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

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Be sure to mark your calendars for the following events:
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0 replies
jlacosta
Mar 2, 2025
0 replies
k i Adding contests to the Contest Collections
dcouchman   1
N Apr 5, 2023 by v_Enhance
Want to help AoPS remain a valuable Olympiad resource? Help us add contests to AoPS's Contest Collections.

Find instructions and a list of contests to add here: https://artofproblemsolving.com/community/c40244h1064480_contests_to_add
1 reply
dcouchman
Sep 9, 2019
v_Enhance
Apr 5, 2023
k i Zero tolerance
ZetaX   49
N May 4, 2019 by NoDealsHere
Source: Use your common sense! (enough is enough)
Some users don't want to learn, some other simply ignore advises.
But please follow the following guideline:


To make it short: ALWAYS USE YOUR COMMON SENSE IF POSTING!
If you don't have common sense, don't post.


More specifically:

For new threads:


a) Good, meaningful title:
The title has to say what the problem is about in best way possible.
If that title occured already, it's definitely bad. And contest names aren't good either.
That's in fact a requirement for being able to search old problems.

Examples:
Bad titles:
- "Hard"/"Medium"/"Easy" (if you find it so cool how hard/easy it is, tell it in the post and use a title that tells us the problem)
- "Number Theory" (hey guy, guess why this forum's named that way¿ and is it the only such problem on earth¿)
- "Fibonacci" (there are millions of Fibonacci problems out there, all posted and named the same...)
- "Chinese TST 2003" (does this say anything about the problem¿)
Good titles:
- "On divisors of a³+2b³+4c³-6abc"
- "Number of solutions to x²+y²=6z²"
- "Fibonacci numbers are never squares"


b) Use search function:
Before posting a "new" problem spend at least two, better five, minutes to look if this problem was posted before. If it was, don't repost it. If you have anything important to say on topic, post it in one of the older threads.
If the thread is locked cause of this, use search function.

Update (by Amir Hossein). The best way to search for two keywords in AoPS is to input
[code]+"first keyword" +"second keyword"[/code]
so that any post containing both strings "first word" and "second form".


c) Good problem statement:
Some recent really bad post was:
[quote]$lim_{n\to 1}^{+\infty}\frac{1}{n}-lnn$[/quote]
It contains no question and no answer.
If you do this, too, you are on the best way to get your thread deleted. Write everything clearly, define where your variables come from (and define the "natural" numbers if used). Additionally read your post at least twice before submitting. After you sent it, read it again and use the Edit-Button if necessary to correct errors.


For answers to already existing threads:


d) Of any interest and with content:
Don't post things that are more trivial than completely obvious. For example, if the question is to solve $x^{3}+y^{3}=z^{3}$, do not answer with "$x=y=z=0$ is a solution" only. Either you post any kind of proof or at least something unexpected (like "$x=1337, y=481, z=42$ is the smallest solution). Someone that does not see that $x=y=z=0$ is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
Similar, posting "I have solved this problem" but not posting anything else is not welcome; it even looks that you just want to show off what a genius you are.

e) Well written and checked answers:
Like c) for new threads, check your solutions at least twice for mistakes. And after sending, read it again and use the Edit-Button if necessary to correct errors.



To repeat it: ALWAYS USE YOUR COMMON SENSE IF POSTING!


Everything definitely out of range of common sense will be locked or deleted (exept for new users having less than about 42 posts, they are newbies and need/get some time to learn).

The above rules will be applied from next monday (5. march of 2007).
Feel free to discuss on this here.
49 replies
ZetaX
Feb 27, 2007
NoDealsHere
May 4, 2019
Inequality => square
Rushil   12
N 27 minutes ago by ohiorizzler1434
Source: INMO 1998 Problem 4
Suppose $ABCD$ is a cyclic quadrilateral inscribed in a circle of radius one unit. If $AB \cdot BC \cdot CD \cdot DA \geq 4$, prove that $ABCD$ is a square.
12 replies
Rushil
Oct 7, 2005
ohiorizzler1434
27 minutes ago
p + q + r + s = 9 and p^2 + q^2 + r^2 + s^2 = 21
who   28
N 43 minutes ago by asdf334
Source: IMO Shortlist 2005 problem A3
Four real numbers $ p$, $ q$, $ r$, $ s$ satisfy $ p+q+r+s = 9$ and $ p^{2}+q^{2}+r^{2}+s^{2}= 21$. Prove that there exists a permutation $ \left(a,b,c,d\right)$ of $ \left(p,q,r,s\right)$ such that $ ab-cd \geq 2$.
28 replies
who
Jul 8, 2006
asdf334
43 minutes ago
H not needed
dchenmathcounts   44
N an hour ago by Ilikeminecraft
Source: USEMO 2019/1
Let $ABCD$ be a cyclic quadrilateral. A circle centered at $O$ passes through $B$ and $D$ and meets lines $BA$ and $BC$ again at points $E$ and $F$ (distinct from $A,B,C$). Let $H$ denote the orthocenter of triangle $DEF.$ Prove that if lines $AC,$ $DO,$ $EF$ are concurrent, then triangle $ABC$ and $EHF$ are similar.

