red cards, 
white cards, and 
and 
the minimal total penalty a player can amass and all the ways in which this minimum can be achieved.
and a set 
consisting of 
disting positive integers smaller than 
are given. Prove that there exists a positive integer 
that can be written in the form 
,
in at least 
different ways.
lies on side 
and point 
lies on side 
and 
,
is parallel to 
be a point on line 
,
.
be the point on line 
,
.
,
.
meet at point 
on line 
be 
.
meet 
respectively. Let 
be the midpoint of 
lie on a circle.
satisfy 
?
such that 
is a multiple of 
?
where 
and 
are constants, 
is the time in minutes, 
is the length of the trail in miles, and 
minutes to hike to the top if a trail is 
miles long and ascends 
feet, as well as if a trail is 
miles long and ascends 
feet. How many minutes does the model estimate it will take to hike to the top if the trail is 
miles long and ascends 
feet?
distinct primes?
index this yr (118.5 on amc10+ 9 on AIME) i’m in 7th rn btw first year comp math also. I will grind so hard until like 30 hrs/week. I’m ok at proofs. made mc nats
and 
meet on 
and 
meet on 
Y by nguyendangkhoa17112003, megarnie, jhu08, tiendung2006, Adventure10, Rounak_iitr, ehuseyinyigit, Narwhal234
Let 
be an acute triangle with circumcircle 
and orthocenter 
. Points 
and 
lie on segments 
and 
respectively, such that 
. The lines through 
and 
parallel to 
intersect again at 
and 
, respectively. Denote by 
the circumcircle of 
.
be an acute triangle with circumcircle 
and orthocenter 
. Points 
and 
lie on segments 
and 
respectively, such that 
. The lines through 
and 
parallel to 
intersect again at 
and 
, respectively. Denote by 
the circumcircle of 
.
- Show that lines
and 
meet on . - Prove that if passes through , then lines
and 
meet on as well.
This post has been edited 2 times. Last edited by pinetree1, Jun 25, 2019, 5:34 PM
.
is cyclic.
since they are both subtended by arc 
.
is cyclic and 
,
,
and 
.
is also cyclic. It immediately follows that 
,
![[asy]
size(8cm,0);
defaultpen(fontsize(10pt));
pair A = dir(110);
pair B = dir(220);
pair C = dir(320);
pair H = A + B + C;
pair L = dir(270);
pair H1 = 2*foot(H,A,L)-H;
pair S = circumcenter(A,H,H1);
pair D = 2*foot(S,A,B)-A;
pair E = 2*foot(S,A,C)-A;
pair P = 2*foot(B,A,L)-A-B-L;
pair Q = 2*foot(C,A,L)-A-C-L;
pair M = (P+Q)/2;
pair O = (0,0);
pair K = (B+C)/2;
pair X = extension(P,E,Q,D);
draw(circumcircle(A,B,C));
draw(A--B--C--cycle,linewidth(1.5));
draw(circumcircle(A,D,E));
draw(P--X--Q,dashed);
draw(P--Q);
draw(K--O);
draw(A--M);
draw(B--P);
draw(C--Q);
draw(D--E);
dot("$A$",A,dir(A));
dot("$B$",B,dir(B));
dot("$C$",C,dir(C));
dot("$D$",D,dir(200));
dot("$E$",E,dir(20));
dot("$H$",H,dir(270));
dot("$P$",P,dir(P));
dot("$Q$",Q,dir(Q));
dot("$M$",M,1.5*dir(70));
dot("$K$",K,NW);
dot("$O$",O,NE);
dot("$H_1$",H1,dir(330));
dot("$X$",X,dir(X));
[/asy]](http://latex.artofproblemsolving.com/b/1/6/b163b639ab8281ea237b34b6de460b1cf9459116.png)
so 
so we are done by spiral lemma.
,
and 
pass through circumcenter 
of 
.
be the midpoint of 
intersect 
,
hence 
or 
so we are done.
