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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
Inspired by Kazakhstan 2017
sqing   1
N a few seconds ago by sqing
Source: Own
Let $a,b,c\ge \frac{1}{2}$ and $a^2+b^2+c^2=2. $ Prove that
$$\left(\frac{2}{a}+\frac{1}{b}-\frac{1}{c}\right)\left(\frac{2}{a}-\frac{1}{b}+\frac{1}{c}\right)\ge \frac{8}{3}$$$$  \left( 2a^2+\frac{2}{a}+\frac{1}{b}-\frac{1}{c}\right)\left( 2a^2+\frac{2}{a}-\frac{1}{b}+\frac{1}{c}\right)  \ge 9\sqrt[3]{4}$$
1 reply
1 viewing
sqing
a minute ago
sqing
a few seconds ago
Number theory
Ecrin_eren   0
12 minutes ago
Show that there are no prime numbers satisfying the equation

(p + r)^q + (q + r)^p = (p + q)^r.

0 replies
Ecrin_eren
12 minutes ago
0 replies
Wait wasn't it the reciprocal in the paper?
Supercali   7
N 14 minutes ago by kes0716
Source: India TST 2023 Day 2 P1
Let $\mathbb{Z}_{\ge 0}$ be the set of non-negative integers and $\mathbb{R}^+$ be the set of positive real numbers. Let $f: \mathbb{Z}_{\ge 0}^2 \rightarrow \mathbb{R}^+$ be a function such that $f(0, k) = 2^k$ and $f(k, 0) = 1$ for all integers $k \ge 0$, and $$f(m, n) = \frac{2f(m-1, n) \cdot f(m, n-1)}{f(m-1, n)+f(m, n-1)}$$for all integers $m, n \ge 1$. Prove that $f(99, 99)<1.99$.

Proposed by Navilarekallu Tejaswi
7 replies
Supercali
Jul 9, 2023
kes0716
14 minutes ago
Functional equation
Ecrin_eren   0
18 minutes ago
Find all functions f:R R satisfying the equation

f(f(x)y) + f(x+ f(y)) = x f(y) + f(x+y)

for all x,y real numbers

0 replies
Ecrin_eren
18 minutes ago
0 replies
Spheres and a point source of light
mofidy   3
N Yesterday at 9:22 PM by kiyoras_2001
How many spheres are needed to shield a point source of light?
Unfortunately, I didn't find a suitable solution for it on the page below:
https://artofproblemsolving.com/community/c4h1469498p8521602
Here too, two different solutions are given:
https://math.stackexchange.com/questions/2791186
3 replies
mofidy
Mar 11, 2025
kiyoras_2001
Yesterday at 9:22 PM
Real Analysis
rljmano   4
N Yesterday at 4:38 PM by alexheinis
In [0 1], {f(t)}*{f’(t)-1} =0 and f is continuously differentiable. How do we conclude that either f is identically zero or f’(t) is identically 1 in [0 1]?
4 replies
rljmano
Yesterday at 6:32 AM
alexheinis
Yesterday at 4:38 PM
find the isomorphism
nguyenalex   14
N Yesterday at 1:49 PM by Royrik123456
I have the following exercise:

Let $E$ be an algebraic extension of $K$, and let $F$ be an algebraic closure of $K$ containing $E$. Prove that if $\sigma : E \to F$ is an embedding such that $\sigma(c) = c$ for all $c \in K$, then $\sigma$ extends to an automorphism of $F$.

My attempt:

Theorem (*): Suppose that $E$ is an algebraic extension of the field $K$, $F$ is an algebraically closed field, and $\sigma: K \to F$ is an embedding. Then, there exists an embedding $\tau: E \to F$ that extends $\sigma$. Moreover, if $E$ is an algebraic closure of $K$ and $F$ is an algebraic extension of $\sigma(K)$, then $\tau$ is an isomorphism.

Back to our main problem:

Since $K \subset E$ and $F$ is an algebraic extension of $K$, it follows that $F$ is an algebraic extension of $E$. Assume that there exists an embedding $\sigma : E \to F$ such that $\sigma(c) = c$ for all $c \in K$. By Theorem (*), there exists an embedding $\tau : F \to F$ that extends $\sigma$. Since $F$ is algebraically closed, $\tau(F)$ is also an algebraically closed field.

