[/quote]
,
and circumscribed about the incircle 
,
.
,
at 
,
,
,
,
,
,
,
intersects 
(distinct from 
be the midpoint of the arc 
of 
and 
at 
,
at 
.
.
but doesn't tell it to anyone. Bernaldo thens pick an arbitrary positive integer 
(not necessarily in 
.
chosen by Arnaldo.
be a set of 2025 positive real numbers. For a subset 
,
as the median of 
when all elements of 
.![\[
P(A) = \sum_{\substack{T \subseteq A \\ |T| \text{ odd}}} M_T, \quad Q(A) = \sum_{\substack{T \subseteq A \\ |T| \text{ even}}} M_T.
\]](http://latex.artofproblemsolving.com/1/d/f/1df2ea8487289111939a53dcc49576fbe6fdef37.png)
Find the smallest real number 
such that for any set ![\[
P(A) - Q(A) \leq C \cdot \max(A),
\]](http://latex.artofproblemsolving.com/c/d/e/cdea343d11c722a423e857aeb01202863405773e.png)
where 
denotes the largest element in 
in 
colors, there exists an arithmetic progression of size 
.
from this?
be circumcenter of a non-isosceles triangle 
.
be foots of perpendicular lines from 
.
is cyclic and 
.
bisects 
such that 
be an integer. Show that, if 
is an integer, then it is a perfect square.[/quote]
.
is at least rational, so that 
must be a perfect square then. Using Conway's topograph method, I have found out that the smallest non-negative pairs 
for which this happens are 
and 
,
,
The eigenvalues of that matrix are irrational, however, so that any calculation which uses powers of that matrix is a little cumbersome. There must be an easier way, but I cannot find it. Can you?
interesting if 
of 
and 
is also a divisor of 
such that 
holds for all real numbers 
and 
.
such that the numbers ![\[ (a_1)^{2018}+a_2, (a_2)^{2018}+a_3, \dots, (a_{2018})^{2018}+a_1 \]](http://latex.artofproblemsolving.com/6/1/a/61ad92b49bd9b4b8839e6f6c4e81c758a11835a3.png)
are all powers of 
be a triangle with 
.
be the midpoint of 
,
be a point on segment 
.
be the point of intersection, different from 
,
and line 
.
and 
as the points of intersection of line 
with 
and 
,
.
with 
be the reflection of 
such that lines 
and 
and 
meet again at 
.
.
Y by sydneymark, Davi-8191, Wizard_32, microsoft_office_word, Flow25, itslumi, somlogan, Abidabi, megarnie, HWenslawski, son7, tiendung2006, Toinfinity, Adventure10, Mango247, Rounak_iitr, Funcshun840
Let 
be an acute triangle with 
. Let 
be its circumcircle, 
its orthocenter, and 
the foot of the altitude from 
. Let 
be the midpoint of 
. Let 
be the point on 
such that 
and let 
be the point on such that 
. Assume that the points , 
, 
, and are all different and lie on in this order.
Prove that the circumcircles of triangles
and 
are tangent to each other.
Proposed by Ukraine
be an acute triangle with 
. Let 
be its circumcircle, 
its orthocenter, and 
the foot of the altitude from 
. Let 
be the midpoint of 
. Let 
be the point on 
such that 
and let 
be the point on such that 
. Assume that the points , 
, 
, and are all different and lie on in this order.Prove that the circumcircles of triangles
and 
are tangent to each other.Proposed by Ukraine
This post has been edited 7 times. Last edited by djmathman, Feb 14, 2020, 4:21 AM
Reason: typo after 4.5 years!
Reason: typo after 4.5 years!
be the inversion with center 
that fixed 
Since 
and the antipode of 
in 
are collinear, so 
is the antipode of 
we get 
is the reflection of 
Clearly, 
is the antipode of 
in 
is the reflection of 
so 
is a rectangle and 
is the perpendicular bisector of 
hence 
is the tangent of 
I.e., 
is tangent to 
be the diameter of 
to meet 
are collinear.
is a parallelogram, so 
are collinear. 
,
is tangent to 
.
be the diameter of 
are collinear. Now 
.
,
is harmonic, so 
.
,
.
,
.
,
is tangent to circumcircle of 
.
and 
is tangent to circumcircle of 
,
at 
.
are the points diametrically opposite 
,
are collinear.
.
,
so 
.
.
from 
.
from 
.
and 
.![$k-m = \frac{(2a+b+c)[(b+c)(a^2-bc)-a(b-c)^2]}{2[(ab+bc+ca)(3a+b+c)+abc]}$](http://latex.artofproblemsolving.com/2/5/f/25fbe6d08381c827994ea5919d86c4881900efd8.png)
and (a+c)}{2a[(ab+bc+ca)(3a+b+c)+abc]}$](http://latex.artofproblemsolving.com/0/5/6/0567483c7f388c4095bff452ff24d8126da02fe3.png)
.
by inspection and luck (mostly luck). Then you can guess the form of ![\begin{align*} & 2a[(a+b+c)(3bc+ab+ac)+abc] - (b+c)[(ab+bc+ca)(3a+b+c)+abc] \\
&= 2a(a+b+c)(ab+bc+ca) + abc(6a+3b+3c) - (ab+bc+ca)(b+c)(3a+b+c) \\
&= 3abc(2a+b+c) - (ab+bc+ca)(b+c)(2a+b+c) - a(ab+bc+ca)(b+c) + 2a(a+b+c)(ab+bc+ca) \\
&= 3abc(2a+b+c) - (ab+bc+ca)(b+c)(2a+b+c) +a(ab+bc+ca)(2a+b+c). \end{align*}](http://latex.artofproblemsolving.com/7/3/3/733670d35c01b998e1b1778f6d676c656d61a90c.png)
The first manipulation was done with the intention of possibly factoring out 
,
appeared and it seemed as if the 
was "pretty close" to 
dear One Plus One. So we must prove that 
is tangent to 
and 
is right and 
,
.
