[/quote]
,
be a function satisfying the following conditions:
,
and 
,
;
is surjective.
.
is 
than the set of numbers 
if among all inequalities of the form 
the number of true inequalities is at least 
times greater than the number of false ones. Prove that there do not exist three sets 
such that 
,
we have 
.
.
and 
.
?
,![\[\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}\geq 4+\frac{8(a-c)^2}{(a+b+c+d)^2}.\]](http://latex.artofproblemsolving.com/7/b/6/7b6e32d32d89ca058a06974e7276f0a5874d90a7.png)
and orthocenter 
intersect 
respectively. Lines through 
intersects 
at 
respectively. Point 
such that 
is a parallelogram. Prove that lines 
and 
are concurrent at a point on 
.
satisfying
Prove
and 
be positive integers. Suppose that the sum of integers between 
.
,
.
be the circumcircle of 
be the midpoint of the arc 
of 
.
meets 
at 
.
at 
.
by 
.
.
Y by narutomath96, spacekid4, Davi-8191, centslordm, megarnie, Adventure10, Mango247, Rounak_iitr
Let 
be a triangle with orthocenter 
and let 
be the second intersection of the circumcircle of triangle 
with the internal bisector of the angle 
. Let 
be the circumcenter of triangle 
and 
the orthocenter of triangle 
. Prove that the length of segment 
is equal to the circumradius of triangle .
be a triangle with orthocenter 
and let 
be the second intersection of the circumcircle of triangle 
with the internal bisector of the angle 
. Let 
be the circumcenter of triangle 
and 
the orthocenter of triangle 
. Prove that the length of segment 
is equal to the circumradius of triangle .This post has been edited 1 time. Last edited by djmathman, Apr 30, 2014, 9:56 PM
be the midpoint of the arc 
be the circumcenter of 
be circumcenter of 
correspond to complex numbers 
.
,
,
,
as desired.
be the center of 
,
.
is the reflection of 
,
.![\[
\angle AYC=180-\angle APC=180-\angle AHC=\angle B
\]](http://latex.artofproblemsolving.com/8/b/9/8b96f6c7fdd8b51222a881f5dbbe8433d20f61b3.png)
are perpendicular to 
and their bisector, 
is isosceles with 
,
.![\[
\angle AOY=2\angle ACY=2(90-\angle PAC)=2(90-\frac{A}{2})=180-\angle A = \angle XOO_1
\]](http://latex.artofproblemsolving.com/f/6/d/f6dca07a72c7753108530576a2a6962acbad7282.png)
as well, and 
,
.
.
took me around 20 minutes, thankfully... I finished the test with five to spare so any more synthetic effort would have finished me.
to meet 
(
)
using the spiral similarity that sends 
to 
at 
,
is a parallelogram so 
.
to 
(these triangles are similar). It turns out 
and so 
,
.
,
is the reflection of 
is isoceles
is a parallelogram, where 
the midpoint of minor arc 
,
to 
,
,
.
,
.
![[asy]
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[/asy]](http://latex.artofproblemsolving.com/0/8/9/089b8d0653590238dcb137f35695cb7e2619cd59.png)
Which means 
and 
.
Which means 
.
will be a parallelogram and we get our desired result.
.
Therefore 
collinear and similarly 
collinear. And another angle chasing gives that 
And we are done. 
Notice 
and since the reflection of 
lies on 
we see 
Notice 
and 
Hence, 
But 
so 
and 
Therefore, 
is the length of the circumradius.
is cyclic by angle chasing.
.
is an isosceles trapezoid.
(where 
)
.
is just 
degrees, 
is just 
is just 
rotated 
.
,
,
(here 
is a directed angle). Then 
,
,
.
,
,
.
.
,
,
,
and 
.
.
,
,
.
.
,
respectively, the circumcenter of 
and 
respectively, 
,
.
