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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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[*]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
Inequality
Sadigly   5
N a few seconds ago by User141208
Source: Azerbaijan Senior NMO 2019
Prove that for any $a;b;c\in\mathbb{R^+}$, we have $$(a+b)^2+(a+b+4c)^2\geq \frac{100abc}{a+b+c}$$When does the equality hold?
5 replies
1 viewing
Sadigly
Yesterday at 8:47 PM
User141208
a few seconds ago
A strong inequality problem
hn111009   4
N 3 minutes ago by hn111009
Source: Somewhere
Let $a,b,c$ be the positive number satisfied $a^2+b^2+c^2=3.$ Find the minimum of $$P=\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}+\dfrac{3abc}{2(ab+bc+ca)}.$$
4 replies
1 viewing
hn111009
Yesterday at 2:02 AM
hn111009
3 minutes ago
Help me this problem. Thank you
illybest   1
N 4 minutes ago by GreekIdiot
Find f: R->R such that
f( xy + f(z) ) = (( xf(y) + yf(x) )/2) + z
1 reply
+1 w
illybest
2 hours ago
GreekIdiot
4 minutes ago
Interesting inequalities
sqing   1
N 16 minutes ago by pooh123
Source: Own
Let $ a,b>0  $ . Prove that
$$ \frac{a^2+b^2}{ab+1}+ \frac{4}{ (\sqrt{a}+\sqrt{b})^2} \geq 2$$$$ \frac{a^2+b^2}{ab+1}+ \frac{3}{a+\sqrt{ab}+b} \geq 2$$$$  \frac{a^3+b^3}{ab+1}+ \frac{4}{(a+b)^2}  \geq 2$$$$  \frac{a^3+b^3}{ab+1}+ \frac{3}{a^2+ab+b^2}  \geq 2$$$$\frac{a^2+b^2}{ab+2}+ \frac{1}{2\sqrt{ab}}  \geq \frac{2+3\sqrt{2}-2\sqrt{2(\sqrt{2}-1)}}{4} $$
1 reply
sqing
Today at 4:34 AM
pooh123
16 minutes ago
Diophantine involving cube
Sadigly   2
N 18 minutes ago by Adywastaken
Source: Azerbaijan Senior NMO 2020
$a;b;c;d\in\mathbb{Z^+}$. Solve the equation: $$2^{a!}+2^{b!}+2^{c!}=d^3$$
2 replies
Sadigly
Yesterday at 10:13 PM
Adywastaken
18 minutes ago
Nice R+ FE
math_comb01   4
N 20 minutes ago by mkultra42
Source: XOOK 2025 P3
Find all functions $f : \mathbb{R}_{>0} \to \mathbb{R}_{>0}$ so that for any $x, y \in \mathbb{R}_{>0}$, we have \[ f(xf(x^2)+yf(x))=xf(y+xf(x)). \]
Proposed by Anmol Tiwari and MathLuis
4 replies
math_comb01
Feb 9, 2025
mkultra42
20 minutes ago
2018 JBMO TST- Macedonia, problem 4
Lukaluce   4
N 29 minutes ago by Erto2011_
Source: 2018 JBMO TST- Macedonia
Determine all pairs $(p, q)$, $p, q \in \mathbb {N}$, such that

$(p + 1)^{p - 1} + (p - 1)^{p + 1} = q^q$.
4 replies
Lukaluce
May 28, 2019
Erto2011_
29 minutes ago
solve in positive integers: 3 \cdot 2^x +4 =n^2
parmenides51   4
N an hour ago by AylyGayypow009
Source: Greece JBMO TST 2019 p2
Find all pairs of positive integers $(x,n) $ that are solutions of the equation $3 \cdot 2^x +4 =n^2$.
4 replies
parmenides51
Apr 29, 2019
AylyGayypow009
an hour ago
ISI UGB 2025 P7
SomeonecoolLovesMaths   10
N an hour ago by Mathworld314
Source: ISI UGB 2025 P7
Consider a ball that moves inside an acute-angled triangle along a straight line, unit it hits the boundary, which is when it changes direction according to the mirror law, just like a ray of light (angle of incidence = angle of reflection). Prove that there exists a triangular periodic path for the ball, as pictured below.

