Let $ABC$ be a triangle with incentre $I$ and circumcircle $\omega$. Let $D$ and $E$ be the second intersection points of $\omega$ with $AI$ and $BI$, respectively. The chord $DE$ meets $AC$ at a point $F$, and $BC$ at a point $G$. Let $P$ be the intersection point of the line through $F$ parallel to $AD$ and the line through $G$ parallel to $BE$. Suppose that the tangents to $\omega$ at $A$ and $B$ meet at a point $K$. Prove that the three lines $AE,BD$ and $KP$ are either parallel or concurrent. Proposed by Irena Majcen and Kris Stopar, Slovenia
Problem
Source: IMO Shortlist 2011, G5
Tags: geometry, incenter, circumcircle, geometric transformation, homothety, IMO Shortlist, Angle Chasing
13.07.2012 15:05
Graph: File:G5.PNG Solution:
13.07.2012 16:05
Easy with bary centre. Hint: It is easy to show that $F=(a,0,b+c)$.We know K,we can easily compute $P$.We can show that $D=(-\frac{a^2}{b+c},b,c)$ and like wise E.
13.07.2012 16:20
After showing $K=AF' \cap BG'$, you can also finish like this: Let $X=AE \cap BD$. Pascal's Theorem on $AAEBBD$ gives $K \in XI$, so $K$ lies on the radical axis of $AEFI$ and $BDGI$. Then $KA\cdot KF'=KB \cdot KG' \Longrightarrow KF'=KG'$. Then the dilation centred at $K$ sending $AB$ to $F'G'$ sends $I$ to $P$, so $K,I,P,X$ are collinear.
14.07.2012 00:01
Babai wrote: Easy with bary centre. Hint: It is easy to show that $F=(a,0,b+c)$.We know K,we can easily compute $P$.We can show that $D=(-\frac{a^2}{b+c},b,c)$ and like wise E. The calculations actually get pretty nasty (at least, the way I did it). Unfortunately, during the actual MOP test, I somehow did not simplify $F$ and had something of much larger degree. As a result I did not finish.
15.07.2012 12:58
Claim:- Let $BD\cap AE= X$. Then $X,I,P$ are collinear. Proof:- By simple angle chasing we can prove that $AIFE, BIGD$ are cyclic. Now the triangles $\triangle IDE, \triangle PFG$ are homothetic. Let $PI\cap DE=Y$. Then $Y$ is the centre of homothety of these two triangles. So we have $\frac {YG}{YF}=\frac {YE}{YD}\Longrightarrow YG.YD=YE.YF\Longrightarrow Y$ lies on the radical axis of $\odot AIE$ and $\odot BID$. Hence $A,I,Y$ are collinear. But $I,P,Y$ are collinear. So $I,P,X$ are collinear. Main Proof:- Let $AB\cap DE=Z$. Then $IZ$ is the polar of $X$ wrt $\odot ABC$. $K$ is the pole of $AB$ wrt $\odot ABC$. $XZ$ is the polar of $I$. Since $X,I,P$ are collinear, polar of $P$ passes through $Z$. Hence we conclude $K,X,P$ are collinear. So done.
23.07.2012 23:14
Observation $\Rightarrow \angle IAF=\angle IEF=\frac {1}{2}\angle A$,so $AIFE$ is cyclic.Let the circle be $\alpha$. $\angle IBG=\angle IDG=\frac {1}{2}\angle B$,so $BIGD$ is cyclic.Let the circle be $\beta$. Let $Q=\alpha \cap \beta \neq I$. Claim1 $\Rightarrow AKBQ$ is cyclic. Proof:$\angle AKB+\angle AQB=180^{\circ}-2\angle C+\angle AQI+\angle BQI$ $=180^{\circ}-2\angle C+2\angle C=180^{\circ}$ Claim2 $\Rightarrow K,I,Q$ are collinear. Proof:$\angle IQB=\angle IDB=\angle ADB=\angle KAB=\angle KQB$ implying $K,I,Q$ are collinear. Claim3 $\Rightarrow QGPF$ is cyclic and $Q,P,I$ are collinear. Let $P'=IQ \cap \odot QGF \neq Q$. now $\angle QP'G=\angle QFG=\angle QIE \rightarrow BE \parallel P'G$ similarly $AD \parallel P'F$ implying $P=P'$. So we get $K,P \in IQ$ i.e. the radical axis of $\alpha,\beta$. Now the radical axis of $\omega,\alpha$ is $AE$,the radical axis of $\omega,\beta$ is $BD$ and the radical axis of $\alpha,\beta$ is $PK$ concur at the radical centre($O'$) of the three circles. If $\alpha,\beta,\omega$ are co-axial then $O'=\infty$ meaning $AE \parallel BD \parallel PK$.
16.08.2012 21:31
We use the following well-known Lemma: $ABDE$ - a cyclic quadrilateral, $ AD$ and $ BE$ intersect at $ I$, $ KB$ and $ KA$ are tangents to the circumcircle at $ A$ and $ B$. Then $ AE$, $ BD$, $ KI$ are either parallel or concurrent. Let $ K'$ be the point of intersection of tangents at points $ D$, $ E$. We need to proof that $ K'$, $ P$, $ I$ are collinear. By simple angle chasing we obtain that triangles $ DIE$ and $ FPG$, $ GCF$ and $ DK'E$ are similar. Also triangles $ DIE$ and $ DCE$ are equal. Let $ K'I$ intersect $ DE$ at $ T$ and $ PI$ intersect $ GF$ at $ S$. $ IGPF$ and $ K'EID$ are similar, therefore $\frac{ET}{TD} = \frac{GS}{SF}$. $ SFP$ and $ SDI$, $ SGP$ and $ SEI$, $ GPF$ and $ EID$ are similar, so $ \frac{SF}{SD} = \frac{PF}{DI}$, $ \frac{GS}{SE} = \frac{GP}{IE}$, $ \frac{DI}{PF} = \frac{IE}{GP}$. From this we get that $ \frac{ES}{SD} = \frac{GS}{SF}$ . So $ S$ = $ T$ and $ K'$, $ P$, $ I$ are collinear. This finishes the proof.
