Let $ABC$ be an acute triangle with $AB<AC<BC$ and let $D$ be a point on it's extension of $BC$ towards $C$. Circle $c_1$, with center $A$ and radius $AD$, intersects lines $AC,AB$ and $CB$ at points $E,F$, and $G$ respectively. Circumscribed circle $c_2$ of triangle $AFG$ intersects again lines $FE,BC,GE$ and $DF$ at points $J,H,H' $ and $J'$ respectively. Circumscribed circle $c_3$ of triangle $ADE$ intersects again lines $FE,BC,GE$ and $DF$ at points $I,K,K' $ and $I' $ respectively. Prove that the quadrilaterals $HIJK$ and $H'I'J'K '$ are cyclic and the centers of their circumscribed circles coincide. by Evangelos Psychas, Greece
Problem
Source: BMO Shortlist 2018 G5
Tags: circumcircle, geometry, cyclic quadrilateral, Circumcenter, IMO Shortlist
05.05.2019 22:48
Feels a lot easier than G6. We will start with a lemma: Let $WXYZ$ be cyclic with center $O$. Let $(OWX)$ meet $XY$, $ZW$ at $P,Q$, and let $(OYZ)$ meet $XY, ZW$ at $R,S$. Then, $PQRS$ is cyclic. To see this, first note that if $XY \cap ZW = T$ exists, then we have the directed length equalities: \[ TP\cdot TR = \frac{TP \cdot TR\cdot TX\cdot TY}{TZ\cdot TW} = \frac{(TP\cdot TX) (TR\cdot TY)}{TZ\cdot TW} = \frac{(TQ\cdot TW) (TS\cdot TZ)}{TZ\cdot TW} = TQ\cdot TS \]so we are done. Otherwise, $XY\parallel ZW$, so there exists a line $\ell\ni O$ that perpendicularly bisects $XY, ZW$, and across which $WX, YZ$ are reflections. Then, $P,R$ and $Q,S$ are reflections over $\ell$, so $PRSQ$ is an isosceles trapezoid and thus cyclic. $\Box$ Applying this lemma to cyclic quads $DEFG$ and $DEGF$, we get the desired cyclic quadrilaterals. Next we claim that $H'J'\parallel HJ$. Indeed, we have \[ \measuredangle HJ'H' = \measuredangle HGH' = \measuredangle DGE = \measuredangle DFE = \measuredangle J'FJ = \measuredangle J'HJ, \]so we are done. Similarly, $K'I'\parallel KI$. So, if $O,O'$ are the centers of $(HIJK)$ and $(H'I'J'K')$, and $X,Y$ are the centers of $(AFG)$, $(ADE)$, then we have: \[ XO\perp HJ \parallel H'J' \perp XO', \]so $X,O,O'$ are collinear. Similarly, since $K'I'\parallel KI$, $YO\parallel YO'$ and so $Y,O,O'$ are collinear. This basically means $O' = XO\cap YO = O$, but we'll dispel some configuration issues first. Suppose that $O\neq O'$, then $XYO$ are collinear. In particular, $JH\parallel KI$, which implies that $\measuredangle KIE = \measuredangle HJF$, or $\measuredangle GDE = \measuredangle DGF$. WLOG $BC$ is horizontal with $B$ to the left of $C$. Then, $\measuredangle GDE = \angle GDE$, and $\measuredangle DGF = 180 - \angle DGF$, so $\angle DGF + \angle GDE = 180$. But \[ \angle GDE < \angle ADE < 90, \angle DGF < \angle AGF < 90, \]so this is a contradiction. Thus, $XYO$ cannot be collinear, so $O' = O$ indeed. $\blacksquare$
15.08.2019 21:36
From EAI'=EDI'=H'GF=H'AF=H'AE (H'EA=HGA=HFA so H'F=H'E) so {A,H',I'} collinear, same for {A,H,I}, {A,K,J}, {A',K',J'}. Observe by angle chase JJ'=HH', KK'=II' so HIJK, H'I'J'K' share perp bisector hence same center
