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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.

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0 replies
jlacosta
May 1, 2025
0 replies
Serbian selection contest for the IMO 2025 - P6
OgnjenTesic   15
N 4 minutes ago by math90
Source: Serbian selection contest for the IMO 2025
For an $n \times n$ table filled with natural numbers, we say it is a divisor table if:
- the numbers in the $i$-th row are exactly all the divisors of some natural number $r_i$,
- the numbers in the $j$-th column are exactly all the divisors of some natural number $c_j$,
- $r_i \ne r_j$ for every $i \ne j$.

A prime number $p$ is given. Determine the smallest natural number $n$, divisible by $p$, such that there exists an $n \times n$ divisor table, or prove that such $n$ does not exist.

Proposed by Pavle Martinović
15 replies
OgnjenTesic
May 22, 2025
math90
4 minutes ago
Easy Number Theory
math_comb01   39
N 21 minutes ago by Adywastaken
Source: INMO 2024/3
Let $p$ be an odd prime and $a,b,c$ be integers so that the integers $$a^{2023}+b^{2023},\quad b^{2024}+c^{2024},\quad a^{2025}+c^{2025}$$are divisible by $p$.
Prove that $p$ divides each of $a,b,c$.
$\quad$
Proposed by Navilarekallu Tejaswi
39 replies
math_comb01
Jan 21, 2024
Adywastaken
21 minutes ago
Painting Beads on Necklace
amuthup   46
N 28 minutes ago by quantam13
Source: 2021 ISL C2
Let $n\ge 3$ be a fixed integer. There are $m\ge n+1$ beads on a circular necklace. You wish to paint the beads using $n$ colors, such that among any $n+1$ consecutive beads every color appears at least once. Find the largest value of $m$ for which this task is $\emph{not}$ possible.

Carl Schildkraut, USA
46 replies
amuthup
Jul 12, 2022
quantam13
28 minutes ago
Iran geometry
Dadgarnia   38
N 36 minutes ago by cursed_tangent1434
Source: Iranian TST 2018, first exam day 2, problem 4
Let $ABC$ be a triangle ($\angle A\neq 90^\circ$). $BE,CF$ are the altitudes of the triangle. The bisector of $\angle A$ intersects $EF,BC$ at $M,N$. Let $P$ be a point such that $MP\perp EF$ and $NP\perp BC$. Prove that $AP$ passes through the midpoint of $BC$.

