[/quote]
,
be a fixed integer greater than 
.
,
,
,
is defined as follows:![\[x_i = \begin{cases}2^i&\text{if }0\leq i \leq m - 1;\\\sum_{j=1}^mx_{i-j}&\text{if }i\geq m.\end{cases}\]](http://latex.artofproblemsolving.com/3/c/6/3c6879be01c394c27712d3acf43e22f1be17db49.png)
Find the greatest 
for which the sequence contains 
be non-negative real numbers such that 
Prove that
.
be the reflection of 
in 
and let 
the reflection of 
.
be the midpoint of segment 
.
.
,
,
be positive real numbers. Prove the inequality![\[ \frac{1}{a\left(b+1\right)}+\frac{1}{b\left(c+1\right)}+\frac{1}{c\left(a+1\right)}\geq \frac{3}{1+abc}. \]](http://latex.artofproblemsolving.com/d/a/5/da5096e4c732244baf08effc43b5d086444e2c62.png)
as follows: 
,
,
,
is the least positive integer such that there is exactly one way to represent 
where 
.
,
,![\[\min_{1\le k\le n}\frac{a_{k+1}}{a_k}\le 1+\frac{c\ln n}n\text{ }?\]](http://latex.artofproblemsolving.com/a/3/2/a32c044b4bb701674aa885f1b0737f2a4eaa2d85.png)
and 
be respectively the midpoints of the arcs 
and 
of 
is not less than the inradius of triangle 
of the form 
with 
,
Determine all ordered pairs of polynomials 
with integer coefficients such that 
.
be a concave function and 
be a continuous function . If 
for all 
prove that 
is a second degree polynomial.
is a square of an integer.
where 
with orthocenter 
.
intersects 
at 
intersects 
.
intersect 
.
be the reflection of 
.
intersects 
different from 
and 
intersect on the circumcircle of triangle 
.
and 
is a circle tangent to 
The tangent
If 
are the contact points of 
respectively, prove that 
Y by Carpemath, Purple_Planet, itslumi, Muaaz.SY, 554183, Adventure10, Mango247, Rounak_iitr
Let 
be a scalene triangle with orthocenter 
and circumcenter 
. Let 
be the midpoint of 
and let 
be on line 
with 
. Let 
be the foot of the altitude from onto line 
. Prove that the midpoint of 
lies on the nine-point circle* of 
.
*The nine-point circle of is the unique circle passing through the following nine points: the midpoint of the sides, the feet of the altitudes, and the midpoints of , 
, and 
.
Proposed by Zack Chroman
be a scalene triangle with orthocenter 
and circumcenter 
. Let 
be the midpoint of 
and let 
be on line 
with 
. Let 
be the foot of the altitude from onto line 
. Prove that the midpoint of 
lies on the nine-point circle* of 
.*The nine-point circle of
is the unique circle passing through the following nine points: the midpoint of the sides, the feet of the altitudes, and the midpoints of , 
, and 
.Proposed by Zack Chroman
This post has been edited 1 time. Last edited by whatshisbucket, Jun 29, 2018, 1:04 AM
denotes a directed angle considered modulo 
.
denote the nine-point circle of 
is the midpoint of 
.
are concyclic. Call this circle 
and 
.
both lie on 
,
bisects 
.
is a parallelogram. Vectorially, this is equivalent to 
.
as desired. 
.
and 
and 
.
of 
is bisected by 
and 
of this triangle. Hence 
is isosceles. In particular, 
.
.
.
,
.
.
,
.
.
is a diamater of the 
,
,
,
.
:
is sent to line 
,
,
of 
,
,
and we are done.
,
,
we have that 
is clearly the 
.
. We have that 
. Thus 
is an isosceles trapezium (since 
is cyclic). However, 
.