,
and 
hit 
.![[asy]
defaultpen(fontsize(10pt));
size(250);
pair A, B, C, H, D, E, P, Q, M, N, S, T, L;
A = dir(115);
B = dir(210);
C = dir(330);
L = dir(90);
H = orthocenter(A, B, C);
M = 2*foot(H, C, A) - H;
N = 2*foot(H, A, B) - H;
P = IP(Line(B, B+L-A, 20), circumcircle(A, B, C), 1);
Q = IP(Line(C, C+A-L, 20), circumcircle(A, B, C), 1);
D = extension(N, P, A, B);
E = extension(M, Q, A, C);
S = extension(M, E, N, D);
T = extension(P, E, Q, D);
draw(A--B--C--cycle, orange);
draw(circumcircle(A, B, C), red);
draw(circumcircle(A, D, E), heavygreen);
draw(B--M^^C--N, lightblue);
draw(M--Q^^N--P, purple+dotted);
draw(Q--T--P, heavycyan+dashed);
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$D$", D, dir(0));
dot("$E$", E, dir(0));
dot("$H$", H, dir(270));
dot("$M$", M, dir(M));
dot("$N$", N, dir(N));
dot("$P$", P, dir(P));
dot("$Q$", Q, dir(Q));
dot("$S$", S, dir(290));
dot("$T$", T, dir(T));
[/asy]](http://latex.artofproblemsolving.com/1/c/7/1c71bb45275a636d246e5c2d7b914bd5a87d3a28.png)
and 
collinear.
.
,
.
and 
collinear.
.
Thus, we must have 
,
and hence 
,
.
to be the max degree of 
.
in the same way.
have degree 
,
values of 
for which 
.
has degree at most 
.
have degrees 
,
has degree at most 
and 
is given by cross product 
;![\[(Q_1 R_2-Q_2 R_1)x + (R_1 P_2-R_2 P_1) y +(P_1 Q_2-P_2 Q_1)z=0.\]](http://latex.artofproblemsolving.com/7/7/c/77cdade478fb6b8069eb9dbb2382fd4a4a2b9444.png)
This is true as the equation must determine a line, and 
and 
both lie on it, so it is line 
for which 
,
factors out of each term. So the degree of the line is at most 
and moving 
are fixed, we get that 
.
passes through a point at infinity which is a constant rotation of the point at infinity on line 
,
.
have degree at most 
.
move linearly in 
have degree at most 
and one for which 
.
,
as well and thus 
,
.
denote the line at infinity. I claim that the points 
are projective in 
and 
.
,
go to infinity. It follows that 
are projective in 
is constant. It suffices then to show that there's a constant rotation sending 
.
,
,
,
.
are reflections of 
Hence, 
and similarly, 
.
is cyclic. Now apply Reim's Theorem again and conclude![[asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(10cm);
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 = -13.397755016958634, xmax = 17.88608307664881, ymin = -9.761581949056891, ymax = 8.82922347542287; /* image dimensions */
pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961);
draw((-3.8,5.57)--(-8.04,-5.49)--(7.86,-5.47)--cycle, linewidth(2) + rvwvcq);
/* draw figures */
draw((-3.8,5.57)--(-8.04,-5.49), linewidth(2) + rvwvcq);
draw((-8.04,-5.49)--(7.86,-5.47), linewidth(2) + rvwvcq);
draw((7.86,-5.47)--(-3.8,5.57), linewidth(2) + rvwvcq);
draw(circle((-0.09413405761646475,-2.1934241949101434), 8.6025692448846), linewidth(0.8));
draw(circle((-3.0536785593388696,2.283662608654139), 3.3700161932175376), linewidth(0.4));
draw((-5.805595844270751,0.3384221609352558)--(0.2684971821854436,1.7178380024590658), linewidth(0.8) + dtsfsf);
draw((-5.784843170218619,4.257934375870217)--(-5.8528383222773375,-8.584161461079916), linewidth(0.8) + dtsfsf);
draw((-5.784843170218619,4.257934375870217)--(8.496193791072775,-1.7346594784502916), linewidth(0.8) + dtsfsf);
draw((7.86,-5.47)--(-8.20064211278282,0.6855903525622036), linewidth(0.4));
draw((-8.04,-5.49)--(2.7638032538325117,5.920538581493395), linewidth(0.4));
draw((-5.805595844270751,0.3384221609352558)--(-3.791731884767069,-1.0031516101797133), linewidth(0.4));
draw((-3.791731884767069,-1.0031516101797133)--(0.2684971821854436,1.7178380024590658), linewidth(0.4));
draw((-5.805595844270751,0.3384221609352558)--(-0.9673541489235893,-0.3628935681639076), linewidth(0.4));
