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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.

Summer camps are starting next month at the Virtual Campus in math and language arts that are 2 - to 4 - weeks in duration. Spaces are still available - don’t miss your chance to have an enriching summer experience. There are middle and high school competition math camps as well as Math Beasts camps that review key topics coupled with fun explorations covering areas such as graph theory (Math Beasts Camp 6), cryptography (Math Beasts Camp 7-8), and topology (Math Beasts Camp 8-9)!

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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
Conditional geo with centroid
a_507_bc   6
N 27 minutes ago by LeYohan
Source: Singapore Open MO Round 2 2023 P1
In a scalene triangle $ABC$ with centroid $G$ and circumcircle $\omega$ centred at $O$, the extension of $AG$ meets $\omega$ at $M$; lines $AB$ and $CM$ intersect at $P$; and lines $AC$ and $BM$ intersect at $Q$. Suppose the circumcentre $S$ of the triangle $APQ$ lies on $\omega$ and $A, O, S$ are collinear. Prove that $\angle AGO = 90^{o}$.
6 replies
a_507_bc
Jul 1, 2023
LeYohan
27 minutes ago
Channel name changed
Plane_geometry_youtuber   0
28 minutes ago
Hi,

Due to the search handle issue in youtube. My channel is renamed to Olympiad Geometry Club. And the new link is as following:

https://www.youtube.com/@OlympiadGeometryClub

Recently I introduced the concept of harmonic bundle. I will move on to the conjugate median soon. In the future, I will discuss more than a thousand theorems on plane geometry and hopefully it can help to the students preparing for the Olympiad competition.

Please share this to the people may need it.

Thank you!
0 replies
Plane_geometry_youtuber
28 minutes ago
0 replies
IMO Shortlist 2010 - Problem G1
Amir Hossein   134
N an hour ago by happypi31415
Let $ABC$ be an acute triangle with $D, E, F$ the feet of the altitudes lying on $BC, CA, AB$ respectively. One of the intersection points of the line $EF$ and the circumcircle is $P.$ The lines $BP$ and $DF$ meet at point $Q.$ Prove that $AP = AQ.$

Proposed by Christopher Bradley, United Kingdom
134 replies
Amir Hossein
Jul 17, 2011
happypi31415
an hour ago
Divisors on number
RagvaloD   34
N an hour ago by cubres
Source: All Russian Olympiad 2017,Day1,grade 10,P5
$n$ is composite. $1<a_1<a_2<...<a_k<n$ - all divisors of $n$. It is known, that $a_1+1,...,a_k+1$ are all divisors for some $m$ (except $1,m$). Find all such $n$.
34 replies
RagvaloD
May 3, 2017
cubres
an hour ago
IMO ShortList 2002, number theory problem 2
orl   59
N an hour ago by cubres
Source: IMO ShortList 2002, number theory problem 2
Let $n\geq2$ be a positive integer, with divisors $1=d_1<d_2<\,\ldots<d_k=n$. Prove that $d_1d_2+d_2d_3+\,\ldots\,+d_{k-1}d_k$ is always less than $n^2$, and determine when it is a divisor of $n^2$.
59 replies
orl
Sep 28, 2004
cubres
an hour ago
None of the circles contains the pentagon - ILL 1970, P34
Amir Hossein   1
N an hour ago by legogubbe
In connection with a convex pentagon $ABCDE$ we consider the set of ten circles, each of which contains three of the vertices of the pentagon on its circumference. Is it possible that none of these circles contains the pentagon? Prove your answer.
1 reply
Amir Hossein
May 21, 2011
legogubbe
an hour ago
interesting incenter/tangent circle config
LeYohan   0
2 hours ago
Source: 2022 St. Mary's Canossian College F4 Final Exam Mathematics Paper 1, Q 18d of 18 (modified)
$BC$ is tangent to the circle $AFDE$ at $D$. $AB$ and $AC$ cut the circle at $F$ and $E$ respectively. $I$ is the in-centre of $\triangle ABC$, and $D$ is on the line $AI$. $CI$ and $DE$ intersect at $G$, while $BI$ and $FD$ intersect at $P$. Prove that the points $P, F, G, E$ lie on a circle.
0 replies
LeYohan
2 hours ago
0 replies
interesting geo config (2/3)
Royal_mhyasd   5
N 2 hours ago by Royal_mhyasd
Source: own
Let $\triangle ABC$ be an acute triangle and $H$ its orthocenter. Let $P$ be a point on the parallel through $A$ to $BC$ such that $\angle APH = |\angle ABC-\angle ACB|$. Define $Q$ and $R$ as points on the parallels through $B$ to $AC$ and through $C$ to $AB$ similarly. If $P,Q,R$ are positioned around the sides of $\triangle ABC$ as in the given configuration, prove that $P,Q,R$ are collinear.
5 replies
Royal_mhyasd
Yesterday at 11:36 PM
Royal_mhyasd
2 hours ago
interesting geometry config (3/3)
Royal_mhyasd   2
N 2 hours ago by Royal_mhyasd
Let $\triangle ABC$ be an acute triangle, $H$ its orthocenter and $E$ the center of its nine point circle. Let $P$ be a point on the parallel through $C$ to $AB$ such that $\angle CPH = |\angle BAC-\angle ABC|$ and $P$ and $A$ are on different sides of $BC$ and $Q$ a point on the parallel through $B$ to $AC$ such that $\angle BQH = |\angle BAC - \angle ACB|$ and $C$ and $Q$ are on different sides of $AB$. If $B'$ and $C'$ are the reflections of $H$ over $AC$ and $AB$ respectively, $S$ and $T$ are the intersections of $B'Q$ and $C'P$ respectively with the circumcircle of $\triangle ABC$, prove that the intersection of lines $CT$ and $BS$ lies on $HE$.

