We are given an acute triangle $ABC$. The angle bisector of $\angle BAC$ cuts $BC$ at $P$. Points $D$ and $E$ lie on segments $AB$ and $AC$, respectively, so that $BC \parallel DE$. Points $K$ and $L$ lie on segments $PD$ and $PE$, respectively, so that points $A$, $D$, $E$, $K$, $L$ are concyclic. Prove that points $B$, $C$, $K$, $L$ are also concyclic. Proposed by Patrik Bak, Slovakia
Problem
Source: 2023 IGO Adcanced P1
Tags: geometry, Concyclic
starchan
18.01.2024 17:17
Let line $AP$ meet $\odot(ADE)$ again at some point $M$. Note that $AKPC$ is cyclic, as \[\angle AKD = \angle AED = \angle ACB\]Similarily one obtains that $ALPB$ is cyclic. Now note that \[\angle AKM = \angle ADM + \angle MDE = B+A/2 = \angle APC = \angle AKC\]which implies that $C, M, K$ are collinear. Similarily one obtains that $B, M, L$ are collinear. As a result, lines $AP, BL, CK$ concur at $M$. The cyclicity of $BLCK$ now follows readily from radical axis theorem on $\odot(APB), \odot(APC)$ and the desired circle. $\square$
VicKmath7
18.01.2024 18:45
We have $\angle PCA=\angle AED=\angle AKD$, so $ACPK$ is cyclic and similarly $ABPL$ is cyclic. Now, we have $$\angle BLK=\angle BLA-\angle ALK=\angle APB-\angle BDP=\angle BAP+\gamma-(\angle BAP+\angle APD)=\angle AKD-\angle APD=\angle KAP=\angle KCB, $$so $BCLK$ is cyclic and we are done.
AlanLG
18.01.2024 21:51
[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -1.479460452652362, xmax = 14.617927525035164, ymin = -2.717936109338099, ymax = 5.4516898426101275; /* image dimensions */ draw((5.2,4.82)--(0,0)--(6.7,0)--cycle, linewidth(0.4) + red); /* draw figures */ draw((5.2,4.82)--(0,0), linewidth(0.4) + red); draw((0,0)--(6.7,0), linewidth(0.4) + red); draw((6.7,0)--(5.2,4.82), linewidth(0.4) + red); draw((5.2,4.82)--(3.9136437398312522,0), linewidth(0.4)); draw((2.46088333069495,2.2810495488364726)--(3.9136437398312522,0), linewidth(0.4)); draw((3.9136437398312522,0)--(5.99012980845338,2.281049548836472), linewidth(0.4)); draw(circle((4.225506569574165,3.1243136744226376), 1.9557580526980252), linewidth(0.4)); draw(circle((3.35,1.600871369294606), 3.7128545811851006), linewidth(0.4)); draw((5.2,4.82)--(13.587135135135135,0), linewidth(0.4)); draw((0,0)--(13.587135135135135,0), linewidth(0.4)); draw(circle((3.35,-1.9338582550615318), 3.868114237024242), linewidth(0.4) + linetype("4 4")); draw(circle((3.913643739831255,1.322550139944132), 1.3225501399441324), linewidth(0.4) + blue); draw((13.587135135135135,0)--(2.715155648035838,1.88180425614787), linewidth(0.4) + linetype("2 2")); draw((2.46088333069495,2.2810495488364726)--(5.99012980845338,2.281049548836472), linewidth(0.4)); /* dots and labels */ dot((5.2,4.82),linewidth(2pt) + dotstyle); label("$A$", (5.252418826831319,4.87390379666642), NE * labelscalefactor); dot((0,0),linewidth(2pt) + dotstyle); label("$B$", (0.05234441333793678,0.05006215727593835), NE * labelscalefactor); dot((6.7,0),linewidth(2pt) + dotstyle); label("$C$", (6.757349923242841,0.05006215727593835), NE * labelscalefactor); dot((3.9136437398312522,0),linewidth(2pt) + dotstyle); label("$P$", (3.9624778870500155,0.05006215727593835), NE * labelscalefactor); dot((2.46088333069495,2.2810495488364726),linewidth(2pt) + dotstyle); label("$D$", (2.511294329796048,2.3343325714719882), NE * labelscalefactor); dot((5.99012980845338,2.281049548836472),linewidth(2pt) + dotstyle); label("$E$", (6.045195029405246,2.3343325714719882), NE * labelscalefactor); dot((2.715155648035838,1.88180425614787),linewidth(2pt) + dotstyle); label("$K$", (2.7665951407944314,1.9312260277903326), NE * labelscalefactor); dot((5.230377397516128,1.4464506943660396),linewidth(2pt) + dotstyle); label("$L$", (5.279292596410096,1.5012457145298996), NE * labelscalefactor); dot((13.587135135135135,0),linewidth(2pt) + dotstyle); label("$F$", (13.637034935409797,0.05006215727593835), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Add $F$ as the intersection of $BC$ with the tangent to $(ABC)$ at $A$ and consider $(KLP)$ Claim.$(KLP)$ is tangent to $BC$ at $P$ Proof: $$\angle KLP=\angle KDE=\angle DPB$$Claim.$FA=FP$ Proof: $$\angle PAF=\angle PAC+\angle CAF=\angle BAP+\angle PBA=\angle FPA$$Claim. $K,L$ and $F$ are collinear Proof: $FA^2=FP^2$ therefore $F$ is on the radical axis of $(ADE)$ and $(KLP)$ which is $KL$ Now we have $BC, KL$ and the tangent to $(ABC)$ at $A$ concur at $F$ so by radical axis on $(ABC),(ADE)$ and $(BKLC)$ we get the latter should be cyclic
IAmTheHazard
18.01.2024 22:31
Invert about $A$ with radius $\sqrt{AD\cdot AC}=\sqrt{AE\cdot AB}$ and reflect about the angle bisector, denoting images with $\bullet'$. This swaps pairs $(B,E),(C,D)$ and swaps $\overline{BC}$ with $(ADE)$. Thus $K'$ and $L'$ are the second intersections of $(AP'C)$ and $(AP'B)$ with $\overline{BC}$. By Reim's, $K',P',D$ and $L',P',E$ are collinear, so $\measuredangle K'L'E=\measuredangle BAP'=\measuredangle P'AC=\measuredangle DK'L'$. Since $\overline{K'L'} \parallel \overline{DE}$ this implies $DEK'L'$ is an isosceles trapezoid, hence cyclic, so $BCKL$ is cyclic as well. $\blacksquare$
lelouchvigeo
19.01.2024 12:10
Spent 1 hr on it Let M be midpoint of arc DE Spiral similarity sends $AME$ to$ ABP$ and $ADM$ to $APC$. After this just angle chase. We are done
Joider
27.01.2024 21:43
When will the results have uploaded in the site??
GrantStar
27.01.2024 23:51
sus Let $T=AA\cap BC$. Note that angle chasing gives $TP^2=TA^2=TB\cdot TC$ by $TAP$ isosceles and PoP. Claim: $(PKL)$ is tangent to $BC$. Proof. Angle chase \[\measuredangle CPL=\measuredangle CPE=\measuredangle DEP=\measuredangle DEL=\measuredangle PKL\]as desired. $\square$ Then, the radical axis of $(PKL)$ and $(AEDKL)$ is clearly line $KL$, but $TP^2=TA^2$ so $T$ lies on this radical axis. Thus $TK\cdot TL=TP^2=TA^2=TB\cdot TC$ which finishes.
sami1618
16.02.2024 21:24
Let $AP$ intersect $(ADE)$ at $Q$. Through simple angle chasing we get that $ACPK$ and $ABPL$ are both concyclic. After that it is not hard to show that $KQC$ and $LQB$ are collinear and you can just finish with radical axes.
Attachments:
Littlelame
10.03.2024 02:27
My solution (because why not)
We first notice that $(KLP)$ is tangent to $BC$, as $\angle KLP=\angle KPB$ by parallelism
Inverting in $A$, we have the following diagram (se attachments):
We get that $B^*P^*=C^*P^*$, and that the thesis is true iff $B^*C^*K^*L^*$ is an isosceles trapezoid ($BC\parallel DE$).
Then, as $(KLP)^*$ is now tangent to $(AB^*C^*)\equiv [BC]^*$, we get that if $K^*P^*=P^*L^*$ we have the thesis; in fact, called $l$ the common tangent at $P$, we have $\angle (L^*P^*,l)=\angle (K^*P^*,l)$ and $\angle (C^*P^*,l)=\angle (B^*P^*,l)$ then we get $\angle L^*P^*C^*=\angle K^*P^*B^*$ that gives $\triangle K^*P^*B^*\cong\triangle L^*P^*C^*$ that implies above.
