thkim1011 wrote: It's pretty easy to see that $B_1C_1BC$ is cyclic. This seems false? In particular it's not possible for $A_1$, $B_1$, $C_1$ to lie on a line parallel to the tangent at $A$, because then the same would be true of the tangents at $B$ and $C$.

[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("$B_1$", (6.070812780000017,3.512074160000003), NE * labelscalefactor); dot((4.070664462530569,-0.8430880791798930),dotstyle); label("$C_1$", (4.228389340000013,-0.6333785800000035), NE * labelscalefactor); dot((7.015358650379657,5.678460892144614),dotstyle); label("$A_1$", (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.

I think we're not doing the same problem. $B_1$ is defined as the point on line $CA$ such that $BA$, $BC$, $B_1K$, $CA$ are the sides of a cyclic quadrilateral.

I'm still not understanding. I'm pretty sure the sides mentioned form a cyclic quadrilateral. Can you post a diagram?

The construction of point $X_1$ 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 $A_1K\cap AB,AC=A_C,A_B$, similarly define $B_A,B_C,C_A,C_B$. It's well known that these points lie on the Lemoine circle with center $K$. Now $A_1,B_1,C_1$ are collinear $\iff$ their polars wrt $(K)$ concur. Let $A_BB_A\cap A_CC_A=A_2$, similarly define $B_2,C_2$, where by the obvious parallels $A_2B_2C_2$ is homothetic(congruent as well) with $ABC$. Moreover, the construction of the polar of $X$ wrt $(K)$ is the line through $X_2$ perpendicular to $X_1K$ or parallel to $XO$. By the correspondence of the homothety the polars concur at the circumcenter of $A_2B_2C_2$ and we done.

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 $A_1$. Then define $B_1$ and $C_1$ similarly. Sorry about that.

Construct the tangents to the circumcircle of $ABC$ at $A$, $B$, $C$ and let it intersect at the three points $A_3$, $B_3$, $C_3$ where each point is opposite of their corresponding point. Apply pascals on degenerate hexagon $AABBCC$. This creates a new set of 3 collinear points $A_2$, $B_2$, $C_2$. 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 $A_1C_1$ is parallel to $A_2C_2$. Similarly, we can find that $A_1B_1$ is parallel to $A_2B_2$ and $B_1C_1$ is parallel to $B_2C_2$. Thus $A_1$, $B_1$, and $C_1$ are collinear.

Another finish from xml's lemma that $6$ points lying on the lemoine circle centered at $K$ note that $A_1,B_1,C_1$ 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

Let $AO$ meet the line through $K$ at $A_2$. Easy to see that under inversion about the second lemoine circle and a flip accros the lemoine point that $A_1 \mapsto A_2$. Hence, the circles with diametres $AA_1$ have common radical axis $HK$. So, $A_1B_1C_1$ are collinear. But, furthermore, we have $HK$ is perpendicular to the isotomic line of $A_1B_1C_1$, and even more $A_1B_1C_1$ 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

Here is a generalisation found with Telv

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What points other than $K$ have the given property? Experimentation suggests that the locus is the Jerabek hyperbola.

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 $\mathcal{H}$. Let the perpendicular from $P$ to $AQ$ meet $BC$ at $A_1$ and define $B_1, C_1$ similarly. Then $A_1B_1C_1$ 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, $XY \perp PQ$. Now, let the orthocentre of $BQC$ be $H^*$, and note $H^* \in \mathcal{H}$. We have $$(\infty, P; X, B_1) = (H^*, P; Q, A) = (\infty, P; Y, C_1) \implies XY \parallel B_1C_1$$ $$\implies B_1C_1 \perp PQ.$$ So, $A_1B_1C_1 \perp PQ$.