19.05.2014 11:11
Suppose $X=BD\cap AE$ (might be at the infinity) Obviously $AIFE(=\omega_1)$ and $BIGD(=\omega_2)$ are concyclic. A little angle chasing shows $IDG\sim EIF\implies \angle IGF=\angle IFG=A/2+B/2$. So $\angle GIF=\angle C$. Let $PF\cap \omega_1=M$ and $PG\cap \omega_2=N$. $\angle DIG=\angle DBG=B/2$ and $\angle NID=\angle NGD=\angle BED=B/2$. So it follows $\angle NIG=B$ Similarly $\angle MIF=A$. So $\angle NIM=A+C+B=180$. Also note that $\angle NMF=\angle IEF=\angle FGP\implies FMNG$ concyclic. Hence $PF\cdot PM=PG\cdot PN\implies IP$ is the radical axis of $\omega_1$ and $\omega_2$. Hence $X$ is the radical center of $\omega_1,\omega_2$ and $ABC$. Thus $IPX$ collinear. It is very well known that in cyclic quad $ABDE$, $K, I, X$ are collinear. Hence $AE,BD,KP$ are concurrent at $X$. [asy][asy]/* File unicodetex not found. */ /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(8.66cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(12); defaultpen(dps); /* default pen style */ real xmin = -0.23, xmax = 15.78, ymin = -2.02, ymax = 7.24; /* image dimensions */ pen uququq = rgb(0.25,0.25,0.25); /* draw figures */ draw(shift((4.78,2.53)) * scale(3.13, 3.13)*unitcircle); draw((3.26,5.26)--(1.74,1.78)); draw((3.26,5.26)--(7.9,2.4)); draw((1.74,1.78)--(7.9,2.4)); draw((6.42,5.19)--(5.09,-0.58)); draw(shift((3.81,1.16)) * scale(2.16, 2.16)*unitcircle); draw(shift((4.83,4.69)) * scale(1.67, 1.67)*unitcircle); draw((5.17,6.32)--(2.26,-0.34)); draw((6.33,2.62)--(2.26,-0.34)); draw((5.17,6.32)--(6.33,2.62)); draw((3.26,5.26)--(5.09,-0.58)); draw((1.74,1.78)--(6.42,5.19)); draw((3.86,3.33)--(5.73,2.18)); draw((3.86,3.33)--(6.04,3.55)); draw((1.74,1.78)--(5.09,-0.58)); draw((3.26,5.26)--(6.42,5.19)); /* dots and labels */ dot((3.26,5.26),blue); label("$A$", (3.14,5.46), NE * labelscalefactor,blue); dot((1.74,1.78),blue); label("$B$", (1.49,1.81), NE * labelscalefactor,blue); dot((7.9,2.4),blue); label("$C$", (8.05,2.31), NE * labelscalefactor,blue); dot((5.09,-0.58),uququq); label("$D$", (5.23,-0.94), NE * labelscalefactor,uququq); dot((6.42,5.19),uququq); label("$E$", (6.47,5.28), NE * labelscalefactor,uququq); dot((6.04,3.55),uququq); label("$F$", (6.32,3.48), NE * labelscalefactor,uququq); dot((5.73,2.18),uququq); label("$G$", (5.95,1.95), NE * labelscalefactor,uququq); dot((3.86,3.33),uququq); label("$I$", (3.44,3.42), NE * labelscalefactor,uququq); dot((6.33,2.62),uququq); label("$P$", (6.38,2.71), NE * labelscalefactor,uququq); dot((5.17,6.32),uququq); label("$M$", (5.23,6.41), NE * labelscalefactor,uququq); dot((2.26,-0.34),uququq); label("$N$", (2.01,-0.7), NE * labelscalefactor,uququq); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy]
23.05.2014 12:26
29.05.2014 23:33
27.09.2014 19:50
First,notice that $AE,BD$ and $IP$ are concurent,cause if $X$ is the second intersection point of the circumcircles of cyclic $BGDI$ and $AEIF$ we have that $PFGX$ is a cyclic and now is an easy angle chase.Now,we will prove that $IK,AD$ and $BE$ are concurent.Now,let $S$ be the second intersection point of the circumcircles of $AEK$ and $BDK$,now we easy obtain that $DIES$ is cyclic so similary like in the previous case we finish with an easy angle chase,so we get $IP,IK,AD,BE$ are concurent,or $PK,AD$ and $BE$ are concurent,so we are finished.
30.09.2014 04:16
A very motivated solution to an extremely nice problem: Define $Q=DB\cap AE, H=(BID)\cap (AIE), Y=BK\cap GP, Z=AK\cap FP$.
Now, by Pascal's on $AADBBE$, we get $K,I,Q$ are collinear. By radical axis on $(BIHD), (AIHE), (ABED)$ we get $Q,I,H$ are collinear. Hence $Q,K,I,H$ are collinear. Now since $\angle DIY =\angle AIZ$ we have either $Z,I,Y$ collinear or a weird case (because we haven't used directed angles). Fortunately this case is $Y=G$ which is obviously false since the only common point of $YK, BG$ is $B\neq G,Y$. So $Z,I,Y$ are collinear. Hence $\angle ZYG =\angle IYG =\angle IBG =\tfrac{ABC}{2}=\angle ZAE =\angle ZFE=\pi -\angle ZFG$, so $ZYFG$ is cyclic. Finally by radical axis on $(ZYFG), (YGHI),(ZFHI)$ we get $P\in HI\equiv KQ$ and we are done.
05.10.2014 18:36
malcolm wrote: Let $X=AE \cap BD$. Pascal's Theorem on $AAEBBD$ gives $K \in XI$... saturzo wrote: Again applying Pascal on $AAEBBD$ shows that $I, K, L$ are collinear. infiniteturtle wrote: Now, by Pascal's on $AADBBE$, we get $K,I,Q$ are collinear. Why does everyone like to call this Pascal? This result has a name, Brokard's Theorem (see page 3 here). Really we're just saying the pole of $\overline{AB}$ (namely $K$) is collinear with $I$ and $L$.
22.10.2014 13:23
I think its an easy problem for a G5. Note that the triangles $IED$ and $PGE$ are homothetic with center of homothety placed at $IP \cap DE=V$.Then we have $\frac{VF}{VD}=\frac{PV}{VI}=\frac{VG}{VE} \implies VF \cdot VE=VG \cdot VD$.Also note that $AIFE$ and $BIGD$ are cyclic quadrilaterals.Thus $IP$ is the radical axis of the two circles.Now the radical axis of $\odot{AIFE}$ and $\odot{ABC}$ is $AE$ and that of $\odot{BIGD}$ and $\odot{ABC}$ is $BD$.Now I asume that $\angle{C} \neq 60^{\circ}$. So $J=AE \cap BD$ lies on the radical axes of $\odot{AIFE}$ and $\odot{BIGD}$.In other words $J,I,P$ are collinear. Now if $AB \cap ED=L$ then note that the polar of $K$ (i.e, $AB$) passes through $L$.Thus the polar of $L$ must pass through $K$.Since the polar of $L$ is $IJ$ this means that $I,J,K$ are collinear. Combining the results of the two paragraphs we obtain that $K,P,J$ are collinear or $AE,BD,KP$ are concurrent. Now if $\angle{C}=60^{\circ}$ then $AE$ and $BD$ are clearly parallel and triangles ${AIE}$ and $BID$ are equilateral.Now since $IP$ is the radical axis of the two circles we have $IO_{AEI}O_{BID} \perp IP$ and also it is perpendicular to each of $AE$ and $BD$.So $AE \parallel BD \parallel KP$ in this case.(or one can also say that $AE,BD,KP$ concur at infinity,so $KP$ is parallel to $AE$ and $BD$)
28.11.2014 07:20
WakeUp wrote: Let $ABC$ be a triangle with incentre $I$ and circumcircle $\omega$. Let $D$ and $E$ be the second intersection points of $\omega$ with $AI$ and $BI$, respectively. The chord $DE$ meets $AC$ at a point $F$, and $BC$ at a point $G$. Let $P$ be the intersection point of the line through $F$ parallel to $AD$ and the line through $G$ parallel to $BE$. Suppose that the tangents to $\omega$ at $A$ and $B$ meet at a point $K$. Prove that the three lines $AE,BD$ and $KP$ are either parallel or concurrent Outline: We first define two new points, $X$ and $Y.$ We then show that two pentagons are cyclic, and discover another cyclic quadrilateral. The radical axis theorem on the two cyclic pentagons and $\omega$ then gives the desired result. While this solution is long, it is quite direct, using only angle chasing and radical axes. Proof. Let $X = KB \cap PG$ and, similarly, let $Y = KA \cap PF.