and 
.
with circumcircle 
is given. Let 
be the foot of altitude from 
be the intersection point of 
and 
.
lies on the arc 
.
.
with diameter 
meets 
and 
(where 
at 
.
is perpendicular to 
.
Y by anantmudgal09, NJOY, a_simple_guy, GeoMetrix, Yaser-, megarnie, son7, Assassino9931, HWenslawski, mistakesinsolutions, Lamboreghini, Adventure10, Rounak_iitr, Tastymooncake2
Two circles 
and 
have common external tangents 
and 
meeting at 
. Suppose touches at 
and touches at 
. A circle 
through and intersects again at 
and again at 
, such that quadrilateral 
is convex.
Suppose lines
and 
meet at point 
, while lines 
and 
meet at point 
. Show that , , are collinear.
Merlijn Staps
and 
have common external tangents 
and 
meeting at 
. Suppose touches at 
and touches at 
. A circle 
through and intersects again at 
and again at 
, such that quadrilateral 
is convex.Suppose lines
and 
meet at point 
, while lines 
and 
meet at point 
. Show that , , are collinear.Merlijn Staps
This post has been edited 1 time. Last edited by a1267ab, Dec 16, 2019, 5:12 PM
![[asy]
size(10cm);
pair T = (0,0);
pair A = 3*dir(15);
pair B = 4*dir(-15);
pair E = 3*dir(-15);
pair F = 4*dir(15);
pair G = incenter(T,A,E);
pair H = incenter(T,B,F);
pair P = intersectionpoints(circumcircle(A,E,G),circumcircle(B,F,H))[0];
pair Q = intersectionpoints(circumcircle(A,E,G),circumcircle(B,F,H))[1];
pair X = 0.6*P + 0.4*Q;
pair C = 2*foot(circumcenter(A,E,G),A,X)-A;
pair D = 2*foot(circumcenter(B,F,H),B,X)-B;
pair Y = extension(A,D,B,C);
dot('$T$', T, dir(180));
dot('$A$', A, dir(120));
dot('$B$', B, dir(240));
dot('$X$', X, dir(225));
dot('$C$', C, dir(C));
dot('$D$', D, dir(D));
dot('$Y$', Y, dir(Y));
draw(circumcircle(A,E,G),deepgreen);
draw(circumcircle(B,F,H),deepgreen);
draw(T--B,deepgreen);
draw(T--F,deepgreen);
draw(A--Y,mediumred);
draw(B--Y,mediumred);
draw(A--C,mediumred);
draw(B--D,mediumred);
draw(circumcircle(A,B,C),lightred);
draw(1.2*P-0.2*Q--2.3*Q-1.3*P,heavygreen);
[/asy]](http://latex.artofproblemsolving.com/b/9/f/b9f72f94068ff54f02fe93156bb75ae2cd3a2b78.png)
be the radical axis of 
by the Radical Axis Theorem. We restate the problem as follows:
,
and 
meet 
.
)
,
has degree at most 
,
.
has degree at most 
:
:
.
:
.
.
for which 
,
.
,
,
.![\[\frac{[TAC]}{[TBD]}=\frac{[YAC]}{[YBD]}=\frac{YA\cdot YC}{YB\cdot YD}=\frac{AC^2}{BD^2},\]](http://latex.artofproblemsolving.com/3/8/8/38809b643c78bccb7bdf7eb7d810c31e1c30eb46.png)
so![\[\frac{AC^2}{TA\cdot AC\sin\angle TAC}=\frac{BD^2}{TB\cdot BD\sin\angle TBD}.\]](http://latex.artofproblemsolving.com/4/e/0/4e0b888142a623e28873814a5e161cf718719eb0.png)
The radii of 
and 
.![\[\frac{r_1}{TA}=\frac{r_2}{TB},\]](http://latex.artofproblemsolving.com/7/f/c/7fc21c99474a55ede9633ea965625d0c5c3625db.png)
which is exactly what we want to prove.![[asy]
size(8cm);
pair A = dir(170);
pair B = dir(5);
pair C = dir(100);
pair D = dir(45);
pair X = extension(A, C, B, D);
pair Y = extension(A, D, B, C);
pair T = X+2.2*(Y-X);
fill(circumcircle(A, C, C+2*A-2*foot(C, A, T)), rgb(66, 160, 255)+white);
