[/quote]
,
and circumcenter 
.
is a parallelogram. Prove that 
.
be an integer. Show that, if 
is an integer, then it is a perfect square.[/quote]
.
is at least rational, so that 
must be a perfect square then. Using Conway's topograph method, I have found out that the smallest non-negative pairs 
for which this happens are 
and 
,
,
The eigenvalues of that matrix are irrational, however, so that any calculation which uses powers of that matrix is a little cumbersome. There must be an easier way, but I cannot find it. Can you?
that satisfy 
for all 
and 
satisfies the following recursion formulas: 
,
.
interesting if 
of 
and 
is also a divisor of 
and 
be points on segment 
denote the circle passing through 
.
denote the circle passing through 
.
)
.
the feet of the altitudes lying on 
respectively. One of the intersection points of the line 
The lines 
and 
meet at point 
Prove that 
airline companies in a country. Between any two cities, there is exactly one 
and circumscribed about the incircle 
,
.
,
,
,
,
,
,
,
intersects 
(distinct from 
be the midpoint of the arc 
of 
at 
,
at 
.
but doesn't tell it to anyone. Bernaldo thens pick an arbitrary positive integer 
(not necessarily in 
.
chosen by Arnaldo.
be a set of 2025 positive real numbers. For a subset 
,
as the median of 
when all elements of 
.![\[
P(A) = \sum_{\substack{T \subseteq A \\ |T| \text{ odd}}} M_T, \quad Q(A) = \sum_{\substack{T \subseteq A \\ |T| \text{ even}}} M_T.
\]](http://latex.artofproblemsolving.com/1/d/f/1df2ea8487289111939a53dcc49576fbe6fdef37.png)
Find the smallest real number 
such that for any set ![\[
P(A) - Q(A) \leq C \cdot \max(A),
\]](http://latex.artofproblemsolving.com/c/d/e/cdea343d11c722a423e857aeb01202863405773e.png)
where 
denotes the largest element in 
in 
colors, there exists an arithmetic progression of size 
.
from this?
.
be foots of perpendicular lines from 
.
is cyclic and 
.
bisects 
such that 
Y by anantmudgal09, a28546, mijail, PROF65, Adventure10, Mango247, Rounak_iitr
Let 
be a fixed acute-angled triangle. Consider some points 
and 
lying on the sides 
and 
, respectively, and let 
be the midpoint of 
. Let the perpendicular bisector of intersect the line 
at 
, and let the perpendicular bisector of 
intersect the lines and at 
and 
, respectively. We call the pair 
, if the quadrilateral 
is cyclic.
Suppose that the pairs
and 
are interesting. Prove that 
Proposed by Ali Zamani, Iran
be a fixed acute-angled triangle. Consider some points 
and 
lying on the sides 
and 
, respectively, and let 
be the midpoint of 
. Let the perpendicular bisector of intersect the line 
at 
, and let the perpendicular bisector of 
intersect the lines and at 
and 
, respectively. We call the pair 
, if the quadrilateral 
is cyclic.Suppose that the pairs
and 
are interesting. Prove that 
Proposed by Ali Zamani, Iran
This post has been edited 4 times. Last edited by hajimbrak, Jul 27, 2015, 6:11 AM
Reason: Changed format
Reason: Changed format
that is interesting.
is cyclic.
to be the associated complex number, and 
its conjugate.
where 
is the argument of 
is its distance to the origin (absolute value)
if and only if 
(this follows from 1 and 2)
where 
(this follows from 4)
(from 
in positive direction) if and only if 
(this follows from 1)
where 
and 
if and only if 
(this follows from 8)
if and only if 
(this follows from 8)
are colinear if and only if 
(this follows from 9)
belong to a circle if and only if 
(this follows from 7)
(this follows from 3 and 12. It also follows directly from 8)
(this follows from 6)
(this follows from 6 and 11)
is the the unit circle in the complex plane.
in terms of 
in terms of 
since 
and therefore 
can be expressed in terms of 
too.
in terms of 
using 15
,
and their conjugates. All of those can be expressed in terms of 
be the middle point of 
and therefore in terms of 
too.
in terms of 
using 15
,
and their conjugates. All of those can be expressed in terms of 
and we get 
and 
expressed in terms of 
and 
with 
.
and their conjugates. All of them we have been able to express in terms of 
and their conjugates; all of which we have expressed also in terms of 
.
be the points on 
,
be the midpoint of 
and 
be the point s.t. 
