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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)!

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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
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
Inspired by qrxz17
sqing   2
N 10 minutes ago by MathsII-enjoy
Source: Own
Let $a, b,c>0 ,(a^2+b^2+c^2)^2 - 2(a^4+b^4+c^4) = 27 $. Prove that $$a+b+c\geq 3\sqrt {3}$$
2 replies
sqing
an hour ago
MathsII-enjoy
10 minutes ago
D,E,F are collinear.
TUAN2k8   1
N 12 minutes ago by Beelzebub
Source: Own
Help me with this:
1 reply
TUAN2k8
Yesterday at 1:07 AM
Beelzebub
12 minutes ago
Inspired by qrxz17
sqing   2
N 24 minutes ago by MathsII-enjoy
Source: Own
Let $ a,b,c $ be reals such that $ (a^2+b^2)^2 + (b^2+c^2)^2 +(c^2+a^2)^2 = 28 $ and $  (a^2+b^2+c^2)^2 =16. $ Find the value of $ a^2(a^2-1) + b^2(b^2-1)+c^2(c^2-1).$
2 replies
sqing
2 hours ago
MathsII-enjoy
24 minutes ago
Prove DK and BC are perpendicular.
yunxiu   63
N 26 minutes ago by sknsdkvnkdvf
Source: 2012 European Girls’ Mathematical Olympiad P1
Let $ABC$ be a triangle with circumcentre $O$. The points $D,E,F$ lie in the interiors of the sides $BC,CA,AB$ respectively, such that $DE$ is perpendicular to $CO$ and $DF$ is perpendicular to $BO$. (By interior we mean, for example, that the point $D$ lies on the line $BC$ and $D$ is between $B$ and $C$ on that line.)
Let $K$ be the circumcentre of triangle $AFE$. Prove that the lines $DK$ and $BC$ are perpendicular.

Netherlands (Merlijn Staps)
63 replies
yunxiu
Apr 13, 2012
sknsdkvnkdvf
26 minutes ago
Geometry with fix circle
falantrng   34
N 28 minutes ago by Aiden-1089
Source: RMM 2018 Problem 6
Fix a circle $\Gamma$, a line $\ell$ to tangent $\Gamma$, and another circle $\Omega$ disjoint from $\ell$ such that $\Gamma$ and $\Omega$ lie on opposite sides of $\ell$. The tangents to $\Gamma$ from a variable point $X$ on $\Omega$ meet $\ell$ at $Y$ and $Z$. Prove that, as $X$ varies over $\Omega$, the circumcircle of $XYZ$ is tangent to two fixed circles.
34 replies
falantrng
Feb 25, 2018
Aiden-1089
28 minutes ago
Set of perfect powers is irreducible
Assassino9931   2
N 32 minutes ago by navid
Source: Al-Khwarizmi International Junior Olympiad 2025 P4
For two sets of integers $X$ and $Y$ we define $X\cdot Y$ as the set of all products of an element of $X$ and an element of $Y$. For example, if $X=\{1, 2, 4\}$ and $Y=\{3, 4, 6\}$ then $X\cdot Y=\{3, 4, 6, 8, 12, 16, 24\}.$ We call a set $S$ of positive integers good if there do not exist sets $A,B$ of positive integers, each with at least two elements and such that the sets $A\cdot B$ and $S$ are the same. Prove that the set of perfect powers greater than or equal to $2025$ is good.

(In any of the sets $A$, $B$, $A\cdot B$ no two elements are equal, but any two or three of these sets may have common elements. A perfect power is an integer of the form $n^k$, where $n>1$ and $k > 1$ are integers.)

Lajos Hajdu and Andras Sarkozy, Hungary
2 replies
Assassino9931
May 9, 2025
navid
32 minutes ago
Stop Projecting your insecurities
naman12   53
N 35 minutes ago by EeEeRUT
Source: 2022 USA TST #2
Let $ABC$ be an acute triangle. Let $M$ be the midpoint of side $BC$, and let $E$ and $F$ be the feet of the altitudes from $B$ and $C$, respectively. Suppose that the common external tangents to the circumcircles of triangles $BME$ and $CMF$ intersect at a point $K$, and that $K$ lies on the circumcircle of $ABC$. Prove that line $AK$ is perpendicular to line $BC$.

