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k a July Highlights and 2025 AoPS Online Class Information
jwelsh   0
Jul 1, 2025
We are halfway through summer, so be sure to carve out some time to keep your skills sharp and explore challenging topics at AoPS Online and our AoPS Academies (including the Virtual Campus)!

[list][*]Over 60 summer classes are starting at the Virtual Campus on July 7th - check out the math and language arts options for middle through high school levels.
[*]At AoPS Online, we have accelerated sections where you can complete a course in half the time by meeting twice/week instead of once/week, starting on July 8th:
[list][*]MATHCOUNTS/AMC 8 Basics
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[*]Plus, AoPS Online has a special seminar July 14 - 17 that is outside the standard fare: Paradoxes and Infinity
[*]We are expanding our in-person AoPS Academy locations - are you looking for a strong community of problem solvers, exemplary instruction, and math and language arts options? Look to see if we have a location near you and enroll in summer camps or academic year classes today! New locations include campuses in California, Georgia, New York, Illinois, and Oregon and more coming soon![/list]

MOP (Math Olympiad Summer Program) just ended and the IMO (International Mathematical Olympiad) is right around the corner! This year’s IMO will be held in Australia, July 10th - 20th. Congratulations to all the MOP students for reaching this incredible level and best of luck to all selected to represent their countries at this year’s IMO! Did you know that, in the last 10 years, 59 USA International Math Olympiad team members have medaled and have taken over 360 AoPS Online courses. Take advantage of our Worldwide Online Olympiad Training (WOOT) courses
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0 replies
jwelsh
Jul 1, 2025
0 replies
The refinement of GMA 567
mihaig   4
N 3 minutes ago by mihaig
Source: Own
Let $a_1,\ldots, a_{n}\geq0~~(n\geq4)$ be real numbers such that
$$\sum_{i=1}^{n}{a_i^2}+(n^2-3n+1)\prod_{i=1}^{n}{a_i}\geq(n-1)^2.$$Prove
$$\left(\sum_{i=1}^{n}{a_i}\right)^2+\frac{2n-1}{(n-1)^3}\cdot\sum_{1\leq i<j\leq n}{\left(a_i-a_j\right)^2}\geq n^2.$$
4 replies
mihaig
Yesterday at 11:22 AM
mihaig
3 minutes ago
Integer-Valued FE comes again
lminsl   216
N 24 minutes ago by TwentyIQ
Source: IMO 2019 Problem 1
Let $\mathbb{Z}$ be the set of integers. Determine all functions $f: \mathbb{Z} \rightarrow \mathbb{Z}$ such that, for all integers $a$ and $b$, $$f(2a)+2f(b)=f(f(a+b)).$$Proposed by Liam Baker, South Africa
216 replies
+2 w
lminsl
Jul 16, 2019
TwentyIQ
24 minutes ago
Peru IMO TST 2022
diegoca1   0
34 minutes ago
Source: Peru IMO TST 2022 D2 P1
Let $N$ be a positive integer. Determine all positive integers $n$ that satisfy the following condition:

For any list $d_1, d_2, \ldots, d_k$ of divisors of $n$ (not necessarily distinct) such that
\[
\frac{1}{d_1} + \frac{1}{d_2} + \cdots + \frac{1}{d_k} > N,
\]there exists a subset of the fractions $\frac{1}{d_1}, \frac{1}{d_2}, \ldots, \frac{1}{d_k}$ whose sum is exactly $N$.
0 replies
diegoca1
34 minutes ago
0 replies
A nice property of triangle with incircle (I)
TUAN2k8   0
38 minutes ago
Source: own
Let \( ABC \) be a non-isosceles triangle with incircle (\( I \)). Denote by \( D, E, F \) the points where (\( I \)) touches \( BC, CA, AB \), respectively. The A-excircle of \( ABC \) is tangent to \( BC \) at \( G \). The lines \( IB \) and \( IC \) meet \( AG \) at \( M \) and \( N \), respectively.

