[/quote]
,
be real numbers such that ![\[ab^2+bc^2+ca^2=6\sqrt{3}+ac^2+cb^2+ba^2.\]](http://latex.artofproblemsolving.com/3/d/0/3d015292c249a69a44d5f923092201d7a271b896.png)
Find the smallest possible value of 
.
such that the following statement holds: Suppose real numbers 
,
,
,
,
,
,
satisfy 
for all 
.
,
,
,
or 
,![\[ \left| \sum_{i=1}^n \varepsilon_i a_i \right| + \left| \sum_{i=1}^n \varepsilon_i b_i \right| \le 1. \]](http://latex.artofproblemsolving.com/6/5/8/658ddc5e8626e7d5ef29809c8c3ac641e3a87757.png)
vertices is given. To each vertex, a natural number from the set 
is assigned. Prove that there are 
vertices 
such that if the numbers 
are assigned to them respectively, then 
and 
is a parallelogram.
such that 
.
.
.
has 
obtuse. Let 
,
,
,
,
respectively. Let 
,
,
be arbitrary points on 
,
,
,
,
are drawn. Show that the lengths of the tangents from the orthocentre of 
Prove that
Where 
Prove that
Let 
Prove that
Let 
Prove that
such that 
.
is inscribed in a circle. Given 
and 
.
and 
to 
.
of a triangle 
respectively. The lines 
and 
meet 
at points 
and 
.
and points 
such that the polynomial 
has 8 linear factors of the form 
with 
for 
.
.
Y by Davi-8191, anantmudgal09, mijail, Adventure10, Mango247
The incircle of triangle 
touches sides 
and 
at points 
and 
, respectively. Let 
be the intersection of lines 
and 
. The reflections of with respect to 
and 
are 
and 
, respectively. Prove that lines 
and 
are concurrent at a point on line 
, where 
and 
are the incenter and circumcenter of triangle .
touches sides 
and 
at points 
and 
, respectively. Let 
be the intersection of lines 
and 
. The reflections of with respect to 
and 
are 
and 
, respectively. Prove that lines 
and 
are concurrent at a point on line 
, where 
and 
are the incenter and circumcenter of triangle .
are the altitudes of 
.
lies on 
on 
on 
.
are collinear. Let 
,
and 
is the angle bisector of 
,
.
and 
concur is equivalent to show that 
concur. This can be done by trig Ceva. Easy to see that:
and so on …
,
indeed concur at a point, say 
.
are collinear. Notice that 
,
,
and 
,
and so 
,
,
goes through 
.
are collinear, once again by simple Trig-Ceva on triangle 
,
are collinear, where 
,
,
concur on 
bisects 
hence 
is a line 
similarly, then 
and 
then 
are collinear, and since 
are corresponding we are done.
and 
then the lines 
and 
are concurrent at the Gergonne Point of 
![\[ux+vy+wz = 0\]](http://latex.artofproblemsolving.com/1/3/d/13d72685675c022d430b098e72dfca13bdee0ebf.png)
will be denoted as ![\[(u \, ; \, v \, ; \, w ).\]](http://latex.artofproblemsolving.com/4/5/a/45acf17235acf70c6ee672359d2cb1216e126b9f.png)
concur at the Gergonne point 
of 
.![\[(\frac{b-c}{a}(-a+b+c) \, ; \, \frac{c-a}{b}(a-b+c) \, ; \, \frac{a-b}{c} (a+b-c))\]](http://latex.artofproblemsolving.com/7/9/2/792112793d0f638561e1cf0c56aa9670e69a0cd4.png)
.![\[\frac{\sin FAX}{\sin XAE} = \frac{FX \cdot \frac{\sin AFX}{AX}}{EX \cdot \frac{\sin AEX}{AX}} = \frac{FG}{EG} \cdot \frac{\sin AFX}{\sin AEX}.\]](http://latex.artofproblemsolving.com/7/e/c/7ecdd62a6a058ccf7c40460b34baaeb86e7f9d4d.png)
A bit of angle chasing yields ![\[\measuredangle AFX = \measuredangle AFE + \measuredangle EFX = \measuredangle FEA + \measuredangle CFE = \measuredangle FEC + \measuredangle CFE = \measuredangle FCE \]](http://latex.artofproblemsolving.com/8/3/0/830687413e6f79491e120388a9093a0245b9c521.png)
(in directed angles) and thus ![\[\frac{\sin FAX}{\sin XAE} = \frac{FG}{EG} \cdot \frac{\sin GCE}{\sin FBG} = \frac{(\frac{\sin GCE}{EG})}{(\frac{\sin FBG}{FG})} = \frac{(\frac{\sin EGC}{CE})}{(\frac{\sin BGF}{BF})} = \frac{BF}{CE} = \frac{a-b+c}{a+b-c}.\]](http://latex.artofproblemsolving.com/5/d/9/5d9bf2adea8a91760404895f587b2dce03aedb86.png)
It follows that ![\[(x : \frac{b}{a-b+c} : \frac{c}{a+b-c}).\]](http://latex.artofproblemsolving.com/0/f/f/0ffeea6e376b3fe167409df1024bc3c80e3caf25.png)
Thus 
concur at the point ![\[Q = (\frac{a}{-a+b+c}:\frac{b}{a-b+c} : \frac{c}{a+b-c}) \]](http://latex.artofproblemsolving.com/d/6/d/d6d2df5beba30d5b02e515cb869d4b405dd6ebf8.png)
which is on line 
.
analogously. First we prove the
concur at a point 
Now 
so we compute 
and similarly we have 
.
