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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.

Are you interested in working towards MATHCOUNTS and don’t know where to start? We have you covered! If you have taken Prealgebra, then you are ready for MATHCOUNTS/AMC 8 Basics. Already aiming for State or National MATHCOUNTS and harder AMC 8 problems? Then our MATHCOUNTS/AMC 8 Advanced course is for you.

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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
Diophantine
TheUltimate123   32
N a few seconds ago by Kempu33334
Source: CJMO 2023/1 (https://aops.com/community/c594864h3031323p27271877)
Find all triples of positive integers \((a,b,p)\) with \(p\) prime and \[a^p+b^p=p!.\]
Proposed by IndoMathXdZ
32 replies
TheUltimate123
Mar 29, 2023
Kempu33334
a few seconds ago
Parallel lines in incircle configuration
GeorgeRP   4
N 2 minutes ago by VicKmath7
Source: Bulgaria IMO TST 2025 P1
Let $I$ be the incenter of triangle $\triangle ABC$. Let $H_A$, $H_B$, and $H_C$ be the orthocenters of triangles $\triangle BCI$, $\triangle ACI$, and $\triangle ABI$, respectively. Prove that the lines through $H_A$, $H_B$, and $H_C$, parallel to $AI$, $BI$, and $CI$, respectively, are concurrent.
4 replies
1 viewing
GeorgeRP
May 14, 2025
VicKmath7
2 minutes ago
Geometry
shactal   1
N 7 minutes ago by ricarlos
Two intersecting circles $C_1$ and $C_2$ have a common tangent that meets $C_1$ in $P$ and $C_2$ in $Q$. The two circles intersect at $M$ and $N$ where $N$ is closer to $PQ$ than $M$ . Line $PN$ meets circle $C_2$ a second time in $R$. Prove that $MQ$ bisects angle $\widehat{PMR}$.
1 reply
shactal
Today at 3:04 PM
ricarlos
7 minutes ago
Curious inequality
produit   1
N 26 minutes ago by arqady
Positive real numbers x_1, x_2, . . . x_n satisfy x_1 + x_2 + . . . + x_n = 1.
Prove that
1/(1 −√x_1)+1/(1 −√x_2)+ . . . +1/(1 −√x_n)⩾ n + 4.
1 reply
produit
May 10, 2025
arqady
26 minutes ago
Simson lines on OH circle
DVDTSB   3
N 4 hours ago by Funcshun840
Source: Romania TST 2025 Day 2 P4
Let \( ABC \) and \( DEF \) be two triangles inscribed in the same circle, centered at \( O \), and sharing the same orthocenter \( H \ne O \). The Simson lines of the points \( D, E, F \) with respect to triangle \( ABC \) form a non-degenerate triangle \( \Delta \).
Prove that the orthocenter of \( \Delta \) lies on the circle with diameter \( OH \).

Note. Assume that the points \( A, F, B, D, C, E \) lie in this order on the circle and form a convex, non-degenerate hexagon.

Proposed by Andrei Chiriță
3 replies
DVDTSB
May 13, 2025
Funcshun840
4 hours ago
IMO Shortlist 2012, Geometry 8
lyukhson   33
N 5 hours ago by awesomeming327.
Source: IMO Shortlist 2012, Geometry 8
Let $ABC$ be a triangle with circumcircle $\omega$ and $\ell$ a line without common points with $\omega$. Denote by $P$ the foot of the perpendicular from the center of $\omega$ to $\ell$. The side-lines $BC,CA,AB$ intersect $\ell$ at the points $X,Y,Z$ different from $P$. Prove that the circumcircles of the triangles $AXP$, $BYP$ and $CZP$ have a common point different from $P$ or are mutually tangent at $P$.

Proposed by Cosmin Pohoata, Romania
33 replies
lyukhson
Jul 29, 2013
awesomeming327.
5 hours ago
IMO 2012 P5
mathmdmb   123
N 6 hours ago by SimplisticFormulas
Source: IMO 2012 P5
Let $ABC$ be a triangle with $\angle BCA=90^{\circ}$, and let $D$ be the foot of the altitude from $C$. Let $X$ be a point in the interior of the segment $CD$. Let $K$ be the point on the segment $AX$ such that $BK=BC$. Similarly, let $L$ be the point on the segment $BX$ such that $AL=AC$. Let $M$ be the point of intersection of $AL$ and $BK$.

Show that $MK=ML$.

Proposed by Josef Tkadlec, Czech Republic
123 replies
mathmdmb
Jul 11, 2012
SimplisticFormulas
6 hours ago
Fixed line
TheUltimate123   14
N 6 hours ago by amirhsz
Source: ELMO Shortlist 2023 G4
Let \(D\) be a point on segment \(PQ\). Let \(\omega\) be a fixed circle passing through \(D\), and let \(A\) be a variable point on \(\omega\). Let \(X\) be the intersection of the tangent to the circumcircle of \(\triangle ADP\) at \(P\) and the tangent to the circumcircle of \(\triangle ADQ\) at \(Q\). Show that as \(A\) varies, \(X\) lies on a fixed line.

