[/quote]
,
such that 
and 
are coprime if and only if 
?
,
be the projection from 
be a point on the extension of 
past 
.
be on the perpendicular bisector of 
such that ![\[\frac12\angle ALE=1.4\angle ABE+3.4\angle ACE-558^\circ\]](http://latex.artofproblemsolving.com/c/6/3/c63eeb053f1fa8a29bef93516814ae121382219e.png)
Let the reflection of 
be 
and 
,
and 
such that 
.
and 
intersect the circumcircles of 
and 
at 
and 
,
and 
intersect 
and 
at 
and 
.
intersect 
.
.
chessboard into 
non-overlapping rectangles subject to the following conditions:
is the number of white squares in the 
-
.
.
denote the determinant of a square matrix 
.![$g:\left[0, \frac{\pi}{2}\right] \rightarrow R$](http://latex.artofproblemsolving.com/5/3/4/534fcbbc20e62de2dfa2579cd0642663c9fc3380.png)
be the function defined by
where 
.
be a quadratic polynomial whose roots are the maximum and minimum values of the function 
and 
.
and orthocenter 
,
and that line 
is not parallel to line 
is perpendicular to 
.
be the points on the circumcircle of triangle 
is a diameter, and 
and 
are parallel. Let 
.
and 
are perpendicular.
and 
are perpendicular.
such that![\[(a^2-b^2)(b^2-c^2) = abc.\]](http://latex.artofproblemsolving.com/7/1/8/7188261c40a45f5e15533100432d36abbf515efd.png)
.
that 
is a prime.
sastifying 
such that 
be a monic polynomial with integer coefficient. And suppose there exist 4 distinct integer 
such that 
.
such that 
such that ![\[f(2x+y) + f(x+f(2y)) = f(x)f(y) - xy\]](http://latex.artofproblemsolving.com/0/6/2/0623b7c328af6d088cfc5dd288afe3e17a94ff29.png)
for all reals 
and 
.
to 
.
)
Y by Davi-8191, anantmudgal09, HWenslawski, megarnie, son7, Adventure10, Funcshun840
Let the excircle of triangle 
opposite the vertex 
be tangent to the side 
at the point 
. Define the points 
on 
and 
on 
analogously, using the excircles opposite 
and 
, respectively. Suppose that the circumcentre of triangle 
lies on the circumcircle of triangle . Prove that triangle is right-angled.
Proposed by Alexander A. Polyansky, Russia
opposite the vertex 
be tangent to the side 
at the point 
. Define the points 
on 
and 
on 
analogously, using the excircles opposite 
and 
, respectively. Suppose that the circumcentre of triangle 
lies on the circumcircle of triangle . Prove that triangle is right-angled.Proposed by Alexander A. Polyansky, Russia
lies on 
,
.
be the incircle of triangle 
are 
respectively.
cut each other at the Nagel point, so 
.
be the midpoint of arc 
,
,
.
be respectively the centers of 
are concurrent at a point 
.
and 
.
.
,
.
.
