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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
prove that a_50 + b_50 > 20
kamatadu   8
N a minute ago by Markas
Source: Canada Training Camp
The sequences $a_n$ and $b_n$ are such that, for every positive integer $n$,
\[ a_n > 0,\qquad\ b_n>0,\qquad\ a_{n+1}=a_n+\dfrac{1}{b_n},\qquad\ b_{n+1} = b_n+\dfrac{1}{a_n}. \]Prove that $a_{50} + b_{50} > 20$.
8 replies
kamatadu
Dec 30, 2023
Markas
a minute ago
EGMO P4 infinite sequence
aditya21   29
N 2 minutes ago by Markas
Source: EGMO 2015, Problem 4
Determine whether there exists an infinite sequence $a_1, a_2, a_3, \dots$ of positive integers
which satisfies the equality \[a_{n+2}=a_{n+1}+\sqrt{a_{n+1}+a_{n}} \] for every positive integer $n$.
29 replies
+1 w
aditya21
Apr 17, 2015
Markas
2 minutes ago
IMO 2014 Problem 1
Amir Hossein   133
N 3 minutes ago by Markas
Let $a_0 < a_1 < a_2 < \dots$ be an infinite sequence of positive integers. Prove that there exists a unique integer $n\geq 1$ such that
\[a_n < \frac{a_0+a_1+a_2+\cdots+a_n}{n} \leq a_{n+1}.\]
Proposed by Gerhard Wöginger, Austria.
133 replies
Amir Hossein
Jul 8, 2014
Markas
3 minutes ago
Sequences and limit
lehungvietbao   16
N 4 minutes ago by Markas
Source: Vietnam Mathematical OLympiad 2014
Let $({{x}_{n}}),({{y}_{n}})$ be two positive sequences defined by ${{x}_{1}}=1,{{y}_{1}}=\sqrt{3}$ and
\[ \begin{cases}  {{x}_{n+1}}{{y}_{n+1}}-{{x}_{n}}=0 \\   x_{n+1}^{2}+{{y}_{n}}=2 \end{cases} \] for all $n=1,2,3,\ldots$.
Prove that they are converges and find their limits.
16 replies
lehungvietbao
Jan 3, 2014
Markas
4 minutes ago
Real triples
juckter   67
N 4 minutes ago by Markas
Source: EGMO 2019 Problem 1
Find all triples $(a, b, c)$ of real numbers such that $ab + bc + ca = 1$ and

$$a^2b + c = b^2c + a = c^2a + b.$$
67 replies
juckter
Apr 9, 2019
Markas
4 minutes ago
Social Club with 2k+1 Members
v_Enhance   24
N 16 minutes ago by mathwiz_1207
Source: USA December TST for IMO 2013, Problem 1
A social club has $2k+1$ members, each of whom is fluent in the same $k$ languages. Any pair of members always talk to each other in only one language. Suppose that there were no three members such that they use only one language among them. Let $A$ be the number of three-member subsets such that the three distinct pairs among them use different languages. Find the maximum possible value of $A$.
24 replies
v_Enhance
Jul 30, 2013
mathwiz_1207
16 minutes ago
A strong inequality problem
hn111009   1
N 22 minutes ago by Tung-CHL
Source: Somewhere
Let $a,b,c$ be the positive number satisfied $a^2+b^2+c^2=3.$ Find the minimum of $$P=\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}+\dfrac{3abc}{2(ab+bc+ca)}.$$
1 reply
hn111009
Today at 2:02 AM
Tung-CHL
22 minutes ago
Altitude configuration with two touching circles
Tintarn   3
N 22 minutes ago by NumberzAndStuff
Source: Austrian MO 2024, Final Round P2
Let $ABC$ be an acute triangle with $AB>AC$. Let $D,E,F$ denote the feet of its altitudes on $BC,AC$ and $AB$, respectively. Let $S$ denote the intersection of lines $EF$ and $BC$. Prove that the circumcircles $k_1$ and $k_2$ of the two triangles $AEF$ and $DES$ touch in $E$.