This, however, is true by the inversive distance formula: in fact \[\frac{K^*P^*}{L^*P^*}=\frac{PK}{PL}\cdot\frac{AL}{AK}\]That is true by similarity of triangles: in fact we have (as $\angle AKE=\angle ABC$ and $\angle KAE=\angle BPD$) $\triangle KEA\sim\triangle PEC$ and similarly $\triangle LDA\sim \triangle PDA$ and in particular \[\frac{LD}{LA}=\frac{CE}{CP}\]\[\frac{KE}{KA}=\frac{BD}{BP}\]furthermore, we get by internal bisector theorem and thales' theorem \[\frac{BP}{PC}=\frac{AB}{AC}=\frac{BD}{CE}\]that with above implies $\frac{LD}{LA}=\frac{CE}{CP}=\frac{BD}{BP}=\frac{KE}{KA}$.
We have then that $\angle PDL=\angle PEK$ that implies $\triangle KEP\sim \triangle DLP$ and in particular \[\frac{KP}{PL}=\frac{KE}{LD}=\frac{KA}{LA}\]Then we finally have $\frac{K^*P^*}{L^*P^*}=\frac{PK}{PL}\cdot\frac{AL}{AK}=1$, and from this follows the thesis.
Attachments:
Amirmahdi2023
18.03.2024 21:40
lelouchvigeo wrote: Spent 1 hr on it Let M be midpoint of arc DE Spiral similarity sends $AME$ to$ ABP$ and $ADM$ to $APC$. After this just angle chase. We are done Thanks a lot, I was looking for a solution using M and yours was great, but I think it was better to mention after wr get ADM~APC,we get ADP~AMC and then angle chase
G.G.T
02.05.2024 13:09
KAE=KDE(ADKLE cyclic) EDP=DPB(DE//BC)(well known) DPC=a => DPB=180⁰-a=PDE=KAE Hence, AKPC is cyclic. LAD=LED(ADKLE cyclic) DEP=EPC(DE//BC) EPB=b => EPC=180⁰-b=PED=LAD Hence, ALPB is cyclic. DAK=x; DAP=y KAP=DAP-DAK=y-x KCP=y-x(AKPC cyclic) BLP=BAP=y(ALPB cyclic) 180⁰-KLE=KLP=KDE=KAE=DAE(2y(bisector))-DAK= 2y-x(ADKLE cyclic) KCB=KLB <- Enough to prove KCB=y-x KLB=KLP-BLP=2y-x-y=y-x So, KCB=KLB=y-x Hence BCKL is cyclic
One_more_fractal
14.06.2024 07:15
AKPC & BPLA are cyclic. Let AP & DE intersect at G and AP and KC meet at H and AP meets circle ADE at I. Since AKPC is concyclic, ∠AKC= ∠APC= ∠AGE. ∠AGE= ∠ADE+1/2∠A= ∠ADI= ∠AKH Which means H and I are the same point. Similarly we can prove this for BL. ∠KAH=∠KLH ∠KAP=∠KLB ∠KCB=∠KLB.
SomeonesPenguin
10.08.2024 00:05
Just invert. Let $F$ be the midpoint of the arc $DE$ in $(ADE)$, $\{M\} = BC \cap (AFB)$, $\{N\} = BC \cap (AFC)$. Consider the inversion with center $A$ and radius $\sqrt{AD\cdot AC}$ composed with the reflection over $AP$. Now we need to prove that $DENM$ is cyclic. Claim: $D$, $F$ and $N$ are collinear. (this is just Reim but here is the angle chase) $$\measuredangle EFD=\measuredangle EDA + \measuredangle AED = \measuredangle EFA + \measuredangle ACB = \measuredangle EFA +\measuredangle AFN = \measuredangle EFN. \square$$ So $F$ is the intersection point of the diagonals of the trapezoid $DENM$ but we know that $FD=FE$ so $DENM$ is indeed a isosceles trapezoid, hence cyclic. Done!
