be positive real numbers. Prove that:![$$\sqrt[3]{a^3+b^3}+\sqrt[3]{b^3+c^3}+\sqrt[3]{c^3+a^3}\geq \sqrt[3]{2}(a+b+c)$$](http://latex.artofproblemsolving.com/2/6/e/26ebcdae35153bee73abbcfaf4a767d3a32b8912.png)
and 
such that
intersects circumcircle of 
,
.
are points chosen on 
.
be a positive integer. In an 
table, an upright path is a sequence of adjacent cells starting from the southwest corner to the northeast corner such that the next cell is either on the top or on the right of the previous cell. Find the smallest number of grids one needs to color in an 
can be partitioned into 
pairs such that the sum of the elements in each pair is a power of 
.
is stylish because the set 
can be partitioned as 
,
,
.
Doing this, we get that 
Now, we can solve this by factoring. By using RRT, we get that 
is a factor. Upon division, we get 
is the factored form. Now, we know that our only positive solutions are 
so our values of 
are 
Summing these gives our answer of 
to 
in or something like that idk my friend explained it pretty wonkily idk if he even solved it right but anyhow we press on. Transformations suck.
be positive real numbers such that 
and 
Determine the value of 
Furthermore, note that 
Thus, the answer is just 
is on point 
such that 
and 
and 
is a point on 
such that 
and 
Given that 
calculate 
so 
Cross multiplying and solving this quadratic, we get that our solution is 
Thus, 
given that 
We know from 
that 
so substituting this gives us 
Now, plugging this in to 
and 
gives us 
Solving these equations gives us 
Thus, by Vieta's, the sum of the roots is 
thus by median properties (the 2:1 ratio split thingy), we know that 
(For reference, I missed this part in-contest) Now, we can calculate 
by the Pythagorean Theorem to be 
and thus 
Now, by Pythagorean theorem, we know that the altitude from 
so our area is just 
Thus, simplifying and substituting 
we get that 
Simple guess and check gives us 
as our solutions. My dumb self forgot the former solution. Thus, 
giving us 
as our sum.![\[f(x) =
\begin{cases}
x - 9, & \text{if } x > 100 \\
f(f(x + 10)), & \text{if } x \leq 100
\end{cases}\]](http://latex.artofproblemsolving.com/4/e/5/4e5b4486030a59e74df648819a1a59f82622b3b7.png)
.
we get that 
Now if we try finding 
we will realize that 
this 
pattern continues until we once again reach 
thus 
Now return to the problem statement; 
but each inner function will just become 
hence the answer is 
be real numbers such that 
Find 
Thus, simplifying, we get that 
or that 
Plugging this in to the first equation, we get that 
so this means that 
Now, 
thus 
are on circle 
and 
Determine the path length from 
formed by segment 
is the diameter, thus we can draw 
and so by Special Right Triangles we do NOT in fact get 
but rather 
is also an integer.
.
to be an integer. Plugging in factors of 
we see that there are 
integers that work.
is a multiple of 
Plugging in numbers gives us 
as our smallest value.
are real numbers such that 
and 
Calculate 
Furthermore, 
so 
meaning that 
.
and thus 
giving us 
will follow the rules such that![\[
p_1 = (0,0), \quad p_{i+1} = (x_i + 1, y_i) \text{ or } (x_i, y_i + 1), \quad p_{10} = (4,5).
\]](http://latex.artofproblemsolving.com/6/0/3/6036ef2619ffb1e91db0ab310388217e472490f2.png)
How many sequences 
are possible such that 
is the only point with equal coordinates?
Furthermore, we know that 
so our answer is just 
in the summation. Essentially, we have 
.
In other words, we want to find 
.
.
in such a polynomial is 
.
,
.
.
,
.
and 
,
and divide by 
.
and 
in our sum, subtract by 
.
term look kind of weird? Well, we actually notice 
,
.
,
,
is our answer!
,
terms. Drawing it out on the unit circle (which I heavily advise), it cycles between 
and 
.
,
,
.
