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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
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0 replies
jlacosta
May 1, 2025
0 replies
Interesting inequality
sqing   0
a minute ago
Source: Own
Let $a,b\geq 0, 2a+2b+ab=5.$ Prove that
$$a+b^3+a^3b+\frac{101}{8}ab\leq\frac{125}{8}$$
0 replies
1 viewing
sqing
a minute ago
0 replies
Inequality em981
oldbeginner   19
N 6 minutes ago by sqing
Source: Own
Let $a, b, c>0, a+b+c=3$. Prove that
\[\sqrt{a+\frac{9}{b+2c}}+\sqrt{b+\frac{9}{c+2a}}+\sqrt{c+\frac{9}{a+2b}}+\frac{2(ab+bc+ca)}{9}\ge\frac{20}{3}\]
19 replies
oldbeginner
Sep 22, 2016
sqing
6 minutes ago
Inspired by RMO 2006
sqing   4
N 9 minutes ago by sqing
Source: Own
Let $ a,b >0  . $ Prove that
$$  \frac {a^{2}+1}{b+k}+\frac { b^{2}+1}{ka+1}+\frac {2}{a+kb}  \geq \frac {6}{k+1}  $$Where $k\geq 0.03 $
$$  \frac {a^{2}+1}{b+1}+\frac { b^{2}+1}{a+1}+\frac {2}{a+b}  \geq 3  $$
4 replies
sqing
Saturday at 3:24 PM
sqing
9 minutes ago
Inspired by 2025 Beijing
sqing   11
N 18 minutes ago by sqing
Source: Own
Let $ a,b,c,d >0  $ and $ (a^2+b^2+c^2)(b^2+c^2+d^2)=36. $ Prove that
$$ab^2c^2d \leq 8$$$$a^2bcd^2 \leq 16$$$$ ab^3c^3d \leq \frac{2187}{128}$$$$ a^3bcd^3 \leq \frac{2187}{32}$$
11 replies
sqing
Saturday at 4:56 PM
sqing
18 minutes ago
nice ecuation
MihaiT   1
N Yesterday at 7:24 PM by Hello_Kitty
Find real values $m$ , s.t. ecuation: $x+1=me^{|x-1|}$ have 2 real solutions .
1 reply
MihaiT
Yesterday at 2:03 PM
Hello_Kitty
Yesterday at 7:24 PM
Linear algebra problem
Feynmann123   1
N Yesterday at 3:51 PM by Etkan
Let A \in \mathbb{R}^{n \times n} be a matrix such that A^2 = A and A \neq I and A \neq 0.

Problem:
a) Show that the only possible eigenvalues of A are 0 and 1.
b) What kind of matrix is A? (Hint: Think projection.)
c) Give a 2×2 example of such a matrix.
1 reply
Feynmann123
Yesterday at 9:33 AM
Etkan
Yesterday at 3:51 PM
Linear algebra
Feynmann123   6
N Yesterday at 1:09 PM by OGMATH
Hi everyone,

I was wondering whether when I tried to compute e^(2x2 matrix) and got the expansions of sinx and cosx with the method of discounting the constant junk whether it plays any significance. I am a UK student and none of this is in my School syllabus so I was just wondering…


6 replies
Feynmann123
Saturday at 6:44 PM
OGMATH
Yesterday at 1:09 PM
Local extrema of a function
MrBridges   2
N Yesterday at 11:36 AM by Mathzeus1024
Calculate the local extrema of the function $f:\mathbb{R}^2 \rightarrow \mathbb{R}$, $(x,y)\mapsto x^4+x^5+y^6$. Are they isolated?
2 replies
MrBridges
Jun 28, 2020
Mathzeus1024
Yesterday at 11:36 AM
Integral
Martin.s   3
N Yesterday at 10:52 AM by Figaro
$$\int_0^{\pi/6}\arcsin\Bigl(\sqrt{\cos(3\psi)\cos\psi}\Bigr)\,d\psi.$$
3 replies
Martin.s
May 14, 2025
Figaro
Yesterday at 10:52 AM
Reducing the exponents for good
RobertRogo   1
N Yesterday at 9:29 AM by RobertRogo
Source: The national Algebra contest (Romania), 2025, Problem 3/Abstract Algebra (a bit generalized)
Let $A$ be a ring with unity such that for every $x \in A$ there exist $t_x, n_x \in \mathbb{N}^*$ such that $x^{t_x+n_x}=x^{n_x}$. Prove that
a) If $t_x \cdot 1 \in U(A), \forall x \in A$ then $x^{t_x+1}=x, \forall x \in A$
b) If there is an $x \in A$ such that $t_x \cdot 1 \notin U(A)$ then the result from a) may no longer hold.

Authors: Laurențiu Panaitopol, Dorel Miheț, Mihai Opincariu, me, Filip Munteanu
1 reply
RobertRogo
May 20, 2025
RobertRogo
Yesterday at 9:29 AM
Sequence divisible by infinite primes - Brazil Undergrad MO
rodamaral   5
N Yesterday at 8:01 AM by cursed_tangent1434
Source: Brazil Undergrad MO 2017 - Problem 2
Let $a$ and $b$ be fixed positive integers. Show that the set of primes that divide at least one of the terms of the sequence $a_n = a \cdot 2017^n + b \cdot 2016^n$ is infinite.
5 replies
rodamaral
Nov 1, 2017
cursed_tangent1434
Yesterday at 8:01 AM
Reduction coefficient
zolfmark   2
N Yesterday at 7:42 AM by wh0nix

find Reduction coefficient of x^10

in(1+x-x^2)^9
2 replies
zolfmark
Jul 17, 2016
wh0nix
Yesterday at 7:42 AM
a^2=3a+2imatrix 2*2
zolfmark   4
N Yesterday at 2:44 AM by RenheMiResembleRice
A
matrix 2*2

A^2=3A+2i
A^3=mA+Li


i means identity matrix,

find constant m ، L
4 replies
zolfmark
Feb 23, 2019
RenheMiResembleRice
Yesterday at 2:44 AM
Find solution of IVP
neerajbhauryal   3
N Yesterday at 12:47 AM by MathIQ.
Show that the initial value problem \[y''+by'+cy=g(t)\] with $y(t_o)=0=y'(t_o)$, where $b,c$ are constants has the form \[y(t)=\int^{t}_{t_0}K(t-s)g(s)ds\,\]

What I did
3 replies
neerajbhauryal
Sep 23, 2014
MathIQ.
Yesterday at 12:47 AM
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
6877 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
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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.

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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
642 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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