,
and 
denote the midpoints of 
and 
.
be the projection of 
onto 
be the reflection of 
.
.
be given as: 
,
(where 
can be 0), and let 
such that 
and 
is considered the same as 
)
is 
,
,
,
are positive integers, and 
are odd primes.
be positive integers, and let 
all be 
-![\[
\alpha_1^j + \alpha_2^j + \cdots + \alpha_n^j = 0 \quad \text{for any } j (0 < j < k).
\]](http://latex.artofproblemsolving.com/3/3/b/33be8eef29cd506bf1132a86def9992ec8d21cb4.png)
Prove that among 
be the orthocenter of acute triangle 
,
,
be the reflection of 
.
intersects line 
and 
.
.
Y by Pirkuliyev Rovsen, bobcats4thewin, claserken, Davi-8191, anantmudgal09, Tumon2001, tenplusten, buratinogigle, Tawan, Ultroid999OCPN, woypat, nguyendangkhoa17112003, AlastorMoody, Centralorbit, RudraRockstar, mathematicsy, samrocksnature, Mahmood.sy, MathLuis, Lamboreghini, Adventure10, Aryan_Agarwal, deplasmanyollari, ItsBesi, Rounak_iitr
Let 
and 
be different points on a circle 
such that 
is not a diameter. Let 
be the tangent line to at . Point 
is such that is the midpoint of the line segment 
. Point 
is chosen on the shorter arc of so that the circumcircle 
of triangle 
intersects at two distinct points. Let 
be the common point of and that is closer to . Line 
meets again at 
. Prove that the line 
is tangent to .
Proposed by Charles Leytem, Luxembourg
and 
be different points on a circle 
such that 
is not a diameter. Let 
be the tangent line to at . Point 
is such that is the midpoint of the line segment 
. Point 
is chosen on the shorter arc of so that the circumcircle 
of triangle 
intersects at two distinct points. Let 
be the common point of and that is closer to . Line 
meets again at 
. Prove that the line 
is tangent to .Proposed by Charles Leytem, Luxembourg
This post has been edited 1 time. Last edited by djmathman, Jun 16, 2020, 4:13 AM
.
.
,
meets 
again at 
are collinear. To prove this, we will instead show that 
is a parallelogram; we already know that 
,
.
,
,
is cyclic we have 
,
.
is a parallelogram, so that 
and 
are collinear. Now since 
we have 
,![\[\dfrac{RK}{RS} = \dfrac{2RS}{AT}\quad\implies\quad RS\cdot RT = RK\cdot AT = RK\cdot RP.\]](http://latex.artofproblemsolving.com/4/1/a/41a7beed447b717c37c08e11d39ff5d097da4087.png)
Thus 
is a cyclic quadrilateral, which means that ![\[\angle KTS = \angle KPS = \angle SAT,\]](http://latex.artofproblemsolving.com/3/f/5/3f5770c6f998a87ddd1ac18e19db4c410562c8fb.png)
implying the tangency.
be the midpoint of 
By Reim's theorem 
so combining 
we get 
hence 
i.e. 
is tangent to 
![\[ \measuredangle RKA = \measuredangle RKJ = \measuredangle RSJ = \measuredangle TSJ = \measuredangle TAJ = \measuredangle TAK \]](http://latex.artofproblemsolving.com/7/0/d/70d0c8c4932fd142802123484ba72ff1dc9150fe.png)
so 
.
is tangent at 
(oppositely), because both equate to 
.
is tangent at 
.
the SAS gives 
.![[asy] size(10cm); pair T = dir(100); pair S = dir(165); pair R = 2*S-T; pair E = dir(0); pair A = IP(unitcircle, R--(R+100*E)); pair B = OP(unitcircle, R--(R+100*E)); pair K = extension(T, T+dir(90)*T, R, R+T-A); draw(R--B, blue);
draw(unitcircle, blue); draw(circumcircle(R, S, K), heavycyan); pair J = -A+2*foot(origin, A, K);
draw(K--R--A--T--cycle, red); draw(A--K, orange); draw(R--T, orange);
// invert
pair Js = extension(T, T+A-B, R, J); pair Ks = extension(T, T+A-B, R, K); draw(K--Ks, heavygreen); draw(Ks--Js, heavygreen); draw(Js--B, dotted+blue); draw(R--Js, dotted+blue);
dot("$T$", T, dir(T)); dot("$S$", S, dir(355)); dot("$R$", R, dir(270)); dot("$A$", A, dir(225)); dot("$B$", B, dir(315)); dot("$K$", K, dir(200)); dot("$J$", J, dir(180)); dot("$J^\ast$", Js, dir(Js)); dot("$K^\ast$", Ks, dir(Ks));
/* TSQ Source:
!size(10cm); T = dir 100 S = dir 165 R355 R = 2*S-T R270 E := dir 0 A = IP unitcircle R--(R+100*E) R225 B = OP unitcircle R--(R+100*E) R315 K = extension T T+dir(90)*T R R+T-A R200 T--K blue R--B blue
unitcircle 0.05 lightcyan / blue circumcircle R S K 0.05 paleblue / heavycyan J = -A+2*foot origin A K R180
K--R--A--T--cycle 0.1 lightred / red A--K orange R--T orange
// invert
J* = extension T T+A-B R J K* = extension T T+A-B R K K--Ks heavygreen Ks--Js heavygreen Js--B dotted blue R--Js dotted blue
*/ [/asy]](http://latex.artofproblemsolving.com/4/f/3/4f368562b8c2a11c4899a1c037c4b3ab8e807ba6.png)
.