,
for when 
terms! Anyways, the answer is 
.
so our coefficient is just 
Find 
Thus, our answer is just 
on the positive integers using the rule that for 
For all prime 
and for all other 
Find the smallest possible value of 
can be written as the sum of two distinct, non-negative integer powers of 
etc.
we get that it is 
Thus, we note that for an integer 
Since 
we know that all values of 
satisfy our conditions. Thus, our answer is just 
be the set of positive integers of 
for some other positive integer 
Find the only three-digit value of 
is our solution. For these Pell Equations, all solutions are of the form 
for 
Thus, we plug in 
giving us 
as our answer.
Find the value of 
and let 
be our three-digit number that we remove. Thus, we can write 
By our condition, we know that 
or that 
Furthermore, we know that 
does not work, so we try 
We know that 
by the fact that it must be divisible by a 
Noting that 
doesn't work, we try 
or that 
This, miraculously, works, and thus our value of 
and both 
and 
are powers of 2. Prove that 
Y by babu2001, JustKeepRunning, Illuzion, Wizard_32, TheMathLife, anantmudgal09, Mr.Chagol, e_plus_pi, char2539, TheDarkPrince, MathInfinite, PCM_noob, FadingMoonlight, MathPassionForever, Euler1728, rmtf1111, Davrbek, Generic_Username, Fermat_Theorem, samuel, Carpemath, Mathman12334, wx_family2, Kayak, fjm30, buratinogigle, v_Enhance, bel.jad5, rkm0959, Satyanweshi_math1729, HolyMath, GeoMetrix, Drunken_Master, SHREYAS333, mathleticguyyy, Arhaan, amar_04, Supercali, EulersTurban, lilavati_2005, Polynom_Efendi, Aryan-23, Feridimo, NJOY, Pi-is-3, AmirKhusrau, kc5170, Purple_Planet, donfluffles1412, Atpar, Inconsistent, IAmTheHazard, psi241, srijonrick, Bassiskicking, OlympusHero, Kanep, TheMathCruncher_007, sotpidot, megarnie, centslordm, tiendung2006, rayfish, magicarrow, crazyeyemoody907, GeoKing, Jalil_Huseynov, mathmax12, EpicBird08, Lamboreghini, Adventure10, NO_SQUARES, CrimsonBlade273, Rounak_iitr, Eka01, Sedro, Funcshun840, iamnotgentle, wizixez
Let 
be the incentre of acute triangle 
with 
. The incircle 
of is tangent to sides 
, and 
at 
and 
, respectively. The line through 
perpendicular to 
meets at 
. Line 
meets again at 
. The circumcircles of triangle 
and 
meet again at 
.
Prove that lines
and 
meet on the line through 
perpendicular to 
.
Proposed by Anant Mudgal, India
be the incentre of acute triangle 
with 
. The incircle 
of is tangent to sides 
, and 
at 
and 
, respectively. The line through 
perpendicular to 
meets at 
. Line 
meets again at 
. The circumcircles of triangle 
and 
meet again at 
.Prove that lines
and 
meet on the line through 
perpendicular to 
.Proposed by Anant Mudgal, India
This post has been edited 2 times. Last edited by djmathman, Jul 17, 2019, 12:31 PM
so 
is cyclic. Now let 
.
are similar. Both quadrilaterals are cyclic, so it suffices to show the measures of the corresponding arcs on the two circles are equal. Clearly 
are both equal to 
,
are also equal. Also, 
and similarly 
,
with 
;
are collinear. Let 
be the antipode to 
be the midpoints of 
.
is obviously harmonic, so similarity implies 
is too. It's well-known that 
are collinear (say, by projecting harmonic bundle 
to 
and sends 
to the midpoint 
of 
.
be the foot of the projection from 
to 
correspond in the inversion.
are concyclic. Let 
;
is cyclic with diameter 
,
.
are midpoints of 
this implies 
so 
.
so 
,
and we're done.
,
,
.
