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 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 $BAC$.

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 $\odot (MIP)$

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.

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

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

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.

Nice and easy! The problem obviously follows from the fact that $MI\Vert AE$. But that's obvious as they both perpendicular to $XD$.

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.

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$\blacksquare$

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.

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, $\angle AIP= \angle ANP=\angle FNP=\angle FXN=\angle IMN=\angle IMP$.

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.

Solved it yay!!

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 $BC$.

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 $\angle AIO = \frac{\pi}{2} \implies AI \perp OI$ . $\blacksquare$

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 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)

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 $PI$.

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.

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).

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.$\blacksquare$

Headsolved 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.$\blacksquare$