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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.

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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
Functional equations
mathematical-forest   2
N 3 minutes ago by jasperE3
Find all funtion $f:C\to C$, s.t.$\forall x \in C$
$$xf(x)=\overline{x} f(\overline{x})$$
2 replies
mathematical-forest
2 hours ago
jasperE3
3 minutes ago
22 light bulbs
dangerousliri   2
N 4 minutes ago by Dontknow4608
Source: Kosovo Mathematical Olympiad 2022, Grade 11, Problem 1
$22$ light bulbs are given. Each light bulb is connected to exactly one switch, but a switch can be connected to one or more light bulbs. Find the least number of switches we should have such that we can turn on whatever number of light bulbs.
2 replies
dangerousliri
Mar 6, 2022
Dontknow4608
4 minutes ago
Interesting inequalities
sqing   2
N 5 minutes ago by SunnyEvan
Source: Own
Let $ a,b,c,d\geq  0 , a+b+c+d \leq 4.$ Prove that
$$a(kbc+bd+cd)  \leq \frac{64k}{27}$$$$a (b+c) (kb c+  b d+  c d) \leq \frac{27k}{4}$$Where $ k\geq 2. $
2 replies
sqing
2 hours ago
SunnyEvan
5 minutes ago
Circles, Tangents, Variable point on BC
SerdarBozdag   4
N 6 minutes ago by ErTeeEs06
Source: DAMO P5
In triangle $ABC$, $D$ is an arbitrary point on $BC$. $(ADC), (ADB)$ cuts $AB$, $AC$ at $F$ and $E$ respectively. Tangents at $B$ and $C$ intersect at $X$. $Z=EF \cap BX$ and $Y=EF \cap CX$. $P$ is a point on $(ABC)$ such that $AP, YZ, BC$ are concurrent. Prove that $P$ lies on $(XYZ)$

Proposed by SerdarBozdag and k12byda5h
4 replies
SerdarBozdag
Mar 19, 2022
ErTeeEs06
6 minutes ago
functional equation with exponentials
produit   4
N 9 minutes ago by jasperE3
Find all solutions of the real valued functional equation:
f(\sqrt{x^2+y^2})=f(x)f(y).
Here we do not assume f is continuous
4 replies
produit
2 hours ago
jasperE3
9 minutes ago
functional inequality with equality
miiirz30   1
N 11 minutes ago by MR.1
Source: 2025 Euler Olympiad, Round 2
Find all functions \( f : \mathbb{R} \to \mathbb{R} \) such that the following two conditions hold:

1. For all real numbers $a$ and $b$ satisfying $a^2 + b^2 = 1$, We have $f(x) + f(y) \geq f(ax + by)$ for all real numbers $x, y$.

2. For all real numbers $x$ and $y$, there exist real numbers $a$ and $b$, such that $a^2 + b^2 = 1$ and $f(x) + f(y) = f(ax + by)$.

Proposed by Zaza Melikidze, Georgia
1 reply
miiirz30
5 hours ago
MR.1
11 minutes ago
Moving stones on an infinite row
miiirz30   1
N 14 minutes ago by genius_007
Source: 2025 Euler Olympiad, Round 2
We are given an infinite row of cells extending infinitely in both directions. Some cells contain one or more stones. The total number of stones is finite. At each move, the player performs one of the following three operations:

1. Take three stones from some cell, and add one stone to the cells located one cell to the left and one cell to the right, each skipping one cell in between.

2. Take two stones from some cell, and add one stone to the cell one cell to the left, skipping one cell and one stone to the adjacent cell to the right.

3. Take one stone from each of two adjacent cells, and add one stone to the cell to the right of these two cells.

The process ends when no moves are possible. Prove that the process always terminates and the final distribution of stones does not depend on the choices of moves made by the player.