Proposed by Iman Maghsoudi, Hooman Fattahi
38 replies
Dadgarnia
Apr 8, 2018
cursed_tangent1434
36 minutes ago
Quadruple Binomial Coefficient Sum
P162008   4
N Yesterday at 8:40 PM by vmene
Source: Self made by my Elder brother
$\sum_{p=0}^{\infty} \sum_{r=0}^{\infty} \sum_{q=1}^{\infty} \sum_{s=0}^{p+q - 1} \frac{((-1)^{p+r+s+1})(2^{p+q-1}) \binom{p + q - s - 1}{p + q - 2s - 1}}{4^s(2p^2q + 2pqr + pq + qr)(2p + 2q + 2r + 3)}.$
4 replies
P162008
Thursday at 8:04 PM
vmene
Yesterday at 8:40 PM
IMC 1994 D1 P5
j___d   5
N Yesterday at 5:39 PM by krigger
a) Let $f\in C[0,b]$, $g\in C(\mathbb R)$ and let $g$ be periodic with period $b$. Prove that $\int_0^b f(x) g(nx)\,\mathrm dx$ has a limit as $n\to\infty$ and
$$\lim_{n\to\infty}\int_0^b f(x)g(nx)\,\mathrm dx=\frac 1b \int_0^b f(x)\,\mathrm dx\cdot\int_0^b g(x)\,\mathrm dx$$
b) Find
$$\lim_{n\to\infty}\int_0^\pi \frac{\sin x}{1+3\cos^2nx}\,\mathrm dx$$
5 replies
j___d
Mar 6, 2017
krigger
Yesterday at 5:39 PM
2023 Putnam A2
giginori   21
N Yesterday at 3:32 PM by pie854
Let $n$ be an even positive integer. Let $p$ be a monic, real polynomial of degree $2 n$; that is to say, $p(x)=$ $x^{2 n}+a_{2 n-1} x^{2 n-1}+\cdots+a_1 x+a_0$ for some real coefficients $a_0, \ldots, a_{2 n-1}$. Suppose that $p(1 / k)=k^2$ for all integers $k$ such that $1 \leq|k| \leq n$. Find all other real numbers $x$ for which $p(1 / x)=x^2$.
21 replies
giginori
Dec 3, 2023
pie854
Yesterday at 3:32 PM
Putnam 2019 A1
awesomemathlete   33
N Yesterday at 3:25 PM by cursed_tangent1434
Source: 2019 William Lowell Putnam Competition
Determine all possible values of $A^3+B^3+C^3-3ABC$ where $A$, $B$, and $C$ are nonnegative integers.
33 replies
awesomemathlete
Dec 10, 2019
cursed_tangent1434
Yesterday at 3:25 PM
IMC 1994 D1 P2
j___d   5
N Yesterday at 3:11 PM by krigger
Let $f\in C^1(a,b)$, $\lim_{x\to a^+}f(x)=\infty$, $\lim_{x\to b^-}f(x)=-\infty$ and $f'(x)+f^2(x)\geq -1$ for $x\in (a,b)$. Prove that $b-a\geq\pi$ and give an example where $b-a=\pi$.
5 replies
j___d
Mar 6, 2017
krigger
Yesterday at 3:11 PM
A Construction in Multivariable Analysis
MrOrange   0
Yesterday at 2:11 PM
Source: Garling's A COURSE IN MATHEMATICAL ANALYSIS
Construct a continuous real valued function \( f \) on \( \mathbb{R}^2 \) for which
\[
\lim_{R \to \infty} \int_{\|x\|_2 \leq R} f(x) \, dx = 0
\]and for which
\[
\lim_{R \to \infty} \int_{\|x\|_\infty \leq R} f(x) \, dx \text{ does not exist.}
\]
0 replies
MrOrange
Yesterday at 2:11 PM
0 replies
Possible values of determinant of 0-1 matrices
mathematics2004   4
N Yesterday at 1:56 PM by loup blanc
Source: 2021 Simon Marais, A3
Let $\mathcal{M}$ be the set of all $2021 \times 2021$ matrices with at most two entries in each row equal to $1$ and all other entries equal to $0$.
Determine the size of the set $\{ \det A : A \in M \}$.
Here $\det A$ denotes the determinant of the matrix $A$.
4 replies
mathematics2004
Nov 2, 2021
loup blanc
Yesterday at 1:56 PM
ISI UGB 2025
Entrepreneur   1
N Yesterday at 1:49 PM by Knight2E4
Source: ISI UGB 2025
1.)
Suppose $f:\mathbb R\to\mathbb R$ is differentiable and $|f'(x)|<\frac 12\;\forall\;x\in\mathbb R.$ Show that for some $x_0\in\mathbb R,f(x_0)=x_0.$

3.)
Suppose $f:[0,1]\to\mathbb R$ is differentiable with $f(0)=0.$ If $|f'(x)|\le f(x)\;\forall\;x\in[0,1],$ then show that $f(x)=0\;\forall\;x.$

4.)
Let $S^1=\{z\in\mathbb C:|z|=1\}$ be the unit circle in the complex plane. Let $f:S^1\to S^1$ be the map given by $f(z)=z^2.$ We define $f^{(1)}:=f$ and $f^{(k+1)}=f\circ f^{(k)}$ for $k\ge 1.$ The smallest positive integer $n$ such that $f^n(z)=z$ is called period of $z.$ Determine the total number of points $S^1$ of period $2025.$

6.)
Let $\mathbb N$ denote the set of natural numbers, and let $(a_i,b_i), 1\le i\le 9,$ be nine distinct tuples in $\mathbb N\times\mathbb N.$ Show that there are $3$ distinct elements in the set $\{2^{a_i}3^{b_i}:1\le i\le 9\}$ whose product is a perfect cube.

