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k a March Highlights and 2025 AoPS Online Class Information
jlacosta   0
Mar 2, 2025
March is the month for State MATHCOUNTS competitions! Kudos to everyone who participated in their local chapter competitions and best of luck to all going to State! Join us on March 11th for a Math Jam devoted to our favorite Chapter competition problems! Are you interested in training for MATHCOUNTS? Be sure to check out our AMC 8/MATHCOUNTS Basics and Advanced courses.

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Be sure to mark your calendars for the following events:
[list][*]March 5th (Wednesday), 4:30pm PT/7:30pm ET, HCSSiM Math Jam 2025. Amber Verser, Assistant Director of the Hampshire College Summer Studies in Mathematics, will host an information session about HCSSiM, a summer program for high school students.
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0 replies
jlacosta
Mar 2, 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
Wait wasn't it the reciprocal in the paper?
Supercali   6
N 27 minutes ago by kes0716
Source: India TST 2023 Day 2 P1
Let $\mathbb{Z}_{\ge 0}$ be the set of non-negative integers and $\mathbb{R}^+$ be the set of positive real numbers. Let $f: \mathbb{Z}_{\ge 0}^2 \rightarrow \mathbb{R}^+$ be a function such that $f(0, k) = 2^k$ and $f(k, 0) = 1$ for all integers $k \ge 0$, and $$f(m, n) = \frac{2f(m-1, n) \cdot f(m, n-1)}{f(m-1, n)+f(m, n-1)}$$for all integers $m, n \ge 1$. Prove that $f(99, 99)<1.99$.

Proposed by Navilarekallu Tejaswi
6 replies
+1 w
Supercali
Jul 9, 2023
kes0716
27 minutes ago
Inspired by Kazakhstan 2017
sqing   0
38 minutes ago
Source: Own
Let $a,b,c\ge \frac{1}{2}$ and $a+b+c=2. $ Prove that
$$\left(\frac{1}{a}+\frac{1}{b}-\frac{1}{c}\right)\left(\frac{1}{a}-\frac{1}{b}+\frac{1}{c}\right)\ge 1$$Let $a,b,c\ge \frac{1}{3}$ and $a+b+c=1. $ Prove that
$$\left(\frac{1}{a}+\frac{1}{b}-\frac{1}{c}\right)\left(\frac{1}{a}-\frac{1}{b}+\frac{1}{c}\right)\ge 9$$
0 replies
1 viewing
sqing
38 minutes ago
0 replies
About old Inequality
perfect_square   0
44 minutes ago
Source: Arqady
This is: $a,b,c>0$ which satisfy $abc=1$
Prove that: $ \frac{a+b+c}{3} \ge \sqrt[10]{\frac{a^3+b^3+c^3}{3}}$
By $  uvw $ method, I can assum $b=c=x,a=\frac{1}{x^2}$
But I can't prove:
$ \frac{2x+\frac{1}{x^2}}{3} \ge \sqrt[10]{ \frac{2x^3+ \frac{1}{x^6}}{3}} $
Is there an another way?
0 replies
perfect_square
44 minutes ago
0 replies
inquality
karasuno   1
N an hour ago by sqing
The real numbers $x,y,z \ge \frac{1}{2}$ are given such that $x^{2}+y^{2}+z^{2}=1$. Prove the inequality $$(\frac{1}{x}+\frac{1}{y}-\frac{1}{z})(\frac{1}{x}-\frac{1}{y}+\frac{1}{z})\ge 2 .$$
1 reply
karasuno
2 hours ago
sqing
an hour ago
Number Theory
karasuno   0
2 hours ago
Solve the equation $$n!+10^{2014}=m^{4}$$in natural numbers m and n.
0 replies
karasuno
2 hours ago
0 replies
Minimum number of values in the union of sets
bnumbertheory   5
N 2 hours ago by Rohit-2006
Source: Simon Marais Mathematics Competition 2023 Paper A Problem 3
For each positive integer $n$, let $f(n)$ denote the smallest possible value of $$|A_1 \cup A_2 \cup \dots \cup A_n|$$where $A_1, A_2, A_3 \dots A_n$ are sets such that $A_i \not\subseteq A_j$ and $|A_i| \neq |A_j|$ whenever $i \neq j$. Determine $f(n)$ for each positive integer $n$.
5 replies
bnumbertheory
Oct 14, 2023
Rohit-2006
2 hours ago
two triangles have equal circumradii
littletush   5
N 3 hours ago by Taha.kh
Source: Italy TST 2009 p5
Two circles $O_1$ and $O_2$ intersect at $M,N$. The common tangent line nearer to $M$ of the two circles touches $O_1,O_2$ at $A,B$ respectively. Let $C,D$ be the symmetric points of $A,B$ with respect to $M$ respectively. The circumcircle of triangle $DCM$ intersects circles $O_1$ and $O_2$ at points $E,F$ respectively which are distinct from $M$. Prove that the circumradii of the triangles $MEF$ and $NEF$ are equal.
5 replies
littletush
Mar 10, 2012
Taha.kh
3 hours ago
Equal lengths in cyclic quadrilateral
LoloChen   4
N 3 hours ago by Nari_Tom
Source: All-Russian MO 2024 9.4
In cyclic quadrilateral $ABCD$, $\angle A+ \angle D=\frac{\pi}{2}$. $AC$ intersects $BD$ at ${E}$. A line ${l}$ cuts segment $AB, CD, AE, DE$ at $X, Y, Z, T$ respectively. If $AZ=CE$ and $BE=DT$, prove that the diameter of the circumcircle of $\triangle EZT$ equals $XY$.
4 replies
LoloChen
Apr 22, 2024
Nari_Tom
3 hours ago
Two circles and many points
CHN_Lucas   5
N 4 hours ago by Captainscrubz
Source: 2022 China Second Round A2
$A,B,C,D,E$ are points on a circle $\omega$, satisfying $AB=BD$, $BC=CE$. $AC$ meets $BE$ at $P$. $Q$ is on $DE$ such that $BE//AQ$. Suppose $\odot(APQ)$ intersects $\omega$ again at $T$. $A'$ is the reflection of $A$ wrt $BC$. Prove that $A'BPT$ lies on the same circle.
5 replies
CHN_Lucas
Dec 22, 2022
Captainscrubz
4 hours ago
Angle QRP = 90°
orl   12
N 4 hours ago by YaoAOPS
Source: IMO ShortList, Netherlands 1, IMO 1975, Day 1, Problem 3
In the plane of a triangle $ABC,$ in its exterior$,$ we draw the triangles $ABR, BCP, CAQ$ so that $\angle PBC = \angle CAQ = 45^{\circ}$, $\angle BCP = \angle QCA = 30^{\circ}$, $\angle ABR = \angle RAB = 15^{\circ}$.