Furthermore, because $\sigma(c) = c$ for all $c \in K$ and $\tau$ is an extension of $\sigma$, we have
$$K = \sigma(K) \subset K \subset \sigma(E) \subset \tau(F) \subset F.$$
This implies that $F$ is an algebraic extension of $\tau(F)$. We conclude that $F = \tau(F)$, meaning that $\tau$ is an automorphism. (Finished!!)

Let choose $F = A$ be the field of algebraic numbers, $K=\mathbb{Q}$. Consider the embedding $\sigma: \mathbb{Q}(\sqrt{2}) \to \mathbb{Q}(\sqrt{2}) \subset A$ defined by
$$
a + b\sqrt{2} \mapsto a - b\sqrt{2}.
$$Then, according to the exercise above, $\sigma$ extends to an isomorphism
$$
\bar{\sigma}: A \to A.
$$How should we interpret $\bar{\sigma}$?
14 replies
nguyenalex
Mar 12, 2025
Royrik123456
Yesterday at 1:49 PM
find the convex sets satisfied
nguyenalex   2
N Yesterday at 9:54 AM by ILOVEMYFAMILY
In $\mathbb{R}^2$, let $B = \{(x, y) \mid x \geq 0\}$. Find all convex sets $C$ such that

\[\mathcal{E}(B \cup C) = B \cup C.\]
2 replies
nguyenalex
Yesterday at 8:57 AM
ILOVEMYFAMILY
Yesterday at 9:54 AM
A great result
steven_zhang123   10
N Yesterday at 9:35 AM by teomihai
Show that $\lim_{n \to \infty} \sum_{k=1}^{n} (\sqrt{1+\frac{k}{n^{2} } } -1)=\frac{1}{4} $.
10 replies
steven_zhang123
Oct 31, 2024
teomihai
Yesterday at 9:35 AM
Weird family of sequences
AndreiVila   9
N Yesterday at 8:40 AM by Fibonacci_math
Source: Romanian District Olympiad 2025 12.3
[list=a]
[*] Let $a<b$ and $f:[a,b]\rightarrow\mathbb{R}$ be a strictly monotonous function such that $\int_a^b f(x) dx=0$. Show that $f(a)\cdot f(b)<0$.
[*] Find all convergent sequences $(a_n)_{n\geq 1}$ for which there exists a scrictly monotonous function $f:\mathbb{R}\rightarrow\mathbb{R}$ such that $$\int_{a_{n-1}}^{a_n} f(x)dx = \int_{a_n}^{a_{n+1}} f(x)dx,\text{ for all }n\geq 2.$$
9 replies
AndreiVila
Mar 8, 2025
Fibonacci_math
Yesterday at 8:40 AM
Monster Integral
Entrepreneur   1
N Yesterday at 7:41 AM by RezerdPrime
$$\color{blue}{\int_{0}^{\pi} \frac{\tan^{-1}\left(\frac{\ln(\sin(x))}{x}\right)dx}{\ln^2\left(x^2 + \ln^2(\sin(x))\right) + 4\arctan^2\left(\frac{\ln(\sin(x))}{x}\right)}= -\frac{\pi \tan^{-1}\left(\frac{2\ln(2)}{\pi}\right)}{\ln^2\left(\frac{\pi^2}{4} + \ln^2(2)\right) + 4\arctan^2\left(\frac{2\ln(2)}{\pi}\right)}.}$$
1 reply
Entrepreneur
Jan 10, 2025
RezerdPrime
Yesterday at 7:41 AM
convex closed set with a nonempty interior
ILOVEMYFAMILY   0
Yesterday at 6:11 AM
a) When $n = 2$, prove that a convex closed set with a nonempty interior that contains exactly one extreme point must contain a ray. (SOLVED)

b) When $n = 2$, find all convex closed sets with a nonempty interior that contain exactly one extreme point.
0 replies
ILOVEMYFAMILY
Yesterday at 6:11 AM
0 replies
Differentiation Marathon!
LawofCosine   178
N Yesterday at 5:26 AM by awzhang10
Hello, everybody!