a) Prove that the circumcircles of triangles \( MBF \), \( NCE \), and \( BIC \) are concurrent at a point.

b) Let \( L \) and \( K \) be the midpoints of \( AG \) and \( BC \), respectively, and let \( J \) be the orthocenter of triangle \( IMN \). Show that the points \( L, K, J \) are collinear.
0 replies
TUAN2k8
38 minutes ago
0 replies
Structure of the group $(\mathbb{Z}/p\mathbb{Z})^{\times}$ and its application t
nayr   1
N Yesterday at 10:05 AM by GreenKeeper
Let $\mathbb{F}_p^{\times} = (\mathbb{Z} / p\mathbb{Z})^{\times}$ be the unit group of $\mathbb{F}_p$. It is well known that this group is cyclic. Let $g$ be a generator of this group and consider the map $\varphi : \mathbb{F}_p^{\times} \rightarrow \mathbb{F}_p^{\times}, x\mapsto x^k$ for a fixed positive integer $k$. I know that the kernel $\ker \varphi$ has oder $d:= (p-1, k)$. By the first isomorphism theorem, $\mathbb{F}_p^{\times} / \ker \varphi \cong \operatorname{im} \varphi$. Since $\mathbb{F}_p^{\times}$ is cyclic, so are its subgroups and hence $\operatorname{im} \varphi$ is cyclic of oder $\frac{p-1}{d}$. Let $H = \operatorname{im} \varphi$. Then $\mathbb{F}_p^{\times}/H$ is cyclic too and hence we have the partition:

$$\mathbb{F}_p = \{0\} \sqcup H \sqcup s^2H \sqcup \cdots \sqcup s^{d-1}H$$
for any $s\notin H$ (for example $g$).

I am trying to use this fact to solve the following question: Show that $3x^3+4y^3+5z^3 \equiv 0 \pmod{p}$ have non-trivial solution for all primes $p$. Here is my attempt:

For simplicity, we rewrite the original equation for $p>3$, as $x^3+Ay^3+Bz^3\equiv 0 \pmod{p}$ (the case $p=2,3$ is easy).

If $p\equiv 2\pmod{3}$, then everything is a cube (since the cubing map $x\,mapsto x^3$ is an anutomorphism by above) and the equation is solvable.

If $p\equiv 1\pmod{3}$, let $H:=\{x^3|x\in \mathbb{F}_p^{\times}\}$ and $sH, s^2H$ be the cosets where $s \notin H$, then we have the following cases:

Case 1: $A \in H$ or $B\in H$, Without loss of generality, assume $A=4/3$ is a cube, then $4/3=a^3$ or $4=3a^3$ and we may take $(x,y,z)=(a,-1,0)$ as our solution.

Case 2: $A \in sH$ and $B\in sH$, then $A=sa^3$ and $B=sb^3$ and we may take $(x,y,z)=(0,b,-a)$ as our solution.

Case 3: $A \in s^2H$ and $B\in s^2H$, then $A=s^2a^3$ and $B=s^2b^3$ and we may take $(x,y,z)=(0,b,-a)$ as our solution.

Case 4: $A \in sH$ and $B\in s^2H$, then $A=sa^3$ and $B=s^2b^3$. This is the case I am stuck with. If we have $s^3=1$, then we may take $(x,y,z)=(ab,b,a)$ as our solution since $1+s+s^2=0$ for $s^3=1$ and $s$ is not $1$). But it is not always possible to have both $s^3=1$ and $s\notin H$. For example, I can take $s=g^{\frac{p-1}{3}}$, then $s^3=1$, but $g^{\frac{p-1}{3}}\notin H$ iff $9\nmid p-1$.