draw((-5.8528383222773375,-8.584161461079916)--(7.86,-5.47), linewidth(0.8) + dtsfsf);
draw((-8.04,-5.49)--(8.496193791072775,-1.7346594784502916), linewidth(0.8) + dtsfsf);
draw((-3.8,5.57)--(-0.09413405761646475,-2.193424194910143), linewidth(0.8) + dtsfsf);
draw((-3.791731884767069,-1.0031516101797133)--(-0.9673541489235893,-0.3628935681639076), linewidth(0.8) + dtsfsf);
draw((-8.20064211278282,0.6855903525622036)--(-5.805595844270751,0.3384221609352558), linewidth(0.4) + linetype("4 4"));
draw((2.7638032538325117,5.920538581493395)--(0.2684971821854436,1.7178380024590658), linewidth(0.4) + linetype("4 4"));
draw((0.2684971821854436,1.7178380024590658)--(-5.8528383222773375,-8.584161461079916), linewidth(0.4) + linetype("4 4"));
draw((-0.9673541489235893,-0.3628935681639076)--(8.496193791072775,-1.7346594784502916), linewidth(0.4) + linetype("4 4"));
draw(circle((-3.7988456534698876,5.577963613649507), 6.581119068594783), linewidth(0.8) + linetype("4 4"));
/* dots and labels */
dot((-3.8,5.57),dotstyle);
label("$A$", (-3.935676118154415,6.111119509912726), NE * labelscalefactor);
dot((-8.04,-5.49),dotstyle);
label("$B$", (-9.115459146768107,-5.376716117903734), NE * labelscalefactor);
dot((7.86,-5.47),dotstyle);
label("$C$", (8.603501609529223,-5.633141020310352), NE * labelscalefactor);
dot((-5.805595844270751,0.3384221609352558),dotstyle);
label("$D$", (-6.65378008366457,0.72619655937376), NE * labelscalefactor);
dot((0.2684971821854436,1.7178380024590658),dotstyle);
label("$E$", (1.0133244982933194,1.726253678759568), NE * labelscalefactor);
dot((8.496193791072775,-1.7346594784502916),dotstyle);
label("$P$", (8.988138963139152,-1.63291254276712), NE * labelscalefactor);
dot((-5.8528383222773375,-8.584161461079916),dotstyle);
label("$Q$", (-7.243557359199793,-8.889737280874392), NE * labelscalefactor);
dot((-5.784843170218619,4.257934375870217),dotstyle);
label("$K$", (-6.37171269101729,4.880279978360962), NE * labelscalefactor);
dot((-3.791731884767069,-1.0031516101797133),dotstyle);
label("$H$", (-3.986961098635739,-2.017549896377046), NE * labelscalefactor);
dot((-0.9673541489235893,-0.3628935681639076),dotstyle);
label("$L$", (-1.6022095062541881,-0.04307814784609218), NE * labelscalefactor);
dot((2.7638032538325117,5.920538581493395),dotstyle);
label("$M$", (3.1929361687495756,6.03419203919074), NE * labelscalefactor);
dot((-8.20064211278282,0.6855903525622036),dotstyle);
label("$N$", (-9.3205990686934,0.21334675456052524), NE * labelscalefactor);
dot((-0.09413405761646475,-2.193424194910143),dotstyle);
label("$O$", (0.013267378907507787,-1.940622425655061), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]](http://latex.artofproblemsolving.com/9/3/4/934bd78ee6b35c4e16cefe39a652faee608f9087.png)
denotes a directed angle modulo 
.![[asy]
size(8cm);
defaultpen(fontsize(10pt));
pair O, A, B, C, H, L, P, Q, Y, D, EE, X, SS, T;
O=(0,0);
A=dir(110);
B=dir(220);
C=dir(320);
H=A+B+C;
L=dir(270);
P=2*foot(O,B,foot(B,A,L))-B;
Q=2*foot(O,C,foot(C,A,L))-C;
Y=A+P+Q;
D=extension(A,B,P,Y);
EE=extension(A,C,Q,Y);
X=extension(P,EE,Q,D);
SS=foot(P,A,Q);
T=foot(Q,A,P);
draw(circle(O,1));
draw(A--B--C--A);
draw(A--L,gray);
draw(B--P);
draw(C--Q);
draw(circumcircle(A,D,EE),gray);
draw(D--EE);
draw(P--X--Q,gray);
draw(P--D,gray);
draw(Q--T,gray);
draw(A--P--Q--A);
draw(H--A--Y,dashed);
dot("$A$",A,N);
dot("$B$",B,B);
dot("$C$",C,C);
dot("$H$",H,S);
dot("$L$",L,S);
dot("$P$",P,P);
dot("$Q$",Q,Q);
dot("$Y$",Y,unit(Y-A));
dot("$D$",D,dir(210));
dot("$E$",EE,dir(18));
dot("$X$",X,NW);
dot("$S$",SS,unit(D-P)*dir(45));
dot("$T$",T,NE);
[/asy]](http://latex.artofproblemsolving.com/1/5/b/15b51044d315976c26ad671f0ca765b8e7698130.png)
.