Robin Son
44 replies
dchenmathcounts
May 23, 2020
Ilikeminecraft
an hour ago
IZHO 2017 Functional equations
user01   51
N 2 hours ago by lksb
Source: IZHO 2017 Day 1 Problem 2
Find all functions $f:R \rightarrow R$ such that $$(x+y^2)f(yf(x))=xyf(y^2+f(x))$$, where $x,y \in \mathbb{R}$
51 replies
user01
Jan 14, 2017
lksb
2 hours ago
chat gpt
fuv870   2
N 2 hours ago by fuv870
The chat gpt alreadly knows how to solve the problem of IMO USAMO and AMC?
2 replies
fuv870
2 hours ago
fuv870
2 hours ago
Inequality with wx + xy + yz + zw = 1
Fermat -Euler   23
N 2 hours ago by hgomamogh
Source: IMO ShortList 1990, Problem 24 (THA 2)
Let $ w, x, y, z$ are non-negative reals such that $ wx + xy + yz + zw = 1$.
Show that $ \frac {w^3}{x + y + z} + \frac {x^3}{w + y + z} + \frac {y^3}{w + x + z} + \frac {z^3}{w + x + y}\geq \frac {1}{3}$.
23 replies
Fermat -Euler
Nov 2, 2005
hgomamogh
2 hours ago
Waiting for a dm saying me again "old geometry"
drmzjoseph   0
2 hours ago
Source: Idk easy
Given $ABCD$ a tangencial quadrilateral that is not a rhombus, let $a,b,c,d$ be lengths of tangents from $A,B,C,D$ to the incircle of the quadrilateral which center is $I$. Let $M,N$ be the midpoints of $AC,BD$ resp. Prove that
\[ \frac{MI}{IN}=\frac{a+c}{b+d} \]
0 replies
drmzjoseph
2 hours ago
0 replies
Finally hard NT on UKR MO from NT master
mshtand1   2
N 2 hours ago by IAmTheHazard
Source: Ukrainian Mathematical Olympiad 2025. Day 1, Problem 11.4
A pair of positive integer numbers \((a, b)\) is given. It turns out that for every positive integer number \(n\), for which the numbers \((n - a)(n + b)\) and \(n^2 - ab\) are positive, they have the same number of divisors. Is it necessarily true that \(a = b\)?

Proposed by Oleksii Masalitin
2 replies
mshtand1
Mar 13, 2025
IAmTheHazard
2 hours ago
IMOC 2017 G5 (<A=120 => E, F, Y,Z are concyclic, incenter related)
parmenides51   4
N 2 hours ago by ehuseyinyigit
Source: https://artofproblemsolving.com/community/c6h1740077p11309077
We have $\vartriangle ABC$ with $I$ as its incenter. Let $D$ be the intersection of $AI$ and $BC$ and define $E, F$ in a similar way. Furthermore, let $Y = CI \cap DE, Z = BI \cap DF$. Prove that if $\angle BAC = 120^o$, then $E, F, Y,Z$ are concyclic.
IMAGE
4 replies
parmenides51
Mar 20, 2020
ehuseyinyigit
2 hours ago
Bosnia and Herzegovina JBMO TST 2013 Problem 1
gobathegreat   3
N 3 hours ago by DensSv
Source: Bosnia and Herzegovina Junior Balkan Mathematical Olympiad TST 2013
It is given $n$ positive integers. Product of any one of them with sum of remaining numbers increased by $1$ is divisible with sum of all $n$ numbers. Prove that sum of squares of all $n$ numbers is divisible with sum of all $n$ numbers
3 replies
gobathegreat
Sep 16, 2018
DensSv
3 hours ago
D1015 : A strange EF for polynomials
Dattier   0
3 hours ago
Source: les dattes à Dattier
Find all $P \in \mathbb R[x,y]$ with $P \not\in \mathbb R[x] \cup \mathbb R[y]$ and $\forall g,f$ homeomorphismes of $\mathbb R$, $P(f,g)$ is an homoemorphisme too.
0 replies
Dattier
3 hours ago
0 replies
P, Q,R collinear and U, R, O, V concyclic wanted, cyclic ABCD, circumcenters
parmenides51   2
N 3 hours ago by DensSv
Source: 2012 Romania JBMO TST2 P4
The quadrilateral $ABCD$ is inscribed in a circle centered at $O$, and $\{P\} = AC \cap BD, \{Q\} = AB \cap CD$. Let $R$ be the second intersection point of the circumcircles of the triangles $ABP$ and $CDP$.
a) Prove that the points $P, Q$, and $R$ are collinear.
b) If $U$ and $V$ are the circumcenters of the triangles $ABP$, and $CDP$, respectively, prove that the points $U, R, O, V$ are concyclic.
2 replies
parmenides51
May 29, 2020
DensSv
3 hours ago
Unsolved Diophantine(I think)
Nuran2010   1
N 4 hours ago by Nuran2010
Find all solutions for the equation $2^n=p+3^p$ where $n$ is a positive integer and $p$ is a prime.(Don't get mad at me,I've used the search function and did not see a correct and complete solution anywhere.)
1 reply
Nuran2010
Mar 14, 2025
Nuran2010
4 hours ago
2^a + 3^b + 1 = 6^c
togrulhamidli2011   1
N 4 hours ago by CM1910
Find all positive integers (a, b, c) such that:

\[
2^a + 3^b + 1 = 6^c
\]
1 reply
togrulhamidli2011
Today at 12:34 PM
CM1910
4 hours ago
circumcenter of BJK lies on line AC, median, right angle, circumcircle related
parmenides51   22
N Today at 5:21 AM by Ilikeminecraft
Source: 2019 RMM Shortlist G1
Let $BM$ be a median in an acute-angled triangle $ABC$. A point $K$ is chosen on the line through $C$ tangent to the circumcircle of $\vartriangle BMC$ so that $\angle KBC = 90^\circ$. The segments $AK$ and $BM$ meet at $J$. Prove that the circumcenter of $\triangle BJK$ lies on the line $AC$.

