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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
Nice function question
srnjbr   1
N 3 minutes ago by Mathzeus1024
Find all functions f:R+--R+ such that for all a,b>0, f(af(b)+a)(f(bf(a))+a)=1
1 reply
srnjbr
5 hours ago
Mathzeus1024
3 minutes ago
Inequality and function
srnjbr   5
N 17 minutes ago by pco
Find all f:R--R such that for all x,y, yf(x)+f(y)>=f(xy)
5 replies
srnjbr
Yesterday at 4:26 PM
pco
17 minutes ago
Difficult factorization
Dakernew192   1
N 22 minutes ago by Thursday
x^5-2x+6
1 reply
Dakernew192
Jan 8, 2024
Thursday
22 minutes ago
every point is colored red or blue
Sayan   8
N an hour ago by Mathworld314
Source: ISI(BS) 2005 #9
Suppose that to every point of the plane a colour, either red or blue, is associated.

(a) Show that if there is no equilateral triangle with all vertices of the same colour then there must exist three points $A,B$ and $C$ of the same colour such that $B$ is the midpoint of $AC$.

(b) Show that there must be an equilateral triangle with all vertices of the same colour.
8 replies
Sayan
Jun 23, 2012
Mathworld314
an hour ago
No more topics!
Line through incenter tangent to a circle
Kayak   32
N Yesterday at 10:22 AM by L13832
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
32 replies
Kayak
Jul 17, 2019
L13832
Yesterday at 10:22 AM
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 • 5 Y
Y by sameer_chahar12, ashraful7525, abhisruta03, Adventure10, MS_asdfgzxcvb
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
429 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
587 posts
#18
Y by
Solved it yay!!
sol
This post has been edited 5 times. Last edited by HoRI_DA_GRe8, Jan 18, 2022, 10:29 AM
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TheCollatzConjecture
153 posts
#19
Y by
yay a tst geo problem , finally!!!
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BVKRB-
322 posts
#20
Y by
Solution
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lneis1
243 posts
#21
Y by
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
Y by
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
Reason: .
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REYNA_MAIN
41 posts
#23 • 1 Y
Y by SPHS1234
Shortage
Attachments:
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tsiannis
5 posts
#24
Y by
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
Y by
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
Y by
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
18 posts
#27
Y by
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
Y by
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
81 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
663 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
550 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
49 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
164 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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L13832
250 posts
#34
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solution
This post has been edited 1 time. Last edited by L13832, Yesterday at 10:22 AM
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