and 
,
,
.
and set 
and 
.
and 
.
and 
,
.
as desired. 
,
,
,
lie on the perpendicular bisector of 
.
,
and 
are isogonal, so 
not containing 
,
,
,
.
and 
.
as desired. 
,
and similarly 
.
,
,
.
Now, we have that
Similaly, 
Notice that 
Finally, we want 
to be real, but 
and we are done. 
![\[\angle AQP = \angle ABP = \angle ADE,\]](http://latex.artofproblemsolving.com/b/7/8/b78941cd93df3d5f05f3e6f366420ec2b0ef30ca.png)
and similarly 
.
have spiral center 
is the second intersection of 
and 
be the reflections of 
,
are collinear. Let 
so the claim is proved. Now define ![\[\angle DEX = \angle EQC = \angle H_BBC = \angle HAE = \angle HDE,\]](http://latex.artofproblemsolving.com/c/8/d/c8d91525a2b3d6cd76ca3b66d99e9259412ca873.png)
so it follows that 
.
as the second intersection of 
with 
and we conclude that 
,
.
meet 
.
from part (a), and we know the center of the circle is on the angle bisector, so we can compute every length that we need to and the rest of the problem is a relatively quick trig bash.
![[asy] size(10cm);
pair A = dir(110); pair B = dir(200); pair C = dir(340); pair H = A+B+C; pair R = A*(1-2cos(7pi/18)); pair M = circumcenter(A,H,R); pair D = 2*foot(M,A,B)-A; pair E = 2*foot(M,A,C)-A; pair S = dir(150); pair T = dir(70); pair O = 0; pair P = 2*foot(O,S,R)-S; pair Q = 2*foot(O,T,R)-T; pair X = intersectionpoints(circumcircle(A,B,C), circumcircle(A,R,H))
[1];
dot('$A$', A, dir(A)); dot('$B$', B, dir(B)); dot('$C$', C, dir(C));draw(B--A--C--B); draw(circumcircle(A,B,C)); dot('$R$', R, dir(R)); dot('$H$', H, dir(H)); draw(circumcircle(A, H, R)); dot('$D$', D, dir(D)); dot('$E$', E, dir(E)); dot('$S$', S, dir(S)); dot('$T$', T, dir(T)); dot('$P$', P, dir(P)); dot('$Q$', Q, dir(Q)); dot('$X$', X, dir(X)); draw(X--P); draw(X--Q); draw(S--P); draw(T--Q); draw(circumcircle(S,H,T));
[/asy]](http://latex.artofproblemsolving.com/9/b/0/9b06f1b83c717c86466e484676605d4551f790af.png)
;
Since 
,
.
as well.
and 
,
.
.
Similarly, 
,
is directly similar to 
.
.
,
and 
;
.
and 
,
meets 
,
.
and we're done.
,
and thus 
is the second intersection point of 
.
.
,
is an isosceles trapezoid. Construct 
.
so 
but 
we conclude that 
,
thus, since 
,
and 
and 
intersects 
.
,
and 
are collinear.