Aleksandr Kuznetsov, Russia
22 replies
parmenides51
Jun 18, 2020
Ilikeminecraft
Today at 5:21 AM
circumcenter of BJK lies on line AC, median, right angle, circumcircle related
G H J
G H BBookmark kLocked kLocked NReply
Source: 2019 RMM Shortlist G1
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parmenides51
30628 posts
#1
Y by
Let $BM$ be a median in an acute-angled triangle $ABC$. A point $K$ is chosen on the line through $C$ tangent to the circumcircle of $\vartriangle BMC$ so that $\angle KBC = 90^\circ$. The segments $AK$ and $BM$ meet at $J$. Prove that the circumcenter of $\triangle BJK$ lies on the line $AC$.

Aleksandr Kuznetsov, Russia
This post has been edited 3 times. Last edited by parmenides51, Oct 9, 2020, 12:31 PM
Reason: source edit
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WizardMath
2487 posts
#2 • 1 Y
Y by Sourorange
Redefine $J$ as the $B-$Humpty point of $ABC$, and $K$ as the intersection of $AJ$ and the circle through $B, J$ and the reflection of $B$ in $AC$ (say $B'$). Note that the reflection of $J$ in $AC$ (say $X$), which is the endpoint of the $B-$symmedian chord, is also on this circle. This circle is in fact the $B-$Apollonian circle of $ABC$, so its center is on $AC$. Now we show that our definitions of $J, K$ are consistent with the previous definitions, i.e., we need to show that $\angle KBC = 90^\circ$ and $KC$ is tangent to $(BMC)$.

For the first part, note that $\angle KBJ = 180^\circ - \angle BJK - \angle JKB = 180^\circ - 180^\circ + \angle BAC - \angle JB'B = \angle BAC - \angle XBB' = \angle BAC - \angle OBM$, where $O$ is the circumcenter of $ABC$. So $\angle KBC = \angle KBJ + \angle JBC = 90^\circ$.

For the second part, we claim that $C, K, X$ are collinear, from where it would follow that $\angle KCM = \angle XCA = \angle XBA = \angle CBJ$, which would finish the proof. To this end, note that $\angle KXC = \angle KXB + \angle BXC = \angle AJB + \angle BAC = 180^\circ$, whence we are done. (The other configurations can be handled similarly)

Note
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Aryan-23
558 posts
#3 • 2 Y
Y by AlastorMoody, Han1728
Beautiful problem ! :love:
Here is a solution
Remark
This post has been edited 2 times. Last edited by Aryan-23, Sep 20, 2020, 8:26 AM
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mmathss
282 posts
#4 • 7 Y
Y by AlastorMoody, anser, jelena_ivanchic, SnowPanda, Aryan-23, srijonrick, tigerzhang
An easy solution which does not invoke HM points.
Let $N$ be the midpoint of $KC$.Clearly $NB=NK$ Let the perpendicular bisector of $BK$ intersect $AC$ at point $O$.It is also easy to see that $AK$ is parallel to $MN$
Since $\angle NOC=\angle BCA=180-\angle MBN \Rightarrow  MBNO$ is cyclic.Therefore $0.5\times \angle BOK=\angle BON=\angle BMN=\angle AJM=\angle BJK$.Thus $O$ is the circumcenter of $BNK$ and we are done$\blacksquare$
This post has been edited 1 time. Last edited by mmathss, Jul 1, 2020, 6:18 PM
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itslumi
284 posts
#5
Y by
Why the shortlist is not in the contest collection?
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v_Enhance
6857 posts
#6 • 3 Y
Y by jelena_ivanchic, Mathematicsislovely, v4913
The following solution was found jointly with Serena An, Sanjana Das, and Sunaina Pati.

Let $D$ be the antipode of $C$ on $(BMC)$, so we have right triangle $DMC$. Complete it to cyclic kite $DCAT$ inscribed in a circle with diameter $\overline{DT}$ (i.e.\ with $\angle DCT = \angle DAT = 90^{\circ}$). We define $P = \overline{BM} \cap \overline{AT}$ and then construct parallelogram $APCQ$.

