Lines intersecting on the circumcircle (BxMO 2023, Problem 3)

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Lines intersecting on the circumcircle (BxMO 2023, Problem 3)

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Source: BxMO 2023, Problem 3

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#1 h May 6, 2023, 1:21 PM

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Let $ABC$ be a triangle with incentre $I$ and circumcircle $\omega$ . Let $N$ denote the second point of intersection of line $AI$ and $\omega$ . The line through $I$ perpendicular to $AI$ intersects line $BC$ , segment $[AB]$ , and segment $[AC]$ at the points $D$ , $E$ , and $F$ , respectively. The circumcircle of triangle $AEF$ meets $\omega$ again at $P$ , and lines $PN$ and $BC$ intersect at $Q$ . Prove that lines $IQ$ and $DN$ intersect on $\omega$ .

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PNT

#2 h May 6, 2023, 6:22 PM • 1 Y

Y by Funcshun840

Hint : Prove that the intersection point is the point of tangency of the mixtilinear incircle.

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#3 h May 6, 2023, 9:29 PM • 1 Y

Y by MS_asdfgzxcvb

This was quite classical. Let $M$ be the midpoint of major arc $BC$ . We split the proof into two Claims.

Claim 1: Lines $MI$ and $DN$ meet on $\omega$ .
Proof: Let $DN$ intersect $\omega$ at point $S$ . Since

$\angle DIB=\angle AIB-\angle AIE=(90^\circ+\dfrac{\angle C}{2})-90^\circ=\dfrac{\angle C}{2}=\angle ICB,$

we obtain that $DI$ is tangent to circle $(BIC)$ . Thus,

$DI^2=DB \cdot DC=DS \cdot DN,$

and so $IS$ is perpendicular to $DN$ , that is $\angle ISN=90^\circ=\angle MSN$ , hence points $M,I,S$ are collinear, as desired $\blacksquare$

Claim 2: Points $M,I,Q$ are collinear.
Proof: Let $MI$ intersect $BC$ at point $Q'$ . Then, by spiral similarity and the fact that $PQ$ bisects $\angle BPC$ , we obtain

$\dfrac{BQ}{QC}=\dfrac{BP}{PC}=\dfrac{BE}{CF}$

Now, by Menelaus' theorem and the fact that $DI$ is tangent to $(BIC)$ , we easily obtain that

$\dfrac{BE}{CF}=\dfrac{DB}{DC}=\dfrac{BI^2}{CI^2}$

Now, note that by an application of the ratio lemma and some sine laws,

$\dfrac{BQ'}{Q'C}=\dfrac{BM}{MC} \cdot \dfrac{\sin \angle BMI}{\sin \angle CMI}=\dfrac{BI^2}{CI^2},$

hence the two ratios $\dfrac{BQ}{QC}$ and $\dfrac{BQ'}{Q'C}$ are equal, and so $Q \equiv Q'$ , as desired $\blacksquare$

To the problem, combining both claims we are done.

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#4 h May 9, 2023, 9:01 PM

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Similar to https://artofproblemsolving.com/community/c6h476508p2667975, Iran TST 2012 Day 1Q2

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PNT

#5 h May 10, 2023, 5:00 PM

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PNT wrote:

Hint : Prove that the intersection point is the point of tangency of the mixtilinear incircle.

First let's prove that $DN\cap \omega =T$ where $T$ is the point of tangency of the mixtilinear incircle.
Let $DN\cap \omega =T'$ and $M$ the midpoint of arc $BAC$ then we want to prove that $M,I$ and $T'$ are collinear or equivalently $\angle BT'I=90^{\circ}$ , but $\angle DIN=90^{\circ}$ so it suffices to prove $DT'\cdot DN=DI^2$ . We have
$DT'\cdot DN=DB\cdot DC=DI^2$ That's because $DI$ is tangent to $(BIC)$ . Therefore $T'=T$ .
Note that $\angle DPN=\angle DPT+\angle TPN=\angle MNT+\angle TMN=90^{\circ}$ (We proved this above) that means $D,P$ and $M$ are collinear. Now define $Q'=MT\cap BC$ , proving $Q=Q'$ will finish the problem.
Since
$-1=(B,C;N,M)\overset{P}{=}(D,Q;B,C)\overset{M}{=}(P,T'';B,C)$ That means $PT''$ and $PN$ meet on $BC$ so $Q=Q'$ .

Attachments:

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#6 h Jun 29, 2023, 11:46 AM

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PNT wrote:

PNT wrote:

Hint : Prove that the intersection point is the point of tangency of the mixtilinear incircle.

