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

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Be sure to mark your calendars for the following events:
[list][*]March 5th (Wednesday), 4:30pm PT/7:30pm ET, HCSSiM Math Jam 2025. Amber Verser, Assistant Director of the Hampshire College Summer Studies in Mathematics, will host an information session about HCSSiM, a summer program for high school students.
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[*]March 11th (Tuesday), 4:30pm PT/7:30pm ET, 2025 MATHCOUNTS Chapter Discussion MATH JAM. AoPS instructors will discuss some of their favorite problems from the MATHCOUNTS Chapter Competition. All are welcome!
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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
Where is the equality?
AndreiVila   2
N 19 minutes ago by MihaiT
Source: Romanian District Olympiad 2025 9.3
Determine all positive real numbers $a,b,c,d$ such that $a+b+c+d=80$ and $$a+\frac{b}{1+a}+\frac{c}{1+a+b}+\frac{d}{1+a+b+c}=8.$$
2 replies
AndreiVila
Mar 8, 2025
MihaiT
19 minutes ago
Truth or lie at a table
SinaQane   7
N 26 minutes ago by Oksutok
Source: 239 2019 S4
There are $n>1000$ people at a round table. Some of them are knights who always tell the truth, and the rest are liars who always tell lies. Each of those sitting said the phrase: “among the $20$ people sitting clockwise from where I sit there are as many knights as among the $20$ people seated counterclockwise from where I sit”. For what $n$ could this happen?
7 replies
SinaQane
Jul 31, 2020
Oksutok
26 minutes ago
2 var inquality
sqing   4
N 27 minutes ago by sqing
Source: Own
Let $ a ,  b\geq 0 $ and $ \frac{1}{a^2+1}+\frac{1}{b^2+1}\le   \frac{3}{2}. $ Show that$$ a+b+ab\geq1$$Let $ a ,  b\geq 0 $ and $ \frac{1}{a^2+1}+\frac{1}{b^2+1}\le   \frac{5}{6}. $ Show that$$ a+b+ab\geq2$$
4 replies
sqing
Today at 4:06 AM
sqing
27 minutes ago
Functional inequality on N
BartSimpsons   21
N 28 minutes ago by Tony_stark0094
Source: European Mathematical Cup 2017 Problem 1
Find all functions $f:\mathbb{N}\rightarrow \mathbb{N}$ such that the inequality $$f(x)+yf(f(x))\le x(1+f(y))$$holds for all positive integers $x, y$.

Proposed by Adrian Beker.
21 replies
BartSimpsons
Dec 27, 2017
Tony_stark0094
28 minutes ago
A colouring game on a rectangular frame
Tintarn   0
33 minutes ago
Source: Bundeswettbewerb Mathematik 2025, Round 1 - Problem 4
For integers $m,n \ge 3$ we consider a $m \times n$ rectangular frame, consisting of the $2m+2n-4$ boundary squares of a $m \times n$ rectangle.

Renate and Erhard play the following game on this frame, with Renate to start the game. In a move, a player colours a rectangular area consisting of a single or several white squares. If there are any more white squares, they have to form a connected region. The player who moves last wins the game.