24.02.2022 20:46
Really easy for G5... Claim1 : $A,H,I$ and $A,H',I'$ and $A,K',J'$ and $A,K,J$ are collinear. Proof : we will prove $A,H,I$ and will use the same approach for other three. $\angle FAH = \angle FGD = \angle FAD/2$ so $AH \perp DF$. $\angle DAI = \angle 180 - \angle DEF = \angle DGF = \angle DAF/2$ so $A,H,I$ is collinear and perpendicular to $DF$. Claim2 : $JIHK$ and $J'I'H'K'$ are cyclic. Proof : $\angle KJA = \angle AGF = \angle AFG = \angle AHK$ so $JIHK$ is cyclic. $\angle K'J'I' = \angle AGF = \angle AFG = \angle AH'K'$ so $J'I'H'K'$ is cyclic. Claim3 : $JIHK$ and $J'I'H'K'$ have same center. Proof : Note that $AK' \perp BC$ and $AI \perp DF$. $\angle KAK' = \angle 90 - \angle AKD = \angle 90 - \angle AI'D = \angle I'AI$ so $KII'K'$ is isosceles trapezoid so $KI$ and $K'I'$ have same perpendicular bisector. with same approach for $HJ$ and $H'J'$ we have $JIHK$ and $J'I'H'K'$ have same center. we're Done.
19.10.2024 04:02
Omg this problem is actually so crazy (but reading what others have said about this problem is implying I’m actually extremely washed — took like two hours to solve this at least I have a better appreciation of how to prove two circles have the same centre given only four points on it). So the first step is showing that: $A-K’-J’$ on the perpendicular bisector of $GD$, $A-K-J$ on the perpendicular bisector of $GE$, $A-H-I$ on the perpendicular bisector of $FD$, and $A-H’-I’$ on the perpendicular bisector of $FE$. We are only going to prove that $GK’=K’D$ however, since for all of the other seven points this proof is going to be similar. Notice that $\measuredangle GK’A=\measuredangle EK’A=\measuredangle EDA=\measuredangle AED=\measuredangle AK’D$. Now recall $AG=AD$, so by the extended sine rule, $(AK’G)$ and $(ADK’)$ have the same circumradius, hence $(AK’G)$ is the reflection of $AK’D$ over $\overline{AK’}$. Coupled with the fact that $\measuredangle GK’A=\measuredangle AK’D$ we must have that $G$ and $D$ are reflections of each other around $AK’$, implying the result. Once we have this we can now start angle chasing to get the two main claims: $HIJK$ and $H’I’J’K’$ cyclic. $J’H’$ and $JH$ parallel; similarly $K’I’$ and $KI$ parallel (since each pair are chords on the same circle their perpendicular bisectors all pass through the same point and so are equal) We prove $HIJK$ cyclic as the other one follows by similar logic. However $\measuredangle HIJ=\measuredangle (\overline{AI},\overline{IJ})=90^{\circ}-\measuredangle IFI’=90^{\circ}-\measuredangle EFD=90^{\circ}-\measuredangle EGD=\measuredangle(\overline{GK},\overline{AK})=\measuredangle GKA=\measuredangle HKJ$ as desired. For the second point we just prove $\overline{J’H’}\parallel\overline{JH}$ for similar reasons. However $\measuredangle JHJ’=\measuredangle JFJ’=\measuredangle EFD=\measuredangle EGD=\measuredangle H’GH=\measuredangle H’J’H$ so we have our parallelism, done.