[asy] size(9cm); pair D = dir(90); pair T = dir(-90); pair C = dir(-20); pair A = -conj(C); pair P = 0.4*T+0.6*A; pair M = extension(A, C, D, T); pair B = conj(abs(A-C)**2/(4*M-4*P))+M; pair Q = A+C-P; pair E = extension(C, Q, D, A); pair K = extension(T, C, B, D);
filldraw(circumcircle(M, B, C), invisible, blue);
filldraw(A--T--C--D--cycle, invisible, red); filldraw(A--P--C--Q--cycle, invisible, orange); draw(P--B, orange); draw(A--T, red); draw(D--T, red); draw(D--K, blue); draw(A--C, red); draw(A--K, dotted+red); pair N = extension(Q, foot(Q, A, C), C, T); draw(C--K, red); draw(C--E, orange+dashed); draw(P--N--Q, blue);
dot("$D$", D, dir(D)); dot("$T$", T, dir(T)); dot("$C$", C, dir(C)); dot("$A$", A, dir(A)); dot("$P$", P, dir(P)); dot("$M$", M, dir(M)); dot("$B$", B, dir(B)); dot("$Q$", Q, dir(Q)); dot("$E$", E, dir(E)); dot("$K$", K, dir(K)); dot("$N$", N, dir(N));
/* TSQ Source:
!size(9cm); D = dir 90 T = dir -90 C = dir -20 A = -conj(C) P = 0.4*T+0.6*A M = extension A C D T B = conj(abs(A-C)**2/(4*M-4*P))+M Q = A+C-P E = extension C Q D A K = extension T C B D
circumcircle M B C 0.1 blue / blue
A--T--C--D--cycle 0.1 lightred / red A--P--C--Q--cycle 0.1 yellow / orange P--B orange A--T red D--T red D--K blue A--C red A--K dotted red N = extension Q foot Q A C C T C--K red C--E orange dashed P--N--Q blue
*/ [/asy]

Claim: Lines $CQ$ and $AD$ meet on $(BMC)$, say at $E$.
Proof. Because $\measuredangle QCM = \measuredangle PAM = \measuredangle TAM = \measuredangle ADM$. $\blacksquare$

Claim: The points $A$, $Q$, $K$ are collinear, so $Q=J$.
Proof. Pascal on $DECCMB$. $\blacksquare$

Claim: Reflect $Q$ across $\overline{AC}$ to obtain $N$. Then we have $BKNQ$ concyclic.
Proof. Because $\measuredangle KNQ = 90^{\circ} - \measuredangle PNT = 90^{\circ} - \measuredangle ACT = \measuredangle 	DCM = \measuredangle DBM = \measuredangle KBQ$. $\blacksquare$

Since $\overline{AC}$ is obviously the perpendicular bisector of $\overline{QN}$, the proof is complete.
This post has been edited 1 time. Last edited by v_Enhance, Sep 13, 2020, 4:33 AM
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jj_ca888
2726 posts
#7
Y by
Let the tangent to $(BMC)$ hit $AC$ at $T$. Since $\triangle KBC$ is a right triangle, $T$ must be the midpoint of $KC$. Let $N$ be the midpoint of $BK$. The perpendicular bisector $NT$ of $BK$ hits $AC$ at $O$. It suffices to show that $O$ is the center of $(BJK)$.

First off, note that\[\angle NTB = \angle TBC = \angle TCB = \angle CMB = \angle OMB\]so $BTOM$ is cyclic. This is in fact, quite nice. It tells us that\[\angle BMT = \angle BDT = \frac12\angle BOK.\]Furthermore, by midpoints, $MT \parallel AK$ so $\angle BMT = \angle BJK$ hence $\angle BJK = \frac12\angle BOK$ and thus $J$ lies on the circle centered at $O$ with radius $OB = OK$, as desired. $\blacksquare$
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niyu
830 posts
#8
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Main Claim: $J$ is the $B$-HM point of $\triangle ABC$.

Proof: Let $K'$ be the $\sqrt{ac}$ inverse of $K$; this inversion swaps $A$ and $C$ as well. Note
\[\angle BK'A = \angle BCK = \angle BMC = 180^\circ - \angle BMA,\]so $K'BMA$ is cyclic. As
\[ \angle K'MA = \angle K'BA = \angle CBK = 90^\circ, \]so $K'$ lies on the perpendicular bisector of $\overline{AC}$.
Now,
\begin{align*}
	\angle BKA &= \angle BCK' \\
	&= C - \angle K'CA \\
	&= C - \angle K'AM \\
	&= C - (180^\circ - \angle K'BM) \\
	&= C - (180^\circ - (90^\circ + \angle ABM)) \\
	&= C + \angle ABM - 90^\circ.
\end{align*}As a result,
\begin{align*}
	\angle AKC &= \angle BKC - \angle BKA \\
	&= (90^\circ - \angle BCK) - (C + \angle ABM - 90^\circ) \\
	&= (90^\circ - \angle BMC) - (C + \angle ABM - 90^\circ) \\
	&= (90^\circ - (180^\circ - \angle MBC - C)) - (C + \angle ABM - 90^\circ) \\
	&= \angle MBC - \angle ABM.
\end{align*}In particular,
\begin{align*}
	\angle CAJ &= 180^\circ - \angle AKC - \angle KCA \\
	&= 180^\circ - (\angle MBC - \angle ABM) - (180^\circ - \angle MBC) \\
	&= \angle ABM,
\end{align*}so $\overline{AC}$ is tangent to $(ABJ)$, implying the claim. $\blacksquare$

Now, let $T$ be the foot of the $B$-internal angle bisector, let $T'$ be the foot of the $B$-external angle bisector, and let $L = \overline{BB} \cap \overline{AC}$. We claim that $L$ is the circumcenter of $(TJBK)$, which is enough to solve the problem.

