Concyclic points and a bisector

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Concyclic points and a bisector

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Source: Saint Petersburg MO 2020 Grade 11 Problem 3

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XbenX

#1 h May 7, 2020, 6:32 PM

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$BB_1$ is the angle bisector of $\triangle ABC$ , and $I$ is its incenter. The perpendicular bisector of segment $AC$ intersects the circumcircle of $\triangle AIC$ at $D$ and $E$ . Point $F$ is on the segment $B_1C$ such that $AB_1=CF$ .Prove that the four points $B, D, E$ and $F$ are concyclic.

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#2 h May 7, 2020, 7:27 PM • 6 Y

Y by amar_04, zuss77, SatisfiedMagma, Mango247, Mango247, Mango247

Let $X \in \odot (ABC)$ with $BX \parallel AC$ . By reflection about the perpendicular bisector of $BC$ it suffices to show that $XDB_1E$ is cyclic. Invert about $B$ with radius $\sqrt{BA \cdot BC}$ and reflect the inverted figure in the angle bisector of $\angle ABC$ to get the following problem-

Inverted problem wrote:

Let $M$ be the midpoint of arc $\widehat{AC}$ not containing $B$ , and suppose the tangent at $B$ to $\odot (ABC)$ meets $AC$ at $X$ . Suppose $\odot (M,MA)$ meets the $B$ -Apollonius circle $\omega$ at $D,E$ . Then $MDXE$ is cyclic.

Since $X$ is the center of $\omega$ , so the result directly follows from the fact that $\omega$ and $\odot (M,MA)$ are orthogonal.

EDIT: To show that $\omega$ and $\odot (M,MA)$ are orthogonal, just invert about $\odot (M,MA)$ and invoke Shooting Lemma to see that $\omega$ remains fixed.

This post has been edited 1 time. Last edited by math_pi_rate, May 7, 2020, 7:32 PM

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#3 h May 7, 2020, 7:36 PM • 3 Y

Y by amar_04, MathsLion, zuss77

Let $N$ be the midpoint of smaller arc $AC$ .
Lemma:
$NB_1 \cdot BN = ND^2=NE^2=NI^2$ because $\triangle AB_1N \sim \triangle BAN$

From this it follows $\angle B_1EN =\angle NBE$ and $\angle B_1DN= \angle NBD$ so basically $B_1$ is the Humpty point of triangle $EDB$ . (key insight)
Since $F$ is the reflection of the Humpty point it follows $F$ is the intersection of the symmedian with the circumcircle of triangle $EDB$ so $E,D,B,F$ are concyclic points and are harmonic at that!

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Mathematicsislovely

#4 h May 7, 2020, 8:14 PM • 1 Y

Y by amar_04

Let $X$ be the midpoint of arc $AC$ of $(ABC)$ not containing $B$ .Note that $D,X,F$ are collinear.Easy to see that $XB.XB_1=XD^2=XE^2$ .[inversion in $(AIC)$ swaps $B,B_1$ because it swaps $AC$ and $(ABC)$ ].
So $\angle XEF=\angle XEB_1=\angle XBE$ [from $XB_1.XB=XE^2$ ]
and $\angle XDF=\angle XDB_1=\angle XBD$ [from $XD^2=XB.XB_1$ ]
SO $\angle EBD= \angle XBE+\angle XBD= \angle XEF+\angle XDF=180^\circ-\angle DFE$ .So $E,B,D,F$ are concyclic.

This post has been edited 6 times. Last edited by Mathematicsislovely, Jun 5, 2020, 11:00 AM

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#5 h Jun 6, 2020, 6:10 AM • 2 Y

Y by zuss77, hakN

Cute

XbenX wrote:

$BB_1$ is the angle bisector of $\triangle ABC$ , and $I$ is its incenter. The perpendicular bisector of segment $AC$ intersects the circumcircle of $\triangle AIC$ at $D$ and $E$ . Point $F$ is on the segment $B_1C$ such that $AB_1=CF$ .Prove that the four points $B, D, E$ and $F$ are concyclic.

Let $\overline{BB_1}$ meet $\odot(ABC)$ again at $M \ne B$ . Note that $\overline{MB}, \overline{MF}$ are symmetric in $\overline{DE}$ hence the reflection $K$ of $F$ in the perpendicular bisector of $\overline{DE}$ (whose midpoint is $M$ ) lies on $\overline{BB_1}$ . Since $MB \cdot MK=MB \cdot MF=MI^2=MD \cdot ME$ , the conclusion follows.

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#6 h Aug 23, 2020, 9:09 AM

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Let $O$ be the circumcenter of $\triangle AIC$ . It is well known that $O$ lies on $BB_1$ . Let $J$ be the reflection of $I$ over $O$ , and let $AB$ hit the circumcircle of $\triangle AIC$ a second time at $K$ . Using directed angles, it is easy to see that $\measuredangle IAC = \measuredangle KAI$ .