Kevin Cong
53 replies
naman12
Dec 12, 2022
EeEeRUT
35 minutes ago
Roots of unity
Henryfamz   1
N 41 minutes ago by Mathzeus1024
Compute $$\sec^4\frac\pi7+\sec^4\frac{2\pi}7+\sec^4\frac{3\pi}7$$
1 reply
Henryfamz
May 13, 2025
Mathzeus1024
41 minutes ago
Shortest number theory you might've seen in your life
AlperenINAN   11
N 41 minutes ago by Assassino9931
Source: Turkey JBMO TST 2025 P4
Let $p$ and $q$ be prime numbers. Prove that if $pq(p+1)(q+1)+1$ is a perfect square, then $pq + 1$ is also a perfect square.
11 replies
AlperenINAN
May 11, 2025
Assassino9931
41 minutes ago
AZE JBMO TST
IstekOlympiadTeam   10
N an hour ago by Assassino9931
Source: AZE JBMO TST
Prove that there are not intgers $a$ and $b$ with conditions,
i) $16a-9b$ is a prime number.
ii) $ab$ is a perfect square.
iii) $a+b$ is also perfect square.
10 replies
IstekOlympiadTeam
May 2, 2015
Assassino9931
an hour ago
Iran TST Starter
M11100111001Y1R   3
N an hour ago by dgrozev
Source: Iran TST 2025 Test 1 Problem 1
Let \( a_n \) be a sequence of positive real numbers such that for every \( n > 2025 \), we have:
\[
a_n = \max_{1 \leq i \leq 2025} a_{n-i} - \min_{1 \leq i \leq 2025} a_{n-i}
\]Prove that there exists a natural number \( M \) such that for all \( n > M \), the following holds:
\[
a_n < \frac{1}{1404}
\]
3 replies
M11100111001Y1R
May 27, 2025
dgrozev
an hour ago
An interesting functional equation
giannis2006   3
N an hour ago by GreekIdiot
Source: Own
Find all functions $f:R^+->R^+$ such that:
$f(xf(y))=xy-xf(x)+f(x)^2$ for all $x,y>0$

The most difficult version of this problem is the following:
Find all functions $f:R^+->R^+$ such that:
$f(xf(y+f(x)))=xy+f(x)^2$ for all $x,y>0$
3 replies
giannis2006
Jun 8, 2023
GreekIdiot
an hour ago
A long non-classical problem
M11100111001Y1R   1
N an hour ago by dgrozev
Source: Iran TST 2025 Test 3 Problem 2
Suppose \( n \in \mathbb{N} \) is a natural number. A function \( f(x, y) \) is called \textit{\( n \)-friendly} if for fewer than 1\% of the integers \( k \) with \( -n \leq k \leq n \), the equation \( f(x, y) = k \) has a solution in natural numbers \( (x, y) \) such that \( \frac{y_0}{x_0} \in \left[\frac{1}{100}, 100\right] \), where \( (x_0, y_0) \) is a solution. Suppose \( f(x, y) \leq g(x, y) \), where \( g(x, y) \) is a polynomial with real coefficients, negative leading coefficients, and total degree greater than 2, and for every real number \( x \), we have \( g(x, y) \to \infty \) as \( \frac{y}{x} \in \left[\frac{1}{100}, 100\right] \). Prove that for sufficiently large \( n \), the function \( f \) is not \( n \)-friendly.
1 reply
M11100111001Y1R
May 27, 2025
dgrozev
an hour ago
Another FE
M11100111001Y1R   1
N an hour ago by Mathzeus1024
Source: Iran TST 2025 Test 2 Problem 3
Find all functions $f: \mathbb{R}^+ \to \mathbb{R}^+$ such that for all $x,y>0$ we have:
$$f(f(f(xy))+x^2)=f(y)(f(x)-f(x+y))$$
1 reply
M11100111001Y1R
2 hours ago
Mathzeus1024
an hour ago
Another orthocenter problem <3
Guendabiaani   17
N Dec 29, 2024 by HamstPan38825
Source: Mexico National Olympiad 2019 P6
Let $ABC$ be a triangle such that $\angle BAC = 45^{\circ}$. Let $H,O$ be the orthocenter and circumcenter of $ABC$, respectively. Let $\omega$ be the circumcircle of $ABC$ and $P$ the point on $\omega$ such that the circumcircle of $PBH$ is tangent to $BC$. Let $X$ and $Y$ be the circumcenters of $PHB$ and $PHC$ respectively. Let $O_1,O_2$ be the circumcenters of $PXO$ and $PYO$ respectively. Prove that $O_1$ and $O_2$ lie on $AB$ and $AC$, respectively.
17 replies
Guendabiaani
Nov 12, 2019
HamstPan38825
Dec 29, 2024
Another orthocenter problem <3
G H J
G H BBookmark kLocked kLocked NReply
Source: Mexico National Olympiad 2019 P6
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Guendabiaani
778 posts
#1 • 1 Y
Y by Adventure10
Let $ABC$ be a triangle such that $\angle BAC = 45^{\circ}$. Let $H,O$ be the orthocenter and circumcenter of $ABC$, respectively. Let $\omega$ be the circumcircle of $ABC$ and $P$ the point on $\omega$ such that the circumcircle of $PBH$ is tangent to $BC$. Let $X$ and $Y$ be the circumcenters of $PHB$ and $PHC$ respectively. Let $O_1,O_2$ be the circumcenters of $PXO$ and $PYO$ respectively. Prove that $O_1$ and $O_2$ lie on $AB$ and $AC$, respectively.
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Al3jandro0000
804 posts
#2 • 1 Y
Y by Adventure10
hint
This post has been edited 1 time. Last edited by Al3jandro0000, Nov 13, 2019, 4:04 AM
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MarkBcc168
1595 posts
#3 • 8 Y
Y by ShinyDitto, Bassiskicking, Professor-Mom, ILOVEMYFAMILY, Adventure10, math_comb01, soryn, Hu_cosine
I have a very different and much shorter solution.