.
be the perpendicular bisector of 
and 
.
are concyclic, then 
is A-symmedian of 
.
is A-median of 
,
are symmetry WRT 
we gte 
in the midpoint of 
is an Isosceles trapezoid 
is A-symmedian of 
cuts 
at 
)
be the orthocenter of 
and 
.
be the midpoint of 
(i.e. 
is a harmonic quadrilateral) .
is the Miquel point of 
WRT 
are the tangents of 
(well-knwon) 
.
is A-symmedian of 
,
are the tangents of 
,
.
,
is the center of spiral similarity that maps 
,
.
is intersection of the tangents of 
at 
and similarly 
is intersection of the tangents of 
at 
This is also the lemma in Telv's previous solution.
center of the spiral similarity carrying 
into 
Since the isosceles 
and 
are directly similar, then 
is the image of 
under the referred spiral similarity 
is cyclic 
touches 
touches 
is radical axis of 
bisecting their common tangent 
i.e. 
Therefore 
be the intersections of 
and 
and 
are collinear then using the converse of Pascal theorem for 6 points 
we get 
.
Therefore 
,
.
Thus 
intersect each other at 
are tangents to 
is polar of 
and 
.
so 
are collinear.
and let 
where 
is orthogonal to cyclic quadrilateral 
(well known) we get that 
passes throw 
with center 
)
passes throw 
.
passes throw 
be the reflection of 
.
,
,
,
,
,
.
.
,
are collinear. Let 
.
,
are concyclic. Then if 
(signed lengths), we have that 
,
,
.
and 
be to interesting pairs and 
the corresponding points 
,
and 
.
,
and 
are cyclic quadrilaterals, so sine law yields
.
and 
.
and 
are tangents to 
if and only if 
are on a circle.
.
is a median in triangle 
as well. Now, the condition is equivalent to saying that reflection of the median 
is a symmedian in triangle 
.
and 
meet on 
are on a line. For this, we apply Pascal's Theorem on the degenerate cyclic hexagon 
we get that 
,
,
are collinear. Therefore, we get 
which proves the claim.
.
followed by reflection in the bisector of angle 
to 
is constant and so, we only want to prove that 
where points 
are the feet of altitudes from 
onto sides 
is a harmonic quadrilateral and so 
is an interesting pair then 
and let 
it follows that 
is parallel to 
and hence 
and thus 
and 
are tangent to 
then 
.
is cyclic. The lemma then follows by straightforward angle chasing 
.
.
and 
and 
lie on segments 
and 
respectively then 
which is a contradiction. Similarly 
and 
can't lie on segments 
and 
respectively. Now assume 
it follows that 
and hence triangles 
and 
are congruent. Thus 
and thus both of these angles must be right. Thus 
is a cyclic quadrilateral which is impossible as 
and 
are tangent to 
and 
be directly similar triangles and let points 
on lines 
,
satisfy 
.
is also directly similar to the two first triangles (here the segments are directed).
then you can construct this by letting 
the intersections of the parallel to 
let 
the projections of 
and 
with the reflection of 
.
and the corresponding point 
.
and 
on lines 
and 
are directly similar, triangle 
is directly similar to both of these, and thus 
.
by Lemma 1, it follows that 
and 
by Lemma 2. It then follows that
follows easily by using 
are similar. Let 
satisfy 
and 
,
.
and thus we wish to show that
.
is a harmonic bundle and hence
which follows from the fact that the symmedians cut opposite sides at points which make ratios equal to the squares of the ratios between the adjacent sides.
.
.
,
is cyclic. Similarly 
is cyclic. Hence
are tangent to 
meet 
,
is cyclic. Similarly 
is cyclic, so by ELMO SL 2013 
and 
,![\[
\frac{E_1E_2}{F_1F_2} = \frac{RE_2}{RF_2},
\]](http://latex.artofproblemsolving.com/3/4/a/34ae76acd131c43f76bb33f66f6d44ba2542f40e.png)
but 
,
as desired. 
.
.
.
is isosceles trapezoid thus lines 
are isogonal w.r.t. 
and 
or 
.