fill(T--A--2*foot(A, circumcenter(A, C, C+2*A-2*foot(C, A, T)), T)-A--cycle, rgb(66, 160, 255)+white);
fill(circumcircle(B, D, D+2*B-2*foot(D, B, T)), rgb(66, 160, 255)+white);
fill(T--B--2*foot(B, circumcenter(B, D, D+2*B-2*foot(D, B, T)), T)-B--cycle, rgb(66, 160, 255)+white);
draw(A--T--B);
draw(unitcircle);
draw(circumcircle(A, C, C+2*A-2*foot(C, A, T)), blue);
draw(circumcircle(B, D, D+2*B-2*foot(D, B, T)), blue);
draw(A--X--B);
draw(A--D);
draw(B--C);
draw(X--T, dotted);
draw(2*foot(B, circumcenter(B, D, D+2*B-2*foot(D, B, T)), T)-B--T--2*foot(A, circumcenter(A, C, C+2*A-2*foot(C, A, T)), T)-A, blue);
string[] names = {"$A$", "$B$", "$C$", "$D$", "$T$", "$X$", "$Y$"};
pair[] pts = {A, B, C, D, T, X, Y};
pair[] labels = {dir(210), dir(-45), dir(80), D, dir(-90), X, Y};
for(int i=0; i<names.length; ++i){
dot(names[i], pts[i], dir(labels[i]));
}
[/asy]](http://latex.artofproblemsolving.com/a/7/9/a79ad2f0a0ad52c27fb776b1b0e8a613bdd86806.png)
![[asy]
size(9cm); defaultpen(fontsize(9pt));
pair T,A,B,O1,O2,P,Q,X,C,D,EE,F,Y,Z;
T=(0,0);
A=dir(40);
B=0.4/A;
draw(A--T--B);
O1=2*circumcenter(T,A,1/A);
O2=2*circumcenter(T,B,reflect(T,O1)*B);
P=intersectionpoint(circle(O1,length(O1-A)),circle(O2,length(O2-B)));
Q=reflect(T,O1)*P;
real x=0.55;
X=x*P+(1-x)*Q;
C=2*foot(O1,X,A)-A;
D=2*foot(O2,X,B)-B;
EE=2*foot(circumcenter(A,B,C),T,A)-A;
F=2*foot(circumcenter(A,B,C),T,B)-B;
Y=extension(B,C,A,D);
Z=extension(A,B,C,D);
draw(circumcircle(A,B,C),gray);
draw(C--A--Y--B--D,gray+dashed);
draw(A--Z--D,gray);
draw(circle(O1,length(O1-A)));
draw(circle(O2,length(O2-B)));
draw(A--T--(1/A));
draw(EE--T--F);
draw(T--Y);
draw(Z--F);
real y=1.1;
clip( (100,y)--(-100,y)--(-100,-100)--(100,-100)--cycle);
dot("$T$",T,N);
dot("$A$",A,dir(150));
dot("$B$",B,SW);
dot("$P$",P,dir(-75));
dot("$Q$",Q,dir(70));
dot("$X$",X,N);
dot("$C$",C,dir(-20));
dot("$D$",D,dir(20));
dot("$E$",EE,SW);
dot("$F$",F,W);
dot("$Y$",Y,E);
dot("$Z$",Z,S);
[/asy]](http://latex.artofproblemsolving.com/6/9/2/692519957043fd09093d073ff262d59cd93a55a1.png)
,
,
concur.
to denote the inverse, to obtain the following picture.![[asy]
size(5cm); defaultpen(fontsize(9pt));
pair O,B,U,T,A,C,P,D,Q;
O=(0,0);
B=dir(10);
U=1.8*B;
A=intersectionpoint( (0,1)--(20,1),circle(U,length(B-U)));
T=2*foot(U,(0,1),(20,1))-A;
C=U+length(B-U)*dir(270)-(0.4,0);
P=extension(A,T,B,C);
D=2*foot(O,B,C)-B;
Q=2*foot(U,B,C)-B;
real x=1.8, y=1.3;
draw(circle(O,1));
draw(circle(U,length(B-U)));
draw(P--C);
draw(foot(P+(-x,0),C,C+(1,0))--(C+(y,0)));
draw(foot(C+(y,0),A,T)--(P+(-x,0)));
label("$\ell_1'$",P+(-x,0),N);
label("$\Gamma_1'$",foot(P+(-x,0),C,C+(1,0)),N);
label("$\Gamma_2'$",dir(160),dir(160));
label("$\ell_2'$",U+(length(B-U),0),E);
dot("$B'$",B,E);
dot("$T'$",T,N);
dot("$A$",A,N);
dot("$C'$",C,S);
dot("$E'$",P,N);
dot("$D'$",D,SW);
dot("$F'$",Q,NE);
[/asy]](http://latex.artofproblemsolving.com/0/1/6/016f0f4ba8c837ec72e62dde38d441ff700a980e.png)
sending 
to 
sends 
to 
,![\[\frac{B'C'}{B'F'}=\frac{B'E'}{B'D'}\implies B'C'\cdot B'D'=B'E'\cdot B'F',\]](http://latex.artofproblemsolving.com/d/c/1/dc1399fd218b671803aa7c6b2b425ba72275db4b.png)
whence 
and 
.