This is a differentiation marathon. It is just like an ordinary marathon, where you can post problems and provide solutions to the problem posted by the previous user. You can only post differentiation problems (not including integration and differential equations) and please don't make it too hard!

Have fun!

(Sorry about the bad english)
178 replies
LawofCosine
Feb 1, 2025
awzhang10
Yesterday at 5:26 AM
Integrals problems and inequality
tkd23112006   1
N Yesterday at 3:55 AM by removablesingularity
Let f be a continuous function on [0,1] such that f(x) ≥ 0 for all x ∈[0,1] and
$\int_x^1 f(t) dt \geq \frac{1-x^2}{2}$ , ∀x∈[0,1].
Prove that:
$\int_0^1 (f(x))^{2021} dx \geq \int_0^1 x^{2020} f(x) dx$
1 reply
tkd23112006
Feb 16, 2025
removablesingularity
Yesterday at 3:55 AM
Line through incenter tangent to a circle
Kayak   31
N Yesterday at 9:16 PM by ihategeo_1969
Source: Indian TST D1 P1
In an acute angled triangle $ABC$ with $AB < AC$, let $I$ denote the incenter and $M$ the midpoint of side $BC$. The line through $A$ perpendicular to $AI$ intersects the tangent from $M$ to the incircle (different from line $BC$) at a point $P$> Show that $AI$ is tangent to the circumcircle of triangle $MIP$.

Proposed by Tejaswi Navilarekallu
31 replies
Kayak
Jul 17, 2019
ihategeo_1969
Yesterday at 9:16 PM
Line through incenter tangent to a circle
G H J
G H BBookmark kLocked kLocked NReply
Source: Indian TST D1 P1
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Kayak
1298 posts
#1 • 6 Y
Y by FadingMoonlight, noob_mathematician, TFIRSTMGMEDALIST, tiendung2006, Adventure10, Mango247
In an acute angled triangle $ABC$ with $AB < AC$, let $I$ denote the incenter and $M$ the midpoint of side $BC$. The line through $A$ perpendicular to $AI$ intersects the tangent from $M$ to the incircle (different from line $BC$) at a point $P$> Show that $AI$ is tangent to the circumcircle of triangle $MIP$.

Proposed by Tejaswi Navilarekallu
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MathPassionForever
1663 posts
#2 • 1 Y
Y by Adventure10
Kayak wrote:
In an acute angled triangle $ABC$ with $AB < AC$, let $I$ denote the incenter and $M$ the midpoint of side $BC$. The line through $A$ perpendicular to $AI$ intersects the tangent from $M$ to the incircle (different from line $BC$) at a point $P$. Show that $AI$ is tangent to the circumcircle of triangle $MIP$.

Proposed by Tejaswi Navilarekallu
FTFY
@below, did you LaTeX it all in...... 10 minutes ?!
This post has been edited 1 time. Last edited by MathPassionForever, Jul 17, 2019, 12:35 PM
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Ankoganit
3070 posts
#3 • 4 Y
Y by sameer_chahar12, ashraful7525, abhisruta03, Adventure10
Solution from the official packet
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Pluto1708
1107 posts
#4 • 3 Y
Y by sameer_chahar12, Adventure10, Mango247
Solution
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Kayak
1298 posts
#5 • 3 Y
Y by FadingMoonlight, Adventure10, Mango247
This is pretty much the only geo I've solved synthetically ever in any contest; my solution is by proving the circle $MIP$ passes through the midpoint of arc $BAC$.
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AlastorMoody
2125 posts
#6 • 2 Y
Y by Adventure10, Mango247
India TST 2019 Day 1 P1 wrote:
In an acute angled triangle $ABC$ with $AB < AC$, let $I$ denote the incenter and $M$ the midpoint of side $BC$. The line through $A$ perpendicular to $AI$ intersects the tangent from $M$ to the incircle (different from line $BC$) at a point $P$> Show that $AI$ is tangent to the circumcircle of triangle $MIP$.