8.)
Let $n\ge 2$ and let $a_1\le a_2\le\cdots\le a_n$ be positive integers such that $$\sum_{i=1}^n a_i=\prod_{i=1}^n a_i.$$Prove that $$\sum_{i=1}^n a_i\le 2n$$and determine when equality holds.
1 reply
Entrepreneur
May 27, 2025
Knight2E4
Yesterday at 1:49 PM
Recurrence trouble
SomeonecoolLovesMaths   3
N Yesterday at 1:44 PM by Knight2E4
Let $0 < x_0 < y_0$ be real numbers. Define $x_{n+1} = \frac{x_n + y_n}{2}$ and $y_{n+1} = \sqrt{x_{n+1}y_n}$.
Prove that $\lim_{n \to \infty} x_n = \lim_{n \to \infty} y_n$ and hence find the limit.
3 replies
SomeonecoolLovesMaths
May 28, 2025
Knight2E4
Yesterday at 1:44 PM
Trigo or Complex no.?
hzbrl   5
N Yesterday at 9:20 AM by GreenKeeper
(a) Let $y=\cos \phi+\cos 2 \phi$, where $\phi=\frac{2 \pi}{5}$. Verify by direct substitution that $y$ satisfies the quadratic equation $2 y^2=3 y+2$ and deduce that the value of $y$ is $-\frac{1}{2}$.
(b) Let $\theta=\frac{2 \pi}{17}$. Show that $\sum_{k=0}^{16} \cos k \theta=0$
(c) If $z=\cos \theta+\cos 2 \theta+\cos 4 \theta+\cos 8 \theta$, show that the value of $z$ is $-(1-\sqrt{17}) / 4$.



I could solve (a) and (b). Can anyone help me with the 3rd part please?
5 replies
hzbrl
May 27, 2025
GreenKeeper
Yesterday at 9:20 AM
Stop Projecting your insecurities
naman12   53
N May 29, 2025 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
May 29, 2025
Stop Projecting your insecurities
G H J
G H BBookmark kLocked kLocked NReply
Source: 2022 USA TST #2
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Leo.Euler
577 posts
#44 • 1 Y
Y by sixoneeight
Here is a computational solution for anyone who doesn't want to do synthetic geometry on Day 1 of the TST.

