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k a March Highlights and 2025 AoPS Online Class Information
jlacosta   0
Mar 2, 2025
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0 replies
jlacosta
Mar 2, 2025
0 replies
Sharygin 2025 CR P2
Gengar_in_Galar   4
N a few seconds ago by FKcosX
Source: Sharygin 2025
Four points on the plane are not concyclic, and any three of them are not collinear. Prove that there exists a point $Z$ such that the reflection of each of these four points about $Z$ lies on the circle passing through three remaining points.
Proposed by:A Kuznetsov
4 replies
Gengar_in_Galar
Mar 10, 2025
FKcosX
a few seconds ago
2 var inquality
sqing   5
N 2 minutes ago by ionbursuc
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$$ a+b+ab\geq2$$
5 replies
+1 w
sqing
Today at 4:06 AM
ionbursuc
2 minutes ago
100 Selected Problems Handout
Asjmaj   32
N 5 minutes ago by John_Mgr
Happy New Year to all AoPSers!
 :clap2:

Here’s my modest gift to you all. Although I haven’t been very active in the forums, the AoPS community contributed to an immense part of my preparation and left a huge impact on me as a person. Consider this my way of giving back. I also want to take this opportunity to thank Evan Chen—his work has consistently inspired me throughout my olympiad journey, and this handout is no exception.



With 2025 drawing near, my High School Olympiad career will soon be over, so I want to share a compilation of the problems that I liked the most over the years and their respective detailed write-ups. Originally, I intended it just as a personal record, but I decided to give it some “textbook value” by not repeating the topics so that the selection would span many different approaches, adding hints, and including my motivations and thought process.

While IMHO it turned out to be quite instructive, I cannot call it a textbook by any means. I recommend solving it if you are confident enough and want to test your skills on miscellaneous, unordered, challenging, high-quality problems. Hints will allow you to not be stuck for too long, and the fully motivated solutions (often with multiple approaches) should help broaden your perspective. 



This is my first experience of writing anything in this format, and I’m not a writer by any means, so please forgive any mistakes or nonsense that may be written here. If you spot any typos, inconsistencies, or flawed arguments whatsoever (no one is immune :blush: ), feel free to DM me. In fact, I welcome any feedback or suggestions.

I left some authors/sources blank simply because I don’t know them, so if you happen to recognize where and by whom a problem originated, please let me know. And quoting the legend: “The ideas of the solution are a mix of my own work, the solutions provided by the competition organizers, and solutions found by the community. However, all the writing is maintained by me.” 



I’ll likely keep a separate file to track all the typos, and when there’s enough, I will update the main file. Some problems need polishing (at least aesthetically), and I also have more remarks to add.

This content is only for educational purposes and is not meant for commercial usage.



This is it! Good luck in 45^2, and I hope you enjoy working through these problems as much as I did!

Here's a link to Google Drive because of AoPS file size constraints: Selected Problems
32 replies
Asjmaj
Dec 31, 2024
John_Mgr
5 minutes ago
Where is the equality?
AndreiVila   2
N an hour 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
an hour ago
Inequality with roots of cubic
Michael Niland   0
Today at 7:23 AM
The equation $x^3+px+q =0$ has real roots $ a_1 ,a_2 , a_3 $ where $a_1 \leq a_2 \leq a_3$.

Similarly the equation $x^3 +rx +s=0$ has real roots $ b_1, b_2, b_3 $ where $b_1 \leq b_2,\leq b_3$.