How should I resolve case 4?
1 reply
nayr
Yesterday at 8:43 AM
GreenKeeper
Yesterday at 10:05 AM
Group Theory resources
JerryZYang   3
N Yesterday at 4:22 AM by JerryZYang
Can someone give me some resources for group theory. ;)
3 replies
JerryZYang
Wednesday at 8:38 PM
JerryZYang
Yesterday at 4:22 AM
Find max(a+√b+∛c) where 0< a, b, c < 1= a+b+c.
elim   7
N Yesterday at 2:25 AM by sqing
Find $\max_{a,\,b,\,c>0\atop a+b+c=1}(a+\sqrt{b}+\sqrt[3]{c})$
7 replies
elim
Feb 7, 2020
sqing
Yesterday at 2:25 AM
Are all solutions normal ?
loup blanc   11
N Wednesday at 9:02 PM by GreenKeeper
This post is linked to this one
https://artofproblemsolving.com/community/c7t290f7h3608120_matrix_equation
Let $Z=\{A\in M_n(\mathbb{C}) ; (AA^*)^2=A^4\}$.
If $A\in Z$ is a normal matrix, then $A$ is unitarily similar to $diag(H_p,S_{n-p})$,
where $H$ is hermitian and $S$ is skew-hermitian.
But are there other solutions? In other words, is $A$ necessarily normal?
I don't know the answer.
11 replies
loup blanc
Jul 17, 2025
GreenKeeper
Wednesday at 9:02 PM
Axiomatic real numbers x^0
Safal   2
N Wednesday at 6:50 PM by Safal
Source: Discussion
Here is an interesting question for you all:

Assume $\mathbb{R}$ is an ordered field with all field axioms holding.

$\textbf{Question:}$ Suppose $x>0$ is a real number and $0\in\mathbb{R}$(Set of real numbers). Prove that if $x^0\in\mathbb{R}$. Then it (that is $x^0$) must be $1$.

Also show same thing is true if $x<0$ is a real number.

Proof

$\textbf{Question:}$ If $z\neq 0$ be any complex number and $0\in\mathbb{C}$. Show that if $z^0\in\mathbb{C}$ , then $z^0=1$.

$\textbf{Remark:}$ Order property is not true for all complex numbers.

Hint
2 replies
Safal
Wednesday at 3:20 PM
Safal
Wednesday at 6:50 PM
D1054 : A measure porblem
Dattier   1
N Wednesday at 5:59 PM by greenturtle3141
Source: les dattes à Dattier
$M=\bigcup\limits_{n\in\mathbb N^*} \{0,1\}^n$, for $m \in M$, $\overline m=\{mx : x\in \{0,1\}^{\mathbb N} \}$

Let $A \subset M$ with $\forall (a,b) \in A,a\neq b$, $\overline a \cap \overline b=\emptyset$.

Is it true that $\sum\limits_{a\in A} 2^{-|a|}\leq 1$ ?

PS : for $m \in M$, $|m|$ is the length of $m$, hence $|0011|=4$
1 reply
Dattier
Wednesday at 4:36 PM
greenturtle3141
Wednesday at 5:59 PM
Analytic Number Theory
EthanWYX2009   0
Wednesday at 2:11 PM
Source: 2024 Jan 谜之竞赛-7
For positive integer \( n \), define \(\lambda(n)\) as the smallest positive integer satisfying the following property: for any integer \( a \) coprime with \( n \), we have \( a^{\lambda(n)} \equiv 1 \pmod{n} \).

Given an integer \( m \geq \lambda(n) \left( 1 + \ln \frac{n}{\lambda(n)} \right) \), and integers \( a_1, a_2, \cdots, a_m \) all coprime with \( n \), prove that there exists a non-empty subset \( I \) of \(\{1, 2, \cdots, m\}\) such that
\[\prod_{i \in I} a_i \equiv 1 \pmod{n}.\]Proposed by Zhenqian Peng from High School Affiliated to Renmin University of China
0 replies
EthanWYX2009
Wednesday at 2:11 PM
0 replies
OMOUS-2025 (Team Competition) P10
enter16180   4
N Wednesday at 8:33 AM by enter16180
Source: Open Mathematical Olympiad for University Students (OMOUS-2025)
Let $f: \mathbb{N} \rightarrow \mathbb{N}$ and $g: \mathbb{N} \rightarrow\{A, G\}$ functions are given with following properties:
(a) $f$ is strict increasing and for each $n \in \mathbb{N}$ there holds $f(n)=\frac{f(n-1)+f(n+1)}{2}$ or $f(n)=\sqrt{f(n-1) \cdot f(n+1)}$.
(b) $g(n)=A$ if $f(n)=\frac{f(n-1)+f(n+1)}{2}$ holds and $g(n)=G$ if $f(n)=\sqrt{f(n-1) \cdot f(n+1)}$ holds.