Proposed by Elliott Liu and Anthony Wang
14 replies
TheUltimate123
Jun 29, 2023
amirhsz
6 hours ago
KJMO 2001 P1
RL_parkgong_0106   1
N Today at 2:05 PM by JH_K2IMO
Source: KJMO 2001
A right triangle of the following condition is given: the three side lengths are all positive integers and the length of the shortest segment is $141$. For the triangle that has the minimum area while satisfying the condition, find the lengths of the other two sides.
1 reply
RL_parkgong_0106
Jun 29, 2024
JH_K2IMO
Today at 2:05 PM
Cotangential circels
CountingSimplex   5
N Today at 2:02 PM by rong2020
Let $ABC$ be a triangle with circumcenter $O$ and let the angle bisector of $\angle{BAC}$ intersect $BC$
at $D$. The point $M$ is such that $\angle{MCB}=90^o$ and $\angle{MAD}=90^o$. Lines $BM$ and $OA$ intersect at
the point $P$. Show that the circle centered at $P$ and passing through $A$ is tangent to segment
$BC$.
5 replies
CountingSimplex
Jun 23, 2020
rong2020
Today at 2:02 PM
Midpoint in a weird configuration
Gimbrint   0
Today at 1:39 PM
Source: Own
Let $ABC$ be an acute triangle ($AB<BC$) with circumcircle $\omega$. Point $L$ is chosen on arc $AC$, not containing $B$, so that, letting $BL$ intersect $AC$ at $S$, one has $AS<CS$. Points $D$ and $E$ lie on lines $AB$ and $BC$ respectively, such that $BELD$ is a parallelogram. Point $P$ is chosen on arc $BC$, not containing $A$, such that $\angle CBP=\angle BDE$. Line $AP$ intersects $EL$ at $X$, and line $CP$ intersects $DL$ at $Y$. Line $XY$ intersects $AB$, $BC$ and $BP$ at points $M$, $N$ and $T$ respectively.

Prove that $TN=TM$.
0 replies
Gimbrint
Today at 1:39 PM
0 replies
Quadrangle, nine-point conic, Steiner line
kosmonauten3114   0
Today at 1:23 PM
Source: My own
Let $P_1P_2P_3P_4$ be a general quadrangle which does not form an orthocentric system. Let $P$, $I$, $M$, $T$ be the Euler-Poncelet point ($\text{QA-P2}$), isogonal center ($\text{QA-P4}$), midray homothetic center ($\text{QA-P8}$), inscribed square axes crosspoint ($\text{QA-P23}$) of $P_1P_2P_3P_4$, respectively.
Let $H_1$ be the orthocenter of $\triangle{P_2P_3P_4}$, and define $H_2$, $H_3$, $H_4$ cyclically.
Let $A_{ij}=P_iP_j \cap H_iH_j$ ($\{i, j\} \in \{1, 2, 3, 4\}, i<j$).
Let $B_{ij}=P_iP_j \cap H_kH_l$ ($\{i, j, k, l\} \in \{1, 2, 3, 4\}, i<j$).
Then, the 12 points $A_{12}$, $A_{13}$, $A_{14}$, $A_{23}$, $A_{24}$, $A_{34}$, $B_{12}$, $B_{13}$, $B_{14}$, $B_{23}$, $B_{24}$, $B_{34}$ lie on the same conic, here denoted by $\mathcal{C}_1$.
Let $\mathcal{C}_2$ be the nine-point conic of $P_1P_2P_3P_4$.
Suppose that $\mathcal{C}_1$ and $\mathcal{C}_2$ have 4 distinct real intersection points, and let $U$, $V$, $W$ be the intersections, other than $P$, of $\mathcal{C}_1$ and $\mathcal{C}_2$.

Prove that the Steiner line of $P$ with respect to $\triangle{UVW}$ passes through $I$ and $M$, and show that the center of $\mathcal{C}_1$ and the orthocenter of $\triangle{UVW}$ coincide with $T$.
0 replies
kosmonauten3114
Today at 1:23 PM
0 replies
My Unsolved Problem
ZeltaQN2008   1
N Today at 12:58 PM by Funcshun840
Source: IDK
Let triangle \(ABC\) be inscribed in circle \((O)\). Let \((I_a)\) be the \(A\)-excircle of triangle \(ABC\), which is tangent to \(BC\), the extension of \(AB\), and the extension of \(AC\). Let \(BE\) and \(CF\) be the angle bisectors of triangle \(ABC\). Let \(EF\) intersect \((O)\) at two points \(S\) and \(T\).

a) Prove that circle \((O)\) bisects the segments \(I_aT\) and \(I_aS\).
b) Prove that \(S\) and \(T\) are the points of tangency of the common external tangents of circles \((O)\) and \((I_a)\) .

1 reply
ZeltaQN2008
Yesterday at 3:07 PM
Funcshun840
Today at 12:58 PM
Geometry Concurrence
KHOMNYO2   1
N Today at 12:32 PM by Funcshun840
Given triangle $XYZ$ such that $XY \neq XZ$. Let excircle-$X$ be tangent with $YZ, ZX, XY$ at points $U, V, W$ respectively. Let $R$ and $S$ be points on the segment $XZ, XY$ respectively such that $RS$ is parallel to $YZ$. Lastly, let $\gamma$ be the circle that is externally tangent with the excircle-$X$ on point $T$. Prove that $VW, UT$, and $RS$ concur at a point.
1 reply
KHOMNYO2
Today at 10:47 AM
Funcshun840
Today at 12:32 PM
variable point on the line BC
orl   27
N May 11, 2025 by Markas
Source: IMO Shortlist 2004 geometry problem G7
For a given triangle $ ABC$, let $ X$ be a variable point on the line $ BC$ such that $ C$ lies between $ B$ and $ X$ and the incircles of the triangles $ ABX$ and $ ACX$ intersect at two distinct points $ P$ and $ Q.$ Prove that the line $ PQ$ passes through a point independent of $ X$.
27 replies
orl
Jun 14, 2005
Markas
May 11, 2025
variable point on the line BC
G H J
Source: IMO Shortlist 2004 geometry problem G7
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orl
3647 posts
#1 • 4 Y
Y by doxuanlong15052000, Adventure10, Mango247, cubres
For a given triangle $ ABC$, let $ X$ be a variable point on the line $ BC$ such that $ C$ lies between $ B$ and $ X$ and the incircles of the triangles $ ABX$ and $ ACX$ intersect at two distinct points $ P$ and $ Q.$ Prove that the line $ PQ$ passes through a point independent of $ X$.
This post has been edited 2 times. Last edited by djmathman, Sep 27, 2015, 2:19 PM
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pohoatza
1145 posts
#2 • 12 Y
Y by doxuanlong15052000, thczarif, Adventure10, aritrads, Soundricio, myh2910, easonlamb, Mango247, and 4 other users
The problem follows nicely by a "lemma" proposed by Virgil Nicula here: http://www.mathlinks.ro/Forum/viewtopic.php?t=132338

Now, returning to the problem:
Let the incircles of $\triangle{ABX}$ and $\triangle{ACX}$ touch $BX$ at $D$ and $F$, respectively, and let them touch $AX$ at $E$ and $G$, respectively.