,
which gives 
.
be a point such that 
is an isosceles trapezoid. We claim that 
and angle chase to get that 
are cyclic. Then angle chase more to get 
are concyclic. We finish by, again, angle chase (or Reim’s on 
and 
)
,
and 
are collinear.
be the reflection of 
,
are collinear.
and 
concur on 
as well as
Since 
,
not lying on 
hence 
,
passes through 
passes through 
and 
so 
.
so 
,
,
and 
.
which is enough to imply that 
![[asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(12.5cm);
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 = -2.1454561278177926, xmax = 2.518586158120346, ymin = -2.6140594649782853, ymax = 2.0882649720471256; /* image dimensions */
pen qqttzz = rgb(0,0.2,0.6); pen qqwuqq = rgb(0,0.39215686274509803,0); pen qqwwzz = rgb(0,0.4,0.6); pen cczzqq = rgb(0.8,0.6,0);
/* draw figures */
draw((-0.9482521532234824,1.2101509391074468)--(-1.6139400016108443,-0.9944591929592713), linewidth(0.7) + blue);
draw((-1.6139400016108443,-0.9944591929592713)--(1.6863454116149246,-0.911274648628153), linewidth(0.7) + blue);
draw((1.6863454116149246,-0.911274648628153)--(-0.9482521532234824,1.2101509391074468), linewidth(0.7) + blue);
draw(circle((0.019364704923429694,-0.28483169268530617), 1.7808019130526807), linewidth(0.7) + qqttzz);
draw((-0.9482521532234824,1.2101509391074468)--(-0.9145761530662614,-0.12591951710936553), linewidth(0.7) + qqwuqq);
draw(circle((-0.7432049066362212,0.5468595735592252), 0.6942620607118334), linewidth(0.7) + qqwwzz);
draw((-0.05144053204733724,0.4880220090830196)--(-1.2810809819276803,0.10789586597863488), linewidth(0.7) + cczzqq);
draw((-1.0032705075450343,-1.7427317838272427)--(-0.05144053204733724,0.4880220090830196), linewidth(0.7) + qqwuqq);
draw((-1.2810809819276803,0.10789586597863488)--(1.7681979773515861,0.051081449034861004), linewidth(0.7) + qqwuqq);
draw((-1.0032705075450343,-1.7427317838272427)--(1.7681979773515861,0.051081449034861004), linewidth(0.7) + qqwuqq);
draw((-1.6139400016108443,-0.9944591929592713)--(1.7681979773515861,0.051081449034861004), linewidth(0.7) + cczzqq);
draw((1.6863454116149246,-0.911274648628153)--(-1.0032705075450343,-1.7427317838272427), linewidth(0.7) + cczzqq);
draw((-1.0032705075450343,-1.7427317838272427)--(-1.3862603305379009,0.8085455126818173), linewidth(0.7) + qqwuqq);
draw((-1.2810809819276803,0.10789586597863488)--(-1.7157415795272901,0.11599449806537983), linewidth(0.7) + qqwuqq);
draw((-0.9482521532234824,1.2101509391074468)--(1.7681979773515861,0.051081449034861004), linewidth(0.7) + qqwuqq);
draw((-0.05144053204733724,0.4880220090830196)--(0.3642501238270317,1.4622543101662346), linewidth(0.7) + qqwuqq);
draw((-1.3862603305379009,0.8085455126818173)--(-0.05144053204733724,0.4880220090830196), linewidth(0.7) + qqwuqq);
draw((-0.05144053204733724,0.4880220090830196)--(1.7681979773515861,0.051081449034861004), linewidth(0.7) + qqwuqq);
draw((-1.7157415795272901,0.11599449806537983)--(1.6863454116149246,-0.911274648628153), linewidth(0.7) + qqwuqq);
/* dots and labels */
dot((-0.9482521532234824,1.2101509391074468),dotstyle);
label("$A$", (-1.0097522606536493,1.2971004806073005), NE * labelscalefactor);
dot((-1.6139400016108443,-0.9944591929592713),dotstyle);
label("$B$", (-1.7179720879301654,-1.1146751465237783), NE * labelscalefactor);
dot((1.6863454116149246,-0.911274648628153),dotstyle);
label("$C$", (1.7465627428009007,-0.9806876116831629), NE * labelscalefactor);
dot((-0.9145761530662614,-0.12591951710936553),linewidth(4pt) + dotstyle);
label("$H$", (-0.9714701078278916,-0.27884814347041503), NE * labelscalefactor);
dot((-0.05144053204733724,0.4880220090830196),linewidth(4pt) + dotstyle);
label("$E$", (0.023865865641806986,0.5186967067713438), NE * labelscalefactor);