Prove that there exist $n_{0} \in \mathbb{N}$ and $d \in \mathbb{N}$ such that for all $n \geq n_{0}$ we have $g(n+d)=g(n)$
4 replies
enter16180
Apr 18, 2025
enter16180
Wednesday at 8:33 AM
Putnam 2012 A3
Kent Merryfield   9
N Wednesday at 6:29 AM by AngryKnot
Let $f:[-1,1]\to\mathbb{R}$ be a continuous function such that

(i) $f(x)=\frac{2-x^2}{2}f\left(\frac{x^2}{2-x^2}\right)$ for every $x$ in $[-1,1],$

(ii) $ f(0)=1,$ and

(iii) $\lim_{x\to 1^-}\frac{f(x)}{\sqrt{1-x}}$ exists and is finite.

Prove that $f$ is unique, and express $f(x)$ in closed form.
9 replies
Kent Merryfield
Dec 3, 2012
AngryKnot
Wednesday at 6:29 AM
AMM problem section
Khalifakhalifa   1
N Wednesday at 4:41 AM by Khalifakhalifa
Does anyone have access to the current AMM edition? I’d like to see the problems section. If so, could someone please share it with me via PM?
1 reply
Khalifakhalifa
Jul 22, 2025
Khalifakhalifa
Wednesday at 4:41 AM
Problem3
samithayohan   120
N Jul 20, 2025 by LHE96
Source: IMO 2015 problem 3
Let $ABC$ be an acute triangle with $AB > AC$. Let $\Gamma $ be its circumcircle, $H$ its orthocenter, and $F$ the foot of the altitude from $A$. Let $M$ be the midpoint of $BC$. Let $Q$ be the point on $\Gamma$ such that $\angle HQA = 90^{\circ}$ and let $K$ be the point on $\Gamma$ such that $\angle HKQ = 90^{\circ}$. Assume that the points $A$, $B$, $C$, $K$ and $Q$ are all different and lie on $\Gamma$ in this order.

Prove that the circumcircles of triangles $KQH$ and $FKM$ are tangent to each other.

Proposed by Ukraine
120 replies
samithayohan
Jul 10, 2015
LHE96
Jul 20, 2025
Problem3
G H J
G H BBookmark kLocked kLocked NReply
Source: IMO 2015 problem 3
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Shreyasharma
688 posts
#135
Y by
20 minute solve, took time to find the right inversion.
Note that $Q$, $H$ and $M$ are collinear from reflecting $H$ about $M$ to $H'$ and noting that $AH'$ is a diameter of $(ABC)$. Now consider the inversion with power $AH \cdot HF$, which swaps the following pairs of points:
  • The pairs $(A, F)$, $(B, E)$ and $(C, D)$ where $D$ and $E$ are the feet from $B$ and $C$ to $AC$ and $AB$ respectively.
  • The pair $(Q, M)$, as $\angle HQA = \angle HFM = 90$.
  • The pair $(K, N)$, where $N$ denotes the point on $(DEF)$ such that $\angle HMN = 90$.
Hence we wish to show $MN$ is tangent to $(NAQ)$ - however noting that $MN \parallel AQ$ as they are both perpendicular to $QM$, it follows that it suffices to show $NA = NQ$. This would follow immediately if we could show $2MN = AQ$, yet this is clear from the homothety at $H$ with scale factor $2$ sending $(DEF)$ to $(ABC)$, finishing. $\square$
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Mathandski
776 posts
#136
Y by
I inverted at H
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SomeonesPenguin
129 posts
#137 • 1 Y
Y by zzSpartan
So this took about 10 minutes.

