such that 
and 
with another positive integer 
,
is a perfect square. For example, if the number on the blackboard is 
,
,
,
,
.
,
with circumcenter 
and orthocenter 
,
be an arbitrary point on the circumcircle of triangle 
and that line 
is not parallel to line 
.
be the point on the circumcircle of triangle 
is perpendicular to 
be the point on line 
such that 
.
and 
be the points on the circumcircle of triangle 
is a diameter, and 
and 
are parallel. Let 
be the midpoint of 
.
and 
are perpendicular.
and 
are perpendicular.
Y by Davi-8191, artsolver, Adventure10, megarnie, Mango247, Novmath, and 5 other users
Triangle 
is inscribed in circle 
. The tangent lines to at 
and 
meet at 
. Point 
lies on ray 
such that 
. Points 
and 
lie on ray 
(with in between and ) such that 
. Prove that triangles and 
are similar to each other.
is inscribed in circle 
. The tangent lines to at 
and 
meet at 
. Point 
lies on ray 
such that 
. Points 
and 
lie on ray 
(with in between and ) such that 
. Prove that triangles and 
are similar to each other.
and 
so that 
bisects 
,
,
(so that 
and 
to be 
.
and 
and 
.
and that 
.
.
,
is cyclic. Similarly, 
is cyclic. Thus, 
and 
.
.
.
,
and thus 
is cyclic since it has center 
is the diameter of the circumcircle of 
.
.
and 
.
.![\[ \angle CSB' + \angle SCB' = \angle CB'T\implies \beta + \angle CSB' = \alpha + \beta\implies \angle CSB' = \alpha = \angle BAB'
\]](http://latex.artofproblemsolving.com/c/d/8/cd8adc0e5592266f8021083f16b14188d5c96247.png)
thus giving us cyclic quaderilateral 
.![\[ \angle B'AS = \angle B'BS = 180 - \angle B'BC = \angle TC'C = \frac {180 - \angle CTC'}{2} = \frac {180 - (2\alpha + 2\beta)}{2} = 90 - \alpha - \beta
\]](http://latex.artofproblemsolving.com/3/2/f/32fbdc0d4529cb9c46aa82cb01fa0b72c288565f.png)
.
,
,
.
and we know that 
.
at 
and 
.
and 
.
and 
.
.
.
.
is cyclic. Thus, 
.
and thus 
.
!
be the midpoint of 
![\[ \frac{BT}{AT}=\frac{\sin{\angle{BAT}}}{\sin{\angle{TBA}}}=\frac{\sin{\angle{CAM}}}{\sin{\angle{BCA}}}\]](http://latex.artofproblemsolving.com/d/b/9/db985bc04b1196adecd657029c97049e76385ba1.png)
![\[ =\frac{MC}{AM}\]](http://latex.artofproblemsolving.com/d/0/c/d0c58d7c2c5a5058f254cf36271efe1758ffb7d5.png)
,
.
,
is cyclic. So 
.
.
,
similarity that 
,
:
,
,
,
,
.
meet the circle again at 
,
meet the circle again at 
.
are collinear.
.
;
,
are collinear reduces to proving that their images lie on a circle through 
is cyclic, or equivalantly 
.
,
,
intersect 
again at 
and 
again at 
.
are collinear, and moreover, 
is similar to triangle 
.
be the reflections of 
about the line 
is similar to 
,
,
are parallel. Let 
intersect our circle (the circle centre 
)
intersect 
at a point 
.
are collinear. But from Pascal applied to the 
points 
,
points, 
,
,
,
and 
are collinear. If 
, we then have 
.
in this diagram is 
in this diagram, is 
is similar to 
Could anyone tell me the author of this problem?
.
.
be the circle centered at 
with radius 
.
and 
and 
,
and 
are directly similar. Since 
are antiparallel with respect to 
is a symmedian of 
.
and 
,
and 
are directly similar, 
at 
(which may be infinite.) Because 
,
,
is the Miquel point of 
,
.