,
and so ![\[ P = \frac{A-R}{1-R\overline{A}} = \frac{\frac{2yz}{y+z} + \frac{yz}{x}} {1 + \frac{yz}{x} \cdot \frac{2}{y+z}} = \frac{yz(2x+y+z)}{2yz+x(y+z)}. \]](http://latex.artofproblemsolving.com/3/3/4/334af39b32b7748631b7bcd7cdcbb12f74e30fd5.png)
We now compute 
Similarly ![\[ O_C = \frac{x-y}{x+y} \cdot \frac{yz(2x+y+z)}{(z-y)(x+z)}. \]](http://latex.artofproblemsolving.com/2/1/f/21f7cb46c9a09d01deddba9c98c46019e0c94ef5.png)
Therefore, subtraction gives ![\[ O_B-O_C = \frac{yz(2x+y+z)}{(x+y)(x+z)(y-z)} \left[ (x-z)+(x-y) \right] = \frac{yz(2x+y+z)(2x-y-z)}{(x+y)(x+z)(z-y)}. \]](http://latex.artofproblemsolving.com/5/1/7/517f3a25ea01049617e918f3c1aa9d31c6538180.png)
It remains to compute 
we have 
so 
.
Thus 
This gives 
as needed.
is cyclic. It allows us to make next
are any points on sides 
.
,
.
,
.
.
pass through mixtilinear touchpoint 
.
be the reflection of 
is 
are concyclic.
again at 
.
so 
.
so 
.
,
.
and 
which means 
.
.
.
, it suffices to show that
We compute the first term first, note that
For the second term, we use Power of Point twice.
And the last term note that 
so,
Thus multiplying all together, the resulting terms
as 
,
![\[ f(\bullet) := \text{Pow}(\bullet, (PBF)) - \text{Pow}(\bullet, (PCE)) \]](http://latex.artofproblemsolving.com/a/b/7/ab704513119ef7baebe1c683bd8bd6fa8212f61c.png)
and ![\[ g(\bullet) := \text{Pow}(\bullet, (BDIF)) - \text{Pow}(\bullet, (CDIE)). \]](http://latex.artofproblemsolving.com/a/d/f/adf8f3f8637294a454b3fc38563e8be222c95c2d.png)
Let 
.
and 
.
is the external angle bisector 
of 
,
-
-
.
.
intersect 
and 
intersect 
.
,![\[ \frac{DB_1}{DC_1} = \frac{KC}{KB} = \frac{AC}{AB}. \]](http://latex.artofproblemsolving.com/4/3/a/43a3d596bf7187ffbe336ece44f49f5dc6058705.png)
Fortunately, this is something very much in our hands.![\[ \frac{DB_1}{DC_1} = \frac{AC}{AB}. \]](http://latex.artofproblemsolving.com/f/7/4/f74e2e38d7c34e02e14ac0021882266d968aadb6.png)
Proof. First, we need to understand what 
,
with 
.![\[ \frac{PE}{PF} = \frac{RF}{RE} = \frac{\cos(\angle F)}{\cos(\angle E)}. \]](http://latex.artofproblemsolving.com/d/4/b/d4bd52969dd675899e1bc277f388c76e86a10698.png)
Second, redefine 
is 
,![\[ \frac{PB_1}{PC_1} = \frac{PE}{PF} = \frac{\cos(\angle F)}{\cos(\angle E)} = \frac{\sin(\angle C/2)}{\sin(\angle B/2)}. \]](http://latex.artofproblemsolving.com/7/8/a/78a839453537b93aee9dee27d0e68b187996300e.png)
Finally, we apply LoS on 
and 
to involve ![\[ \angle DPB_1 = 90^{\circ} + \angle B/2. \]](http://latex.artofproblemsolving.com/8/e/a/8eafad303175984f1ef66104e9abb75d50c7af4f.png)
This is easy angle chasing, in that we write ![\[ \angle DPB_1 = 360^{\circ} - (\angle B_1PF + \angle DPF) = 360^{\circ} - ((180^{\circ}-\angle B)+(90^{\circ}+\angle B/2)) = 90^{\circ}+\angle B/2, \]](http://latex.artofproblemsolving.com/9/5/b/95bc62478d11c32f7ed1432f4c1fbda36ff29418.png)
as desired.
intersect 
,
intersect 
,
,
,
,
and 
is isoceles, and we can prove 
,
,
,
is a rhombus
is and isoceles trapezoid, and that 
and 
are congruent, and from that we can prove that 
the unit circle, where 
represent 
.
and cyclic permutations. Let 
for some real constant 
must be pure imaginary,
So 
.