Proposed by Tejaswi Navilarekallu
Solution: Let $MG$ be tangent to incircle $\implies$ $DIGM$ is cyclic. Since, $\angle IAP=90^{\circ}$ $=$ $\angle IGM$ $\implies$ $AIGP$ is cyclic. Let $T_A$ be the $A-$extouch point $\implies$ $MD$ $=$ $MG$ $=$ $MT_A$ $\implies$ $\angle DGT_A$ $=$ $90^{\circ}$. Also, If $D'$ is the reflection of $D$ over $I$ $\implies$ $A,D',T_A$ collinear. Hence, $G$ $\in$ $T_AD'$. Now,
$$ \angle AIP  =90^{\circ}-\angle API  =90^{\circ}-\angle AGI =90^{\circ}-\angle D'GI  =90^{\circ}-\frac12 \angle DIG =90^{\circ}-\angle MIG =\angle IMG $$Hence, $AI$ is tangent to $\odot (MIP)$
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zuss77
520 posts
#7 • 1 Y
Y by Adventure10
I'll use diagram in post #3.
$A,D',X,E$ are collinear (proof in #3 or @above).
$IM$ - midline in $\triangle EDD'$ , so $IM \parallel AX$.
$AIXP$ - cyclic.
$\angle AIP = \angle AXP = \angle IMP$, which proves needed tangency.
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jayme
9767 posts
#8 • 1 Y
Y by Adventure10
Dear Mathlinkers,
1 AIXP cyclic OK.
2. (I) goes through X
3. IM//AX
4. By a converse of the Reim's theorem we are done...

Sincerely
Jean-Louis
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e_plus_pi
756 posts
#9 • 1 Y
Y by Adventure10
IMO, a bit too easy for a TST 1
Kayak wrote:
In an acute angled triangle $ABC$ with $AB < AC$, let $I$ denote the incenter and $M$ the midpoint of side $BC$. The line through $A$ perpendicular to $AI$ intersects the tangent from $M$ to the incircle (different from line $BC$) at a point $P$> Show that $AI$ is tangent to the circumcircle of triangle $MIP$.

Proposed by Tejaswi Navilarekallu

Let the tangent from $M$ onto $\odot(I)$ meet it at $X$. Then from USA TS 2016 we have that $ A - X - D'$ where $D'$ is the ex-touch point of side $BC$. Now it is well known that $IM \parallel AD'$. Hence $180^{\circ} - \angle MIA = \angle IAX = \angle IPM$ since, evidently, $AIXP$ is cyclic.
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mashina
158 posts
#10 • 2 Y
Y by Adventure10, Mango247
Nice and easy! The problem obviously follows from the fact that $MI\Vert AE$. But that's obvious as they both perpendicular to $XD$.
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ubermensch
820 posts
#11 • 1 Y
Y by Adventure10
Note that the problem reduces to proving $IM \Vert AD$, where $D$ is $MP$ $ \cap$ incircle, as we need to prove $ \angle IMP = \angle AIP = \angle ADP$ (as $AIPD$ is cyclic) $<=> IM \Vert AD$.

But as $IM \perp DF$, and moreover is the perpendicular bisector of $DF$ (as $IFMD$ is cyclic and $IF=ID$, $MD=MF$, $F$ being the intouch point of incircle on $BC$), the problem further reduces to proving $\angle ADF=90^{\circ} <=>$ circumcenter of $ADF$ is midpoint of $AF$. But, as $IM$ is the perpendicular bisector of $DF$, all we need to show is $IM$ bisects $AF$.