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28.12.2024 06:41
idk what to do so imma post this solution lol. We'll prove these two claims first. Claim 1. $\overline{H,A,I}$ and $\overline{J,A,K}$. Proof. Observe that $\angle{IFD}=180^{\circ}-\angle{EFD}=\angle{EGD}=\frac{1}{2}\angle{EAD}=\frac{1}{2}(180^{\circ}-\angle{EID})=90^{\circ}-\frac{\angle{FID}}{2}$ $\Longrightarrow$ $\triangle{FID}$ isosceles. Since that $\triangle{FAD}$ is also isosceles, then we know that $AI$ is the perpendicular bisector of $DF$. Then, notice that $\angle{FHD}=\angle{FHG}=180^{\circ}-\angle{FAG}=180^{\circ}-2\angle{FDG}=180^{\circ}-2\angle{FDH}$ $\Longrightarrow$ $\triangle{FDH}$ is isosceles. Since that $\angle{FHA}=\angle{AHG}=\angle{AHD}$, then $AH$ must be the perpendicular bisector of $DF$. Hence, $H$, $A$, $I$ are collinear. By the similar reasoning, we obtain that $J$, $A$, $K$ are also collinear. $\square$ Claim 2. $\overline{I',H',A}$ and $\overline{K',J',A}$. Proof. We'll prove this claim for $K'$, $J'$, and $A'$ only. For $I'$, $H'$, $A'$, analogous. Notice that $\angle{K'GD}=\angle{EGD}=\frac{1}{2}\angle{EAD}=\frac{1}{2}(180^{\circ}-\angle{EK'D})=90^{\circ}-\frac{\angle{EK'D}}{2}$, so $\triangle{K'GD}$ must be isosceles. Similarly, $\angle{GJ'D}=\angle{GJ'F}=180^{\circ}-\angle{FAG}=180^{\circ}-2\angle{FDG}=180^{\circ}-2\angle{J'DG}$ $\Longrightarrow$ $\triangle{J'DG}$ is isosceles. After that, we can have $\angle{AJ'D}=\angle{AJ'F}=\angle{AJ'G}$ and $\angle{GK'A}=\angle{EK'A}=\angle{AK'D}$. Since $A$ lies on the perpendicular bisector of $DG$, we conclude that both $J'$ and $K'$ also lie on that line. Thus, $K'$, $J'$, $A'$ and $I'$, $H'$, $A$ are collinear. $\square$ Finally, observe that $\angle{K'I'J'}=\angle{K'I'D}=\angle{K'AD}=\angle{J'AD}=\angle{J'AG}=180^{\circ}-\angle{GH'J'}=\angle{K'H'J'}$ and $\angle{JHI}=\angle{JHA}=180^{\circ}-\angle{JFA}=180^{\circ}-\angle{EFA}=180^{\circ}-\angle{FEA}=180^{\circ}-\angle{IEA}=\angle{AKI}=\angle{JKI}$ $\Longleftrightarrow$ $H'I'J'K$ and $HIJK$ are cyclic quadrilaterals. $\blacksquare$ We'll introduce our third claim to prove that the center of these two circles is the same. Claim 3. $H'J'\parallel HJ$ and $K'I'\parallel KI$ Proof. Notice that $H'J'$ and $DE$ are parallel since that $\angle{K'H'J'}=\angle{K'I'J'}=\angle{K'I'D}=\angle{K'ED}$. Next, we have $\angle{IJH}=\angle{FJH}=180^{\circ}-\angle{FGH}=\angle{FGD}=\angle{FED}=\angle{IED}$ $\Longrightarrow$ $DE$ is parallel with $HJ$. Therefore, $H'J$ $\parallel$ $DE$ $\parallel$ $HJ$.Then, notice again that $\angle{I'K'G}=\angle{I'K'H'}=\angle{I'J'H'}=180^{\circ}-\angle{H'J'F}=\angle{HGF}=\angle{K'GH}$ $\Longrightarrow$ $K'I'$ $\parallel$ $GF$. Once again, $\angle{JFG}=\angle{EFG}=\angle{EDG}=180^{\circ}-\angle{EDF}=\angle{EIK}=\angle{JIK}$ $\Longrightarrow$ $GF$ $\parallel$ $KI$, thus $K'I'$ $\parallel$ $GF$ $\parallel$ $KI$. $\square$ From Claim 3., we can conclude that $H'J'HJ$ and $K'I'KI$ are both isosceles trapezoids. Therefore, pairs of ($HJ$,$H'J'$) and ($KI$,$K'I'$) has same perpendicular bisectors. Hence, centers of ($H'I'J'K'$) and ($HIJK$) coincide. $\blacksquare$ Q.E.D.
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