It follows that the circumcenter 
of 
is located at
Intermediate steps are omitted so the reader does not need to suffer the factorization process like I did. Then we have 
It will suffice to show that 
is pure imaginary. In particular, 
So it suffices that the ratio 
which is clear as the expression equals the negative of its own conjugate.
.
and 
are similar. Since 
symmedian of 
,
symmedian of 
,
.
.
.
to 
Queue point of 
,
![[asy]
import geometry;
size(10cm);
defaultpen(fontsize(9pt));
pen pri; pri=RGB(31, 191, 184);
pen sec; sec=RGB(5, 113, 108);
pen tri; tri=RGB(25, 120, 165);
pen qua; qua=RGB(3, 17, 99);
pen fil=invisible;
pen sfil=invisible;
pen tfil=invisible;
pen qfil=invisible;
pair O, A, B, C, I, D, E, F, R, P, Q, K, D_prime;
O=(0, 0);
A=dir(125);
B=dir(220);
C=dir(320);
I=incenter(A, B, C);
D=foot(I,B,C);
E=foot(I,A,C);
F=foot(I,A,B);
R=2*foot(I,D,foot(D,E,F))-D;
P=2*foot(I,A,R)-R;
Q=intersectionpoints(circle(B,F,P), circle(C,E,P))[1];
K=extension(D,I,P,Q);
D_prime=2I-D;
fill(D--E--F--cycle,sfil);
filldraw((path)(A--B--C--cycle), white+0.1*pri, pri);
filldraw(circumcircle(B, P, F), tfil, tri);
filldraw(circumcircle(C, E, P), tfil, tri);
filldraw(circumcircle(B, I, C), sfil, sec);
filldraw(circumcircle(A, B, C), fil,pri);
filldraw(incircle(A, B, C), fil, pri);
draw(R--D,tri+dashed);
draw(P--K,tri+dashed);
draw(D--K,tri+dashed);
draw(A--K,tri+dashed);
dot("$A$",A,dir(110));
dot("$B$",B,dir(200));
dot("$C$",C,dir(-15));
dot("$K$",K,dir(-5));
dot("$I$",I,dir(-30));
dot("$D$",D,S);
dot("$E$",E,dir(75));
dot("$F$",F,dir(130));
dot("$R$",R,NW);
dot("$P$",P,dir(120));
dot("$Q$",Q,dir(-20));
dot("$D'$",D_prime,dir(45));
[/asy]](http://latex.artofproblemsolving.com/a/1/4/a1476ee173037fd0cbdfb02c95b44d0784b4bc5e.png)
which proves the claim. 
and 
are isogonal.
and 
where 
lies on 
.
we observe 
,![[asy]
import geometry;
size(7cm);
defaultpen(fontsize(9pt));
pen pri; pri=RGB(31, 191, 184);
pen sec; sec=RGB(5, 113, 108);
pen tri; tri=RGB(25, 120, 165);
pen qua; qua=RGB(3, 17, 99);
pen fil=invisible;
pen sfil=invisible;
pen tfil=invisible;
pen qfil=invisible;
pair I, D, E, F, A, B, C, X, Y, P, Q, D_prime, K;
I=(0, 0);
D=dir(110);
E=dir(220);
F=dir(320);
A=((E+F)/2);
B=((D+F)/2);
C=((D+E)/2);
X=((D+A)/2);
Y=intersectionpoints(circle(D,E,F),line(D, A))[0];
P=intersectionpoints(circle(X,abs(X-D)), circle(D, E, F))[1];
Q=intersectionpoints(circle(P, E, C), circle(B, P, F))[0];
D_prime=2I-D;
K=foot(A,D,D_prime);
filldraw((path)(D--E--F--cycle), white+0.1*pri, pri);
filldraw(circumcircle(D, E, F), sfil, pri);
filldraw(circumcircle(P,A,D), sfil, tri);
filldraw(circumcircle(P,Q,Y), sfil, tri);
draw(circumcircle(P,Q,C), dashed+sec);
clipdraw(circumcircle(P,Q,B), dashed+sec);
draw(D--Y,tri);
draw(Q--B,tri);
draw(D--D_prime,tri);
draw(A--K,tri);
draw(D_prime--P,tri+dashed);
dot("$A^*$",A,dir(-120));
dot("$B^*$",B,dir(40));
dot("$C^*$",C,dir(-110));
dot("$I$",I,dir(20));
dot("$D$",D,N);
dot("$E$",E,dir(-100));
dot("$F$",F,dir(-40));
dot("$P$",P,dir(130));
dot("$Q^*$",Q,dir(W));
dot("$X$",X,dir(240));
dot("$Y$",Y,dir(-40));
dot("$D'$",D_prime,dir(-50));
dot("$K$",K,dir(-15));
[/asy]](http://latex.artofproblemsolving.com/d/2/5/d25b23e501c5932b5b5849bfcd2d5dc7a923646b.png)
.