.
are interesting, then 
and 
so
hence the result follows.
.
is harmonic, as 
is a symmedian of 
,
is similar to a triangle of fixed shape 
,
,
both be inscribed in 
and 
coincide at the center of spiral similarity sending 
to 
.
,
.
,
,
,
.
is the midpoint of 
we also have 
,
.
and 
are given as in the problem, then ![\[ \frac{E_1E_2}{F_1F_2} = \frac{\operatorname{dist}(S,\overline{AC})}{\operatorname{dist}(S,\overline{AB})} = \frac{AB}{AC} \]](http://latex.artofproblemsolving.com/4/9/7/4977cd7fbf007b1d52d1f37c29c292fb3f4705f3.png)
with the last equality since 
be the point on the circumcircle of 
is parallel to 
intersect the circumcircle of 
is cyclic. Note that this gives us that:
.
and let 
be the point on 
are collinear. Angle chase gives 
.
so 
we can take 
we can take 
and 
,
.
![[asy]
size(8cm); real labelscalefactor = 0.5; /* changes label-to-point distance */pen dps = linewidth(0.7) + fontsize(0); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -35.52923006253961, xmax = 9.606596061817754, ymin = -16.259434463888752, ymax = 22.17778849006929; /* image dimensions */pen qqwuqq = rgb(0,0.39215686274509803,0); pen fuqqzz = rgb(0.9568627450980393,0,0.6); pen ccqqqq = rgb(0.8,0,0); pen zzttff = rgb(0.6,0.2,1); /* draw figures */draw(circle((-8.518985135430142,8.738475867737094), 4.251059521670271), linewidth(0.8) + blue); draw(circle((-6.069937159026376,5.954033836015797), 7.959287321946787), linewidth(0.8) + blue); draw(circle((-17.85255294334901,5.277216173516793), 9.319620113551203), linewidth(0.8) + qqwuqq); draw((-12.647483832307847,7.725063036427688)--(-5.217400295075848,6.060595250770354), linewidth(0.8) + fuqqzz); draw((-25.284462723147833,-0.3461312357971944)--(1.9111769434423325,-0.788850951299825), linewidth(0.8)); draw((-6.1587811750061725,-2.0047576183526017)--(-11.32653884568014,11.93052054487508), linewidth(0.8) + ccqqqq); draw((-25.284462723147833,-0.3461312357971944)--(-11.32653884568014,11.93052054487508), linewidth(0.8) + zzttff); draw((-11.32653884568014,11.93052054487508)--(-13.7997530247477,4.056614178864074), linewidth(0.8)); draw((-13.7997530247477,4.056614178864074)--(-15.234057361746412,-0.5097424858665198), linewidth(0.8)); draw((-11.32653884568014,11.93052054487508)--(1.9111769434423325,-0.788850951299825), linewidth(0.8)); draw((-13.7997530247477,4.056614178864074)--(0.11164233385230922,0.9402199536978176), linewidth(0.8) + fuqqzz); draw((-11.32653884568014,11.93052054487508)--(-6.844055345447695,2.498417066280946), linewidth(0.8)); draw((-6.844055345447695,2.498417066280946)--(-10.607618375973532,-0.5850566088907295), linewidth(0.8)); /* dots and labels */dot((-11.32653884568014,11.93052054487508),dotstyle); label("$A$", (-11.525902035275358,12.847591217013086), NE * labelscalefactor); dot((-15.234057361746412,-0.5097424858665198),dotstyle); label("$B$", (-15.473293189260675,-1.4666857959235688), NE * labelscalefactor); dot((1.9111769434423325,-0.788850951299825),dotstyle); label("$C$", (2.070667495118512,-0.39012457210939155), NE * labelscalefactor); dot((-8.566436770833844,4.487681189199803),linewidth(4pt) + dotstyle); label("$D$", (-9.65188805308031,4.833190995285321), NE * labelscalefactor); dot((-12.647483832307847,7.725063036427688),dotstyle); label("$F$", (-12.044246328222926,8.102747304646897), NE * labelscalefactor); dot((-5.217400295075848,6.060595250770354),linewidth(4pt) + dotstyle); label("$E$", (-4.508317761523683,5.7901343053423675), NE * labelscalefactor); dot((-8.932442063691848,6.892829143599021),linewidth(4pt) + dotstyle); label("$M$", (-8.535454191347087,7.105931356670807), NE * labelscalefactor); dot((-10.607618375973532,-0.5850566088907295),linewidth(4pt) + dotstyle); label("$K$", (-10.688576638975443,-1.5065584338426126), NE * labelscalefactor); dot((-13.7997530247477,4.056614178864074),linewidth(4pt) + dotstyle); label("$T$", (-15.114439447989282,4.195228788580623), NE * labelscalefactor); dot((0.11164233385230922,0.9402199536978176),linewidth(4pt) + dotstyle); label("$S$", (-0.08245495250984282,1.7231252375999198), NE * labelscalefactor); dot((-25.284462723147833,-0.3461312357971944),linewidth(4pt) + dotstyle); label("$K'$", (-26.518013892835754,-0.9882141408950457), NE * labelscalefactor); dot((-6.1587811750061725,-2.0047576183526017),linewidth(4pt) + dotstyle); label("$G$", (-6.342459105799689,-3.0217186747662694), NE * labelscalefactor); dot((-6.844055345447695,2.498417066280946),linewidth(4pt) + dotstyle); label("$I$", (-6.501949657475864,2.799686461414097), NE * labelscalefactor); dot((-6.661440209152045,-0.6492967185831722),linewidth(4pt) + dotstyle); label("$N$", (-6.501949657475864,-0.31037929627130434), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
[/asy]](http://latex.artofproblemsolving.com/5/2/4/524f33f52a803fa7bd5479d4c2b5fc4afaec5a29.png)
are concyclic.
lies on the 
and the 
symmedian in 
.
Moreover, 
intersect 
.
to 
,
On the other hand,
For each point 
,
or it lies below 
by ratio lemma, 
where 
from the harmonic conditions, hence
as desired. ![[asy]
size(8cm); real labelscalefactor = 0.5; /* changes label-to-point distance */pen dps = linewidth(0.7) + fontsize(0); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -25.760433772373943, xmax = 19.375392351983418, ymin = -27.005110383071003, ymax = 11.432112570887039; /* image dimensions */pen zzttqq = rgb(0.6,0.2,0); draw((-2.5744297462234242,-5.259500087187417)--(-7.088141678673,-5.076104623147327)--(-6.12886969510659,-2.768890110948807)--cycle, linewidth(0.8) + zzttqq); draw((-2.5744297462234242,-5.259500087187417)--(0.12615653932945348,-3.642990092573564)--(-1.1176249307501425,-2.325277556090636)--cycle, linewidth(0.8) + zzttqq); /* draw figures */draw(circle((-2.8060639557129097,-1.8084335226740398), 3.4588314268995517), linewidth(0.8) + blue); draw(circle((-4.717476877953287,-2.366750011988238), 3.600090833073584), linewidth(0.8) + blue); draw((-6.12886969510659,-2.768890110948807)--(0.12615653932945348,-3.642990092573564), linewidth(0.8)); draw((-7.088141678673,-5.076104623147327)--(-1.1176249307501425,-2.325277556090636), linewidth(0.8)); draw((-4.468445123604657,1.2247172636118744)--(-9.093671122213715,-9.899748715801291), linewidth(0.8)); draw((-9.093671122213715,-9.899748715801291)--(6.935129321241813,-10.856692025858337), linewidth(0.8)); draw((6.935129321241813,-10.856692025858337)--(-4.468445123604657,1.2247172636118744), linewidth(0.8)); draw((-2.5744297462234242,-5.259500087187417)--(-7.088141678673,-5.076104623147327), linewidth(0.8) + zzttqq); draw((-7.088141678673,-5.076104623147327)--(-6.12886969510659,-2.768890110948807), linewidth(0.8) + zzttqq); draw((-6.12886969510659,-2.768890110948807)--(-2.5744297462234242,-5.259500087187417), linewidth(0.8) + zzttqq); draw((-2.5744297462234242,-5.259500087187417)--(0.12615653932945348,-3.642990092573564), linewidth(0.8) + zzttqq); draw((0.12615653932945348,-3.642990092573564)--(-1.1176249307501425,-2.325277556090636), linewidth(0.8) + zzttqq); draw((-1.1176249307501425,-2.325277556090636)--(-2.5744297462234242,-5.259500087187417), linewidth(0.8) + zzttqq); /* dots