Determine all pairs $(m,n)$ for which Renate has a winning strategy.
0 replies
Tintarn
33 minutes ago
0 replies
Find the value
sqing   5
N 33 minutes ago by sqing
Source: Own
Let $a,b,c$ be distinct real numbers such that $ \frac{a^2}{(a-b)^2}+ \frac{b^2}{(b-c)^2}+ \frac{c^2}{(c-a)^2} =1. $ Find the value of $\frac{a}{a-b}+ \frac{b}{b-c}+ \frac{c}{c-a}.$
Let $a,b,c$ be distinct real numbers such that $\frac{a^2}{(b-c)^2}+ \frac{b^2}{(c-a)^2}+ \frac{c^2}{(a-b)^2}=2. $ Find the value of $\frac{a}{b-c}+ \frac{b}{c-a}+ \frac{c}{a-b}.$
Let $a,b,c$ be distinct real numbers such that $\frac{(a+b)^2}{(a-b)^2}+ \frac{(b+c)^2}{(b-c)^2}+ \frac{(c+a)^2}{(c-a)^2}=2. $ Find the value of $\frac{a+b}{a-b}+\frac{b+c}{b-c}+ \frac{c+a}{c-a}.$
5 replies
sqing
Today at 5:02 AM
sqing
33 minutes ago
A touching question on perpendicular lines
Tintarn   0
34 minutes ago
Source: Bundeswettbewerb Mathematik 2025, Round 1 - Problem 3
Let $k$ be a semicircle with diameter $AB$ and midpoint $M$. Let $P$ be a point on $k$ different from $A$ and $B$.

The circle $k_A$ touches $k$ in a point $C$, the segment $MA$ in a point $D$, and additionally the segment $MP$. The circle $k_B$ touches $k$ in a point $E$ and additionally the segments $MB$ and $MP$.

Show that the lines $AE$ and $CD$ are perpendicular.
0 replies
Tintarn
34 minutes ago
0 replies
The last nonzero digit of factorials
Tintarn   0
36 minutes ago
Source: Bundeswettbewerb Mathematik 2025, Round 1 - Problem 2
For each integer $n \ge 2$ we consider the last digit different from zero in the decimal expansion of $n!$. The infinite sequence of these digits starts with $2,6,4,2,2$. Determine all digits which occur at least once in this sequence, and show that each of those digits occurs in fact infinitely often.
0 replies
Tintarn
36 minutes ago
0 replies
Fridolin just can't get enough from jumping on the number line
Tintarn   0
38 minutes ago
Source: Bundeswettbewerb Mathematik 2025, Round 1 - Problem 1
Fridolin the frog jumps on the number line: He starts at $0$, then jumps in some order on each of the numbers $1,2,\dots,9$ exactly once and finally returns with his last jump to $0$. Can the total distance he travelled with these $10$ jumps be a) $20$, b) $25$?
0 replies
Tintarn
38 minutes ago
0 replies
D1016 : A strange result about the palindrom polynomials
Dattier   0
an hour ago
Source: les dattes à Dattier
Let $Q\in \{-1,1,0\}[x]$ with $Q(1)=0$.