Part 1: Talking about the "wannabe" point
Let $H'$ be the reflection of the orthocenter $H$ of $\triangle ABC$ with respect to line $BC$. Let $D$ be the foot of the altitude of $A$ onto $BC$. Define \[ f(X) := \text{Pow}(X, (BME)) - \text{Pow}(X, (ABC)). \]Compute $f(H) = BH \cdot HE = BD \cdot BC - BH^2$ and $f(D) = \tfrac{BD \cdot BC}{2}$, so \[ f(H') = f(2D-H) = 2f(D) - f(H) = BH^2 = 4R^2 \cdot (\cos B)^2 \]by linearity. Thus, \[ \text{Pow}(H', (BME)) = 4R^2 \cdot (\cos B)^2. \]Using a similar function with $B$ swapped with $C$, we can compute \[ \text{Pow}(H', (CMF)) = 4R^2 \cdot (\cos C)^2, \]so we conclude that \[ \frac{\text{Pow}(H', (BME))}{\text{Pow}(H', (CMF))} = \left(\frac{\cos B}{\cos C}\right)^2. \]
Part 2: Seeing reality
A quick LoS computation shows that $\text{rad}(BME)/\text{rad}(CMF)=\tfrac{\cos B}{\cos C}$. Thus \[ \frac{\text{Pow}(K, (BME))}{\text{Pow}(K, (CMF))}=\left(\frac{\cos B}{\cos C}\right)^2. \]
Part 3: The magical and forgotten step to stardom
Suppose that $K$ lies on $(ABC)$. Then \[ \frac{\text{Pow}(K, (BME))}{\text{Pow}(K, (CMF))} = \frac{\text{Pow}(H', (BME))}{\text{Pow}(H', (CMF))}. \]Denote $Q$ by the intersection of $(BME)$ and $(CMF)$ distinct from $M$. Leveraging the Unforgotten Coaxiality Lemma, we have that $K$, $H'$, $Q$, and $M$ are concyclic. Since $K$ is external to $(BME)$ and $(CMF)$, it must be $H'$, and we are done.
This post has been edited 1 time. Last edited by Leo.Euler, Dec 5, 2023, 4:46 AM
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Dukejukem
695 posts
#45
Y by
Note that points $B, C, E, F$ are inscribed inside the circle of diameter $\overline{BC}$ with center $M$. Let $Q$ be the Miquel point of quadrilateral $BCEF$. By the standard facts, $Q$ lies on $\odot(BME)$ and $\odot(CMF)$ and $\angle MQA = 90^{\circ}$. Therefore, line $QM$ meets $\odot(ABC)$ for a second time at the antipode $A'$ of $A$.


[asy]
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draw(circle((9.522388430690755,6.177083213710761), 2.482680032776302), linewidth(1.2) + linetype("2 2") + blue); 
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[/asy]
Consider the inversion with pole $M$ fixing $\odot(ABC)$. This inversion swaps $\{B, C\}, \{Q, A'\}$, and say $\{K, L\}$. Since $K$ is the intersection of the two tangent lines to $\odot(BMQ), \odot(CMQ)$, it follows under inversion that $L$ is the second intersection of the two circles through $M$ tangent to lines $CA', BA'$. By symmetry, $L$ must be the reflection of $M$ in the angle bisector of $\angle BA'C$. Thus, arcs $\widehat{QB}$ and $\widehat{LC}$ on $\odot(ABC)$ are equal. It follows that points $Q, L$ are symmetric in the perpendicular bisector of $\overline{BC}$. Therefore, points $A', K$ are also symmetric in the perpendicular bisector of $\overline{BC}$. We conclude that $A'K \parallel BC$, hence $AK \perp BC$.
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wu2481632
4239 posts
#46
Y by
Note that proving $AK$ is perpendicular to $BC$ is equivalent to proving that $K$ is the reflection of the orthocenter $H$ of triangle $ABC$ in line $BC$. We establish furthermore the following well-known properties:
  • $(AEHF)$, $MH$, and $(ABC)$ concur at some point $K'$;
  • $K'BKC$ is a harmonic quadrilateral;
  • $MK' \cdot MH = ME^2 = MF^2$.
Invert about the circle with diameter $BC$. We rephrase the problem as follows:
Inverted wrote:
Let $BHC$ be an obtuse triangle with $\angle H > 90^{\circ}$, and let $M$ be the midpoint of $BC$. Two circles $\omega_1$ and $\omega_2$ pass through $M$ and are tangent to lines $BH$ and $CH$. Suppose they meet at $T \neq M$. Show that if $T$ lies on $(BHC)$, then quadrilateral $TBHC$ is harmonic.
This turns out to be quite simple. Note that $T$ and $M$ are symmetric with respect to $\angle BHC$. Thus $\angle BHT = \angle CHM$ and we are done.
This post has been edited 1 time. Last edited by wu2481632, Mar 20, 2024, 5:02 PM
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ihatemath123
3449 posts
#47 • 1 Y
Y by OronSH
We invert about $(BFEC)$. This sends $(BME)$ to line $BE$, sends $(CMF)$ to line $CF$, fixes $M$, sends $A$ to the $A$-humpty point, sends $H$ to the $A$-queue point and sends $K$ to $K^*$, the reflection of $M$ across the angle bisector of $\angle BHC$. So, $(ABC)$ is sent to $(BHC)$.