and labels */dot((-4.468445123604657,1.2247172636118744),dotstyle); label("$A$", (-4.8671715027950935,1.6601322286974445), NE * labelscalefactor); dot((-9.093671122213715,-9.899748715801291),dotstyle); label("$B$", (-9.25316167388989,-11.255418405048774), NE * labelscalefactor); dot((6.935129321241813,-10.856692025858337),dotstyle); label("$C$", (7.232582079622686,-10.979493991639378), NE * labelscalefactor); dot((-2.5744297462234242,-5.259500087187417),linewidth(4pt) + dotstyle); label("$D$", (-2.315322675976303,-6.0700650443587595), NE * labelscalefactor); dot((-6.12886969510659,-2.768890110948807),dotstyle); label("$F_{1}$", (-7.658256157128147,-2.5233107007782247), NE * labelscalefactor); dot((-7.088141678673,-5.076104623147327),dotstyle); label("$F_{2}$", (-8.57532682926615,-5.23465007927319), NE * labelscalefactor); dot((0.12615653932945348,-3.642990092573564),linewidth(4pt) + dotstyle); label("$E_1$", (1.1137241850614472,-3.79923511418762), NE * labelscalefactor); dot((-1.1176249307501425,-2.325277556090636),linewidth(4pt) + dotstyle); label("$E_2$", (-0.88118133170029703,-1.9252211319925705), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
[/asy]](http://latex.artofproblemsolving.com/c/9/5/c95a34409ca9d798efb1f8e2375b6e847def0cb0.png)
to 
,
as desired.
is the distance from the point to the line, and 
is the distance between two parallel lines. First, we characterize interesting pairs as follows:
are on the inner circle such that 
are tangent to the inner circle. Let 
cut the outer circle at 
.
.
,
,
are collinear, which implies 
.
,
,
)
implies that 
at 
)
cut 
,
.
,
.
,
.
are similar: they are isosceles with base angle equal to 
.
clockwise 
units and intersect it with line 
.
units around 
intersect 
around 
also passes through 
are collinear.
,
are concyclic. Then by the converse of Reim's theorem 
cutting the circles 
and 
)
are collinear. 
.
,
concyclic, so 
.
,![\[\triangle ELE' \sim \triangle ACB\]](http://latex.artofproblemsolving.com/b/c/8/bc877e60274d64ba3a3e55ba314c319f4057ce61.png)
so![\[\frac{EE'}{AB} = \frac{EL}{AC} = \frac{FF'}{AC}.\]](http://latex.artofproblemsolving.com/8/5/5/855261541c0db62c1b311c76ec1190259a862727.png)
and 
respectively. We will first show the following claim which has been used in other solutions as well.
.
are collinear. Therefore, 
are symmetric over 
is a convex quadrilateral, it follows that 
is a rhombus, and angle chasing gives us![\[\measuredangle EFK=\measuredangle MFK=\measuredangle MSK=\measuredangle TKS=\measuredangle EAF. \blacksquare\]](http://latex.artofproblemsolving.com/0/5/0/050e98e39ba10d821071c728d6286df57f1a942c.png)
This lets us completely remove 
and 
from the diagram.
.
centered at 
.
and 
are directly similar. Therefore, by spiral similarity, we have 
and hence,![\[\measuredangle (F_1K_1,K_1K_2')=\measuredangle F_1E_1E_2'=\measuredangle (K_1F_1,F_1A)\]](http://latex.artofproblemsolving.com/0/c/5/0c57cc8ade5fddeb4a537797fd0b426711551746.png)
and so 
.![\[\frac{AF_1}{F_1F_2}=\frac{AK_2'}{K_2K_2'}=\frac{BK_1}{K_1K_2}\]](http://latex.artofproblemsolving.com/e/3/a/e3abd38452d575c529e1a9137406fed7b8a637f4.png)
as desired. 