Noting that $(J, B)$ are inverses under $(AC)$ inversion, it follows that $\frac{AT}{TC} = \frac{AB}{BC} = \frac{AJ}{JC}$, implying that $\overline{JT}$ bisects $\angle AJC$. As $(AC;TT') = -1$, it follows that $\angle TJT' = 90^\circ$ (by right angles/bisectors lemma), so $TJBT'$ is cyclic with diameter $TT'$. Noting that $AJHC$ is cyclic (this follows from $(AC)$ inversion), we find
\begin{align*}
	\angle TJK &= 180^\circ - \angle AJT \\
	&= 180^\circ - \frac{1}{2}\angle AJC \\
	&= 180^\circ - \frac{1}{2}\angle AHC \\
	&= 90^\circ + \frac{B}{2} \\
	&= \angle TBK.
\end{align*}Hence, $TJBKT'$ is cyclic.

Yet \[ \angle LBT = \angle LBA - \angle TBA = 180^\circ - C - \frac{B}{2} = \angle LTB, \]so $L$ is the circumcenter of $TBT'$. Thus, $L$ is the circumcenter of $\triangle BJK$, and we are done! $\Box$
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srijonrick
168 posts
#10 • 1 Y
Y by Mango247
Storage :D
parmenides51 wrote:
Let $BM$ be a median in an acute-angled triangle $ABC$. A point $K$ is chosen on the line through $C$ tangent to the circumcircle of $\vartriangle BMC$ so that $\angle KBC = 90^\circ$. The segments $AK$ and $BM$ meet at $J$. Prove that the circumcenter of $\triangle BJK$ lies on the line $AC$.

Aleksandr Kuznetsov, Russia

Solution. Let $N$ be the midpoint of $KC.$ We know that $NB=NK$. Let the perpendicular bisector of $BK$, through $N$, intersect $AC$ at $O.$ Now, as $CB \perp KB$, and also $ON \perp BK$, so $BC \parallel ON$, and thus $$\angle NOC = \angle BCA \overset{(\spadesuit)}{=} 180^{\circ} - \angle MBN,$$thus $M,O,N,B$ are concyclic; and $(\spadesuit)$ follows as $\angle NBC = \angle NCB$ and $\measuredangle (\overline{AMC}, \overline{CC}) = \angle MBC$ by Alternate segment theorem, so $\angle MBN = \angle NCB + \measuredangle (\overline{AMC}, \overline{CC}) = 180^{\circ} - \angle BCA.\quad(\clubsuit)$

Now, by Midpoint theorem in $\triangle ACK$, we have $MN \parallel AK-(*).$ Lastly, we note that $$\frac{1}{2} \angle BOK = \angle BON \overset{(\clubsuit)}{=} \angle BMN \overset{(*)}{=} \angle AJM \overset{\text{ver.opp.}}{=} \angle BJK.\ \blacksquare$$
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rcorreaa
238 posts
#11 • 3 Y
Y by Mango247, Mango247, Mango247
Amazing problem! :trampoline:

We prove that $J$ is actually the $B$-humpty point WRT $ABC$.

Let $X_B$ the $B$-humpty point WRT $ABC$, and let $K'$ the intersection of $AX_B$ with the line through $C$ tangent to $(BMC)$. Observe that from the ratio lemma, we have that $$\frac{sin \angle X_BBA}{sin \angle X_BBC}=\frac{BA}{BC} \qquad (\spadesuit)$$Now, notice that since $CK'$ is tangent to $(BMC)$, then $\measuredangle X_BAC= \measuredangle X_BBA \implies \measuredangle K'AC= \measuredangle X_BBA$.

Hence, by Law of Sines on $K'AC$, $$\frac{K'A}{K'C}= \frac{sin \angle K'AC}{sin \angle ACK'}=\frac{sin \angle X_BBA}{sin \angle X_BBC} \underbrace{=}_{(\spadesuit)} \frac{BA}{BC}$$$\implies K'$ lies on the $B$-Apollonius circle of $ABC$.

Let $L$ the centerof the $B$-Apollonius circle of $ABC \implies LK'^2=LB^2=LA.LC$ (it's well known that $L$ is the intersection of the tangent through $B$ to $(ABC)$ with $AC$) $\implies \angle LK'C= \angle CAK'= \angle CAX_B= \angle X_BBA (\star)$, and since $\angle ACK'= 180º- \angle MBC= 180º- \angle X_BBC \implies \angle LK'C= \angle X_BBC- \angle X_BBA \implies$ from $(\star)$, $\angle LK'X_B= \angle  X_BB_C \implies$ if $P= K'L \cap BC \implies P \in (BX_BK')$, and since $PK$ is diameter of $(BX_BK')$, $\angle PBK'=90º \implies \angle CBK'= 90º \implies K=K'$ and $J=X_B$, so $(BJK)$ is the $B$-Apollonius circle of $ABC$, whose center lies on $AC$ (it's well known that the center of the $B$-Apollonius circle of $ABC$ is the intersection of the tangent through $B$ to $(ABC)$ with $AC$), as desired.