Is it true that $\exists Q \in \mathbb Z[x]$ palindrom with $Q | P$ ?
0 replies
Dattier
an hour ago
0 replies
EMC last problem
GreekIdiot   1
N an hour ago by Tintarn
Source: EMC 2024 P4, Ioannis Galamatis
Find all functions $f: \mathbb{R_+} \rightarrow \mathbb{R_+}$ such that $f(x+yf(x))=xf(y+1) \: \forall\: x, \:y \in \mathbb{R_+}$.
Note: With the symbol $\mathbb{R_+}$ we denote the set of positive real numbers.
1 reply
GreekIdiot
Yesterday at 7:02 PM
Tintarn
an hour ago
Functional equation with a parameter
Tintarn   8
N an hour ago by NicoN9
Source: Baltic Way 2024, Problem 1
Let $\alpha$ be a non-zero real number. Find all functions $f: \mathbb{R}\to\mathbb{R}$ such that
\[
xf(x+y)=(x+\alpha y)f(x)+xf(y)
\]for all $x,y\in\mathbb{R}$.
8 replies
Tintarn
Nov 16, 2024
NicoN9
an hour ago
D1015 : A strange EF for polynomials
Dattier   1
N an hour ago by Dattier
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.
1 reply
Dattier
Yesterday at 8:37 PM
Dattier
an hour ago
Functional Equations Marathon March 2025
Levieee   19
N an hour ago by Jupiterballs
1. before posting another problem please try your best to provide the solution to the previous solution because we don't want a backlog of many problems
2.one is welcome to send functional equations involving calculus (mainly basic real analysis type of proofs) as long it is of the form $\text{"find all functions:"}$
19 replies
Levieee
Today at 1:03 AM
Jupiterballs
an hour ago
IMO 2012 P5
mathmdmb   122
N Today at 2:51 AM by KevinYang2.71
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
122 replies
mathmdmb
Jul 11, 2012
KevinYang2.71
Today at 2:51 AM
IMO 2012 P5
G H J
Source: IMO 2012 P5
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mathmdmb
1547 posts
#1 • 30 Y
Y by Pirshtuk, siavosh, buratinogigle, Potla, KOSNITA, Amir Hossein, samuelbf, Mathematicalx, Scorpion.k48, tenplusten, Davi-8191, anantmudgal09, Wizard_32, nguyendangkhoa17112003, XbenX, Gumnaami_1945, Richangles, amar_04, Adventure10, Mango247, home245, GeoKing, Rounak_iitr, Tastymooncake2, crazyeyemoody907, ItsBesi, Sedro, and 3 other users
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
This post has been edited 3 times. Last edited by Eternica, Jun 19, 2024, 10:03 AM
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Concyclicboy
49 posts
#2 • 17 Y
Y by siavosh, buratinogigle, Swistak, armann, ZacPower123, myh2910, Jalil_Huseynov, Adventure10, Mango247, Tastymooncake2, Sedro, and 6 other users
I think:
1° AL^2=AD*AB then <ALD=<LBA and a tangent to circuncircles is obviously
2° Let F the intersection of CD and the perpendicular to AL through L, then FLDA is cyclic and <DFA=<DLA=<LBA
3 If we call T the intersection of BX with AF then BFTD is cyclic and X is orthocenter with AX perpendicular to BF

By similar way the perpendicular for BK through K passes through F, and if U is the intersection of AX with BF we have: FL^2=FT*FA=FU*FB=FK^2, impliying that LM=KM
This post has been edited 1 time. Last edited by Concyclicboy, Jul 12, 2012, 1:19 AM
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Aryan5000
2 posts
#3 • 13 Y
Y by siavosh, ropro01, buratinogigle, Swistak, Algie, myh2910, Adventure10, Tastymooncake2, and 5 other users
MK and ML are both tangents to a circle
let BX meets the circle(A,AC)at J And AX meets circle(B,BC)at I(circle (c,r)means circle with center c and radius r)
easily we can find that JKLI is cyclic Let name this that have these four points on itself circle C2
easily we can find that BK is tangent to C2 (because we have BC*BC=Bl*BJ and we have BC =BK
so BK*BK=BL*BJ)
with a similar work we can see that AL is tangent to the C2 too.
So MK and ML are both tangents to the C2 SO MK=ML
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Potla
1886 posts
#4 • 14 Y
Y by buratinogigle, diablo901, mathbuzz, Sx763_, biomathematics, Gumnaami_1945, Adventure10, Tastymooncake2, MS_asdfgzxcvb, and 5 other users
Obviously the perpendiculars through $K, L$ and to $BK, AL$ intersect at a point $E$ which lies on $CD,$ which is also the intersection of $\odot BDK$ and $\odot ADL.$ So, by some angle chasing, it follows that $\angle EKD=\angle AXE;$ meaning that $EK$ is tangent to $\odot KDX.$ So, by power of $E$, it follows that $EK^2=ED\cdot EX=EL^2,$ because $EL$ is also tangent to $\odot LDX.$ Therefore $EK=EL,$ and $MKE\cong MLE,$ leading to $MK=ML.\Box$
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buratinogigle
2314 posts
#5 • 10 Y
Y by silly_mouse, phuongtheong, p1a, PARISsaintGERMAIN, Adventure10, Mango247, Tastymooncake2, and 3 other users
Let me propose this general problem