Since $(ABCK)$ is cyclic, $BHCK^*$ is cyclic. Now, letting $X$ be the midpoint of arc $BK^*C$, we have
\[\angle K^*MB = \angle HXM = \angle HBX - 90^{\circ} = \angle HMX - 90^{\circ} = \angle HMB,\]so line $BC$ bisects $\angle HMK^{*}$. So, line $BC$ also bisects $\angle HMK$ which finishes.
This post has been edited 1 time. Last edited by ihatemath123, Aug 22, 2024, 11:30 PM
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Pyramix
419 posts
#48
Y by
We first prove a bunch of claims for any general acute triangle.

Let $AD,BE,CF$ be the altitudes and $H$ be the orthocenter of $ABC$. Let $Q$ be the $A-$Queue Point. Note that $\angle EMB=2\angle ECB=2C$ and since $\angle BEM=\angle MBE$, we have \[\measuredangle MBE=90^\circ-C=\measuredangle HAE=\measuredangle HQE=\measuredangle MQE\](it is well-known that $M,Q,H$ are collinear - it follows by $\sqrt{-HA\cdot HD}$ inversion). Since $\angle MBE=\angle MQE$, we have that $Q\in(BME)$. Similarly, $Q\in(CMF)$. So, $MQ$ is the radical axis of the two circles.

Let $A,K_1$ be the intersections of $AD$ with $ABC$. Let $O_1,O_2$ be the centres and $r_1,r_2$ be the radii of circles $(BME),(CMF)$, respectively.

Claim: $\frac{K_1B}{K_1C}=\frac{r_1}{r_2}$ and $\overline{K_1B},\overline{K_1C}$ are tangents to $(BME),(CMF)$.
Proof. Note that \[\frac{r_1}{r_2}=\frac{\frac{MB}{2\sin(\measuredangle BEM)}}{\frac{MC}{2\sin(\measuredangle MFC)}}=\frac{\sin(90^\circ-B)}{\sin(90^\circ-C)}=\frac{\sin(\measuredangle HCB)}{\sin(\measuredangle CBH)}=\frac{HB}{HC}=\frac{K_1B}{K_1C},\]as desired. Moreover, $\measuredangle BEM=90^\circ-C=\measuredangle CBH=\measuredangle K_1BC$, which means $\overline{K_1B}$ is tangent to $(BME)$ by alternate segment theorem. The claim is therefore true. $\blacksquare$

Claim: Call $(BME)$ as $\tau_1$ and $(CMF)=\tau_2$. Then, the locus of all points $X$ such that
\[\frac{\text{length of tangent from }X\text{ to }\tau_1}{\text{length of tangent from }X\text{ to }\tau_2}=\sqrt{\frac{Pow_{\tau_1}(X)}{Pow_{\tau_2}(X)}}=\frac{r_1}{r_2} \ \ \ (\star)\]is $(QMK_1)$.
Proof. Given equation is simply:
\[\frac{O_1X^2-r_1^2}{O_2X^2-r_2^2}=\frac{r_1^2}{r_2^2}\Longrightarrow \frac{O_1X}{O_2X}=\frac{r_1}{r_2}\]which means the equation is that of an Apollonius circle for some harmonic bundle. In particular, points $M,Q,K_1$ lie on this circle - $M,Q$ lie because they are intersection points of the two circles, and $K_1$ lies on the circle by the previous claim. $\blacksquare$

We now return to the original problem. If $K$ is the meeting point of the common external tangents, then it satisfies the equation $(\star)$ in the above claim. Also, from the above claim, we have that $K\in(MQK_1)$ and $K\in(ABC)$, which forces $K\in\{Q,K_1\}$. However, note that $K=Q$ is impossible as $Q$ itself is an intersection point of $\tau_1,\tau_2$. So, $K=K_1$ must be true. Hence, $AK\perp BC$, as claimed. $\blacksquare$
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This post has been edited 1 time. Last edited by Pyramix, Apr 8, 2024, 6:46 PM
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shendrew7
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#49
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Define $Q$ as the $A$-queue point. First notice that $(ABC)$, $(BME)$, $(CMF)$ all pass through $Q$, which can be proved by inverting about $(BC)$.