First let's prove that $DN\cap \omega =T$ where $T$ is the point of tangency of the mixtilinear incircle.
Let $DN\cap \omega =T'$ and $M$ the midpoint of arc $BAC$ then we want to prove that $M,I$ and $T'$ are collinear or equivalently $\angle BT'I=90^{\circ}$ , but $\angle DIN=90^{\circ}$ so it suffices to prove $DT'\cdot DN=DI^2$ . We have
$DT'\cdot DN=DB\cdot DC=DI^2$ That's because $DI$ is tangent to $(BIC)$ . Therefore $T'=T$ .
Note that $\angle DPN=\angle DPT+\angle TPN=\angle MNT+\angle TMN=90^{\circ}$ (We proved this above) that means $D,P$ and $M$ are collinear. Now define $Q'=MT\cap BC$ , proving $Q=Q'$ will finish the problem.
Since
$-1=(B,C;N,M)\overset{P}{=}(D,Q;B,C)\overset{M}{=}(P,T'';B,C)$ That means $PT''$ and $PN$ meet on $BC$ so $Q=Q'$ .

You assumed the collinearity when you stated that $\angle DPT = \angle MNT$ , here's a proof for the collinearity:
Notice that $P$ is the center of the homothety sending $BE$ to $CF$ , and since $D$ is the intersection of $(EF)$ and $(BC)$ then $PEBD$ IS cyclic, thus $\angle PDI= \angle PDE = \angle PBE = \angle PBA = \angle PNA = \angle PNI$ thus, $PIND$ is cyclic and this gives us $\angle DPN = 90$ .

This post has been edited 2 times. Last edited by Cusofay, Apr 5, 2024, 8:56 PM

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#7 h Aug 10, 2023, 9:02 AM • 1 Y

Y by erkosfobiladol

$\angle DIB=\angle \frac{C}{2}=\angle ICB \implies DI^2=DB.DC=DK.DN \implies \angle IKD=90$
Let's prove that $QK \perp DN$
$\angle APE=180-\frac{B+C}{2}$ and $\angle APB=180-C \implies \angle EPB=\frac{B-C}{2}=\angle EDB$
$\implies P,E,D,B$ are cyclic.
$\angle PNI=\angle PNA=\angle PBA= \angle PBE=\angle PDE=\angle PDI \implies P,D,N,I$ are cyclic. So $\angle DPN=90$
Also $\angle BPN=\angle \frac{A}{2}=\angle BPN,$ $NK.ND=NI^2=NB^2=NQ.NP$ which means $P,D,K,Q$ are cyclic.
$\implies QK \perp DN$

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Om245

#8 h Jan 12, 2024, 5:07 PM

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We redefines everything

Let $T$ be mixtilinear Incircles touch point.Note that circle pass though $E$ and $F$

Define $Y$ be antipode of $N$ then we know $\overline{Y-I-T}$ and $\angle YTN = 90$
Let $D = NT \cap BC$ and $M$ be midpoint of $BC$ . Redefine $Q = DM \cap YT$
Now note $Q$ is orthocenter of $\triangle YND$ .

Redefine $P = DY \cap (ABC)$ now we get $NP \perp DY$
Now if we prove $P$ lie on $(AEF)$ we are done

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from $\angle DPN = \angle DIN = \angle DMN = 90 \implies D,P,I,M,N$ are cyclic.
which give us $\angle IPM = \angle IDM$

Now as $I$ and $M$ are midpoint of $EF$ and $BC$ and $P$ lie on $(ABC)$
$P$ is miqual point of complete quadrilateral $BCFE$ and hence from $A = BE \cap CF$ we get $P$ lie on $(AEF)$ $\blacksquare$

This post has been edited 1 time. Last edited by Om245, Jan 12, 2024, 5:07 PM
Reason: image

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#9 h Jan 13, 2024, 7:43 AM

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This question is basically like a lemma

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#10 h Aug 31, 2024, 1:39 PM

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Take $IR \cap (ABC)$ at $R$ . $\angle PAF=\angle PEF=\angle PBD$ so, $PEBD$ is cyclic $\implies$ $\angle EPB=\angle EDB=\angle FPC$ so, $PFCD$ cyclic. $\angle QPD=90$ , $\angle BPQ=\angle QPC$ that means $(D,Q;C,B)=-1$ .Take pencil from $P$ w.r.t $(ABC)$ so, $PBNC$ is harmonic quadrilateral.Then, take pencil from $N$ w.r.t line $BD$ . So, $B$ fixes $B$ , $P$ fixes $Q$ , $C$ fixes $C$ and $N$ fixes $NR \cap BD$ (call it $D'$ ).
Thus, $(B,C;Q,D')=-1$ also $(B,C;Q,D)=-1$ and that means $D'=D$ .