$ Then we claim that $BIGDX$ is a cyclic pentagon. Moreover, quadrilateral $BIGX$ is an isoceles trapezoid. For a proof of the first part, we first show that $GDXB$ is cyclic, and we then show that $BIGD$ is cyclic. Note that since $KB = KA$ by equal tangents, it follows that \[\angle KBA = 90^{\circ} - \frac{1}{2}\angle BKA = 90^{\circ} - \frac{1}{2}\left(180^{\circ} - \angle BOA\right) = \frac{1}{2}\angle BOA = C\] where $O$ is the circumcenter of $\triangle ABC.$ We use this fact twice throughout the proof. Now, for the first cyclic quad, note that since $GX \parallel IB$, it follows that \[\angle GXB = \angle IBK = \angle IBA + \angle ABK = \frac{B}{2} + C.\] In addition, we have that \[\angle GDB = \angle EDB = \angle EDA + \angle ADB = \frac{B}{2} + C.\] Therefore, $\angle GXB = \angle GDB$, which implies that quadrilateral $GDXB$ is cyclic. For the second cyclic quad, we have that \[\angle BID = 180^{\circ} - \angle BIA = 180^{\circ} - \left(90^{\circ} + \frac{C}{2}\right) = 90^{\circ} - \frac{C}{2}.\] In addition, we have that \[\angle BGD = \frac{\overarc{BD} + \overarc{EC}}{2} = \frac{A}{2} + \frac{B}{2} = 90^{\circ} - \frac{C}{2}.\] Therefore $\angle BID = \angle BGD$, which implies that that quadrilateral $BIGD$ is cyclic. Therefore, $BIGDX$ is one cyclic pentagon. This concludes the proof of the first part. $\blacksquare$ The second part follows immediately from the first: Since $BIGX$ is a cyclic quadrilateral, and $GX \parallel IB$, it follows that $BIGX$ is a isoceles trapezoid. $\blacksquare$ Analagously, we find that $AIFEY$ is a cyclic pentagon, and $AIFY$ is an isoceles trapezoid. Now, we claim that $I \in \overline{XY}.$ Moreover, quadrilateral $XGFY$ is cyclic. For the first part, due to cyclic pentagons $BIGDX$ and $AIFEY$, and since $ID \parallel PF; IE \parallel PG$, it follows that \[\angle XIY = \angle XID + \angle DIE + \angle EIY = \angle XGD + \angle DIE + \angle EFY\]\[= \angle FGP + \angle FPG + \angle GFP = 180^{\circ}.\] Therefore, $I \in \overline{XY}$, as desired. $\blacksquare$ For the second part, we have \[\angle XGF = 180^{\circ} - \angle XGD = 180^{\circ} - \angle XBD = \angle KBD\]\[= \angle KBA + \angle ABC + \angle CBD = C + B + \frac{A}{2} = 180^{\circ} - \frac{A}{2}.\] In addition, we have \[\angle XYF = \angle IYF = \angle IAF = \frac{A}{2}.\] Therefore, $\angle XGF + \angle XYF = 180^{\circ}$, which implies that quadrilateral $XGFY$ is cyclic. $\blacksquare$ Now, we claim that $\triangle IFG$ is isoceles with $IF = IG.$ This follows from \[\angle FGI = 180^{\circ} - \angle IGD = \angle IBD = \angle EBD\]\[= \angle EAD = \angle EAI = 180^{\circ} - \angle EFI = \angle IFG.\] Therefore $IF = IG = AY = BX$ where the last steps follow from isoceles trapezoids $BIGX$ and $AIFY.$ Combining this equality with $KA = KB$, it follows that $KX = KY.$ Therefore, $KB \cdot KX = KA \cdot KY$, which implies that $K$ lies on the radical axis of $\odot (BIGDX)$ and $\odot (AIFEY).$ In addition, by Power of a Point on cyclic quadrilateral $XGFY$, it follows that $PG \cdot PX = PF \cdot PY.$ Therefore $P$ also lies on the radical axis of $\odot (BIGDX)$ and $\odot (AIFEY).$ It follows that $KP$ is the radical axis of $\odot (BIGDX)$ and $\odot (AIFEY).$ Then, by the Radical Axis Theorem on circles $\omega, \odot (BIGDX), \odot (AIFEY)$ it follows that $AE, BD, KP$ all either concur, or are parallel. The proof is complete. $%Error. "qedsymbol" is a bad command. $
28.11.2014 10:10
My solution: Let $ X=PF \cap (AEF), Y=PG \cap (BDG), Z=AE \cap BD $ . Since $ \angle FAI=\angle IAB=\angle FEI $ , so we get $ A, E, F, I $ are concyclic . Similarly, we can prove $ B, D, G, I $ are concyclic . Since $ \angle AIX=\angle CFP=\angle CAI=\angle IAB $ , so we get $ IX \parallel AB $ . Similarly, we can prove $ IY \parallel AB $ , so we get $ X, I, Y $ are collinear and parallel to $ AB $ . Since $ \angle PFG=\angle IDF =\angle IYG $ , so we get $ F, G, X, Y $ are concyclic and $ PF \cdot PX=PG \cdot PY $ , hence $ PI $ is the radical axis of $ (AEFI) $ and $ (BDGI) $ . Since $ ZA \cdot ZE=ZB \cdot ZD $ , so $ Z $ lie on the radical axis of $ (AEFI) $ and $ (BDGI) $ . ie. $ P, I, Z $ are collinear ... $ (*) $ From Maclaurin Theorem we get $ K, I, Z $ are collinear , so combine with $ (*) $ we get $ PK $ is the radical axis of $ (AEFI) $ and $ (BDGI) $ , hence from radical center theorem we get $ PK, AE, BD $ are concurrent or parallel . Q.E.D
27.11.2015 00:26
First we claim that $BIGD$ is cyclic. Indeed, $\angle IBG=\angle EBC=\angle EBA=\angle EDA=\angle IDG$ as desired. Then by symmetry, $AIFE$ is cyclic. Thus by the radical axis theorem on $(AIFE), (BIGD)$ and $\gamma$, lines $AE$, $BD$ and the radical axis of $(AIFE)$ and $(BIGD)$ are concurrent (possibly at the point at infinity). Thus, it suffices to show that $K$ and $P$ have equal power with respect to both circles. Let $KB$ intersect $(BIGD)$ again at $R$ and $(AIFE)$ again at $S$. Then we claim that $RS \parallel AB$, implying that $KS=KR\longrightarrow KA\cdot KS=KB\cdot KR$, so that $K$ has equal power with respect to both circles. First, $\angle BRI=\angle BDI=\angle DBA$ and $\angle ASI=\angle AEB=\angle ADB$ which implies that $R,I,S$ are collinear, so we get the conclusion. Now we want to show that $P$ has equal power with respect to both circles. Let $GP$ hit $(BIGD)$ again at $R'$; then $\angle BR'G=\angle BDG=\angle BDE=\angle KBE=\angle KBI$, so $R'\equiv R$. Then $\angle PGF=\angle PGE=\angle RGD=\angle RID=\angle RSP$; thus $\triangle PGF \sim \triangle PSR$, so that $PG \cdot PR=PF \cdot PS$, and we're done$.\:\blacksquare\:$
16.01.2016 18:36
Ok here's an angle-chasing-free solution . First , in what follows $(UVW)$ denotes the circle passing through $U,V$ and $W$ . Let $I_A, I_B , I_C $ be the excenters . Let the tangents to the circumcircle at $D$ and $E$ intersect at $X$. Some facts we'll use a lot : $E$ and $D$ are the midpoints of $[II_A]$ and $[II_B]$ because $ABC$ is the orthic triangle of $I_AI_BI_C$ and $(ABC)$ is the euler circle of $I_AI_BI_C$ . By Pascal on $AADBBE$ and $DDAEEB$ , $(AE)$ and $(BD)$ intersect on $(XIK)$ , so proving that $P$ lies on the line through $X,I$ and $K$ would be enough . Since $PFG$ and $II_AI_B$ are homothetic , $(IP)$ goes through the center of homothety sending $F$ to $I_A$ and $G$ to $,I_B$ . Let's look for the image of $I$ by that homothety ; it is the point $Y$ satisfying $(FI_) // (YI_A)$ and $(GI)//(YI_B)$ . Claim : The reflection of $I$ over $X$ is that point . Note that the reflection of $I$ over $X$ is the image of $X$ by the homothety centered at $I$ sending $D$ to $I_A$ , showing that $(EX)//(IG)$ and $(DX)//(IF)$ will prove the claim and solve the problem, but $(EX)//(AC)$ and $(DX)//(BC)$. So It , again , suffices to show that $(IF)//(BC)$ and $(IG)//(AC)$ . Now if $(IF)$ intersects $(I_BI_C)$ at $U$ , by homothety $F$ is the midpoint of $[IU]$ , hence proving the parallelism is equivalent to $(CF,CG;CI,CU)=-1 <=> (CA,CB;CI_C,CI_B)=-1$ wich is true (because $ABC$ is the orthic triangle of $(I_AI_BI_C$ ) . Do the same for $(IG)$ and $(AC)$ , and QED.