is the midpoint of 
)
are cocyclic. However, one can easily prove that 
be the midpoint of 
be the orthic triangle of ![[asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(20cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.5) + fontsize(15); defaultpen(dps); /* default pen style */
pen dotstyle = green; /* point style */
real xmin = -16.62, xmax = 9.58, ymin = -4.7, ymax = 7.34; /* image dimensions */
pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451);
draw((0.6012190533078552,6.245840367777759)--(0.7153786855300965,5.987059421085614)--(0.974159632222241,6.101219053307855)--(0.86,6.36)--cycle, linewidth(1.6));
draw((2.1473823505951684,4.17200291842619)--(2.215332324429749,3.8974436332874793)--(2.48989160956846,3.96539360712206)--(2.421941635733879,4.239952892260771)--cycle, linewidth(1.6));
draw((2.8721950772975338,-1.3203223722714752)--(2.592517449569009,-1.2781272949739417)--(2.5503223722714754,-1.5578049227024662)--(2.83,-1.6)--cycle, linewidth(1.6));
/* draw figures */
draw((0.86,6.36)--(3.222841081768753,1.0038312727493026), linewidth(2) + wrwrwr);
draw((-12.083581619768571,0.6500181167155676)--(-2.34,-0.82), linewidth(2) + wrwrwr);
draw((-1.2065428110900374,1.7231945676167293)--(8,-2.38), linewidth(2) + wrwrwr);
draw((-12.083581619768571,0.6500181167155676)--(0.86,6.36), linewidth(2) + wrwrwr);
draw((0.86,6.36)--(-0.2703584479813297,-1.132247661619838), linewidth(2) + wrwrwr);
draw((-12.083581619768571,0.6500181167155676)--(2.421941635733879,4.239952892260771), linewidth(2) + wrwrwr);
draw((2.421941635733879,4.239952892260771)--(3.222841081768753,1.0038312727493026), linewidth(2) + wrwrwr);
draw((3.0970212111124376,3.621685520290938)--(-2.34,-0.82), linewidth(2) + wrwrwr);
draw((0.86,6.36)--(2.83,-1.6), linewidth(2) + wrwrwr);
draw((2.83,-1.6)--(3.222841081768753,1.0038312727493026), linewidth(2) + wrwrwr);
draw((0.86,6.36)--(-2.34,-0.82), linewidth(2) + wrwrwr);
draw((-2.34,-0.82)--(8,-2.38), linewidth(2) + wrwrwr);
draw((8,-2.38)--(0.86,6.36), linewidth(2) + wrwrwr);
/* dots and labels */
dot((0.86,6.36),dotstyle);
label("$A$", (0.56,6.78), NE * labelscalefactor);
dot((-2.34,-0.82),dotstyle);
label("$B$", (-2.7,-1.36), NE * labelscalefactor);
dot((8,-2.38),dotstyle);
label("$C$", (7.92,-2.9), NE * labelscalefactor);
dot((3.222841081768753,1.0038312727493026),linewidth(4pt) + dotstyle);
label("$O$", (3.46,1.12), NE * labelscalefactor);
dot((-12.083581619768571,0.6500181167155676),linewidth(4pt) + dotstyle);
label("$T$", (-12.22,0.86), NE * labelscalefactor);
dot((-0.2703584479813297,-1.132247661619838),linewidth(4pt) + dotstyle);
label("$D$", (-0.3,-1.82), NE * labelscalefactor);
dot((3.0970212111124376,3.621685520290938),linewidth(4pt) + dotstyle);
label("$E$", (3.3,3.7), NE * labelscalefactor);
dot((-1.2065428110900374,1.7231945676167293),linewidth(4pt) + dotstyle);
label("$F$", (-1.66,1.74), NE * labelscalefactor);
dot((0.0743178364624929,1.152337454501396),linewidth(4pt) + dotstyle);
label("$H$", (0.32,0.46), NE * labelscalefactor);
dot((0.46715891823124644,3.756168727250698),linewidth(4pt) + dotstyle);
label("$P$", (0.62,3.26), NE * labelscalefactor);
dot((2.421941635733879,4.239952892260771),linewidth(4pt) + dotstyle);
label("$X$", (1.82,4.48), NE * labelscalefactor);
dot((1.4445502769825629,3.998060809755734),linewidth(4pt) + dotstyle);
label("$M$", (0.76,4.1), NE * labelscalefactor);
dot((2.83,-1.6),linewidth(4pt) + dotstyle);
label("$M_A$", (2.78,-2.18), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]](http://latex.artofproblemsolving.com/5/7/b/57b509ea47e09d433a34e8eb226967694ba3fa4d.png)
,
is tangent to 
.
lie on a circle.
We can redefine 
.
and 
is the perpendicular bisector of 
is perpendicular to 
is the orthocenter of 
,
lies on 
lies on the nine-point circle of 
be the midpoint of 
It is well known that 
and 
so 
Also, 
so 
Note that pentagon 
and 
so 
Then 
is the perpendicular bisector of 
so 
as desired.