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Eka01

#11 h Nov 8, 2024, 5:19 PM

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Let $T$ be the $A$ mixtilinear touch point. It is well known that $DN \cap \omega=T$ . We prove that $\overline{T-Q-I}$ . Since $(PT;BC) \stackrel{Q}{=} (N,TQ \cap \omega ;C,B)=-1$ $\implies TQ \cap \omega =M$ where $M$ is antipode of $N$ in $\omega$ . But $I$ lies on $TM$ and so the required is proved.

This post has been edited 3 times. Last edited by Eka01, Nov 8, 2024, 5:36 PM

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#12 h Nov 8, 2024, 8:47 PM

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Let $T_A$ the A-mixtilinear intouch point of $\triangle ABC$ , by $\sqrt{bc}$ invert we have $M,I,T_A$ colinear, where $M$ is midpoint of arc $BAC$ in $\omega$ . Let $L$ midpoint of $BC$ .
Claim 1: $D,P,M$ are colinear.
Proof: Note that $P$ is miquel of $BEFC$ so by homologue points we have it's miquel of $BEIL$ as well so $DPIL$ is cyclic, but clearly $DILN$ is cyclic with diameter $DN$ therefore both combined give $\angle DPN=90$ which concludes.
Finish: Note that $-1=(M, N; B, C) \overset{P}{=} (D, Q; B, C)$ and also that $DI$ is trivially tangent to $(BIC)$ from I-E Lemma, which means that in $\triangle BIC$ , $IQ$ is the I-symedian but from trivial angle case we have $BN, NC$ tangent to $(BIC)$ therefore $N,I,Q,T_A$ are colinear, now projecting from $T_A$ gives $D,T_A,N$ colinear which finishes as $\angle NT_AM=90$ thus we are done :cool:

:cool:

.

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SimplisticFormulas

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SimplisticFormulas

#13 h Nov 8, 2024, 9:09 PM • 1 Y

Y by MathLuis

I have not done mixtilinear circles or whatever, so heres a solution based purely on angle chase:
Let $DN$ meet $\omega$ again in $L$ . Let $IL$ meet $BC$ in $Q*$ and $QN$ meet $\omega$ in $P*$ . We claim that $P$ and $P*$ are the same by showing that $AFEP$ is cyclic.
Note that $\angle EDB=180-\angle EBD-\angle DEB=\frac{B-C}{2}$ . Also, note that

$\angle DIB=180-\angle IDB-\angle DBI= \frac{B-C}{2}$ . Hence, $DB$ is tangent to $(BIC)$ at $I$ .
$\implies DI^2=DB \cdot DC=DL \cdot DN$
$\implies DI$ is tangent to $(LIN)$ in $I$ .
$\implies \angle DIN= \angle ILN=90$
$\implies \angle ILD=90$
Also note that
$\angle LDQ=\angle BDL=180-\angle DBL-\angle DLB=180-\angle LNC-\angle BCN=180-\angle LNI-\angle B-\angle \frac{A}{2}=\angle C+\angle \frac {A}{2}-\angle LNI=\angle BCN +\angle ACB-\angle LNI=\angle ACN-\angle LNI=\angle ALD-\angle ANL=\angle LAN=\angle LP*N=\angle LP*Q*$ ,
giving us the fact that $LDP*Q*$ is cyclic.
$\implies \angle NP*D=\angle Q*P*D=\angle Q*LD=90$
$\implies \angle DP*C=\angle DP*N +\angle NP*C=90+\angle NAC=90+\angle \frac{A}{2}=180-(90-\angle \frac{A}{2})=\angle DFC$
$\implies DP*FC$ is cyclic
$\implies \angle P*AC=\angle P*BA=\angle P*DF$
$\implies \angle P*DE=\angle P*BE$
$\implies EP*DB$ is cyclic
$\implies \angle P*EA=\angle P*DB=\angle P*DC=\angle P*FA$ , hence $P* \in (AEF)$ , as required

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#14 h Nov 19, 2024, 9:49 AM

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Om245 wrote:

We redefines everything

Let $T$ be mixtilinear Incircles touch point.Note that circle pass though $E$ and $F$

Define $Y$ be antipode of $N$ then we know $\overline{Y-I-T}$ and $\angle YTN = 90$
Let $D = NT \cap BC$ and $M$ be midpoint of $BC$ . Redefine $Q = DM \cap YT$
Now note $Q$ is orthocenter of $\triangle YND$ .