dot((-1.2810809819276803,0.10789586597863488),linewidth(4pt) + dotstyle);
label("$D$", (-1.4818988121713268,0.1550162550611018), NE * labelscalefactor);
dot((1.7681979773515861,0.051081449034861004),linewidth(4pt) + dotstyle);
label("$P$", (1.8358877660610018,0.03378943782435445), NE * labelscalefactor);
dot((-1.0032705075450343,-1.7427317838272427),linewidth(4pt) + dotstyle);
label("$Q$", (-1.1182183603266291,-1.8675574851519987), NE * labelscalefactor);
dot((-1.3862603305379009,0.8085455126818173),linewidth(4pt) + dotstyle);
label("$T$", (-1.5329416826056703,0.8313342880661133), NE * labelscalefactor);
dot((-0.2220540932637905,0.08816398968567697),linewidth(4pt) + dotstyle);
label("$F$", (-0.1994466925084459,-0.030014150194986263), NE * labelscalefactor);
dot((-1.7157415795272901,0.11599449806537983),linewidth(4pt) + dotstyle);
label("$M$", (-1.9157632108632467,0.10397338464562923), NE * labelscalefactor);
dot((0.3642501238270317,1.4622543101662346),linewidth(4pt) + dotstyle);
label("$N$", (0.35564452346503983,1.5267933974769272), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]](http://latex.artofproblemsolving.com/0/f/7/0f7779041a366f33a3836e67de41ee0f583158e8.png)
minutes.
is the Miquel point of 
.
as the Miquel point of 
.
.
and hence proven. 
.
are collinear.![\[\angle AMD = \angle AHD = \angle AED = \angle ADE = \angle ABP = \angle AMP\]](http://latex.artofproblemsolving.com/3/1/4/31413caa578f903c109eccb3ecafebfc6a8fa5e9.png)
hence proven. 
collinear. Then we wish to show that 
lies on 
.![\[\angle PFQ = 180 - (\angle MPQ + \angle NQP) = 180 - (\angle MPB + \angle BPQ + \angle NQC - \angle PQC)\]](http://latex.artofproblemsolving.com/b/0/c/b0c5015551c614f809ca8892c56d51ef95446f29.png)
However noting that as 
is an isosceles trapezoid we find 
.
which is what we wished to show. 
.
,
.
are collinear. Then we have
are also collinear and we're done.
,
,
,
.
.![\[\angle LED=\angle LCB=\angle LPB,\]](http://latex.artofproblemsolving.com/b/2/2/b2207d77ff2962de8efaca2f5517d32f6308d3aa.png)
so 
are collinear. We can use a similar argument to show that 
are collinear, proving part (a) 
and 
are collinear
.
We can use a similar method to show that ![\[\angle HDF=\angle HDP=2\angle CRP,\]](http://latex.artofproblemsolving.com/a/1/a/a1a48b4dfc286af25c28cf2e057f1eaf1ee9f246.png)
![\[\angle HEF=\angle HEQ=2\angle BSQ=2\angle CRP,\]](http://latex.artofproblemsolving.com/8/0/e/80e11f3eb30d9dcc57246e86f50a8702fea1d848.png)
so 
Denote 
and 
The map 
is evidently projective, and the map 
is also projective, where the first map is induced by a rotation about 
It suffices to check these maps coincide for three cases of 
we have 
at which the result is obvious.
we have 
with 
at which the result is obvious.
we have 
from which we may deduce 
implying the result.
and 
be the reflections of 
and 
.
where 
because 
is constant, 
the center of 
because it's the intersection of the perpendicular bisector of 
respectively. As these altitudes always go through the same point at infinity, both of them different of course, we have that the line is 
Note that the 
so it has 
As 
we have that 
as well. Now, the maximum degree of 
where the 
is when 
then 
by an angle chase by showing that 
the reflection of 
This has 
We will work on line 
The cases are when 
and when 
cases, so we're don
(b) Let 
be the midpoint of 
,
,
so we have, 
and similarly 
.
and because 
,
after some simplification we get, 
and similarly we also get 
or
we use the complex number angle between two lines formula
and 
and by symmetry we also have 
we get 
and 
and we are done 
.
,
.
.
.
.
Therefore 
where the last step is due to arc 
equals arc 
which is 
.
.
and 
to be collinear.
which implies 
are collinear.
is an isosceles trapezium. Then: 
which implies 
