A_1, M,Q,P lie on a circle

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A_1, M,Q,P lie on a circle

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ts0_9

#1 h May 25, 2012, 2:34 PM • 2 Y

Y by mathematicsy, Adventure10

Given a non-degenerate triangle $ABC$ , let $A_{1}, B_{1}, C_{1}$ be the point of tangency of the incircle with the sides $BC, AC, AB$ . Let $Q$ and $L$ be the intersection of the segment $AA_{1}$ with the incircle and the segment $B_{1}C_{1}$ respectively. Let $M$ be the midpoint of $B_{1}C_{1}$ . Let $T$ be the point of intersection of $BC$ and $B_{1}C_{1}$ . Let $P$ be the foot of the perpendicular from the point $L$ on the line $AT$ . Prove that the points $A_{1}, M, Q, P$ lie on a circle.

This post has been edited 3 times. Last edited by v_Enhance, May 26, 2012, 1:15 PM
Reason: Repaired English. Hope you don't mind.

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#2 h May 27, 2012, 6:24 AM • 5 Y

Y by Tima95, JustN, HamstPan38825, Adventure10, Mango247

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yunxiu

#3 h May 28, 2012, 6:29 AM • 2 Y

Y by Adventure10, Mango247

Because the polar of ${A_1}, A$ about $\left( I \right)$ intersect at $T$ , $A{A_1}$ is the polar of $T$ about $\left( I \right)$ , so $TQ$ touch $\left( I \right)$ at $Q$ .
So $M,Q,{A_1},I,T$ are all on the circle $\omega$ with diameter $IT$ .
Because the polar of $A,T$ about $\left( I \right)$ intersect at $L$ , $AT$ is the polar of $L$ about $\left( I \right)$ , so $IL \bot AT$ , hence $I,L,P$ are collinear, so $P$ is on the circle $\omega$ , we done.

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gold46

#4 h May 28, 2012, 10:00 AM • 2 Y

Y by Adventure10, Mango247

Let $I$ be incenter. $AQ\cdot AA_1=AB_1^2=AM\cdot AI \implies A_1, I , M,Q$ lie on a circle (*). On the other hand , we've $AB_1IC_1P$ is cyclic. (All points lie a circle with AB diameter.) Thus $LI\cdot LP=LC_1\cdot LB_1=LQ\cdot LA_1 \implies P,Q,I,A_1$ points lie on a circle (**). Combining (*) , (**) results , we've $P , M \in ( A_1IQ )$ as desired. We're done

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#5 h Jun 8, 2020, 7:41 PM

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We claim that $A_1,M,Q,P$ lie on a circle with diameter IT. Since $IM \perp B'C'$ , $IA' \perp BC$ , we only need to prove that $IL \perp AT$

We claim L is the pole of AT.
Proof:
a: We see that (BC,A'T)=-1 and thus (C',B';L,T)=-1. Thus, A'L is the polar of T. Thus, T lies on the polar of L.
b: L lies on B'C' the polar of A, so A lies on the polar of L.

Thus, AT is the polar of L and $IL \perp AT$ so I,L,P are collinear.

Thus $A_1,M,Q,P$ lie on a circle with diameter IT

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algebra_star1234

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algebra_star1234

#6 h Jun 13, 2020, 11:32 PM

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Note that $A_{1}C_{1}QB_{1}$ is harmonic, so this means that $TQ$ is also tangent to the incircle. Also, since $AA_{1}$ , $BB_{1}$ , and $CC_{1}$ intersect at the Gergonne point, we have that $-1= (TA_{1};BC) \stackrel A = (TL;C_{1}B_{1})$ . . Therefore, we have $TQ^2 = TC_{1}\cdot TB_{1} = TL \cdot TM.$ This means that $TQ$ is tangent to $(QLM)$ so $\angle TA_1Q = \angle TQA_1 = \angle TMQA_1$ , and $TQMA$ is cyclic. We also have that $PAML$ is cyclic since $\angle APL = \angle AML = 90^{\circ}$ , so $TP \cdot TA = TM \cdot TL = TC_{1}\cdot TB_{1}$ . Therefore, $PAB_{1}C_{1}$ is cyclic. We can now find $\angle TPC_{1} = \angle PC_{1}B_{1} = 90 -\frac{A}{2}$ , which implies $\angle ATC_{1}=\angle PC_{1}A$ . Therefore, $AC_{1}$ is tangent to $(APC_{1}T)$ . We have $AQ \cdot AA_{1}= AC_{1}^2 = AP \cdot AT$ , so $TPQA_{1}$ is cyclic and the result follows.