,
with respect to 
,
,
.
and 
;
,
.
concur at a point 
,
each have degree 
.
,
values of 
.
.
,
,
follows analogously.
.
,
,
follows analogously.
of 
,
and 
are projective, so this reduces the problem to three cases.
,
.
,
lies on 
lie on 
respectively, such that 
lie on a circle 
,
are collinear. Point 
concur. Lines 
and 
meet 
concur;
to the line at infinity, preserving 
is now a rectangle, and 
and 
,
and 
,
(unless double roots). We prove 
,
is projective; since 
by definition. Note that 
is projective; so 
is a projective map 
.
is projective; since 
is always perpendicular to the 
.
cyclic forces 
projective; so 
is a projective map 
:
both coincide with the other intersection 
,
of 
:
;
.
(this is really nice if you haven't seen it before).
intersect 
and 
intersect 
.![\[AN\cdot AB=AB_1^2=BA_1^2=BM\cdot BA\]](http://latex.artofproblemsolving.com/f/3/6/f36e9a80c32e85657f2e27e9a2ea14996865a40a.png)
And the conclusion follows
intersect. We prove this by Trig Ceva's.
to 
,
.
,![\[\frac{AP\cdot CB}{CP}=\frac{BQ\cdot DA}{DQ}\]](http://latex.artofproblemsolving.com/1/5/4/154b7aec752552b64ac583dbdc4fcbcc4e13db70.png)
![\[AP\cdot CB\cdot DQ = BQ\cdot DA\cdot CP\]](http://latex.artofproblemsolving.com/1/c/f/1cfa1a744b0dc29333dab9507909f6dec1d33800.png)
And so 
.
by the radical axis theorem on 
.
is sent to the line at infinity (and hence 
lies on the line at infinity, implying 
)
be the image of 
under the homothety sending 
.
.![[asy]
size(10cm);
real d = 5/2;
real s = 3/2;
real c = 3/4;
path g1p = circle((0, -d), 1);
path g1 = circle((c, -d), 1);
path g2 = scale(s) * g1p;
draw(g1 ^^ g2);
//draw(g1p, dotted);
pair O = origin; dot("$O$", O, dir(90));
pair T = extension((0, 0), (1, 0), s * (0, -d), (c, -d)); dot("$T'$", T, dir(90));
real a = abs(T - (c, -d));
pair A = intersectionpoints(g1, circle(T, sqrt(a^2 - 1)))[0]; dot("$A'$", A, dir(135));
pair B = intersectionpoints(g2, circle(T, s*sqrt(a^2 - 1)))[1]; dot("$B'$", B, dir(-30));
draw(A + (s-1 + 0.2) * (A - T) -- T -- B + 0.2 * (B - T));
pair C = (2 * c - A.x, A.y); dot("$C'$", C, dir(45));
pair D = (- B.x, B.y); dot("$D'$", D, dir(180));
draw(A -- C ^^ B -- D);
pair Y = extension(A, D, B, C); dot("$Y'$", Y, dir(90));
draw(D -- Y -- B);
//draw(T -- Y, dotted);
pair M = s * intersectionpoints(O -- D, g1p)[0]; dot("$M$", M, dir(120));
pair N = s * intersectionpoints(O -- B, g1p)[0]; dot("$N$", N, dir(60));
//draw(B -- O -- D, dotted);
draw(M -- N);
draw(O -- T, dotted);
draw(D -- O -- B, dashed);
[/asy]](http://latex.artofproblemsolving.com/a/f/1/af17b728424d7bf315b255972802199bbf84edb5.png)
and 
,
are collinear. Finally, by Pascal's theorem on hexagon 
,
are collinear. Combining the previous two results, 
is collinear, so 
is collinear.
and 
.
are concurrent.
and let the inverse of any point or inverted image of any cline 
this happens 
which is 
to 
respectively. Let an arbitary line 
passing through 
intersect 
at 
respectively.
and 
.
at 
,
and 
.
So power of 
and 
of 
we get that 
Polar of 
.
respectively.
?