IMAGE

Proposed by Luka Tsulaia, Georgia
1 reply
miiirz30
4 hours ago
genius_007
14 minutes ago
Interesting inequalities
sqing   8
N 24 minutes ago by sqing
Source: Own
Let $ a,b,c,d\geq  0 , a+b+c+d \leq 4.$ Prove that
$$a(bc+bd+cd)  \leq \frac{256}{81}$$$$ ab(a+2c+2d ) \leq \frac{256}{27}$$$$  ab(a+3c+3d )  \leq \frac{32}{3}$$$$ ab(c+d ) \leq \frac{64}{27}$$
8 replies
sqing
Yesterday at 1:25 PM
sqing
24 minutes ago
2000 KJMO P1 easy euclidean lemma
RL_parkgong_0106   4
N 29 minutes ago by JH_K2IMO
Source: KJMO 2000
For arbitrary natural number $a$, show that $\gcd(a^3+1, a^7+1)=a+1$.
4 replies
RL_parkgong_0106
Jun 29, 2024
JH_K2IMO
29 minutes ago
Sum floor 2^k/3 from k=0 to 100
v_Enhance   8
N 29 minutes ago by MathIQ.
Source: All-Russian MO 2000
Evaluate the sum \[ \left\lfloor \frac{2^0}{3} \right\rfloor  + \left\lfloor \frac{2^1}{3} \right\rfloor + \left\lfloor \frac{2^2}{3} \right\rfloor  + \cdots + \left\lfloor \frac{2^{1000}}{3} \right\rfloor. \]
8 replies
v_Enhance
Dec 30, 2012
MathIQ.
29 minutes ago
Points on a lattice path lies on a line
navi_09220114   3
N 30 minutes ago by Ianis
Source: TASIMO 2025 Day 1 Problem 3
Let $S$ be a nonempty subset of the points in the Cartesian plane such that for each $x\in S$ exactly one of $x+(0,1)$ or $x+(1,0)$ also belongs to $S$. Prove that for each positive integer $k$ there is a line in the plane (possibly different lines for different $k$) which contains at least $k$ points of $S$.
3 replies
navi_09220114
May 19, 2025
Ianis
30 minutes ago
Graph Theory
ABCD1728   1
N 36 minutes ago by ABCD1728
Can anyone provide the PDF version of "Graphs: an introduction" by Radio Bumbacea (XYZ press), thanks!
1 reply
ABCD1728
Today at 5:32 AM
ABCD1728
36 minutes ago
Inspired by Crux 4975
sqing   4
N an hour ago by sqing
Source: Own
Let $ a,b,c\geq 0 $ and $ a(b +c +bc)=1.$ Prove that
$$a^2+b^2+c^2+ 3abc(a+b+c)\geq 2 $$
4 replies
sqing
Oct 18, 2024
sqing
an hour ago
CMI Entrance 19#6
bubu_2001   7
N an hour ago by Apple_maths60
$(a)$ Compute -
\begin{align*}
\frac{\mathrm{d}}{\mathrm{d}x} \bigg[ \int_{0}^{e^x} \log ( t ) \cos^4 ( t ) \mathrm{d}t \bigg]
\end{align*}$(b)$ For $x > 0 $ define $F ( x ) = \int_{1}^{x} t \log ( t ) \mathrm{d}t . $

$1.$ Determine the open interval(s) (if any) where $F ( x )$ is decreasing and all the open interval(s) (if any) where $F ( x )$ is increasing.

$2.$ Determine all the local minima of $F ( x )$ (if any) and all the local maxima of $F ( x )$ (if any) $.$
7 replies
bubu_2001
Nov 1, 2019
Apple_maths60
an hour ago
angle BAS = angle CAM
Ovchinnikov Denis   14
N Nov 8, 2024 by mcmp
Source: All-Russian Olympiad 2010 grade 10 P-3
Let $O$ be the circumcentre of the acute non-isosceles triangle $ABC$. Let $P$ and $Q$ be points on the altitude $AD$ such that $OP$ and $OQ$ are perpendicular to $AB$ and $AC$ respectively. Let $M$ be the midpoint of $BC$ and $S$ be the circumcentre of triangle $OPQ$. Prove that $\angle BAS =\angle CAM$.
14 replies
Ovchinnikov Denis
Sep 9, 2010
mcmp
Nov 8, 2024
angle BAS = angle CAM
G H J
Source: All-Russian Olympiad 2010 grade 10 P-3
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Ovchinnikov Denis
470 posts
#1 • 3 Y
Y by HWenslawski, Adventure10, Mango247
Let $O$ be the circumcentre of the acute non-isosceles triangle $ABC$. Let $P$ and $Q$ be points on the altitude $AD$ such that $OP$ and $OQ$ are perpendicular to $AB$ and $AC$ respectively. Let $M$ be the midpoint of $BC$ and $S$ be the circumcentre of triangle $OPQ$. Prove that $\angle BAS =\angle CAM$.
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aboojiga
43 posts
#2 • 4 Y
Y by NHN, Everything999, Adventure10, and 1 other user
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kyyuanmathcount
1973 posts
#3 • 4 Y
Y by Adventure10, Mango247, and 2 other users
Easily with angles, we see that OPQ is similar to ABC in that orientation --> Let perpendicular from S to AD be ST.