is cyclic. This follows as,
which proves the claim. 
,
and 
.
by homothety.
.
as,
where we have 
as both 
because 
because,
so we have proven the claim. 
from the picture.
lies on 
.
and thus the claim is proven. 
implies 
are concurrent.
,
,
.
and so on. Define 
,
is power of point 
.
.
.
.
.
and 
.
and similar equality with segment 
.
,
,
is harmonic quadrilateral, so 
and it ie enough to prove that 
or 
,
)
as the intersection of the external bisector of 
and 
is orthogonal 
are concyclic
are concyclic, 
are concyclic
are concyclic
are concyclic
and 
are concyclic (this is obvious)
are concyclic
be 
,
as 
.
(so the antipode of 
,
,
.
and 
.
collinear.
we get 
.
and 
are symmetric about line 
is a median so 
is a symmedian. This implies that 
by projecting 
onto 
is tangent to 
is the bisector, and 
from 
,
.
and 
denote the inverses of 
,
and the foot from 
be the midpoints of 
,
,
and 
are cyclic. Then, ![\[\measuredangle PQB'=\measuredangle PFB'=\measuredangle PFD=\measuredangle PED=\measuredangle PEC'=\measuredangle PQC'\]](http://latex.artofproblemsolving.com/7/d/5/7d53afe464cf8f392bc13c4b1d1241c20291df72.png)
so 
collinear. Also, 
so 
collinear. A homothety at 
to 
as both are perpendicular to 
as desired.
. We now invert with respect to the incircle and restate in terms of 
with center 
.
such that 
,
is cyclic.
is a cyclic quadrilateral with diameter 
are cyclic and finally even more angle chasing shows that 
is cyclic.
,
are collinear.
is median on 
and 
is 
s![\[\measuredangle KPE=\measuredangle KDE=\frac{\measuredangle C}{2}=\measuredangle FPA=\measuredangle MPE\]](http://latex.artofproblemsolving.com/0/5/b/05b00df055c33cbfb4b882400cc257b0ddcde346.png)
Hence 
.
respectively. 
is the foot of the altitude from 
are the midpoints of 
respectively and 
intersects 
intersects 
.
then 
are concyclic.
.![\[\measuredangle PQF=\measuredangle PBF=\measuredangle PBA=\measuredangle PCA=\measuredangle PCE=\measuredangle PQE\]](http://latex.artofproblemsolving.com/3/a/b/3ab100d48e1465aefe067c10faa4a634e00f7093.png)
Thus, 
are collinear
are collinear.
.![\[\measuredangle T'QF=\measuredangle PQF=\measuredangle PBA=\measuredangle PT'A\]](http://latex.artofproblemsolving.com/3/4/4/344fe4afd13bd5161925118bf501e06784fc8762.png)
This yields 
.
is the angle bisector of 
.
and 
.![\[-1\overset{?}{=}(Q(XY)_{\infty},\overline{QFE};\overline{QPT},QH)\overset{XY}{=}(XY_{\infty},Y;X,H)\]](http://latex.artofproblemsolving.com/8/3/c/83ca7efc560668cd246233c272a5deeee7b3796c.png)
.
with respect to 
.
and 
are collinear.