)" It is only if and when projective properties of real and of imaginary points have been shown to be identical,that the process of projecting real points to the circular points at infinity is justifiable as a means of obtaining a theorem for real points".
and 
.
and 
.
are concyclic.
cut 
and 
cut 
.
thus arc 
of 
of 
defined by homothety at 
followed by reflection across the angle bisector of 
,
and 
.
which means 
.
which implies the concyclicity.
and 
where mixed up.
touch 
at 
respectively. Clearly 
is cyclic. Let 
and 
.
we get 
is cyclic. Hence from simple angle chasing, 
is cyclic. Also 
thus 
are collinear.
we have 
are collinear. Moreover, applying Pascal's on 
means that 
are collinear. Combining this results gives us that 
are collinear.
,
concur via the projective transformation mapping 
through 
.
,
by Reim's theorem. Now notice 
and 
.
,
.
,
,
.
,
,
and 
.
,
,
and 
.![\[\angle T_2BT = 180 - \angle BDT_2 = 180 - \angle ACT_1 = \angle TAT_1.\]](http://latex.artofproblemsolving.com/1/6/2/162212c71462b71cc07d578a8087e3a2f1b9426d.png)
Furthermore, since 
,
,
and 
are isogonal wrt 
.
and 
are isogonal wrt 
,
,
and 
are isogonal wrt 
is isogonal with 
,
.
respectively. Now,![\[\frac{AR_1}{BR_2} = \frac{AT}{BT} = \frac{AT_1}{BT_2}\]](http://latex.artofproblemsolving.com/5/a/f/5af18737e9d28c5d6439f5e70514c989e4d49d07.png)
so showing 
suffices. We have![\[(T_1A;G_1R_1) \overset{B}{=} (CA;X\ell_2\cap AG_2) \overset{A'}{=} (CA;C'A') = (C'A';CA)\]](http://latex.artofproblemsolving.com/7/b/0/7b0773b67f8a7c176885b52a2a85029a81cb30c1.png)
Similarly, 
.
is cyclic by radical axis, and so is 
,
and 
are collinear. By homothety through ![\[(C'A';CA) = (TD\cap \Gamma_2 B; TD'\cap \Gamma_2 B') \overset{T}{=} (DB;D'B')\]](http://latex.artofproblemsolving.com/b/0/0/b000df537a520785c4c948d5af6567b1e3ce0628.png)
Therefore, 
and 
meet 
is cyclic 
are antihomologous points so 
are collinear
.
are midpoints of 
.
as 
maps 
say 
.
we get 
are collinear. By 
on quad 
with point 
hence by 
with point 
collinear.
and 
was an isosceles trapezium and hence cyclic,in the inverted picture points 
should be collinear.
and by homothety reasons 
.
This implies 
and 
[In inverted diagram ofc].
have a common point.
concurrent.Let they all intersect at 
.
concur at point 
and 
.
,
,
:
are all collinear, implying the conclusion.
:
.
:
Lemma 2: if 
and 
then 
and 
& point 
We have :
now by Lemma 1 in 
We are Done:)
![[asy]
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draw(circle((9.299520820490239,0.8869011879778688), 1.5428849251131946), linewidth(0.5));
draw((4.817527272727283,0.4536)--(8.633834637842673,2.2787904789682334), linewidth(0.5));
draw((4.817527272727283,0.4536)--(8.912795304303,-0.6067310516551885), linewidth(0.5));
draw((7.34,1.66)--(8.731841990792773,0.37394423298667867), linewidth(0.5));
draw((7.850427725877604,1.4166398927064516)--(8.912795304303,-0.6067310516551885), linewidth(0.5));
draw((7.34,1.66)--(8.605907646213748,1.0564445797363042), linewidth(0.5));
draw((8.605907646213748,1.0564445797363042)--(8.912795304303,-0.6067310516551885), linewidth(0.5));
draw((8.009713508732649,1.7335953421316597)--(8.195794072343277,-0.1911902541393934), linewidth(0.5));
draw((4.817527272727283,0.4536)--(8.605907646213748,1.0564445797363042), linewidth(0.5) + linetype("2 2"));
draw(circle((7.893684292592922,2.1643370218071647), 0.7489473462329945), linewidth(0.5));
draw(circle((8.272753306502535,-0.21779870730589768), 0.7489473462329939), linewidth(0.5));
draw((8.633834637842673,2.2787904789682334)--(7.838118503380405,0.392131383232714), linewidth(0.5));
draw((4.817527272727283,0.4536)--(10.739584596250433,0.3330869466614837), linewidth(0.5));
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label("$A$", (7.377293775285266,1.7550732363857555), NE * labelscalefactor);
label("$\Gamma_1$", (6.9586594848748025,1.1461506321523538), NE * labelscalefactor);
dot((4.817527272727283,0.4536),dotstyle);