Let $O$ be the circumcenter, $A'$ be the $A$ antipode, $N$ be the midpoint of $AH$ and $T$ lies on the nine-point circle such that $TM\perp HM$.

It is known that $\overline{A'-M-H-Q}$. Now invert at $H$ with radius $\sqrt{-HA\cdot HF}$. It's easy to see that $Q\mapsto M$, $K\mapsto T$ and $A\mapsto F$.

After inversion, it suffices to prove that $MT$ is tangent to $(AQT)$. From middle lines, $ON$ is parallel to $A'Q$, hence $ON\perp AQ$ so $ON$ is the perpendicular bisector of $AQ$. Now notice that $NM$ is a diameter in the nine-point circle so $\angle NTM=\angle TMH=90^\circ$ , or $NT\parallel MH$. Therefore, we get $\overline{T-O-N}$ and since $ON$ is the perpendicular bisector of $AQ$, we get the desired conclusion. $\blacksquare$
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ihatemath123
3495 posts
#138 • 1 Y
Y by OronSH
Invert at $H$.

Let $N^*$ be the minor arc midpoint of arc $B^{*}C^*$ in $(A^*B^*C^*)$. Since $N^*$ and $Q^*$ are antipodal arc midpoints in $(A^*B^*C^*)$, it follows that $\angle N^* K^* Q^* = 90^{\circ}$. We also have $\angle K^* Q^* H = 90^{\circ}$ by definition. It's well known that the uninverted $M$, $H$ and $Q$ are collinear, so $M^*$, $H$ and $Q^*$ are collinear – since $M^*$ lies on $(HB^*C^*)$, it follows that $M^*$ is the foot from $F^*$ to line $HQ^*$.

Now, what we want to show is clearly true: $\overline{Q^*K^*} \parallel \overline{M^* F^*}$ and $M^* K^* = F^* K^*$ since line $N^* K^*$ is the perpendicular bisector of $\overline{M^* F^*}$. Therefore, $(F^* K^* M^* )$ is tangent to line $KQ$.
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iStud
271 posts
#140
Y by
reim's spam lol

Note that it suffices to prove that $\angle{MFK}-\angle{QHK}=\angle{MKQ}=90^\circ+\angle{HKM}$.

Let $O$ be the circumcenter of $\triangle{ABC}$. Define $A',Q'$ to be the antipodes of $A,Q$, respectively .Notice that $Q$ is the $A$-queue point of $\triangle{ABC}$, so it's well known that $\overline{Q,H,M,A'}$. Moreover, $\overline{K,H,Q'}$. Let $J$ be the midpoint of $Q'H$. Note that $AF\cap(ABC)$ is the reflection of $H$ over $BC$ and $A'$ is the reflection of $H$ over $M$. Using Reim's Theorem 2 times, we conclude that $AJFK$ and $MJQK$ are cyclic.

Let $\angle{HQK}=\alpha$ and $\angle{JQM}=\beta$. Note that $J$ must be lying on the mid parallel line of $AQ$ and $A'Q'$ (which passes through $O$), so $\angle{AJQ}=2\angle{JQM}=2\beta$. We have $JO\parallel HQ$, so $\angle{AFK}=\angle{AJK}=\angle{AJO}+\angle{OJK}=\frac{\angle{AJQ}}{2}+\angle{QHK}=90^\circ-\alpha-\beta$, then $\angle{MFK}=90^\circ+\angle{AFK}=180^\circ-\alpha+\beta$. Lastly, using information that $\angle{QHK}=90^\circ-\alpha$, we can have $\angle{MFK}-\angle{QHK}=90+\beta=90^\circ+\angle{JQM}=90^\circ+\angle{JKM}=90^\circ+\angle{HKM}$, as desired. $\blacksquare$