,
and 
'
,
is cyclic and 
.
.
.
.
are similar. Since 
.
is isosceles, 
which implies that 
is cyclic. Hence 
which since 
implies that triangles 
and 
are similar. By the same argument, triangles 
and 
are similar. This implies that 
are similar.
very nice.
then 
,
are concyclic. So 
and 
;
,![[asy]/* DRAGON 0.0.9.6
Homemade Script by v_Enhance. */
import olympiad;
import cse5;
size(12cm);
real lsf=0.8000;
real lisf=2011.0;
defaultpen(fontsize(10pt));
/* Initialize Objects */
pair B_1 = (3.0, -3.0);
pair C_1 = (11.0, -3.0);
pair T = midpoint(B_1--C_1);
pair B = (6.451128850753037, 0.9621636086265188);
pair C = (10.082123836968163, -0.45038970554465996);
pair S = IntersectionPoint(Line(B,C,lisf),Line(B_1,C_1,lisf));
pair K = IntersectionPoint(Line(B,B_1,lisf),Line(C,C_1,lisf));
pair A = foot(T,K,S);
/* Draw objects */
draw(CirclebyPoint(T,B_1), rgb(0.0,0.6,0.6));
draw(circumcircle(A,B,C), rgb(0.0,0.6,0.8) + linetype("4 4"));
draw(T--B, rgb(0.0,0.8,0.8));
draw(T--C, rgb(0.0,0.8,0.8));
draw(K--S, rgb(0.0,0.6,0.8));
draw(circumcircle(K,B_1,C_1), rgb(0.0,0.8,0.8) + linewidth(1.0) + dotted);
draw(T--A, rgb(0.0,0.8,0.8) + linetype("4 4"));
draw((abs(dot(unit(T-A),unit(S-A))) < 1/2011) ? rightanglemark(T,A,S) : anglemark(T,A,S), rgb(0.0,0.8,0.8));
draw(B--B_1, rgb(1.0,0.6,0.0) + linewidth(1.0));
draw(B_1--C_1, rgb(1.0,0.6,0.0) + linewidth(1.0));
draw(C_1--C, rgb(1.0,0.6,0.0) + linewidth(1.0));
draw(C--B, rgb(1.0,0.6,0.0) + linewidth(1.0));
draw(K--B, rgb(0.0,0.8,0.8) + dashed);
draw(K--C, rgb(0.0,0.8,0.8) + dashed);
draw(C--S, rgb(0.0,0.8,0.8) + dashed);
draw(C_1--S, rgb(0.0,0.8,0.8) + dashed);
/* Place dots on each point */
dot(B_1);
dot(C_1);
dot(T);
dot(B);
dot(C);
dot(S);
dot(K);
dot(A);
/* Label points */
label("$B_1$", B_1, lsf * dir(225));
label("$C_1$", C_1, lsf * dir(-45));
label("$T$", T, lsf * dir(-90));
label("$B$", B, lsf * dir(140));
label("$C$", C, lsf * dir(70) * 1.414);
label("$S$", S, lsf * dir(-45));
label("$K$", K, lsf * dir(60));
label("$A$", A, lsf * dir(45));
[/asy]](http://latex.artofproblemsolving.com/4/8/e/48e4fcca54a23cceacfb4a93e062dec2bb5a40c7.png)
and 
are collinear?
was meant rather than 
.
be the intersection of 
and 
intersect 
at 
and 
intersect 
and let the foot of perpendicular from 
is cyclic 
is cyclic
i
is cyclic. 
and 
and we are done.
and 
meet at 
.
so 
is cyclic so 
.
,
,
and 
which implies that 
also shows that 
,
by Brocard we get that 
.
,
we get that 
.
,
,
.
we conclude 
,
is the orthocenter of 
,
and 
.
,
which implies the desired result.
lies on 
.
,
are collinear, implying that 
.
and 
.
and 
,
.
,
is cyclic, implying that 
,
.