final problem for this "points on parallels forming strange angles with the orthocenter" config, for now. personally i think its pretty cool :D
2 replies
Royal_mhyasd
Today at 7:06 AM
Royal_mhyasd
2 hours ago
Convex Quadrilateral with Bisector Diagonal
matinyousefi   8
N 3 hours ago by lpieleanu
Source: Germany TST 2017
In a convex quadrilateral $ABCD$, $BD$ is the angle bisector of $\angle{ABC}$. The circumcircle of $ABC$ intersects $CD,AD$ in $P,Q$ respectively and the line through $D$ parallel to $AC$ cuts $AB,AC$ in $R,S$ respectively. Prove that point $P,Q,R,S$ lie on a circle.
8 replies
matinyousefi
Apr 11, 2020
lpieleanu
3 hours ago
Kids in clubs
atdaotlohbh   0
3 hours ago
There are $6k-3$ kids in a class. Is it true that for all positive integers $k$ it is possible to create several clubs each with 3 kids such that any pair of kids are both present in exactly one club?
0 replies
atdaotlohbh
3 hours ago
0 replies
Turbo's en route to visit each cell of the board
Lukaluce   22
N 3 hours ago by HamstPan38825
Source: EGMO 2025 P5
Let $n > 1$ be an integer. In a configuration of an $n \times n$ board, each of the $n^2$ cells contains an arrow, either pointing up, down, left, or right. Given a starting configuration, Turbo the snail starts in one of the cells of the board and travels from cell to cell. In each move, Turbo moves one square unit in the direction indicated by the arrow in her cell (possibly leaving the board). After each move, the arrows in all of the cells rotate $90^{\circ}$ counterclockwise. We call a cell good if, starting from that cell, Turbo visits each cell of the board exactly once, without leaving the board, and returns to her initial cell at the end. Determine, in terms of $n$, the maximum number of good cells over all possible starting configurations.

Proposed by Melek Güngör, Turkey
22 replies
Lukaluce
Apr 14, 2025
HamstPan38825
3 hours ago
n lamps
pohoatza   47
N 3 hours ago by yayyayyay
Source: IMO Shortlist 2006, Combinatorics 1, AIMO 2007, TST 2, P1
We have $ n \geq 2$ lamps $ L_{1}, . . . ,L_{n}$ in a row, each of them being either on or off. Every second we simultaneously modify the state of each lamp as follows: if the lamp $ L_{i}$ and its neighbours (only one neighbour for $ i = 1$ or $ i = n$, two neighbours for other $ i$) are in the same state, then $ L_{i}$ is switched off; – otherwise, $ L_{i}$ is switched on.
Initially all the lamps are off except the leftmost one which is on.

