Then 
is
and whose roots are all real.
.
with a finitely generated k-algebra with one generator such that make 
there exist 
such that 
.
then 
then the result from a) may no longer hold.
invertible, for an infinite number of 
:
such that 
is a scalar
![\[
\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}
\frac{1 - e^{-2} \cos\left(2\left(u + \tan u\right)\right)}
{1 - 2e^{-2} \cos\left(2\left(u + \tan u\right)\right) + e^{-4}}
\, \mathrm{d}u
\]](http://latex.artofproblemsolving.com/9/1/1/91165e6b531233c56b799ed32ded982d57b6c87a.png)
and 
be positive integers. Prove that![\[
a^2 + \left\lceil \frac{4a^2}{b} \right\rceil
\]](http://latex.artofproblemsolving.com/5/6/2/562d003cf06ca46031cec9e4e968f8b67784225f.png)
is not a perfect square.
matrices
matrices
,
Y by farhad.fritl, Frd_19_Hsnzde, ehuseyinyigit, pomodor_ap, Nuran2010, Rounak_iitr, user4747
In an acute-angled triangle 
, 
be the orthocenter of it and 
be any point on the side 
. The points 
are on the segments 
, respectively, such that the points 
and 
are cyclic. The segments 
and 
intersect at 
is a point on 
such that 
is tangent to the circumcircle of triangle 
at 
and 
intersect at 
. Prove that the points 
and lie on the same line.
Proposed by Theoklitos Parayiou, Cyprus
, 
be the orthocenter of it and 
be any point on the side 
. The points 
are on the segments 
, respectively, such that the points 
and 
are cyclic. The segments 
and 
intersect at 
is a point on 
such that 
is tangent to the circumcircle of triangle 
at 
and 
intersect at 
. Prove that the points 
and lie on the same line.Proposed by Theoklitos Parayiou, Cyprus
This post has been edited 1 time. Last edited by falantrng, Apr 27, 2025, 4:38 PM
.
we have
Suppose 
.
,
is the foot of the altitude, we have
All segments on the left-hand sides are effectively computable (Sine Law/Ratio Lemma seems enough) via 
and the variable 
,
from the tangency to 
.
are all cyclic. Let 
be reflection of 
.
.
so 
are collinear. Now Pascal on 
solves.
lies on 
.
.
.
.
gives 
and similarly 
as well.
is the external angle bisector of 
.
.
,
are collinear, and 
are concyclic.
be the intersection of 
and the tangent from 
to 
and 
be the intersection of 
and 
,
are collinear, qed.
and 
.
be the second intersection of 
then 
and 
are reflections over 
,
is cyclic, implying that 
,
now.
,
,
,
,
,
,
denote 
,
,
,
,
,
,
,
respectively. (I know that's a lot but they're just common notation anyway)
.
.
so 
.
.
.
,
and 
.
so we're done.
,
are cyclic and 
,
is cyclic. Now, we claim that 
lies on the radical axis of 
and 
,
due to 
.
and 
.
and 
,
,
are collinear. Then 
,
is cyclic, i.e. 
and thus 
lie on the same line is equivalent to saying 
are concurrent lines.
and 
.
be the feet of the heights from 
are collinear,
,
is parallel to these two lines as well.
.
.
,
and the conclusion follows.
,
Let lines 
intersect the unit circle again at 
respectively. Now since 
are concyclic, we have
So
and similarly
Then
(So 
,
And since 
,
Now we find the coordinate of 
.
Since 
,
Now we find the vectors
and
Then
which is real. 
are concyclic, 
.
hence projecting DDIT at 
.
thus, 
are collinear. Pascal at 
gives the result as desired
,
.
.