Note from simple angles that <SOA = 90, so AOST is cyclic and <SAT = <SOT = <OAM.

It follows that <BAS = <BAD + <SAT = <OAC + <OAM = <CAM.
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erfan_Ashorion
102 posts
#4 • 2 Y
Y by Adventure10, Mango247
oh its really nice problem but....!
suppose $op$ intersect the $AB$ at $Y$ and $OQ$ intersect $AC$ at $x$.
it is really easy to show that $XY$ is parallel to $BC$ so if midpoint of $XY$ is $N$ we know that $XAN=CAM$
lemma:
in the triangle $AXY$ if the tangent to the circumcircle of it on $X$ and $Y$ intersect each other on $Z$ than $AZ$ concides with a symmedian of $AXY$ ..!
proof lemma:
on attachment!!
so it easy to show that $S$ lie on $AZ$..!(not so easy :D if you need i will say it :lol: )
Attachments:
lemma.pdf (196kb)
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sahadian
112 posts
#5 • 1 Y
Y by Adventure10
It is enough if we prove that $\angle DAM=\angle OAS$ SO we prove that the triangles $DAM$ and $AOS$ are similar .
for proving this first we prove $\angle AOS=\angle ADM=90^{\circ}$
then it is easy to prove that $\frac{AD}{DM}=\frac{AO}{OS}$ by using this:
$DM=\frac{AC^2-AB^2}{2BC}$
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junioragd
314 posts
#6 • 2 Y
Y by Adventure10, Mango247
Let the tangent at $A$ to the circumcircle of $ABC$ intersects $BC$ at $D$.Now,by an easy angle chase and using that $MOAD$ is cyclic,it is enough to prove that $DBO$ is similar to $APS$ and we have that angles $DBO$ and $APS$ are equal,so it is enough to prove that $AP/PS=BD/OB$ but this is easy cause $AOP$ is similar to $ADB$ and $OAB$ is similar to $OPS$,so done.
This post has been edited 1 time. Last edited by junioragd, Jan 27, 2015, 5:36 PM
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TelvCohl
2312 posts
#7 • 3 Y
Y by amar_04, Adventure10, Mango247
My solution:

Let $ N $ be the projection of $ O $ on $ PQ $ .

Since $ OP \perp AB, OQ \perp AC, PQ \perp BC $ ,
so we get $ \triangle OPQ \cup S \cup N \sim \triangle ABC \cup O \cup D $ and $ \angle AOS=90^{\circ} $ ,
hence from $ AO:OS=AD:ON=AD:DM \Longrightarrow Rt \triangle AOS \sim Rt \triangle ADM $ ,
so we get $ \angle DAM=\angle OAS \Longrightarrow \angle BAS=\angle CAM $ .

Q.E.D
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djmathman
7938 posts
#8 • 1 Y
Y by Adventure10
WLOG let $AB<AC$, so that $O$ is in the interior of $\triangle ADC$. Let $M_C=OP\cap AB$ and $M_B=OQ\cap AC$. Note that by angle-chasing $\angle QPO=\angle APM_C=\angle ABD$ and $\angle PQO\equiv\angle AQM_B=\angle ACD$, so $\triangle ABC\sim\triangle OPQ$. Hence there exists a spiral similarity sending $\triangle ABC$ to $\triangle OPQ$. Note that this sends $PQ$ to $BC$, so the angle of rotation of this spiral similarity is $90^\circ$. Thus, since $A$ is sent to $O$ and $O$ is sent to $S$, we have $AO\perp OS$.