19.03.2016 05:41
I shall use the diagram of Particle. Lemma 1. The quaderilaterals $BIGD$ and $AIFE$ are cyclic. Proof. Easy angle chasing gives $\angle GPF = 90 - \frac{C}{2}$ and $\angle FGP = \frac{A}{2}.$ Also $\angle FGC= \angle GFC$ by angle chasing. Now note that $\angle BGD = \angle BID = 90 - \frac{C}{2}$ to finish. Lemma 2. If $PG,PF$ meet $\odot (AIFE),\odot (BIGD)$ at $M,N$ then $M,I,N$ are collinear. Proof. Easy angle chase again, with cyclic quads and angle sum property. Lemma 3. $P$ lies on the radical axis of $\odot (BIGD),\odot (AIFE).$ Proof. Since $MNGF$ is cyclic (trivial angle chase again) by power of point we have $PG \cdot PN= PF \cdot PM.$ The result follows. Lemma 4. $K$ lies on the radical axis of $\odot (BIGD), \odot (AIFE).$ Proof. First note that $\angle MAF = \angle MAE + \angle CAE = \angle MFE + \angle CBE = \frac{B}{2} + \frac{B}{2} = B.$ Thus $MA$ and similarly $BN$ are tangent to $\odot (ABC).$ Hence $K = NB \cap MA.$ But also $\angle AMI = \angle AEB = \angle ACB = C,$ and $\angle NBA = A +B = 180 - C$ so $BAMN$ is cyclic. By power of point we have $KB \cdot KN = KA \cdot KM.$ Now we are ready to finish off. Note that $AE,BD,KP$ are the radical axes of the circles $ \odot (AIFEM), \odot (ABDCE), \odot (BIGDN)$ taken pairwise. Hence the required result follows by radical axes theorem. A note, and some motivation. See the way the problem is worded? The lines are concurrent or parallel. When the solution just involves, say, like Ceva, Menelaus, Desargues, Pappus etc. we don't usually have that bold part. The only way this can enter is by virtue of radical axes. Hence we wish to interpret the lines as the radical axes of three circles. Hence the construction of the two new points and two new circles.
18.06.2022 15:06
WakeUp wrote: Let $ABC$ be a triangle with incentre $I$ and circumcircle $\omega$. Let $D$ and $E$ be the second intersection points of $\omega$ with $AI$ and $BI$, respectively. The chord $DE$ meets $AC$ at a point $F$, and $BC$ at a point $G$. Let $P$ be the intersection point of the line through $F$ parallel to $AD$ and the line through $G$ parallel to $BE$. Suppose that the tangents to $\omega$ at $A$ and $B$ meet at a point $K$. Prove that the three lines $AE,BD$ and $KP$ are either parallel or concurrent. Let $(IGDB)\cap (IFEA)=S$. $\measuredangle PEG=\measuredangle CED=\measuredangle CAI =\measuredangle FSI$ and $\measuredangle PGF=\measuredangle CDE=\measuredangle CBI=\measuredangle GSI$. So we get $PESG$ is cyclic and $P,I,S$ are collinear. From radical axis theorem on $(AEDB),(AEIS),(ISBD)$ we get $AE,DB,IS$ are concurrent and since $P\in IS$ we only need to prove $K\in IS$. Now let $KA\cap (AEI)=R$ and $KB\cap (BDI)=L$. $\angle LIB=\angle DIB-\angle DIL=\angle DIB-\angle DBL=\angle DIB-\angle DAB=\angle IBA$. So $IL\parallel AB$. Similarly $IR\parallel AB$,which means $R,I,L$ are collinear. $\angle ARL=\angle AEI=\angle ACB=\angle ABK \implies ABLR$ is cyclic. From radical axis theorem on $(RAIS),(ISBL),(BLAR)$ we get $RA,BL,IS$ are concurrent, which means $IS$ passes through $K$, as desired.
09.08.2022 21:45
Nice problem, solved with walkthrough. By Pascal on $AAEBBD$ (or Brokard), we find that $\overline{KI},\overline{AE},\overline{BD}$ concur (possibly at infinity). Hence it suffices to show that $\overline{PI},\overline{AE},\overline{BD}$ concur, since then $P,I,K$ are collinear and the conclusion follows. Note that since $$\angle FEI=\angle DEB=\angle DAB=\angle CAD=\angle FAI,$$$AEFI$ is cyclic, and so is $BDGI$. Further, from the parallelisms, letting $T=\overline{PI} \cap \overline{FG}$, we have $\triangle ITE \sim \triangle PIG$ and $\triangle ITG \sim \triangle PIF$. This means that $\frac{TF}{TG}=\frac{TD}{TE} \implies TF \cdot TE=TG \cdot TD$, so $T$ lies on the radical axis of $(AEFI)$ and $(BDGI)$, hence $P$ does as well (since $I$ also obviously lies on the radical axis). Then, by radical center on $(AEFI),(BDGI),(ABC)$, it follows that $\overline{PI},\overline{AE},\overline{BD}$ concur. $\blacksquare$
11.08.2022 00:39