Let $A' = PH\cap\omega$. Then using Reim's theorem with $\overline{PHA'}$ and $\overline{BBC}$ gives $CA'\parallel BH$ or $A'$ is the $A$-antipode in $\omega$. This means $\odot(PCH)$ is also tangent to $BC$ which means that it suffices to show that $O_1\in AB$.

Now we use a common trick. We will reflect $O$ across $AB$ to get $O'$ and show that $P,X,O,O'$ are concyclic instead. Notice that $O'X$ and $OO'$ are perpendicular bisectors of $BH, BA$ respectively. Thus
$$\angle XO'O = \angle ABH = 45^{\circ}$$On the other hand, we can easily chase $\angle XPO$ Since $OX$ is the perpendicular bisector of $BP$, we get
$$\angle XPO = \angle XBO = 90^{\circ} - \angle OBC = 45^{\circ}$$so we are done.
This post has been edited 1 time. Last edited by MarkBcc168, Nov 13, 2019, 10:21 AM
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Pathological
578 posts
#4 • 2 Y
Y by Adventure10, soryn
Notice that since $\angle BPC + \angle BHC = 180$, we know from the condition of the problem that $H$ is the $P-$HM point of $\triangle BPC.$ This means that $HP$ bisects $BC$, and so $P \in HA'.$

We have $\angle PYC = 2 (180 - \angle PHC) = 2 \angle A'HC = 2 \angle PCB = \angle POB.$

Analogously $\angle PXB = \angle POC.$ This means that $PXBO$ and $POCY$ are similar kites. Hence, we've that $\angle PO_1O = 2 \angle PXO = 2 \angle POY = \angle POC.$

Similarly, $\angle PO_2O = \angle POB$, so we get that $\triangle PO_1O \sim \triangle POC$ and $\triangle PO_2O \sim \triangle POB.$

We are now set up to begin the complex bash. WLOG let $B = 1, C = i, A = a$ for some complex number $a$ on the unit circle. If we let $E, F$ be the feet of $B, C$ onto $AC, AB$ respectively, then we know that $\triangle PEF \sim \triangle PCB.$ We have $F = \frac{a+i+1-ai}{2}$ and $E = \frac{a+i+1+ai}{2}$. This means that $P = \frac{EB - CF}{E + B - F - C} = \frac{ai - 1 - a}{ai - 1 + i}$.

Hence, we have from $\triangle PO_1O \sim \triangle POC$ that $O_2 = P + \frac{P^2}{1-P} = \frac{P}{1-P} = \frac{ai-1-a}{a+i}.$

To show that $A, O_2, C$ are collinear, we need to show that:

$$\frac{a - \frac{ai-1-a}{a+i}}{a - i} \in \mathbb{R}.$$
This simplifies to $\frac{a^2+a+1}{a^2+1}$, and this is clearly equal to its complex conjugate, hence it's real.

Therefore, $O_2 \in AC$. Analogously, $O_1 \in AB$ and we're done.

$\square$
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ShinyDitto
63 posts
#6 • 3 Y
Y by Adventure10, Mango247, soryn
I am only going to prove $O_1 \in AB$ since the other case is anologous. Once you get $POO_1 \sim PCO$, write: \[ \frac{PC}{CO} = \frac{PO}{OO_1} = \frac{AO}{OO_1}. \]Let $P'$ be the point where the parallel to $AB$ through $P$ cuts $\omega$ and let $A'$ be the point where $AH$ cuts $\omega$. Notice that $HO \perp CP'$, thus $HP'=HC$. Furthermore $CHP' \sim COP$. Thus we write: \[ \frac{PC}{CO} = \frac{P'C}{CH} = \frac{P'C}{CA'}. \]Finally $\angle AOO_1 = \angle P'CA' \implies AOO_1 \sim P'CA'  \implies \angle O_1AO= \angle A'P'C = \angle BAO \implies O_1 \in AB$.
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juckter
324 posts
#7 • 2 Y
Y by Adventure10, Mango247
Bonus problem: In the same configuration, let $B'$ and $C'$ be the $B$ and $C$-antipodes respectively. Show that the circumcenter of $AO_1O_2$ lies on $B'C'$.