Let two circles $(O_1)$ and $(O_2)$ intersect at $A,B$. $C,D$ are on line $O_1O_2$ such that $AC\perp O_1A$ and $AD\perp O_2A$. $P$ is a point on $AB$. $CP$ cut $(O_1)$ at $L$ such that $C,L$ don't have same side with $AB$. $DP$ cut $(O_2)$ at $K$ such that $D,K$ don't have same side with $AB$. $LO_1$ cuts $KO_2$ at $M$. Prove that $MK=ML$.
Attachments:
Figure856.pdf (6kb)
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Goutham
3130 posts
#6 • 12 Y
Y by buratinogigle, p1a, Wizard_32, Adventure10, Mango247, Tastymooncake2, and 6 other users
Let circumcircle of $ADL$ intersect $DC$ at $U$. Then $\angle AUD=\angle ALD$. Also, $AL^2=AC^2=AD\cdot AB$ and so, $LA$ is antiparallel to $BD$ which means, $\angle AUD=\angle LBD=\angle XBD$. Now, this implies $UAD\sim BXD$. This implies $\frac{XD}{AD}=\frac{BD}{UD}\Longrightarrow \frac{UD}{BD}=\frac{AD}{XD}$. So, $UDB\sim AXD$. This means $\angle BUD=\angle DAX$ but by similar reasonings, $\angle DAX=\angle DKB$. So, $BDKU$ is cyclic. Since in circles $ADLU$ and $BDKU$, we have $AU, BU$ as diameters, we also have $\angle ALU=\angle BKU=\frac{\pi}{2}$. But $U$ also lies on $CD$. So, the perpendiculars from $K, L, D$ to $BM, AM, AB$ concur at a point, $U$. By carnot's theorem, $BK^2-KM^2+ML^2-LA^2+AD^2-DB^2=0$. Using $AL=AC, BK=BC, AD^2-DB^2=(AD^2+DC^2)-(DC^2+DB^2)=AC^2-CB^2$, we have $MK=ML$ as required.
This post has been edited 1 time. Last edited by Goutham, Jul 11, 2012, 9:56 PM
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vladimir92
212 posts
#7 • 7 Y
Y by diablo901, buratinogigle, p1a, Adventure10, Mango247, Tastymooncake2, and 1 other user
Let AX and BX meet the circumcircle of ABC at V and U respectively, AU and BV meet at W, clearly X is the orthocenter of WAB. See that AL²=AD.AB=AX.AV, then A,D,L,V,W are concyclic, similarly B,D,K,U,W are concyclic too. See that WLV~WBL and WKU~WAK, it follows that WL=WK, but WLKM cyclic, hence ML=MK, As desired.
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Goutham
3130 posts
#8 • 9 Y
Y by buratinogigle, p1a, PARISsaintGERMAIN, IFA, Adventure10, Mango247, Tastymooncake2, and 2 other users
buratinogigle wrote:
Let me propose this general problem

