Another "triangle BIC" configuration

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Another "triangle BIC" configuration

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Source: Greece National Olympiad 2023, Problem 3

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#1 h Feb 19, 2023, 2:16 PM

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A triangle $ABC$ with $AB>AC$ is given, $AD$ is the A-angle bisector with point $D$ on $BC$ and point $I$ is the incenter of triangle $ABC$ . Point M is the midpoint of segment $AD$ and point $F$ is the second intersection of $MB$ with the circumcirle of triangle $BIC$ . Prove that $AF\bot FC$ .

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#2 h Feb 19, 2023, 2:53 PM

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Obviously $AD \cap (BIC)=I_A$ is the $A$ -excenter. We want $\angle AFC=90^{o}$ and we have that $\angle MFC=90^{o}-\frac{\alpha}{2}$ , so we want $\angle AFM =\angle BAM \iff MD^2=MA^2=MF \cdot MB=MI \cdot MI_A$ , but the last one easily follows since $(AD; II_A)=-1$ .

This post has been edited 2 times. Last edited by VicKmath7, Feb 19, 2023, 2:55 PM

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#3 h Mar 3, 2023, 11:18 AM • 1 Y

Y by Ismayil_Orucov

Let $L$ be the midpoint of the arc $BC$ not containing $A$ in $(ABC)$ and let $N$ be the midpoint of $AC$ .

It is easy to check that $\triangle ALC\sim \triangle ABD$ , as $LN$ and $BM$ are medians we have $\angle NLC=\angle MBD$ .

On the other hand $\angle FLC = 2\angle FBD$ so $LN$ bisects $\angle FLC$ , this implies that $\triangle NLC = \triangle NLF$ .

Finally we have $NA = NC = NF$ which implies the result.

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#5 h Jul 6, 2023, 1:47 AM

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Let the centre of the circumcirle of triangle $BIC$ be $T$ ,then because $BT$ = $IT$ = $CT$ , $T$ is on the circumcirle of triangle $BAC$ ,easily we have $MF \cdot MB$ = $MN^2-BT^2$ = $MN^2-DT \cdot TA$ = $MN^2-(MN-MT) \cdot(MN+AM)$ = $MN^2$ then $\triangle ABM$ is similar to $\triangle AFM$ just angle chasing we can know that $AF\bot FC$

This post has been edited 3 times. Last edited by how_to_what_to, Jul 6, 2023, 1:51 AM
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#6 h Jul 6, 2023, 4:52 PM • 1 Y

Y by AndresSchwepps

$J$ is the excentre wrt $A$ of $\Delta ABC$
its welknow $(ADIJ)=-1$ so $MA^2 =MF.MB \rightarrow \angle MFA = \angle BAM$
$\rightarrow \angle AFC = \angle CFM + \frac{ \angle A}{2} = 180^{\circ} - \angle BIC + \frac{ \angle A}{2} =90^{\circ}$

This post has been edited 1 time. Last edited by tungthanh, Jul 6, 2023, 4:53 PM

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#7 h Jul 8, 2023, 5:40 AM

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Let $AD\cap \odot(BIC)=J$ ,suppose $T\in BC,AT\bot BC$
Note that $\dfrac{JD}{DI}=\dfrac{JA}{AI}$
So $MI^{2}=MI\cdot MJ=MF\cdot MB,MT^{2}=MF\cdot MB=MA^{2}$
Which means that $\angle AFM=\dfrac{1}{2}\angle A,\angle MTF=\angle MDF$
So $M,F,T,D$ are cyclic
Hence $\angle MFT=\angle ADC=180^{\circ}-\dfrac{1}{2}\angle A-\angle C$
So $\angle AFT=180^{\circ}-C$
Which means that $AFTC$ are cyclic
So $\angle AFC=\angle ATC=90^{\circ}\Rightarrow AF\bot FC$

This post has been edited 1 time. Last edited by SQTHUSH, Jul 8, 2023, 5:40 AM

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#8 h Jun 22, 2024, 2:32 PM • 2 Y

Y by Om245, GeoKing

Note that the condition that $AF \perp FC$ simply says that $F$ lies on the circle with diameter $AC$ . Thus, we wish to show that points $A$ , $F$ , $H$ and $C$ are concyclic, where $H$ is the foot of the $A-$ altitude. Let $N_A$ denote the $BC$ arc midpoint which is well known to be the center of $(BIC)$ .