Another proof to the previous generalization: As $\mathcal{H}$ is rectangular, the orthocenters $H_B$ and $H_C$ of $\triangle BPQ$ and $\triangle CPQ$ are the second intersections of $\mathcal{H}$ with $PB_1$ and $PC_1,$ respectively. By Pascal theorem for $PH_CCABH_B,$ the intersections $C_1 \equiv PH_C \cap AB,$ $B_1 \equiv PH_B \cap AC$ and $CH_C \cap BH_B$ (point at infinity of $\perp PQ$) are collinear $\Longrightarrow$ $(B_1C_1 \parallel BH_B \parallel CH_C) \perp PQ.$ Hence, we conclude that $A_1,B_1,C_1$ lie on a perpendicular to $PQ.$

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

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 $\mathcal{H}$ passing through $P$ also contain $P^*$ , hence from the generalization above we get $A_1, B_1, C_1$ are collinear .

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

Please see: https://groups.yahoo.com/neo/groups/AdvancedPlaneGeometry/conversations/messages/1327

[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("$A_1$",A1,dir(A1)); dot("$B_1$",B1,dir(B1)); dot("$C_1$",C1,dir(C1)); [/asy][/asy] Let $XYZ$ be the triangle such that $ABC$ is its contact triangle. We see that $KA_1\parallel YZ$ and the cyclic variations thereof. We now use barycentric coordinates with respect to $\triangle XYZ$. We see that $$K=\left(\frac{1}{s-x}:\frac{1}{s-y}:\frac{1}{s-z}\right)$$ and $\infty_{YZ}=(0:1:-1)$. Thus, $$K\infty_{YZ}=K\times\infty_{YZ}=\begin{vmatrix} \star & \star & \star \\ 0 & 1 & -1 \\ \frac{1}{s-x} & \frac{1}{s-y} & \frac{1}{s-z}\end{vmatrix}=\left(\frac{1}{s-z}+\frac{1}{s-y}:-\frac{1}{s-x}:\frac{1}{s-x}\right)$$ and $$BC=B\times C=\begin{vmatrix}\star & \star & \star \\ s-y & s-x & 0 \\ s-z & 0 & s-x\end{vmatrix}=(s-x:y-s:z-s).$$ Thus, $$A_1=K\infty_{YZ}\times BC=\begin{vmatrix}\star & \star & \star \\ s-x & y-s & z-s \\ \frac{1}{s-z}+\frac{1}{s-y} & -\frac{1}{s-x} & -\frac{1}{s-x}\end{vmatrix}=\left(\frac{z-y}{s-x}:\frac{s-z}{s-y}:-\frac{s-y}{s-z}\right).$$ Thus, the problem is equivalent to verifying $$\begin{vmatrix}\frac{z-y}{s-x} & \frac{s-z}{s-y} & -\frac{s-y}{s-z} \\ -\frac{s-z}{s-x} & \frac{x-z}{s-y} & \frac{s-x}{s-z} \\ \frac{s-y}{s-x} & -\frac{s-x}{s-y} &\frac{y-x}{s-z}\end{vmatrix}=0.$$ But $(s-x)\text{row}_1+(s-y)\text{row}_2+(s-z)\text{row}_3=0$, so this is true, as desired.

If we bary with $ABC$ as reference triangle we get $$A_1=(0,b^2(b^2+a^2-c^2),c^2(c^2+a^2-b^2))$$

Let $\triangle XYZ$ be the triangle with incenter $(ABC)$ ($X$ opposite $A$, etc.) Also let $T_a$ be the intersection of the tangent to $(ABC)$ at $A$ with $BC$, and define $T_b$ and $T_c$ similarly. Claim: $T_a-T_b-T_c$ are colinear Proof: Pascal on $AABBCC$ $\blacksquare$ Call this line $r$. Claim $A_1B_1\parallel to r$ and so on. Proof: Notice that $KA_1\parallel YZ$ and $KB_1\parallel XZ$. This means $\triangle T_aT_bZ$ and $\triangle A_1B_1K$ are homothetic from $C$, and thus $r=T_aT_b\parallel A_1B_1$. $\blacksquare$ This solves the problem.