(This was actually included in the original proposal for this problem, but it was deemed too difficult for the National Olympiad)
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ShinyDitto
63 posts
#8 • 1 Y
Y by Adventure10
juckter wrote:
Bonus problem: In the same configuration, let $B'$ and $C'$ be the $B$ and $C$-antipodes respectively. Show that the circumcenter of $AO_1O_2$ lies on $B'C'$.

(This was actually included in the original proposal for this problem, but it was deemed too difficult for the National Olympiad)

We can use MarkBcc168's trick. Reflect $A$ across $B'C'$ to get $K$. We may easily check that $\angle O_1C'K = \angle O_2B'K$. Also: \[ \frac{O_2B'}{O_1C'} = \frac{O_2P}{O_1P} = \frac{CP}{BP} = \frac{BA'}{CA'} = \frac{B'K}{C'K}. \]$\implies  O_1C'K \sim O_2B'K \implies O_1KO_2 \sim C'KB' \implies \angle O_1KO_2=135^{\circ} \implies AO_1KO_2 \text{ cyclic}$.
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ShinyDitto
63 posts
#9 • 1 Y
Y by Adventure10
juckter wrote:
Bonus problem: In the same configuration, let $B'$ and $C'$ be the $B$ and $C$-antipodes respectively. Show that the circumcenter of $AO_1O_2$ lies on $B'C'$.

(This was actually included in the original proposal for this problem, but it was deemed too difficult for the National Olympiad)

We might also let $O_3$ be said circumcenter. Check that $O_3$ lies on the circumcircles of $PC'O_1$ and $PB'O_2$. Furthermore $O_3$ lies on $B'C'$.
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v_Enhance
6878 posts
#10 • 3 Y
Y by v4913, ChCar, Rounak_iitr
Solution from Twitch Solves ISL:

We let $\overline{BK}$ and $\overline{CL}$ be the altitudes of $\triangle ABC$. The circle with diameter $\overline{BC}$ will be denoted by $\gamma$; and we'll denote the center by $M$.

Claim: The circle $(PHC)$ is also tangent to line $BC$.
Proof. We are given $\measuredangle PBC = \measuredangle PHB$. Since $\measuredangle BHC = -\measuredangle BAC = -\measuredangle BPC$, it follows $\measuredangle PCB = \measuredangle PHC$ too. $\blacksquare$

Claim: Points $P$, $H$, $M$ are collinear. Actually, $P$ is the inverse of $H$ with respect to $\gamma$.
Proof. Line $PH$ bisects $\overline{BC}$ by radical axis on $(BPH)$ and $(CPH)$. Also, $MH \cdot MP = MB^2 = MC^2$. $\blacksquare$

[asy]size(10cm); pair B = dir(180); pair C = dir(0); pair K = dir(65); pair L = dir(90)*K; filldraw(unitcircle, invisible, blue); filldraw(B--C--K--L--cycle, invisible, blue); pair A = extension(B, L, C, K); pair H = extension(B, K, C, L); draw(B--K, blue); draw(C--L, blue); pair O = circumcenter(A, B, C);
pair P = 1/conj(H); pair M = origin; filldraw(circumcircle(P, B, H), invisible, orange); pair X = circumcenter(P, H, B); pair O_1 = circumcenter(P, X, O); draw(L--A--K, dotted+deepcyan);
draw(circumcircle(X, O, L), dashed+deepgreen); draw(P--M, blue); draw(CP(O_1, P), dotted+grey);
dot("$B$", B, dir(225)); dot("$C$", C, dir(315)); dot("$K$", K, dir(K)); dot("$L$", L, dir(L)); dot("$A$", A, dir(A)); dot("$H$", H, dir(110)); dot("$O$", O, dir(60)); dot("$P$", P, dir(P)); dot("$M$", M, dir(-90)); dot("$X$", X, dir(X)); dot("$O_1$", O_1, dir(O_1));
/* TSQ Source:
B = dir 180 R225 C = dir 0 R315 K = dir 65 L = dir(90)*K unitcircle 0.1 lightcyan / blue B--C--K--L--cycle 0.1 lightcyan / blue A = extension B L C K H = extension B K C L R110 B--K blue C--L blue O = circumcenter A B C R60
P = 1/conj(H) M = origin R-90 circumcircle P B H 0.1 yellow / orange X = circumcenter P H B O_1 = circumcenter P X O L--A--K dotted deepcyan
circumcircle X O L dashed deepgreen P--M blue CP O_1 P dotted grey
*/ [/asy]
We now actually use the condition that $\measuredangle BAC = 45^{\circ}$, which is equivalent to $\measuredangle BHC = 135^{\circ}$ and $\measuredangle BOC = 90^{\circ}$. This means that $O$ is the arc midpoint of $(BC)$.