H
the midpoint of 
.
.
and 
,
.
is the midpoint of 
.
be the reflection of 
.
is a parallelogram. We use complex numbers: let 
,
,
,
.
.
.
is cyclic. From claim #1, 
(as 
is perpendicular to 
,
.
and 
,
is cyclic. As 
,
is cyclic. Thus the points 
;
lie on a single circle, as desired. 
in the plane are concyclic, on a circle with center 
,
is an isosceles trapezoid. Obviously their perpendicular bisectors must be the same line, or 
and 
so we get 
.
is an isosceles trapezoid. We have 
(see claim #3), but also 
(from claim #1). Thus 
,
.
,
.
has length equal to the circumradius of 
.
is isosceles, with base 
.
-- but as 
in the plane, 
if and only if 
and 
be the center of the nine point circle
be the reflection of 
and that 
and ![[asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(13cm);
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 = -11.72, xmax = 11.4, ymin = -7.6, ymax = 8.48; /* image dimensions */
pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen sexdts = rgb(0.1803921568627451,0.49019607843137253,0.19607843137254902); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961);
/* draw figures */
draw((-2.26,4.08)--(-3.2,-2.12), linewidth(0.5) + wrwrwr);
draw((-2.26,4.08)--(6.13,-2.34), linewidth(0.5) + wrwrwr);
draw((xmin, -0.023579849946409406*xmin-2.19545551982851)--(xmax, -0.023579849946409406*xmax-2.19545551982851), linewidth(0.5) + wrwrwr); /* line */
draw(circle((1.5254779442162991,0.33481463426398006), 5.325059313272605), linewidth(0.5) + wrwrwr);
draw((xmin, 42.409090909090956*xmin + 99.92454545454557)--(xmax, 42.409090909090956*xmax + 99.92454545454557), linewidth(0.5) + wrwrwr); /* line */
draw((1.5254779442162991,0.33481463426398006)--(-2.26,4.08), linewidth(0.5) + wrwrwr);
draw((xmin, 1.0107585004600594*xmin + 6.364314211039734)--(xmax, 1.0107585004600594*xmax + 6.364314211039734), linewidth(0.5) + wrwrwr); /* line */
draw((xmin, 0.5903328050713154*xmin + 2.885039619651347)--(xmax, 0.5903328050713154*xmax + 2.885039619651347), linewidth(0.5) + wrwrwr); /* line */
draw((0.014828758983379346,2.893793522537732)--(1.5254779442162991,0.33481463426398006), linewidth(0.5) + wrwrwr);
draw((1.5254779442162991,0.33481463426398006)--(1.465,-2.23), linewidth(0.5) + rvwvcq);
draw((-8.27559930220558,-2.0003181300658923)--(-2.4927944304726974,-5.792600165046676), linewidth(0.5) + rvwvcq);
draw((-2.4927944304726974,-5.792600165046676)--(1.465,-2.23), linewidth(0.5) + rvwvcq);
draw((-2.346158243128199,0.4261072346086519)--(1.5254779442162991,0.33481463426398006), linewidth(0.5) + wrwrwr);
draw((-2.3204779442162993,1.51518536573602)--(1.465,-2.23), linewidth(0.5) + wrwrwr);
draw((-2.380955888432599,-1.0496292685279605)--(1.5254779442162991,0.33481463426398006), linewidth(0.5) + linetype("4 4") + wrwrwr);
draw((-1.15282459261646,2.204489444136876)--(1.465,-2.23), linewidth(0.5) + linetype("4 4") + wrwrwr);
draw((-8.27559930220558,-2.0003181300658923)--(1.5254779442162991,0.33481463426398006), linewidth(0.5) + rvwvcq);
draw(circle((-0.4277389721081497,-0.35740731713199014), 2.6625296566363024), linewidth(0.8) + linetype("4 4") + sexdts);
draw(circle((-3.3750606789946396,-0.8327517479009561), 5.037706824969461), linewidth(0.8) + linetype("4 4") + dtsfsf);
/* dots and labels */
dot((-2.26,4.08),dotstyle);