is a parallelogram.But note that 
so it is actually a rectangle, and the conclusion follows.![[asy]
unitsize(2.5cm);
void b() {
pointpen=black; pathpen=rgb(0.4,0.6,0.8); pointfontpen=fontsize(10); pen dd=linetype("4 8");
/* Define the excenter */
pair excenter(pair A=(0,0), pair B=(0,0), pair C=(0,0)) {
return extension(A,bisectorpoint(C,A,B),B,rotate(90,B)*bisectorpoint(A,B,C));
}
/* Draw points */
pair A=D("A",dir(115),N),
B=D("B",(-1,0),W),
C=D("C",(1,0),E),
I=D(incenter(A,B,C)),
Ia=D("I_a",excenter(A,B,C),S),
Ib=D("I_b",excenter(B,C,A),E),
Ic=D("I_c",excenter(C,A,B),W),
A1=D("A_1",foot(Ia,B,C),SSW),
B1=D("B_1",foot(Ib,C,A),E),
C1=D("C_1",foot(Ic,A,B),N),
Ma=D("M_a",midpoint(Ib--Ic),N),
Mb=D("M_b",midpoint(Ic--Ia),SW),
Mc=D("M_c",midpoint(Ia--Ib),SE),
V=D("V",circumcenter(Ia,Ib,Ic),SE);
/* Draw paths */
D(unitcircle,heavyblue);
D(circumcircle(B,C,V),linetype("2 2")+rgb(0.6,0,1));
D(circumcircle(B,C1,V),linetype("2 2")+rgb(0.6,0,1));
D(A--B--C--cycle);
D(Ia--Ib--Ic--cycle,gray(0.3)+linewidth(1));
D(A1--B1--C1--cycle);
D(I--Ia,dd+red); D(I--Ib,dd+red); D(I--Ic,dd+red);
D(V--Ia,dd+heavygreen); D(V--Ib,dd+heavygreen); D(V--Ic,dd+heavygreen);
}
b(); pathflag=false; b();
[/asy]](http://latex.artofproblemsolving.com/a/5/7/a57080c45ed62201c373ce31e09a5d7dc1774469.png)
be 
be 
,
,
;
.
and 
,
(SAS), thus 
,
is one of 
;
concur at the Bevan point 
of 
.
is the common perpendicular bisector of 
and 
,
.
is the circle on diameter 
,
.
,
![[asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
import graph; size(4.22cm);
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 = 2.24, xmax = 6.46, ymin = -0.88, ymax = 3.1; /* image dimensions */
/* draw figures */
draw((3.5,-0.32)--(3.5,2.62));
draw((3.5,-0.32)--(5.3,-0.32));
draw((3.5,2.62)--(5.3,-0.32));
draw(circle((4.4,1.15), 1.723629890666787));
draw((3.5,1.9736298906667877)--(4.102372300006038,1.6361252433234716));
draw((4.102372300006038,1.6361252433234716)--(4.653629890666787,-0.32));
draw((4.653629890666787,-0.32)--(3.5,1.9736298906667877));
draw((2.93,0.25)--(4.653629890666787,-0.32));
draw((2.93,0.25)--(3.5,1.9736298906667877));
draw((2.93,0.25)--(4.4,1.15));
draw((2.93,0.25)--(3.5,-0.32));
draw((2.93,0.25)--(5.3,-0.32));
draw((2.93,0.25)--(3.5,2.62));
draw((2.93,0.25)--(2.93,-0.32));
draw((2.93,-0.32)--(3.5,-0.32));
draw((2.93,0.25)--(3.5,0.25));
/* dots and labels */
dot((3.5,-0.32),linewidth(3.pt) + dotstyle);
label("$A$", (3.22,-0.62), NE * labelscalefactor);
dot((3.5,2.62),linewidth(3.pt) + dotstyle);
label("$B$", (3.22,2.78), NE * labelscalefactor);
dot((5.3,-0.32),linewidth(3.pt) + dotstyle);
label("$C$", (5.42,-0.62), NE * labelscalefactor);
dot((4.4,1.15),linewidth(3.pt) + dotstyle);
label("$O$", (4.48,1.26), NE * labelscalefactor);
dot((4.102372300006038,1.6361252433234716),linewidth(3.pt) + dotstyle);
label("$A_1$", (4.18,1.76), NE * labelscalefactor);
dot((4.653629890666787,-0.32),linewidth(3.pt) + dotstyle);
label("$B_1$", (4.58,-0.78), NE * labelscalefactor);
dot((3.5,1.9736298906667877),linewidth(3.pt) + dotstyle);
label("$C_1$", (3.08,1.84), NE * labelscalefactor);
dot((2.93,0.25),linewidth(3.pt) + dotstyle);
label("$I$", (2.78,-0.08), NE * labelscalefactor);
dot((3.5,0.25),linewidth(3.pt) + dotstyle);
label("$H$", (3.58,0.36), NE * labelscalefactor);
dot((2.93,-0.32),linewidth(3.pt) + dotstyle);
label("$F$", (2.76,-0.64), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]](http://latex.artofproblemsolving.com/6/1/6/616881e7aadfd019b56fced4717ae74a4a65650e.png)
be the circumcenter of 
and 
(set 
)
hence 
(s.a.s), thus 
,
w.r.t 
,![\[ IC^2\cdot BA_1+IB^2\cdot CA_1=BC(IA_1^2+BA_1\cdot CA_1) \]](http://latex.artofproblemsolving.com/0/c/a/0caf3a073d3a928a907036e20a5a4ad63bc3927a.png)
or 
(since 
,![\[ IB^2-IA_1^2=\frac{a^2-(b-c)^2}{4}\quad (1) \]](http://latex.artofproblemsolving.com/d/9/5/d95a90e563ff38cf178c1945698740789bae9ca9.png)
Let 
be the perpendicular projection of 
,
,
.