Claim: $LOKH$ is a parallelogram.
Proof. Since arcs $KL$ and $OB$ of $\gamma$ measure $90^{\circ}$, the arcs $BL$ and $OK$ are equal, hence $\overline{LO} \parallel \overline{BK}$. Similarly $\overline{KO} \parallel \overline{CL}$. $\blacksquare$

Claim: We have $\measuredangle XPO = 45^{\circ}$.
Proof. Since $\triangle HLK \sim \triangle HBC$, and $\overline{HO}$ bisects $\overline{LK}$ by previous claim, it follows \[ \measuredangle BHM = \measuredangle OHL. \]Now \begin{align*} 		\measuredangle XPO &= \measuredangle XPH + \measuredangle MPO 		= (90^{\circ} - \measuredangle HBP) + \measuredangle HOM \\ 		&= 90^{\circ} - \measuredangle MBP + \measuredangle MBH + \measuredangle HOM \\ 		&= 90^{\circ} - \measuredangle BHM + \measuredangle MBH + \measuredangle HOM \\ 		&= 90^{\circ} - \measuredangle OHL + \measuredangle MBH + \measuredangle HOM \\ 		&= 90^{\circ} + \measuredangle(LH, HO) + \measuredangle(MB, BH) + \measuredangle(HO,MO) \\ 		&= 90^{\circ} + \measuredangle(LH, BH) + \measuredangle(MB, MO) = 45^{\circ}.  	\end{align*}$\blacksquare$

Claim: We have $X$, $O_1$, $O$, $L$, $H$ are concyclic, in the circle with diameter $\overline{XO}$.
Proof. First, since $\triangle LBH$ is a $45^{\circ}$-$45^{\circ}$-$90^{\circ}$ triangle and $XB=XH$, it follows that $\overline{XL}$ is the perpendicular bisector of $\overline{BH}$. Hence, $\measuredangle XLO = 90^{\circ}$.
On the other hand, since $\measuredangle XPO = 45^{\circ}$, we have $\measuredangle XO_1O = 90^{\circ}$. This implies concyclic. $\blacksquare$

Finally, $\measuredangle BLO = 135^{\circ}$ and \[ \measuredangle OLO_1 = \measuredangle OXO_1 = 90^{\circ} - \measuredangle XPO = 45^{\circ} \]this implies $O_1$ lies on line $BL$, ergo lies on line $AB$.
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L567
1184 posts
#12 • 1 Y
Y by geometry6
Solved with geometry6, mueller.25

Let $P'$ be the $H$ humpty point in $\triangle HBC$. So, $\angle BP'C = \angle BP'H + \angle CP'H = \angle HBC + \angle HCB = \angle BAC$ and so $P' \in (ABC)$ and since $(P'BH)$ is tangent to $BC$, we have $P' = P$, which gives $(PHC)$ is tangent to $BC$ too. So, it suffices to show $O_1 \in AB$.

Let $BX \cap (ABC) = Q$. Since $\angle XQP = \angle BQP = \frac{\angle BOP}{2} = \angle XOP$, we have $PQOX$ is cyclic. So, $\angle XPO = \angle XQO = \angle BQO = 90 - \angle OBC = 45^\circ$.

Define $O'$ to be the reflection of $O$ across line $AB$. It suffices to show that $O' \in (PXO)$, or that $\angle XO'O = 45^\circ$. But note that $O'$ is the center of $(AHB)$, so $\angle XO'O = \angle HBA = 45^\circ$, and so we are done. $\blacksquare$
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Eyed
1065 posts
#13 • 1 Y
Y by hyc2021
Complex cuz

Let $A', C'$ be the antipode of $A, C,$ and $M$ the midpoint of $BC$. Note that
\[\angle BPA' = 90 - \angle C = \angle HBC = \angle BPH\]so $P, H, A'$ are collinear. It is well known that $M$ lies on this line. Next, note that $OX$ perpendicularly bisects $PB$ so
\[\angle PXO = 90 + \angle PHB = 90 + \overarc{PB} + 45 = 135 + \overarc{PB} = 180 - (45 + \overarc{PB}) = 180 - \angle PC'O\]so $(PXOC')$ concyclic. Now, we set complex numbers with the circumcenter as the origin, $C = -1, B = i, C' = 1, A' = a$. We have $M = \frac{i-1}{2}, A = -a$, and
\[ap\overline{m} + m = a + p \implies p\left(a\frac{1+i}{2} + 1\right) = \frac{i-1}{2} -a \implies p = \frac{i-1-2a}{a(i+1)+2}\]$O_1$ is the circumcenter of $OPC'$. Since $C' = 1$, then $o_1 = \frac{p}{p+1} = \frac{i-1-2a}{i+ai + 1 - a}$. Now, we want to show that $O_1\in AB$, or $-ai\overline{o_1} + o_1 = -a + i$. However, this resolves to
\[-ai\overline{o_1} + o_1 = \frac{-ai(-i-1-\frac{2}{a})(-a)}{(-a)(-i - \frac{1}{a}i + 1 - \frac{1}{a})} + \frac{i-1-2a}{i+ai+1-a} = \frac{ai(-ai-a-2) + i-1-2a}{i+ai + 1 -a} = i-a.\]We conclude $O_1\in AB$. Similarly, because $\angle HPC = \angle A'PC = 90 - \angle B = \angle HCB$, we have $(PHC)$ is tangent to $BC$ as well, so by symmetry, $O_2\in AC$.
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gvole
201 posts
#14
Y by
Like everyone else, we can show $P$ is the $A$ - Queue point, also $(PHC)$ tangent to $BC$. We just want to show $O_1\in AB$. Notice that the antipode $C'$ of $C$ is on $(POX)$ due to some angles. We will now use moving points.