$\blacksquare$
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rafaello
1079 posts
#12
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I present my solution which is a bit different from others (idea is still the same ig).
We want to claim that $J$ is the $B$-humpty point of $\triangle ABC$.

So let $I$ be the $B$-humpty point of $\triangle ABC$ (Nota bene: $I$ is not a incenter lol).
Let $I'$ be the reflection of $I$ over $AC$ and let $K=I'C\cap AI$. Let $D$ be the centre of $(MBC)$.

We have $\measuredangle IKI'=\measuredangle ACI'-\measuredangle CAK=\measuredangle ICA-\measuredangle CAI$ and
$\measuredangle IBI'=\measuredangle IBC-\measuredangle I'BC=\measuredangle ICA-\measuredangle ABJ=\measuredangle ICA-\measuredangle  CAI$, since it is kind of well-known that $BI'$ is the symmedian (can be proven by simple angle chase using properties of a humpty point).

Hence, $I'IBK$ is cyclic.
Now, $$\measuredangle IBK=\measuredangle II'K=\measuredangle II'C=\measuredangle CII'=\measuredangle ICA-90^\circ= \measuredangle IBC-90^\circ\implies \measuredangle CBK=90^\circ.$$
We have $$\measuredangle CMB+\measuredangle BCD=\frac{\measuredangle CDB}{2} +\measuredangle BCD=90^\circ$$Also, $$\measuredangle KCB=90^\circ-\measuredangle BKC=90^\circ+\measuredangle I'KB=90^\circ+\measuredangle I'IC+\measuredangle CIB=90^\circ+90^\circ-\measuredangle ICA+\measuredangle CIB=\measuredangle ACI+\measuredangle CIM=\measuredangle CMI=\measuredangle CMB,$$thus $\measuredangle KCD=90^\circ$.
Thus, $K$ coincides with original $K$ and $J=I$.
Since $AC$ is the perpendicular bisector of $JJ'$ ($J'$ is the reflection of $J$ over $AC$). And since $J'$ lies on $JBK$, we have $O$ lying on $AC$.
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Kimchiks926
256 posts
#13
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Took 1.5 bloody hours. WLOG $BA<BC$. Define $CK \cap \odot(ABC) = X$.
Claim: $BX$ is symmedian of $\triangle ABC$.
Proof: This is simple angle chasing $\angle ACX = \angle MBC =\angle ABX$ $\blacksquare$

The most difficult part of the problem is to realise that $\odot(BJK)$ is $B$ Apollonian circle. This simply motivates to add point $F$, which is interaction of $\angle ABC$ bisector and side $BC$. Then $\odot(BFX)$ is $B$ Apollonian circle (well-known fact).

Claim: $XF$ intersects $\odot(ABC)$ at point $N$, which is midpoint of arc $ABC$.
Proof: Well -known.Click to reveal hidden text

Consequently this immediately gives us $\angle FXC =90^{\circ} - 0.5 \angle B = \angle KBF$. Thus $K$ lies on $\odot(BFX)$. From Apollonian circle definition it follows:
$$ \frac{KA}{KC}=\frac{BA}{BC} = \frac{AF}{FC} $$This reveals that $KF$ is angle bisector $\angle AKC$. We conclude by:
$$ \angle AKF = \angle FKX = \angle FBX = \angle JBF $$since this simply says that $J$ lies on $\odot(BFK)$. Now we are done since both points $J,K$ lies on $B$ Apollonian circle.
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amuthup
779 posts
#14
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Invert around $B$ with radius $\sqrt{BA\cdot BC}$ and then reflect the diagram in the angle bisector of $\angle ABC.$

Let $X=\overline{CK}\cap(ABC),$ and let $T$ be the intersection of the tangents to $(ABC)$ at $A$ and $C.$

$\textbf{Claim: }$ $M^*=X$.

$\emph{Proof: }$ Since $\angle ABX=\angle ACX=\angle MBC,$ we know $\overline{BM},\overline{BX}$ are isogonal with respect to $\angle ABC.$ Therefore, since the inverse of $\overline{AC}$ is $(ABC),$ we are done. $\blacksquare$

$\textbf{Claim: }$ $J^*=T$

$\emph{Proof: }$ By definition, $K\in\overline{CX},$ so $K^*\in (BMA).$ Furthermore, $\angle KBC=90^\circ,$ so $\angle K^*BA=90^\circ.$ This implies that $K^*$ is the antipode of $A$ in $(BMA).$ In particular, $K^*$ lies on the perpendicular bisector of $\overline{AC}.$

Since $J\in\overline{KA},$ quadrilateral $BK^{*}CJ^*$ is cyclic. Therefore, \begin{align*}
\angle CJ^*B
&=180^\circ-\angle BK^*C\\
&=180^\circ-(\angle BK^*A+2\angle AK^*M)\\
&= 180^\circ-(\angle BMA+2\angle ABM)\\
&= \angle CAB-\angle ABM.\end{align*}But we also know that $J\in\overline{BM},$ so $J^*\in\overline{BX}$ and thus $\angle J^*BC=\angle ABM.$ It follows that $\angle BCJ^*=180^\circ-\angle CAB,$ which implies the desired conclusion. $\blacksquare$