From here, we use complex numbers with $(AIC)$ as the unit circle. Let $I = x^2$ , $A = y^2$ and $C = z^2$ . Because $DE$ is the perpendicular bisector of $AC$ , we have that $\{ D, E \} = \{ yz, -yz \}$ . WLOG, let $D = yz$ and $E = -yz$ . Since $B_1 = IJ \cap AC$ , we obtain that
$B_1 = \frac{ AC(I+J) - IJ(A+C)}{AC - IJ} = \frac{ x^4(y^2+z^2)}{x^4 + y^2z^2}.$ Because $F$ is the reflection of $B_1$ over the midpoint of $AC$ , we have
$F = A + C - B_1 = \frac{ y^2z^2(y^2+z^2)}{x^4 + y^2z^2}.$
Because $\measuredangle IKC = \measuredangle KCI$ , we get $K = \frac{x^4}{z^2}$ . Then $B = IJ \cap AK$ implies
$B = \frac{ AK(I+J) - IJ(A+K)}{AC - IJ} = \frac{ x^4(y^2 + \frac{x^4}{z^2})}{y^2 \frac{x^4}{z^2} + x^4}= \frac{x^4 + y^2z^2}{y^2 + z^2}$ To check that $B,D,E,F$ are concyclic, we just have to check that the quantity
$\frac{B - D}{F-D} \div \frac{B-E}{F-E} \in \mathbb{R}$ We just simplify:
$\begin{align*} \frac{B - D}{F-D} \div \frac{B-E}{F-E} &= \frac{ \frac{x^4 + y^2z^2}{y^2+z^2} - yz}{\frac{y^2z^2(y^2+z^2)}{x^4+y^2z^2} - yz} \cdot \frac{ \frac{y^2z^2(y^2+z^2)}{x^4 + y^2z^2} + yz}{\frac{x^4+y^2z^2}{y^2+z^2}+ yz} \\ &= \frac{ (x^4+ y^2z^2 - yz(y^2 + z^2))(yz(y^2+z^2) + (x^4 + y^2z^2))}{(yz(y^2+z^2) - (x^4+y^2z^2) ) (x^4 + y^2z^2 + yz(y^2+z^2))} \\ &= 1, \end{align*}$ which is obviously a real number.

This post has been edited 3 times. Last edited by HKIS200543, Aug 23, 2020, 9:12 AM

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#7 h Aug 23, 2020, 6:09 PM • 2 Y

Y by Mango247, Mango247

See my solution on my Youtube channel in the link below. It is similar to anantmudgal09's solution, but somewhat different in the end.

https://www.youtube.com/watch?v=Bv_HFRmVBio

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zuss77

#8 h Oct 30, 2020, 7:07 PM • 2 Y

Y by Mango247, Mango247

Let $BI$ meet $\odot (ABC)$ at $M$ . Inversion $(M; MI)$ followed by a reflection about $MD$ swaps $(ABC)\mapsto AC$ , $B\mapsto F$ , $D\mapsto D$ . Hence $\odot (BDF)$ is fixed. So, second point of intersecion $\odot (BDF)$ with $\odot (M)$ lays on $MD$ , which is $E$ .

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#9 h Oct 30, 2020, 9:14 PM • 2 Y

Y by KST2003, Muaaz.SY

Let $N = BI \cap (ABC)$ , which is the circumcenter of $IAC$ . Let $B_2$ be the reflexion of $B_1$ w.r.t $N$ .
On one hand $NB_2 \cdot NB= NB_1 \cdot NB = NI^2 = ND \cdot NE$ , implying that $B,D,B_2,E$ are concyclic.
On the other hand, $DEB_2F$ is an isosceles trapezoid, which are concyclic, which completes the proof.

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zuss77

#10 h Nov 6, 2020, 2:24 PM • 3 Y

Y by Mango247, Mango247, Mango247

I'll explain math_pi_rate's solution, because in its essence it's more simple than it looks and quite inversion-instructive. We don't even need to construct inversive diagram to understand it.

Let $\odot (AIC)$ be $\omega$ and $M$ is where $BI$ hit $(ABC)$ . Let $X\in (ABC), BX\parallel AC$ .
Perform inversion $B_1\mapsto M$ with center $B$ (bisector-flip actually don't even needed here).

$\omega$ is fixed.
Since $DE$ orthogonal to $\omega$ it goes to circle (say, $\Omega$ ) orthogonal to $\omega$ .
Since $X$ is a reflection of $B$ over $DE$ it goes to circumcenter of $\Omega$ (say, $X'$ ).
$D,E$ obviously go to $\omega \cap \Omega=[D',E']$ .