Define $P$ as the intersection of $BC$ with the tangent to $(ABC)$ at $Q$. We claim $PKMQ$ is cyclic, which follows from Coaxiality Lemma and LOS:
\[\frac{\operatorname{pow}(K,(BME))}{\operatorname{pow}(K,(CMF))} = \left(\frac{R_{(BME)}}{R_{(CMF)}}\right)^2 = \left(\frac{QB}{QC}\right)^2 = \frac{PB}{PC} = \frac{\operatorname{pow}(P,(BME))}{\operatorname{pow}(P,(CMF))}.\]
Since $O \in (MPQ)$, it follows that $PK$ is also tangent to $(ABC)$, giving the desired harmonic quadrilateral $QBKC$. $\blacksquare$
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Ywgh1
139 posts
#50
Y by
USA TST 2023 p2

Let $Q$ be the queue point of $\triangle ABC$.
And let $Q'$ be the intersection of $(BEM)$ and $(CFM)$.
Let $K'$ be the intersection of the $A$-altitude with $(ABC)$
And let line $KM$ intersect $(ABC)$ again at $R$. We first start off with the following claim.

Claim 1: $Q=Q'$

Proof: Since $M$ is the center $(BCEF)$, and $BCEF$ is cyclic, this gives us $Q=Q'$.$\blacksquare$

A well know property is that $(QK'BC)$ is harmonic. Hence we want to show that $(QKBC)$ is harmonic.

Claim 2: $QR \| BC$

Proof: long angle chase. $\blacksquare$

Now this basically implies that $(QK'BC)$ is harmonic so $K=K'$, hence we done.
This post has been edited 2 times. Last edited by Ywgh1, Aug 14, 2024, 7:41 PM
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ironball
110 posts
#52
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BlizzardWizard wrote:
To show that $k=\frac1s$, it now suffices to show that $s^2\overline s^2-s^2-\overline s^2=0$.

Could you please explain why we have $|s^2-1|=1$?
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InterLoop
279 posts
#53
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solution
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cursed_tangent1434
653 posts
#54
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Solved with stillwater_25. I personally found this problem very challenging and it was pretty hard to avoid accidentally circular reasoning. It turns out the natural way to eliminate the headaching common tangents is via inversion (even though it seems counterintuitive since they will now become circles tangent to two lines). Let $H$ denote the orthocenter, $Q_A$ denote the $A-$Queue Point, $H_A$ the $A-$Humpty Point and $X_A$ the $A-$Ex Point. We first make a minor observation.

Claim : $Q_A$ lies on circles $(BME)$ and $(CMF)$.
Proof : This is a direct angle chase. It is well known that $M$ is the center of cyclic quadrilateral $BCEF$, $M$ is the intersection of the tangents to $(AEF)$ at $E$ and $F$ and points $Q_A$ , $H$ and $M$ are collinear. Thus,
\[\measuredangle MQ_AF =\measuredangle HQ_AF = \measuredangle HFM = \measuredangle CFM = \measuredangle MCF\]which implies that $CMFQ_A$ is cyclic. Similarly we can show that $BMEQ_A$ is also cyclic, proving the claim.