label("$A$", (-2.18,4.28), NE * labelscalefactor);
dot((-3.2,-2.12),dotstyle);
label("$B$", (-3.12,-1.92), NE * labelscalefactor);
dot((6.13,-2.34),dotstyle);
label("$C$", (6.22,-2.14), NE * labelscalefactor);
dot((-2.4066361873444984,-2.138707399655328),linewidth(4pt) + dotstyle);
label("$D$", (-2.32,-1.98), NE * labelscalefactor);
dot((1.5254779442162991,0.33481463426398006),linewidth(4pt) + dotstyle);
label("$O$", (1.6,0.5), NE * labelscalefactor);
dot((-8.27559930220558,-2.0003181300658923),linewidth(4pt) + dotstyle);
label("$T$", (-8.2,-1.84), NE * labelscalefactor);
dot((-2.380955888432599,-1.0496292685279605),linewidth(4pt) + dotstyle);
label("$H$", (-2.3,-0.88), NE * labelscalefactor);
dot((-2.3204779442162993,1.51518536573602),linewidth(4pt) + dotstyle);
label("$P$", (-2.24,1.68), NE * labelscalefactor);
dot((0.014828758983379346,2.893793522537732),linewidth(4pt) + dotstyle);
label("$X$", (0.1,3.06), NE * labelscalefactor);
dot((-1.15282459261646,2.204489444136876),linewidth(4pt) + dotstyle);
label("$K$", (-1.08,2.36), NE * labelscalefactor);
dot((1.465,-2.23),linewidth(4pt) + dotstyle);
label("$M$", (1.54,-2.08), NE * labelscalefactor);
dot((-2.4927944304726974,-5.792600165046676),linewidth(4pt) + dotstyle);
label("$P'$", (-2.42,-5.64), NE * labelscalefactor);
dot((-2.346158243128199,0.4261072346086519),linewidth(4pt) + dotstyle);
label("$L$", (-2.26,0.58), NE * labelscalefactor);
dot((-0.42773897210815004,-0.3574073171319902),linewidth(4pt) + dotstyle);
label("$N_9$", (-0.34,-0.2), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]](http://latex.artofproblemsolving.com/9/e/b/9ebbda1c16925d1d312df62e072f508603b25a7c.png)
Therefore, 
is cyclic, but as 
,
is cyclic as well also 
so 
is cyclic too. Hence, the points 
are all concyclic.
Which means 
And we are done. 
![[asy]
import geometry;
import olympiad;
unitsize(3cm);
pair A = dir(135), B = dir(210), C = -1/B;
triangle tri = triangle((point) A, (point) B, (point) C);
line t = tangents(circumcircle(tri), A)[0];
pair T = intersectionpoint(t, tri.BC);
pair O = circumcenter(tri);
pair H = orthocentercenter(tri);
pair P = (A+H)/2;
pair X = foot(O, P, T);
pair M = (B+C)/2;
pair A_ = -A;
pair N = (O + H)/2;
pair Y = (P+X)/2;
draw(A--T--B); draw(tri); draw(A--M); draw(circle(M, abs(M-P)), 0.8blue); draw(T--X);
draw(A--H--A_, dotted); draw(M--P^^A--A_, linetype("0 2"));
draw(circumcircle(tri)); draw(circumcircle((point) O, (point) T, (point) A), 0.9red);
draw(circle(N, abs(N - P)), 0.8green);
dot(A^^B^^C); dot(T^^P^^X); dot(O^^H); dot(M); dot(Y); dot(N);
label("A", A, A);
label("B", B, B*dir(10));
label("C", C, C);
label("T", T, dir(180));
label("P", P, 2dir(-10));
label("H", H, dir(45));
label("O", O, dir(180));
label("X", X, dir(60));
label("M", M, dir(-60));
label("N$_9$", N, dir(-90));
label("Y", Y, dir(-30));
[/asy]](http://latex.artofproblemsolving.com/c/c/7/cc7ed1a4a3c9552b0fe06d2faed7a9fe65c9ecbd.png)
are concyclic.![\[\measuredangle OAT = \measuredangle OXT = \measuredangle OMT = 90^{\circ}\]](http://latex.artofproblemsolving.com/3/2/b/32b99ec59554b994a030b82093f7be6803eaaabd.png)
.
and 
,
,
and 
,
refers to the circle with center 
.
.
,
.
.
and 
;
and 
so the conclusion follows readily.
and 
,
.
and 
.
.
where the second and third equalities follow from the previous lemmas.
.