(Karl Czakler)
3 replies
Tintarn
Jun 1, 2024
NumberzAndStuff
22 minutes ago
IMO Shortlist 2009 - Problem G3
April   48
N 30 minutes ago by Markas
Let $ABC$ be a triangle. The incircle of $ABC$ touches the sides $AB$ and $AC$ at the points $Z$ and $Y$, respectively. Let $G$ be the point where the lines $BY$ and $CZ$ meet, and let $R$ and $S$ be points such that the two quadrilaterals $BCYR$ and $BCSZ$ are parallelogram.
Prove that $GR=GS$.

Proposed by Hossein Karke Abadi, Iran
48 replies
April
Jul 5, 2010
Markas
30 minutes ago
Mount Inequality erupts in all directions!
BR1F1SZ   3
N 31 minutes ago by NumberzAndStuff
Source: Austria National MO Part 1 Problem 1
Let $a$, $b$ and $c$ be pairwise distinct nonnegative real numbers. Prove that
\[
(a + b + c) \left( \frac{a}{(b - c)^2} + \frac{b}{(c - a)^2} + \frac{c}{(a - b)^2} \right) > 4.
\](Karl Czakler)
3 replies
BR1F1SZ
May 5, 2025
NumberzAndStuff
31 minutes ago
Show that (DEN) passes through the midpoint of BC
v_Enhance   24
N 31 minutes ago by Markas
Source: Sharygin First Round 2013, Problem 21
Chords $BC$ and $DE$ of circle $\omega$ meet at point $A$. The line through $D$ parallel to $BC$ meets $\omega$ again at $F$, and $FA$ meets $\omega$ again at $T$. Let $M = ET \cap BC$ and let $N$ be the reflection of $A$ over $M$. Show that $(DEN)$ passes through the midpoint of $BC$.
24 replies
v_Enhance
Apr 7, 2013
Markas
31 minutes ago
2020 EGMO P5: P is the incentre of CDE
alifenix-   49
N 31 minutes ago by Markas
Source: 2020 EGMO P5
Consider the triangle $ABC$ with $\angle BCA > 90^{\circ}$. The circumcircle $\Gamma$ of $ABC$ has radius $R$. There is a point $P$ in the interior of the line segment $AB$ such that $PB = PC$ and the length of $PA$ is $R$. The perpendicular bisector of $PB$ intersects $\Gamma$ at the points $D$ and $E$.

Prove $P$ is the incentre of triangle $CDE$.
49 replies
alifenix-
Apr 18, 2020
Markas
31 minutes ago
The reflection of AD intersect (ABC) lies on (AEF)
alifenix-   61
N 32 minutes ago by Markas
Source: USA TST for EGMO 2020, Problem 4
Let $ABC$ be a triangle. Distinct points $D$, $E$, $F$ lie on sides $BC$, $AC$, and $AB$, respectively, such that quadrilaterals $ABDE$ and $ACDF$ are cyclic. Line $AD$ meets the circumcircle of $\triangle ABC$ again at $P$. Let $Q$ denote the reflection of $P$ across $BC$. Show that $Q$ lies on the circumcircle of $\triangle AEF$.

Proposed by Ankan Bhattacharya
61 replies
alifenix-
Jan 27, 2020
Markas
32 minutes ago
JBMO 2013 Problem 2
Igor   44
N 32 minutes ago by Markas
Source: Proposed by Macedonia
Let $ABC$ be an acute-angled triangle with $AB<AC$ and let $O$ be the centre of its circumcircle $\omega$. Let $D$ be a point on the line segment $BC$ such that $\angle BAD = \angle CAO$. Let $E$ be the second point of intersection of $\omega$ and the line $AD$. If $M$, $N$ and $P$ are the midpoints of the line segments $BE$, $OD$ and $AC$, respectively, show that the points $M$, $N$ and $P$ are collinear.
44 replies
Igor
Jun 23, 2013
Markas
32 minutes ago
The best computer problems of the year
NT_G   9
N Apr 10, 2025 by bin_sherlo
Source: https://t.me/NeuroGeometry
1. I is incenter of triangle $ABC$. $K$ is the midpoint of arc $BC$ not containing $A$. $L$ is the midpoint of arc $(BAC)$ $M$ is the midpoint of $IK$ and $N$ is the midpoint of $LM$.  $D$ is projection of $I$ onto $BC$. $S$ is the second intersection of $KD$ and $(ABC)$
Prove that $(ISD)$ is tangent to $(AMN)$.