Prove that

a.) $\angle QRP = 90\,^{\circ},$ and

b.) $QR = RP.$
12 replies
orl
Nov 12, 2005
YaoAOPS
4 hours ago
IMO 2014 Problem 4
ipaper   166
N 4 hours ago by hgomamogh
Let $P$ and $Q$ be on segment $BC$ of an acute triangle $ABC$ such that $\angle PAB=\angle BCA$ and $\angle CAQ=\angle ABC$. Let $M$ and $N$ be the points on $AP$ and $AQ$, respectively, such that $P$ is the midpoint of $AM$ and $Q$ is the midpoint of $AN$. Prove that the intersection of $BM$ and $CN$ is on the circumference of triangle $ABC$.

Proposed by Giorgi Arabidze, Georgia.
166 replies
ipaper
Jul 9, 2014
hgomamogh
4 hours ago
IMO 2016 Problem 1
quangminhltv99   146
N 4 hours ago by Ilikeminecraft
Source: IMO 2016
Triangle $BCF$ has a right angle at $B$. Let $A$ be the point on line $CF$ such that $FA=FB$ and $F$ lies between $A$ and $C$. Point $D$ is chosen so that $DA=DC$ and $AC$ is the bisector of $\angle{DAB}$. Point $E$ is chosen so that $EA=ED$ and $AD$ is the bisector of $\angle{EAC}$. Let $M$ be the midpoint of $CF$. Let $X$ be the point such that $AMXE$ is a parallelogram. Prove that $BD,FX$ and $ME$ are concurrent.
146 replies
quangminhltv99
Jul 11, 2016
Ilikeminecraft
4 hours ago
A function equation
YaWNeeT   8
N 5 hours ago by HamstPan38825
Source: 2017 Taiwan TST 2nd round day 2 P4
Find all integer $c\in\{0,1,...,2016\}$ such that the number of $f:\mathbb{Z}\rightarrow\{0,1,...,2016\}$ which satisfy the following condition is minimal:
(1) $f$ has periodic $2017$
(2) $f(f(x)+f(y)+1)-f(f(x)+f(y))\equiv c\pmod{2017}$

