circle of $\triangle ABC$

$\Rightarrow PR\perp OH.$
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Pyramix
419 posts
#62
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Let $S$ be the mid-point of $\overline BC$, $D$ be the feet of altitudes from $A$ in $\triangle ABC$. Then, $M,N,D,E,F,S,$ are cyclic as they form the nine-point circle.

Note that $R$ is the radical center of circles $(AQEF)$, $(MNDS)$, $(ABC)$. So, $R,B,C$ are collinear. Since $(BCEF)$ is cyclic, we have $RB\cdot RC=RE\cdot RF$. Since $(EFDS)$ is cyclic, we have $RE\cdot RF=RD\cdot RS$. Hence, we have $RD\cdot RS=RB\cdot RC$, which means $Pow_{(ABC)}(R)=Pow_{(MNDS)}(R)$ and hence $R$ lies on the radical axis of the circumcircle and the nine-point circle.

Similarly, since $\overline{PA}$ is tangent to $(ABC)$ which is also tangent to $(AMN)$ because $(AMN)$ is just $(ABC)$ with homothety at $A$ with factor $\frac12$, we have $PA^2=PM\cdot PN$ which means $Pow_{(ABC)}(P)=Pow_{(MNDS)}(P)$ and hence $P$ lies on the radical axis of the circumcircle and the nine-point circle. So, $PR$ is the radical axis, which means $PR\perp ON_9$ and hence $PR\perp OH$. $\blacksquare$

Remark. If $U,S$ are the intersections of the $A-$median and the nine-point circle, then $\overline{PU},\overline{PD}$ are the tangents from $P$ to nine-point circle. So, $DU$ is the pole of $P$ in nine-point circle. Also, the mid-point of $\overline{AH}$, say $L$, lies on the pole of $R$.
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ihatemath123
3430 posts
#63 • 1 Y
Y by GeoKing
Applying the radical axis theorem on $(ABC)$, $(AMN)$ and $(MNEF)$, it suffices to show that $R$ has equal power WRT $(ABC)$ and $(MNEF)$. This is true because $RQ \cdot RA = RF \cdot RE$.
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OronSH
1720 posts
#64 • 1 Y
Y by GeoKing
Since $(AMN)$ is tangent to $(ABC)$ by homothety, $PA^2=PB\cdot PC,$ so $P$ lies on the radical axis of the circumcircle and the nine point circle. Since $(AEFQ)$ is cyclic, $RQ\cdot RA=RE\cdot RF,$ so $R$ lies on the radical axis of the circumcircle and the nine point circle. Thus $PR$ is perpendicular to the line through their centers which is $OH.$
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Eka01
204 posts
#65 • 1 Y
Y by AaruPhyMath
It is well known that $R$ lies on the radical axis of the nine point circle and the circumcircle, also called the orthic axis.
By radax on $(ABC),(AMN),(MNEF)$, we see that $P$ also lies on this radical axis and the line $OH$ is the line joining the centers of the $NPC$ and $(ABC)$ so it trivially follows that $PR \perp OH$.
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ohiorizzler1434
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#67
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Bro what the sigma? This is triv! This is trivialised by radical axis because P and R have equal power wrt the NPC and the circumcircle!
This post has been edited 2 times. Last edited by Luis González, Mar 17, 2025, 1:22 PM
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