[/quote]
,
be real numbers such that 
.
is the interval ![$\left[-\frac{2}{3}, \frac{2}{3}\right]$](http://latex.artofproblemsolving.com/4/f/2/4f22df2b391cf34b9b4f86634575e3e0ad23a386.png)
.
selects a graph with 
vertices such that each vertex is incident to at least one edge. Then 
deletes some of the edges (possibly none) from the chosen graph. Finally, 
of a table 
is written the number 
.
such that there are exactly 
rows not containing a perfect square.
be an acute scalene triangle. A point 
varies on its side 
.
and 
are the midpoints of the arcs 
and 
(not containing 
and 
,
passes through a fixed point, independent of the choice of 
such that 
divides 
for any positive integers 
and 
.
.
intersects 
and 
.
is the midpoint of 
.
be the altitude in 
,
intersects 
be the midpoint of 
be the reflection of 
and 
are similar.
and 
Prove that 
and 
.
Prove 
Find maximum 
that is not isosceles at 
,
be its incircle, which is tangent to 
at 
,
and 
intersect the line passing through 
at 
and 
,
and perpendicular to 
intersect 
and 
at 
and 
,
are collinear and that 
are concurrent.
,
.
such that 
.
be non-negative real numbers such that 
Prove that
Y by doxuanlong15052000, mathmaths, Davi-8191, Rarvic, thczarif, lilavati_2005, Purple_Planet, samrocksnature, megarnie, Adventure10, Mango247, ItsBesi, Rounak_iitr, MS_asdfgzxcvb
Let 
be an acute triangle with orthocenter 
. Let 
be the point such that the quadrilateral 
is a parallelogram. Let 
be the point on the line 
such that 
bisects 
. Suppose that the line intersects the circumcircle of the triangle 
at 
and 
. Prove that 
.
be an acute triangle with orthocenter 
. Let 
be the point such that the quadrilateral 
is a parallelogram. Let 
be the point on the line 
such that 
bisects 
. Suppose that the line intersects the circumcircle of the triangle 
at 
and 
. Prove that 
.This post has been edited 2 times. Last edited by v_Enhance, Jul 7, 2016, 7:46 PM
Reason: Use official SL wording
Reason: Use official SL wording
be the intersection of 
.
.
.
are concyclic. Let 
be the second intersection of this circle with 
,
is a parallelogram.
,
is 
as desired.
.
on the other hand we have 
,
.
and we have 
.
and it's not hard to find 
with 
and 
.
and we are done.
.
is cyclic. Furthermore 
.
and 
:
be the midpoint of 
.
and 
,
,
.
,
is cyclic, so ![\[ \measuredangle CAH = \measuredangle CBH = \measuredangle CGH = \measuredangle CGI = \measuredangle CJI = \measuredangle MJI. \]](http://latex.artofproblemsolving.com/f/c/a/fca46f1eaf4e3d9a1797b1e6ff21fcfb5c8593e1.png)
Moreover, 
,
,
is cyclic. Define 
,
to 
.
and 
are cyclic. Then 
which, combined with 
gives us that 
is a median and an angle bisector in 
,
;
bisect each other, so 
is a parallelogram. Furthermore, 
.
is cyclic, 
.
is cyclic because 
is perpendicular to 
and 
degrees. Then 
.
.
.
is 
,
,
.
and a well-known result: 
,
we have 
But by Law of Sines, 
and 
we have 
and consequently 
cyclic. But now 
where the last equality is derived from 
intercepting both angles.
so 
but since 
and similar for 
,
is a parallelogram. Note 
lie on circle with diameter 
.
be te reflection of 
apone 
.
i
and 
.
wrt circumcircle. Hence
is a parallelogram 
. This fact together with above result implies that
be the intersection of 
and 
.
. Now since 
are collinear, by applying sine rule to triangles 
and 
we can get that 
![[asy]
import olympiad; import cse5;
defaultpen(fontsize(10pt));
usepackage("amsmath"); usepackage("amssymb"); usepackage("gensymb"); usepackage("textcomp");
size(8cm);
pair MRA (pair B, pair A, pair C, real r, pen q=anglepen) {
r=r/2;
pair Bp=unit(B-A)*r+A;
pair Cp=unit(C-A)*r+A;
pair P=Bp+Cp-A;
D(Bp--P--Cp,q);
return A;
}
pointpen=black+linewidth(2);
pen polyline=linewidth(pathpen)+rgb(0.6,0.2,0);
pen polyfill=polyline+opacity(0.1);
pen angleline=linewidth(pathpen)+rgb(0,0.4,0);
pen anglefill=angleline+opacity(0.4);
markscalefactor=0.01;
size(12cm);
pair A=dir(130),B=dir(210),C=dir(330);
// filldraw(A--B--C--cycle,polyfill,polyline);
D(A--B--C--cycle,polyline);
pair H=orthocenter(A,B,C);
pair G=B+H-A;
pair P=extension(G,H,A,C);
pair I=2*P-H;
pair J=2*foot(circumcenter(G,C,I),A,C)-C;
pair X=H+I-J;
D(B--G--H--A,polyline);
D(H--I--J--A); D(circumcircle(G,C,I));
D(H--X);
D("A",D(A),dir(65));
D("B",D(B),B);
D("C",D(C),C);
D("H",D(H),dir(-41));
D("G",D(G),S);
D("P",D(P),dir(190));
D("I",D(I),unit(I-H));
D("J",D(J),unit(J-H));
D("X",D(X),unit(A+C));
[/asy]](http://latex.artofproblemsolving.com/0/f/e/0fef6a7271fbad7aed89af150b5c5e9d5b91fc05.png)
.