Since 
,
and 
,
Done.
as the circumcircle of 
as the circumcircle of 
.
.
as the 
and 
,
.
.
.
is the external angle bisector of 
colinear.
.
.
and the 
.
lies on 
so 
as desired.
and 
the midpoints of the sides 
resp. of the excentral triangle 
which are in the same time the midpoints of the arcs 
of the circumcircle of 
is rotation with center ,the second intersection of 
and on the bisector of 
so 
similarly we get 
hence ,since the center of the circumcircle of 
is on the circumcircle of 
but 
thus 
similarly 
thus 
so 
.
are the midpoints of arcs 
respectively and 
to 
and thus, points 
are concyclic. Similarly for the other two vertices. This gives that in fact 
is the perpendicular bisector of 
and 
of 
.
,
and similarly 
.
and so 
and so, 
be the incenter, circumcenter of 
be the complex numbers corresponding to points 
respectively. Scale the diagram such that 
on the unit circle such that 
and such that 
.
be the tangency point of the incircle with 
.
.
and 
respectively.
.
.
denote the midpoint of line segment 
be the intersection of 
and 
is parallel to 
;
.
and 
.
are collinear, we get that 
.
are collinear, we get a similar equation. We find 
.
;
.
.
is never equal to 0. Note that 
.
,
.
,
.
is a nonnegative real number for any complex number 
.
be the intersection points of 
with 
is the center of the spiral similarity that takes 
to 
and 
,
and 
.
.
and on 
is the perpendicular bisector of 
and 
is the perpendicular bisector of 
.
which implies 
.![[asy]
unitsize(100);
pen n_purple = rgb(0.7,0.4,1),
n_blue = rgb(0,0.6,1),
n_green = rgb(0,0.4,0),
n_orange = rgb(1,0.4,0.1),
n_red = rgb(1,0.2,0.4);
pair A = dir(130), B = dir(180), C = dir(0), V = -incenter(A, B, C), A1 = foot(V, B, C), B1 = foot(V, C, A), C1 = foot(V, A, B), O = circumcenter(A1, B1, C1);
draw(unitcircle, n_blue);
draw(circumcircle(A, B1, C1), dashed + gray(0.6));
draw(O--A^^B1--C1, gray(0.6));
draw(A--B--C--cycle);
draw(O--A1^^O--B1^^O--C1, n_orange);
draw(O--B^^O--C, n_green);
dot(A^^B^^C^^A1^^B1^^C1^^O);
label("$A$", A, dir(A));
label("$B$", B, dir(B));
label("$C$", C, dir(C));
label("$A_1$", A1, dir(270));
label("$B_1$", B1, dir(80));
label("$C_1$", C1, rotate(90) * dir(A - B));
label("$O$", O, dir(O));
[/asy]](http://latex.artofproblemsolving.com/4/4/b/44bb33ff00cf9acdf5ebf3daf554cccb4b1ee42d.png)
,
,
.