Here is a complete bary bash which has surprisingly not been posted yet. Not a trivial, but definetly doable. Employ barycentric coordinates on $\triangle ABC$ with $A=(1,0,0),B=(0,1,0),C=(0,0,1)$. By Fact 5, $D$ is the midpoint of $II_a$ so $$D=\frac{I+I_a}2=\frac12\cdot\frac{1}{(a+b+c)(-a+b+c)}[(-a^2+ab+ac,-ab+b^2+bc,-ac+bc+c^2)+(-a^2-ab-ac,ab+b^2+bc,ac+bc+c^2)].$$Thus,$$D=(-a^2:b^2+bc:c^2+bc).$$Similarly, $$E=(a^2+ac:-b^2:c^2+ac).$$ Since $F$ is on cevian $AC$, it can be parameterized by $F=(t:0:1)$. However, $D,E,F$ are colinear so we have $$\begin{vmatrix}t&0&1\\ -a^2&b^2+bc&c^2+bc\\ a^2+ac&-b^2&c^2+ac\end{vmatrix}=0 \iff t=\frac{a}{b+c}$$so $$F=(a:0:b+c).$$Similarly, $$G=(0:b:a+c).$$ Let $P_a=(-b-c:b:c)$ be the point at infinity along the $A$-angle bisector. Then the line through $F$ parallel to $AD$ has equation $$FP_a: \begin{vmatrix}x&y&z\\ a&0&b+c\\ -b-c&b&c\end{vmatrix}=0\iff x[b(b+c)]+y[(b+c)^2+ac]+z[-ab]=0.$$Similarly, the line through $G$ parallel to $BE$ has equation $$x[(a+c)^2+bc]+y[a(a+c)]+z[-ab]=0.$$ Their point of intersection, $P$, is the solution to the system$$\begin{bmatrix}1&1&1\\ b(b+c)&(b+c)^2+ac&-ab\\ (a+c)^2+bc&a(a+c)&-ab\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}1\\0\\0\end{bmatrix}.$$We can solve this using Cramer's Rule (we don't need the denominator since we only care about unhomogenized coordinates). This gives $$x=\begin{vmatrix} (b+c)^2+ac&-ab\\ a(a+c)&-ab\end{vmatrix}=-ab(a+b+c)(-a+b+c)$$and $$y=\begin{vmatrix} -ab&b(b+c)\\-ab&(a+c)^2+bc\end{vmatrix}=-ab(a+b+c)(a-b+c).$$The $z$ component is a little harder to compute. Cramer's Rule gives \begin{align*}z&=\begin{vmatrix}b(b+c)&(b+c)^2+ac\\ (a+c)^2+bc&a(a+c)\end{vmatrix}\\&=-[(b+c)^2(a+c)^2+ac(a+c)^2+bc(b+c)^2+abc^2-ab(ab+bc+ca+c^2)]\\&=-[(ab+bc+ca+c^2)(c)(a+b+c)+(c)[a(a+c)^2+b(b+c)^2+abc]]\\&=-[(ab+bc+ca+c^2)(c)(a+b+c)+(c)[a^3+2a^2c+ac^2+b^3+2b^2c+bc^2+abc]]\\&=-[(ab+bc+ca+c^2)(c)(a+b+c)+(c)[(a+b+c)(-ab+bc+ca+a^2+b^2]]\\&= -(a+b+c)(c)(2ac+2bc+a^2+b^2+c^2).\end{align*}Factoring $-(a+b+c)$, we get $$P=(ab(-a+b+c):ab(a-b+c):c(2ac+2bc+a^2+b^2+c^2)).$$ Now, instead of showing that $AE,BD,PK$ are either parallel or concurrent, we will show that $T=AE\cap BD,K,P$ are colinear. Since $T$ lies on cevian $BD$, it can be parameterized by $T=(-a^2:t:c^2+bc)$. However, since $A,T,E$ are colinear, we must have $$\begin{vmatrix}1&0&0\\ -a^2&t&c^2+bc\\ a^2+ac&-b^2&c^2+ac\end{vmatrix}=0\iff t=-\frac{b^2(b+c)}{a+c}$$so $$T=(a^2(a+c):b^2(b+c):-c(b+c)(a+c)).$$The tangents to the circumcircle at $A$ and $B$ have equations $$b^2z+c^2y=0$$and $$a^2z+c^2x=0$$respectively, so it is easy to find $$K=(a^2:b^2-c^2).$$Finally, to show that $K,T,P$ are colinear, we just need to check that the determinant $$\begin{vmatrix}a^2&b^2&-c^2\\ a^2(a+c)&b^2(b+c)&-c(a+c)(b+c)\\ ab(-a+b+c)&ab(a-b+c)&c(2ac+2bc+a^2+b^2+c^2) \end{vmatrix}$$vanishes. However, observe \begin{align*}\begin{vmatrix}a^2&b^2&-c^2\\a^2(a+c)&b^2(b+c)&-c(a+c)(b+c)\\ ab(-a+b+c)&ab(a-b+c)&c(2ac+2bc+a^2+b^2+c^2) \end{vmatrix}&=abc\cdot \begin{vmatrix}a&b&-c\\a(a+c)&b(b+c)&-(a+c)(b+c)\\ b(-a+b+c)&a(a-b+c)&(2ac+2bc+a^2+b^2+c^2) \end{vmatrix}\\&= a(b+c)[b(2ac+2bc+a^2+b^2+c^2)+a(a-b+c)(a+c)]-b(a+c)[a(2ac+2bc+a^2+b^2+c^2)+b(-a+b+c)(b+c)]-c[a^2(a-b+c)(a+c)-b^2(-a+b+c)(b+c)]\\&=a(b+c)[(2abc+2b^2c+a^2b+b^3+bc^2)+(a^3+2a^2c-a^2b-abc+ac^2)]-b(a+c)[(2a^2c+2abc+a^3+ab^2+ac^2)+(-ab^2-abc+b^3+2b^2c+bc^2)]-c[a(a^3+2a^2c-a^2b-abc+ac^2)-b(-ab^2-abc+b^3+2b^2c+bc^2)]\\&=a(b+c)[a^3+b^3+2a^2c+2b^2c+ac^2+bc^2+abc]-b(a+c)[a^3+b^3+2a^2c+2b^2c+ac^2+bc^2+abc]-c[a^4+2a^3c-a^3b-a^2bc+a^2c^2-b^4-2b^3c+b^3a+ab^2c-b^2c^2]\\&=c(a-b)(a^3+b^3+2a^2c+2b^2c+ac^2+bc^2+abc)-c[a^4+2a^3c-a^3b-a^2bc+a^2c^2-b^4-2b^3c+b^3a+ab^2c-b^2c^2]\\&=0\end{align*}so $K,T,P$ are colinear and we are done.
14.09.2022 13:22
Quote: Let $ABC$ be a triangle with incentre $I$ and circumcircle $\omega$. Let $D$ and $E$ be the second intersection points of $\omega$ with $AI$ and $BI$, respectively. The chord $DE$ meets $AC$ at a point $F$, and $BC$ at a point $G$. Let $P$ be the intersection point of the line through $F$ parallel to $AD$ and the line through $G$ parallel to $BE$. Suppose that the tangents to $\omega$ at $A$ and $B$ meet at a point $K$. Prove that the three lines $AE,BD$ and $KP$ are either parallel or concurrent. Proposed by Irena Majcen and Kris Stopar, Slovenia Let $X = AE \cap BD$. By Pascal's theorem on $AAEBBD$ we have points $K, I, X$ collinear. Thus, it suffices to prove that $P \in IX$. Denote by $\ell _1$ and $\ell _2$ the lines through $F$ parallel to $AD$ and the line through $G$ parallel to $BE$, respectively. Let $A' = AE \cap \ell _1 , E' = AE \cap \ell _2, D' = BD \cap \ell _1, B' = BD \cap \ell_2.$ Notice that triangles $AIE$ and $A'PE'$ are homothetic, say, with centre $X'$. Now, triangles $BID$ and $B'PD'$ are also homothetic. Clearly both of these homotheties have the same ratio, maps $P$ to $I$, and hence have the same centre. Therefore, $X' = AE \cap BD = X$, i. e. $X, I, P$ are collinear. $\blacksquare$.