$ (a)$ Prove that there are infinitely many integers $ n$ for which all the lamps will eventually be off.
$ (b)$ Prove that there are infinitely many integers $ n$ for which the lamps will never be all off.
47 replies
pohoatza
Jun 28, 2007
yayyayyay
3 hours ago
prove that at least one of them is divisible by some other member of the set.
Martin.s   0
3 hours ago
Given \( n + 1 \) integers \( a_1, a_2, \ldots, a_{n+1} \), each less than or equal to \( 2n \), prove that at least one of them is divisible by some other member of the set.
0 replies
Martin.s
3 hours ago
0 replies
IMO ShortList 2002, geometry problem 7
orl   110
N May 30, 2025 by SimplisticFormulas
Source: IMO ShortList 2002, geometry problem 7
The incircle $ \Omega$ of the acute-angled triangle $ ABC$ is tangent to its side $ BC$ at a point $ K$. Let $ AD$ be an altitude of triangle $ ABC$, and let $ M$ be the midpoint of the segment $ AD$. If $ N$ is the common point of the circle $ \Omega$ and the line $ KM$ (distinct from $ K$), then prove that the incircle $ \Omega$ and the circumcircle of triangle $ BCN$ are tangent to each other at the point $ N$.
110 replies
orl
Sep 28, 2004
SimplisticFormulas
May 30, 2025
IMO ShortList 2002, geometry problem 7
G H J
Source: IMO ShortList 2002, geometry problem 7
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Pyramix
419 posts
#113
Y by
Let $S$ be the antipode of $K$ in $\Omega$ and $T=EF\cap BC$ where $E,F$ are the $B-,C-$ touchpoints of $\Omega$.

Claim. $S,T,N$ are collinear.
Proof. Projecting $(A,D;M,\infty_{AD})=-1$ from $K$ onto $\Omega$ gives $(AK\cap\Omega,K;N,S)=-1$. Let $L=AK\cap\Omega$. So, $LNKS$ is harmonic quadrilateral. So, tangents at $L,K$ meet on line $SN$. But since tangents to $E,F$ meet on line $LK$, it means $LEKF$ is harmonic quadrilateral. Hence, $T=EF\cap BC$ is the meeting point of tangents at $L,K$. So, $T\in SN$, as claimed. $\blacksquare$

Since $KN\perp NS$, it means $N$ is the foot from $K$ to $ST$. Extend $NK$ to meet $(BCN)$ again at $P$. Note that $(T,K;B,C)=-1$ which means $(BCN)$ is the Apollonius circle for $KT$ as $\angle TNK=90^\circ$. Hence, $\frac{BN}{NC}=\frac{BK}{KC}$, which means $NK$ is the angle-bisector of $\angle BNC$. So, $NK$ extended to meet the circle $(BNC)$ is the midpoint of arc $BC$ not containing $N$. Since $\Omega$ is tangent to $BC$, $\Omega$ is tangent to $(BNC)$ by Shooting Lemma.
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fearsum_fyz
56 posts
#114 • 1 Y
Y by GeoKing
We can remove $AD$ by replacing $KM$ with $KI_A$, where $I_A$ is the $A$-excenter.

By shooting lemma, it would suffice to show that $TK \cdot TN = TB^2 = TC^2$ where $T$ is the midpoint of arc $\widehat{BC}$ of $(BCN)$. We will show this using phantom points.

Let $T'$ be the intersection of the perpendicular bisector of $BC$ and line $KI_A$. Let $A'$ be the midpoint of $BC$ and $X$ be the $A$-extouch point. Since $T'A'$ and $IK$ are both perpendicular to $BC$, they are parallel, and hence $\angle{NKI} = \angle{KT'A'} = \angle{T'I_AX} = x$ (say).
By the converse of midpoint theorem in $\Delta{KI_AX}$, $T'$ is the midpoint of $KI_A$. This yields:

$KT' \cdot KN = \frac{1}{2} KI_A \cdot KN =  \frac{1}{2} \cdot KI_A \cdot \frac{\sin{(180^{\circ} - 2x)}}{\sin{x}} \cdot r = KI_A \cdot \cos{x} \cdot r = r_a \cdot r = (s - b) \cdot (s - c) = KB \cdot KC$

implying that $N, B, C, T'$ are concyclic as desired.
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bjump
1035 posts
#115 • 1 Y
Y by OronSH
Xoink phone write ups
Let $MK$ meet the perpendicular bisector of $BC$ at $E$ note that by the shooting lemma $E \in (BCN)$. Let $O$ be the circumcenter of $(BCN)$. Then note that since $ON=OE$ and $IK=IN$ and $\angle NKI = \angle NEO$. Since $\triangle KNI \sim \triangle ENO$ and $N$, $K$, and $E$ are collinear. This implies $O$, $I$, $N$ collinear, and we are finished.
This post has been edited 1 time. Last edited by bjump, Jul 3, 2024, 10:43 PM
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MELSSATIMOV40
29 posts
#116
Y by
#attachments[url][/url]
Attachments:
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MELSSATIMOV40
29 posts
#117
Y by
MELSSATIMOV40 wrote:
#attachments[url][/url]
2 nd page
Attachments:
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Reason: 2 nd page
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Eka01
204 posts
#118 • 2 Y
Y by Sammy27, SuperBarsh
By the midpoints of altitudes lemma, $KN$ passes through the $A$ excenter $I_A$ so now we remove the extraneous altitude. Now we take the tangent to the incircle at $N$ and show it is tangent to $(BCN)$ as well. Let the tangent intersect $BC$ at $T$ and let $IT \cap KN=X$. Since $IT$ is the perpendicular bisector of $KN$, X is the midpoint of $KN$ and it lies on the circle with diameter $II_A$ which is $(BIC)$.
Since $\Delta IKT$ is right angled at $T$ and $X$ is foot of altitude from $K$ to the hypotenuse, it follows by similarity that $TK^2=TX.TI$.
Now by Power of Point,
$$TN^2=TK^2=TX.TI=TB.TC$$So $TN$ is indeed tangent to the circle $(BCN)$.
Attachments:
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mcmp
53 posts
#119
Y by
Someone PLEASE stop me from becoming too reliant on projective stuff :noo:

Relabel several of the points to match with the diagram.
Funny diagram
Let $A’=(DEF)\cap\overline{AD}\neq D$, $T’$ the pole of $\overline{A’D}$ and $T$ the midpoint of $AD$. I claim that $\overline{KT}$ tangent to $(DEF)$, and $KT^2=TB\cdot TC$ which will finish. First notice that $\overline{KD’}$ passes through $T$, since $(A’D;KD’)\stackrel{D}{=}(AX;M\infty_{AX})=-1$, however since $A$ is the pole of $\overline{EF}$, the polar of $A$ which is just $\overline{EF}$ also passes through $T’$, the pole of $\overline{A’D}$. Hence by the midpoints of harmonic bundles lemma $TD^2=TD\cdot TT’=TB\cdot TC$. So it suffices to show $KT=TD$.

Now $\measuredangle D’KD=90^{\circ}$, so $\triangle T’KD$ has circumcentre $T$. Hence we indeed have $KT=DT$. Since $K\in(DEF)$, we must also have that $KT$ tangent to $(DEF)$ and $(BKC)$ as desired. :yoda:
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Aiden-1089
302 posts
#120
Y by
Let $I$ be the incentre, $\Delta KPQ$ be the intouch triangle, and let $R=PQ \cap BC$.
Let $K'$ be the antipode of $K$ in $\Omega$.
Let $\Omega_A$ be the $A$-excircle with centre $I_A$, $L$ be the $A$-extouch point, $L'$ be the antipode of $L$ in $\Omega_A$.
Recall that $A-K-L'$ and $A-K'-L$. Since $I_A$ is the midpoint of $LL'$, there exists negative homothety centred at $K$ takes $D$ to $L$, $A$ to $L'$, and so $M$ to $I_A$. Thus $M-K-I_A$.