.
on triangle 
it is enough to prove: ![\[\frac{LH}{LH_A} \cdot\frac{H_AD}{DC} \cdot \frac{CX}{XH} = 1 \]](http://latex.artofproblemsolving.com/c/a/5/ca5a5fbc872bbbe5053a4c254dc952119b2c08ae.png)
Using the Ratio Lemma on triangle 
we get: ![\[\frac{LH}{LH_A} = \frac{\cos(\beta)}{\sin(\gamma)\sin(\alpha)}\]](http://latex.artofproblemsolving.com/c/6/b/c6b89a205981fb3614471caa45100f846abf9ede.png)
Using the Ratio Lemma on triangle 
we get: ![\[ \frac{CX}{CH} = \frac{\sin(\alpha)\sin(x)}{\cos(\beta)\cos(\gamma + x)} \]](http://latex.artofproblemsolving.com/f/b/b/fbb1710f7ba004243867ac4e5289368adb29ef17.png)
Using the Ratio Lemma on triangle 
we get: ![\[ \frac{H_AD}{DC} = \frac{\sin(\gamma)\cos(\gamma + x)}{\sin(x)} \]](http://latex.artofproblemsolving.com/1/c/1/1c12889bc42c4ae75f5e19990aa222c97ca0fc2d.png)
Multiplying the three expressions, we get:![\[ \frac{LH}{LH_A} \cdot \frac{H_AD}{DC} \cdot \frac{CX}{XH} = \frac{\cos(\beta)}{\sin(\gamma)\sin(\alpha)} \cdot \frac{\sin(\gamma)\cos(\gamma + x)}{\sin(x)} \cdot \frac{\sin(\alpha)\sin(x)}{\cos(\beta)\cos(\gamma + x)} = 1 \]](http://latex.artofproblemsolving.com/3/6/9/369a31e4ab0808652255bcd39573fe1fd1de4586.png)
cyclic and additional trivially from the angles we have 
,
is the tangent to 
at 
.
and 
.
and 
are concurrent at 
and 
which is 
,
and 
which is 
are collinear. However, since 
,
and 
also works.
.![\[\angle BDP = \angle AEC = \angle ADC = \angle CDQ.\]](http://latex.artofproblemsolving.com/1/0/7/10760fcdfeedf54b41023f3410dd8fcdbb095652.png)
Also, since 
,
is isosceles.
be the image of 
are concyclic,
are collinear. Note that 
,
.
,
,
,
and 
.
cases. Pick 
,
.
.
Let 
.
thus we're done 
which means that 
.
defined by 
is projective since 
,
is the tangent line to 
.
so by prism lemma (or Steiner conic) the line 
we get 
,
.
that map 
and 
.
are collinear has deg 
.
concyclic has a deg 
:
.
,
is tangent to 
are concyclic.
holds. 
and 
are concyclic.
One can similiarly prove that 
,
are collinear.
.
hold. This is enough to prove the claim. 
implies 
are collinear. 
inscribed 
.
,
and 
lie on the same straight line. This means that 
,
,
and 
.
at 
and connect 
then do ratio lemma with menelaus.
move on BC, similar to other solutions we know 
cyclic so 
if we let 
so 
since 
,
where 
is the foot from 
to 
(
are concylic points.
,
is collinear.
and sides 
and 
and therefore 
.
.
and so 
.
,
,
,
and 
.
.
and therefore 
by applying the Sine theorem to 
.
and 
we obtain 
and 
.
Q.E.D.
is the 
be the second intersection of the circles 
the intersection of 
and 
is a harmonic.
be on the circle 
.
with respect to 
.
be the midpoint of 
,
with 
.
be the center of the circle 
.
.
and 
,
colinear and 
,
,
colinear.
be the intersection of 
and 
.
and 
.
,
and 
is cyclic, by Power of a Point we have 
,
is the PoP of U with respect to 
is the PoP of U in respect to 
are colinear. Now, its easy to see that 
,
from 
are all cyclic. Let ![\[\measuredangle PDB = \measuredangle AEP = \measuredangle ADC = -\measuredangle ADB = -\measuredangle BDA'.\]](http://latex.artofproblemsolving.com/2/0/4/2047cf43305fe5cfb9de955727629a4293eef7fe.png)
are collinear. Finally, Pascal's on hexagon 
implies 
collinear, as needed.
and 
are cyclic.
and 
( Due to 
and 
being cyclic)
.
.
( due to 
being tangent to 
)
.
are not on the same line we get an obvious contraction )
.
and 
because 
and 
.
Which implies the claim. 
.
Which implies the claim. 
Now by applying Pascal's theorem on cyclic hexagon 
,
,
,
must be collinear, thus finishing the problem. 