Clearly, $DE \| FG$.
If the line $PQ$ intersects $BX$ and $AX$ at $M$ and $N$, respectively, then $MD^{2}= MP\cdot MQ = MF^{2}$, i.e., $MD = MF$ and analogously $NE = NG$.

It follows that $PQ \| DE$ and $FG$ and equidistant from them. The midpoints of $AB, AC$, and $AX$ lie on the same line $m$, parallel to $BC$.

Applying the "lemma" to $\triangle{ABX}$, we conclude that $DE$ passes through the common point $U$ of $m$ and the bisector of $\angle{ABX}$.
Analogously, $FG$ passes through the common point $V$ of $m$ and the bisector of $\angle{ACX}$. Therefore $PQ$ passes through the midpoint $W$ of the line segment $UV$ . Since $U, V$ do not depend on $X$, neither does $W$.
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math154
4302 posts
#3 • 3 Y
Y by Adventure10, Mango247, and 1 other user
Let the incircle of $\triangle{ABC}$ meet $BC,CA,AB$ at $D_1,E_1,F_1$ and have inradius $r$. Then setting $u=AE_1=AF_1$ and so on, we have $r^2=uvw/(u+v+w)$. Similarly, let $\triangle{ABX}$'s incircle meet $BX,XA,AB$ at $D_2,E_2,F_2$ and let $\triangle{ACX}$'s incircle meet $CX,XA,AC$ at $D_3,E_3,F_3$. By simple length chasing, we find that $D_2D_3=E_2E_3=CD_1=CE_1=w$. Now define $M=BX\cap PQ$ and $N=AX\cap PQ$. Then $MD_2^2=MQ\cdot MP=MD_3^2$, and so $MD_2=MD_3=NE_2=NE_3=w/2$. Clearly we also have $D_3E_3 \| MN \| D_2E_2$, so
\[\angle{NMX}=\frac\pi2-\frac{\angle{AXB}}{2}=\frac{\angle{D_1BF_1}}{2}+\frac{\angle{E_2AF_2}}{2},\]and the tangent addition formula combined with some simple ratios and lengths gives us
\begin{align*}
\tan\angle{NMX}=\frac{\frac{r}{v}+\frac{s}{u+v-sv/r}}{1-\frac{r}{v}\frac{s}{u+v-sv/r}}=\frac{r(u+v)-sv+sv}{v(u+v-sv/r)-rs}
=\frac{r^2(u+v)}{v(r(u+v)-sv)-r^2s}
&=\frac{uvw(u+v)}{v(u+v+w)(r(u+v)-sv)-uvws}\\
&=\frac{uw(u+v)}{r(u+v)(u+v+w)-s(v+u)(v+w)}\\
&=\frac{uw}{r(u+v+w)-s(v+w)}.
\end{align*}Also,
\[BM=BD_2+MD_2=\frac{sv}{r}+\frac{w}{2}.\]Now consider the coordinate plane with $x$-axis $BM$ and origin $B=0$. The line $MPQN$ is
\[y=\frac{uw}{r(u+v+w)-s(v+w)}\left(x-\frac{sv}{r}-\frac{w}{2}\right).\]But it's easy to check that the point
\[\left(\frac{w}{2}+\frac{v(u+v+w)}{v+w},\frac{uvw}{r(v+w)}\right)\]always lies on this line, so we're done.
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GoldenFrog1618
667 posts
#4 • 3 Y
Y by Adventure10, Mango247, and 1 other user
Diagram

Let the incircle of $ABX$ be tangent to $BX$ at $B_1$ and $AX$ and $B_2$. Also, let the incircle of $ACX$ touch $CX$ at $C_1$ and $AX$ at $C_2$. Let $M_1$ be the midpoint of $B_1$ and $C_1$, and $M_2$ be the midpoint of $B_2$ and $C_2$. Since $PQ$ is a radical axis, it passes through $M_1$ and $M_2$. Thus, it is sufficient to prove that, as $X$ varies, there is a constant point which $M_1M_2$ always passes through.

Let $D$, $E$, and $F$ be the midpoints of $AB$, $AC$, and $AX$ respectively. Since $B$, $C$, and $X$ are collinear, the points $D$, $E$, and $F$ are collinear also. I will prove that $K=DE\cap M_1M_2$ does not change as $X$ changes, this will show that all $M_1M_2$ pass through a common point. It is sufficient to prove that $DK$ is a constant.