Note that $AK$ is the polar of $R$ wrt $\Omega$, so $AK \perp IR$. It follows that $\Delta IKR \sim \Delta KLL'$.
Let $T$ be the midpoint of $KR$, then we have $\Delta IKT \sim \Delta KLI_A$, so $IT \perp KI_A$, and this implies that $KI_A$ is the polar of $T$ wrt $\Omega$.
Now $N$ lies on $KI_A$, so $TN$ is tangent to $\Omega$ at $N$. Then since $(B,C;K,R)=-1$, we have $TB \cdot TC = TK^2 = TN^2$.
Thus $TN$ is also tangent to $(BCN)$, so we are done. $\square$
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Thelink_20
69 posts
#121
Y by
[asy]  /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(25cm); 
real labelscalefactor = 0.5; /* changes label-to-point distance */
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Lemma 1: If $P'$ is the midpoint of the arc $\overarc{BC}$ in $(BNC)$, then $(BNC)$ is tangent to $\Omega\iff N-K-P'$.
Proof: Inverting by $P'$ fixing $B$ and $C$ sends $BC\longleftrightarrow (BNC)$ and $\Omega$ is tangent to $(BNC)\iff\Omega\longleftrightarrow\Omega\iff K\longleftrightarrow N\iff N-K-P' \ $ $_{\blacksquare}$

Lemma 2: $M-K-I_A$.
Proof: If $K''$ is the antipode of $K'$ on the excircle, then the homothety from $A$ sending the incircle to th excircle yields $A-K-K''$ thus the homothety from $K$ mapping $AD\mapsto K'K''$ maps $M\mapsto I_A\Rightarrow M-K-I_A \ $ $_{\blacksquare}$

Lemma 3: If $P=(BNC)\cap MK$, then $KP=I_AP$ .
Proof: Let $Q=(BIC)\cap MK$. $\angle IQK=\angle IQI_A=\angle IBI_A=90^{\circ}\implies NQ=KQ$. But $NK\cdot KP=BK\cdot CK=KQ\cdot KI_A=\frac{NK}{2} \cdot KI_A\implies KP=\frac{KI_A}{2}=I_AP \ $ $_{\blacksquare}$

Now notice that $P$ lies on the perp. bissector of $KK'$ because $K'I_A\perp BC$, thus it lies on the perp. bissector of $BC\implies \boxed{P=P'}$, but that means we are done because clearly $N-K-P' \ $ $_{\blacksquare}$.
This post has been edited 3 times. Last edited by Thelink_20, Nov 19, 2024, 5:52 PM
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Scilyse
388 posts
#122
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If $AB = AC$ then the conclusion is obvious since $N$ lies on the perpendicular bisector of $\overline{BC}$; thus without loss of generality assume that $AB < AC$. Let $I_A$ be the $A$-excentre, which is known to lie on line $MK$, let $I$ be the incentre, and let the tangent to $\Omega$ at $N$ intersect line $BC$ at $T$. Further let $M_A$ be the midpoint of arc $BC$ of $(ABC)$ not containing $A$, let $\ell$ be the tangent to $(ABC)$ at $M_A$, let $K'$ be the foot from $I_A$ to $BC$ and let $L$ be the midpoint of $\overline{AC}$. As $T$ is the pole of line $NK$ in $\Omega$, we have $TI \perp KI_A$.