Although I presume this is well known, here's a short barycentric proof:
We know $F= (0:s-c:s-b) \implies M_{AF}=(a:s-c:s-b)$. As we want to show $M-I-M_{AF}$, this is equivalent to showing:
$0 \cdot (b(s-b)-c(s-c)) -a(s-b-(s-c))+a(c-b) =0$, which is trivially true.
This post has been edited 1 time. Last edited by ubermensch, Jan 12, 2020, 7:13 AM
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aops29
452 posts
#12 • 2 Y
Y by Adventure10, Mango247
Simple
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arzhang2001
248 posts
#13
Y by
let $S$ be the intersection of $A$ -exercle with segment $BC$ and $L$ be the tangency point of $MP$ and incircle . easy to see $A,L,S$ collinear. and $LA || IM$. From this two result we are done$\blacksquare$
This post has been edited 2 times. Last edited by arzhang2001, Aug 12, 2020, 5:01 PM
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hakN
428 posts
#15
Y by
Let $D$ be the tangency point of the incircle with $BC$. Let the other tangent from $M$ to incircle cut incircle at $E$. Let $F$ be the antipode of $D$ in the incircle. By a well known lemma, we have $A,F,E$ are collinear. Also, since $90 = \angle IAP = \angle IEP$, we have $AIEP$ is cyclic.
Finally, since $IDME$ is also cyclic, we have $\angle IMP = \angle IME = \angle IDE = \angle FDE = \angle FEP = \angle AEP = \angle AIP$, implying that $AI$ is tangent to $(MIP)$, as needed.
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primesarespecial
364 posts
#16
Y by
Meh,
Let $N$ be the tangency point mentioned.
Let the incircle be tangent to $BC$ at $X$.
It is well known that $AN \cap incircle $ is the $X-antipode$ which we call $F$.
$AINP$ is cyclic .
Thus, $\angle AIP= \angle ANP=\angle FNP=\angle FXN=\angle IMN=\angle IMP$.
This post has been edited 1 time. Last edited by primesarespecial, Aug 23, 2021, 1:25 PM
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rafaello
1079 posts
#17
Y by
Probably overcomplicated this but whatever.

Let $N$ be the midpoint of arc $BAC$. Let $T=NI\cap (ABC)$, let $E$ be the reflection of $D$ over $M$, where $D$ is the point on $BC$ such that $ID\perp BC$. Let $F$ be the point on $(I)$ such that $MP$ is tangent to $(I)$. Let $K$ be the intersection of $AI$ and $(ABC)$, let $L$ be the point on $(ABC)$ such that it is reflection of $K$ over the circumcenter of $\triangle ABC$.

Firstly note that $A,F,E$ are collinear as $\angle DFE=90^\circ$ and $AE$ passes through the antipode of $D$ wrt $(I)$. It is also well-known that $T$ is the intersection of mixtilinear incircle and the circumcircle and furthermore $AT,AE$ are isogonal. Hence,
$$\measuredangle KIM=\measuredangle ILK=\measuredangle TAK=\measuredangle KAE=\measuredangle IAF=\measuredangle IPM,$$where we used the fact that $AIFP$ is cyclic due to proper right angles. Done.
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HoRI_DA_GRe8
584 posts
#18
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Solved it yay!!
sol
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TheCollatzConjecture
153 posts
#19
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yay a tst geo problem , finally!!!
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BVKRB-
322 posts
#20
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Solution
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lneis1
243 posts
#21
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smh it was really easy, idk why I wasted so much time.

Let $MD$ be the tangent to incircle, where $D \in (I)$, then clearly $ID \perp MP$

Hence $\square PAID$ is cyclic.
Note that it suffices to show that $\angle AIP=\angle IMP$ due to tangent secant theorem.