P2 of GOWACO 2021 was definitely much harder than this one :P
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Ilikeminecraft
734 posts
#141
Y by
rename appropriately first($F \to D, Q\to D$).
Invert about $H$ first with radius $-\sqrt{AH\cdot HD}.$ Then, the circumcircle maps onto the 9-point circle, $K$ maps to intersection of 9 point circle and the line perpendicular to $GM$ that passes through $M.$ Clearly, $G\leftrightarrow M$. Problem simplifies to proving that $(AGK^*)$ is tangent to $MK^*.$
Then, reflect $H$ across $K^*$ and $M,$ and they both land on the circumcircle. In fact, they form a rectangle. This finishes.
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cj13609517288
1939 posts
#142 • 1 Y
Y by MS_asdfgzxcvb
This works????????

Let $\ell$ be the line through $M$ perpendicular to $HM$, and let $S$ be the point on $\ell$ such that $(QAS)$ is tangent to $\ell$. Since $QA\perp QM$, $S$ is the projection of the midpoint of $QA$ onto $\ell$. So in complex coordinates with $(ABC)$ as the unit circle,
\[s=m+\frac{a-q}{2}\Longrightarrow 2s-h=2m+a-q-h=-q.\]Therefore, the reflection of $H$ over $S$ is exactly the $Q$-antipode, meaning that $K$ lies on $HS$.

Now consider the negative inversion centered at $H$ swapping $A$ and $F$. Then it also swaps $Q$ and $M$, and $K$ and $S$ because they are collinear with $H$ and also $\ell$ swaps with $(QH)$. So since $(QAS)$ is tangent to $\ell$, we get $(MFK)$ is tangent to $(QH)$. $\blacksquare$
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VideoCake
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#143
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Solution. Let \(A'\) be the antipode of \(A\) on \(\Gamma\). Observe that \(H, M, A'\) are collinear (as \(M\) is known to be the midpoint of \(HA'\)) and observe that \(Q, H, A'\) are collinear (as \(AQ \perp QH\)). Thus, \(Q, M, H\) are collinear. Let \(D, E\) be the foots of the altitudes from \(B, C\).
Perform an inversion around a circle with center \(H\). Denote \(X^*\) to be the inverted image of \(X\). We get that
  • \(F^*D^*E^*\) is a triangle with orthic triangle \(A^*B^*C^*\) and orthocenter \(H^*\).
  • \(Q^*\) lies on the line \(A^*D^*\) and on the circle \((A^*B^*C^*)\), as \(AQDH\) is cyclic.
  • \(K^*\) is on the perpendicular line to \(H^*Q^*\) through \(Q^*\) and on \((A^*B^*C^*)\), as \(K\) lies on the circle with diameter \(QH\).
  • \(M^*\) is the intersection of \((H^*B^*C^*F^*)\) and \(Q^*M^*\), as line \(BC\) goes to the circle \((H^*B^*C^*)\).
It is enough to show that \((M^*F^*K^*)\) is tangent to \(Q^*K^*\). For simplicity, we remove the \(*\) symbols from the points and solve the remaining problem. Let \(O\) be the circumcenter of \((BMFCH)\).
First, it is known that \(H\) is the incenter of \(\Delta ABC\), and that \(D\), \(E\) and \(F\) are the three excenters of \(\Delta ABC\). This means that \(O\) is the \(A\)-southpole and lies on the circle \((ABC)\). In particular, \(O\) is the midpoint of \(HF\). As \(FM \perp MH\), we have \(OMF\) isosceles. As \(AQ\) is the exterior angle bisector of \(\angle BAC\), we know that \(Q\) is the northpole. Therefore, \(OQ\) is the diameter of \(ABC\), meaning \(QK \perp KO\). Notice that \(KO\) is also the perpendicular bisector of \(MF\), so \(KMF\) is isosceles. Notice that \(FM \perp MQ \perp QK \implies MF \parallel QK\). Thus, \(\angle QKM = \angle FMK = \angle KFM\), and \((MFK)\) is indeed tangent to \(QK\), as desired.
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Retemoeg
62 posts
#144
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Let $AF$ intersect $(O)$ twice at $H'$. Denote $L$ the orthogonal projection of $Q$ onto $AD$. We'll call the midpoints of segments $QL, QA, QH$, $Y, X, I$ resp. Let $Z$ be the center of $(KDM)$. Denote $A'$ the antipodal point of $A$ in $(O)$.