.
is a parallelogram as its diagonals bisect each other. Therefore it suffices to prove 
.
is cyclic, and that 
,![\[\angle HAX = 90^\circ - \angle ACB = \angle HBC = 180^\circ - \angle HXC = \angle AXH\]](http://latex.artofproblemsolving.com/9/2/7/92779c4e5f0ccd73fc81a2a72f23caec3a1013fd.png)
so we are done.
,
and 
.
,![\[\angle CGH = \angle CBH = \angle DAC.\]](http://latex.artofproblemsolving.com/8/d/2/8d242b09fb82d78d1002ee721cc88f1d401ddbec.png)
Now we have that by the sine law;
as desired. 
with ratio 
.
,
.
the 
and 
.
,
be a point on 
is the midpoint of 
.
is a parallelogram.
,
,
However, 
is cyclic, so 
Thus, 
is isoceles, and 
,
is perpendicular to 
,
is cyclic, 
and 
,
and 
so 
,
cyclic 
due to equality of all angles with 
and Conclusion follows
,
so 
has length 
where 
is the radius of the circumcircle. Let 
and 
,
.
by reflecting the orthocenter, and we are done.
![\[
\textbf{IMOSL 2015/G1}
\]](http://latex.artofproblemsolving.com/0/8/8/088f3516af46e8ebf74e12238fa3d27fdf658707.png)
.
to 
at 
in 
to get 
![\[
AH = HM = BG \quad \text{and} \quad \angle BMH = \angle BMA - \angle HMA= \angle A-\angle 90- \angle C=90-\angle B=\angle BGH\]](http://latex.artofproblemsolving.com/8/3/e/83e3b066ae51951d13b514f74d78b4c6b6a083dd.png)
imply that 
is an isosceles trapezium.
and the claim follows.
and 
are cyclic, we get:![\[
CK \cdot KJ = IK \cdot KG= KH\cdot KG
\]](http://latex.artofproblemsolving.com/9/b/f/9bf8e59b20ad852600497bebc60577b2b49b98ae.png)
![\[
\implies KJ = \frac{KH \cdot KG}{CK} =KG \frac{KH}{CK}=KG \cos A=KM
\]](http://latex.artofproblemsolving.com/8/4/6/8467376630b34c57f84d147c0724ffe1fe5f9cf5.png)
is a parallelogram.![\[
\angle IJL = \angle IJM = \angle JMH = 90 - \angle C
\]](http://latex.artofproblemsolving.com/e/1/4/e14f539e9fac5f75bc16fdf8b7d36ccd06e35850.png)
![\[
=\angle HAA = \angle JL
\]](http://latex.artofproblemsolving.com/1/8/2/1826b3ed61fd807b54c6a8ec81872d887c21644a.png)
,
is a parallelogram.
.![\[\angle IJX = \angle IGC = \angle HBC = \angle CAH,\]](http://latex.artofproblemsolving.com/4/0/1/40196d4c0bdbc32bd3ca285d0b88c799839f8dd8.png)
where the second step follows since 
,
.
we get 
Addditionally, 
,
.
with 
.
is the 
.![\[\measuredangle IJP=\measuredangle IGC=\measuredangle ABZ=\measuredangle ACZ=\measuredangle JPI,\]](http://latex.artofproblemsolving.com/f/1/3/f1379f8503365e4764c6b7215af37b63cfa85bb8.png)
so 
as desired.
![[asy]
import olympiad; import geometry;
size(6cm);
pair A,B,C,H,G,K,I,J,P;
A=dir(120); B=dir(210); C=dir(330); H=orthocenter(A,B,C); G=(B+H-A); K=intersectionpoint(line(G,H),line(A,C)); I=(2K-H);J=intersectionpoints(circle(G,C,I),line(A,C))[1];P=(2K-J);
draw(A--B--C--cycle,blue); draw(B--G--H--A,blue); draw(H--I,blue);draw(circle(G,C,I),purple);draw(A--J,blue); draw(H--P--I--J--cycle,orange); draw(arc(circumcenter(G,B,C),C,G),deepred);
draw(G--C--H,blue);
dot(A);dot(B); dot(C);dot(H); dot(G);dot(K); dot(I);dot(J);dot(P);
label(A,"$A$",N); label(B,"$B$",NW); label(C,"$C$",NE); label(G,"$G$",SW); label(H,"$H$",W); label(P,"$P$",NE); label(K,"$K$",E); label(I,"$I$",NE); label(J,"$J$",NW);
[/asy]](http://latex.artofproblemsolving.com/0/e/1/0e1af03defef23caba45f6c306a432bd116a45c9.png)
is parallelogram and since 
is cyclic, we have 
thus 
is cyclic and so 
and since 
we have that 
and thus 
.
we have 
which implies that 
is an equilateral trapezoid and thus 
.
and we are done.
(also this is India TST 2016 P1)
as feet of altitudes from 
to 
respectively.
.
.
implying that, 
.
is a cyclic quadrilateral.
.
.
.
thus, 
or 
.
and 
.
we have, 
.
and 
.
.
,
as desired. 
)
.
Therefore 
is cyclic. Reims gives 
this combined with 
means that 
is a parallelogram so 
.