But because of concyclicity $\angle AIP=\angle ADP$, hence it reduces to proving $AD || IM$, which well known and trivial due to the following
EGMO wrote:
Let $ABC$ be a triangle whose incircle is tangent to $BC$ at $D$. If $DE$ is a diameter of the incircle and ray $AE$ meets $BC$ at $X$, then $BD = CX$ and $X$ is the tangency point of the $A-excircle$ to $BC$.
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Hedra
110 posts
#22
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Inverting around the incircle we see , Let $O$ be the circumcenter of $\triangle PIM$,$P'M'||A'I$ and $P'M'$ passes through the intersection points of $(PIM)$ and incircle , Thus $P'M'$ is the radical axis of these 2 circles $\implies P'M' \perp OI$ , say $P'N' \cap OI = N$ and since $O , O' , I$ are collinear we can say $\angle ON'P' = \angle O'NP' = \frac{\pi}{2}$ but $P'M'||A'I \implies \angle O'N'P' = \angle O'IA' = \frac{\pi}{2}$ , Inverting back we get $\angle AIO = \frac{\pi}{2} \implies AI \perp OI$ . $\blacksquare$
This post has been edited 2 times. Last edited by Hedra, Dec 2, 2021, 1:10 PM
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REYNA_MAIN
41 posts
#23 • 1 Y
Y by SPHS1234
Shortage
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tsiannis
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#24
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Let $D$ the tangency point of incircle and $BC$. We consider $D'$ the reflection of $D$ through $I$ and $E$ the reflection of $D$ through $M$. It is a well-known fact that $A,D',E$ are collinear and let $X$ the intersection of $AE$ and incircle. $\angle DXD'=90^\circ$ , hence $\angle DXE=90^\circ$. So, $DM=ME=MX$ and $X$ is the tangency point of the line $MP$ and incircle. $APXI$ is cyclic so $\angle AIP= \angle AXP= \angle MXE= \angle IMX$ because $IM$ and $AE$ are parallel. So, $AI$ is tangent to $(MIP)$ since $\angle AIP= \angle IMP$
This post has been edited 1 time. Last edited by tsiannis, Mar 18, 2022, 9:46 PM
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maolus
75 posts
#25
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This my ideas:
MP intersect (I) at K.
It's clearly that when you proved: AK//IM it's will be done. My hint for is AK intersect (I) at S and prove SD is perpendicular to BC (use the common lemma: S' is the intersection of the line through I perpendicular to BC, AS' intersect BC at N then BD = NC. Then prove S' is S)
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VicKmath7
1385 posts
#26
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Cute.
Let $D'$ be the antipode of $D$, $AD' \cap BC =E$ and $AD' \cap (I)=E'$. It is well-known that $ME'$ is tangent to the incircle and that $IM \parallel AE$. The rest is angle chasing: $\angle IME'= \angle ME'E= \angle AE'P=\angle AIP$, where the last equality follows from $AIE'P$ being cyclic with diameter $PI$.
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Sanjana42
17 posts
#27
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Let $D$ and $K$ be the intouch and extouch points respectively on $BC$. Let $Q$ be the point other than $D$ such that $MQ$ is tangent to the incircle. (So $Q$ lies on $MP$). $\angle IQP = 90^\circ = \angle AIP \implies AIQP$ cyclic.

Claim: $\angle AQD = 90^\circ$.
Proof: Let $AK$ intersect the incircle at $Q'$. Let $L$ be the antipode of $D$ on the circumcircle, let $N$ be a point on $AK$ on the opposite side of $K$ w.r.t. $Q'$, and let $I_A$ be the excenter. Note that $L$ lies on $AQ'KN$, by homothety. If $\angle DLQ' = \theta$, then $\angle Q'DK$ and $\angle I_AKN$ are also $\theta$, because of tangency and homothety respectively. Also $\angle DKI_A=90^\circ$ (tangency).
So $\angle DKQ' = 180^\circ - \theta - 90^\circ = 90^\circ - \theta$. Therefore in $\triangle Q'DK$, $\angle DQ'K = 180^\circ - \theta - (90^\circ - \theta) = 90^\circ$. This implies $\angle AQ'D = 90^\circ$ and $MD=MQ' \implies Q=Q'$, proving our claim.

Note that $\angle AQD = 90^\circ = \angle IQM \implies \angle DQM = \angle IQA = \angle IPA$. So $\angle IMP = 90^\circ -\angle DQM = 90^\circ - \angle IPA = \angle AIP \implies AI$ is tangent to $(MIP)$ as desired.
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mannshah1211
651 posts
#28
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Cute, and easy problem. Still took me like 30 minutes though. Let the tangent from point $M$ meet the incircle at point $Q$, let $D$ be the tangency point of incircle with $\overline{BC},$ and let $T$ be the tangency point of the $A$-excircle with $\overline{BC}$.
Claim: $A, Q, T$ are collinear.
Proof. Let $\overline{AT}$ meet the incircle at points $X', Q'$ where $X'$ is the one further from $T$. Then, we have by Diameter Of Incircle Lemma that $P', I, D$ are collinear. Thus, $\angle DQ'X' = 90^{\circ},$ which means that $\angle DQ'T = 90^{\circ},$ which means that the circle with diameter $\overline{DT}$ meets the incircle at point $Q'$, which means that $MD = MQ'$ (since $M$ is the midpoint of $\overline{DT}$, it is the center of this circle), which means that $MQ'$ is tangent to the incircle, and thus $Q \equiv Q',$ completing the proof.