,
is the second intersection of 
is the intersection of 
;
is an isosceles trapezoid. In particular, one reads 
from this, hence 
is a parallelogram.
is tangent to 
is cyclic, from 
collinear.
so 
is parallel to 
.
at 
then 
is a parallelogram.
so 
are cyclic.
,
is parallelogram.
collinear since 
Then 
As desired.
. Define point 
.
, but 
is midline in 
.Therefore 
)then three minutes.
and 
thus 
so 
and 
Now these give 
so 
thus 
,
,
.
are collinear. From angle chasing,we get 
,
,
,
we get 
which gives us the desired result.
.
,
are collinear, we have 
becomes 
so we just have to calculate 
and 
,
,
,
,
which is indeed symmetric to 
.
and 
is parallel to 
and 
so triangles 
and 
are similar and 
,
.
so triangles 
and 
are similar.
so 
.
.
implies 
is a parallelogram.
is cyclic.
implies 
.
,
![\[\angle JAT=180-\angle JST=\angle RSJ=\angle RKJ\]](http://latex.artofproblemsolving.com/1/9/0/190eb23f2ee865a72d0bd6f9168fe6adf405d217.png)
.
is a parallelogram. Therefore, 
is cyclic![\[\angle ARS=\angle RKS=180-\angle SK'A\]](http://latex.artofproblemsolving.com/8/4/2/842ebbd5ccdd11ff5122611afb0e896c30e03ce1.png)
![\[\angle RTK=\angle K'RS=\angle TAS,\]](http://latex.artofproblemsolving.com/1/7/1/171b6a49ba3c735cb1ca73531939abaf873380e1.png)
as desired 
similar to 
so 
is cyclic
is a parallelogram. By Reim's, 
so 
is cyclic. To finish, note that
we have:
and 
we get by 
chriter that triangles 
and 
are simmilar 
C
we get: 
Combining with 
we get by 
chriter that triangles 
and 
are simmilar 
-
.![\[\angle TAK = 180 - \angle JST = \angle JSR = \angle JKR\]](http://latex.artofproblemsolving.com/1/7/5/1752e81029aa6cfd132fb3383579519f934c6abb.png)
so 
.
.![\[\angle RAK' = 180 - \angle ARK = 180-\angle ARS - \angle SRK\]](http://latex.artofproblemsolving.com/3/a/7/3a7a1fc0ac0c6c7a5a904c0acef4d77a00a19ed8.png)
Notice that 
.
,![\[\angle RTK = \angle TRK' = \angle K'AS = \angle TAS = \angle TJS\]](http://latex.artofproblemsolving.com/2/a/e/2ae9a1d9de06f834b600d906fa2bbaafa6f1bb57.png)
so 
since 
(from the tangency) and 
(from the paralle lines). So,![\[ \frac{AT}{TR} = \frac{SR}{RK} = \frac{TS}{RK} \implies \frac{AT}{TS} = \frac{TR}{RK}.\]](http://latex.artofproblemsolving.com/3/f/7/3f7cfb213dbf547e0b65b79d9a261b1b471b8389.png)
Combined with the fact that 
,
from SAS similarity. So, it follows that 
,
so 
Furthermore, since 
the reflection 
are collinear and 
In particular, we have that 
since 
is tangent to 
,
and we conclude.
,
intersect line 
at 
with 
passes through the other intersection 
and 
.
symmedian with 
,
and 
are 
inverses. It follows that 
and we're done by converse Reim.
.
and hence 
are collinear. From the tagnency and the parallel lines we get that: 
is cyclic. From the cyclic quadrilaterals and the parallel lines we get that : 
and, hence, 
![\[\measuredangle AKR = \measuredangle JKR = \measuredangle JSR = \measuredangle JAT = \measuredangle KAT\]](http://latex.artofproblemsolving.com/8/a/7/8a74f3ca535deef48087271cce92a94cc6fd75c2.png)
from which it follows that 
.
,
is a parallelogram. We can further note that,![\[\measuredangle SRA = \measuredangle SKR = \measuredangle MKR = \measuredangle KMT\]](http://latex.artofproblemsolving.com/5/c/a/5cad1c56e724e702f44f24acdb564d08f5d1a5ad.png)
so quadrilateral 
must be cyclic. Finally note that, 
and 
,
,![\[\measuredangle ATK = \measuredangle MTK = \measuredangle KRM = \measuredangle ASR\]](http://latex.artofproblemsolving.com/9/3/c/93cd2c5eacd6668fa451eb7d8d6d69a049c27b16.png)
which implies that 
.
map to each other in this inversion.
are collinear along with 
.
is a cyclic trapezium.
were collinear which implies 
is a cyclic trapezium. Therefore 
is a parallelogram. It suffice to show that 
are tangent to each other.
is a cyclic quadrilateral which implies 
to be collinear. Therefore 
which implies 