we get that 
are concyclic. Also 
and 
are isosceles trapezoids. By 
we have![\[\measuredangle PTA=\measuredangle PSA=\measuredangle DSH=\measuredangle DT'H=\measuredangle H'TD'=\measuredangle H'TA\]](http://latex.artofproblemsolving.com/b/6/2/b623e0b22f4d9986b77af2edd01161b53b7ddfab.png)
This yields 
are collinear. Hence 
.![\[\measuredangle PQH=2\measuredangle PQF=2\measuredangle PBA=\measuredangle POA=180-\measuredangle HOP\]](http://latex.artofproblemsolving.com/f/4/c/f4c85ad767bfceb98998cd575ecaeea57255cb33.png)
As desired
is cyclic. Points 
,![\[\angle PQC = \angle PEC = \angle PFB = \angle PQB.\]](http://latex.artofproblemsolving.com/c/c/d/ccddb8c57355e35d227a450ab2f6f3bb697f1186.png)
Now, 
finishes.
meet 
and we will also prove 
.
are isogonal so it suffices for ![\[\frac{\sin(\angle BAR)}{\sin(\angle CAR)}=\frac{\sin(\angle BAT)}{\sin(\angle CAT)}=\frac{\sin(\angle CAS)}{\sin(\angle BAS)}\]](http://latex.artofproblemsolving.com/4/b/4/4b4a27edf7a58841aa19db32fd02dbe628b14798.png)
By ratio lemma ![\[\frac{\sin(\angle CAS)}{\sin(\angle BAS)}=\frac{CS}{BS}\cdot\frac{AB}{AC}=\frac{p-b}{p-c}\cdot\frac{AB}{AC}\]](http://latex.artofproblemsolving.com/0/3/a/03adbb3514a34778a73cad428ba35bb8e588c2c2.png)
By trig ceva ![\[\frac{\sin(\angle BAR)}{\sin(\angle CAR)}=\frac{\sin(\angle AFR)}{\sin(\angle RFE)}\cdot\frac{\sin(\angle FER)}{\sin(\angle REA)}=\left(\frac{\sin(\angle FDR)}{\sin(\angle EDR)}\right)^2=\left(\frac{\sin C/2}{\sin B/2}\right)^2\]](http://latex.artofproblemsolving.com/f/8/8/f88c5c46cc6cad9322b8c9169f2513c3a20264f6.png)
One can easily check that these are equal so we get ![\[\measuredangle BQC=\measuredangle BQP+\measuredangle PQC=\measuredangle BFP+\measuredangle PEC=\measuredangle EDF= \measuredangle BIC\]](http://latex.artofproblemsolving.com/9/a/1/9a151055dc9fcb16dbd1396072cd9b4768353466.png)
So 
,
and 
are the midpoints of 
,
is the midpoint of 
and 
are the intersections of 
with the circle with diameter 
and 
,
and 
is cyclic. Now look at the problem from the perspective of 
is the nine-point circle. The problem translates to 
being cyclic.
.
.
(which is centered at 
and since 
is a parallelogram, we get that 
we have 
so 
is an isosceles trapezoid, which suffices.
so 
is actually the perpendicular bisector of 
.![\[\measuredangle DIM'=\measuredangle M'IP= \measuredangle K'IN'\]](http://latex.artofproblemsolving.com/1/0/b/10bca1fa7c33aafa0df0ba52aa5feecccf6a4750.png)
And since 
we get that 
is cyclic (well known lemma). 
.
and ![\[\measuredangle PDK'=\measuredangle PDE+\measuredangle EDK'=\measuredangle PDE+90^\circ - \measuredangle DFE=90^\circ-\measuredangle PED\]](http://latex.artofproblemsolving.com/6/7/c/67ca8eb9d62cb4626cf8e1ffa7eb42fa3cd4d7e2.png)
Therefore, using that 
we get ![\[\measuredangle M'K'P=90^\circ-\measuredangle PDK'=\measuredangle PED=\measuredangle C'Q'P. \ \ \ \blacksquare\]](http://latex.artofproblemsolving.com/7/9/6/796dcf4d72f6b93d2114c454169357d475de2b19.png)
be the reflection of 
be the feet from 
be the intersections of 
with 
be the second intersections of 
with 
and 
.
are collinear. Next, 
by Iran lemma, and by symmetry 
.
