.
to the equation 
.
and 
are located within ![\[\angle Q_1BC=\angle ABP,\quad\angle Q_1CB=\angle DCP,\quad\angle Q_2AD=\angle BAP,\quad\angle Q_2DA=\angle CDP.\]](http://latex.artofproblemsolving.com/d/b/3/db31d11ec24322724c0e0f1485aadf8864eabf7a.png)
Prove that 
if and only if 
.
be an integer. Find, with proof, all sequences 
of positive integers with the following three properties:
;
for all 
;
and 
(not necessarily distinct) for which 
,
such that 
.
![$[n] = \{1, 2, . . . , n\}$](http://latex.artofproblemsolving.com/4/f/f/4ff58e6e89a225ffa8db0ab50038fdadd9d4e2bc.png)
is a sequence 
such that each element of ![$[n]$](http://latex.artofproblemsolving.com/0/d/e/0deff9086fd7840ef23860f5bc924f1d4d2d2310.png)
appears precisely one time as a term of the sequence. For example, 
is a permutation of ![$[5]$](http://latex.artofproblemsolving.com/c/4/2/c4216c69e041fca0904e8c5366d4480d1b2ca8b8.png)
.
be the number of permutations of 
is a perfect square for all 
.
such that 
.
has diameter 
of length 
.
lies tangent to 
and 
.
and 
,
is 
,
and 
are relatively prime positive integers, and 
is a positive integer not divisible by the square of any prime. What is 
?
of side length 
is given. Suppose that 
is drawn below.)![\[n \leq \frac{2}{3} L^{2}.\]](http://latex.artofproblemsolving.com/9/5/f/95fe9c1b574a4e693e5ad5190a56fcb62aa26701.png)
be positive real numbers such that 
.![\[\frac{ab+1}{(a+b)^2}+\frac{bc+1}{(b+c)^2}+\frac{ca+1}{(c+a)^2}\geq 3.\]](http://latex.artofproblemsolving.com/5/c/3/5c3cc6f6ed19fc776dbf6ed4f3db9d728c174fa4.png)
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 
,
.
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 
.
.
,
,
.
.
,
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 
.
.
= 
,
,
thus 
.
be on 
are collinear
,
.
,
which is also the perp. bisector of 
(
so 
.
is perpendicular to 
so it's sufficient to show 
.
and 
are similar and it follows that 
and 
are similar and you can just compute 
and conclude that 
