Line through incenter tangent to a circle

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Kayak

1298 posts

Kayak

#1 h Jul 17, 2019, 12:26 PM • 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

MathPassionForever

#2 h Jul 17, 2019, 12:32 PM • 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

Ankoganit

#3 h Jul 17, 2019, 12:34 PM • 5 Y

Y by sameer_chahar12, ashraful7525, abhisruta03, Adventure10, MS_asdfgzxcvb

Solution from the official packet

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Pluto1708

1107 posts

Pluto1708

#4 h Jul 17, 2019, 12:45 PM • 3 Y

Y by sameer_chahar12, Adventure10, Mango247

Solution

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Kayak

1298 posts

Kayak

#5 h Jul 17, 2019, 12:47 PM • 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

AlastorMoody

#6 h Jul 18, 2019, 9:50 AM • 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

zuss77

#7 h Jul 18, 2019, 11:43 AM • 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

jayme

#8 h Sep 12, 2019, 8:50 AM • 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

e_plus_pi

#9 h Sep 17, 2019, 3:03 PM • 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

mashina

#10 h Nov 18, 2019, 7:04 AM • 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

ubermensch

#11 h Jan 12, 2020, 7:13 AM • 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

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aops29

#12 h Feb 13, 2020, 7:03 PM • 2 Y

Y by Adventure10, Mango247

Simple

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arzhang2001

248 posts

arzhang2001

#13 h Aug 12, 2020, 5:00 PM

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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

hakN

#15 h Jul 5, 2021, 3:16 PM

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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

primesarespecial

#16 h Aug 23, 2021, 1:22 PM

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

rafaello

#17 h Aug 25, 2021, 12:49 PM

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

HoRI_DA_GRe8

#18 h Oct 21, 2021, 5:48 PM

Y by

Solved it yay!!
sol

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TheCollatzConjecture

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TheCollatzConjecture

#19 h Oct 21, 2021, 6:31 PM

Y by

yay a tst geo problem , finally!!!

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BVKRB-

322 posts

BVKRB-

#20 h Nov 13, 2021, 8:18 AM

Y by

Solution

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lneis1

243 posts

lneis1

#21 h Nov 13, 2021, 2:40 PM

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

Hedra

#22 h Dec 1, 2021, 9:15 AM

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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
Reason: .

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REYNA_MAIN

41 posts

REYNA_MAIN

#23 h Dec 22, 2021, 2:05 PM • 1 Y

Y by SPHS1234

Shortage

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tsiannis

5 posts

tsiannis

#24 h Mar 18, 2022, 9:44 PM

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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

maolus

#25 h Mar 19, 2022, 10:25 AM

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

VicKmath7

#26 h Oct 7, 2022, 5:56 PM

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

Sanjana42

#27 h Nov 20, 2023, 9:01 PM

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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

mannshah1211

#28 h Jan 16, 2024, 8:11 PM

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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

81 posts

SimplisticFormulas

#29 h Dec 28, 2024, 6:32 AM • 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

bin_sherlo

#30 h Dec 28, 2024, 8:44 AM

Y by

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

cursed_tangent1434

#31 h Feb 27, 2025, 12:35 AM

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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

Retemoeg

#32 h Feb 27, 2025, 4:54 PM

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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

ihategeo_1969

#33 h Mar 14, 2025, 9:16 PM

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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 $(NIM)$ .