Now comes the crux of this solution. We perform an inversion about circle $(BC)$ (with center $M$). Clearly $Q_A$ goes to $H$ and $A$ goes to $H_A$ under this inversion. Further, circles $(BME)$ and $(CMF)$ become lines $\overline{BE}$ and $\overline{CF}$ respectively. Thus, the common external tangents to these circles become the two circles tangent to $\overline{BE}$ and $\overline{CF}$ passing through $M$. Also, $K$ goes to the second intersection of these two circles. Since it is well known that the radical axis of two intersecting circles is bisected by the line passing through its centers, $K$ must be the reflection of the internal $\angle BHC$-bisector. Now, it suffices to prove the following problem.
Inverted and Rephrased Version wrote:
Let $\triangle ABC$ be a triangle with orthocenter $H$ , $A-$Queue point $Q_A$, $A-$Humpty Point $H_A$ and $M$ the midpoint of side $BC$. If the reflection of $K$ across the internal $\angle BHC$-bisector lies on $(BHC)$, show that it also lies on circle $(MH_AQ_A)$.
Note that $MH$ is the $H-$ median of $\triangle BHC$, and its reflection across the $\angle BHC$-bisector must be the $H-$symmedian of $\triangle BHC$. Since $K$ lies on $(BHC)$, this means it is the intersection of the $H-$symmedian of $\triangle BHC$ with $(BHC)$. Now, $X_A$ is also the $H-$Ex Point of $\triangle BHC$ quite clearly, and thus it is a well known lemma that $K$ lies on the circle $(MX_A)$ which is precisely what we needed to show.
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bjump
1035 posts
#55
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Recall that $(ABC) \cap (EMB)  \cap (CMF) \cap (AEF) = Q_A$ the A-Queue point.

Now when we invert about $(BEFC)$ the following things happen:
  • $(MFC)$ is mapped to $FC$
  • $(BEM)$ is mapped to $EB$
  • $Q_A$ is mapped to $H$ the orthocenter of $\triangle ABC$
  • The external tangents are mapped to the circle through $B$, and $M$ tangent to $CF$, and $EB$, and the circle through $C$, and $M$ tangent to $CF$, and $EB$.

Now
$$90^\circ - C = \measuredangle CBH = \measuredangle MBH = \measuredangle MK'B =\measuredangle MBK$$Finishing.
This post has been edited 1 time. Last edited by bjump, Dec 20, 2024, 7:19 PM
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HamstPan38825
8869 posts
#56
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me when I forget the forgotten coaxiality lemma :maybe:

Because $HQ \cdot HM = HD \cdot HA = HB \cdot HE$, the queue point $Q$ lies on both $(BEM)$ and $(CFM)$. Let $P$ be the tangent intersection now, and $K$ the point on the circumcircle such that $\overline{AK} \perp \overline{BC}$. The key claim is the following:

Claim: $MKPQ$ is cyclic.

Proof: Note that $\overline{KB}$ and $\overline{KC}$ are tangent to their respective circles. Now use the forgotten coaxiality lemma: let $k$ be the homothety ratio between the two circles. Then \[\frac{\operatorname{pow}(K, (BEM))}{\operatorname{pow}(K, (CFM))} = \frac{BK^2}{CK^2} = \frac{\cos^2 B}{\cos^2 C} = k^2 = \frac{\operatorname{pow}(P, (BEM))}{\operatorname{pow}(P, (CFM))} \]as $EM=FM$ by tangents lemma. $\blacksquare$

Hence when $P$ lies on $(ABC)$, we must have $P = K$ as $P \neq Q$. The result follows.
This post has been edited 5 times. Last edited by HamstPan38825, Dec 29, 2024, 3:17 AM
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cj13609517288
1926 posts
#57
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P2???????

Invert around $M$ with radius $MB$, then $K$ goes to the reflection of $M$ over the angle bisector of angle $BHC$. Therefore, $HK^\ast$ is the $H$-symmedian of triangle $BHC$, so since $K^\ast$ lies on $(BHC)$, we get that $(BC;HK^\ast)=-1$. Therefore, $K^\ast$ is the reflection of the $A$-queue point over $BC$, so $K$ is the reflection of $H$ over $BC$. $\blacksquare$
This post has been edited 1 time. Last edited by cj13609517288, Mar 11, 2025, 10:38 PM
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ihategeo_1969
243 posts
#58 • 1 Y
Y by Aaronjudgeisgoat
Let $Q_A$ and $H_A$ be $A$-Queue and $A$-Humpty points respectively.