,
and 
,
![[asy] /* coordinates from geogebra */
import geometry; size(10cm); pair A,B,C,T,S,B_1,C_1,D; A=(-0.89,1.11); B=(1.96,-1.44); C=(-1.44,-1.06); T=(-0.1,-4.5); S=(-9.76,-0.13); B_1=(3.26,-6.02); C_1=(-3.47,-2.98); D=(1.16,1.4); fill(A--B--C--cycle,white+white); fill(B--C--C_1--B_1--cycle,white+white+white+white); draw(circle(A,B,C)); draw(circle(B_1,C_1)); draw(circle(A,C_1,B_1));draw(C_1--D--B_1--cycle,green); draw(A--B--C--cycle,red); draw(A--B_1--C_1--cycle,red); draw(D--S); draw(S--C); draw(D--B); draw(D--C); draw(B--T); draw(C--T); draw(A--T); draw(S--C);draw(S--C_1);dot(A^^B^^C^^T^^S^^B_1^^C_1^^D); label("$A$",A,NE); label("$B$",B,2W); label("$C$",C,NE); label("$T$",T,SW); label("$S$",S,2E); label("$B_1$",B_1,SW); label("$C_1$",C_1,SE); label("$D$",D,2N); [/asy]](http://latex.artofproblemsolving.com/0/d/b/0dbdf5e1e404846d04361fd0c083e97aa730b8af.png)
. Chase, 
.
is cyclic. As 
are collinear by radical axes ,we use 
.
![[asy]
import graph; size(12cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.5) + fontsize(13); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -2, xmax = 4, ymin = -4.7, ymax = 2; /* image dimensions */
draw((0.33709926477356134,0.6527195501031977)--(0.2983270446038727,0.5051548519589105)--(0.44589174274816,0.4663826317892219)--(0.4846639629178486,0.6139473299335092)--cycle, linewidth(0.75) + blue);
/* draw figures */
draw((0.4846639629178486,0.6139473299335092)--(-0.6435760050714912,-0.10350482168263458), linewidth(0.5));
draw((-0.6435760050714912,-0.10350482168263458)--(0.5736726499363176,-0.10910136722290036), linewidth(0.5));
draw((0.5736726499363176,-0.10910136722290036)--(0.4846639629178486,0.6139473299335092), linewidth(0.5));
draw(circle((-0.033620893008616766,0.18314254712320893), 0.673952485546976), linewidth(0.5));
draw((-0.6435760050714912,-0.10350482168263458)--(-0.04083567317250101,-1.386072138521609), linewidth(0.5));
draw((-0.04083567317250101,-1.386072138521609)--(0.5736726499363176,-0.10910136722290036), linewidth(0.5));
draw((-0.04083567317250101,-1.386072138521609)--(0.4846639629178486,0.6139473299335092), linewidth(0.5));
draw((-0.2732001930556983,0.8130740606736085)--(3.283969388430581,-0.12156250165275902), linewidth(0.5));
draw((3.283969388430581,-0.12156250165275902)--(0.5736726499363176,-0.10910136722290036), linewidth(0.5));
draw(circle((-0.04083567317250101,-1.386072138521609), 1.4171361366930244), linewidth(0.5));
draw((-0.2732001930556983,0.8130740606736085)--(-1.3654078509418188,-1.8898413076860952), linewidth(0.5));
draw((1.2837365045968172,-0.8823029693571232)--(-0.2732001930556983,0.8130740606736085), linewidth(0.5));
draw((-1.3654078509418188,-1.8898413076860952)--(3.283969388430581,-0.12156250165275902), linewidth(0.5));
draw((-0.2732001930556983,0.8130740606736085)--(0.48686186746905835,-1.1853750711357969), linewidth(0.5));
draw((-0.6435760050714912,-0.10350482168263458)--(1.2837365045968172,-0.8823029693571232), linewidth(0.5) + linetype("4 4"));
draw((0.5736726499363176,-0.10910136722290036)--(-1.3654078509418188,-1.8898413076860952), linewidth(0.5) + linetype("4 4"));
draw(circle((1.9313054361793436,0.42502876216911906), 1.4589248703776139), linewidth(0.5) + blue);
