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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
solve in positive integers: 3 \cdot 2^x +4 =n^2
parmenides51   4
N 20 minutes 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
20 minutes ago
A strong inequality problem
hn111009   3
N 20 minutes ago by ZeltaQN2008
Source: Somewhere
Let $a,b,c$ be the positive number satisfied $a^2+b^2+c^2=3.$ Find the minimum of $$P=\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}+\dfrac{3abc}{2(ab+bc+ca)}.$$
3 replies
hn111009
Yesterday at 2:02 AM
ZeltaQN2008
20 minutes ago
ISI UGB 2025 P7
SomeonecoolLovesMaths   10
N 22 minutes 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
SomeonecoolLovesMaths
Yesterday at 11:28 AM
Mathworld314
22 minutes ago
2018 JBMO TST- Macedonia, problem 4
Lukaluce   3
N 24 minutes ago by Erto2011_
Source: 2018 JBMO TST- Macedonia
Determine all pairs $(p, q)$, $p, q \in \mathbb {N}$, such that

$(p + 1)^{p - 1} + (p - 1)^{p + 1} = q^q$.
3 replies
Lukaluce
May 28, 2019
Erto2011_
24 minutes ago
Trigo relation in a right angled. ISIBS2011P10
Sayan   12
N 31 minutes 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
31 minutes ago
Six variables
Nguyenhuyen_AG   4
N 42 minutes 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
42 minutes ago
Expressing polynomial as product of two polynomials
Sadigly   3
N an hour 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
an hour ago
Help me this problem. Thank you
illybest   0
an hour ago
Find f: R->R such that
f( xy + f(z) ) = (( xf(y) + yf(x) )/2) + z
0 replies
illybest
an hour ago
0 replies
Product of consecutive terms divisible by a prime number
BR1F1SZ   2
N an hour 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
an hour ago
Pentagon with given diameter, ratio desired
bin_sherlo   2
N an hour 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
+1 w
bin_sherlo
Yesterday at 7:21 PM
Umudlu
an hour ago
ISI UGB 2025 P4
SomeonecoolLovesMaths   7
N 2 hours ago by SomeonecoolLovesMaths
Source: ISI UGB 2025 P4
Let $S^1 = \{ z \in \mathbb{C} \mid |z| =1 \}$ be the unit circle in the complex plane. Let $f \colon S^1 \longrightarrow S^2$ be the map given by $f(z) = z^2$. We define $f^{(1)} \colon = f$ and $f^{(k+1)} \colon = f \circ f^{(k)}$ for $k \geq 1$. The smallest positive integer $n$ such that $f^{(n)}(z) = z$ is called the period of $z$. Determine the total number of points in $S^1$ of period $2025$.
(Hint : $2025 = 3^4 \times 5^2$)
7 replies
1 viewing
SomeonecoolLovesMaths
Yesterday at 11:24 AM
SomeonecoolLovesMaths
2 hours ago
Hard geometry
Lukariman   0
2 hours ago
Given triangle ABC, any line d intersects AB at D, intersects AC at E, intersects BC at F. Let O1,O2,O3 be the centers of the circles circumscribing triangles ADE, BDF, CFE. Prove that the orthocenter of triangle O1O2O3 lies on line d.
0 replies
Lukariman
2 hours ago
0 replies
Shortest number theory you might've seen in your life
AlperenINAN   6
N 2 hours ago by AlperenINAN
Source: Turkey JBMO TST 2025 P4
Let $p$ and $q$ be prime numbers. Prove that if $pq(p+1)(q+1)+1$ is a perfect square, then $pq + 1$ is also a perfect square.
6 replies
+1 w
AlperenINAN
Yesterday at 7:51 PM
AlperenINAN
2 hours ago