$\textbf{Claim: }$ $X\in (BJK)$

$\emph{Proof: }$ Just note that $J^*,X^*,K^*$ all lie on the perpendicular bisector of $\overline{BC}.$ $\blacksquare$

Now write
\begin{align*}
    \angle CAJ
    &=\angle CAB-\angle JAB\\
    &=\angle CAB-\angle CJ^*B\\
    &=\angle ABM\\
    &=\angle XBC\\
    &=\angle XAC,
\end{align*}and similarly, $\angle JCA=\angle ACX.$

Thus, $\overline{AC}$ is the perpendicular bisector of $\overline{JX},$ so it passes through the center of $(BJXK).$
This post has been edited 1 time. Last edited by amuthup, May 11, 2021, 3:15 PM
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sriraamster
1492 posts
#15
Y by
nice problem

Let $N$ denote the midpoint of $BK$ and $O$ the circumcenter of $BJK.$ The key claim is as follows:

Claim: $BONM$ is cyclic.
Proof:

Note that $ON \parallel BC$ and $NM \parallel KA,$ so $\measuredangle ONM$ is equal to the complement of the angle formed by lines $KA$ and $BC.$ Let $\measuredangle JBC = \alpha$ and $\measuredangle JKB = \theta.$ Then $\measuredangle ONM = 90 + \theta.$ A quick angle chase gives $\measuredangle OBK = \alpha - \theta,$ so \[ \measuredangle OBJ = \measuredangle OBM = \measuredangle OBK + \measuredangle KBM = 90 - \alpha + \alpha - \theta = 90 - \theta. \]Therefore, $\measuredangle OBM + \measuredangle ONM = 180$ so the points are concyclic.

Furthermore, note that $BN$ happens to be tangent to $(BMC)$ since $BN = NC$ (circumcenter of right triangle). Then, \[ \measuredangle OMB = \measuredangle ONB = \measuredangle CBN = \measuredangle NCB = \alpha + \measuredangle C  = \measuredangle AMB \]by exterior angles. Hence, since $\measuredangle OMB = \measuredangle AMB,$ $O,A,$ and $M$ are collinear which implies that $O \in AC$ as desired.
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L567
1184 posts
#16 • 1 Y
Y by everythingpi3141592
Let $N$ be midpoint of $KC$ and $(BJK)$ meet $KC$ at $Z$. Let the circumcenter of $(BJK)$ be $O$

Since $\angle NZB = 180 - \angle KZB = 180 - \angle KJB = 180 - \angle NMB$, we have $BMNZ$ is cyclic

Since $\angle ZOB = 2 \angle BKC = \angle BNC$, we also have $BMNZO$ is cyclic

But now we have $\angle OMB = \angle ONB = \angle NBC = \angle AMB$ and so $O$ lies on $AC$, as desired. $\blacksquare$
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Fakesolver19
106 posts
#17
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Storage
This post has been edited 1 time. Last edited by Fakesolver19, Oct 9, 2021, 12:48 PM
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tigerzhang
351 posts
#18 • 2 Y
Y by centslordm, Bradygho
my job is to overkill geo

Let $C'$ be the antipode of $C$ with respect to the circumcircle of $\triangle BMC$. Notice that $AK$ passes through $C'$. We claim that $J$ is the $B$-humpty point of $\triangle ABC$, after which we are done by easy angle chasing. Rewrite the problem as the following:

Let $M$ be the midpoint of $\overline{BC}$ in $\triangle ABC$. Let $B'$ be the antipode of $B$ with respect to the circumcircle of $\triangle ABM$. Let $J$ be the $A$-humpty point of $\triangle ABC$. Prove that $\overline{AB'}$, $\overline{CJ}$, and the line passing through $B$ tangent to the circumcircle of $\triangle ABM$ are concurrent.

Let the circumcircle of $\triangle ABM$ intersect the circle centered at $B'$ passing through $B$ at a point $L \neq B$.

Claim: $C$, $J$, and $L$ are collinear.
Proof: We have $$\measuredangle BLJ=\measuredangle BAM=\measuredangle BB'M=\measuredangle BLC.$$
Claim: $AB'CL$ is cyclic.
Proof: More angle chasing.

Now, we use the radical axis theorem on the circumcircles of $\triangle ABM$, $\triangle BCL$, and $AB'CL$ to finish.
This post has been edited 1 time. Last edited by tigerzhang, Nov 4, 2021, 3:20 AM
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OronSH
1718 posts
#19
Y by
solved with orz @GrantStar!!!

Let $N,P,O$ be the midpoint of $BC,$ the midpoint of $CK$ and the circumcenter of $BKJ.$ First we notice $BP$ is also tangent to $(BMC)$ so $PM$ is the $M$ symmedian of $\triangle BMC.$ Now we have $\angle CAJ=180-\angle CMP=\angle NMB=\angle ABJ$ so $J$ is the $B$ humpty point. Then by the tangency $\angle KCM=\angle CBM=\angle JCM,$ and we have $\angle JCK=2\angle MBC=2(90-\angle KBJ)=180-\angle KOJ$ so $KOJC$ is cyclic. However $O$ is on the perpendicular bisector of $JK$ so it is the fact $5$ point so it lies on $AC$ which bisects $\angle JCK.$
This post has been edited 1 time. Last edited by OronSH, Feb 1, 2024, 11:16 PM
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HamstPan38825
8846 posts
#20
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The HM point ends up being sort of a red herring...

