and 
. Prove that
Let 
. Prove that
and 
. Prove that
contestant of EGMO are named 
.
with 
.
has at least 
numbers 
and 
numbers 
are written in some order.
.
Y by GeoKing, LoloChen, aansc1729, HoripodoKrishno, Rounak_iitr
Let 
be a scalene triangle, 
be the midpoint of 
be the common point of 
and the incircle of closest to 
, and 
be the common point of the ray and the excircle farthest from . The tangent to the incircle at 
meets 
at point 
, and the tangent to the excircle at meets at 
. Prove that 
.
be a scalene triangle, 
be the midpoint of 
be the common point of 
and the incircle of closest to 
, and 
be the common point of the ray and the excircle farthest from . The tangent to the incircle at 
meets 
at point 
, and the tangent to the excircle at meets at 
. Prove that 
.
? @above noo sniped 
. @below lmao ded 
I
and 
denote the incircle and 
e
i.e., 
.
be the points where 
touches 
.
is the contact triangle, then 
concur which I refer to as Incircle Concurrency Lemma.
be the incenter and 
denote the incircle and 
be the point where 
.
be the points where the 
be the line through 
.
.
.
and 
is cyclic.
T
,
.
centred at 
also maps 
and 
,
and 
.
.
and so 
and similarly 
.
is cyclic.
and so we get 
are tangent to each other at 
be the tangents at 
.
and 
is the angle bisector of 
,
.
are collinear 
and 
.
are the same line, we conclude the circles are tangent at 
.
is the midpoint of 
.
and 
be the point where 
be the centre of 
.
are collinear 
.
and 
are collinear and 
,
centred at 
.
and so their centers also get mapped, i.e. 
.
.
and 
which together with the fact 
.
from which we can conclude that 
.
and 
.
as the chords, we get 
so 
.
as the chords, we get 
so 
a little (because yeah, I messed them up and changing the definitions here would have needed to rewrite the entire Length Bash below which I am too lazy to do 
and 
.
are tangents to 
and 
.
and
.
and 
.
,
And we are done 
.
.
and hence 
.
where 
are the intouch and extouch points to BC respectively.
![[asy]
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.
are isotomic conjugates in 
and we have 
.
meets the incircle again at 
and the excircle at points 
in that order along the line 
.
be the touch points of 
with the incircle and 
the corresponding touch points with the excircle. We'll prove a short claim before proceeding:
and 
concur at some point 
.
contains the polar of 
.
is tangent to the excircle, which in particular will imply that the polar of 
to 
,
,
.
is right angled at 
.
,
and then because 
is tangent to the excircle, it follows that 
.![\[(XD;BC) = (DP, DD; DF, DE) \overset{D}{ = } (PD;FE) \overset{A}{ = } (Q'U;FE)\]](http://latex.artofproblemsolving.com/9/b/b/9bb327064601085217ddb99e553e8e5d496457c8.png)
where 
is defined as the tangent to the incircle at 
.![\[(QD';E'F') = (D'Q, D'D'; D'E', D'F')\]](http://latex.artofproblemsolving.com/5/a/0/5a09611f3e99027b67aab1024453ced1be08f024.png)
where 
is the tangent to the excircle at ![\[(D'Q, D'D'; D'E', D'F') = (YD';CB)\]](http://latex.artofproblemsolving.com/6/1/b/61bc2db13bf26c891aea1e016f0733ba7155b8f5.png)
Putting all the above equalities together, we have obtained that 
.![\[\dfrac{\dfrac{XB}{XC}}{\dfrac{DB}{DC}} = \dfrac{\dfrac{YC}{YB}}{\dfrac{D'C}{D'B}}\]](http://latex.artofproblemsolving.com/4/3/7/437f1f60c383ccf4873955be2616cb29a48b12c0.png)
Because 
.
or that 
.
,
so 
.
similarly 
so 
.
to 
so 
is radax of the incircle and excircle .Since the radax of incircle and excircle is also midline of trapezoid 
which bisects 
so 
we get 
and hence 
.
and 
,
,
.
,
.
to the incircle meet 
and thus 
.
are parallelograms, so 
.
,
.
and the excircle of 
.
Let 
be a triangle with 
are two different circles with centers 
,
where 
are orthogonal projections of 
with 
.
.
be midpooint of 
.
.
which are midpoints of 
and 
and 
is radax of the 
circles
is simson line of 
be the center of the incircle of 
be the center of the 
be the intersection of 
and 
.
.
be the second intersection of 
and 
are parallel, so 
so 
and our lemma is proved.
centered at 
and 
both internally tangent to 
respectively. If 
passes through the midpoint 
,
,
so triangles 
and 
are congruent and so 
.
and so 
and 
,
.
intersect 
and 
intersect 
.
does not pass through this midpoint. Let 
be the midpoint of 
.
and 
at 
and 
.
(because 
.
and 
(because 
and 
,
and 
must have a common point (by assumption 
)
,
goes to the tangent of excircle through 
at 
and since 
,
.
is degenerate quadriple. Apply DDIT from 
is an involution.
is an involution.
is refletcion of 
is also involution which implies 
since any involution on a line has 
respectivily.
hit the incircle again at 
,
.
Hence 