We readily notice that $\angle AQH = \angle AQA' = 90^{\circ}$, thus $Q, H, A'$ are collinear. From a common result: $H, M, A'$ are collinear. Hence all these four points lies on a straight line. We'll show that $\angle MKI = \angle ZKM$, i.e $\angle MKI = \angle KFA = \angle H'KF + \angle KH'F$. A spiral similarity $\phi$ centered at $K$ sends $Q$ to $A$ and $H$ to $A'$, thus:
\[ \angle MKI = \angle MKH + \angle HKI = \angle MKH + \angle QHK = \angle MKH + \angle AA'K = \angle MKH + \angle AH'K \]So we'll aim to show that $\angle MKH = \angle H'KF$.
Observe that $\phi$ also sends $F$ to $Y$, $H'$ to $L$, $M$ to $X$. Thus, $\angle MKH = \angle XKQ$ and $\angle H'KF = \angle LKY$. And to finish off the problem, $X$ is the intersection of the two tangents from $Q$ and $L$ w.r.t $(I)$. Thus $KX$ is the $K$-symmedian w.r.t $\triangle QKL$, hence $\angle QKX = \angle LKY$ and we are done.

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This post has been edited 2 times. Last edited by Retemoeg, Apr 20, 2025, 5:26 AM
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bjump
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#145
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Consider a Negative inversion at $H$ such that the circumcircle of $\triangle ABC$ is mapped to the Nine Point Circle of $\triangle ABC$. $F$ is swapped with $A$ under this inversion. $Q$ is mapped to $M$ and vice versa. $K$ is mapped to the point on $(A'B'C')$ such that $\angle HQ'K' = 90^\circ$. Since $M' \in (HB'C')$ we have $\angle F'M'H = \angle F'M'Q' = \angle M'Q'K' = 90^\circ$, so $Q'K' \parallel M'F'$, Note that the $Q'$-antipode in $(A'B'C')$ is the midpoint of $HF'$, $K'$ lies on the perpendicular bisector of $F'M'$ therefore $Q'K'$ is tangent to $(F'M'K')$. Uninverting gives the desired conclusion.
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fearsum_fyz
85 posts
#146
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Consider the composition $\mathcal{T}$ of inversion centered at $H$ with radius $\sqrt{HA \cdot HF}$ and reflection over $H$.
It is easy to see that $\mathcal{T}$ swaps the circumcircle and nine point circle, sending $A$ to $F$ and $Q$ to $M$.
Let $K^*$ be the image of $K$ under $\mathcal{T}$.
Then $\measuredangle{QMK^*} = \measuredangle{HMK^*} = \measuredangle{HKQ} = 90^{\circ}$, so $AQ \parallel MK^*$. Hence it would suffice to show that $A K^* = Q K^*$.
Let $Q'$ be the antipode of $Q$. By homothety at $H$, $K^*$ is the midpoint of $HQ'$.
By Apollonius's theorem in $\Delta{AQ'H}$ and $\Delta{QQ'H}$, we have
$2 A{K^*}^2 + 2 H{K^*}^2 = AQ'^2 + AH^2$
$2 Q{K^*}^2 + 2 H{K^*}^2 = QQ'^2 + QH^2$
Subtracting and using the Pythagorean theorem, we get
$2 A{K^*}^2 - 2 Q{K^*}^2 = AQ^2 - AQ^2 = 0$
as desired.
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zuat.e
96 posts
#147 • 1 Y
Y by Kingsbane2139
Consider the inversion at $H$ with radius $r=\sqrt{AH\cdot AF}$ followed by a reflection w.r.t $H$.

It suffices to show that $K'M$ is tangent to $(AQK')$, where $K'$ is the inverse of $K$.