fasttrust_12-mn
29.08.2024 08:33
let $\Omega$ be the circumcircle of $\triangle ADE$, let $F=\Omega\cap AP$, we have that $F$ is the mid of $\overarc{DE}$ we have $\angle ADE=\angle ABC=\angle AFE$ we also have $\angle BAP=\angle PAC$ thus $\triangle FAE \sim \triangle BAP$ by the same way we have that $\triangle APC \sim \triangle ADF$, we know that $\angle FAE=\angle FLP$ , so $$\angle ABP+\angle FLP+\angle ALF=\angle FAE+\angle EFA+\angle FEA=180^{\circ}$$thus $(ALPB)$ is a cyclic, by the same way we have $(AKPC)$ is a cyclic, since $(AKPC),(ALPB)$ are cyclic we have $B-F-L$ and $C-F-K$ since $BL\cap CK \cap AP=F$ and $(AKPC),(ALPB)$ are cyclic by radical axis we have $(BKLC)$ is cyclic $\square$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(13cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -25.00590274104705, xmax = 35.14367552830164, ymin = -13.551433924933864, ymax = 15.158320856629341; /* image dimensions */ /* draw figures */ draw((-2.577204079854346,9.542443375108213)--(18.890064796518487,-9.67182500532846), linewidth(0.4)); draw((18.890064796518487,-9.67182500532846)--(-7.423280618283065,-9.67182500532846), linewidth(0.4)); draw((-7.423280618283065,-9.67182500532846)--(-2.577204079854346,9.542443375108213), linewidth(0.4)); draw((9.849190330090408,-1.5797947898324374)--(3.299827075486145,-9.67182500532846), linewidth(0.4)); draw((3.299827075486145,-9.67182500532846)--(-5.38237034269336,-1.5797947898324374), linewidth(0.4)); draw(circle((2.2334099936985248,2.4142789798793682), 8.599577629748635), linewidth(0.4)); draw((3.299827075486145,-9.67182500532846)--(-2.577204079854346,9.542443375108213), linewidth(0.4)); draw((-5.38237034269336,-1.5797947898324374)--(9.849190330090408,-1.5797947898324374), linewidth(0.4)); draw((2.2334099936985243,-6.185298649869268)--(-5.38237034269336,-1.5797947898324374), linewidth(0.4)); draw((2.2334099936985243,-6.185298649869268)--(9.849190330090408,-1.5797947898324374), linewidth(0.4)); draw(circle((5.733392089117715,-22.39609249742666), 18.30314235152345), linewidth(0.4)); draw((18.890064796518487,-9.67182500532846)--(7.727005897682741,-4.201848244849687), linewidth(0.4)); draw((-1.215558875922951,-5.4633697268385895)--(-7.423280618283065,-9.67182500532846), linewidth(0.4)); draw((-7.423280618283065,-9.67182500532846)--(7.727005897682741,-4.201848244849687), linewidth(0.4)); draw((-1.215558875922951,-5.4633697268385895)--(18.890064796518487,-9.67182500532846), linewidth(0.4)); draw((-1.215558875922951,-5.4633697268385895)--(-2.577204079854346,9.542443375108213), linewidth(0.4)); draw((7.727005897682741,-4.201848244849687)--(-2.577204079854346,9.542443375108213), linewidth(0.4)); draw(circle((11.094945936002317,3.218385217463934), 15.06391043644898), linewidth(0.4)); draw(circle((-2.061726771398459,-0.8058208501862476), 10.361094987116086), linewidth(0.4)); /* dots and labels */ dot((-2.577204079854346,9.542443375108213),dotstyle); label("$A$", (-3.3855199224158583,10.138517106478721), NE * labelscalefactor); dot((18.890064796518487,-9.67182500532846),dotstyle); label("$B$", (19.07149685457377,-9.236164034453504), NE * labelscalefactor); dot((-7.423280618283065,-9.67182500532846),dotstyle); label("$C$", (-7.260456150602304,-9.236164034453504), NE * labelscalefactor); dot((3.299827075486145,-9.67182500532846),linewidth(4pt) + dotstyle); label("$P$", (3.483685209369204,-9.324230766912287), NE * labelscalefactor); dot((-5.38237034269336,-1.5797947898324374),linewidth(4pt) + dotstyle); label("$E$", (-6.20365536109691,-1.8825918741451357), NE * labelscalefactor); dot((9.849190330090408,-1.5797947898324374),linewidth(4pt) + dotstyle); label("$D$", (10.308856974924876,-1.6624250429981786), NE * labelscalefactor); dot((7.727005897682741,-4.201848244849687),linewidth(4pt) + dotstyle); label("$K$", (7.8870218323083465,-3.86409335446775), NE * labelscalefactor); dot((-1.215558875922951,-5.4633697268385895),linewidth(4pt) + dotstyle); label("$L$", (-1.0517515122581127,-5.097027608890709), NE * labelscalefactor); dot((2.2334099936985243,-6.185298649869268),linewidth(4pt) + dotstyle); label("$F$", (2.4268844198638098,-5.845594834790363), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]
