Prove that
Let 
Prove that
for which 
and 
are both perfect squares.
for every real 
and 
has four distinct real roots, then the sum of the roots is
can be defined as a sequence such that 
,
,
.
is such that 
.
for the function 
are perfect squares?
and 
are the roots of the cubic equation 
,
can be written in the form 
,
and 
are relatively prime positive integers. Compute 
.
where 
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 
that doesn't contain 
.
be the circumcenter of 
be circumcenter of 
correspond to complex numbers 
.
are on the unit circle, 
,
,
,
as desired.
be the center of 
,
.
is the reflection of 
across 
,
.![\[
\angle AYC=180-\angle APC=180-\angle AHC=\angle B
\]](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 
,
.
![[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 
are reflections over 
and since the reflection of 
lies on 
we see 
Notice 
and 
Hence, 
But 
so 
and 
Therefore, 
be the circumcentre of 
,
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 
,
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 
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 
.
.
= 
,
,
thus 
.
be on 
are collinear
,
.
,
which is also the perp. bisector of 