Orientate $BC$ parallel to the $x$-axis; now $\operatorname{slope}(TI) = \frac{IK}{TK} = \frac{r}{TK}$ and \[\operatorname{slope}(KI_A) = \frac{I_A K}{-KK'} = \frac{\operatorname{dist}(I_A, \ell) + M_A L}{-KK'} = -\frac{\operatorname{dist}(I, \ell) + M_A L}{2KL} = -\frac{IK + 2M_A L}{2(BL - BK)} = -\frac{r + 2BL \tan \angle M_ABC}{2(\frac a2 - (s - b))} = -\frac{r + a \tan(\frac{\alpha}{2})}{b - c}.\]Since $\operatorname{slope}(TI) \cdot \operatorname{slope}(KI_A) = -1$, we have \[-\frac{r}{TK} \cdot \frac{r + a \tan(\frac{\alpha}{2})}{b - c} = -1 \implies TK = \frac{r(r + a \tan(\frac{\alpha}{2}))}{b - c}.\]Let $E$ be the foot from $I$ to $AC$; now $\tan\left(\frac{\alpha}{2}\right) = \tan(\angle IAE) = \frac{IE}{AE} = \frac{r}{s - a}$. As such,
\begin{align*}
TK &= \frac{r(r + a \tan(\frac{\alpha}{2}))}{b - c} \\
&= \frac{r^2(1 + \frac{a}{s - a})}{b - c} \\
&= \frac{(\frac{A}{s})^2(\frac{s}{s - a})}{b - c} \\
&= \frac{\frac{s(s - a)(s - b)(s - c)}{s^2} \cdot \frac{s}{s - a}}{b - c} \\
&= \frac{(s - b)(s - c)}{b - c}.
\end{align*}Now we can calculate
\begin{align*}
TK &= \frac{(s - b)(s - c)}{b - c} \\
TK((s - c) - (s - b)) &= (s - b)(s - c) \\
TK^2 &= TK^2 + TK((s - c) - (s - b)) - (s - b)(s - c) \\
TK^2 &= (TK - (s - b))(TK + (s - c)) \\
TK^2 &= (TK - BK)(TK + CK) \\
TK^2 &= TB \cdot TC.
\end{align*}But $TN = TK$, so $TN^2 = TB \cdot TC$, and therefore $TN$ is tangent to $(NBC)$. Since $TN$ is by definition also tangent to $\Omega$, we're done.
This post has been edited 2 times. Last edited by Scilyse, Apr 15, 2025, 10:08 AM
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Captainscrubz
84 posts
#123 • 1 Y
Y by MrdiuryPeter
hehe nice problem :-D
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cursed_tangent1434
656 posts
#124 • 1 Y
Y by ihategeo_1969
Extremely dissatisfying solution but it's the best I've got. We denote by $I_a$ the $A-$Excenter and by $D$ , $E$ and $F$ the intouch points of $\triangle ABC$ (this overrides a point definition in the problem). It is well known by the Midpoint of the Altitude Lemma that the midpoint of the $A-$altitude , $D$ and $I_a$ are collinear. Thus, it suffices to show that points $D$ , $N$ and $I_a$ lie on the same line which we shall now do.

Let $P$ and $Q$ be the reflections of $D$ across points $B$ and $C$ respectively. We start off with the following key claim.

Claim : Quadrilaterals $BFQI_a$ and $CEPI_a$ are cyclic.

Proof : We only provide the proof for one since the other is entirely similar. Let $X= PD \cap AI$. Note,
\[\measuredangle I_aXP = \measuredangle CPF + \measuredangle (BC,AI) = 90 + \measuredangle IBA + \measuredangle  ACB + \measuredangle IAC = 90 + \measuredangle ICB = \measuredangle PCI_a\]which implies that points $P$ , $X$ , $C$ and $I_a$ are concyclic. Further note that since $PF \perp FD \perp BI$ lines $PF$ and $BI$ are parallel. Thus, $\triangle AFX \sim \triangle ABI$. This then gives us,
\[AX \cdot AI_a = \frac{AX\cdot AC \cdot AB}{AI} = AC \cdot AF = AC \cdot AE\]which implies that points $X$ , $E$ , $C$ and $I_a$ are also concyclic, which in conjunction with the previous observation will imply the claim.

Now, let $N = (I_aBF) \cap (I_aCE) \ne I_a$. We first note that,
\[\measuredangle ENF = \measuredangle I_aNF+ \measuredangle ENI_a = \measuredangle NI_aB + \measuredangle EI_aN = \measuredangle EI_aF = \measuredangle EDF\]which implies that $N$ lies on the incircle of $\triangle ABC$. Further,
\[\text{Pow}_{(CEI_a)}(D)=DP \cdot DC = 2DB \cdot DC = DQ \cdot DB = \text{Pow}_{(BFI_a)}(D)\]which indicates that $D$ lies on the radical axis of circles $(BFI_a)$ and $(CEI_a)$ and hence points $N$ , $D$ and $I_a$ are collinear. We now simply show that this point $N$ is the desired point of tangency.

Claim : Circle $(BNC)$ is tangent to the incircle at $N$.