Since $B_1,B_2, C_1,$ and $C_2$ are points of tangency of incircles, $B_1X=B_2X=\frac{BX+AX-AB}{2}$ and $C_1X=C_2X=\frac{CX+AX-AC}{2}$. Thus, \[M_1X=M_2X=\frac{BX+CX+2AX-AC-AB}{4}\]Since $DF\|BX$, $M_2FK\sim M_2XM_1$. Since $M_1X=M_2X$, $M_2F=KF$. We can now compute the length of $DK$ (notice the signed lengths):
\begin{align*}
DK&=DF-KF\\
    &=\frac{BX}{2}-M_2F\\
    &=\frac{BX}{2}-M_2X+FX\\
    &=\frac{BX}{2}-\frac{BX+CX+2AX-AC-AB}{4}+\frac{AX}{2}\\
    &=\frac{BX-CX+AC+AB}{4}\\
    &=\frac{BC+AC+AB}{4}
\end{align*}
This is independent of $X$, so all $M_1M_2$ (and so $PQ$) intersect at a common point, as desired.
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skytin
418 posts
#5 • 2 Y
Y by Adventure10, Mango247
another solution :
Let , U is exenter of triangle ABC opposite to A , I is incenter of ACX , incircle of CXA is tangent to XB at point T , incircle of ABX is tangent to BX at point K , incircle of BCA is tangent to BC at point L
Easy to see that KT = LC
A - Excircle of ABC is tangent to XB at point H
Let line thru L' = midpont of LC and || to CI intersect line thru H' midpoint of BH and perpendicular to BU at point D , let line thru point D and perpendicular to XI intersect XB at point N
Easy to see that angle NDL' = IAC , angle L'DH' = CAU , angle DH'N = AUI , so H'L'ND ~ UCIA , UC/CI = H'L'/L'N = HC/CT = H'L'/CT , L'N = CT , so N is midpoint of KT , line ND = line PQ , D is fixed . done
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aZpElr68Cb51U51qy9OM
1600 posts
#6 • 4 Y
Y by proglote, Lcz, Adventure10, Mango247
Let $\omega_1$ and $\omega_2$ be the incircles of $\triangle ABX$ and $\triangle ACX$, respectively. Denote by $(P, \omega)$ the power of $P$ with respect to circle $\omega$. Define a function $f: \mathbb{R}^2 \to \mathbb{R}$ by\[f(P) = (P, \omega_1) - (P, \omega_2).\]This function is linear. We now use barycentrics with respect to $\triangle ABC$. Let $R = (x:y:z)$ be the constant point that lies on $PQ$. Since $R$ lies on the radical axis of $\omega_1$ and $\omega_2$, we have $f(R) = 0$. Let $BC = a$, $CA = b$, $AB = c$, $AX = p$, and $CX = q$. We claim that the point $R = (2a: a-b-c: a+b+c)$ works, which is independent of the position of $X$. We can compute
\begin{align*}
f(A) &= \left(\frac{a+c+p+q}{2}-a-q\right)^2 - \left(\frac{b+p+q}{2}-q\right)^2 \\&= \frac{1}{4}(-a+b+c+2p-2q)(-a-b+c), \\ f(B) &= \left(\frac{a+c+p+q}{2}-p\right)^2 - \left(\frac{b+p+q}{2}-p+a\right)^2 \\&= \frac{1}{4}(3a+b+c-2p+2q)(-a-b+c), \\ f(C) &= \left(\frac{a+c+p+q}{2}-c-q\right)^2 - \left(\frac{b+p+q}{2}-p\right)^2 \\&= \frac{1}{4}(a+b-c)(a-b-c+2p-2q).
\end{align*}
By linearity, we have $f(R) = xf(A)+yf(B)+zf(C)$. But now it's straightforward to check that indeed $2af(A)+(a-b-c)f(B)+(a+b+c)f(C) = 0$, implying that $R$ always lies on the radical axis of $\omega_1$ and $\omega_2$, as desired.

Remark
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JuanOrtiz
366 posts
#7 • 2 Y
Y by Adventure10, Mango247
Short Solution:

Take $B'$ and $C'$ fixed points that lie on $MN$ (line connecting midpoints of $AB$ and $AC$) such that $AB'B = AC'C = 90$. Let $K$ be the midpoint of $B'C'$. I claim $K$ is the desired point. First I prove a lemma

LEMMA In triangle $XYZ$, let $X'$ and $Z'$ be the points of tangency of the incircle (centered at $I$) with $YX$ and $YZ$ respectively and let $W$ be the intersection of the bisector of angle $Z$ and $X'Z'$. Then, if $M$ and $N$ are the midpoints of $XY$ and $XZ$, we have $MNW$ are collinear and $XWZ=90$.
Proof: Angle chasing

Therefore, $B'$ lies on the line of the points if tangency of the incircle of $ABC$ with $AX$ and $BC$, and analogously $C'$. From this we see clearly $K$ lies on the radical axis, and we're done.
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pi37
2079 posts
#8 • 2 Y
Y by Adventure10, Mango247
Let $\omega$ be the circle centered at the midpoint of the two incircles with radius the average of the other two. Let $Y$ be the point on $BC$ such that $\omega$ is the incircle of $AYX$. Note that
\[
AY+AX-YX=\frac{AX+AC-CX}{2}+\frac{AX+AB-BX}{2}
\]
so
\[
2AY+BY-CY=AB+AC
\]
There are finitely many points $Y$ on $BC$ satisfying this, and $Y$ as a function of $X$ is continuous, so $Y$ must be the same for all $X$. Note that $PQ$ meets $XY,XA$ at their tangency points with $\omega$, so it suffices to show that for all $X$ varying on ray $BY$ past $Y$, there exists a constant point on all $X$-touch chords of $AYX$. But by a well known lemma, the intersection of the bisector of $AYX$ and the circle with diameter $AY$ suffices.
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anantmudgal09
1980 posts
#9 • 3 Y
Y by Davsch, Adventure10, Mango247
Let the $A$-midline of $\triangle ABC$ intersect the internal bisector of $\angle B$ at $V$ and the external bisector of $\angle C$ at $W$. Let $L$ be the midpoint of $\overline{VW}.$ We will show that $L \in \overline{PQ}$ to prove the result.

Note that $\overline{PQ}$ is the mid-parallel of the $X$-touch chord of the incircles of $\triangle ABX$ and $\triangle ACX$. Point $V, W$ lie on the $X$ touch chord of $\triangle ABX$ and $\triangle ACX$, respectively (from the "right angles on the intouch chord" lemma). Evidently, $L$ being the midpoint of $\overline{VW}$ lies on $\overline{PQ}$ as desired.
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nikolapavlovic
1246 posts
#10 • 3 Y
Y by neyft, Adventure10, Mango247
Lemma:In $\triangle ABC$ the $B$-bisector $A$-midline and the $C$-touch chord of the incircle are concurrent
Proof:
Take the polar dual and it suffices to show that the line connecting $C$ and orthocenter of $\triangle CIB$ is perpendicular to $AI$ which is immediate.