Another config geo with concurrent lines
a_507_bc   16
N 2 hours ago by Tkn
Source: BMO SL 2023 G5
Let $ABC$ be a triangle with circumcenter $O$. Point $X$ is the intersection of the parallel line from $O$ to $AB$ with the perpendicular line to $AC$ from $C$. Let $Y$ be the point where the external bisector of $\angle BXC$ intersects with $AC$. Let $K$ be the projection of $X$ onto $BY$. Prove that the lines $AK, XO, BC$ have a common point.
16 replies
a_507_bc
May 3, 2024
Tkn
2 hours ago
Common chord bisects segment
mofumofu   11
N Apr 19, 2025 by sttsmet
Source: China TSTST 3 Day 1 Q2
Let $ABCD$ be a non-cyclic convex quadrilateral. The feet of perpendiculars from $A$ to $BC,BD,CD$ are $P,Q,R$ respectively, where $P,Q$ lie on segments $BC,BD$ and $R$ lies on $CD$ extended. The feet of perpendiculars from $D$ to $AC,BC,AB$ are $X,Y,Z$ respectively, where $X,Y$ lie on segments $AC,BC$ and $Z$ lies on $BA$ extended. Let the orthocenter of $\triangle ABD$ be $H$. Prove that the common chord of circumcircles of $\triangle PQR$ and $\triangle XYZ$ bisects $BH$.
11 replies
mofumofu
Mar 18, 2017
sttsmet
Apr 19, 2025
Common chord bisects segment
G H J
Source: China TSTST 3 Day 1 Q2
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mofumofu
179 posts
#1 • 10 Y
Y by rkm0959, anantmudgal09, baopbc, ThE-dArK-lOrD, guangzhou-2015, buratinogigle, nguyendangkhoa17112003, Ya_pank, Adventure10, Rounak_iitr
Let $ABCD$ be a non-cyclic convex quadrilateral. The feet of perpendiculars from $A$ to $BC,BD,CD$ are $P,Q,R$ respectively, where $P,Q$ lie on segments $BC,BD$ and $R$ lies on $CD$ extended. The feet of perpendiculars from $D$ to $AC,BC,AB$ are $X,Y,Z$ respectively, where $X,Y$ lie on segments $AC,BC$ and $Z$ lies on $BA$ extended. Let the orthocenter of $\triangle ABD$ be $H$. Prove that the common chord of circumcircles of $\triangle PQR$ and $\triangle XYZ$ bisects $BH$.
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TelvCohl
2312 posts
#2 • 16 Y
Y by baopbc, Ankoganit, smy2012, dagezjm, Saro00, guangzhou-2015, zed1969, anantmudgal09, Mosquitall, enhanced, nguyendangkhoa17112003, Gaussian_cyber, AllanTian, JG666, Adventure10, thinkcow
Let $ \widetilde{A}, $ $ \widetilde{D} $ be the isogonal conjugate of $ A, $ $ D $ WRT $ \triangle BCD, $ $ \triangle ABC, $ respectively. Clearly, $ \widetilde{A} $ and $ \widetilde{D} $ are symmetry WRT $ BC, $ so one of the intersection $ U $ of $ \odot (PQR) $ and $ \odot (XYZ) $ lies on $ BC. $ Let $ V $ be the second intersection of these two circles, then note that $ ( A,B,P,Q ) $ and $ ( B,D,Y,Z ) $ are concyclic we get $$ \measuredangle ZVQ = \measuredangle ZYU + \measuredangle UPQ = \measuredangle ZDB + \measuredangle BAQ = 2\measuredangle ABD \ , $$hence $ V $ lies on the 9-point circle of $ \triangle ABD. $ Let $ M $ be the midpoint of $ BH. $ Since $ M, $ $ Q, $ $ V, $ $ Z $ are concyclic (the 9-point circle of $ \triangle ABD $), so we conclude that $$ \measuredangle MVQ = \measuredangle BAQ = \measuredangle UPQ = \measuredangle UVQ \Longrightarrow M \ \text{lies on} \ UV \ . $$
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bobthesmartypants
4337 posts
#3 • 3 Y
Y by anantmudgal09, Adventure10, Mango247
solution