Proof : Let $M$ denote the midpoint of segment $DI_a$ (overrides a point definition in the problem). It is clear that $M$ lies on the perpendicular bisector of segment $BC$ since it is well known that the midpoint of $BC$ is the midpoint of the segment joining the $A-$intouch and $A-$extouch points. Further,
\[DB \cdot DC = \frac{DP \cdot DC}{2} = \frac{DN\cdot DI_a}{2}= DM \cdot DN\]which implies that $M$ lies on $(NBC)$. Thus, $M$ must be the minor $BC-$arc midpoint in $(NBC)$. By Shooting Lemma it now suffices to show that $MD \cdot MN = MB^2$ which is exactly what we shall do next.

We now define the function, $f(\bullet)=\text{Pow}_{(I_aBF)}(\bullet)-\text{Pow}_{(B)}(\bullet)$ which by the Linearity of Power of Point, we know is linear. Let $X_a$ denote the $A-$extouch point and $M_a$ the midpoint of segment $BC$ in $\triangle ABC$. Thus,
\begin{align*}
f(M)&= \frac{f(D)+f(I_a)}{2}\\
&= \frac{-2DB\cdot DC - DB^2 + 0 - I_aB^2}{2}\\
&= \frac{-2DB \cdot DC-DB^2 - I_aX_a^2 - BX_a^2}{2}\\
&= \frac{-2DB\cdot DC - DB^2 - 4MM_a^2-BX_a^2}{2}\\
&= \frac{-2(s-b)(s-c)-(s-b)^2 - 4MM_a^2 - (s-c)^2}{2}\\
&= \frac{-((s-b)+(s-c))^2-4MM_a^2}{2}\\
&= \frac{-a^2 - 4MM_a^2}{2}\\
&= \frac{-4(BM_a^2+MM_a^2)}{2}\\
&= \frac{-4BM^2}{2}\\
&= -2BM^2
\end{align*}Thus,
\[-MN \cdot MI_a-MB^2=-2MB^2\]so
\[MB^2 = MN\cdot MI_a = MN \cdot MD\]as desired.
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ihategeo_1969
245 posts
#125 • 1 Y
Y by cursed_tangent1434
Why don't most solutions do this?

Let $I_A$ be the $A$-Excenter and by Midpoint of Altitude lemma, we have $N=\overline{KI_A} \cap$ incircle.

Now invert about the incircle and so we solve this instead
Quote:
Let $\triangle ABC$ be a triangle with circumcenter $O$ and $\triangle MNP$ as medial triangle. If $T=\overline{MO} \cap \overline{NP}$ then if $Y=(ATO) \cap (ABC)$ then prove that $(PNY)$ is tangent to $(ABC)$.
We will prove $Y$ is $A$-Why point. Now we will phantom point this and use two well known facts about $Y_A$ (the Why point).
  • $Y_A=\overline{DGA'} \cap (ABC)$ where $G$ is centroid.
  • $(Y_APN)$ is tangent to $(ABC)$.
So we just need to prove $Y_A \in (ATO)$. Firstly by $-\frac 12$ homothety at $G$ we get $T \in \overline{DGA'}$ and so \[\measuredangle AY_AT=\measuredangle AY_AA'=\measuredangle ODA=\measuredangle AOM=\measuredangle AOT\]And done.
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Ilikeminecraft
676 posts
#126
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Let $S$ be the second intersection of $AK$ with the incircle.
Let $KEF$ denote the intouch triangle.
Let $K’$ denote the antipode of $K$ in the incircle.
Let $T = EF\cap BC.$
Let $L$ denote the midpoint of $KT.$
Let $S$ be the intersection of $AK$ with the incircle.
Note that $TS$ is tangent to the incircle because $SEKF$ is harmonic as $AF, AE$ are tangent to incircle and $ASK$ collinear.
Observe that $-1=(\infty M;AD) \stackrel K= (K’N;SK)$ which implies $T, N, K’$ are collinear.
Projection of the bundle $(TK;\infty L) = - 1$ through $N$ onto the incircle implies $LN$ is tangent to the incircle.
Finally, note that $(TK;BC) = -1$ due to gergonne point. By well-known configuration properties, it follows that $LB\cdot LC = LN^2,$ which implies $LN$ is tangent to $(BCN).$
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SimplisticFormulas
128 posts
#127
Y by
goofy aah sol
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N Quick Reply
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