denote the distance from point 
.![\[\frac{F_1F_2}{E_1E_2}=\frac{AF_1\cdot K_1K_2}{BK_1}\div\frac{AE_1\cdot K_1K_2}{CK_1}=\frac{AF_1}{AE_1}\cdot \frac{CK_1}{BK_1}.\]](http://latex.artofproblemsolving.com/e/2/2/e22c860e8380dce149e654889aa35fb98d03db9a.png)
Since ![\[\frac{AF_1}{AE_1}\cdot \frac{CK_1}{BK_1} = \frac{d(K_1,AB)}{d(K_1,AC)} \cdot \frac{CK_1}{BK_1}=\frac{[ABK_1]}{[ACK_1]}\cdot \frac{AC}{AB}\cdot \frac{CK_1}{BK_1}=\frac{AC}{AB}\]](http://latex.artofproblemsolving.com/6/4/a/64a4fe2c7f65c7efa398a7be12d18ddb3eed23e2.png)
and we're done.
.
is orthocenter of 
so 
,
from 
is the 
,
.
.
so 
and similarly, 
.
then 
so 
and 
where 
,
and 
and lines 
implies that 
.
and similarly 
.
,
are cyclic. Then the spiral similarity induced by lines 
and 
implies 
so 
where the last step follows from 
.
Then by Law of Sines in 
gives 
and similarly for 
.
are collinear. Then 
Finally, 
so 
as desired.
are cyclic in Step 3 of Solution 1.
corresponding to each pair.
and 
so 
(because 
and 
are cyclic) and similarly, 
so 
as desired.
Obviously 
and since 
we deduce 
which completes 
Converse of Pascal on 
gives 
be 
respectively. Let 
by sqrtbc so 
and 
so by radius SR inversion at R, 
so 
which is fixed so it follows that 
and 
meet again at a point on line ![\[BF \cdot BA + CE \cdot CA = BC^2\]](http://latex.artofproblemsolving.com/9/1/9/919a195b757c0fb7ecf0b393204c9e08862165fe.png)
.
.
we obtain that points 
and 
concur. Since 
are collinear, this implies that 
are collinear. Hence, 
,![\[BF \cdot BA + CE \cdot CA = BP \cdot BC + CP \cdot CB = BC^2\]](http://latex.artofproblemsolving.com/1/1/c/11c257aa2ce6c0ad0afc78b565c119ddae57f1bf.png)
proving the lemma.
,![\[\frac{E_1E_2}{AB} = \frac{F_1F_2}{AC}\]](http://latex.artofproblemsolving.com/f/2/3/f238e1d4163fc1009ff565da492d13ff0364daed.png)
as required. 
and 
such that 
has midpoint 
and 
is the 
and similarly for 
,
work perfectly).
,
is a median of 
and 
intersect at 
.
and 
lies on 
.
,
lies on 
.
,
,![\[ \text{Pow}(\bullet, (AEF))-\text{Pow}(\bullet, (AEB))=\text{Pow}(\bullet, (AFC))-\text{Pow}(\bullet, (ABC)). \]](http://latex.artofproblemsolving.com/a/3/9/a39eb183b5cf16e9eef60663f93e818165067925.png)
Proof. Let 
denote LHS and 
denote the RHS. Note that 
are linear functions. It suffices to show that 
holds for three distinct non-collinear positions of 
.
for each 
.![\[ \frac{E_1E_2}{AB} = \frac{F_1F_2}{AC} \]](http://latex.artofproblemsolving.com/7/e/6/7e63fd14955a86cdfac855f35a968a333b9ac06d.png)
holds if and only if 
.
and 
denote the intersection points in the second claim for ![\[ \theta(B) + \theta(C) = (BX_1 \cdot BC - BX_2 \cdot BC) + (CX_1 \cdot BC - CX_2 \cdot BC) = BC^2-BC^2 = 0, \]](http://latex.artofproblemsolving.com/7/9/3/7931c12b4d8706805e46a1792794a59ffd940411.png)
and we are done.
where 
,
and 
.
)
.
.
,
,
and define 
similarly. By Pascal's on 
we get that 
as well. So then 
and 
.
.
so 
.
which implies the desired.
,
.
.
,
to be collinear. Thus. 
and we are done.
)
denote the foot of perpendiculars from 
upon their opposite sides.
.
to 
.
be two interesting pair of points. Then: 
and we are done..!
.
with 
is cyclic, then 
,
,
.
.
.
