Let ω be the circumcircle of triangle ABC. A point T on the line BC is such that AT touches ω. The bisector of angle BAC meets BC and ω at points L and A0 respectively. The line TA0 meets ω at point P. The point K lies on the segment BC in such a way that BL=CK. Prove that ∠BAP=∠CAK.
Problem
Source: Sharygin Correspondence Round 2024 P10
Tags: geometry
06.03.2024 18:33
Let the parallel line WRT BC through A intersect (ABC) at D. By symmetry, ADKL is cyclic. Applying a √bc inversion through A followed by a reflection through the internal angle bisector of ∠BAC gives us that P is the image of K. Hence, ∠BAP=∠CAK. ◼
06.03.2024 18:40
Note if M is midpoint of BC then K is reflection of L above M. Take point E on (ABC) such that AE∥BC. If O is circumcenter of (ABC) then A is reflection of E and L is reflection of K above OM. AL is reflection of KE and AE∥LK. Hence A,E,K,L cyclic. Take √bc inversions. Its well know T↔E (As T′ should lie on (ABC) and ∡TAI=∡IAT′=∡IAE) and A0↔L Let P′ be image of P then P′ should lie on BC and (AEL) hence P′=K As √bc inversions is reflection about angle bisector we get AP is reflections of AK hence ∡BAP=∡KAC. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(15.741534557788974cm); 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 = -12.081393817971838, xmax = 37.66014073981714, ymin = -18.28892983088801, ymax = 7.701611330231495; /* image dimensions */ pen yqqqyq = rgb(0.5019607843137255,0.,0.5019607843137255); /* draw figures */ draw(circle((11.095712613319792,-6.177893877780872), 10.192679672815137), linewidth(0.4)); draw((xmin, -0.006085192697768812*xmin-8.692576064908721)--(xmax, -0.006085192697768812*xmax-8.692576064908721), linewidth(0.4)); /* line */ draw((xmin, 0.7609797307563224*xmin-1.813196043720326)--(xmax, 0.7609797307563224*xmax-1.813196043720326), linewidth(0.4)); /* line */ draw((xmin, -2.9954813531338718*xmin + 16.680825838893796)--(xmax, -2.9954813531338718*xmax + 16.680825838893796), linewidth(0.4)); /* line */ draw((xmin, -0.38657790910946643*xmin-12.105004285069663)--(xmax, -0.38657790910946643*xmax-12.105004285069663), linewidth(0.4) + yqqqyq); /* line */ draw((4.285776955438827,-13.761790979412739)--(4.923256615628692,1.9333024500850746), linewidth(0.4)); draw((4.923256615628692,1.9333024500850746)--(13.672198269465575,-8.775774025980523), linewidth(0.4)); draw((xmin, -0.006085192697768574*xmin + 1.963261415291745)--(xmax, -0.006085192697768574*xmax + 1.963261415291745), linewidth(0.4)); /* line */ draw(circle((11.121554363944778,-1.9312328584077858), 7.304349945571356), linewidth(0.4) + yqqqyq); draw((17.36642423119154,1.857583377373747)--(11.033689341740999,-16.370384840563304), linewidth(0.4)); draw((4.923256615628692,1.9333024500850746)--(1.22,-8.7), linewidth(0.4)); draw((4.923256615628692,1.9333024500850746)--(20.94,-8.82), linewidth(0.4)); /* dots and labels */ dot((4.923256615628692,1.9333024500850746),dotstyle); label("A", (4.007988805578389,2.4563547240191688), NE * labelscalefactor); dot((1.22,-8.7),dotstyle); label("B", (1.326425034986684,-8.41723902706144), NE * labelscalefactor); dot((20.94,-8.82),dotstyle); label("C", (21.069806642639893,-8.535109962032504), NE * labelscalefactor); dot((-8.968445578518395,-8.638001345363984),linewidth(4.pt) + dotstyle); label("T", (-8.83994310626758,-8.387771293318673), NE * labelscalefactor); dot((8.487801730534423,-8.744225974019479),linewidth(4.pt) + dotstyle); label("L", (8.604955269449881,-8.505642228289737), NE * labelscalefactor); dot((11.033689341740999,-16.370384840563304),linewidth(4.pt) + dotstyle); label("A0", (11.581196377469245,-16.019914332695034), NE * labelscalefactor); dot((4.285776955438827,-13.761790979412739),linewidth(4.pt) + dotstyle); label("P", (4.391069344234347,-13.515156964559935), NE * labelscalefactor); dot((11.08,-8.76),linewidth(4.pt) + dotstyle); label("M", (11.198115838813289,-8.535109962032504), NE * labelscalefactor); dot((13.672198269465575,-8.775774025980523),linewidth(4.pt) + dotstyle); label("K", (13.791276408176696,-8.535109962032504), NE * labelscalefactor); dot((11.095712613319792,-6.177893877780875),linewidth(4.pt) + dotstyle); label("O", (10.196212891559245,-6.76704593746655), NE * labelscalefactor); dot((11.157735884898583,4.014597085001561),linewidth(4.pt) + dotstyle); label("N", (11.286519040041586,4.2538864823278875), NE * labelscalefactor); dot((7.089519596849112,-4.555697916144881),linewidth(4.pt) + dotstyle); label("I", (7.219971783539881,-4.321224036816982), NE * labelscalefactor); dot((17.36642423119154,1.857583377373747),linewidth(4.pt) + dotstyle); label("E", (17.474743126022442,2.1027419191059784), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]
06.03.2024 20:43
OK only inversion solutions here u can also do this easily by sine rule and length chasing
08.03.2024 14:25
This is also easy to do by barycentric coordinates, by setting ABC as the reference triangle and let P′ be the point where the isogonal of AK meets ω. Now one gets P′=(−a2bc:b3(b+c):c3(b+c)), A0=(−a2:b(b+c):c(b+c)) and T=(0:−b2:c2) Now its easy to check that P′, T and A0 and collinear and hence, P≡P′
08.03.2024 20:23
My solution: Claim: A,A0,K,T are concyclic. Proof: Let M be mid-point of BC. It is well known that the angle subtended by the orthocentre and incentre at A is |∠B−∠C|/2, and AH||A0M, so ∠LA0M=|B−C|/2, hence ∠AA0K=∠LA0K=|B−C|. It is easy to observe that ∠TAA0=∠B+∠A/2 (or C instead of B, depending upon T is farther from which) and ∠ALT=∠B+∠A/2 (or ∠C+∠A/2). Hence, ∠ATL=∠ALK=|∠B−∠C|. Thus, the claim follows. Reflect the line AK about AA0, and let it intersect ω at P′. We want to show P=P′. For this, let ∠BAP′=CAK=α and ∠P′AA0=∠KAA0=β. In the configuration where AC>AB, by simple angle chasing in ΔAP′A0, we have ∠AA0P′=∠C+α. But, due to the claim, we also have ∠AA0T=∠AKT=C+α, so T,P′,A0 are collinear. Similar equality holds when AC<AB. ◼
12.03.2024 12:30
Let A′ be the reflection of A across the perpendicular bisector of BC. It suffices to show that P,L,A′ are collinear. By Pascal's theorem on PA0A0AAA′, we have that AA0∩PA′ lies on BC, so we are done.
09.06.2024 23:38
Let the line passing through A and parallel to BC intersect w again at A′. Notice that under √bc-inversion, ϕ(A′)=T and ϕ(A0)=L. Since AKLA′ is concyclic and K∈BC, ϕ(K)=ϕ(AKLA′)∩ϕ(BC)=TA0∩w=P. Thus AP and AK are isogonal lines and the result follows.