Proposed by William Chao
8 replies
YaWNeeT
Apr 15, 2017
HamstPan38825
5 hours ago
Circumcenter lies on altitude
ABCDE   58
N 5 hours ago by cj13609517288
Source: 2016 ELMO Problem 2
Oscar is drawing diagrams with trash can lids and sticks. He draws a triangle $ABC$ and a point $D$ such that $DB$ and $DC$ are tangent to the circumcircle of $ABC$. Let $B'$ be the reflection of $B$ over $AC$ and $C'$ be the reflection of $C$ over $AB$. If $O$ is the circumcenter of $DB'C'$, help Oscar prove that $AO$ is perpendicular to $BC$.

James Lin
58 replies
ABCDE
Jun 24, 2016
cj13609517288
5 hours ago
Immaculate Geometry
anantmudgal09   6
N Oct 1, 2024 by Pyramix
Source: The 1st India-Iran Friendly Competition Problem 6
Let $ABC$ be a triangle with midpoint $M$ of $BC$. A point $X$ is called immaculate if the perpendicular line from $X$ to line $MX$ intersects lines $AB$ and $AC$ at two points that are equidistant from $M$. Suppose $U, V, W$ are three immaculate points on the circumcircle of triangle $ABC$. Prove that $M$ is the incentre of $\triangle UVW$.


Proposed by Pranjal Srivastava and Rohan Goyal
6 replies
anantmudgal09
Jun 13, 2024
Pyramix
Oct 1, 2024
Immaculate Geometry
G H J
G H BBookmark kLocked kLocked NReply
Source: The 1st India-Iran Friendly Competition Problem 6
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anantmudgal09
1979 posts
#1 • 4 Y
Y by GeoKing, Supercali, Rg230403, kiyoras_2001
Let $ABC$ be a triangle with midpoint $M$ of $BC$. A point $X$ is called immaculate if the perpendicular line from $X$ to line $MX$ intersects lines $AB$ and $AC$ at two points that are equidistant from $M$. Suppose $U, V, W$ are three immaculate points on the circumcircle of triangle $ABC$. Prove that $M$ is the incentre of $\triangle UVW$.


Proposed by Pranjal Srivastava and Rohan Goyal
This post has been edited 1 time. Last edited by anantmudgal09, Jun 13, 2024, 3:42 PM
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anantmudgal09
1979 posts
#2 • 3 Y
Y by GeoKing, Rijul saini, Supercali
anantmudgal09 wrote:
Let $ABC$ be a triangle with midpoint $M$ of $BC$. A point $X$ is called immaculate if the perpendicular line from $X$ to line $MX$ intersects lines $AB$ and $AC$ at two points that are equidistant from $M$. Suppose $U, V, W$ are three immaculate points on the circumcircle of triangle $ABC$. Prove that $M$ is the incentre of $\triangle UVW$.


Proposed by Pranjal Srivastava and Rohan Goyal

Let $X$ be an immaculate point and $P, Q$ be the points where lines $AB, AC$ meet the perpendicular to line $MX$ at $X$. Since $XP=XQ$, we get $$-1 = A(PQ; X \infty) = (AB, AC; AX, \ell)$$where $\ell$ is the perpendicular to $MX$ from $A$. Let $D$ be the antipode of $A$ in $BC$. Look at lines perpendicular to lines of this pencil passing through $D$, we get $(DB, DC; DX, DK) = -1$ where $K$ lies on the line through $D$ parallel to line $XM$. Let tangents to $B$ and $C$ meet at $T$; then $T, K, X$ are collinear so we redefine $K$ as such and compute the conditions on $X$ with $XM \parallel DK$. We now use complex numbers with $(ABC)$ as the unit circle. Let $X$ be an immaculate point with coordinate $x^2$, with an appropriate sign to be chosen later.