2. Circle $w$ with center $I$ is incircle of triangle $ABC$. $D$ is projection of $I$ onto $BC$ Points $E, F$ on segments $BI$, $CI$ are following condition: $IE = IF$. $M$ is the midpoint of $EF$. $P$ is the second intersection of $(IMD)$ and $w$. $Q$ is the second intersection of $AP$ and $(IMD)$.
Prove, that $I,E,F,Q$ are concyclic.

I am grateful to Savva Chuev for making the diagrams.
9 replies
NT_G
Dec 31, 2023
bin_sherlo
Apr 10, 2025
The best computer problems of the year
G H J
Source: https://t.me/NeuroGeometry
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NT_G
45 posts
#1 • 2 Y
Y by GeoKing, SBYT
1. I is incenter of triangle $ABC$. $K$ is the midpoint of arc $BC$ not containing $A$. $L$ is the midpoint of arc $(BAC)$ $M$ is the midpoint of $IK$ and $N$ is the midpoint of $LM$.  $D$ is projection of $I$ onto $BC$. $S$ is the second intersection of $KD$ and $(ABC)$
Prove that $(ISD)$ is tangent to $(AMN)$.

2. Circle $w$ with center $I$ is incircle of triangle $ABC$. $D$ is projection of $I$ onto $BC$ Points $E, F$ on segments $BI$, $CI$ are following condition: $IE = IF$. $M$ is the midpoint of $EF$. $P$ is the second intersection of $(IMD)$ and $w$. $Q$ is the second intersection of $AP$ and $(IMD)$.
Prove, that $I,E,F,Q$ are concyclic.

I am grateful to Savva Chuev for making the diagrams.
Attachments:
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NT_G
45 posts
#2 • 1 Y
Y by GeoKing
P1 has a generalisation:
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SBYT
196 posts
#3 • 3 Y
Y by GeoKing, vuanhnshn, Om245
Proof of part 1

It is well know that $\angle ASI=\frac{\pi}{2}$ (Sharky Devil).Let $AS$ meets $BC$ at $E$,so $EI$ is the diameter of $\odot ISD$.
By $KI^2=KB^2=KD\cdot KS$, we can know that $KI$ is a tangent line of $\odot IDS$,so $AK\perp IE$.
Let $\odot ISD$ meets $\odot ABC$ at $S,F$,$\angle SFI=\angle SEI=\frac{\pi}{2}-\angle SAI=\frac{\pi}{2}-\angle SLK=\angle SKL=\angle SFL$,so $F,I,L$ are conlinear.
By $MF=MI=MK$, we can get $MF$ is another tangent line of $\odot IDS$.
It means that $FI$ is the polar of $M$ to $\odot ISD$ ,and $L$ lies on this line.
By $N$ is the midpoint of $ML$, we know $N$ lies on the radical axis of $\odot ISD$ and $\odot M$.
Let $NI$ meets $\odot ISD$ at $G$ again,then $NI\cdot NG=NM^2=NL^2=NA^2$.
So $A,N,M,G$ are concyclic.
Let $GP$ is a tangent line of $\odot ISD$,then $\angle IGP=\angle GIK=\angle AIN=\angle NAG$,so $IG$ tangents $\odot AMN$,too.$\Box$.
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Om245
164 posts
#4 • 2 Y
Y by ATGY, GeoKing
Solution of frist part

It well known that $S$ is Sharky-Devil point (as if $S$ is Sharky-Devil point then $SD$ is angle bisector so $S - D - K$)
$$\angle SIA = \angle AES = \angle AEF - \angle SEF$$
Spiral similarity at $S$ sends $\overline{EF}$ to $\overline{BC} \implies \angle SEF = \angle SCB$
so $\angle SIA = 90 - \frac{\angle A}{2} - \angle SCB$. From $S - D - K \implies \angle KSC = \angle KAC$

So $\angle SIA = 90 - \angle SDA$ from $\overline{ID} \perp \overline{BC}$ we get $\angle SIA = \angle SDI$ hence we get $\overline{AK}$ tangent to $(SID)$.