EDIT: 4321st post woo
This post has been edited 1 time. Last edited by bobthesmartypants, Aug 31, 2017, 7:08 PM
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anantmudgal09
1980 posts
#5 • 2 Y
Y by Adventure10, Mango247
I think that this solution is a bit different from others posted here. Not that hard a problem with the knowledge of circumrectangular hyperbolas, especially with geogebra by your side :D
mofumofu wrote:
Let $ABCD$ be a non-cyclic convex quadrilateral. The feet of perpendiculars from $A$ to $BC,BD,CD$ are $P,Q,R$ respectively, where $P,Q$ lie on segments $BC,BD$ and $R$ lies on $CD$ extended. The feet of perpendiculars from $D$ to $AC,BC,AB$ are $X,Y,Z$ respectively, where $X,Y$ lie on segments $AC,BC$ and $Z$ lies on $BA$ extended. Let the orthocenter of $\triangle ABD$ be $H$. Prove that the common chord of circumcircles of $\triangle PQR$ and $\triangle XYZ$ bisects $BH$.

First we clear up our notation a bit. Consider the equivalent problem:

China TST 2017 #2, morally correct notation wrote:
Let $ABCP$ be a non-cyclic quadrilateral and $\mathcal{H}$ denote the rectangular hyperbola circumscribing them. Let $H_B$ be the orthocenter of triangle $ABP$. Let $\omega_1$ be the pedal circle of $P$ wrt $\triangle ABC$ and $\omega_2$ be the pedal circle of $A$ wrt $\triangle BCP$. Prove that the radical axis (or common chord) of $\omega_1, \omega_2$ passes through the mid-point of $\overline{BH_B}$.

For convenience, suppose $P$ lies in the interior of $\triangle ABC$. The general case can be accounted for by directing angles accordingly. Now let $Z$ be the center of $\mathcal{H}$ and $P_A, P_B, P_C$ be the projections of $P$ on $\overline{BC}, \overline{CA}, \overline{AB}$ respectively. Let $\odot(P_AP_BP_C)$ meet the sides $\overline{BC}, \overline{CA}, \overline{AB}$ again at $Q_A,Q_B,Q_C$ respectively.

Let $M$ be the midpoint of $\overline{BH_B}$ and $P_B'=\overline{AP} \cap \overline{BH_B}$. Then $H_B \in \mathcal{H} \implies Z \in \odot(P_BMP_B')$.

Claim. $\overline{ZQ_A}$ is the common chord of $\omega_1, \omega_2$ (in fact these two points lie on both the circles).

(Proof) Observe that $Z$ is a common point due to the fundamental theorem and $Q_A \in \omega_1$. So let $A'$ be the isogonal conjugate of $A$ wrt $\triangle BPC$; and $P'$ be the isogonal conjugate of $P$ wrt $\triangle ABC$. Then $A', P'$ are symmetric in $\overline{BC}$. Consequently, $Q_A$ lies on the pedal circle $\omega_2$ of $A'$ wrt $\triangle BPC$, as desired. $\blacksquare$

Finally, we see $$\angle P_BZQ_A=\angle P_BP_AB=90^{\circ}-\angle ABP$$and $$\angle P_BZM=\angle P_BP_B'B=90^{\circ}-\angle ABP$$hence $M$ lies on $\overline{ZQ_A}$, as desired. $\blacksquare$

P.S. 1400th post! :)
This post has been edited 1 time. Last edited by anantmudgal09, Oct 27, 2017, 4:42 PM
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nguyenhaan2209
111 posts
#6 • 3 Y
Y by top1csp2020, JG666, Adventure10
ZR-BC=U then XZU=XAR=XYC so UXYZ cyclic, similarly QXU collinear so RQU=RZX=RAC=RPC so UPQR cyclic hence ZVQ=ZVU-UVQ=ZYU-UPQ=180-ZDB-QB=ZMQ so MVQ+QVU=180-MZQ+QAB=180 so q.e.d
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mathaddiction
308 posts
#7
Y by
Notice that $Z,Q,X,R$ all lies on the circle with diameter $AD$. Denote this circle by $\omega$.Let $N$ be the midpoint of $AD$. Let $M$ be the midpoint of $BH$. Suppose $ZX$ and $QR$ meet at $G$, and $ZR$ and $QX$ meet at $I$.
CLAIM. Both $I$ and $G$ lies on the common chord
Proof. Applying radical axis theorem to $(PQR)$, $(XYZ)$ and $\omega$ we see that $G$ is the radical center of the three circles hence lying on the common chord. Now
$$\angle QIR=\angle ZRQ-\angle XQR=\angle BAQ-(90^{\circ}-\angle ACD)=\angle QPC-\angle RPC=\angle QPR$$hence $I$ lies on $(PQR)$. By symmetry it lies on $(XYZ)$ as well. This justfies our claim.

Now notice that $\angle MQB+\angle NQD=\angle MBQ+\angle NQD=90^{\circ}$. Therefore $MZ$ and $MQ$ are both tangent to $\omega$. Let $ZQ$ and $DX$ meet at $J$. Then $J$ lies on $ZQ$, the polar of $M$ w.r.t. $\omega$. Hence by La Hire's theorem, $M$ lies on the polar of $J$. w.r.t. $\omega$, that is, $GI$ by Brokard's theorem.
This shows that the common chord of the two circles bisect $BH$.
This post has been edited 2 times. Last edited by mathaddiction, Jul 18, 2020, 11:47 PM
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Plops
946 posts
#8
Y by
Pascal's on $\odot (AZRDXQ)$, $ZR \cap QX \cap PY=I'_2$. Let $U=AD \cap BC$, $V=ZQ \cap RX$, and assume $AB<CD$. By simple angle chasing, we see