.![\[x^2+y^2=2\lambda x(b-a) + 2\lambda y(ab+1).\]](http://latex.artofproblemsolving.com/3/9/2/392b8aa52a5270c0a39b55f51d3afb4415457862.png)
By the definition of 
and 
.![\[4\cdot (2\lambda-1)^2 = (2\lambda - 1) \cdot (a-b)^2 + 4\cdot (2\lambda-1) \cdot \lambda \cdot (ab+1).\]](http://latex.artofproblemsolving.com/a/4/4/a448d63df42872b403128230a6a044e2abf4d102.png)
Solving this, ![\[y_k - 2 = 4(\lambda-1) = \frac{(a+b)^2}{1-ab}.\]](http://latex.artofproblemsolving.com/7/1/a/71a766b48d3af7a1ca4d0a057a4e8eb3f24eec81.png)
so it has the same 
,
,
.
and 
,![\[\tan(\alpha)=-\tan(\beta+\gamma)=-\frac{1/a+1/b}{1-1/ab} = \frac{a+b}{1-ab}.\]](http://latex.artofproblemsolving.com/2/5/3/253ac4482115ad0efd762ba5cc2b3e81c978c6d4.png)
,![\[y_{k'}-2 = (a+b)\tan(\alpha) = \frac{(a+b)^2}{1-ab} = y_k-2,\]](http://latex.artofproblemsolving.com/9/2/0/920f9df63678275870d9380beeaec9168ace8a4d.png)
concluding the proof.
is a harmonic bundle.
.
at 
.
.
.
and 
.
,
.
is also a harmonic bundle, so 
.
.
,
.
is 
.
.
and 
.
-
-
and 
.
be the reflection of 
.
.
.
would have 
.
with 
.
.
.![\[\frac{c^2x_{t'}}{z_k} + \frac{b^2x_{s'}}{y_k} = 2a^2.\]](http://latex.artofproblemsolving.com/3/d/e/3debc9ffdae5ef1654e915ef055a05b56a544ae0.png)
Since 
,![\[\frac{c^2x_{t'}}{z_k} = \frac{c^2(1-2y_m)}{z_k}+c^2-\frac{c^2y_m}{z_m} = \frac{b^2z_m}{y_m}-2(b^2z_m+c^2y_m)+2c^2-\frac{c^2y_m}{z_m}.\]](http://latex.artofproblemsolving.com/9/9/c/99c743e54100328c2c002e7ba8c16867c369cd71.png)
With a similar computation, ![\[\frac{b^2x_{s'}}{y_k} = -\frac{b^2z_m}{y_m}-2(b^2z_m+c^2y_m)+2b^2+\frac{c^2y_m}{z_m}.\]](http://latex.artofproblemsolving.com/1/c/f/1cf631662f74ace89a46dcdd35313d72ac6ed166.png)
Therefore, ![\[2b^2+2c^2-2a^2 = 4b^2z_m+4c^2y_m.\]](http://latex.artofproblemsolving.com/2/a/9/2a91117fd35e787eff346b6acf581dd241313807.png)
Therefore, ![\[c^2y+b^2z = \frac{b^2+c^2-a^2}{2},\]](http://latex.artofproblemsolving.com/7/3/f/73fea2ac437a7106a5b27e7772d6014c9f2e11a6.png)
which is a line.
(which is 1), we can rearrange to ![\[-\frac{b^2+c^2-a^2}{2}x + \frac{a^2+c^2-b^2}{2}y + \frac{a^2+b^2-c^2}{2}z = 0.\]](http://latex.artofproblemsolving.com/1/0/0/1007035d60714cd1d15b4c78ee3cd4125ec279eb.png)
and 
be the feet of the altitudes from 
,
.
.![$\left[BC_0:AC_0:0\right]$](http://latex.artofproblemsolving.com/6/1/8/618a86652257cde3138ccc08d7bda16dc2ca20ab.png)
,![$\left[\frac{a^2+c^2-b^2}{2}:\frac{b^2+c^2-a^2}{2}:0\right]$](http://latex.artofproblemsolving.com/3/c/c/3ccbf2c0d7fbda0ed8161c7ec8c8add0bdc192a3.png)
.
.
is cyclic, by Power of a Point 
.
.
.
,
,
,
.
,
,
.
and 
,