Animate $A$ on $(BAC)$ projectively. Let $\ell$ be the bisector of $C'O$.

Clearly $A\to AB\to \ell \cap AB$ is projective.

Also $A\to A' \to A'M \cap (ABC)=P\to P'\to P'O\cap \ell$ is projective, where $A'$ is the antipode of $A$, $M$ is the midpoint of $BC$, and $P'$ is the midpoint of $C'P$.

We want to show those two bisectors intersect on $AB$, so it's enough to check for 3 cases of $A$. It's easy to verify what happens when $A$ is the antipode of $B$ or $C$, and also when $A$ is the midpoint of major arc $BC$.
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CT17
1481 posts
#15
Y by
Let $A'$ be the $A-$ antipode. Observe that

$$\measuredangle BPH = \measuredangle CBH = \measuredangle BAA' = \measuredangle BPA'$$
so $P$ is actually the $A-$ orthic Miquel point. Now let $O'$ be the reflection of $O$ over $AB$. As $OX$ is the perpendicular bisector of $BH$, so we have

$$\measuredangle XO'O = \measuredangle HBX = 90^\circ - \measuredangle BAC$$
and

$$\measuredangle XPO = \measuredangle BPO + \measuredangle XPB = (90^\circ + \measuredangle BA'P) - (90^\circ + \measuredangle PHB = \measuredangle BA'H + \measuredangle A'HB = \measuredangle A'BH = \measuredangle BAC$$
so from $\angle BAC = 45^\circ$ it follows follows that $O'$ lies on $(PXO)$. Since $AB$ is the perpendicular bisector of $OO'$, we are done.
This post has been edited 1 time. Last edited by CT17, Jan 29, 2024, 2:37 AM
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GrantStar
821 posts
#16
Y by
Rename $P$ to $Q$ since it is the queue point. Let $O'$ be the reflection of $O$ over $AB$ and $F$ the foot from $C$ to $AB$. Let $H$ be the orthocenter of $ABC$. Let $AB$ hit $(BHQ)$ again at $J$.

Claim: $FXJH$ is cyclic
Proof. First, note that $\angle JHF = 90^{\circ}$. Then, $\angle JXH = 2\angle JBH=2\angle ABH = 90^{\circ}$ as desired. $\blacksquare$
Note by reflection that $\angle XQO=\angle XBO=45 ^{\circ}$, so it suffices to show $\angle XO'O=45^{\circ}$. But by the claim, as $\angle XFB=135 ^{\circ}$ and $\angle BFO=180^{\circ} - \angle OCB = 45 ^{\circ}$ we get that $F,X,O'$ collinear, so \[\angle XO'O=\angle FO'O=\angle FOO'=90^{\circ} - \angle FA = 45 ^{\circ}\]as desired.
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Ywgh1
139 posts
#17
Y by
Mexico 2019/4
Label the points as the following.
We divide the problem into four claims.