Now, it is just angle chasing, $\angle IMQ = \frac{\angle DMQ}{2} = \angle MQT = \angle AQP = \angle AIP$ (where the last equality follows from the fact that $AIQP$ is cyclic).
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SimplisticFormulas
80 posts
#29 • 1 Y
Y by Combe2768
Let $F$ be the diameter of the incircle opposite to $D$. Let $AF \cap BC=N$ and $AF$ met the incircle in $K$.
It is well known that $N$ is the $A$-excentre tangency point, and that $DM=MN$.
The key claim is that $A,I,K,P$ are concyclic.
For this note that $\angle IKM=\angle IKD+\angle DKM=\angle IDK+\angle MDK=\angle FDM=90=\angle IAP$. Since $\angle IKM=90, $ K is the second tangency point of $M$ with the incircle and $P-K-M$.
So $\angle AIP=\angle AKP=\angle IMP$ and we are done.$\blacksquare$
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bin_sherlo
661 posts
#30
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Headsolved :cool:
Invert around the incircle.
New Problem Statement: $ABC$ is a triangle with circumcenter $O$. Let $A'$ be the antipode of $A$ and $D\in (ABC)$ such that $-1=(A',D;B,C)$. If $P$ is the foot of the altitude from $O$ to $MD$, then show that $NP\perp BC$.
Let $DM\cap (ABC)=E$. $-1=(A',D;B,C)=(EA',ED;EB,EC)=(EA'\cap BC,M;B,C)$ thus, $EA'\parallel BC$ which gives $NP\parallel AE\perp BC$ as desired.$\blacksquare$
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cursed_tangent1434
548 posts
#31
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Quite simple. Let $X$ denote the $A-$extouch point and $E$ the second tangency point from $M$ to the incircle. It is well known that points $A$ , $E$ and $X$ are collinear with $AX \parallel MI$. Now, we note that since $AIXP$ is cyclic due to the right angles we have,

\[\measuredangle MPI = \measuredangle EPI = \measuredangle EAI = \measuredangle XAI = \measuredangle MIA\]which implies the desired tangency.
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Retemoeg
47 posts
#32
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Denote $N$ and $Q$ the midpoint of minor arc $BC$ and arc $BAC$. Incircle touches $BC$ at $D$. Common result: $NI = NB = NC$. Now, $NM\cdot NQ = NB^2 = NI^2 \implies \triangle NMI \sim \triangle NIQ$. Thus, $\angle AQI = \angle IMD = \angle IMP$, implying that $IQPM$ is cyclic. The rest is trivial, as now, $\angle AIP = \angle AIQ + \angle PIQ = \angle IMQ + \angle QMP = \angle IMP$, showing that $IA$ is tangent to $(MIP)$, as desired.
This post has been edited 1 time. Last edited by Retemoeg, Feb 27, 2025, 5:00 PM
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ihategeo_1969
158 posts
#33
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We define some new points.

$\bullet$ Let $N$ and $L$ be major and minor arc midpoints of $\widehat{BC}$.
$\bullet$ Let $T$ and $X$ be $A$-mixtilinear intouch and $A$-extouch point.
$\bullet$ Let $D$ be $A$-intouch point.
$\bullet$ $D'$ be $D$-antipode of incircle and $X'=\overline{MP} \cap \text{ incircle}$ and it is well known that $\overline{AD'X'X}$ is a line and is parallel to $\overline{IM}$.

Claim: $(NPMI)$ is cyclic.
Proof: Just one big aah angle chase. \begin{align*}
\angle PMI=\angle X'MI =180 ^{\circ}-\angle D'X'M = \angle D'DX' = 90 ^{\circ}-\angle DD'X &= \angle AXB \overset{\sqrt{bc}}= \angle ACT = \angle ANI=180 ^{\circ}-\angle PNI
\end{align*}And done. $\square$

To finish see that $LM \cdot LN=LI^2$ (by $(BIC)$ inversion) and so $\overline{AIL}$ is tangent to $(NIM)$.
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