,
,
.
,
,
)
,
,
and 
we get![\[\angle BOC_1 = \angle COB_1 \quad \text{or} \quad \angle BOC_1 + \angle COB_1 = 180^{\circ}.\]](http://latex.artofproblemsolving.com/0/0/8/00893beb0d5c3606cd9754158f7d877ef8afe837.png)
Doing a similar thing in 
and 
gives![\[\angle AOB_1 = \angle BOA_1 \quad \text{or} \quad \angle AOB_1 + \angle BOA_1 = 180^{\circ}.\]](http://latex.artofproblemsolving.com/8/d/8/8d8f3ba71ce4cb8a65f5dba2e1cff89cd9740712.png)
Suppose that both 
and 
,![\[\angle BOC_1 + \angle COB_1 + \angle AOB_1 + \angle BOA_1 = 360^{\circ}\]](http://latex.artofproblemsolving.com/f/c/b/fcbc4218b4a663b802034991d3e7c7ba0b219502.png)
or![\[(\angle A_1OB + \angle BOC_1) + (\angle AOB_1 + \angle B_1OC) = \angle A_1OC_1 + \angle AOC = 360^{\circ}.\]](http://latex.artofproblemsolving.com/9/0/8/908c8cb568372b7b814a22019bf2a04ca378c4ca.png)
This is an obvious contradiction since 
.
or 
.
and 
are congruent, so 
.
,
,
.
means that 
,
.
,
,
,
.
,![\[\frac{\sin AB_1C_1}{\sin AC_1B_1} = \frac{AC_1}{AB_1}\]](http://latex.artofproblemsolving.com/e/0/6/e066852d2e98c2b5d8945159413be6d691c3b72d.png)
and by the Ratio Lemma in 
with cevian 
,![\[\frac{AC_1}{AB_1} = \frac{CA_1}{BA_1} = \frac{CO}{BO} \cdot \frac{\sin COA_1}{\sin BOA_1},\]](http://latex.artofproblemsolving.com/0/6/9/0697c1280b9735caf3ee2d82b1ddf84582cd1045.png)
thus 
since 
,![\[\angle AB_1C_1 = \angle COA_1 \quad \text{or} \quad \angle AB_1C_1 + \angle COA_1 = 180^{\circ}.\]](http://latex.artofproblemsolving.com/a/c/9/ac97182195c8e1ecaae664e91562a0d587e676d4.png)
However, if 
,
,
,![\[\angle AOC_1 + \angle COA_1 < \angle AOB + \angle COB = \angle AOB < 180^{\circ}.\]](http://latex.artofproblemsolving.com/3/3/b/33bce852fd1493404c26940d7ae7dd88e3375663.png)
Thus, ![\[\angle BAC = \angle BOC = \angle BOA_1 + \angle COA_1 = \angle AC_1B_1 + \angle AB_1C_1 = 180^{\circ} - \angle B_1AC_1,\]](http://latex.artofproblemsolving.com/9/e/9/9e95bee1cb9d32a04200c29906c8756f436c92a7.png)
from which 
,
be the largest of all thereupon we can easily see that 
is the one which is more than 90. afterward we claim the mid point of arc 
.
,
is concyclic and 
and this the beginning of complex. I just explain it briefly and would go into details. take 
the unit circle.
and 
and then we can calculate 
has to be real
is real and I think the rest is straightforward to give us 
be the orthic triangle of 
.
.
be the medial triangle of 
lie on a circle. All other such results hold cyclically.
lies on 
,
to 
As 
,
.
,
is the perpendicular bisector of 
.
lies on 
.