Let two circles $(O_1)$ and $(O_2)$ intersect at $A,B$. $C,D$ are on line $O_1O_2$ such that $AC\perp O_1A$ and $AD\perp O_2A$. $P$ is a point on $AB$. $CP$ cut $(O_1)$ at $L$ such that $C,L$ don't have same side with $AB$. $DP$ cut $(O_2)$ at $K$ such that $D,K$ don't have same side with $AB$. $LO_1$ cuts $KO_2$ at $M$. Prove that $MK=ML$.
Let $AB$ intersect $CD$ at $S$. Then, circumcircle of $O_1SL$ intersect $AB$ at $U$. So, $\angle O_1US=\angle O_1LS$. Also, $O_1L^2=O_1A^2=O_1S\cdot O_1C$ and so, $O_1L$ is tangent to circumcircle of $LSC\Longrightarrow \angle O_1LS=\angle LCS=\angle PCS$. So, $PCS\sim O_1US\Longrightarrow \frac{PS}{O_1S}=\frac{CS}{US}$. Now, $O_1AS\sim ACS\Longrightarrow AS^2=O_1S\cdot CS$. Similarly, $AS^2=O_2S\cdot DS$. This implies $\frac{O_1S}{O_2S}=\frac{DS}{CS}$. Multiplying this with $ \frac{PS}{O_1S}=\frac{CS}{US}$, we have $\frac{PS}{O_2S}=\frac{DS}{US}\Longrightarrow PDS\sim O_2US$. So, $\angle O_2US=\angle PDS$. We also have $O_2K^2=O_2A^2=O_2S\cdot O_2D$ and hence, $O_2K$ is tangent to cirumcircle of $KSD\Longrightarrow \angle PDS=\angle KDS=\angle O_2KS$. This means $O_2KUS$ is cyclic. In circles $O_1LUS$ and $O_2KUS$, we have $O_1U$ and $O_2U$ as diameters and so, $\angle O_1LU=\angle O_2KU=\frac{\pi}{2}$. $K$ lies on $AB$. The perpendiculars to sides $MO_1, MO_2, O_1O_2$ passing through $L, K, S$ concur at $U$. Hence, by carnot's theorem, $ML^2-LO_1^2+O_1S^2-SO_2^2+O_2K^2-KM^2=0$. We have $O_1S^2-O_2S^2=(O_1S^2+SA^2)-(SA^2+O_2S^2)=O_1A^2-O_2A^2=O_1L^2-O_2K^2$. Using this, we have $MK=ML$ as desired.

When $\angle O_1AO_2=\frac{\pi}{2}$, then $D, C$ coincide with $O_1, O_2$ respectively. In that case, it can be seen that we get the IMO problem.
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pavel kozlov
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#9 • 2 Y
Y by Adventure10, Mango247
Another addition to this problem:
Prove that the bisector of $\angle{ACB}$ and the lines $XM$ and $AB$ meet each other at one point.
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pavel kozlov
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#10 • 3 Y
Y by Goutham, Adventure10, Mango247
I very like the solution of Goutham with Carnot's theorem. Very effective finalization really!
By the way: accurate calculation in Cartesian coordinates together with the sines theorem applied to triangles $AKM, BLM, ALX, BKX$ (that trick excludes calculation of the point $M$) gives desire result in half an hour.
This post has been edited 1 time. Last edited by pavel kozlov, Jul 14, 2012, 10:40 PM
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ropro01
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#11 • 11 Y
Y by dizzy, Hantaehee, TheMathsBoy, Jalil_Huseynov, Adventure10, Mango247, Tastymooncake2, Rounak_iitr, and 3 other users
Very nice problem, my favorite one this year :)

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This post has been edited 1 time. Last edited by ropro01, Jul 13, 2012, 9:44 PM
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Orin
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#12 • 6 Y
Y by numbertheorist17, Adventure10, Mango247, Tastymooncake2, and 2 other users
First we construct three circles:the circumcircles of $\triangle ABC(\Gamma),$,The circle($\alpha$) with centre $A$ and radius $AC$ and the circle($\beta$) with centre $B$ and radius $BC$.
Note that the centre of $\Gamma$ lies on the midpoint of $AB$,so the three circles are co-axial with radical axis $CD$.
Let $AX \cap \Gamma =Y \neq A,BX \cap \Gamma =Z \neq B,AY \cap BZ =P$.
so $\angle AYB =\angle AZB=90^{\circ}$,so $X$ must be the orthocenter of $\triangle ABP$ implying $P$ lies on the radical axis.
let us denote $PWR(P)$ by the power of $P$ wrt $\alpha,\beta,\Gamma$.[which are equal]
From similar triangles $ABC,ACD$ we get $AC^2=AD*AB=AL^2 \Rightarrow \angle ALD = \angle ABL$
similarly $\angle BKD = \angle BAK$
now $\angle APD=\angle ABZ=\angle ALZ$ implying $ADLP$ is cyclic.
so $AL \bot PL$.....(1) meaning $PL^2=PWR(P)$.
similarly we get $BK \bot PK$.....(2) meaning $PK^2=PWR(P)$
hence $PK=PL$.....(3)
so $\triangle MKP \cong MLP$[using (1),(2),(3)]
proving $MK=ML$
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Jeroen
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#13 • 6 Y
Y by buratinogigle, Swistak, Adventure10, Mango247, Tastymooncake2, and 1 other user
This was the NLD6's solution:

$E$ is the reflection of $C$ in $AB$. It lies on the circle $\Gamma$ with centre $B$ en radius $BC$, as does $K$. The line $AK$ intersects $\Gamma$ a second time in $K'$ and the tangent in $K$ to $\Gamma$ intersects $CD$ in $S$. The tangents in $C$ and $E$ to $\Gamma$ intersect in $A$ en $A$ is on $KK'$, so $CEKK'$ is a harmonic quadrilateral. This yields $-1=(CEKK')=K(CEKK')=K(CESX)=(CESX)$.

If we define $T$ to be the point of intersection of $CD$ and the tangent in $L$ to the circle with centre $A$ and radius $AC$, we similarly find $(CETX)=-1$. From this follows $S=T$ and the power of a point now yields $SK^2=SC\cdot SE=SL^2$. So $SK=SL$ and now we have congruent triangles $SKM$ and $SLM$, which yields $KM=LM$.
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proglote
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#14 • 11 Y
Y by buratinogigle, TheStrangeCharm, like123, Roct-7, Adventure10, Mango247, Tastymooncake2, and 4 other users
Denote by $\mathcal{C}_A$ the circle centered at $A$ passing through $C$, and define $\mathcal{C}_B$ similarly. Suppose $AK$ intersects $\mathcal{C}_B$ again at $T_A$, and define $T_B$ similarly. Observe that $X$ is in the radical axis $\ell$ of $\mathcal{C}_A, \mathcal{C}_B$, so $XL \cdot XT_B = XK \cdot XT_A \implies KLT_AT_B$ cyclic with circumcircle $\omega$. Note that $\ell$ is the polar of $A$ w.r.t. $\mathcal{C}_B$, so $(A, X; K, T_A)$ is harmonic $\implies X$ is in the polar $a$ of $A$ w.r.t. $\omega$. But $AL = AT_B \implies a \parallel LT_B \implies a \equiv LT_B.$ Similarly $b \equiv KT_A.$

So $AL, BK$ are tangent to $\omega.$ So $MK, ML$ are tangent to $\omega.$ So done.
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Neerjhor
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#15 • 6 Y
Y by p1a, Jalil_Huseynov, Adventure10, Tastymooncake2, and 2 other users
I found this problem easy. I'm very confused as everyone saying it's too hard. Please tell me if there's a bug. :S

Let $AX$ and $BX$ intersect the circumcircle $O_1$ of $ABC$ at $P$ and $Q$. Let $AQ$ and $BP$ intersect at $S$. By Radical Axis theorem (I mean Radical Centre), $S$ lies on $CD$. Now, triangle $ABC$ and $ACD$ are similar. So, $AC^2=AB.AD=AL^2$. Again, as quadrilateral $SQDB$ is cyclic, so $AD.AB=AQ.AS=AL^2$. This means that $AL$ is a tangent to the circumcircle of triangle $SQL$ at $L$. And so, $AL$ is perpendicular to $SL$ as $SQL=90^0$. Similarly, $BK$ is perpendicular to $SK$. And $SK=SL$ because they lie on the radical axis of circle $(A,AC)$ and circle $(B,BC)$ and are tangents to them. So, if we draw a circle with centre $S$ and radius $SK$ then it would go through $K,L$ and as $<MKS=90^0$ and $<MLS=90^0$, it implies $MK=ML$.

I hope I got it right. :/
This post has been edited 1 time. Last edited by Neerjhor, Jul 13, 2012, 1:33 PM
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