29.01.2023 10:58
Let $X= AE \cap BD$, and $T= (ADX) \cap (BEX)$. We wish to show $X,K,P$ collinear. By Pascal's Theorem on $AADBBE$, $X,K,I$ are collinear. Hence, it would suffice to prove that $X,I,P$ are collinear. Claim 1: $X,I,T$ collinear Proof: The inversion about $X$ with power $XA \cdot XE$ sends $I$ to $T$. Claim 2: $AEFI$ and $BDGI$ are cyclic quadrilaterals, and their circumcircles intersect at $I$ and $T$. Proof: By definition, $T$ is the Miquel point of $(AEDB)$, so it also lies on $(AEI)$ and $(BDI)$. $AEFI$ is cyclic because $\angle FEI = \angle FAI = \frac{1}{2} \angle A$. Similarly, $BDGI$ is cyclic Claim 3: $T \in (PFG)$ Proof: Firstly, \[\angle PFG = 180^{\circ} - \angle PFG - \angle PGF = 180^{\circ} - \angle FDA - \angle GEB = 180^{\circ} - \frac{1}{2} \angle A - \frac{1}{2} \angle B \]Now from Claim 2, \[\angle FTG = 360^{\circ} - \angle FTI - \angle GTI = 360^{\circ} -(180^{\circ}-\angle FAI) - (180^{\circ}-\angle GBI) = \frac{1}{2} \angle A + \frac{1}{2} \angle B = 180^{\circ} - \angle PFG \]as desired. Claim 4: $I,T,P$ collinear Proof: \begin{align*} \angle FTP + \angle FTI = \angle FGP + 180^{\circ} - \angle FAI &= \angle DEB + 180^{\circ}-\frac{1}{2} \angle A \\ &= \frac{1}{2} \angle A + 180^{\circ}-\frac{1}{2} \angle A \\ &=180^{\circ} \end{align*} Combining Claims 1 and 4, $X,I,P$ are collinear so we are done.
02.02.2023 20:31
First, we eliminate the point $K$. Note that the points $K, I, \overline{AE} \cap \overline{BD}$ are collinear by Pascal on $AADBBE$. Thus it suffices to show that $P$ lies on this line. Note that $BDGI$ is cyclic because $\measuredangle IBC = \measuredangle ADE$, and similarly $AIFE$ is cyclic. As a result, we can characterize this line as the radical axis of those two circles. Set $Q = \overline{IP} \cap \overline{FG}$. The two pairs of parallel lines imply $\frac{QF}{QD} = \frac{QP}{QI} = \frac{QG}{QE}$, so $QG \cdot QD = QF \cdot QE$. Thus $Q$ lies on the radical axis, which implies the conclusion.
16.02.2023 20:52
Let $H = AK \cap FP$ and $J = BK \cap GP$. Claim: $(AHEFI)$ and $(BJDGI)$ are cyclic. Proof: We will prove $(AHEFI)$ cyclic; the other proof is similar. We have \[\angle AIE = 180^\circ - \angle AIB = \angle BAI + \angle ABI = \frac12(\angle A + \angle B),\]and \[\angle AFE = 180^\circ - \angle AFD = 180^\circ - (360^\circ - \angle BAF - \angle FDB - \angle DBA) = 180^\circ - (360^\circ - \angle A - \angle C - \frac12 \angle B - \angle B - \frac12 \angle A) = \frac32 \angle A + \frac32 \angle B + \angle C - 180^\circ = \frac12 \angle A + \frac12 \angle B = \angle AIE,\]giving $(AEFI)$ cyclic. Furthermore, \[\angle HAI = \angle ABD = 180^\circ - \angle AED = 180^\circ - \angle AEF = \angle AIF,\]so $AHFI$ is an isosceles trapezoid, giving that $H$ lies on $(AEFI)$ as desired. $\square$ Claim: $H$, $I$, and $J$ are colinear, and $HJ \parallel AB$. Proof: Since \[\angle AHI = \angle AEI = \angle AEB = \angle C = \angle KAB,\]$HI \parallel AB$, so $H$ is on the line through $I$ parallel to $AB$. Similarly, $J$ is also on the same line, giving the desired conclusion. $\square$ Claim: $KP$ is the radical axis of $(AHEFI)$ and $(BJDGI)$. Proof: It suffices to show that $K$ and $P$ lie on the radical axis. Since $AB \parallel HJ$, $\triangle KAB \sim \triangle KHJ$ are isosceles, giving $KA = KB$ and $KH = KJ$, so $KA \cdot KH = KB \cdot KJ$, and $K$ lies on the radical axis. Also note that \[180^\circ - \angle HFG = \angle DFP = \angle ADE = \frac12 \angle B = \angle IBG = \angle IJG = \angle HJG,\]so $(HFGJ)$ is cyclic, giving $PF \cdot PH = PG \cdot PJ$, so $P$ lies on the radical axis as well. $\square$ Now $AE$, $BD$, and $KP$ are concurrent or parallel, from radical axis on $(AHEFI)$, $(BJDGI)$, and $\omega$.
03.06.2023 15:22
OK hear me out. This solution is not necessarily as complicated as it seems.
17.09.2023 08:46
One of my favorite geo problems, though misplaced for g5; this problem splices together several different ideas, so I really like my solution. Note that EIA=180-AIB=90-C/2=CFG=AFE so AEFI is cyclic and analogously BDGI; let H=(AEFI)\cap(BDGI). Then, PFH=180-CFP-AFH=180-CAD-AFH=AIH-A/2, analogously PGH=BIH-B/2, and indeed PFH+PGH=360-AIB-A/2-B/2=180, so PFHG is cyclic; we also have FHP=FGP=FEB=180-FHI, so PHI is a line. Furthermore, in degenerate quad ADBE with AE\cap BD=J, since H is the miquel point of that, with ADBE cyclic, it's well known that HIJ is a line. At this point, the problem is just screaming Pascal's, since it suffices to prove IJK is a line, which are all intersections relating AA,BB,D,E; in particular, Pascal's on AADBBE (say) finishes. \textbf{Remark.} Most of this angle chasing is well motivated by the fact that all these points are well defined in terms of A,B,C,I -- this builds intuition that there is an elementary way.
22.01.2024 05:30
Broke: IAmTheHazard wrote: Nice problem, solved with walkthrough. By Pascal on $AAEBBD$ (or Brokard), we find that $\overline{KI},\overline{AE},\overline{BD}$ concur (possibly at infinity). Hence it suffices to show that $\overline{PI},\overline{AE},\overline{BD}$ concur, since then $P,I,K$ are collinear and the conclusion follows. Note that since $$\angle FEI=\angle DEB=\angle DAB=\angle CAD=\angle FAI,$$$AEFI$ is cyclic, and so is $BDGI$. Further, from the parallelisms, letting $T=\overline{PI} \cap \overline{FG}$, we have $\triangle ITE \sim \triangle PIG$ and $\triangle ITG \sim \triangle PIF$. This means that $\frac{TF}{TG}=\frac{TD}{TE} \implies TF \cdot TE=TG \cdot TD$, so $T$ lies on the radical axis of $(AEFI)$ and $(BDGI)$, hence $P$ does as well (since $I$ also obviously lies on the radical axis). Then, by radical center on $(AEFI),(BDGI),(ABC)$, it follows that $\overline{PI},\overline{AE},\overline{BD}$ concur. $\blacksquare$ Woke: Observe that $I=\overline{AD} \cap \overline{BE}$ and $\overline{AE} \cap \overline{BD}$ both lie on the polar of $Q:=\overline{BC} \cap \overline{DE}$ by Brocard, and $K$ lies on the polar of $Q$ by La Hire. Hence it suffices to show that $\overline{PI} \perp \overline{OQ}$ where $O$ is the circumcenter of $ABC$. To that end, we use complex numbers with $(ABC)$ as the unit circle, assigning $C=x^2,A=y^2,B=z^2$ so that $D=-xz$ and $E=-xy$. Thus $I=-xy-xz-yz$. We compute \begin{align*} Q&=\frac{x^2yz(y^2+z^2)-y^2z^2(-xy-xz)}{x^2yz-y^2z^2}=\frac{x(xy^2+xz^2+y^2z+yz^2)}{x^2-yz},\\ F&=\frac{x^2y^2(-xy-xz)-x^2yz(x^2+y^2)}{x^2y^2-x^2yz}=-\frac{x^2z+xy^2+xyz+y^2z}{y-z}\\ G&=\frac{x^2y+xyz+xz^2+yz^2}{y-z}, \end{align*}where the last line comes from swapping $y$ and $z$ in our expression for $f$. We now compute $P$. Letting $j$ denote the complex coordinate of $I$ (to avoid confusion with the imaginary unit), we have $$\frac{p-f}{p-g}=\frac{j-d}{j-e} \implies p(d-e)=f(j-e)-g(j-d) \implies p=\frac{(f-g)j+(dg-ef)}{d-e} \implies p-j=\frac{((f-g)-(d-e))j+(dg-ef)}{d-e}.