We let $D$ be the midpoint of $\overline{CK}$. It suffices to show that $O =\overline{D\infty_{BC}} \cap \overline{AC}$ satisfies $\angle BOD = \angle BJK$ as $\overline{OD}$ bisects $\angle AOK$.

Observe that as $\angle MOD + \angle MBC + \angle CBD = 180^\circ$, it follows that $MBDO$ is cyclic. Then $\angle BOD = \angle BMD = \angle BJK$ as $\overline{MD} \parallel \overline{JK}$ by midline.

Remark: If we instead wrote $\angle BOD = \angle BMD = \angle AJM$, the configuration actually follows from $J$ being the $B$-HM point. Proving this, however, is much more difficult than the problem itself, hence the irony.
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Marinchoo
407 posts
#21
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Introduce $B'$ as the reflection of $B$ over $\overline{AC}$. The condition then naturally rewrites as $KBJB'$ being cyclic, which is equivalent to $\angle B'BM = \angle B'KA$. Now we complex bash with unit circle $(BCM)$. Then $a = 2m - c$, $k = \frac{3bc-c^2}{b+c}$, and $b' = c + m - \frac{cm}{b}$, so $\frac{b'-k}{a-k} = \frac{b-c}{2b}$ and $\frac{b'-b}{m-b} = \frac{b-c}{b}$, done.
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shendrew7
787 posts
#22
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The bulk of this proof is to show that $J$ is the $B$-Humpty point. Consider $\sqrt{ca}$ inversion about $B$ composed with a reflection over the $B$-bisector.
  • $CK \mapsto (ABM)$, so $K$ maps to the point on $(ABM)$ such that $\angle K^*BA = 90$.
  • Define $T = AA \cap CC$, which lies on the $B$-symmedian. Note
    \begin{align*}
\measuredangle BTC &= \measuredangle BAC + \measuredangle ABM \\
&= \measuredangle BMC + 2 \measuredangle ABM \\
&= \measuredangle BK^*A + 2 \measuredangle AK^*M \\
&= \measuredangle BK^*C,
\end{align*}so $T \in (K^*BC)$, implying $T$ is the inverse of $J$, making $J$ the Humpty Point.

If we reflect $K$ over $AC$ to $L$, showing that $JBKL$ is cyclic suffices. From the Humpty Point classification, we get
\[\measuredangle ACJ = \measuredangle CBM = \measuredangle KCA = \measuredangle ACL,\]
so $J$, $C$, and $L$ are collinear. Finally, we the desired from
\[\measuredangle KBJ = 90 + \measuredangle CBJ = 90 + \measuredangle ACL = \measuredangle KLJ. \quad \blacksquare\]
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Saucepan_man02
1294 posts
#23
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Huh, G1? I guess its bit harder for G1:

Here we go (HM Points):

Re-define $J$ to be $B$ HM point, and let $\omega$ denote the $B$ Apollonius circle. Let $X, B'$ denote the reflections of $J, B$ with respect to $AC$ respectively. Note that $AX$ is $A$-symmedian.
Note that: $X, B' \in \omega$. Refine $K = JA \cap \omega$ with $K \neq J$. Note that the circumcenter of $BJK$ lies on $AC$.
It suffice to show $KC$ to be tangent to $(BMC)$ and $\measuredangle KBC=90^\circ$.

Claim:$KC$ is tangent to $(BMC)$
Note that: $\frac{AJ}{JC}=\frac{AK}{KC}$ as both of them lie of the $B$-Apollonius circle. Thus, by Angle-Bisector theorem, $\angle JCA = \angle KCA$. Notice that: $\angle KCA = \angle JCA = \angle JBC = \angle MBC$, and we are done.

Claim:$\measuredangle KBC=90^\circ$
Pure-Angle-Chase:
$$\measuredangle KBJ =\measuredangle KJB+ \measuredangle BKJ = \measuredangle CBA + \measuredangle XBB' =  \measuredangle CBA + \measuredangle OBM = 90^\circ - \measuredangle JBC.$$
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Ilikeminecraft
283 posts
#25
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SO COOOLLLLLL

Let $L$ be second intersection of $(M)$ and $(BMC)$
Let $R$ be second intersection of $BK$ and $(BMC)$
Reflect $J$ across $AC$ to get $N$
Let $T = RM\cap CK$

Clearly:
1. $\angle RBC = 90 $ since $\angle RBC = 180 - \angle CBK = 90$
2. $ALR$ collinear $\angle ALR = \angle ALC + \angle RLC = 180$
3. $RATC$ is a cyclic kite
Now, let $J' = BM\cap LC.$
By Pascal's on $LRBMCC,$ we get $J'AK$ are collinear. Thus, $J = J'$

Finally, note that $\angle KBM = \angle MCR = 90 - \angle MAR= 90 - \angle LCR = \angle CJN.$
Thus, since $AC$ is perp bisector of $JN,$ we are done.

In fact, one can get $AT\parallel LC$, as $\angle CAT = \angle ACT = \angle ART = \angle LCM$
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