Observe $\angle LAM = 90$ and $N$ is midpoint of $LM$ we get $AN = NM$.

Let $Y =\overline{LI} \cap (ABC)$ from $\angle LYK = 90$. $Y$ also lie on circle $w$ with center $M$ with radius $MI$.
$\overline{MI}$ is tangent to $(SID)$ and $MI = MY$ so $MY$ also tangent to $(SID)$.

Let $ X = \overline{LI} \cap (SID)$ and $ P = \overline{MX} \cap (SID)$
$$(Y,I;X,P)\stackrel{I}{=}(L,M;N,\infty{\overline{LM}}) = -1 \implies \overline{IP} \parallel  \overline{LM}$$
$$\angle IXM = \angle PIM = \angle IMN \implies \angle IXM = \angle XMN$$
Now as $AN=NM$ we get $\angle IXM = \angle XMN = \angle MAN$ so $X,A,N,M$ cyclic points.

Now as tangent to $I$ and $N$ are parallel, Homothety at $X$ sends $(SID)$ to $(ANM)$ hence $X$ is tangent point of $(SID)$ and $(ANM)$


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This post has been edited 3 times. Last edited by Om245, Jan 3, 2024, 4:03 AM
Reason: image
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NT_G
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#5 • 3 Y
Y by Grifon, SBYT, mathlove_13520
My first solution of P1 was similar to SBYT's. I found it accidently while Cartesian bashing the problem for the New Year stream on NeuroGeometry. Trick with proving tangency using radical axis was new for me...

My second solution:
Due to the fact that $ID \parallel KL$, $\angle IDS = \angle LKS$, therefore $LI$ passes through the second intersection of $(SID)$ and $(ABC)$. It means that $(SID)$ contains the point of tangency of circumcircle and mixtilinear circle of $\triangle ABC$.
Now we invert the problem with center in $A$. Problem turns into:
In triangle $ABC$ $I_a$ is the $A$ - excenter. $N$ is the projection of $I_a$ onto $BC$. $L, L_a$ are foots of bisectors of $\angle BAC$. K is a point on $AI_a$ such that $(I,I_a; K, A) = -1$. $A'$ is a point on $\Gamma = (AL_{a}K)$ suiting: $KA = KA'$. $\gamma$ is a circle passing through $N, I_a$ that is tangent to $AI_a$. We have to prove that $KA'$ is tangent to $\gamma$.
Let's spot that due to the fact that $\angle L_{a}AK = \frac{\pi}{2}$, $A'$ is the reflection of $A$ in $KL_a$.Therefore, we have to prove that line $O_{\Gamma}K$ passes through $O_{\gamma}$. $L_a, K, O_{\Gamma}$ are obviously collinear. Let $I_{a}'$ be the reflection of $I_{a}$ over $O_{\Gamma}$. $I_{a}'$ lies on $BC$. Then after spotting that $I_{a}I_{a}' \parallel AL_{a}$ and projecting $(I,I_a; K, A)$ from $L_a$ onto $I_{a}I_{a}' $ we get what we needed.
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SerdarBozdag
892 posts
#6 • 1 Y
Y by GeoKing
First problem. $G=AS \cap BC$ is on $(SID)$. $LI \cap ABC = J$. $A'$ is the antipode of $A$ and $S,I,A'$ are collinear. By radical axis theorem on $(ASFIE), (IJK), (ABC)$, $G$ is on $JK$. $H = IN \cap (SID)$. Let $I'$ be the reflection of $I$ across $N$.

$LIMI'$ is parallelogram so $LI' = MI = MK$ and $I'M \parallel JL \implies LJMI'$ is cyclic.

$\angle I'HJ + \angle JMI' = \angle ISJ + \angle JMI + \angle IMI' = \angle JAA' +\angle JMI + \angle JIM = \angle JAA' + 90 + \angle JKA =  \angle JAA' + 90^{\circ} + \angle JA'A = 180^{\circ} \implies H \in (LJMI')$.