$$Substituting in and clearing denominators, it follows that $$\frac{p-j}{q}=\frac{(x^2-yz)((x^2y+x^2z+xy^2+xz^2+y^2z+yz^2+2xyz+x(y-z)^2)(xy+xz+yz)-x(2x^2yz+xy^3+xy^2z+xyz^2+xz^3+y^3z+yz^3))}{x^2(y-z)^2(xy^2+xz^2+y^2z+yz^2)}.$$Upon expansion and collecting terms, the numerator turns out to equal $$(x^2-yz)(x^3(y^2+z^2)+x^2(y+z)(y^2+yz+z^2)+2xyz(y^2+yz+z^2)+y^2z^2(y+z).$$By performing synthetic division, we may factorize this as $$(x(y^2+z^2)+(y^2z+yz^2))(x^2+(y+z)x+yz)=(x(y^2+z^2)+(y^2z+yz^2))(x+y)(x+z).$$Hence $$\frac{p-j}{q}=\frac{(x^2-yz)(x+y)(x+z)}{x^2(y-z)^2}.$$It is straightforward to verify that the conjugate of the RHS is its negative, so we are done. $\blacksquare$ Remark: It is possible to first realize that $x+y$ (and symmetrically $x+z$) divide the numerator through geometric means. I did not do this however, instead realizing that $x(y^2+z^2)+(y^2z+yz^2)$ had to divide the numerator. Conjugating the denominator and then multiplying by a suitable $x^ay^bz^c$ term can send the $x^2$ and $(y-z)^2$ terms to themselves, but it's impossible to get $x(y^2+z^2)+(y^2z+yz^2)$ (or its negative) through this. But for the problem to be true, we must be able to perform this conjugation, clear denominators, and then get the negative of the original expression. Thus $x(y^2+z^2)+(y^2z+yz^2)$ must divide the numerator as well for the conjugation to have any chance of working, in which case we could just cancel these factors out and be happy.
22.01.2024 21:28
[asy][asy] import olympiad; size(8cm,0); pair A = dir(120), B = dir(210), C = dir(330), D = dir(270), E = dir(45); pair I = incenter(A,B,C); pair F = intersectionpoints((10A-9C)--(10C-9A),(10D-9E)--(10E-9D))[0]; pair G = intersectionpoints((10B-9C)--(10C-9B),(10D-9E)--(10E-9D))[0]; path AA = (A + 10*dir(120+90))--(A + 10*dir(120-90)); path BB = (B + 10*dir(210+90))--(B + 10*dir(210-90)); path FY = (F + 10*(D - A))--(F - 10*(D - A)); path GX = (G + 10*(E - B))--(G - 10*(E - B)); pair Z = intersectionpoints(circumcircle(A,I,F), circumcircle(B,I,G))[1]; pair Y = intersectionpoints(AA,FY)[0], X = intersectionpoints(BB,GX)[0]; pair K = intersectionpoints(AA,BB)[0], P = intersectionpoints(FY,GX)[0]; draw(A--D, deepgreen); draw(B--E, deepgreen); draw(D--E); draw(circumcircle(A,I,F), gray+dashed); draw(circumcircle(B,I,G), gray+dashed); draw(circumcircle(A,B,C)); draw(B--C--A); draw(A--B, blue); draw(Y--K, magenta); draw(X--K, magenta); draw(P--X, deepgreen); draw(P--Y, deepgreen); draw(K--P, orange); draw(X--Y,blue); pair[] points = { A,B,C,D,E,I,F,G,X,Y,K,P }; int[] dirs = { 120,210,330,270,40,130,20,-40,-110,70,90,-30 }; string[] labels = { "$A$", "$B$", "$C$", "$D$", "$E$", "$I$", "$F$", "$G$", "$X$", "$Y$", "$K$", "$P$" }; pen[] pens = { defaultpen, defaultpen, defaultpen, deepgreen, deepgreen, defaultpen, blue, blue, defaultpen, defaultpen, deepmagenta, defaultpen }; for (int i = 0; i < points.length; ++i) { label(points[i], labels[i], dir(dirs[i]), pens[i]); } [/asy][/asy] Let $Y = PF \cap AA$ and $X = BB \cap PG$, where $AA$ and $BB$ are the tangent lines to the circumcircle from those respective points. Claim: $A$, $I$, $G$, $E$, and $Y$ are cyclic. Likewise $I$, $G$, $D$, $X$, and $B$ are cyclic. Proof: $\angle IEF = \angle BED = \angle DAC = \angle IAF$, so $AEIF$ is cyclic. Now, $\angle AYF = \angle KAI = \angle KAD = \angle ACD = AED = \angle AEF$, so $AYFE$ is also cyclic. This finishes the proof, and the same reasoning can be used on $IGDXB$. Claim: $YX \parallel AB$ Proof: $\angle KYI = \angle AYI = \angle AEI = \angle AEB = \angle KAB$. So, $YI \parallel AB$. Using the same reasoning on the other side, $XI \parallel AB$ as well. So, $X$, $I$, and $Y$ are collinear and $XY \parallel AB$. Claim: $K$ is on the radical axis of $(AYEFI)$ and $(BIGDX)$. Proof: Because of the previous claim, their powers of points are $KA \cdot KY$ and $KB \cdot KX$ respectively, where both are clearly equal. Now, since $AB \parallel XY$, $AI \parallel PY$, and $BI \parallel PX$, triangles $\triangle AIB$ and $\triangle APX$ are homothetic; therefore $K,I,P$ are collinear. Now, the problem dies by radax on those two circles and the circumcircle of $\triangle ABC$.
13.06.2024 22:01
Let $ABDE$ be the contact quadrilateral of $KNLM$ with $A$ on $MK$, $B$ on $KN$, $C$ on $NL$, and $D$ on $LM$. Let $AB$ intersect $DE$ at $H$ and $AE$ intersect $BD$ at $J$. Now, $\angle OL^*H=\angle OK^*H=90^\circ$ so $L$, $H^*$, $K$ are collinear. The polar of $H$ with respect to $\omega$ is $IJ$ so $H^*$ is on $IJ$. Note that Brianchon's on $LDNKAM$ and $NBKMEL$ gives $KL,MN,AD,BE$ concur at $I$. Therefore, $H^*$ is also on line $KIL$ which means that $K,I,L,J$ are all collinear. Let $LI$ intersect $DE$ at $Q$. Note that $DE$ is the perpendicular bisector of $CI$. Take the homothety centered at $Q$ that takes $I$ to $L$. Note that $\triangle IFG\sim \triangle LDE$ as well as $FG\parallel DE$ implies that this homothety takes $FG$ to $DE$. Now, it is clear that $\triangle IED\sim\triangle PGF$ so it takes $P$ to $I$ so $P$ is on the line $K,I,L,J$. We are done.
10.07.2024 14:14
Sorry I’m just nine and don’t know latex so i just wrote it out. Enjoy!