It is now clear that the claim must hold. As an immediate corollary, we see 
.
be the isogonal conjugate of 
is a line. Note that 
.
Now we undertake the arduous task of plugging in 
Compute the RHS as 
On the other hand, compute LHS as 
Consequently, it suffices to check 
which is clearly true. 
bisects 
?
from 
you can see.
denote the foot of altitude from 
similarly. We will prove that 
,
,
,
,
bisects 
.
.
,
concur at the homothetic centre of the two triangles, say 
.
are concurrent on the Isogonal Mittenpunkt Point 
of 
which lies on 
.
is the foot of the altitude from 
.
meets 
,
intersect at a point 
and moreover 
,
and 
are reflections about 
is collinear. Now we know that 
to the nine-point circle of 
at 
respectively.
Let 
be the reflection of 
and let 
be the foot of the altitude from 
are collinear.
and let 
so 
collinear. Furthermore, 
collinear and 
to 
It is easy to see that this homothety sends 
so we are done. 
are the feet of the altitudes from 
to 
respectively. Now the concurrency follows from the Cevian Nest Theorem, so let 
so 
Similarily, 
and 
so triangles 
and 
the circumcenter of 
as desired.
its circumcircle and 
the altitude from 
to 
its reflection WRT 
are collinear.
.
are collinear, AKA the 
since 
bissects 
,
,
cuts 
such that 
.
similarly. An easy angle chasing gives us that 
are homothetic 
are concurrent (as well as 
)
.
are collinear. This implies that 
be the orthic triangle of 
,
.
,
,
.
,
,![[asy]
size(12cm);
pair A = dir(130); pair B = dir(210); pair C = dir(330); pair I = incenter(A, B, C); pair D = foot(I, B, C); pair E = foot(I, C, A); pair F = foot(I, A, B); pair G = extension(B, E, C, F); pair X = -G+2*foot(G, E, F); pair Y = -G+2*foot(G, F, D); pair Z = -G+2*foot(G, D, E); pair J = foot(D, E, F); pair K = foot(E, F, D); pair L = foot(F, D, E);
filldraw(A--B--C--cycle, invisible, blue); filldraw(incircle(A, B, C), invisible, blue); filldraw(D--E--F--cycle, invisible, red); draw(D--J, red); draw(E--K, red); draw(F--L, red); draw(A--D, blue); draw(B--E, blue); draw(C--F, blue); draw(A--J, deepgreen); draw(B--K, deepgreen); draw(C--L, deepgreen); draw(G--X, dashed+orange); draw(G--Y, dashed+orange); draw(G--Z, dashed+orange);
dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$I$", I, dir(I)); dot("$D$", D, dir(D)); dot("$E$", E, dir(E)); dot("$F$", F, dir(F)); dot("$G$", G, dir(280)); dot("$X$", X, dir(X)); dot("$Y$", Y, dir(Y)); dot("$Z$", Z, dir(Z)); dot("$J$", J, dir(120)); dot("$K$", K, dir(K)); dot("$L$", L, dir(315));
/* -----------------------------------------------------------------+ | TSQX: by CJ Quines and Evan Chen | | https://github.com/vEnhance/dotfiles/blob/main/py-scripts/tsqx.py | +-------------------------------------------------------------------+ !size(12cm); A = dir 130 B = dir 210 C = dir 330 I = incenter A B C D = foot I B C E = foot I C A F = foot I A B G 280 = extension B E C F X = -G+2*foot G E F Y = -G+2*foot G F D Z = -G+2*foot G D E J 120 = foot D E F K = foot E F D L 315 = foot F D E A--B--C--cycle / 0.1 lightcyan / blue incircle A B C / 0.1 lightcyan / blue D--E--F--cycle / 0.1 yellow / red D--J / red E--K / red F--L / red A--D / blue B--E / blue C--F / blue A--J / deepgreen B--K / deepgreen C--L / deepgreen G--X / dashed orange G--Y / dashed orange G--Z / dashed orange */
[/asy]](http://latex.artofproblemsolving.com/5/2/0/520d0adf0a8d8ce018f9cbb4065037c73a3728bb.png)
![\[ \measuredangle JKF = \measuredangle JKD = \measuredangle JED = \measuredangle FED = \measuredangle BFD = \measuredangle BFK \implies \overline{BA} \parallel \overline{JK}. \]](http://latex.artofproblemsolving.com/0/7/d/07de3eafd6c9a7c251c2eac88c026538e55a100f.png)
Now the incenters of 
and 
are the reflections of 
be the tangential triangle. Then 
,
and 
concur on the Euler line.
a
the reflection of the symmedian point of triangle 
be the reflection of 
the intersection of line 
with the circumcircle of triangle 
with line ![\[-1=(AC;BK_b) \overset{B}{=}(AL;TK)\overset{D}{=}(\infty,M;T,DK \cap MT)\]](http://latex.artofproblemsolving.com/f/5/c/f5c39904bb9125bec7152598d8859356c21cde44.png)
But this implies that 
and 
which proves the claim.
and 
the foot from 
to be the intersection of 
it follows that by right angles and bisectors, 
which implies that 
are collinear.
implying that 
and 
the nine point center of 
are collinear.
are collinear.
