Let ABC be a triangle with symmedian point K. Select a point A1 on line BC such that the lines AB, AC, A1K and BC are the sides of a cyclic quadrilateral. Define B1 and C1 similarly. Prove that A1, B1, and C1 are collinear. Proposed by Sammy Luo
Problem
Source: ELMO 2014 Shortlist G1, by Sammy Luo
Tags: geometry, circumcircle, Asymptote, geometric transformation, homothety, blogs, conics
24.07.2014 18:21
24.07.2014 18:45
thkim1011 wrote: It's pretty easy to see that B1C1BC is cyclic. This seems false? In particular it's not possible for A1, B1, C1 to lie on a line parallel to the tangent at A, because then the same would be true of the tangents at B and C.
24.07.2014 19:06
[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(15cm); 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 = -8.491418640000010, xmax = 32.25448436000006, ymin = -9.207733820000017, ymax = 10.98806158000001; /* image dimensions */ pen zzttqq = rgb(0.6000000000000006,0.2000000000000002,0.000000000000000); draw((5.520000000000006,4.380000000000004)--(-0.1000000000000001,-0.5000000000000006)--(7.680000000000009,-1.140000000000001)--cycle, zzttqq); /* draw figures */ draw((5.520000000000006,4.380000000000004)--(-0.1000000000000001,-0.5000000000000006), zzttqq); draw((-0.1000000000000001,-0.5000000000000006)--(7.680000000000009,-1.140000000000001), zzttqq); draw((7.680000000000009,-1.140000000000001)--(5.520000000000006,4.380000000000004), zzttqq); draw(circle((3.903934634484357,0.5650179004503991), 4.143157694983760)); draw((xmin, 2.214677842689021*xmin-18.14872583185168)--(xmax, 2.214677842689021*xmax-18.14872583185168)); /* line */ draw((xmin, 2.214677842689021*xmin-9.858298469367947)--(xmax, 2.214677842689021*xmax-9.858298469367947)); /* line */ draw(circle((2.244283885012643,2.476337001276425), 3.788700156874010)); draw((xmin, 0.8683274021352324*xmin-0.4131672597864773)--(xmax, 0.8683274021352324*xmax-0.4131672597864773)); /* line */ /* dots and labels */ dot((5.520000000000006,4.380000000000004),dotstyle); label("C", (5.645638140000016,4.575010760000004), NE * labelscalefactor); dot((-0.1000000000000001,-0.5000000000000006),dotstyle); label("B", (0.04750538000000536,-0.2790663800000031), NE * labelscalefactor); dot((7.680000000000009,-1.140000000000001),dotstyle); label("A", (7.984098660000020,-1.129415660000004), NE * labelscalefactor); dot((5.006357207366608,1.229169910373350),dotstyle); label("K", (5.149601060000014,1.457063400000000), NE * labelscalefactor); dot((5.942049952202645,3.301427899926592),dotstyle); label("B1", (6.070812780000017,3.512074160000003), NE * labelscalefactor); dot((4.070664462530569,-0.8430880791798930),dotstyle); label("C1", (4.228389340000013,-0.6333785800000035), NE * labelscalefactor); dot((7.015358650379657,5.678460892144614),dotstyle); label("A1", (6.602281080000018,5.956828340000006), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] It is true for B and C too (doing the same thing with B and C gives you 6 more points). It has to be cyclic from the alternate segment theorem.
24.07.2014 22:14
I think we're not doing the same problem. B1 is defined as the point on line CA such that BA, BC, B1K, CA are the sides of a cyclic quadrilateral.
25.07.2014 00:07
I'm still not understanding. I'm pretty sure the sides mentioned form a cyclic quadrilateral. Can you post a diagram?