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#7 h Sep 28, 2020, 5:48 AM

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Note $TQ, TD$ are tangent to incircle, and $AI\perp EF$ , so $T, Q, M, I, D$ are concyclic. However, since the polar of $A$ is $EF$ and the polar of $T$ is $AD$ , the polar of $L$ is $AT$ , so thus $I, L, P$ are collinear. Hence, $IP\perp AT$ so $TPID$ is cyclic, done.

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pad

#8 h Dec 30, 2020, 9:56 PM

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Diagram

Claim: $T,Q,M,A_1$ are concyclic.

Proof: Note $(TL;C_1B_1) \stackrel{A}{=} (TA_1;BC)=-1$ by Ceva-Menelaus. So $ML\cdot MT=MB_1^2$ , so $LM\cdot LT = ML\cdot MT - ML^2 = MB_1^2-ML^2 = LB_1\cdot LC_1.$ Now $LQ\cdot LA_1=LB_1\cdot LC_1 = LM\cdot LT$ , proving the concyclicity. $\blacksquare$

Claim: $T,P,Q,A_1$ are concyclic.

Proof: Since $\angle AML=\angle APL=90^\circ$ , we have $APLM$ is cyclic. Since $(TL;C_1B_1)=-1$ , we have $TP\cdot TA = TL\cdot TM = TC_1\cdot TB_1$ , so $PAB_1C_1$ is cyclic. Now, we claim $\overline{AC_1}$ is tangent to $(PC_1T)$ . Indeed, $\begin{align*} \angle PC_1A &= \angle PB_1A_1 = 180^\circ - \angle PAB_1 - \angle APB_1 = 180^\circ - \angle TAC - \angle AC_1B_1, \\ \angle ATC_1 &= \angle ATB_1 = 180^\circ - \angle TAC - \angle AB_1C_1, \end{align*}$ which are equal, proving the tangency. Hence $AP\cdot AT = AC_1^2 = AL\cdot AA_1$ , proving the concyclicity. $\blacksquare$

Combining the above two claims implies that $T,A_1,M,Q,P$ are concyclic, implying the desired.

Remarks

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#9 h Jan 1, 2021, 11:02 PM

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Very nice

$\color{black}\rule{25cm}{1pt}$
Let $I$ be the incenter of $ABC$ .

Since $AA_1$ , $BB_1$ and $CC_1$ are concurrent we have that:
$-1=(T,A_1;B,C)\overset{A}{=}(T,L;C_1,B_1)\overset{A_1}{=}(A_1,Q;B_1,C_1)$ thus we have that $TQ$ is a tangent to the incircle.
Since we have that $\angle IA_1T=\angle IQT=\angle IMT = 90$ , we just need to show that $I,L,P$ are colinear.

Since we have that the polar of $T$ is $AA_1$ w.r.t. incircle of $ABC$ , this holds since we have the tangents $TQ$ and $TA_1$ , by La Hire we have that $T$ must be on the polar of $A$ w.r.t the incircle of $ABC$ , but this holds true since the polar of $A$ w.r.t. the incircle is $B_1C_1$ and $T$ is on $B_1C_1$ .

Now since $L$ is the intersection of $AA_1$ and $B_1C_1$ , this would imply by La Hire (again) that $AT$ is the polar of $L$ , w.r.t. to the incenter, is $AT$ . This implies that the pole of $L$ w.r.t. the incircle of $ABC$ is $P$ , this implies that $I,L,P$ are colinear, thus we have that $\angle IPT = 90$ .

Thus we are done.

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#10 h Sep 5, 2021, 2:48 AM

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The points are concyclic along the circle with diameter $\overline{TI}$ . Obviously $A_1$ and $M$ lie on this circle, and $Q$ also lies on it because the tangent at $Q$ passes through $T$ is well-known. Now by La Hire's, $A$ and $T$ both lie on the polar of $L$ wrt the incircle, so $AT$ is the polar of $L$ and it follows that $I, L, P$ are collinear, so $P$ also lies on the circle.