,
is cyclic and 
.
.
lie on the circle with diameter 
.
by Iran lemma, so by symmetry 
and thus ![\[\measuredangle PZP_C=\measuredangle PZE=\measuredangle PFE=\measuredangle PX_CP_C\]](http://latex.artofproblemsolving.com/0/7/4/0744b511398448a78df97c554ff20a074fd74f65.png)
as desired.![\[FR_C\cdot FE=FP\cdot FX_C=FP_B\cdot FZ=EP_B\cdot ED=EQ_C\cdot EF,\]](http://latex.artofproblemsolving.com/5/d/b/5db577bfc3dbb01c65f64fb06fd577a38c236bae.png)
since 
by Iran lemma. This implies 
,
.
.
and line 
and 
we get the involution is reflection over 
must also pass through 
be the circular points at infinity. Construct the cubic 
through 
.![\[((BDIF\mathcal I\mathcal J)\cup\overline{AEC})\cap((CDIE\mathcal I\mathcal J)\cup\overline{AFB})\]](http://latex.artofproblemsolving.com/c/2/5/c25d0bd6a0f20204cc7284f2a0ecabbe865968e0.png)
we see 
.![\[B+F+P+\mathcal I+\mathcal J=P+\mathcal I+\mathcal J-A=C+E+P+\mathcal I+\mathcal J\]](http://latex.artofproblemsolving.com/5/b/f/5bfb8e8551ad9a3c0c3f8a1cc540106739e376f7.png)
we get 
.![\[B+F+\mathcal I+\mathcal J+P+Q=0=B+F+\mathcal I+\mathcal J+D+I,\]](http://latex.artofproblemsolving.com/b/0/4/b0481a4c3a0a99f977680c245612e1e090013cbb.png)
we get 
.![\[B+D+I+F+\mathcal I+\mathcal J=0=B+F+A+D+I+X,\]](http://latex.artofproblemsolving.com/4/2/d/42ddd1fdd440494a7626196f96ef07f23842c51e.png)
so 
.![\[\measuredangle BQC=\measuredangle BQP+\measuredangle PQC=\measuredangle BFP+\measuredangle PEC=\measuredangle FEP+\measuredangle PFE=\measuredangle FPE=\measuredangle FDE=\measuredangle BIC,\]](http://latex.artofproblemsolving.com/1/e/9/1e904e132621d75b7722cbf2060bedf811edf03c.png)
so 
.
are coconic, 
,![\[E+F=\mathcal I+\mathcal J=A+X.\]](http://latex.artofproblemsolving.com/8/5/a/85a571df813ac39063b1512e6d2e3cb903f0956d.png)
Thus 
intersect on the line at infinity, as desired.
Let 
Let 
and let 
meet 
We start with a preliminary claim:
and 
then 
Let 
and let 
It is clear that 
Projecting this harmonic bundle through 
so 
passes through 
are collinear.
and 
say 
gets sent to the second intersection of 
say 
Similarly define 
intersects 
then 
intersects 
at 
then 
are collinear.
and 
are cyclic and 
be the 
are concyclic, and let 
with 
We will show that this 
is cyclic.
Clearly 
Finally, say by the incircle diameter lemma, we see that 
is tangent to 
This proves our claim.
is cyclic.
is tangent to 
The circle 
This if 
then 
Therefore, our inverted claim is as follows:
such that 
Let 
Let 
be the midpoints of 
respectively, Prove that 
is cyclic.
the claim thus follows by Reim's Theorem.
is cyclic. This equates our two definitions of 
are collinear.
so symmedian properties give 
But 
and 
Therefore, 
and we are done!
.
and the antipode of 
.
.
where 
is the reflection of 
and 
then by angle chasing we have 
So 
is cyclic. Also by PoP we have 
Which gives 
is cyclic and so 
.
is cyclic. Let 
and 
.
and 
.
which would imply the result.
and 
which gives 
is a parallelogram and so 
.
.
.
And 
so we must prove that 
and this is true since
And we are done.