Suppose $M$ has coordinate $m \in \mathbb{R}$ by aligning the diameter line through $M$ with the real axis (using a rotation if necessary). Then $m = \overline{m}$. So $T$ has coordinate $\frac{1}{m}$. Now $$K \in \overline{TX} \iff \frac{k-x^2}{x^2-\frac{1}{m}} = \frac{\frac{1}{k}-\frac{1}{x^2}}{\frac{1}{x^2}-\frac{1}{m}} \iff k = \frac{x^2-\frac{1}{m}}{x^2\left(\frac{x^2-m}{mx^2}\right)} = \frac{mx^2-1}{x^2-m}$$and $$XM \parallel DK \iff  \frac{k-d}{x^2-m} = \frac{\frac{1}{k}-\frac{1}{d}}{\frac{1}{x^2}-m} \iff kd = \frac{x^2(x^2-m)}{mx^2-1} \iff d = \frac{x^2(x^2-m)^2}{(mx^2-1)^2}.$$Fix a square root $\lambda$ of $d$. Choosing the value of $x$ such that $$\frac{x(x^2-m)}{(mx^2-1)} = \lambda \iff x^3-\lambda mx^2-mx+\lambda = 0.$$Thus, if $X, Y, Z$ are immaculate points, there exists complex numbers $x, y, z$ such that $X=x^2, Y=y^2, Z=z^2$, and $-m = xy+yz+zx$ by Vieta's formula, hence $m=-(xy+yz+zx)$. Thus, $M$ is either the incentre or one of the three excentres of $\triangle ABC$. Since $M$ lies in the interior of $(XYZ)$, we are done.


Anecdote
This post has been edited 1 time. Last edited by anantmudgal09, Jun 13, 2024, 5:51 PM
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mueller.25
86 posts
#3 • 2 Y
Y by GeoKing, anantmudgal09
Let $U'$ denote the point on $(ABC)$ such that $(UU';BC)=-1$. Define $V'$ and $W'$ similarly. Projecting $(UU';BC)=-1$ from $A$ onto the line through $U$ perpendicular to $MU$, we get that $AU' \perp UM$.

Note that we have the cyclic quadrilaterals $UU'OM$, $VV'OM$, $WW'OM$ where $O$ is the circumcentre of $ABC$. As $V'A \perp MV$ and $W'A \perp MW$,
\[\measuredangle V'AW'=\measuredangle VMW\]\begin{align*}
	2 \measuredangle V'AW'
	&=\measuredangle V'OW'\\
    &= \measuredangle V'OM +\measuredangle MOW'\\
    &= \measuredangle V'VM + \measuredangle MWW'\\
    &= \measuredangle V'VU+\measuredangle UVM+ \measuredangle MWU+ \measuredangle UWW'\\
	&= \measuredangle V'WU+\measuredangle UWW'+\measuredangle UVM+ \measuredangle MWU\\
	&= \measuredangle V'AW'+ \measuredangle UVM+ \measuredangle MWU\\
\end{align*}\[\measuredangle VMW=\measuredangle UVM+ \measuredangle MWU\implies 2 \measuredangle VMW= \measuredangle VUW\]The centre $N_1$ of $(VMW)$ lies on $(UVW)$. Similarly, the centres $N_2,N_3$ of $(UMW)$ and $UMV$ lie on $(UVW)$. If any two of $N_1,N_2,N_3$ are the midpoints of the corresponding arcs not containing the opposite vertex, then $M$ is the incentre. If not, then $N_1,N_2,N_3$ are the arc midpoints of $\widehat{VUW},\widehat{UVW},\widehat{UWV}$. It follows that $M$ is the reflection of $U$ in $N_2N_3$, but then $UM$ internally bisects $\angle U$ and similarly $VM,WM$ are also internal angle bisectors which implies that $M$ must be the incentre.
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math_comb01
659 posts
#4 • 1 Y
Y by GeoKing
Nice Problem!
Let $BB \cap CC = T$ and $TU,TV,TW \cap (ABC) = U^{*},V^{*},V^{*}$.
Claim 1: $AU^{*} \perp MU$
Proof
Claim 2: If $O$ is the circumcenter of $ABC$, then $UU^{*}OM$ is cyclic.
Proof
Claim 3: The center of $(MUV)$ lies on $(ABC)$.
Proof
Claim 3 and cyclic variants imply $M$ is incenter of $UVW$.
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mathscrazy
113 posts
#5 • 2 Y
Y by GeoKing, Om245
Relabel the immaculate points on $(ABC)$ as $X_1,X_2,X_3$. Let $\omega=(ABC)$ with center $O$. Let $Y_i$ be the point on $\omega$ such that $(X_iY_i;BC)=-1$ for $i=1,2,3$.