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#11 h Aug 13, 2022, 3:40 AM • 1 Y

Y by centslordm

We claim all the points lie on the circle with diameter $\overline{TI},$ where $I$ is the incenter. Since $T$ lies on the polar of $A,$ La Hire yields that $A$ is on the polar of $T$ so $\overline{QT}$ is tangent to the incircle at $Q.$ Also, $L$ lies on the polar of $A$ and $T,$ so $\overline{AT}$ is the polar of $L$ so $\overline{AT}\perp\overline{IL}.$ Hence, $I,P,L$ are collinear and $\angle IPT=90.$ It is clear that $\angle IQT=\angle IA'T=\angle IMT=90,$ so we are done. $\square$

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#12 h Sep 3, 2022, 11:43 AM

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Since $B_1C_1=A^*$ , then $T$ lies on the polar of $A$ . Then, $A$ lies on the polar of $T$ . Thus $TQ$ is tangent to the incircle.

Claim: The line $LP$ passes through $I$
Proof: Since $L=A^*\cap T^*$ , then $L^*=AT$ . Thus $IL\perp AT$ , so we have $I\in LP$ . $\blacksquare$

We have that, $IP\perp TP$ , $IQ\perp TQ$ and as $M$ is the midpoint of $B_1C_1$ , $IM\perp TM$ . So $PQMIT$ is a cyclic with $IT$ as the diameter. Since $IA_1\perp TA_1$ , we have $A_1\in (PQM)$ . $\blacksquare$
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#13 h Mar 8, 2023, 5:57 AM • 7 Y

Y by Mattthebat2569, Significant, Taco12, Bluedevils, programmeruser, mahaler, mulberrykid

New favorite problem??

https://media.discordapp.net/attachments/908472255858753536/1082907345979453480/IMG_3639.png?width=787&height=387

By 02IRN, the tangent to the incircle at $Q$ passes through $T$ . Thus, since $L$ lies on the polar of $A$ and the polar of $T$ , $AT$ is the polar of $L$ .

Let the center of the incircle be $I$ . We claim that $TI$ is the diameter of the circumcircle of $A_1MQP$ . By Brokard's on $QB_1A_1C_1$ (the extended theorem claims the center of the incircle is the orthocenter of $PQR$ ), we have that $P$ , $L$ , and $I$ are collinear. Therefore, we have that $TP\perp{}TI$ . Next, since $TQ$ and $TA_1$ are both tangents to the incircle, we have that $TQ\perp{}QI$ and $TA_1\perp{}A_1I$ . Finally, since $M$ is the midpoint of the chord $B_1C_1$ of the incircle, we have that $TM\perp{}MI$ .

Therefore, $A_1$ , $M$ , $Q$ , and $P$ all lie on the circle with diameter $TI$ , and we are done.

This post has been edited 4 times. Last edited by peppapig_, Mar 8, 2023, 6:07 AM

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#14 h Aug 10, 2023, 8:48 AM

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Claim: The polar of $L$ is $AT$ .

Since the polar of $A$ is line $B_1C_1$ , and $T$ lies on it, we have that $A$ also lies on the polar of $T$ . Due to the tangency, the polar of $T$ is just $AA_1$ . Note that $L$ is also on this, so $T$ is on the polar of $L$ also. Clearly, $A$ is also on the polar of $L$ since the polar of $A$ is $B_1C_1$ , hence proven.

This means that $IL\perp AT$ , which also means that $P,L,I$ are collinear.

Then, we have $\angle TNI=90$ from tangency, $\angle TMI=90$ obviously as well, $\angle TQI=90$ since the polar of $T$ is $AA_1$ which contains $Q$ so $TQ$ is tangent to the incircle, and finally $\angle TPI=\angle TPL=90,$ hence they all lie on $(IT)$ and we are done.

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#15 h Aug 10, 2023, 7:31 PM

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We claim that all the points lie on the circle with diameter $\overline{TI}$ . Clearly $A_1$ and $M$ lie on this circle. By La Hire's we find that:

$A$ lies on the polar of $T$ , since line $B_1C_1$ is the polar of $A$ . This means that $AA_1$ is the polar of $T$ , so $\overline{TQ}$ is tangent to the incircle, and $\overline{TQ} \perp \overline{QI} \implies Q$ lies on the desired circle.
$A$ and $T$ both lie on the polar of $L$ , so the polar of $L$ is just line $AT$ . Thus, since $\overline{IL} \perp \overline{AT}$ and $\overline{LP} \perp \overline{AT}$ , points $I$ , $L$ and $P$ must be collinear, and hence $\overline{PT} \perp \overline{PI} \implies P$ lies on the desired circle.