draw(circle((1.3118282790787361,-1.9326634023434865), 2.677578575444292), linewidth(0.5) + blue);
/* dots and labels */
dot((0.4846639629178486,0.6139473299335092),linewidth(3pt) + dotstyle);
label("$A$", (0.5270237491555813,0.7436458119447124), NE * labelscalefactor);
dot((-0.6435760050714912,-0.10350482168263458),linewidth(3pt) + dotstyle);
label("$B$", (-0.961798164673135,-0.08347747351568333), NE * labelscalefactor);
dot((0.5736726499363176,-0.10910136722290036),linewidth(3pt) + dotstyle);
label("$C$", (0.6564865242711218,-0.04032321514383659), NE * labelscalefactor);
dot((-0.04083567317250101,-1.386072138521609),linewidth(3pt) + dotstyle);
label("$T$", (-0.10590537363150583,-1.6298383985068579), NE * labelscalefactor);
dot((-0.2732001930556983,0.8130740606736085),linewidth(3pt) + dotstyle);
label("$Q$", (-0.4511394406062806,0.9090704690367915), NE * labelscalefactor);
dot((3.283969388430581,-0.12156250165275902),linewidth(3pt) + dotstyle);
label("$S$", (3.3680124253021653,-0.07628509712037554), NE * labelscalefactor);
dot((-0.033620893008616766,0.18314254712320893),linewidth(3pt) + dotstyle);
label("$O$", (-0.005212104097196526,0.18264045311070484), NE * labelscalefactor);
dot((0.20595840703846502,-0.4467889664271907),linewidth(3pt) + dotstyle);
label("$P$", (0.23213631694796114,-0.3639801529326871), NE * labelscalefactor);
dot((-1.3654078509418188,-1.8898413076860952),linewidth(3pt) + dotstyle);
label("$B_1$", (-1.7529595681569938,-1.9822648418769395), NE * labelscalefactor);
dot((1.2837365045968172,-0.8823029693571232),linewidth(3pt) + dotstyle);
label("$C_1$", (1.4044936693831338,-1.0616406632775426), NE * labelscalefactor);
dot((0.48686186746905835,-1.1853750711357969),linewidth(3pt) + dotstyle);
label("$S^*$", (0.48386949078373437,-1.4068747302523164), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
[/asy]](http://latex.artofproblemsolving.com/8/e/1/8e1bdb178e6db9163a1f80d7357e6948eca621d5.png)
.
are collinear. Now Pascal on 
gives that 
are collinear and then Pascal on 
gives that 
are collinear and thus we conclude that 
and 
lie on 
and 
.
and similarly that 
.
is the midpoint of 
and 
as in the problem statement.
is cyclic. Similarly, 
is also cyclic.
are orthogonal to each other. So upon performing an Inversion 
,
.
denote the inverse of 
is also cyclic, and similarly 
is also cyclic.
gives that 
are collinear. We finally have 
is also cyclic.
,
,
and thus we are done.
and 
.
so 
and 
.
,
and 
are tangent to 
by three tangents lemma, which is enough to imply that 
with 
and 
.
(as 
,
so 
.
,
so 
,
.
be the foot of the altitude from 
,
which is also just 
are tangent to 
(this can be shown by angle chasing and showing that 
)
.
.
lies on 
from the first claim. 
meets 
.
.
.
being acute,
,
.
intersect 
.
and 
and 
.
we have 
are collinear. But 
then 
hence 
.
and 
.
we have 
and 
or 
and let 
.
is self-polar. From here we have that 
First, we have that 
Thus, we conclude 
is cyclic with diameter 
Next, note that 
where 
However, we also know that 
so 
and thus, 
which implies that 
are both tangent to 
Thus, 
is cyclic with 
as diameter. Since 
we have that either 