.
follows 
,
.
which must actually be a spiral congruence since 
,
.![[asy] pair A = dir(110); pair B = dir(180); pair C = dir(0); pair X = dir(90); pair Y = dir(235); pair Z = dir(325);
pair A_1 = foot(Y+Z-X, B, C); pair B_1 = foot(Z+X-Y, C, A); pair C_1 = foot(X+Y-Z, A, B);
draw(unitcircle, grey); draw(C_1--X--B_1, red); draw(A--B--C--cycle, grey); draw(A_1--B_1--C_1--cycle, heavycyan); draw(Y--X--Z, orange); draw(circumcircle(X, B_1, C_1), dotted+grey); draw(X--A_1, red+dashed);
dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$X$", X, dir(X)); dot("$Y$", Y, dir(Y)); dot("$Z$", Z, dir(Z)); dot("$A_1$", A_1, dir(270)); dot("$B_1$", B_1, dir(B_1)); dot("$C_1$", C_1, dir(C_1));
/* TSQ Source:
A = dir 110 B = dir 180 C = dir 0 X = dir 90 Y = dir 235 Z = dir 325
A_1 = foot Y+Z-X B C R270 B_1 = foot Z+X-Y C A C_1 = foot X+Y-Z A B
unitcircle 0.1 grey / grey C_1--X--B_1 red A--B--C--cycle grey A_1--B_1--C_1--cycle 0.1 cyan / heavycyan Y--X--Z orange circumcircle X B_1 C_1 dotted grey X--A_1 red dashed
*/ [/asy]](http://latex.artofproblemsolving.com/3/0/1/3019331c4211a0e716de795440f4bbe70fd6f56d.png)
.
(since 
)
,
and 
are exactly the perpendicular bisectors of 
which solves the problem.
is cyclic (with diameter 
)
is cyclic; a calculation then gives 
.
.
as the midpoint of 
on 
and 
and 
.![[asy]
size(9cm);
defaultpen(fontsize(9pt));
pair A=(32.5,53.3), B=(0,0), C=(120,0), D=(80,0), E=(100.8,11.7), F=(11.7,19.2), K=(60,60), L=(28.8,-51.2), M=(45.9,9.6);
dot(M);
draw(A--B--C--cycle, linewidth(0.5));
draw(circle((60,0),60), linewidth(0.5));
draw(F--K--B, linewidth(0.5));
draw(E--K--C, linewidth(0.5));
draw(E--F, linewidth(0.4)+grey);
draw(F--D, linewidth(0.4)+grey);
draw(circumcircle(D,E,F), linewidth(0.8)+dashed);
draw(F--L--A, linewidth(0.5));
draw(D--L--C, linewidth(0.5));
real xmin=-100, ymin=-110, xmax=200, ymax=80;
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
draw(K--L, linewidth(0.7)+dotted);
dot("$A$", A, (0,1));
dot("$B$", B, (-1,0));
dot("$C$", C, (1,0));
dot("$A_1$", D, (0.5,-1));
dot("$B_1$", E, (0,-1));
dot("$C_1$", F, (-0.5,0));
dot("$K_A$", K, (0,1));
dot("$K_B$", L, (-0.5,-1));
[/asy]](http://latex.artofproblemsolving.com/7/8/5/785af11bd6caecb321ba2b4898c79c7afd43e81f.png)
and 
,
and 
respectively, so
Now, since 
for some value 
.
or 
(and 
by symmetry). Finally, the desired collinearity occurs iff
which can be shown to be equivalent to 
with a few minutes of omitted expansion and row/column operations.