$NA^2 = NM^2 = NM \cdot NL = NI' \cdot NH = NI \cdot NH \implies HMNA $ is cyclic. If tangents to $(SID)$ at $I$ and $H$ intersect at $Q$. $\angle NHQ = \angle HIQ = \angle HAN \implies HQ$ is tangent to $(HMNA)$.
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SerdarBozdag
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#7 • 1 Y
Y by GeoKing
Second problem. Let $G= EF \cap BC$. $(IMDG) \cap (I) = P$ , $(IMDG) \cap (IEF) = Q$, $(IMDG) \cap AI = L$, $PM \cap GL = S$, $N$ is the antipode of $I$ in $(IEF)$ and $AI \cap (I) = J$. I will prove that $A \in PQ$.

$\angle IQG = 90^{\circ} = \angle IQN \implies N \in QG$.

$\angle EID = 90^{\circ} - \frac{B}{2} = \angle JIF \implies J $ is the reflection of $D$ across $AI$. $\angle IJM = \angle IDM = \angle IPM = \angle IJP \implies J \in PM$. $\angle LGD = \angle DIJ = 2 \cdot \angle DIM = 2 \cdot \angle DGM \implies ML = MD \implies M$ is the center of $(SLJD)$.

$N,S,A$ are collinear $\iff$ (by Menelaus)
$$\frac{NM}{NI} \cdot \frac{IA}{AJ} \cdot \frac{JS}{SM} = 1$$This is true because $\frac{NM}{NI} = \cos^2 (\angle IFE) = \frac{1+\cos (90^{\circ} - \angle A/2) }{2}= \frac{JA}{IA} \cdot \frac{JS}{SM}$.

Applying Pascal on $PQG  LIM$ shows that $PQ \cap LI \in SN \implies A=PQ \cap LI$ as desired.
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NT_G
45 posts
#9 • 1 Y
Y by GeoKing
P2: (R.Prozorov's solution)
Let $R$ be the second intersection of $(IMD)$, $BC$. Obviously, $R$ lies on $EF$, and $IR$ is diameter of $(IMD)$.
$\angle IQD = \angle PQI$. So, using DIT for $ABDC$, we get that $\angle BQI = \angle CQI$.
$K = IQ \cap BC$. $IQ \perp QR$, therefore $(R,K; B,C) = -1 \Rightarrow $ $CE$, $BF$, $IQ$ are concurent. Using isogonal theorem for $\angle BQC$ and points $E$, $F$, we prove that $\angle EQI = \angle IQF$. Therefore, due to the equality: $IE = IF$, we get what we needed.
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SBYT
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#10 • 1 Y
Y by GeoKing
Another solution which is similar to NT_G's:
We get $(R,K;B,C)=-1$ the same way,$IK$ meets $EF$ at $N$.$(R,K;B,C)=-1\implies(IR,IK;IB,IC)=-1\implies(R,N;E,F)=-1\implies(QR,QN;QE,QF)=-1$.
Due to $IQ\perp RQ$,$IQ$ is the angle bisector of $\angle EQF$,and $IE=IF$,so $I,E,Q,F$ are concyclic.$\Box$
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bin_sherlo
722 posts
#11 • 1 Y
Y by MS_asdfgzxcvb
I 'll present a solution to the first problem.
Let $AS\cap BC=R,MR\cap (ABC)=T$. Let $R,W'$ be the feet of the perpendicular from $P,N$ to $IN,PI$ respectively. Let $W,I'$ be the reflections of $I$ over $W',N$. Note that $P,T,I,R,S,D$ are concyclic.
Claim: $A,M,N,R$ are concyclic.
Proof: $A,P,K,W$ are concyclic because $IP.IW=IL.IT=IA.IK$ so $A,P,M,W'$ are concyclic. $IA.IM=IP.IW'=IN.IR$ which gives the result.
Let $PR\cap (AMNR)=V$ which is the antipode of $N$. Since $\measuredangle LAM=90$, we have $NA=NM$ and $NV$ is diameter thus, $NV\perp AI\perp IP$. Hence $(RPI)$ and $(RNV)$ are tangent to each other as desired.$\blacksquare$
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