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#16 h Aug 20, 2023, 2:58 PM • 1 Y

Y by centslordm

I claim they lie on the circle with diameter $\overline{TI}$ , where $I$ is the incircle. Indeed, $\angle TA_1I=\angle TMI=90^\circ$ is clearly true. By La Hire's, since $T$ lies on the polar of $A$ , $A$ lies on the polar of $T$ , and so does $A_1$ , hence $Q$ does too, i.e. $\overline{TQ}$ is tangent to the incircle, so $\angle TQI=90^\circ$ . Finally, by La Hire's, since $L$ lies on the polars of $A$ and $T$ , $A$ and $T$ lie on the polar of $L$ , hence $\overline{IL} \perp \overline{AT}$ , hence $P,L,I$ are collinear so $\angle TPI=90^\circ$ . $\blacksquare$

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#17 h Sep 20, 2023, 7:08 PM

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Let $Q'\neq A_1$ be the unique point lying on the incircle $\omega$ of $\triangle ABC$ such that $TQ'$ is tangent to $\omega$ . Noting that $AB_1$ , $AC_1$ , $TA_1$ , and $TQ'$ are all tangent to $\omega$ , $Q$ lies on $AA_1$ , and $T$ lies on $B_1C_1$ , it follows that $A_1B_1C_1Q$ and $A_1B_1C_1Q'$ are both harmonic, implying that $Q=Q'$ .

It thus follows that $TQ$ is tangent to $\omega$ , which is centered at $I$ , so that $\angle TQI=90^\circ$ . However, we have by definition that $\angle TPI=\angle TA_1I=90^\circ$ , and we also have that $\angle TMI=\angle B_1MI=90^\circ$ , as $\triangle B_1C_1I$ is isosceles with $B_1I=C_1I$ and $M$ being the midpoint of $B_1C_1$ . Thus $\angle TA_1I=\angle TMI=\angle TQI=\angle TPI$ , so that $A_1MQP$ is cyclic with $TI$ being the diameter of its circumcircle, proving the desired. $\blacksquare$

- Jörg

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#18 h Dec 30, 2023, 6:26 PM • 1 Y

Y by GeoKing

Note that both $L$ lies on the polar of $T$ and also on the polar of $A$ . So by using La Hire's Theorem, we get that $A$ and $T$ both lie on the polar of $L$ , that is, the line $AT$ is the polar of $L$ . Thus we have that $IL\perp AT$ , that is, $\overline{I-L-P}$ are collinear.

Now note that since $Q$ lies on the polar of $T$ , we get that $\measuredangle IQT = 90^\circ$ . Finally we have that, $\measuredangle IPT=\measuredangle IA_1T=\measuredangle IMT=\measuredangle IQT = 90^\circ \implies IMQPTA_1$ is cyclic. :yoda:

:yoda:

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#19 h Aug 21, 2024, 5:27 AM • 1 Y

Y by dolphinday

Denote $I$ as the incenter of $\triangle ABC$ .

We claim that the circle is the circle with diameter $\overline{IT}$ . It is easy to see that $A_1$ and $M$ lie on this circle. For point $Q$ , note that $A_1B_1QC_1$ is harmonic, so $\overline{QT}$ is tangent to the incircle.

Now, to get started on point $P$ , we make a claim:

Claim: The polar of $L$ with respect to the incircle is $\overline{AT}$ .

Proof: Note that the polar of $A$ is $\overline{B_1C_1}$ , so $L$ lies on the polar of $A$ . Similarly, $L$ lies on the polar of $T$ as well. Thus, La Hire implies that $A$ and $T$ lie on the polar of $L$ . $\square$

Now, we can determine that $I$ , $L$ , and $P$ are collinear, which implies it lies on the desired circle. $\square$

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#20 h Mar 28, 2025, 4:54 AM

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wut

Let $I$ be the incenter.
With respect to the incircle: $A$ is the pole of $B_1C_1,$ so the polar of $T$ is $AA_1.$
Hence, $TQ$ is tangent to the incircle.
Hence, $\angle IQT = 90.$
Clearly, $\angle IA_1T = 90.$
Obviously, $\angle IMT = 90.$
Define $P’$ to be the second intersection of $(AC_1IB_1)$ with $(QMIA_1T).$ By Radax on $(AC_1IB_1), (QMIA_1T)$ and the incircle, we get $I, L, P’$ are collinear. However, it is obvious that $\angle AP’I = 90.$ Thus, $P’ = P.$
Thus, $\angle IPT = 90.$

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