Claim 1 : $AY_i \perp MX_i$ and $X_iY_iOM$ is cyclic for $i=1,2,3$.
Proof : Projecting the cross ratio $(X_iY_i;BC)=-1$, from $A$ onto the line through $X_i$ perpendicular to $MX_i$, we infer that $AY_i \perp X_iM$.
If $BB\cap CC=T$, then $TX_i\cdot TY_i = TB^2 = TO\cdot TM$ which implies that $X_iY_iOM$ is cyclic. $\blacksquare$

Claim 2 : $M$ is orthocenter of $Y_1Y_2Y_3$.
Proof : Angel chase $\measuredangle Y_1Y_2Y_3 = \measuredangle Y_1AY_3 = \measuredangle X_1MX_3 = \measuredangle Y_3MY_1$.
The second equality follows since $AY_i \perp MX_i$; and the last equality follows by reflection over line $OM$, since it is angle bisector of $\angle X_iMY_i$.
Similarly doing the angle chase for $\measuredangle Y_1Y_3Y_2$ implies the claim. $\blacksquare$

Claim 3 : $M$ lies on angle bisector $X_1X_3X_2$.
Proof : Look at quadrilateral $X_3Y_2Y_3X_2$ with center $O$. Since $M = (X_3Y_3O) \cap (X_2Y_2O)$, $M$ is the spiral center of the quadrilateral.
Hence $\measuredangle MX_3X_2 = \measuredangle MY_2Y_3 = 90 - \measuredangle Y_2Y_3Y_1$ by claim 2.
Hence, $90 - \measuredangle MX_3X_2 =  \measuredangle Y_2Y_3Y_1 = - \measuredangle Y_1Y_3Y_2 = -90 + \measuredangle MX_3X_1$, which implies the claim. $\blacksquare$

Hence $M$ is incentre or excentre of $X_1X_2X_3$. Since $M$ lies inside $(X_1X_2X_3)$, it must be the incentre and we are done!
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sami1618
871 posts
#6 • 1 Y
Y by GeoKing
Seems too simple, but I couldn't find a mistake :)

Claim: $\measuredangle AWM+\measuredangle BMW=90^{\circ}$
Intersect the perpendicular to $MW$ through $W$ with $AB$ and $AC$ at $B_W$ and $C_W$. Let $W'$ be the point such that $AB_WW'C_W$ is a parallelogram. Then triangle $AB_W W'$ and $CWB$ are similar with opposite orientation so $\measuredangle WMB= \measuredangle C_W W A$ and the result follows.

Cyclic variants also hold. Thus $$\angle WMV=-\measuredangle BMW-\measuredangle VMC=\measuredangle AWM+\measuredangle MVA=\measuredangle WAV-\measuredangle  WMV\iff 2\measuredangle WMV=\measuredangle WUV$$Cyclic variants imply $M$ is the incenter of $UVW$ as desired. remark
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This post has been edited 1 time. Last edited by sami1618, Aug 18, 2024, 9:32 PM
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Pyramix
419 posts
#7
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We make the use of complex numbers. Toss $ABC$ onto the unit circle. Let $X$ denote an immaculate point and let $X_1,X_2$ be points of $AB,AC$ such that $X_1,X_2,X$ are collinear and that line is perpendicular to $MX$.

First note that since $MX_1=MX_2$ and $MX\perp X_1X_2$, we must have $X$ to be mid-point of $\overline{X_1X_2}$. So, $2X-X_1$ must lie on line $AC$. We have,
\[x_1+ab\overline{x_1}=a+b, \ \ \ x_2+ac\overline{x_2}=a+c, \ \ \ x_1+x_2=2x\]This systems gives
\[x_1=\frac{2bc+ab+ac-\frac{2abc}{x}-2bx}{c-b},x_2=\frac{2bc+ab+ac-\frac{2abc}{x}-2cx}{b-c}\]Hence, $\frac{x_2-x_1}{\overline{x_2-x_1}}=\frac{ax\left(b+c-\frac{2bc}{x}\right)}{b+c-2x}$
Since $MX\perp X_1X_2$, we want
\[ax\left(\frac{m-\frac{bc}{x}}{m-x}\right)=ax\left(\frac{b+c-\frac{2bc}{x}}{b+c-2x}\right)=\frac{x_2-x_1}{\overline{x_2-x_1}}=\frac{x-m}{\overline{m}-\frac1x}\]\[\Longrightarrow x(x-m)^2+a(mx-bc)(\overline{m}x-1)=0.\]The roots of the cubic are $x_1,x_2,x_3$, then we need to compute $-\left(\sqrt{x_1x_2}+\sqrt{x_2x_3}+\sqrt{x_3x_1}\right)$
The rest follows by Vieta's relations.
This post has been edited 1 time. Last edited by Pyramix, Oct 1, 2024, 2:24 PM
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