![[asy]
import geometry;
size(8cm,0);
defaultpen(fontsize(9pt));
pair B = (-25,0), C = (25,0), N = (0,25), A = (-7,24);
triangle t = triangle(A,B,C);
drawline(t);
circle c1 = excircle(t.BC);
circle c2 = excircle(t.AC);
circle c3 = excircle(t.AB);
clipdraw(c1,bp+green);
clipdraw(c2,bp+green);
clipdraw(c3,bp+green);
dot("$B$", B, (-1,0));
dot("$C$", C, (1,0));
dot("$N_A$", N, (0,1));
dot("$A$", A, (0,1));
draw(circumcircle(A,B,C));
draw(box((-60,-20), (60,50)), invisible);
draw(extouch(t));
dot("$A_1$",extouch(t).A);
dot("$B_1$",extouch(t).B);
dot("$C_1$",extouch(t).C);
point V = circumcenter(excenter(t.AB),excenter(t.BC),excenter(t.CA));
dot("$V$", V);
point NN = (0,25);
line l = perpendicular(NN, line(V,NN));
point VV = reflect(l)*V;
dot("$V'$", VV);
point BB = (-25,0);
point CC = (25,0);
point BBB = reflect(l)*BB;
dot("$B'$", BBB);
draw(circumcircle(BB,CC,V));
dot("$I_C$",excenter(t.AB));
dot("$I_B$",excenter(t.AC));
[/asy]](http://latex.artofproblemsolving.com/b/6/5/b65d43d22861dc8b44ed163c29c8cdba2dd95b20.png)
be the midpoint of major arc 
.
by SAS, it follows that 
and so 
and on the perpendicular bisector of 
,
.
are isogonal conjugates WRT 
is the reflection of 
,
.
are actually reflections across 
)![\[\angle VBC=\angle V'BA=\angle VB'A=\angle VB'C\]](http://latex.artofproblemsolving.com/5/4/3/543c4a2f1f1c32ad5dce4b088c7e3ebce02c3297.png)
so 
is cyclic. This means 
.
has three right angles so it must be a rectangle, done.
denote the midpoints of arcs 
containing the respective points 
.
,
,
.
are the perpendicular bisectors of segments 
respectively. Thus ![\[\measuredangle B_1A_1C_1 = \measuredangle(B_1A_1, A_1C_1) = \measuredangle(N_bP, N_cP) = \measuredangle N_bPN_c = \measuredangle N_bN_aN_c\]](http://latex.artofproblemsolving.com/0/5/8/058cd2ef65dda074d7ed0405268090d7b045730a.png)
Denoting by 
we have ![\[\measuredangle N_bN_aN_c = \measuredangle (N_bN_a, N_aN_c) = \measuredangle(CI, IB) = \measuredangle CIB \]](http://latex.artofproblemsolving.com/6/3/f/63faf2fd7f4eac4e66b91961532ff35385f16711.png)
Thus if we denote by 
the angles of the triangle, we have ![\[\angle B_1A_1C_1 \in \{90^{\circ}+\angle A/2, 90^{\circ}-\angle A/2\}\]](http://latex.artofproblemsolving.com/0/8/1/08192ee683660b2fbc413e560d2face503d6a163.png)
A similar condition holds true for the other angles. Now ![\[180^{\circ} = \angle B_1A_1C_1+\angle A_1C_1B_1+\angle C_1B_1A_1 \geq (90^{\circ}-\angle A/2)+(90^{\circ}-\angle B/2)+(90^{\circ}-\angle C/2) = 180^{\circ}\]](http://latex.artofproblemsolving.com/6/b/c/6bca860d5bdb1cedca070147305e03e8fca9d949.png)
It follows that equality holds everywhere and the angles of 
and 
respectively.
,
,
denote the side lengths of the triangle, and 
the semiperimeter. Note that 
both lie on the perpendicular bisector of 
.
meet ![\[\frac{CB_1}{CA_1} = \frac{CP}{CB}\]](http://latex.artofproblemsolving.com/1/7/0/17043018ad569cc584d340b82bed41205247338f.png)
or ![\[\frac{s-a}{s-b} = \frac{c}{a+b}\]](http://latex.artofproblemsolving.com/2/6/5/26530a15015d1979d386c7896d4791f2a7e6f4f5.png)
Cross multiplication along with simple algebra reduces the above to 
,
,
,
.
as 
.