Prove that if $ \frac{a_1}{b_1} \leq \frac{a_2}{b_2} \leq \frac{a_3}{b_3}$ (s non zero), then $(\frac{p}{r})^3 =(\frac{q}{s})^2$
0 replies
Michael Niland
Today at 7:23 AM
0 replies
Inequalities
nhathhuyyp5c   3
N Today at 6:36 AM by giangtruong13
Let $a,b,c$ be positive reals such that $a+b+c+2=abc$. Find the maximum value of $$\frac{a+1}{a^2+2}+\frac{b+1}{b^2+2}+\frac{c+1}{c^2+2}$$Given $n\ge 2$ non-zero reals $x_1,x_2,\cdots x_n$ such that their sum is $100$. Prove that there exists two numbers $x_i,x_j$ such that $\frac{1}{2}\le \left|\frac{x_i}{x_j}\right|\le 2$
3 replies
nhathhuyyp5c
Jan 10, 2025
giangtruong13
Today at 6:36 AM
Algebra-1
JetFire008   3
N Today at 5:11 AM by BackToSchool
Find real numbers $p$,$q$ if $1+i$ is a root of $x^3+px^2+qx+6=0$ and solve the equation.
3 replies
JetFire008
Yesterday at 3:19 PM
BackToSchool
Today at 5:11 AM
Cool problem
jkim0656   3
N Today at 4:09 AM by RedFireTruck
Hey AoPS!
I came across a problem recently and it goes like this:
Which is greater:
$$2 ^ {100!} $$or $$2^{100}!$$Soo... can u guys help? thx!
:yoda: and may the force be with u!
Notes: I don't exactly know where this problem came from, but if u find that u are the orig maker of this problem feel free to drop me a PM and ill add u to my post :)
3 replies
jkim0656
Today at 3:04 AM
RedFireTruck
Today at 4:09 AM
help me at this pls i'm so confused
myname17042005   12
N Today at 3:07 AM by sqing
if a, b, c are real numbers and a, b, c >0, prove that
pls help meee
12 replies
myname17042005
Jun 12, 2020
sqing
Today at 3:07 AM
Trig Multiplication
szhang7   1
N Today at 2:24 AM by joeym2011
Find the exact value of $(2-\sin^2(\frac{\pi}{7}))(2-\sin^2(\frac{3\pi}{7}))(2-\sin^2(\frac{3\pi}{7}))$.
1 reply
szhang7
Yesterday at 3:50 PM
joeym2011
Today at 2:24 AM
Algebra-2
JetFire008   1
N Today at 2:11 AM by joeym2011
Prove that if $f(x)$ is a polynomial such that $f(x^n)$ is divisible by $x-1$, then $f(x^n)$ is divisible by $x^n-1$.
1 reply
JetFire008
Yesterday at 3:23 PM
joeym2011
Today at 2:11 AM
a/b + b/a never integer ?
MTA_2024   3
N Yesterday at 10:35 PM by ohiorizzler1434
Let $a$ and $b$ be 2 distinct positive integers.
Can $\frac a b +\frac b a $ be in an integer. Prove why ?
3 replies
MTA_2024
Yesterday at 3:08 PM
ohiorizzler1434
Yesterday at 10:35 PM
2014 Community AIME / Marathon ... Algebra Medium #1 quartic
parmenides51   5
N Yesterday at 7:17 PM by CubeAlgo15
Let there be a quartic function $f(x)$ with maximums $(4,5)$ and $(5,5)$. If $f(0) = -195$, and $f(10)$ can be expressed as $-n$ where $n$ is a positive integer, find $n$.

proposed by joshualee2000
5 replies
parmenides51
Jan 21, 2024
CubeAlgo15
Yesterday at 7:17 PM
interesting problem
sausagebun   1
N Yesterday at 4:05 PM by mathprodigy2011
Six points, labeled A, B, C, D, E, and F, are positioned consecutively on a straight line. Let G be a point not located on this line. The following distances are given: AC = 26, BD = 22, CE = 31, DF = 33, AF = 73, CG = 40, and DG = 30. Determine the area of triangle BGE.
I brute forced this with trig, was wondering if theres a more elegant way of doing this
1 reply
sausagebun
Yesterday at 3:21 PM
mathprodigy2011
Yesterday at 4:05 PM
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
611 posts
#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
611 posts
#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
62 posts
#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
22 posts
#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
22 posts
#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
958 posts
#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
12 posts
#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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