Now let $X$ be the orthogonal projection of $O$ onto $PQ\equiv AD$. Remark that by similarity and the fact that $MOXD$ is a rectangle, \[\dfrac{MD}{DA}=\dfrac{OX}{AD}=\dfrac{AO}{OS},\quad\text{so}\quad \triangle AOS\sim\triangle ADM.\]Thus $\angle OAS=\angle MAD$. Combining this with $\angle BAD=\angle CAO$ yields the desired equality. $\blacksquare$
This post has been edited 2 times. Last edited by djmathman, Dec 12, 2015, 10:01 PM
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Kayak
1298 posts
#9 • 1 Y
Y by Adventure10
Let $M_B, M_C$ be the midpoint of $\overline{AC}, \overline{AB}$ respectively. Now $\angle PQO = 90 - \angle QAM_C = \angle ACB$, similarly $\angle QPO = \angle ABC$ and thus $\Delta ABC \sim \Delta OPQ$. Let $X$ be the Miquel point of $\Delta M_BM_CD$ w.r.t $\Delta ABC$, and observe $X$ is the center of spiral similarity $f$ mapping $(A,B,C,O,X) \overset{f}{\mapsto} (O,P,Q,S,X)$. Since $f(\overline{PQ}) = \overline{BC}$, $f$ rotates things by $90$. So $\angle SXO = 90$, and thus $\angle AXS = \angle AXO + \angle SXO = 90 + 90  = 180$, and thsu $A,X,S$ are colinear.

Now let $\Phi(Y)$ denote the image of $Y$ after a $\sqrt{bc}$ inversion, then a reflection along $A$-angle bisector, and then a homothety at $A$ with factor $\frac{1}{2}$. We have $\Phi(M_B) = C, \Phi(M_C) = B, \Phi(B) = M_C, \Phi(C) = M_B, \Phi(D) = O$, and thus clearly $\Phi(E) = M$ ('coz $M_C,O,M,B$ are conyclic).

Thus $AX$ and $AM$ are isogonal and we're done.
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rmtf1111
698 posts
#10 • 10 Y
Y by Kayak, AlastorMoody, RudraRockstar, amar_04, khina, 606234, hakN, IAmTheHazard, Adventure10, Mango247
this problem is the proof that russians have time machines. how could they know about 2017G3 in 2010 otherwise??
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SerdarBozdag
892 posts
#12
Y by
$$\frac{IS}{SJ}=\frac{AI}{JF}  \leftrightarrow  \frac{OM}{OK}=\frac{KJ}{JF} \leftrightarrow      \frac{OK}{OM}=\frac{KF}{KJ}-1    \leftrightarrow  \frac{MK\cdot KJ}{OM}=KF$$We have $t=BF$, $MK=h$, $AP=\frac{c^2}{2h}$ and $AQ=\frac{b^2}{2h} \implies KJ=AI=\frac{b^2+c^2}{4h}$, $OM=R\cdot cos(A)$. Thus it is enough to prove that $$\frac{b^2+c^2}{4R\cdot cos(A)}=h+tsin(A) \leftrightarrow t=\frac{a}{2cos(A)}$$
$x=\angle BAF$, $t=FA\frac{sinx}{sinc}=FA\frac{a}{2AM}=\frac{a}{2cos(A)}$
In the last equation we used sin law in $FBA$ and $MAC$. We also used the fact that $\frac{AM}{AF}=cos(A)$ which can be proved by noticing $AMC\sim AFR$.
Attachments:
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Kingjon
4 posts
#13
Y by
I have solution from sinus. But its easy to chase angels and sinus is very similar to it anyway
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math_comb01
662 posts
#14 • 1 Y
Y by Rounak_iitr
Amazing problem.
[asy]/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(10cm); 
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ 
pen dotstyle = black; /* point style */ 
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draw((-0.68,1.85)--(-2.2,-2.65)--(6.9,-3.07)--cycle, linewidth(2) + zzttqq); 
 /* draw figures *//* special point *//* special point *//* special point */
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draw((-2.2,-2.65)--(6.9,-3.07), linewidth(0.8)); 
draw((6.9,-3.07)--(-0.68,1.85), linewidth(0.8)); 
draw(circle((2.403617451019996,-1.6982885612334209), 4.700962486344153), linewidth(0.8)); /* special point *//* special point *//* special point *//* special point *//* special point */
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label("$O$", (2.48,-1.61), NE * labelscalefactor); 
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label("$Q$", (-1.02,-7.01), NE * labelscalefactor); 
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label("$S$", (-0.04,-3.81), NE * labelscalefactor); 
label("$D$", (-0.82,-2.63), NE * labelscalefactor); 
label("$Q_A$", (-0.22,-1.89), NE * labelscalefactor); 
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 /* end of picture */[/asy]
Let $M$ denote the midpoint of $AB$, $N$ denote the midpoint of $AC$, and let $Q_A$ denote the $A$-dumpty point.
Claim 1: $\triangle OPQ \sim \triangle ABC$
Proof
Since both the triangles are positively similar , therefore there exists a point $Q_A$ such that it takes the former triangle to the latter.
Claim 2:$Q_A$ is the $A$-dumpty point
Proof
Claim 3: $A-Q_A-S$
Proof
Remark
This post has been edited 2 times. Last edited by math_comb01, Nov 14, 2023, 5:12 PM
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Hertz
32 posts
#15
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Let $F$ and $G$ be midpoints of $AB$ and $AC$ respectively. Let $K$ be the midpoint $PQ$.