Z4ADies
02.10.2024 21:06
Firstly, assume that $KL$ is parallel to $DE$ then, by angle chase $\implies$ $BCLK$ cyclic. Next, assume that $KL$ is not parallel to $DE$. Thus, $KL \cap BC$ at $X$. By angle chase $XP$ tangents to $(KLP)$. So, by PoP $XP^2=XK \cdot XL$ if we prove $XP^2=XB \cdot XC$ we are done. From this motivation let's prove $XA$ tangents do $(ADE)$. Let $XA \cap (ABC)$ another point $R$ and $XA \cap (ADE)$ another point $R'$. From angle chasing we deduce $BR$ is parallel to $DR'$ and $CR$ is parallel to $ER'$. So, $\triangle BRC$ and $\triangle DR'E$ are oriented and similar. Thus, $RR', BC,$ and $DE$ intersect at one point but, it is impossible ,since, $DE$ is parallel to $BC$ and they intersect on $P_{\infty}$. Thus, $R$ and $R'$ have to be same point and it implies that there exists unique homothety that takes $DE$ to $BC$ which is point $A$. So, $AX$ tangents to $(ADE)$ and $(ABC)$.
cursed_tangent1434
08.10.2024 06:12
Cool opening problem! Similar to some other solutions above but posting for storage purposes. We start off by proving the following key claim. Claim : Quadrilaterals $AKPC$ and $ALPB$ are cyclic. Proof : Simply note that, \[\measuredangle BPD = \measuredangle EDP = \measuredangle EDK = \measuredangle EAK\]which implies that $AKPC$ is cyclic. Similarly one can show $ALPB$ cyclic finishing the proof of the claim. Now, we can show the following result. Claim :Lines $\overline{BL}$ and $\overline{CK}$ intersect on the internal $\angle A-$bisector. Proof : We first show that $N = \overline{BL} \cap \overline{CK}$ lies on $(ADE)$. To see why, \[\measuredangle KNL = \measuredangle CNB = \measuredangle NCB + \measuredangle CBN = \measuredangle KCP + \measuredangle PBL = \measuredangle KAP + \measuredangle PAL = \measuredangle KAL\]Now, note that \[\measuredangle NAB = \measuredangle ANL + \measuredangle LBA = \measuredangle AEL + \measuredangle LPA = \measuredangle EAP = \measuredangle CAP = \measuredangle PAB\]which implies points $A$ , $N$ and $P$ are collinear, which proves the claim. Now, simply note that \[BN \cdot NL = PN \cdot NA = CN \cdot NK\]from which we conclude that $BKLC$ is cyclic, as desired.
cosdealfa
13.10.2024 10:36
Nice problem! Here is my solution(hopefully I didn’t make any mistakes): Let $T$ be the point where the common tangent of circles $(ADE)$ and $(ABC)$ meets $BC$, $I$=$PE\cap AB$, $J$=$PD\cap AC$. Claim: $I$-$J$-$T$ are collinear. Proof: Fix $\triangle ABC$, then $P$ and $T$ are also fixed. It’s easy to get that: $deg(D)=deg(E)=deg(I)=deg(J)=1$. So in order for the claim to be true we must check $3$ cases. Cases $D \equiv A$ and $D \equiv B$ are trivial. We will now treat the case in which $D$ is such that $PE || AB$. It suffices to prove $JT || AB$. By Thales and Bisector theorem we get $$\frac{BP}{PC}=\frac{AE}{EC}=\frac{AD}{DB}=\frac{AB}{AC}$$. Applying Menelaus in $\triangle ABC$ with transversal $D-P-J$ we get $\frac{CJ}{AJ}=\frac{AB^2}{AC^2}$. We are left to prove $\frac{CT}{BT}= \frac{AB^2}{AC^2}$, and this is true by LOS. $\square$ Now, apply Pascal in $KDAAEL$ to get that: $KD\cap AE$, $DA\cap EL$, $AA\cap KL$ are collinear. Using the claim above, we get that $K$, $L$, $T$ must be collinear. Writing the power of $T$ wrt $(ABC)$ and $(ADE)$ we get $AT^2=BT \cdot CT = KT \cdot LT$ $\blacksquare$