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26.07.2024 20:17
Let $T$ be the intersection of $AE$ and $BD$. By Pascal's on $AEBBDA$, we have that $T,I,K$ are collinear. Thus, it suffices to show that $T,I,P$ are collinear. Note that $EFIA$ and $DGIB$ are cyclic as $\angle GFI=\angle CFG=90-\angle C/2=\angle EAD$. The key observation here is that due to the parallel lines, $\triangle PFG$ and $\triangle IDE$ are negatively homothetic at $PI\cap DE=S$. This means that $SG\cdot SD=SF\cdot SE$. In particular, this means that $S$ lies on the radical axis of $(EFIA)$ and $(DGIB)$. Thus, line $SI$, which is also line $PI$, is the radical axis of $(EFIA)$ and $(DGIB)$. Since $T$ lies on $AE$ and $BD$ and $AEDB$ is cyclic, we have $TE\cdot TA=TD\cdot TB$ as well, so $T$ also lies on the radical axis, as desired. remark: the application of Pascal is used because of the fact that 2 out of the 3 points we wish to show collinear are intersections of chords. We then use this Pascal to eliminate the point $K$, as it seems unrelated to the rest of the diagram (some people call these "floating" points). Once this happens, finding a better description of the line $PI$ as a radical axis finishes.
13.11.2024 20:43
Note that $\odot(GDBI)$ because $\measuredangle GBI=\measuredangle CBE=\measuredangle ADE=\measuredangle IDG$, similarly $\odot(FAEI)$. Let $(GDBI)\cap (FAEI)=\{I,N\}$, by radax on $\odot(I), (GDBI), (FAEI)$, $IN, BD, AE$ concur at say $R$. Let $PF\cap AA=X$ an $PG\cap BB=Y$, so we have \begin{align*} &\measuredangle BYG=\measuredangle KBI=\measuredangle BCE=\measuredangle BDE=\measuredangle BDG\implies Y\in (GDBI)\\ &\measuredangle AXF=\measuredangle KAI=\measuredangle ACD=\measuredangle AED=\measuredangle AEF\implies X\in (FAEI) \end{align*}Clearly $K$ lies on radax of $(FAEI), (GDBI)$ so $\overline{R-K-I-N}$. Now we show $P\in \overline{R-K-I-N}$, for this we prove $\odot(FPGN)$ \begin{align*} &\measuredangle FPG= \measuredangle PFG + \measuredangle PGF=\measuredangle EAX+ \measuredangle DBY= \measuredangle EBA + \measuredangle BAD= \frac{1}{2}(\measuredangle B+\measuredangle A)\\ & \measuredangle FNG= \measuredangle FNI+ \measuredangle ING= \measuredangle FAI+ \measuredangle IBG= \frac{1}{2}(\measuredangle B+\measuredangle A) \end{align*}and finally $$\measuredangle FNP=\measuredangle FGP= \measuredangle DGY= \measuredangle DBY= \measuredangle BAI= \measuredangle FAI= \measuredangle INF \implies \overline{I-N-P}$$So $\overline{R-K-I-N-P}$ and $AE,BD$ and $KP$ are concurrent or parallel. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(12cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(9); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -14.632474284239294, xmax = 3.151279799891681, ymin = -10.01080318645678, ymax = 5.760087249878838; /* image dimensions */ draw((-5,1)--(-7,-4)--(2,-4)--cycle, linewidth(0.2)); /* draw figures */ draw(circle((-2.5,-2.9), 4.632493928760187), linewidth(0.6)); draw(circle((-4.108580229954061,-2.04241351035751), 1.9575864896424906), linewidth(0.6) + red); draw((-5,1)--(-7,-4), linewidth(0.6)); draw((-7,-4)--(2,-4), linewidth(0.6)); draw((2,-4)--(-5,1), linewidth(0.6)); draw((-2.5,-7.5324939287601875)--(0.1925824035672506,0.8696153649941529), linewidth(0.6)); draw((-2.5,-7.5324939287601875)--(-5,1), linewidth(0.6) + blue); draw((-7,-4)--(0.1925824035672506,0.8696153649941529), linewidth(0.6) + blue); draw((-0.36584198949262775,-3.321533654879412)--(-13.645997604571859,1.217099153045522), linewidth(0.6) + red); draw((-13.645997604571859,1.217099153045522)--(0.1925824035672506,0.8696153649941529), linewidth(0.6) + red); draw((-13.645997604571859,1.217099153045522)--(-2.5,-7.5324939287601875), linewidth(0.6) + red); draw(circle((-4.183979572227289,-5.045200594210912), 3.003733565376939), linewidth(0.6) + red); draw(circle((-2.4246006577267747,0.10278708385340259), 2.7272097075732264), linewidth(0.6) + red); draw((-7.785714285714283,-0.7857142857142865)--(-2.1645854554010784,2.8175734260249503), linewidth(0.6)); draw((-7.785714285714283,-0.7857142857142865)--(-6.200267692549023,-7.2716321668449115), linewidth(0.6)); draw((-6.200267692549023,-7.2716321668449115)--(-2.1645854554010784,2.8175734260249503), linewidth(0.6)); draw(circle((-1.367959144454577,-2.920686792894633), 1.079313207105367), linewidth(0.6)); draw((-6.200267692549023,-7.2716321668449115)--(-0.36584198949262775,-3.321533654879412), linewidth(0.6) + blue); draw((-2.1645854554010784,2.8175734260249503)--(-0.36584198949262775,-3.321533654879412), linewidth(0.6) + blue); /* dots and labels */ dot((-5,1),linewidth(3pt) + dotstyle); label("$A$", (-5.335949395452039,1.1533271487386971), NE * labelscalefactor); dot((-7,-4),linewidth(3pt) + dotstyle); label("$B$", (-7.553320073008038,-4.304230989098497), NE * labelscalefactor); dot((2,-4),linewidth(3pt) + dotstyle); label("$C$", (2.107477076687604,-4.262728645844982), NE * labelscalefactor); dot((-4.108580229954061,-2.042413510357509),linewidth(3pt) + dotstyle); label("$I$", (-4.315386127607148,-2.523386087331148), NE * labelscalefactor); dot((-2.5,-7.5324939287601875),linewidth(3pt) + dotstyle); label("$D$", (-2.4930388825862946,-8.060193053541585), NE * labelscalefactor); dot((0.1925824035672506,0.8696153649941529),linewidth(3pt) + dotstyle); label("$E$", (0.26686694377242137,0.987317775724638), NE * labelscalefactor); dot((-0.7406210854994871,-2.0424135103575094),linewidth(3pt) + dotstyle); label("$F$", (-0.4179217199105683,-2.1461091399157284), NE * labelscalefactor); dot((-1.367959144454575,-4),linewidth(3pt) + dotstyle); label("$G$", (-1.3517244431146451,-4.490991533739313), NE * labelscalefactor); dot((-0.36584198949262775,-3.321533654879412),linewidth(3pt) + dotstyle); label("$P$", (-0.2519123468965102,-3.619442325415503), NE * labelscalefactor); dot((-7.785714285714283,-0.7857142857142865),linewidth(3pt) + dotstyle); label("$K$", (-8.278859908317785,-1.229301730204616), NE * labelscalefactor); dot((-13.645997604571859,1.217099153045522),linewidth(3pt) + dotstyle); label("$R$", (-13.926934448929547,0.6815900366330283), NE * labelscalefactor); dot((-2.4057963782299048,-2.624357794614598),linewidth(3pt) + dotstyle); label("$N$", (-2.99043064569876535,-3.0591606914930534), NE * labelscalefactor); dot((-2.1645854554010784,2.8175734260249503),linewidth(3pt) + dotstyle); label("$X$", (-2.0780154500511494,2.937927908639833), NE * labelscalefactor); dot((-6.200267692549023,-7.2716321668449115),linewidth(3pt) + dotstyle); label("$Y$", (-6.539517349803961,-7.748925479140224), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]