[asy]
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import graph; size(11.11960960330805cm); 
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ 
pen dotstyle = black; /* point style */ 
real xmin = -20.76039899260374, xmax = 32.35921061070431, ymin = -9.566482834569173, ymax = 14.026720905139266;  /* image dimensions */
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draw(circle((0.5058327053282743,-1.7350287966840203), 4.489852339053012), linewidth(1.) + ffccww); 
draw(circle((0.5058327053282743,2.754823542368992), 6.349610070941334), linewidth(1.) + ffwwqq); 
draw((-0.7078110158086872,8.98736852585242)--(-3.9840196337247384,-1.7350287966840203), linewidth(1.) + qqzzcc); 
draw((-0.7078110158086872,8.98736852585242)--(4.995685044381286,-1.7350287966840203), linewidth(1.) + qqzzcc); 
draw((-3.9840196337247384,-1.7350287966840203)--(4.995685044381286,-1.7350287966840203), linewidth(1.) + qqzzcc); 
draw((-3.9840196337247384,-1.7350287966840203)--(3.015283336501507,1.988065555626175), linewidth(1.) + afeeee); 
draw((4.995685044381286,-1.7350287966840203)--(-3.217261646981921,0.7744218344892114), linewidth(1.) + afeeee); 
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draw((0.5058327053282743,-1.7350287966840203)--(-4.919015326353041,6.054607154390085), linewidth(1.) + ffzztt); 
draw((-0.7078110158086872,-1.7350287966840203)--(-0.7078110158086872,8.98736852585242), linewidth(1.) + afeeee); 
draw(circle((-1.7390934641982323,2.485363414583891), 2.2610400852463703), linewidth(1.) + dotted + rvwvcq); 
draw((-3.217261646981921,0.7744218344892114)--(0.5058327053282743,2.754823542368992), linewidth(1.) + eqeqeq); 
draw((0.5058327053282743,2.754823542368992)--(-2.0085535919833317,4.7302895841103965), linewidth(1.) + eqeqeq); 
draw((-3.9840196337247384,2.215903286798792)--(-3.217261646981921,0.7744218344892114), linewidth(1.) + eqeqeq); 
draw((-3.9840196337247384,2.215903286798792)--(-2.0085535919833317,4.7302895841103965), linewidth(1.) + eqeqeq); 
 /* dots and labels */
dot((0.5058327053282743,-1.7350287966840203),dotstyle); 
label("$M$", (0.6340298423685562,-1.3996046169777905), NE * labelscalefactor); 
dot((-3.9840196337247384,-1.7350287966840203),dotstyle); 
label("$B$", (-3.833322345031858,-1.3996046169777905), NE * labelscalefactor); 
dot((4.995685044381286,-1.7350287966840203),linewidth(4.pt) + dotstyle); 
label("$C$", (5.136283218733036,-1.4694069949059219), NE * labelscalefactor); 
dot((0.5058327053282743,2.754823542368992),linewidth(4.pt) + dotstyle); 
label("$O$", (0.6340298423685562,3.0328463814585582), NE * labelscalefactor); 
dot((-0.7078110158086872,8.98736852585242),dotstyle); 
label("$A$", (-0.5526105824096789,9.349961583954457), NE * labelscalefactor); 
dot((3.015283336501507,1.988065555626175),linewidth(4.pt) + dotstyle); 
label("$H_1$", (3.1469154477812893,2.265020224249112), NE * labelscalefactor); 
dot((-3.217261646981921,0.7744218344892114),linewidth(4.pt) + dotstyle); 
label("$H_2$", (-3.065496187822412,1.043478610506811), NE * labelscalefactor); 
dot((-0.7078110158086893,0.007663847746394663),linewidth(4.pt) + dotstyle); 
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[/asy]
Claim 1: $P$ is the $A$-queue point of $\triangle ABC$.
Proof: Easy angle chase gives us that $BC$ is tangent to $(PHC)$, then by PoP we have that $MC^2=MB^2=MH \cdot MP$. $\blacksquare$

Claim 2: $HOH_1H_2$ is a parallelogram.
Proof: $\angle BMH_1=\angle OMH_2$ hence $\angle OH_1H_2=\angle BH_1H_2$ hence $OH_2 \| HH_1$. And similarly $OH_1 \| HH_2$. $\blacksquare$

Claim 3: $\angle XO_1O=90$.
Proof: Long angle chase. $\blacksquare$

Finally we how our last claim.
Claim 4: $OH_2O_1X$ is cyclic.
Proof: By our previous claim $\angle XO_1O=90$, and we also have that $\angle XH_2O=90$ since $XH_2$ is perpendicular to $BH$, hence it’s cyclic. $\blacksquare$

Now since $\angle O_1H_2O= \angle H_2BH=45$ and $\angle O_1XO=45$, hence we have that $O_1$ must lie on $AB$. So we are done.
This post has been edited 5 times. Last edited by Ywgh1, Aug 15, 2024, 7:52 PM
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InterLoop
279 posts
#18 • 3 Y
Y by cursed_tangent1434, idkk, ohiorizzler1434
truly rizztastic problem
solution
This post has been edited 1 time. Last edited by InterLoop, Oct 8, 2024, 10:03 AM
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cursed_tangent1434
649 posts
#19 • 3 Y
Y by InterLoop, idkk, stillwater_25
Solved with stillwater_45. Very interesting problem. After I realized that $P$ with the $A-$Queue Point of $\triangle ABC$ I thought this could turn into a orthocenter-configuration problem. Guess my wish was not to be granted.