Let the incircles of $\triangle ABX,\triangle $ touch AX,BC in $E,F,E_1,F_1$.Now let $EF,E_1F_1\cap \text{A-midline}=\{R,S\}$.By lemma $R$ lies on the B-angle bisector and
$S$ on C- outer angle bisector and hence they're fixed.$PQ||EF||E_1F_1$ and $PQ$ bisects $EE_1$ and hence it bisects $RS$ $\implies$ it passes thru the midpoint of $RS$ which is fixed.$\blacksquare$
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MarkBcc168
1595 posts
#11 • 3 Y
Y by Tawan, Adventure10, Mango247
Really nice problem!

Let the incircle of $\Delta ABX, \Delta ACX$ centered at $I_1, I_2$ and touches $BC$ at $E, F$ resp. Let $I_B$ be the center of $B$-excircle of $\Delta ABC$, which touches $BC$ at $T$. Let $K, M$ be the midpoint of $AT, EF$ and finally let $R$ be the point on $BC$ such that $AR\perp I_1I_2$.

Note that $M$ lies on radical axis of those incircles, which is $PQ$. Now we claim that $K$ also lies on $PQ$, which is the fixed point. It suffices to show that $MK\perp I_1I_2$, which is parallel to $AR$. Or equivalently, $M$ is the midpoint of $RT$.

To prove this, note by symmetry that $XR=XA$. So by side-chasing,
\begin{align*} ET &= CT - CF \\
&= \frac{AB + AC - BC}{2} - \frac{AC + CX - AX}{2} \\
&= \frac{AB + AX - BC - CX}{2} \\
&= \frac{AB + AX - BX}{2} \\
\end{align*}and
\begin{align*}FR &= XR - EX \\
&= XA - \frac{XA + XB - AB}{2} \\
&= \frac{XA + AB - XB}{2} \\
\end{align*}so $ET=FR$ which implies $M$ is midpoint of $RT$ and we are done.

Note : The most effective way to claim the fixed point is to plug $X=C$ and $X={\infty}_{BC}$ and intersect the radical axii.
This post has been edited 1 time. Last edited by MarkBcc168, Apr 21, 2018, 1:12 PM
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Th3Numb3rThr33
1247 posts
#12 • 4 Y
Y by eisirrational, rjiangbz, Adventure10, Mango247
Nice algebra problem! Solved with eisirrational.

We use barycentric coordinates on $\triangle ABC$. Let $CX = d$ and $AX = x$. Moreover, let the incircles of $ABX$ and $ACX$ be tangent to line $AX$ at $M_1$, $N_1$, and to line $BX$ at $M_2$, $N_2$, respectively.

I claim that the point $R = \left( \frac{1}{2}, \frac{a-b-c}{4a}, \frac{a+b+c}{4a} \right)$ is the desired point. It suffices to show that $R$ lies on the radical axis of the two incircles, i.e. $R$, the midpoint of $\overline{M_1N_1}$, and the midpoint of $\overline{M_2N_2}$ are collinear. Call these midpoints $S_1$ and $S_2$.

We first compute some lengths. By equal tangents, we have that $N_1X = N_2X = \frac{x+d-b}{2}$ and $M_1X = M_2X = \frac{a+d+x-c}{2}$, so $S_1X = S_2X = \frac{2x+2d+a-b-c}{4}$. The coordinates of $S_2$ is readily calculated to be (with directed ratios)
$$(0:CS_2:BS_2) = (0:CX-XS_2:BX-XS_2) = \left(0,\frac{a-b-c+2x-2d}{4a},\frac{3a+b+c-2x+2d}{4a} \right).$$The coordinates of $S_1$ are a bit harder, but still not too bad. Note the coordinates of $X$ are $(0:CX:BX) = \left( 0, -\frac{d}{a}, \frac{a+d}{a} \right)$, so
$$S_1 = XS_1 \cdot (1,0,0) + AS_1 \cdot \left( 0, -\frac{d}{a}, \frac{a+d}{a} \right) = \left( \frac{2x+2d+a-b-c}{4x}, \frac{-2xd+2d^2+ad-bd-cd}{4ax}, \frac{2ax+2dx-3ad-d^2-a^2+ab+bd+ac+cd}{4ax}\right).$$To show that $X,S_1,S_2$ are collinear, we simply show that the displacement vectors $\overrightarrow{XS_1}$ and $\overrightarrow{XS_2}$ are proportional. In fact, because all of these coordinates are homogenized, we only to verify the proportion for the first two coordinates. Looking at the first coordinates, we receive a ratio of
$$\frac{\frac{1}{2} - \frac{2x+2d+a-b-c}{4x}}{\frac{1}{2}} = \frac{-2d-a+b+c}{2x}.$$But for the second coordinates we also have
$$\frac{\frac{a-b-c}{4a} - \frac{2xd+2d^2+ad-bd-cd}{4ax}}{\frac{a-b-c}{4a} - \frac{a-b-c+2x-2d}{4a}} = \frac{(d-x)(2d+a-b-c)/4ax}{(2x-2d)/4a} = \frac{-2d-a+b+c}{2x}.$$We thus have the desired.
This post has been edited 1 time. Last edited by Th3Numb3rThr33, Nov 3, 2019, 9:54 PM
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yayups
1614 posts
#13 • 1 Y
Y by Adventure10
//cdn.artofproblemsolving.com/images/2/5/4/254309e73ff0023e01fbffb09a87f56abae26ac2.jpg
Let $I_1$ be the incenter of $ABX$ and let $I_2$ be the incenter of $ACX$. Let $D$ and $F$ be the contact points of the incircle of $ABX$ with $BX$ and $AX$ respectively, and let $E$ and $G$ be the contact of the incircle of $ACX$ with $CX$ and $AX$ respectively.