IMAGE
10 replies
1 viewing
SomeonecoolLovesMaths
Yesterday at 11:28 AM
Mathworld314
an hour ago
Trigo relation in a right angled. ISIBS2011P10
Sayan   12
N an hour ago by SomeonecoolLovesMaths
Show that the triangle whose angles satisfy the equality
\[\frac{\sin^2A+\sin^2B+\sin^2C}{\cos^2A+\cos^2B+\cos^2C} = 2\]
is right angled.
12 replies
Sayan
Mar 31, 2013
SomeonecoolLovesMaths
an hour ago
Six variables
Nguyenhuyen_AG   4
N an hour ago by Sunjee
Let $a,\,b,\,c,\,x,\,y,\,z$ be six positive real numbers. Prove that
$$\frac{a}{b+c} \cdot \frac{y+z}{x} + \frac{b}{c+a} \cdot \frac{z+x}{y} + \frac{c}{a+b} \cdot \frac{x+y}{z} \geqslant 2+\sqrt{\frac{8abc}{(a+b)(b+c)(c+a)}}.$$
4 replies
Nguyenhuyen_AG
Yesterday at 5:09 AM
Sunjee
an hour ago
Expressing polynomial as product of two polynomials
Sadigly   3
N 2 hours ago by Sadigly
Source: Azerbaijan Senior NMO 2021
Define $P(x)=((x-a_1)(x-a_2)...(x-a_n))^2 +1$, where $a_1,a_2...,a_n\in\mathbb{Z}$ and $n\in\mathbb{N^+}$. Prove that $P(x)$ couldn't be expressed as product of two non-constant polynomials with integer coefficients.
3 replies
Sadigly
Yesterday at 9:10 PM
Sadigly
2 hours ago
Product of consecutive terms divisible by a prime number
BR1F1SZ   2
N 2 hours ago by bin_sherlo
Source: 2025 Francophone MO Seniors P4
Determine all sequences of strictly positive integers $a_1, a_2, a_3, \ldots$ satisfying the following two conditions:
[list]
[*]There exists an integer $M > 0$ such that, for all indices $n \geqslant 1$, $0 < a_n \leqslant M$.
[*]For any prime number $p$ and for any index $n \geqslant 1$, the number
\[
a_n a_{n+1} \cdots a_{n+p-1} - a_{n+p}
\]is a multiple of $p$.
[/list]


2 replies
BR1F1SZ
Yesterday at 12:09 AM
bin_sherlo
2 hours ago
Pentagon with given diameter, ratio desired
bin_sherlo   2
N 2 hours ago by Umudlu
Source: Türkiye 2025 JBMO TST P7
$ABCDE$ is a pentagon whose vertices lie on circle $\omega$ where $\angle DAB=90^{\circ}$. Let $EB$ and $AC$ intersect at $F$, $EC$ meet $BD$ at $G$. $M$ is the midpoint of arc $AB$ on $\omega$, not containing $C$. If $FG\parallel DE\parallel CM$ holds, then what is the value of $\frac{|GE|}{|GD|}$?
2 replies
bin_sherlo
Yesterday at 7:21 PM
Umudlu
2 hours ago
Stop Projecting your insecurities
naman12   52
N Apr 11, 2025 by ihategeo_1969
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
52 replies
naman12
Dec 12, 2022
ihategeo_1969
Apr 11, 2025
Stop Projecting your insecurities
G H J
G H BBookmark kLocked kLocked NReply
Source: 2022 USA TST #2
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asdf334
7585 posts
#43
Y by
wait I actually forgot that inversion existed that's probably not good. spent like 30 minutes then like 2 minutes after inverting ;-;

Let $Q$ be the $A$-Queue point.

Step 1: Notice by angle chase from Miquel configuration, we have $Q\in (BME),(CMF)$.

Step 2: Invert around $(BFEC)$, since $K\in (ABC)$ we get $K'\in (BHC)$.

Step 3: The external tangents are mapped to two circles passing through $M$ and $K'$ tangent to $BE$ and $CF$. Put another way, $K'$ should be the reflection of $M$ about the angle bisector of $\angle BHC$.

Step 4: $BHCK'$ is harmonic. As a result, let $H'$ be the reflection of $H$ across the perpendicular bisector of $BC$; then $H'\in MK'$. We also define $X\in (ABC)$ with $AX\perp BC$. We get $X\in H'M$ which means that $X\in MK'$.

Step 5: Since $X$ is both on the circle and on $MK'$ it coincides with $K$. Done!
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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((8.546847373658695,10.341098226284037)--(7.679547878066041,4.51346100817611)--(15.07954787806604,4.54546100817611)--cycle, linewidth(1.2)); 
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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
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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
3446 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
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#48
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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. $\blacksquare$
Attachments:
This post has been edited 1 time. Last edited by Pyramix, Apr 8, 2024, 6:46 PM
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shendrew7
796 posts
#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
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#53
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solution
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cursed_tangent1434
632 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
1028 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
8864 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
1916 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
235 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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