,
.
.
we should have
Let 
-
,
,
.
or 
,
be the largest angle in 
.
,
,
.
;
,
and 
similarly to 
instead. Note that 
and 
.![\[\angle B = \angle A_1O_1C_1 = 2 \angle O_2O_1O_3.\]](http://latex.artofproblemsolving.com/f/f/9/ff96fe1e72b0d8ca3b109101cfd1ef48918868b4.png)
denote the measure of arc 
caused by the inscribed angle theorem will be omitted.![\[2 \angle O_2O_1O_3 = 2 m(O_1O_2) = 2( m(AC) - m(AO_2) - m(CO_3)).\]](http://latex.artofproblemsolving.com/1/f/2/1f2f75a117b4ad20f3669088edfefdc4ec3e6572.png)
has measure 
because the angle bisector of 
has measure 
.![\[2 \angle O_2O_1O_3 = 2 \left( \angle B - \frac{\angle A}{2} - \frac{\angle C}{2} \right) = \angle B \]](http://latex.artofproblemsolving.com/d/1/1/d11ce0b98c3942aeac5f8ffa82030611e69e6ab3.png)
![\[\implies \angle A + \angle C = \angle B.\]](http://latex.artofproblemsolving.com/8/6/7/86764c04d76a9db626a2df298b5e876c7e54ae5d.png)
.
and 
respectively, and let the circumcenter of 
intersect 
intersect 
at 
,
.
and 
,
,
,
,
and 
to 
Notice that 
and symmetric. Now construct 
to be the intersection of perpendicular bisectors of 
We have 
directly congruent by SSS so 
so it is on 
Thus it must be the midpoint of arc 
Define 
similarly. Let 
is the perpendicular bisector of 
and define 
lies on 
coincide. However 
are distinct, and if 
impossible. Thus we may assume that 
coincide. This implies that 
are the perpendicular bisectors of 
respectively.
intersect 
Since 
bisects 
we have that 
and since 
share an arc midpoint we have 
Similarly 
so 
However from the spiral similarity we have 
so 
as desired.
are the midpoints of arc 
and 
as well as 
and 
.![\[\triangle PAB_1 \cong \triangle PBA_1, \quad \triangle P'AB_1 \cong \triangle P'BA_1\]](http://latex.artofproblemsolving.com/1/b/9/1b9838ab841d772c8de3736f0c90e1db579b37ce.png)
and 
are the perpendicular bisectors of ![\[\angle C = \angle APB = \angle A_1PB_1 = 2 \angle NPM = 180 - \angle C \implies \angle C = 90. \quad \blacksquare\]](http://latex.artofproblemsolving.com/9/c/7/9c7aa8c67fe4dd2d56b88cab28865793600e8b13.png)
by X being the midpoint, also 
since ABCX is cyclic 
,
is cyclic. Now we have that 
.
.
meets 
.
is the midpoint of arc 
,
as well, so 
and 
are congruent and thus 
,
.
or 
and either 
or 
.
is perpendicular to 
.
,
and we are done.
are parallel to the internal bisectors): we know 
cyclic so 
so 
.
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.
point circle properties. We look at this problem from the excentral triangle, let them be 
Let 
define 
similarly. WLOG let 
we have that 
is the perpendicular bisector of both of 
so 
As 
by Reim we see that 
is cyclic. Therefore we get that 
so we're don
.
(to show variability) of 
due to congruency of two triangles of the form 
where 
and 
.
the midpoint of 
.
lie on the circle with diameter 
so 
°
are meant to say "perpendicular bisector of 
be the point where the 
e
.
.
.
and
Let 
be the points where the incircle touches 
and 
.
Thus, clearly 
Thus, 
which means 
lies on the perpendicular bisector of 
.
we have
We then also have,
Then, note that 
which gives us that 
.
Thus, 
The other cases are entirely similar (with 
and 
being 
)