Let $P$ and $Q$ be the intersections of $DF$ and $EG$ with the $A$-midline respectively. The key claim is that $P$ and $Q$ are fixed as $X$ varies. This follows from the so called Iran lemma, as $Q$ is the intersection of the internal $C$-angle bisector with the midline and $Q$ is the intersection of the external $B$-angle bisector with the midline.

Note that the midpoints of $DE$ and $FG$ are on the radical axis of the two circles, so the intersection of the radical axis with $PQ$ is simply the midpoint of $PQ$, which is a fixed point, as desired.
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KST2003
173 posts
#14
Y by
Let the incircle of $\triangle ABX$ touch $XA$ and $XB$ at $S$ and $T$, and let the incircle of $\triangle ACX$ touch $XA$ and $XC$ at $U$ and $V$. Let the incenters of $\triangle ABX$ and $\triangle ACX$ be $I$ and $J$ respectively, and let $Y$ and $Z$ be the foots of perpendicular from $A$ to $\overline{BI}$ and $\overline{CJ}$. Let $M$ be the midpoint of segment $YZ$. We claim that $M$ is the desired fixed point. By the Iran lemma, it follows that $Y$ and $Z$ lie on $\overline{UV}$ and $\overline{ST}$ respectively. It is also easy to see that $UV\parallel ST$ since both of them are perpendicular to $\overline{IX}$. Moreover, the radical axis passes through the midpoints of segments $US$ and $VT$, so it is actually the midline of $\overline{UV}$ and $\overline{ST}$, and this passes through $M$ as desired.
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nixon0630
31 posts
#15 • 1 Y
Y by Tafi_ak
First ISL G7.
Here's solution using famous Iran Lemma:
Let $I_{1}$ and $I_{2}$ be the incenters of $\triangle{ABX}$ and $\triangle{ACX}$, respectively. Denote by $T_{1}$, $T_{2}$ touchpoints of incircle of $\triangle{ABX}$ with $\overline{BX}$ and $\overline{AX}$, respectively. Similarly, let incircle of $\triangle{ACX}$ touches $\overline{CX}$ and $\overline{AX}$ at $T_{3}$ and $T_{4}$, respectively. Let $M$ and $N$ be the midpoints of $\overline{AB}$ and $\overline{AX}$, respectively.
By Iran Lemma we have $T = T_{1}T_{2} \cup BI_{1} \cup MN$ and $S = T_{3}T_{4} \cup CI_{2} \cup MN$ are fixed, since $MN$, $BI_{1}$, and $CI_{2}$ are fixed.
Simple angle chasing gives us $T_2T_1T_3T_4$ is cyclic. From angle chasing we also get, that $T_1T_2 \parallel T_3T_4$. Hence, $TT_1T_3S$ is parallelogram. It's well-known, that $PQ$ bisects $\overline{T_1T_3}$; since $TT_1T_3S$ is parallelogram , it's also pass through the midpoint of $\overline{TS}$, which is fixed. $\blacksquare$
This post has been edited 5 times. Last edited by nixon0630, Oct 28, 2021, 9:01 AM
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HoRI_DA_GRe8
599 posts
#17
Y by
Solution(with geogebra)
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JAnatolGT_00
559 posts
#18
Y by
Let incircles of $ABX,ACX$ meet $AX$ again at $D_1,D_2$ and $BX$ at $E_1,E_2$ respectively. Let also line homothetic to $BC$ wrt $A$ with coefficient $\frac{1}{2}$ intersect internal bisector of and $ABC$ and external bisector of $ACB$ at $X,Y$ respectively, so by Iran lemma $X\in D_1E_1,Y\in D_2E_2.$ But obviously $D_1XD_2Y$ is an isosceles trapezoid, therefore $PQ$ passes through midpoint of $XY,$ which is fixed.
This post has been edited 1 time. Last edited by JAnatolGT_00, Jul 20, 2022, 7:37 PM
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Jalil_Huseynov
439 posts
#19 • 1 Y
Y by Mango247
Let $M,N$ be midpoints of $AB,AC$. Incircle of triangle $ABX$, which has center $I$ touches $XA,XB$ at $K,L$. Let incircle of triangle $ACX$ touches $AX,CX$ at $J,G$. Let $D=PQ\cap MN, E=PQ\cap BC, F=BI\cap MN$ and $R=PQ\cap AX$. From Iran Lemma $F\in KL$. Since $RK=RJ$ and $El=EG$, we get $KL||RE \implies FDEL$ is parallelogram
$\implies FD=EL=\frac{LG}{2}=\frac{XL-XG}{2}=\frac{(XA+XB-AB)-(XC+XA-AC)}{4}=\frac{BC+AC-AB}{4}$.
Also $MF=MB=\frac{AB}{2} \implies MD=MF+FD=\frac{(BC+AC-AB)+2AB}{4}=\frac{AB+BC+CA}{4}$, which is constant when $ABC$ is fixed. So $D$ lies on fixed line $MN$ and length of $MD$ is constant, which implies $D$ is fixed point as $X$ varies. So we are done.
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awesomeming327.
1732 posts
#20
Y by
Let the incircle of $ABX$ intersect $BX$ at $D$ and $AX$ at $E$, and let $M$, $N$ be the midpoints of $AB$ and $AX$ respectively. Let $I$ be the incenter, then we claim that $BI$, $MN$, $DE$ are concurrent.
https://media.discordapp.net/attachments/1091890356796280852/1095004634659172352/Screenshot_2023-04-10_at_11.09.13.png?width=1692&height=1112
Let $S$ be the intersection point of $DE$ and $MN$. Let $F$ be on $DE$ such that $AF\parallel BX$. Let $G$ be $AS$ intersection with $BX$. Let $R$ be the contact point on $AB$. Clearly, $AFGD$ is a parallelogram with center $S$. We have \[\angle AEF = \angle DEX = \angle EDX = \angle AFE\]so $AE=AF$. Thus, $DG=AF=AE=AR$. Also, $BD=BR$ so $AB=AG$. Since $AS=SG$, $\triangle ABS\cong \triangle GBS$ so $S$ is on $AI$ as desired.
https://media.discordapp.net/attachments/1091890356796280852/1095004634894057572/Screenshot_2023-04-10_at_11.12.41.png?width=1898&height=1048
Now, similarly, if $J$, $K$ are the points of contact of the incircle of $ACX$ on $CX$ and $AX$, respectively then $JK$ passes through $T$, a fixed point on the midline and the $C$-angle bisector. Let $U$ be the midpoint of $ST$, also a fixed point with respect to $ABC$.