25.07.2014 06:02
The construction of point X1 is the intersection of its opposite side with the line through K perpendicular to XO where X=A,B,C and O is the circumcenter of ABC. Let A1K∩AB,AC=AC,AB, similarly define BA,BC,CA,CB. It's well known that these points lie on the Lemoine circle with center K. Now A1,B1,C1 are collinear ⟺ their polars wrt (K) concur. Let ABBA∩ACCA=A2, similarly define B2,C2, where by the obvious parallels A2B2C2 is homothetic(congruent as well) with ABC. Moreover, the construction of the polar of X wrt (K) is the line through X2 perpendicular to X1K or parallel to XO. By the correspondence of the homothety the polars concur at the circumcenter of A2B2C2 and we done.
26.07.2014 19:12
thkim1011 wrote: I'm still not understanding. I'm pretty sure the sides mentioned form a cyclic quadrilateral. Can you post a diagram? Oof yeah, I managed to break the problem while editing it. The original wording was something like Quote: The anti-parallel of BC through K wrt AB and AC meets BC again at A1. Then define B1 and C1 similarly. Sorry about that.
26.07.2014 22:00
Construct the tangents to the circumcircle of ABC at A, B, C and let it intersect at the three points A3, B3, C3 where each point is opposite of their corresponding point. Apply pascals on degenerate hexagon AABBCC. This creates a new set of 3 collinear points A2, B2, C2. We construct the antiparallels by constructing the parallels to the three tangent. We note that there must be a homothety from B mapping two antiparallels to two tangents (this is true because K lies on the line connecting B and the intersection of the tangents at A and C. Thus A1C1 is parallel to A2C2. Similarly, we can find that A1B1 is parallel to A2B2 and B1C1 is parallel to B2C2. Thus A1, B1, and C1 are collinear.
26.07.2014 23:17
Another finish from xml's lemma that 6 points lying on the lemoine circle centered at K note that A1,B1,C1 have same power to the lemoine circle and circumcircle of ABC, so they must be on the radical axis of the two circles, thus collinear
23.10.2014 08:50
01.01.2015 13:17
Let AO meet the line through K at A2. Easy to see that under inversion about the second lemoine circle and a flip accros the lemoine point that A1↦A2. Hence, the circles with diametres AA1 have common radical axis HK. So, A1B1C1 are collinear. But, furthermore, we have HK is perpendicular to the isotomic line of A1B1C1, and even more A1B1C1 is parallel to the polar line of O wrt K. With all this, we appear to have some sort of mapping between isotomic lines; interesting
01.01.2015 18:48
Here is a generalisation found with Telv
01.01.2015 18:50
The requested blog entry does not exist
02.01.2015 16:52
What points other than K have the given property? Experimentation suggests that the locus is the Jerabek hyperbola.
02.01.2015 20:01
Oops! Yes it is. But more generally, here is the result found by Telv and me; Let ABC be a triangle. Let P,Q be two points on a rectangular circum-hyperbola H. Let the perpendicular from P to AQ meet BC at A1 and define B1,C1 similarly. Then A1B1C1 is a line. Found with Telv Cohl Proof: Suppose BQ meets the perpendicular to CQ from P at Y and define X similarly for point on CQ. Of course, XY⊥PQ. Now, let the orthocentre of BQC be H∗, and note H∗∈H. We have (∞,P;X,B1)=(H∗,P;Q,A)=(∞,P;Y,C1)⟹XY∥B1C1 ⟹B1C1⊥PQ. So, A1B1C1⊥PQ.
03.01.2015 07:39
Another proof to the previous generalization: As H is rectangular, the orthocenters HB and HC of △BPQ and △CPQ are the second intersections of H with PB1 and PC1, respectively. By Pascal theorem for PHCCABHB, the intersections C1≡PHC∩AB, B1≡PHB∩AC and CHC∩BHB (point at infinity of ⊥PQ) are collinear ⟹ (B1C1∥BHB∥CHC)⊥PQ. Hence, we conclude that A1,B1,C1 lie on a perpendicular to PQ.