.
cut 
at 
cuts 
.
cuts 
and circle 
are tangent.
cuts 
so 
similarly, 
.
are diameter of 
so 
are collinear. So 
.
is tangent to 
.
of 
and 
,
,
,
,
is common tangents of 
and circle 
or they are tangent. We are done.
.
passing through 
intersects 
at 
,
.
.
.
and 
are tangent.
,
.
be the point on 
such that 
and let 
be the point on 
.
.
and 
are tangent to each other.
)![[asy]size(6cm);
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pair T = midpoint(A--H);
pair N_9 = midpoint(O--H);
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,
.
.
we just need to prove 
.
is a diameter, hence 
is a rectangle, so 
passes through the center of 
be the reflection of 
intersects 
is cyclic. If 
intersects circle 
are collinear. => let 
be line which tangents circle 
at 
are midpoints of segments 
respectively,
and 
are midlines of triangles 
and 
resprectively.
lie on Euler's circle of 
is a line through 
,
are concurrent at radical center 
and 
.
and 
are congruent,
and 
,
.
the second intersection of 
.
be its incenter. Let 
be its A-excenter (we know that 
is cyclic for obvious reasons, we call this circle 
)
be the midpoint of the circumcircle arc containing 
be the second intersection of 
with 
be the intersection between 
and the perpendicular line wrt 
passing through 
is tangent to 
at 
.
is orthogonal to 
.
is a diameter of 
and a point 
.
,
and 
denote the antipode of 
are collinear and 
,
(Note that 
are collinear since 
is cyclic):![[asy]
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is a harmonic pencil and 
cuts 
at its midpoint 
.
,
all lie on the circle with diameter 
.
,
is cyclic. But note that 
,
is tangent to 
and 
,
,
at 
.
.
is tangent to 
.
,
.
is a concyclic and thus a rectangle. This is because 
composed with a reflection about 
,
.
.
,
.
is isosceles, which is equivalent to proving the perpendicular bisector and altitude to 
are the same. Define 
,
as the centers of 
,
,
is a rectangle, which then implies that 
.
,
is a tangent. 
are collinear. Let 
: 
and 
are internally tangent, and also externally tangent to 
.
: 
is tangent to line 
Then 
and 
are collinear. Note that 
is a rectangle. Thus,
Note that 
.
and 
is tangent to 
.
.
Which gives our desired tangency.
be diameters of 
be the midpoint of 
.
is a parallelogram 
are collinear and 
so 
are collinear. Since 
we have 
collinear. We have 
so 
is cyclic. Also 
so 
is cyclic. Since 
and 
,
is the perpendicular bisector of 
(since 
)
and so 
is cyclic. Now we have that 
.
and hence 
,
and 
,
,
the 
.
,
and 
.
tangent to 
.
,
.
= 
(Can easily be proved projecting through T or by ratio lemma)
i.e. prove that line 
as follows:
,
!
,
,
and WLOG the radius of 
in function of 
with 
;
)
,
,
(EZ to get 
)
.
;
,
with 
;
:
,
in fact 
.
at 
swaps 
.
.
.
.
,
with respect to 
is tangent to 
.
to 
.
be the midpoint of 
.
.
is a rectangle, proving that we get that 
,
,
.
causing me to forget what I'd done so this is just for storage basically.
of 
be 
is tangent to 
.
and it is well known that 
is diameter of the nine point circle.
is a rectangle. Now by some easy angle chase we are done.
is a diameter of 
,
,
and 
where 
and 
,
.
,
such that 
.
- however noting that 
as they are both perpendicular to 
,
.
,
be the 
.
.
.
,
and 
.
.
is parallel to 
,
so 
is a diameter in the nine-point circle so 
, or 
.
and since 
be the minor arc midpoint of arc 
in 
.
are antipodal arc midpoints in 
.
by definition. It's well known that the uninverted 
,
,
to line 
.
and 
since line 
is the perpendicular bisector of 
.
is tangent to line 
.
.
.
to be the antipodes of 
.
.
be the midpoint of 
is the reflection of 
and 
are cyclic.
and 
.
(which passes through 
.
,
,
.
,
,
)
Then, the circumcircle maps onto the 9-point circle, 
that passes through 
Clearly, 
.
is tangent to 
and 
and they both land on the circumcircle. In fact, they form a rectangle. This finishes.
be the line through 
is tangent to 
,
onto ![\[s=m+\frac{a-q}{2}\Longrightarrow 2s-h=2m+a-q-h=-q.\]](http://latex.artofproblemsolving.com/2/3/d/23d2e105a7c9f6d8ac53f0836d68bad487a88368.png)
Therefore, the reflection of 
.
.
is tangent to 