With $M,X'$ being centers of orthogonal circles we immediately see that $X'D'ME'$ is cyclic. Inverting back it makes $XDB_1E$ cyclic. And finally reflect it about $DE$ to get what we need.

This post has been edited 1 time. Last edited by zuss77, Nov 7, 2020, 9:25 AM

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#11 h Jan 8, 2021, 7:05 AM

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Let $M$ and $N$ be the midpoint of $BC$ and arc $BC$ . $AM$ meets $(ABC)$ at $X$ . Then as $XM$ and $NB_{1}$ are concurrent on $(ABC)$ , by butterfly theorem, $XF$ and $NM=DE$ are also concurrent on $(ABC)$ . Let this concurrency point be $Y$ . Then by power of a point,
$MX\cdot MB=MA\cdot MC=MD\cdot ME$ and so $B,D,X,E$ are concyclic. Consider an inversion at $Y$ with radius $YA=YC$ . This swaps $AC$ with $(ABC)$ , and preserves the circle $(AIC)$ . Therefore, $D\rightarrow E$ and $F\rightarrow X$ , and consequently,
$YD\cdot YE=YA^2=YF\cdot YX$ which shows that $F$ also lies on $(DEX)$ .

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#12 h Jan 21, 2021, 11:23 PM

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Very nice problem:
Let us consider $N$ the intersection of the median of $AC$ with the circle. Clearly, $N$ (radical axis argument) lies in the circle $(BDE)$ . Now we prove the following Lemma:
Claim: $N,F,L$ are collinear where $L$ is the midpoint of arc $ABC$ .
Proof: Apply Angle Bissector Theorem in $ANC$ and you'll be good to go.
In conclusion, notice that $LF\times LN = LA^2=LB^2$ because $\angle ANF=\angle FAL$ .
However, $LA^2= LE.LD$ which implies that $LE.LD = LF.LN$ $\implies$ $(EDFN)$ is cyclic desired.

This post has been edited 1 time. Last edited by RodSalgDomPort, Jan 21, 2021, 11:30 PM
Reason: typo

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#13 h Jan 27, 2021, 3:30 PM

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let $N,R,M$ be the midpoint of arc $BC$ , the intersection of the external bisector of $\angle ABC$ with $BC$ and the midpoint of $BC$
let $B'$ be the reflection of $B_1$ beyond $N$

$\angle RBB'=90=\angle B_1FB'$ so $RBFB'$ is cyclic
since $(C,A;B_1,R)=-1$ we have $MD.ME=MA.MC=MB_1.MR=MF.MR$ so $DEFR$ is cyclic
$NB'.NB=NB_1.NB=NA^2=ND.NE$ so $DB'EB$ IS cyclic
so $BDFB'ER$ is cyclic

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#14 h Aug 7, 2021, 5:37 PM • 1 Y

Y by SatisfiedMagma

Let $G$ be the midpoint of $AC$ and $H$ be the intersection of $BG$ and $(ABC)$ .
$DG \cdot GE = AG \cdot GC = BG \cdot GH$ So $BDHE$ is cyclic. Let $J$ be the intersection of perpendicular bisector and $HF$ . Since $G$ is midpoint of $B_1F$ by butterfly it follows that $J$ lies on $(ABC)$ .
$\angle JFD = \angle GDF - \angle DJF = \angle B_1BD - \angle B_1BG = \angle GBD = \angle HBD = \angle HED$ Which is the exterior angle of the quadrilateral $DFHE$ so $DFHE$ is cyclic. Combining both results we get that $BDFE$ is cyclic.

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mathuz

#15 h Apr 26, 2023, 9:39 PM

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Let $BB_1$ meets the circumcircle of $ABC$ again at $X$ . We know that $X$ is the center of $(AIC)$ that is the midpoint of $DE$ . Assume that $XF$ meets the circumcircle of $ABC$ again at $P$ . Then, the points $B$ and $P$ , $B_1$ and $F$ are symmetrical with respect to the line $DE$ .

Considering an inversion centred at $X$ with radius $XI$ , we obtain that $B\leftrightarrow B_1$ , $P\leftrightarrow F$ and the points $B,C,D,E$ are fixed. So the circle $(BDF)$ goes to the circle $(PDB_1)$ .

Let $E'$ be the second intersection point of the circles $(BDF)$ and $(PDB_1)$ . As they are mapped each to other by the inversion, we get that $E'$ should be fixed under the inversion, i.e. $E'$ lies on $(AIC)$ . On the other hand, the triangles $BDF$ and $PDB_1$ are symmetrical with respect to the line $DE$ , so their circumcenters are too. That means $DE$ be the radical axis of their circumcircles. Hence $E\equiv E'$ and $B$ , $D$ , $F$ , $E$ are concyclic.

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