$$\angle RI_2B=\angle RAD-\angle ZDA+\angle DUC=\angle ADC-\angle DAB+\angle DAB+\angle CBA-\pi=\angle ADC+\angle CBA-\pi=\pi-\angle AQD+\pi-\angle PQA-\pi=\pi-\angle AQD-\angle PQA=\pi-\angle PQR$$
so $\odot (I'_2PQR)$ is cyclic, and similarly, $\odot (I'_2YXZ)$ is cyclic. Let $M$ be the midpoint of $BH$. Then, $MH=MZ=MQ=MB$, and

$$\angle MZB= \angle MBZ=\frac{\pi}{2}-\angle ZAD=\angle ZDA$$
so $MZ, MQ$ are tangent to circle $\odot (AZRDXQ)$, and $M$ lies on the polar of $V$ w.r.t. $\odot (AZRDXQ)$, which, by Brokard's theorem, is $I_1I_2$, the radical axis of $\odot (XYZ)$ and $\odot (PQR)$.
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KST2003
173 posts
#10
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Let $K$ be the orthocenter of $\triangle ACD$, and let $M$ and $N$ be the midpoints of $BH$ and $CK$.

Claim: $\overline{HK}, \overline{BC}$ and $\overline{ZR}$ are concurrent one of the intersections of $(PQR)$ and $(XYZ)$, say $S$.

Proof. Let $U=\overline{KR}\cap\overline{HZ}$, and $V=\overline{CR}\cap\overline{BZ}$. Then as $D$ is the orthocenter of $\triangle AUV$, $UV\parallel CK\parallel BH$. Hence $\triangle KRC$ and $\triangle HZB$ are perspective and the concurrency follows. Now it is left to show that this point lies on both circles. Since $ZAQR$ and $BAQP$ are cyclic, by Miquel's theorem it follows that $S$ lies on $(PQR)$. Similarly, $S$ lies on $(XYZ)$ so we are done. $\square$

Now as $BH\parallel CK$ it follows by homothety that $\overline{MN}$ passes through $S$. Consider a rectangular circumhyperbola $\mathcal{H}$ passing through $A,B,C,D,H$. Obviously, this passes through $K$ as well. Let $T$ be the center of this hyperbola. Then by the fundamental theorem, this must be the other intersection point of $(PQR)$ and $(XYZ)$ (Configuration issues can be dealt without much difficulty). Since $BH$ and $CK$ are parallel chords of a conic, it follows that $\overline{MN}$ passes through $T$, so it must be the radical axis of $(PQR)$ and $(XYZ)$ as desired. $\square$
This post has been edited 1 time. Last edited by KST2003, Jun 6, 2021, 6:29 AM
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cronus119
74 posts
#11
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casey theorem for distance $H,B$ with radical axis of circles $\odot XYZ$,$\odot PQR$.
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ChanandlerBong
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#12
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Denote by $\Omega$ the circle passing through $A,Z,R,D,X,Q$.
By Pascal's theorem on $(ZRDQXA)$, we have that $L:=ZR \cap QX$ lies on line $BC$. Simple angle chasing indicates that $L$ is one of the intersections of circles $(PQR)$ and $(XYZ)$.
Consider the radical axis of $\Omega$, $(PQR)$ and $(XYZ)$, we have $S:=XZ\cap RQ$ lies on the radical axis of $(PQR)$ and $(XYZ)$, which we denote by $l$. Thus now we conclude that $l$ is line $SL$, so we only have to prove that the $M$ , the midpoint of segment $BH$, lies on $SL$.
To finish, it is well-known that $M$ is the pole of line $QZ$ with respect to $\Omega$, therefore apply Pascal's theorem on $(ZZRQQX)$, and then we are done!:P
This post has been edited 2 times. Last edited by ChanandlerBong, Dec 10, 2022, 3:05 AM
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trinhquockhanh
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#13 • 1 Y
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$\text{a good problem for training the \textit{Pascal theorem}, but it is a bit easy for China TST}$
https://i.ibb.co/340Wpbq/2017-China-TST-R3-D1-P2.png
geogebra solution link
This post has been edited 5 times. Last edited by trinhquockhanh, Aug 9, 2023, 1:49 PM
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sttsmet
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#14
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Allow me to disagree with the above comment! This is a very beautiful problem, well placed as a P2. It's difficulty relies mostly on drawing a good diagram on paper (sth you may haven't noticed through geogebra) as well as keeping it clear from the many lines/circles that seem tempting.
This post has been edited 1 time. Last edited by sttsmet, Apr 19, 2025, 3:40 PM
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