Claim: $Q_A=(BME) \cap (CMF)$.
Proof: Invert about $H$ with radius $-\sqrt{HB \cdot HE}$ and see that $M$ goes to $Q_A$ so done by PoP. $\square$

Now invert about $(M,\overline{MB})$, and see that $MH \cdot MQ_A=ME^2$ and so $Q_A$ swaps with $H$ so $(ABC)$ swaps with $(BHC)$ so $A$ swaps with $H_A$.

Also $K$ swaps with reflection of $M$ over angle bisector of $\angle BHC$. So we need to prove $K^* \in (BHC) \implies (K^*H_AM)$ is tangent to $\overline{M \infty_{\perp BC}}$. We exploit duality between $\triangle BAC$ and $\triangle BHC$ and solve this instead.
Quote:
Let $\triangle ABC$ have midpoint $M$ and Queue point as $Q_A$. If reflection of $M$ over angle bisector of $\angle BAC$ lies on $(ABC)$ then prove that $(M'MQ_A)$ is tangent to $\overline{M \infty_{\perp BC}}$.
So $M'$ basically is point on $(ABC)$ such that $(AM';BC)=-1$, call it $K$ instead. We actually disregard the whole reflection condition and prove that $(KMQ_A)$ has center on $\overline{BC}$.

See that if $X_A$ is $A$-Ex point, then $H_A$ and $Q_A$ lie on $(X_AM)$ by right angles and $K$ also lies on it as $H_A$ and $K$ are reflections of each other over $\overline{BC}$.
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EeEeRUT
84 posts
#59
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Let $Q$ denote the $A$ queue point of $\triangle ABC$.
Claim: $(BME) \cap (CMF)$ at $Q$.
Proof: Perform an inversion at $H$ with radius $-\sqrt{HA\cdot HD}$, where $D$ is foot of altitude from $A$. Notice that the circle $(BME)$ is fixed upon this inversion, hence if $M$ is on this circle, its inversion also lies on this circle, which is $Q$. That is $QMEQ$ is cyclic. $\blacksquare$

Let $\Phi$ be an inversion centered at $K$ with radius $KQ$. Notice that this inversion swap $(BME)$ and $(CMF)$. Let $B_1, C_1$ be images of $B, C$. Note that $B_1 \in (CMF), C_1 \in (BME)$.

Claim: $Q, B_1, C_1$ are collinear.
Proof]: Consider an inversion $\Phi$. Notice that this inversion sends $(ABC)$ to a line, since its center is on $(ABC)$. That is $B_1, C_1$ and an inverse of $Q$ are collinear. But $Q$ is a fixed point, so we have the claim. $\blacksquare$

Claim: $Q$ is midpoint of $B_1, C_1$.
Proof: Consider $2$ circles, $(BC_1MQ)$, $(CB_1MQ)$. By reims, $B_1C \parallel C_1B$. And since $(BB_1CC_1)$, by radical axis, $CB_1 \parallel BC_1 \parallel QM$. That is $QM$ is midline of isosceles trapezoid $BB_1CC_1$, hence $PQ$ bisect $B_1C_1$. $\blacksquare$


Claim:$(Q, K; B, C) = -1$
Proof: Since, $(Q, \infty ; B_1, C_1) = -1$, perform an inversion $\Phi$ on these ratio gives $$(Q, \infty ; B_1, C_1) = (Q, K; B, C)= -1$$$\blacksquare$.

It is well known that $(Q, K_1; B, C) =-1$, where $K_1$ is the reflection of $H$ across $BC$.
Hence, $K = K_1$ and we are done. $\blacksquare$
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