Denote by $B_1$ and $C_1$ the feet of the altitudes from $B$ and $C$. We can start off by noting that $O$ lies on $(BC)$ as well since we have $\measuredangle COB = 2\measuredangle BAC = \frac{\pi}{2}$. Now, we prove the following claim, invoking some symmetry into the problem.

Claim : Circle $(PHC)$ is also tangent to $\overline{BC}$ and in particular, $P$ is the $A-$Queue Point of $\triangle ABC$.
Proof : Let $P'$ denote the intersection of the circles tangent to $\overline{BC}$ passing through points $B$ and $H$ and $C$ and $H$ respectively. Note that,
\[\measuredangle CPB = \measuredangle CPH + \measuredangle HPB = \measuredangle BCH + \measuredangle HBC = \frac{\pi}{4} \]which implies that $P'$ lies on $(ABC)$ and indeed $P'\equiv P$. Thus, the first part of the claim is proved. For the second part simply note that, since the radical axis $\overline{PH}$ of circles $(PHB)$ and $(PHC)$ must bisect the common tangent $\overline{BC}$, $\overline{PH}$ passes through the midpoint of side $BC$. It is well known that this implies that $P$ is the $A-$Queue Point of $\triangle ABC$.

Also note that,
\[\measuredangle OPX = \measuredangle BPX + \measuredangle OPB = \frac{\pi}{2}+\measuredangle PHB + \frac{\pi}{2} + \measuredangle BCP = \measuredangle PHC_1 + \measuredangle C_1HB + \measuredangle BAP = \frac{\pi}{4}\]A similar calculation yields that $\measuredangle YPO = \frac{\pi}{4}$ as well.

With the symmetry now established and the above observations, we can prove the following results.

Claim : Quadrilaterals $XO_1OC_1$ and $YO_2OB_1$ are cyclic.
Proof : We prove the first one, the other follows similarly due to symmetry. First note that,
\[\measuredangle OO_1X = 2\measuredangle OPX = \frac{\pi}{2}\]so it suffices to show that $\measuredangle XC_1O=\frac{\pi}{2}$ as well, which we shall do as follows.

Note that $C_1$ lies on the perpendicular bisector of segment $BH$ (since $\triangle BC_1H$ is a 45-45-90 right triangle). Also, $X$ lies on this perpendicular bisector as it is the center of $(BHP)$. Thus, $\overline{XC_1}$ is the perpendicular bisector of segment $BH$ implying that $XC_1 \perp BH$. Further note that, $O$ and $C_1$ (due the 45-45-90 right triangle $\triangle AC_1B$) lie on the perpendicular bisector of $AC$. Thus, $OC_1 \perp AC \perp BH$, implying that $OC_1 \parallel BH$. Thus, $XC_1 \perp OC_1$ and indeed $\measuredangle XC_1O = \frac{\pi}{2}$ as desired.

Now we are essentially done since,
\[\measuredangle OC_1O_1 = \measuredangle OXO_1 = \frac{\pi}{4} = \measuredangle OC_1A\]which implies that $O_1$ lies on $AB$. A similar argument shows that $O_2$ also lies on $AC$ which finishes the proof of the problem.
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HamstPan38825
8868 posts
#20
Y by
Nice! Let $A'$ be the $A$-antipode and $C'$ be the $C$-antipode.

Claim: $P$ is the orthocenter Miquel point, i.e. it lies on $\overline{HA'}$.

Proof: The actual Miquel point $P'$ satisfies $\measuredangle BP'A' = \measuredangle BAA' = \measuredangle CBH$, i.e. $\overline{CB}$ is tangent to $(P'BH)$. So $P = P'$. $\blacksquare$

Claim: $\overline{BO}$ and $\overline{CP}$ intersect at the intersection point $Q$ of $(PXO)$ and $(PBH)$.

Proof: $Q$ lies on $\overline{BO}$ as \[\angle PQO + \angle PQB = 180^\circ - \frac 12 \angle PXB + \angle PQB = 180^\circ - \angle PQB + \angle PQB = 180^\circ.\]$Q$ lies on $\overline{CP}$ because \[\measuredangle QPH = \measuredangle OBH = \angle C - 45^\circ = \measuredangle A'PC. \ \blacksquare\]Now note that $C'$ lies on $(POQX)$ because $CO \cdot CC' = 2BC^2$ as $\angle BOC = 90^\circ$. In particular, $\angle COQ = 90^\circ$, so it follows that $O_1$ is the midpoint of $\overline{C'Q}$.

To finish, let $N$ be the midpoint of $\overline{AB}$. Note that $\overline{QH} \parallel \overline{AB}$ by the same angle chase as before, so a homothety with ratio $\frac 12$ at $C'$ takes $\overline{QH}$ to $\overline{O_1N}$. It follows that $O_1$ lies on $\overline{AB}$, and the other part follows similarly.
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