Let $ABC$ be a triangle and $I$ its incenter. The lines $BI$ and $CI$ intersect the circumcircle of $ABC$ again at $M$ and $N$, respectively. Let $C_1$ and $C_2$ be the circumferences of diameters $NI$ and $MI$, respectively. The circle $C_1$ intersects $AB$ at $P$ and $Q$, and the circle $C_2$ intersects $AC$ at $R$ and $S$. Show that $P$, $Q$, $R$ and $S$ are concyclic.
Problem
Source: Cono Sur 2021 P2
Tags: geometry, Cyclic Quadrilaterals, triangle -incenter, circumcircle
MathLuis
01.12.2021 00:28
Solution during the test Let $(MI) \cap (NI)=D$ and let $U$ be the midpoint of the minor arc $BC$. Its known that $I$ is ortocenter of $\triangle MNU$ and since $\angle MDI=90=\angle NDI$ we have $M,N,D$ colinear and $ID \perp MN$ hence $D$ lies on $AI$ thus now by PoP we have $AP \cdot AQ=AI \cdot AD=AR \cdot AS$ thus $PQRS$ cyclic as desired, thus we are done Easiest problem of the test ?
DottedCaculator
01.12.2021 00:31
Since $C_1$ and $C_2$ pass through the midpoint of $AI$ by the Incenter-Excenter lemma, this means that $A$ lies on the radical axis of $C_1$ and $C_2$. Therefore, $AP\cdot AQ=AR\cdot AS$, so $PQRS$ is cyclic.
Leo890
14.02.2022 23:31
Let $O_1$ be the center of $C_1$ and $O_2$ the center of $C_2$. We want to show that $AP \cdot AQ = AR \cdot AS \Longleftrightarrow AO_1^2-\frac{NI^2}{4} = AO_2^2-\frac{MI^2}{4}$
Observe that $\angle AIN = 90^{\circ}- B \Longrightarrow cos(\angle AIN) = sen(B)$. Similarly we have $\angle AIM = sen(C)$. By law of cosines in triangles $AO_1I$ and $AO_2I$:
$$AO_1^2 = AI^2+\frac{NI^2}{4}-AI \cdot NIsen(B)$$$$AO_2^2 = AI^2+\frac{MI^2}{4}-AI \cdot MIsen(C)$$Subtracting, is equivalent to prove that $NIsen(B) = MIsen(C)$. By incenter-excenter lemma we have $AN=NI$ and $AI=MI$. By law of cosines in triangle $ANI$ and $AMI$:
$$NI^2 = AN^2 + AI^2-2AN \cdot AI sen(B)$$$$MI^2 = AM^2 + AI^2-2AM \cdot AI sen(C)$$Subtracting again we conclude that $AM sen(C) = AN sen(B) \Longrightarrow NIsen(B) = MIsen(C)$, as desired. $\blacksquare$
Mahdi_Mashayekhi
08.03.2022 21:42
Let $C_1$ and $C_2$ meet at $I,S$. Claim1 : $I,S,A$ are collinear. Proof : It's well known that $NI = NA$ and $MI = MA$ so $NAMI$ is kite and we have $\angle ISN = \angle 90$ so $I,S,A$ are collinear. Now we have $AP.AQ = AS.AI = AS.AR$ so $PQRS$ is cyclic.
PennyLane_31
24.05.2023 14:00
First, notice that if $AI$ is the radical axis of $C_1, C_2$, then, as $AI, PQ, RS$ are concurrent $\Leftrightarrow PQRS$ is cyclic, and the problem is finished.
So, let's prove that $AI$ is the radical axis of $C_1, C_2$. For that, call $B_1$ the second intersection of $C_1, C_2$ (the first one is $I$, clearly). (see attachment).
As $NI$ is the diameter of $C_1\implies \angle NB_1I= \angle NPI= 90º$(1). Similarly, $\angle MB_1I= 90º\implies M, B_1, N$ are collinears.
Also, as $ANBC$ is cyclic $\implies \angle ANI= \angle ANC= \angle ABC= B.$ (2)
And, as $MNCB$ is cyclic $\implies \angle B_1NI= \angle MNC= \angle MBC= \angle IBC= \frac{B}{2}$.(3)
From the Incenter/Excenter Lemma, is known that $AN= NI= NB\implies ANI$ is isosceles.(4)
By (1), (2), (3), and (4), we discover that $NB_1$ is the angle bisector of $\angle ANI$ (because $B= \angle ANI= \angle B_1NI+\angle B_1NA\implies B= \frac{B}{2}+\angle B_1NA\implies \angle B_1NA= \frac{B}{2})$, similarly, we have for $C_2\implies MB_1$ the angle bisector of $\angle AMI$. And, since $\angle NB_1I= 90º\implies NB_1\perp B_1I\implies NB_1I\perp AI$ (notice that the angles are still the same, because, as $ANI$ is isosceles $\implies \angle NAI= \angle AIN= 90º-\frac{B}{2}$, and as $\angle B_1NI= \frac{B}{2}$(by (3)) $\implies \angle B_1IN= 90-\frac{B}{2}=\angle AIN$). Therefore, $NB_1\perp AI$ at $B_1$ ($B_1$ is the projection of $N$ in $B_1I$ and is the intersection of the angle bisector of $\angle ANI$ with $AI$).
So, we conclude that $B_1$ is the midpoint of $AI\implies B_1, A, I$ are collinear $\implies A\in$ the radius axis of $C_1, C_2$, as we wanted.
Attachments:
Thelink_20
18.07.2024 21:19
$PQRS$ is cyclic$\iff AP\cdot AQ=AS\cdot AR\iff Pow_{C_1}A=Pow_{C_2}A$ Let $L$ and $K$ be the midpoints of $NI$ and $MI$. Note that $AL$ is a median of $\Delta ANI$. $Pow_{C_1}A=AL^2-\frac{NI^2}{4}=\frac{2AI^2+2AN^2-NI^2}{4}-\frac{NI^2}{4}=\frac{AI^2}{2}$. ($AN=NI$ by the incenter-excenter lemma) Analogously $Pow_{C_2}A=\frac{AI^2}{2}\Rightarrow PQRS$ is cyclic.
For those who don't know, the measure of the median relative to a vertex $A$ of a triangle $\Delta ABC$ can be expressed as:
$m^2=\frac{2b^2+2c^2-a^2}{4}$ And the proof can be obtained using the Law of Cossines.
zuat.e
19.07.2024 15:22
Let $K=AI \cap (ABC)$, $L=(NI) \cap (MI)$, $L'=AI \cap MN$. Note that, in directed angles, where $(XY)$ stands for the arc XY, $\angle AL'M$ = $\frac{(AM)+ (KN)}{2} = \frac {(AC) + (ABC)}{4} = 90º$, $\angle MLI= 90º$ and $M,L,N$ collinear, since $\angle MLI = \angle ILN = 90º$, hence $L \equiv L'$ and so $A-L-I$ collinear. Therefore, A lies in the radical axis of $(NI)$ and $(MI)$. The conclusion follows by power of the point A.
AshAuktober
30.11.2024 09:15
Note from angle chasing that $AI \perp MN$, so $C_1 \cap C_2 \setminus I = K$ lies on $AI$. But now we have $$AP \cdot AQ = AI \cdot AK = AR \cdot AS,$$so $PQRS$ is cyclic. $\square$