[/quote]
,
such that![\[
f\left(\frac{x^2 - f(x)}{f(x) - 1}\right) = x
\]](http://latex.artofproblemsolving.com/8/0/0/8009a4709ffd6708023d67db2b544e605d1fbe68.png)
for all real numbers 
satisfying 
.
such that 
for all 
?
be a sequence of integers, with the property that every consecutive group of 
'
and 
,![\[\frac{a_n+a_{n+1}+\cdots+a_{n+k-1}}{k}\]](http://latex.artofproblemsolving.com/a/7/b/a7b72c85a43bcba1ebbc642e075c7443ccb6d23c.png)
is always the square of an integer. Prove that the sequence must be constant (all 
be an acute triangle with 
.
be its circumcircle, 
its orthocenter, and 
the foot of the altitude from 
.
be the midpoint of 
.
be the point on 
such that 
and let 
be the point on 
.
,
,
and 
are tangent to each other.
and 
.
(so 
is a function from the positive real numbers) such that![\[ \frac {\left( f(w) \right)^2 + \left( f(x) \right)^2}{f(y^2) + f(z^2) } = \frac {w^2 + x^2}{y^2 + z^2}
\]](http://latex.artofproblemsolving.com/0/4/b/04b955fb570ad384fb7f0e85bf0dacccf4c982a1.png)
satisfying 
and 
of cyclic quadrilateral 
.
,
,
,
such that:
for all 
.
be an integer. Ivan writes the numbers 
each on different cards. He then shuffles these 
cards, and divides them into two piles. Prove that at least one of the piles contains two cards such that the sum of their numbers is a perfect square.
?
be six distinct real numbers satisfying the following condition:
,![\[
x_i^6 + (-1)^{i+1} x_i = -x_1 x_2 x_3 x_4 x_5 x_6.
\]](http://latex.artofproblemsolving.com/3/6/f/36f886654cf5f7af956e2475be94acbe62bb8153.png)
.
| 
with 
are positive integers and 
is prime 
)
,
is the midpoint of 
.
is an arbitrary point on 
.
be the intersection of 
and 
.
intersects with the circumcircle of 
,
.
is the circumcenter of triangle 
.
.
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 
,
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.
.
.
,
are collinear, and 
are concyclic.
be the intersection of 
and the tangent from 
and 
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 
,
.
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 
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.