$\angle POQ = 180 - \angle AOF - \angle AOG = 180 - \angle C - \angle B = \angle A$

Since triangle $AFP$ and triangle $ADB$ are similar we get $\angle QPO = \angle AFP = \angle B$

So we get triangle $OPQ$ and triangle $ABC$ are similar in that orientation.

$\angle SOA = \angle FOA + \angle SOP = \angle C + 90 - \angle C = 90$ So $AO$ is tangent to ($OPQ$), thus $AOST$ is cyclic.

$\angle SAK = \angle SOK = \angle OAM$

$\angle BAS = \angle BAD + \angle SAK = \angle OAC + \angle OAM = \angle CAM$
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mcmp
53 posts
#16
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Woah Russian Geo :love: :love:

I love it when I can just nuke a problem with a technique that appears seven whole years after the problem (in the guise of ISL 2017/G3 :D).

Rename $M\to M_a$, let the midpoint of $PQ$ be $M$, and let the tangent to $(ABC)$ at $A$ intersect $\overline{BC}$ at $A’$. Let $M_b$, $M_c$ be the midpoints of $AC$, $AB$ resp.
Diagram
Imma first show that $AO$ tangent to $(OPQ)$; to do this, notice that it would suffice to show that $AP\cdot AQ=AO^2$. However notice that $M_cPDB$ and $M_bPDC$ are cyclic because of the right angles, so $AP\cdot AD=AM_c\cdot AB$ and $AQ\cdot AD=AM_b\cdot AC$. Hence:
\begin{align*}
AP\cdot AQ&=\frac{AM_c\cdot AM_b\cdot AB\cdot AC}{AD^2}\\
&=\frac{AB^2\cdot AC^2}{4AD^2}\\
&=\frac{A^2}{AD^2\sin^2(\angle BAC)}\\
&=\frac{BC^2}{\sin^2(\angle BAC)}\\
&=AO^2
\end{align*}the last line following from the law of sines. Hence we have the tangency. Now $\measuredangle OPQ=\measuredangle M_cPD=\measuredangle M_cBD=\measuredangle ABC$, and similarly with $\measuredangle OQP=\measuredangle ACB$, so $A’ABCM_a\stackrel{+}{\sim}AOPQM$. The rest is now trivial. $\measuredangle SAD=\measuredangle SAM=\measuredangle OA’M_a=\measuredangle OAM_a$, so $AM_a$ and $AS$ are conjugate in $\angle DAO$; since $AD$ and $AO$ conjugate in $\angle BAC$, they share the same angle bisector and hence $AM_a$ and $AS$ are conjugate in $\angle BAC$ as desired. $\square$
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