Since $XI\perp DE$ and $XI\perp JK$, $DE\parallel JK$. Now, since $PQ$ is the radical axes of the two circles, it bisects $EK$ and $DJ$. Thus, $PQ$ is the line right in the middle of $DE$ and $JK$, so it passes through $U$, as desired.
This post has been edited 1 time. Last edited by awesomeming327., Apr 10, 2023, 3:18 PM
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huashiliao2020
1292 posts
#21
Y by
nice problem! glad i solved it in ~45 minutes? this is the last problem im doing on configgeo im done

Define the points as in the diagram, with all the lines given by Iran Lemma; note that $YJ\perp CD\perp VW\implies YJ\parallel VW\parallel PQ$. On the other hand, since PQ is radax which bisects the common tangent RK, PQ is the midline of JYWV, which passes through the midpoint of YV. We conclude. :surf:

also note that my diagram is extremely overkill i didnt need ALL of those info but its helpful to list out when you dont know what to do
Attachments:
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cursed_tangent1434
639 posts
#22 • 1 Y
Y by Shreyasharma
Consider an arbitrary point $X$ on ray $BC$. Let $I$ be the incenter of $\triangle ABC$ with $\triangle DEF$ being the intouch triangle of $\triangle ABC$. Further, let $M_B$ and $M_C$ be the midpoints of $AC$ and $AB$ respectively. Let $M_{CD}$ and $M_{CE}$ be the midpoints of $CD$ and $CE$. We claim that all lines $PQ$ pass through $Z=\overline{M_{CD}M_{CE}} \cap \overline{M_CB_C}$.

Let $\omega$ be the incircle of $\triangle ABC$ and $\omega_B$ and $\omega_C$ the incircles of $\triangle ABX$ and $\triangle ACX$ respectively. Let $O_B$ and $O_C$ be the centers of $\omega_B$ and $\omega_C$ respectively. Now, let $G$ and $H$ be the tangency points of the incircle of $\triangle ABX$ with sides $AX$ and $BX$. Let $M$ be the midpoint of $AX$.

Clearly, $B-I-O_B$ and $C-O_C-O_B$ due to the common tangents. Further, $M_C-M_B-M$. By Iran Lemma on $\triangle ABC$, we see that $\overline{BI},\overline{M_BM_C}$ and $\overline{DE}$ concur (say at $N$). Now, clearly $N$ also lies on $BI$ and $M_CM_B$ due to the above collinearities.Further notice that by Iran Lemma on $\triangle ABX$, we must have that lines $\overline{GH},\overline{M_CM}$ and $\overline{BO_B}$ concur. But clearly the latter two of these lines intersect at $N$. Thus, $GH$ also must pass through $N$.

Now, simply notice that $\overline{ED}\parallel\overline{M_{CD}M_{CE}}$ by Midpoint Theorem and $\overline{GH} \parallel \overline{PQ}$ since both these lines are perpendicular to $XO_C$. Thus, the intersection of $EF$ and $GH$ and the intersection of $M_{CD}M_{CE}$ and $PQ$ must form two triangles which are similar. Now, we prove the following. Let $D',E'$ be the tangency points of $\omega_C$ with $AX$ and $BX$. Then,

Claim : $GD'=CD$.

Proof :
Note that $XD'$ and $XG$ are tangents to $\omega_B$ and $\omega_C$ respectively. Then, let $s_1$ and $s_2$ denote that semiperimeters of $\triangle ABX$ and $\triangle ACX$ and $s$ denote the semiperimeter of $\triangle ABC$. We have,
\begin{align*}
        GD' &= XG-XD'\\
        &= s_1-AB - (s_2-AC)\\
        &= s_1-s_2 + AC - AB\\
        &= \frac{AX+XB + AB - AX - CX - AC}{2} + AC - AB\\
        &= \frac{BC + AB - AC}{2} + AC - AB\\
        &= \frac{AB+AC+BC}{2}-AB\\
        &= s-AB\\
        &= CD
    \end{align*}Thus, indeed $GD'=CD$ as claimed.

Now, let $R = \overline{PQ} \cap \overline{BX}$. It is well known that the radical axis bisects the common tangent. Thus, $RD'=\frac{GD'}{2}=\frac{CD}{2}=DM_{CD}$
Thus, the previously described similar triangles must infact also be congurent. This means, that the intersections of $EF$ and $GH$ and of $M_{CD}M_{CE}$ and $PQ$ must lie on a line parallel to $BC$. But clearly the former intersection point is $N$ which lies on $\overline{M_CM_B}$ and thus $Z'=  M_{CD}M_{CE} \cap PQ $ must also lie on $\overline{M_CM_B}$. This means, $Z'=Z$.

Thus, all lines $PQ$ must pass through a common point, this point being $Z=\overline{M_{CD}M_{CE}} \cap \overline{M_CB_C}$.
This post has been edited 1 time. Last edited by cursed_tangent1434, Oct 21, 2023, 6:41 AM
Reason: wrong sol
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shendrew7
799 posts
#23
Y by
Define line $\ell$ as the $A$-midline and the touch points of the incircles of $\triangle ABX$ and $\triangle ACX$ to $AX$, $BX$ to be $J_1$, $J_2$ and $K_1$, $K_2$. If we suppose $J = J_1J_2 \cap \ell$ and $K = K_1K_2 \cap \ell$, we know
  • $J$ and $K$ are fixed as they lie on the $B$-angle bisector and $C$-external angle bisector, respectively, through Iran Lemma.
  • $PQ$ bisects both $J_1K_1$ and $J_2K_2$ by power of a point, and $J_1J_2 \parallel PQ \parallel K_1K_2$.