Let $V,W$ be the reflections of $H$ across $K',M$, so as $AQWV$ is cyclic (because $VH\cdot HK=2K'H\cdot HK=2HQ\cdot HM=HQ\cdot HW$ and $W\in (ABC)$) and $\measuredangle AQW=\measuredangle QWV=90º$, we have $ AQWV$ rectangle, yet clearly $K'$ is center of $\triangle VHW$, from where it is clear that $\triangle K'AQ$ is isosceles, consequently $\measuredangle MK'Q=\measuredangle K'QA=\measuredangle K'AQ$, as desired.
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cursed_tangent1434
737 posts
#148
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As a true config main we refrain from indulging in inversion. We start off by noting the following which allows us to adjust our frame of reference.

Claim : Point $H$ is the $Q-$Humpty Point in $\triangle QBC$.

Proof : Simply note that if the $A-$antipode is $A'$,
\[\measuredangle HBC = \measuredangle CAF = \measuredangle A'AB = \measuredangle A'QB = \measuredangle HQB\]a similar argument shows that $\measuredangle BCH = \measuredangle CQH$ which suffices to show the claim.

Now, we rewrite the problem with $\triangle QBC$ as our reference triangle.
Reference Shift wrote:
Let $\triangle ABC$ be a triangle with $A-$Humpty point $H_a$. Let $M$ denote the midpoint of segment $BC$ and $F$ the foot of the altitude from $H_a$ to $BC$. Let $K$ denote the second intersection of $(AH_a)$ and $(ABC)$. Show that circles $(AH_aK)$ and $(KFM)$ are tangent.

The following is the key claim.

Claim : Lines $BK$ and $CH_a$ intersect on $(AH_a)$ (and similarly).

Proof : We show the proof for one as the other is entirely similar. Note,
\[\measuredangle KBC + \measuredangle BCH_a = \measuredangle  KAC + \measuredangle CAH_a = \measuredangle  KAH_a\]which implies that $BK \cap CH_a$ lies on $(AH_a)$ as desired.

Let $X= BK \cap CH_a$ and $Y = CK \cap BH_a$. Now, by Pascal's Theorem on concyclic hexagon $H_aH_aYKKX$ it follows that the tangents to $(AH_a)$ at $K$ and $H_a$ intersect on $\overline{BC}$.

The tangent to $(AH_a)$ at $H_a$ is perpendicular to $AM$ and hence must pass through $H$. Thus, this intersection of tangents is the $A-$Ex point of $\triangle ABC$. Thus, if the tangent intersection point is $T$,
\[TK^2 = TH_a^2 = TF \cdot TM\]which implies that $\overline{TK}$ is also tangent to $(KFM)$ and the result follows.
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ihategeo_1969
284 posts
#149
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Changing labels.
Quote:
Let $\triangle ABC$ have orthic triangle $\triangle DEF$, orthocenter $H$, midpoint of $\overline{BC}$ as $M$. Let $Q_A$ be its $A$-Queue point. Let $K=(Q_AH) \cap (ABC)$. If $AB<AC$ and $A,Q_A,K,B,C$ are cyclic in this order; then prove that $(KQ_AH)$ and $(DKM)$ are tangent to each other.
Invert about $H$ swapping $(ABC)$ and $(DEF)$ which means $K^*=\overline{M\infty_{\perp HM}} \cap (DEF)$. Let $N$ be midpoint of $\overline{AH}$. Now $\measuredangle NK^*M=\measuredangle NDM=90^\circ$; which means $\overline{NK^*} \parallel \overline{HM} \perp \overline{AQ_A} \parallel \overline{MK^*}$ which means $K^*A=K^*Q_A$. This implies $\overline{MK^*}$ is tangent to $(AQ_AK^*)$ and invert back and done.
This post has been edited 2 times. Last edited by ihategeo_1969, Jul 11, 2025, 5:23 PM
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LHE96
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#150
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I think it was too easy for a P3

Synthetic sol
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