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dot((8.633834637842673,2.2787904789682334),linewidth(4.pt) + dotstyle);
label("$E$", (8.671254309281245,2.3544814249280104), NE * labelscalefactor);
label("$\Gamma_2$", (9.156489509529736,2.0593372680165413), NE * labelscalefactor);
dot((8.009713508732649,1.7335953421316597),linewidth(4.pt) + dotstyle);
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label("$X$", (8.186019109032753,1.041492059549738), NE * labelscalefactor);
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label("$B$", (9.013773274162533,-0.5664441922540882), NE * labelscalefactor);
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label("$D$", (7.6768992343134355,1.4886695970336423), NE * labelscalefactor);
dot((8.731841990792773,0.37394423298667867),linewidth(4.pt) + dotstyle);
label("$C$", (8.823484960339595,0.38499737686060187), NE * labelscalefactor);
dot((8.605907646213748,1.0564445797363042),linewidth(4.pt) + dotstyle);
label("$Y$", (8.737855219119274,1.1651794635346477), NE * labelscalefactor);
dot((7.838118503380405,0.392131383232714),linewidth(4.pt) + dotstyle);
label("$H$", (7.47224066171082,0.46111270238977703), NE * labelscalefactor);
dot((6.814053828530863,0.4129709587450996),linewidth(4.pt) + dotstyle);
label("$J$", (6.854000912272186,0.4896559494632177), NE * labelscalefactor);
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label("$I$", (10.62170952596636,0.44208387100748325), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]](http://latex.artofproblemsolving.com/9/9/e/99e254d6db81961194a45c4978a1b9606223713f.png)
.
are congruent. Now Since 
,
to 
,
meet 
.
.
cyclic, which is equivalent to 
collinear (Bcoz 
is cyclic and radical axis). Let 
meet 
and 
which implies 
.
cyclic. Hence by radical axis, 
are collinear. We are done.
,
,
,
.
and 
.
which is valid since 
.
is the midpoint 
,
is a parallelogram. Thus 
.
and 
are homothetic; hence 
.
are collinear.
and 
.
.
are cyclic. Then ![\[\measuredangle(CJ, JXB_1)=\measuredangle CJE=\measuredangle CAB_1=\measuredangle CEA=\measuredangle (CE,AE)=\measuredangle(CE,JXB_1)\]](http://latex.artofproblemsolving.com/0/6/8/06800ab90c00cd928543420e5a9d0620d158a380.png)
so 
or equivalently 
so 
passes through 
is cyclic by radical axis and then ![\[\measuredangle(JI,FB_1I)=\measuredangle JIB_1=\measuredangle FAB_1=\measuredangle ECA=\measuredangle(EC,AC)=\measuredangle(EC,FB_1I)\]](http://latex.artofproblemsolving.com/c/6/2/c62545935bb1eea1607f21f0629a52717d2cf965.png)
.
gives 
lies on the line between 
and 
.
and 
gives that they are perspective from 
must lie on the line between 
and 
.
are collinear and since line 
is line 
by the Pascal, we get 
and 
respectively. Vary the center 
and 
are degree 
has degree 
has degree 
.
,
and 
.
,
,
,
(clearly it can't pass through both).
cases, so the problem always holds when 
and vary 
—
Since 
it follows 
Note 
and 
are projective maps, and in particular, we have 
Then, we see 
so by Zack's Lemma, 
where we subtract the one case when 
is a line). Thus, the hypothetical collinearity 
has degree 
It remains to check five cases of 
all coincide, and 
we have 
so 
we have 
so 
we have 
and evidently we have 
Therefore we can move 
with respect to 
respectively, as 
Thus, we simply need to check that the problem is true for 
cases of 
and 
.
cyclic. An inversion at 
fixes 
and swaps 
passes through 
and 
,
are collinear. By Pascal on 
,
and 
(which obviously passes through 
since it is the radical axis of 
.
,
is cyclic. Similarly, 
is cyclic. We now have![\[XA \cdot XC = XB \cdot XD \implies XE \cdot XC = XF \cdot CD,\]](http://latex.artofproblemsolving.com/a/d/4/ad491e7255845532ab8469c0969d7673bed9cd19.png)
so 
and the angle chase![\[\angle A'CB + \angle ADB' = ( \angle BCA - \angle A'CA) + (360^{\circ} - \angle ADB - \angle BDB') = 180^{\circ}.\]](http://latex.artofproblemsolving.com/0/1/a/01a8f65bb2939a5ac1c10af64be527e2b8788978.png)