Redefine $P = DY \cap (ABC)$ now we get $NP \perp DY$
Now if we prove $P$ lie on $(AEF)$ we are done

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from $\angle DPN = \angle DIN = \angle DMN = 90 \implies D,P,I,M,N$ are cyclic.
which give us $\angle IPM = \angle IDM$

Now as $I$ and $M$ are midpoint of $EF$ and $BC$ and $P$ lie on $(ABC)$
$P$ is miqual point of complete quadrilateral $BCFE$ and hence from $A = BE \cap CF$ we get $P$ lie on $(AEF)$ $\blacksquare$

i'm fanumtaxing this diagram:

Note that $D,P,Y$ collinear because $\angle DPC=\angle DFC$ , $\angle CPD=\angle CNM$ . This is easy to check as they are $90^\circ-\frac{\angle A}{2}$ . Where we note that $DPFC$ is cyclic by Miquel.

Suppose that $IY\cap (ABC)=T'$ . We claim that $D,T',N$ collinear. This is radax on $(IT'N)$ , $(BIC)$ , and $(ABC)$ .

Now, note that $T'=T$ . Note that as $PN\perp DY$ , and $YT\perp DN$ , and $DQ\perp YN$ . Thus $Q$ also lies on $IT$ .

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GeorgeMetrical123

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GeorgeMetrical123

#15 h Mar 31, 2025, 11:41 AM

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Here's a proof I haven't seen mentioned above:
We use the same notation from Om245's picture.
We invert around circle $(BIC)$ ., whose centre is $N$ . Let $T$ be the intersection of $DN$ and $\omega$ . Let $Q$ be the intersection of $PN$ with $BC$ .
From there, we see that:
$\begin{align} &BC \rightarrow \omega \\ & T \rightarrow D \\ & Q \rightarrow P \\ & B \rightarrow B \\ & I \rightarrow I\\ &C \rightarrow C \\ & M \rightarrow Y \end{align}$ We will first prove that $D, P, Y$ are collinear.
$P$ is the Miquel point of $BEFC$ , so $DPEB$ is concyclic. From there:
$\begin{align} \angle DPB &= \angle DEB \\ &= \angle FEA \\ &= 90^\circ - \frac{1}{2}\angle BAC \\ &= \angle YCB \\ &= 180^\circ - \angle YPB \end{align}$ From there, the inversion shows us that $(TQMN)$ is concyclic. Because $\angle QTN = \angle TMN = 90^\circ$ , we see that $T, Q, Y$ are collinear because $Y$ is the antipode of $N$ . Now we finish by remarking that $T, I, Y$ are collinear because $\angle DIN = \angle DMN = 90^\circ$ is a cyclic quad, which the inversion sends to the required line, finishing the proof! Q.E.D.

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GeorgeMetrical123

#16 h Mar 31, 2025, 9:37 PM

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Suppose we already have $T, I, Q$ and $D, P, Y$ collinear following proofs from my previous post.
Let $f(X) = \pm \frac{BX}{CX}$ where the sign is positive if $X$ is strictly above $BC$ .
$\begin{align*} f(T)f(Y) &= f(T) \\ &= \frac{f(D)}{f(N)} \\ &= -f(D) \\ &= -f(P)f(Y) \\ &= -f(P) \\ &= f(P)f(N) \end{align*}$ So $PN$ and $TY$ intersect on $BC$ , so we are finished. Q.E.D.
I present another proof, where we do not assume $D, P, Y$ collinear.
We see that we want to prove that $f(D) = f(P)$ . We also see that $f(D) = f(I)^2$ . From here, we start calculation, eventually showing those two are equal.
$\begin{align*} f(P) = f(I)^2 \\ \frac{BP}{CP} = \left( \frac{\sin \angle ICB}{\sin \angle IBC} \right) ^2 \\ \frac{BE}{CF} = \left( \frac{\sin \angle EIB}{\sin \angle FIC} \right) ^2 \\ \frac{\sin \angle FIC ^2}{CF} = \frac{\sin \angle EIB ^2}{BE} \\ \frac{ \sin \angle FIC \sin \angle IFC}{CI} = \frac{ \sin \angle EIB \sin \angle IEB}{BI} \\ \frac{ \sin \angle FIC}{CI} = \frac{ \sin \angle EIB}{BI} \\ \frac{BI}{CI} = \frac{ \sin \angle EIB}{\sin \angle FIC} \\ \frac{BI}{CI} = \frac{\sin \angle ICB}{ \sin \angle IBC} \end{align*}$ which is obviously true by sine theorem! Q.E.D.

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