,
maps circumcircle to nine-point circle. Suppose 
such that 
or 
.
is a parallelogram. We need to show that 
,![\[-\frac{t-a}{a}=\frac{\overline{t}-\frac{1}{a}}{\frac{1}{a}}=a\overline{t}-1\implies 2a=t+a^2\overline{t}.\]](http://latex.artofproblemsolving.com/0/c/7/0c722f7faab24f7d7be81c0af94991a50777df47.png)
As 
are collinear, ![\[\frac{t-b}{b-c}=\frac{\overline{t}-\frac{1}{b}}{\frac{1}{b}-\frac{1}{c}}\implies b+c=t+bc\overline{t}.\]](http://latex.artofproblemsolving.com/a/e/2/ae2bb39b7ca46d9d71c660a63b70fa6fa393a562.png)
Combining the results gives ![\[\overline{t}=\frac{2a-(b+c)}{a^2-bc},t=\frac{a(a(b+c)-2bc)}{a^2-bc}.\]](http://latex.artofproblemsolving.com/6/d/6/6d69d1aa85572dc812dd4d683f3e9fd15189f3a1.png)
As ![\[p=\frac{a+h}{2}=\frac{a+a+b+c}{2}=\frac{2a+b+c}{2},\overline{p}=\frac{2bc+a(b+c)}{2abc}.\]](http://latex.artofproblemsolving.com/7/a/4/7a466b3fef71559b35b8ad674d37e12f2175d66b.png)
As 
,![\[x=\frac{\overline{p}t-p\overline{t}}{2(\overline{p}-\overline{t})}.\]](http://latex.artofproblemsolving.com/6/1/a/61a0e56e54f0ba64edf4e69f097c14de5241e39a.png)
We have,
Also,
Thus ![\[x=\frac{a(b-c)^2}{2(a(b+c)-2bc)}.\]](http://latex.artofproblemsolving.com/d/6/f/d6f382d99e12a39ef863de356c3f9e0ac902be35.png)
As ![\[m=p+x-h=\frac{2a+b+c}{2}+x-\frac{2a+2b+2c}{2}=x-\frac{b+c}{2}.\]](http://latex.artofproblemsolving.com/6/c/5/6c5220c64b08d8c9d80806be68e257b07eb8f92c.png)
Therefore, ![\[m=\frac{a(b-c)^2}{2(a(b+c)-2bc)}-\frac{b+c}{2}=\frac{bc(b+c-2a)}{a(b+c)-2bc}.\]](http://latex.artofproblemsolving.com/e/b/0/eb05dc323bda0a3fdace3f397861afdd2bbc7ee0.png)
Thus we have ![\[\overline{m}=\frac{\frac{1}{bc}\left(\frac{1}{b}+\frac{1}{c}-\frac{2}{a}\right)}{\frac{1}{ab}+\frac{1}{ac}-\frac{2}{bc}}=\frac{ac+ab-2bc}{b^2c+bc^2-2abc}=\frac{a(b+c)-2bc}{bc(b+c-2a)}=\frac{1}{m}.\]](http://latex.artofproblemsolving.com/0/c/0/0c096c789cf29486e9ff9ba7ba198f5cbf1d24e7.png)
Thus, 
lies on the circumcircle of 
,
,
and 
,
are tangent to 
implies that 
is, in fact, the radical axis of 
.