Thus $PQ$ passes through the midpoint of $JK$, which is fixed. $\blacksquare$
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Leo.Euler
577 posts
#24
Y by
I'm surprised no one found this clean linpop solution!

Let $\omega_1$ and $\omega_2$ denote the incircles of $\triangle ABX$ and $\triangle ACX$ respectively. Define $f(\bullet) = \text{Pow}(\bullet, \omega_1) - \text{Pow}(\bullet, \omega_2)$. We claim that the intersection of the $A$-midline and the radical axis of $\omega_1$ and $\omega_2$ is fixed. In order to prove this, it suffices to show that the powers of $f((A+B)/2)$ and $f((A+C)/2)$ are independent of $X$, because it follows that the point on the $A$-midline that $f$ vanishes on is independent of $X$. This can be done by bashing.

Here's an example of what the bash would look like; consider the example of proving $f((A+B)/2)=(f(A)+f(B))/2$ constant (proving $f((A+C)/2)$ is analogous). Let all lengths be signed. Then using the intouch points, we calculate \[ f(A) = \frac{1}{4} \left[ (AB+AX-BX)^2 - (AC+AX-CX)^2\right] \]and \[ f(B) = \frac{1}{4} \left[ (AB+BX-AX)^2 - (2BC+AC+CX-AX)^2\right]. \]Now we can bash out $f(A)+f(B)$, and utilizing the fact that $BC+CX=BX$, we compute this quantity to be independent of $X$, as desired.
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HamstPan38825
8868 posts
#25
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This problem is so cute!

I claim that the fixed point $K$ lies on the $A$-midline $\overline{MN}$. Let $E$ and $F$ are the tangency points of the incircle of $ABX$ to $\overline{BX}$ and $\overline{AX}$, respectively, and define $G$ and $H$ similarly. Let $I_1$ and $I_2$ be the incenters of triangles $ABX$ and $ACX$, and recall that $R = \overline{BI_1} \cap \overline{MN}$ lies on the tangent chord $\overline{EF}$. Similarly, $S = \overline{CI_1} \cap \overline{MN}$ lies on the tangent chord $\overline{GH}$. Note that $R$ and $S$ are fixed points.

Now, by radical axis, $\overline{PQ}$ is the midline of trapezoid $EGHF$, i.e. $K = \overline{PQ} \cap \overline{MN}$ is the midpoint of $\overline{RS}$. It follows that $K$ is the desired fixed point.
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Saucepan_man02
1356 posts
#26
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Nice Problem: :D

Let $X, Y$ denote the intersection of $A$-midline at the angle bisectors of $B$ (internally) and $C$ (externally) respectively.
Let $Z$ denote the midpoint of $XY$. We claim that $Z \in PQ$.
Note that $PQ$ bisects the common chords of both the incircle.
Let the incircle of $\triangle ABX$ touch $BX, AX$ at $U, V$ respectively, and the incircle of $\triangle ACX$ touch $CX, AX$ at $S, T$ respectively. Thus, it suffice to show $X \in (UV), Y \in (ST)$ which is immediate due to Iran's Lemma.
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InterLoop
279 posts
#27 • 1 Y
Y by ihategeo_1969
solved with ihategeo_1969 and Agrivulture
solution

remarks
This post has been edited 1 time. Last edited by InterLoop, Mar 7, 2025, 11:38 PM
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Ilikeminecraft
658 posts
#28
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Let $U, V$ denote the intersections of the $A$-midline with the $B$-internal angle bisector and $C$ external angle bisector. Let $D, F, E, G$ denote the tangencies of $BX, AX$ with incircle of $ABX,$ incircle of $AX,$ respectively.
By homothety, we have $DF\parallel EG.$ By radax, $PQ$ passes throuhg the midpoint of $FG, DE,$ implying that $PQ\parallel DF\parallel EG.$ Finally, note that $UV\parallel DE,$ implying that $PQ$ passes through the midpoint of $UV.$ Since $UV$ is fixed, this point is also fixed.
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Markas
150 posts
#29
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Let the $\triangle ABX$ incircle be $\omega_1$ and the $\triangle ACX$ incircle be $\omega_2$. Let M be the midpoint of AB, N be the midpoint of AX, D - the contact point of $\omega_1$ with BX, E - the contact point of $\omega_1$ with AX, F - the contact point of $\omega_2$ with CX, G - the contact point of $\omega_2$ with AX. Now from the Iran lemma we have that $BI_1 \cap DE \cap MN = Y$ and $CI_2 \cap FG \cap MN = Z$. We know that $BI_1$, $CI_2$, DE, FG, MN are fixed $\Rightarrow$ Y and Z are fixed too and YZ is fixed. Let L be the midpoint of DF and J be the midpoint of EG. By radical axis properties it follows that L and J lie on the $rad(\omega_1,\omega_2)$ $\Rightarrow$ $L \in PQ$ and $J \in PQ$. We have that $YZ \parallel DF$, since $MN \parallel BX$ and from angle chase we get $YD \parallel ZF$ $\Rightarrow$ YDFZ is a parallelogram. Now since $PQ \cap DF = L$ and L is the midpoint of DF, by the parallelogram, if $PQ \cap YZ = K$, then K should be the midpoint of YZ $\Rightarrow$ $K \in PQ$, where K is midpoint of YZ and since YZ is fixed, K is fixed too $\Rightarrow$ PQ passes trough the fixed point K - midpoint of YZ, independant of X $\Rightarrow$ we are ready.
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