We perform a $\sqrt{AB \cdot AC}$ inversion followed by a reflection across the $A$ -angle bisector. Then, the problem rewrites to

Inverted Problem wrote:

Let $\triangle ABC$ be a triangle with minor arc midpoint of $BC$ , $N_A$ . Let $P$ be the reflection of $A$ across $N_A$ . Now, let $R$ be the second intersection of circles $(ABP)$ and $(BIC)$ . Show that the points $R$ , $A'$ (the $A$ -antipodal point) and $C$ are collinear.

Let $(BIC) = \omega$ . We instead let $R = \overline{AC'} \cap \omega$ and show that points $A$ , $B$ , $R$ and $P$ are concyclic. We first prove the following claim.

Claim : Lines $\overline{BR}$ and $\overline{AI}$ are parallel.
Proof : Let $R'$ be the intersection of the line through $B$ parallel to the $A$ -angle bisector and $\overline{CA'}$ . Clearly,
$\measuredangle BR'C = \measuredangle IAC + \measuredangle ACA' = 90 + \measuredangle IAC = \measuredangle BIC$ from which it follows that $R'$ indeed lies on $\omega$ which proves the claim.

Now, we can observe the following pair of congruent triangles. Note that,
$AN_A = N_AP \text{ and }BN_A=N_AR$ due to reflection and radii respectively. Further,
$\measuredangle PN_AR = \measuredangle BRN_A = \measuredangle N_ABR = \measuredangle BN_AA$ from which it follows that $\triangle ABN_A \cong \triangle N_ARP$ . Now, we already know that $ABRP$ is a trapezoid due to the previous claim. Due to the proved pair of congruent triangles, we further have $RP=AB$ so this trapezoid is in fact isosceles, and in turn cyclic. Thus, points $A$ , $B$ , $R$ and $P$ are concyclic which finishes the proof.

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Om245

#9 h Jun 22, 2024, 2:52 PM • 2 Y

Y by cursed_tangent1434, GeoKing

We do inversion with center $M$ with radius $AM$ .
If $J$ is $A$ -excenter and $E= MC \cap (IBC)$ , then as $(A,D;I,J)=-1$ we have from $AM^2 = MI.MJ \Rightarrow I \leftrightarrow J$ and hence so $F \leftrightarrow B$ and $E \leftrightarrow C$ .

Now proving $\angle CFA= 90 \Leftrightarrow \angle BAM + \angle MEB = 90$ . But as $\angle MEB = \angle CJB = 90 - \angle BAI$ , our statement is true.

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#10 h Jun 23, 2024, 2:14 AM • 1 Y

Y by GeoKing

Let $N$ be midpoint of $CA;$ $AD$ intersects $(ABC)$ again at $S,$ $B'$ is reflection of $B$ in $AD$ . We have $\angle{MQB'} = \angle{ACB} = 180^{\circ} - \angle{MNC}$ and $\angle{MSB'} = \angle{MSB} = \angle{ACB} = 180^{\circ} - \angle{MNC}$ . Then $Q, S \in (MNB')$ . From this, we have $\angle{NCQ} = \angle{QBB'}$ . But $\angle{QMB'} = \angle{QNB'}$ then $\triangle NQC \sim \triangle MBB'$ . Combine with $\triangle MBB'$ is isosceles at $M,$ we have $\triangle NQC$ is isosceles at $N$ or $NC = NQ$ . Hence $\triangle AQC$ is $Q$ - right triangle

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