is one of the diameters.![[asy]
size(12cm);
defaultpen(fontsize(9pt));
pair A=(-40,80), B=(-80,-40), C=(80,-40), D=(-40,-40), E=(0,40), F=(-64,8), H=(-40,0), P=(-40,40), T=(-280,-40), M=(0,-40), O=(0,0), X=(-16,48), N=(-28,44);
dot(A);
dot(B);
dot(C);
dot(D);
dot(E);
dot(F);
dot(H);
dot(P);
dot(T);
dot(M);
dot(O);
dot(X);
dot(N);
draw(circumcircle(A,B,C), linewidth(0.5));
draw(A--B--C--cycle, linewidth(0.5));
draw(A--D, linewidth(0.5));
draw(B--E, linewidth(0.5));
draw(C--F, linewidth(0.5));
draw(circumcircle(B,E,C), linewidth(0.7)+dashed);
draw(A--T--B, linewidth(0.5));
draw(T--X--O, linewidth(0.5));
draw(circumcircle(T,A,O), linewidth(0.8)+dotted);
draw(A--M, linewidth(0.4)+dashed);
label("$A$", A, (0,1));
label("$B$", B, SW);
label("$C$", C, SE);
label("$D$", D, (0,-1));
label("$M$", M, (0.5,-1));
label("$O$", O, (1,0));
label("$T$", T, SW);
label("$E$", E, (0.5,1));
label("$F$", F, (-1,0));
label("$P$", P, (-0.5,0.5));
[/asy]](http://latex.artofproblemsolving.com/6/3/3/6337ee0fd821a24bcde5021a70a2698326540367.png)
must be an isosceles trapezoid based on the parallelism we established above. It follows that 
,
is an isosceles triangle; therefore, 
and 
it follows that 
is a parallelogram, since 
and 
Thus we can redefine point 
Let 
I claim that 
which finishes since 
Since 
the projection from the center of 
is 
which always lies on 
lies on 
![[asy]
size(10cm);
pair A,B,C,F,O,T,M,P,Z,X; A=dir(130); B=dir(220); C=dir(320); M=(B+C)/2; O=(0,0); F=A+dir(220); T=extension(A,F,B,C); P=A+M; Z=extension(P,T,A,M); X=2Z-P;
draw(A--B--C--cycle);draw(unitcircle);draw(A--T--B);draw(A--M);draw(T--X--O);draw(circumcircle((A+B)/2,M,P),dotted);
dot("$A$",A,dir(A));dot("$B$",B,dir(B));dot("$C$",C,dir(C));dot("$O$",O,W);dot("$M$",M,dir(M));dot("$T$",T,dir(T));dot("$P$",P,N);dot("$Z$",Z,dir(Z));dot("$X$",X,dir(X));
[/asy]](http://latex.artofproblemsolving.com/4/4/d/44df4e57e963bef527694f1953cc2d3478554fed.png)
hence 
by perpendicularity lemma. Thus 
.
and 
so 
lie on the circle with diameter 
.
because 
by homothety at 
.
.
.![\[AP\cos\angle PAM=AT\cos\angle TAM\iff OM\cos\angle DAM=OM\cdot\frac{\sin\angle AMT}{\sin\angle OAM}\sin\angle OAM\iff\]](http://latex.artofproblemsolving.com/d/4/5/d45163bd309bafa5769c76b2b00a808028fab3fc.png)
![\[\cos\angle DAM=\sin\angle AMD,\]](http://latex.artofproblemsolving.com/4/a/d/4ad06bd66de36827a94b582f05182d1aff0144ee.png)
which is trivial. Let 
.
.
,
be the foot of perpendicular from 
respectively, 
and let 
be midpoints of 
respectively Now as 
so 
is a parallelogram so 
So 
And now construct 
Now As 
so 
so 
So 
and 
Hence 
so 
is also cyclic and Hence 
nine point circle of 
where 
and 
and 
,
,
,
.
be the point such that 
is harmonic. Then
so
Also, the tangents to 
through 
Clearly
so
Now observe that
so
implying that the midpoint of 
and 
and 
are colinear so 
colinear, now note 
is cyclic so by final angle chase we get 
and 
so 
as desired thus we are done 
.
is cyclic with diameter 
;
is cyclic.![\[\angle P'MT = \angle PMT = 90^{\circ} - \angle PMO = 90^{\circ} - \angle PAO = \angle TAP'.\]](http://latex.artofproblemsolving.com/0/c/e/0ce84b7310ddacddf537f4fa002d450d4ab45967.png)
What is a configuration issue 
is equal to the radius of 
,
.
and 
inscribe an angle of 
in congruent circles 
,
be the intersection of the 
.
,
cyclic. Indeed, we have
where 
is the reflection of 
,
as well. On the other hand, clearly 
is cyclic, so 
.
,
is cyclic. Since 
is the nine-point circle of 
,
be the reflection of 
.
,
.
but this is true as we can easily show that 
by noticing that 
,
be the 
point and 
be the intercetion 
then:
but 
so 
.
,
so if we let 
.
done
be the circle with diameter 
.
.
,
.
,
.
and 
,
.
are altitudes in 
,
and 
,
,
are concyclic. Note that circumcircle of 
is the nine-point circle by definition. Hence, 
is the 
.
be the intersections of 
are collinear and 
.
.
are tangents from 
.
.
and 
.
,
Now Note that 
as,
which proves the claim. However then 
and thus upon noting that 
,
we have 
.
is isosceles.
which proves the claim. 
and hence 
is isosceles, finishing.