03.01.2015 20:40
Variation of the generalization: Let ABC be a triangle and P,P* two isogonal conjugate points. A1 = (Perpendicular to AP* through P) /\ BC Similarly B1, C1. Which is the locus of P such that A1,B1,C1 are collinear? The locus is the McCay cubic + Circumcircle + line at infinity How about a synthetic proof? ie Let P,P* be two isogonal conjugate points collinear with O. Let A1 = (Perpendicular to AP* through P) /\ BC Similarly B1, C1. To prove: A1,B1,C1 are collinear. See also another locus related to circunormal triangle by Cesar Lozada: https://groups.yahoo.com/neo/groups/Hyacinthos/conversations/messages/22977
03.01.2015 20:57
rodinos wrote: Let P,P* be two isogonal conjugate points collinear with O. Let A1 = (Perpendicular to AP* through P) /\ BC Similarly B1, C1. To prove: A1,B1,C1 are collinear. Since P,P∗,O are collinear , so the rectangular circum-hyperbola H passing through P also contain P∗ , hence from the generalization above we get A1,B1,C1 are collinear .
03.01.2015 22:40
Well... that's the problem: if P,P*,O are collinear then P* lies on the r. c/hyperbola through P The isogonal conjugate of the line PP*O contains the points P*,(P*)*, O* = P*, P, H : on the r. c/hyperbola through P
12.05.2015 10:18
Please see: https://groups.yahoo.com/neo/groups/AdvancedPlaneGeometry/conversations/messages/1327
21.07.2019 12:01
[asy][asy] unitsize(1.5inches); pair X=dir(120); pair Y=dir(210); pair Z=dir(-30); pair I=incenter(X,Y,Z); pair A=foot(I,Y,Z); pair B=foot(I,X,Z); pair C=foot(I,X,Y); pair K=extension(A,X,B,Y); pair A1=extension(B,C,K,K+Y-A); pair B1=extension(A,C,K,K+X-B); pair C1=extension(A,B,K,K+X-C); draw(X--Y--Z--cycle); draw(incircle(X,Y,Z)); draw(A--X); draw(B--Y); draw(C--Z); draw(B--A1); draw(B--C1); draw(A--B1); draw(K--A1); draw(K--B1); draw(K--C1); draw(A1--B1--C1,dotted); dot("A",A,dir(-90)); dot("B",B,dir(B)); dot("C",C,1.5*dir(160)); dot("X",X,dir(X)); dot("Y",Y,dir(Y)); dot("Z",Z,dir(Z)); dot("K",K,2*dir(10)); dot("A1",A1,dir(A1)); dot("B1",B1,dir(B1)); dot("C1",C1,dir(C1)); [/asy][/asy] Let XYZ be the triangle such that ABC is its contact triangle. We see that KA1∥YZ and the cyclic variations thereof. We now use barycentric coordinates with respect to △XYZ. We see that K=(1s−x:1s−y:1s−z)and ∞YZ=(0:1:−1). Thus, K∞YZ=K×∞YZ=|⋆⋆⋆01−11s−x1s−y1s−z|=(1s−z+1s−y:−1s−x:1s−x)and BC=B×C=|⋆⋆⋆s−ys−x0s−z0s−x|=(s−x:y−s:z−s).Thus, A1=K∞YZ×BC=|⋆⋆⋆s−xy−sz−s1s−z+1s−y−1s−x−1s−x|=(z−ys−x:s−zs−y:−s−ys−z).Thus, the problem is equivalent to verifying |z−ys−xs−zs−y−s−ys−z−s−zs−xx−zs−ys−xs−zs−ys−x−s−xs−yy−xs−z|=0.But (s−x)row1+(s−y)row2+(s−z)row3=0, so this is true, as desired.
11.09.2019 18:52
If we bary with ABC as reference triangle we get A1=(0,b2(b2+a2−c2),c2(c2+a2−b2))
06.02.2022 23:27
Let △XYZ be the triangle with incenter (ABC) (X opposite A, etc.) Also let Ta be the intersection of the tangent to (ABC) at A with BC, and define Tb and Tc similarly. Claim: Ta−Tb−Tc are colinear Proof: Pascal on AABBCC ◼ Call this line r. Claim A1B1∥tor and so on. Proof: Notice that KA1∥YZ and KB1∥XZ. This means △TaTbZ and △A1B1K are homothetic from C, and thus r=TaTb∥A1B1. ◼ This solves the problem.