,
.
and 
.
gives 
,
and 
which means 
.
be the perpendicular bisector of 
.
and 
follows from the orthocenter condition, we know 
and 
as desired. 
,
,
is cycle, which finally shows that 
is isosceles![\[\angle (OX,XO_1) = \angle (AB,AP) = \angle (AP,AC) = \angle (O_1X, OO_1)\]](http://latex.artofproblemsolving.com/5/7/3/573c29ee258eba4935d4f2c12ba9173147e5cd16.png)
due to perpendicularity conditions 
is an isosceles trapezoid.
.
and we're done. 
is a parallelogram, which finishes.
.
.
,
.
,
since 
,
.
,
.
is isosceles. Since 
is also isosceles, in fact we have 
.
is on 
,
and hence their angle between them is equal to 
.
and hence 
and we are done since 
.
is the reflection of the circumcircle of 
be the centers of 
,
and 
,
,
is cyclic. Letting 
be the intersection of 
,
due to symmetry, so 
,
is cyclic. Thus,![\[ \angle O_1X O_2 = \angle MXN = \angle MAN = \angle CAN = \angle TAN = \angle TO_2 N = \angle O_1 O_2 X\]](http://latex.artofproblemsolving.com/c/0/a/c0a649de9a1ae6e62f4483e323cb9f3a01c74750.png)
so 
is isosceles with 
and 
.
.
(Aware of config issues here but also too lazy to do anything about it.)
is a parallelogram.
There's probably a faster way we could do this, but whatever.
.
,
and 
by SAS congruence. Thus, 
.
not too easy, not too hard, not too bashy, not too clean
.![\[\angle AYC = 180^\circ - \angle APC = 180^\circ - \angle AHC = \angle ABC,\]](http://latex.artofproblemsolving.com/0/4/0/04075d58067ffcfa7e4765a4acbd1009466a2a6d.png)
and 
are perpendicular to 
is cyclic, which means 
.![\[\angle AOY = 2 \angle ACY = 2 (90^\circ-\angle PAC) = 180^\circ - 2 \angle PAC = 180^\circ - \angle XOO'.\]](http://latex.artofproblemsolving.com/5/e/0/5e07a4b973462759b94d7d943d00e90783a90ae7.png)
and 
and 
,
.![\[\angle OXO' = \angle BAP = \angle CAP = \angle O'XO,\]](http://latex.artofproblemsolving.com/1/6/2/162c43154e4d9862d064139f9e56dbdd6d11a592.png)
is isosceles with 
.
,
;
.
so 
cyclic. Now note that 
.
from the three points 
map to three points 
along the perpendicular bisector of 
is the midpoint of 
,
,
.
be the midpoint of 
,
.
,
,
(say by considering the arc midpoint). Now, let 
and 
represent the complex numbers associated with the reflections of 
,
by the circumcenter formula, where 
is the reflection of 
over 
This clearly has magnitude 
,
since 
.
,
,
.
then 
implies 
passes through 
,
and thus 
,
to 
have distance 
,
.
is a parallelogram and 
as desired.
is the reflection of 
with 
.
to suffice. We angle chase to get 
and 
as desired. 
.
and 
.
from 
and 
as well meaning there is a spiral similarity at 
to ![\[\angle (BC, XY) = \angle BCM + \angle (BM, XY) \]](http://latex.artofproblemsolving.com/b/5/a/b5aa567fd1752336d19573a083b9f4f3de738291.png)
![\[\angle \frac A2 + \angle (PM, PY) = \angle \frac A2 + 90 - \angle (PM, AC) = 90\]](http://latex.artofproblemsolving.com/e/f/6/ef614728e2ad514ee7814f12856efd388588d9c1.png)
as desired 
is a parallelogram meaning 
as desired.
,
.
,
is obviously equal to the circumradius
the reference, not 
be the midpoint of arc 
.
is a rhombus. Let 
is a rhombus. First, note that 
,
,
and 
is a parallelogram then 
,
by the Pythagorean theorem, so 
