[/quote]
,
of positive integers that satisfy the equation![\[
x + xy + xyz = 31.
\]](http://latex.artofproblemsolving.com/a/7/8/a78d1e9d66831f11e7de08c109855e7727ec8535.png)
be real numbers such that 
,![\[
a + \frac{1}{a} + b + \frac{1}{b} + c + \frac{1}{c} + d + \frac{1}{d} = 0.
\]](http://latex.artofproblemsolving.com/a/9/0/a90c94d82e3650560635616dacaf392d300b8c91.png)
Prove that one of the numbers 
or 
is equal to 
.
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?
numbers placed in a circular way is either 
or 
.
consecutive numbers.
Find the minimal possible value of this sum.
Find the maximal possible value of this sum.
people who do not know each other. Prove that it is possible to introduce them in such a way that no three of them have the same number of acquaintances.
,
at points 
,
and 
.
be the common point of 
and the incircle, closest to 
be the common point of 
and 
.
.
be the 
-
and 
for 
)
,
be the smallest integer 
.
,
to 
: 
interesting if 
of 
and 
is also a divisor of 
distinct points on a circle. Consider the set of distances between any two out of the 
such that for any positive integers 
,
we have 
.
,
is divisible by 
.
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 
Y by nguyendangkhoa17112003, Adventure10, mathematicsy, Mango247, ehuseyinyigit
Let 
be the circumcenter and 
the orthocenter of an acute-angled triangle 
such that 
. Let 
be the foot of the altitude 
of triangle . The perpendicular to the line 
at the point intersects the line 
at 
. Prove that 
.
be the circumcenter and 
the orthocenter of an acute-angled triangle 
such that 
. Let 
be the foot of the altitude 
of triangle . The perpendicular to the line 
at the point intersects the line 
at 
. Prove that 
.
,
,
concur if and only if the lines 
,
,
concur. Hereby, for any point P and any line g, the notion 
means the perpendicular from the point P to the line g.
.
.
.
.
and AC. Since OF || C'Z, we can rewrite 
,
.
,
,
concur. In order to prove this, we denote by S the point of intersection of the lines 
.
,
,
,
,
,
.
,
.
from 
again at a point D and consider the cyclic quadrilateral ADBC with the diagonal intersection F. Let the perpendicular to OF at F meet AC at P, DB at P', the circumcircle arc DC opposite to the vertex B at X, and the circumcircle arc DC opposite to the vertex A at X'. Since 
,
.
has right angle 
.
,
are midlines of this right angle triangle, i.e., 
.
are (oppositely) congruent and 
.
,
,
intersects the circumcircle of 
at 
.
are concyclic because 
and we are asked to show 
.
.
,
,
.
.
.
,
.
and 
.
.
.
.
and 
it follows that 
are isogonal conjugates with respect to 
Consequently, if 
denote the midpoints of 
then 
is the pedal triangle of 
Since 
is N-isosceles, then 
is cyclic, so 
but 
.
,
nor pedal triangle of 
,
,
and 
,
,
,
.
.
so that 
,
cuts 
at 
and cuts the circumcirle at 
.
is a well known property, 
.
then 
and 
,
and we are done.
and 
,
.
,
and 
,
at 
respectively, thus 
and 
.
,
,
are corresponding points concerning similar triangles 
and 
.
be the point that corresponds to 
and 
we now just need to prove 
.
and 
,
be the orthic triangle of 
and we already have that 
as isogonal wrt 
,
are isogonal conjugates wrt 
.
(since they are the feet of the pedals from 
so 
so 
as desired.
,
is perpendicular to the sides of the pedal triangles.
instead of 
.
![[asy]
size(8cm);
pair A=(0,0), B=(120,0), C=(34.5,140.9), F=(34.5,0), O=(60,60), P=(3.3,13.3), Q=(136,97.7), H=(34.5,20.9), X=(19.4,-14.5), Y=(34.4,-20.9);
label("$A$", A, SW);
label("$B$", B, SE);
label("$C$", C, NW);
label("$Q$", Q, NE);
label("$H$", H, NW);
label("$O$", O, NW);
label("$P$", P, NW);
label("$H_1$", X, SW);
label("$H_2$", Y, S);
draw(A--B--C--cycle, linewidth(0.5)+blue);
draw(circle(O, 60sqrt(2)), linewidth(0.4)+blue);
draw(C--F, linewidth(0.5)+blue);
draw(F--P, linewidth(0.5)+blue);
draw(X--O--F--cycle, linewidth(0.5)+red);
draw(F--Y--O, linewidth(0.5)+red);
draw(F--Q--C, linewidth(0.4)+dashed+blue);
draw(circle((17.2,2.8), 17.42), linewidth(0.4)+dashed+grey);
dot(A);
dot(B);
dot(C);
dot(F);
dot(O);
dot(P);
dot(Q);
dot(H);
dot(X);
dot(Y);
[/asy]](http://latex.artofproblemsolving.com/6/0/3/6034660f955b984745b16a09de43eb32c60af7ae.png)
and 
be the reflections of 
and 
respectively, and let 
.
too.
,
.
,
,
so 
,
and 
.
as well. This yields that
or that 
.
so 
is cyclic and
Yep...
.
.
.
and so, 
and so, 
.
gives power of 
power of 
is isosceles and thus, 
.
be the circle with center 
,
at 
,
.
at F
is cyclic quadrilateral of 
.
.
meet 
at 
.
by SAS congruency.
.
.
such that 
.
,
.
.
and together with 
it follows that 
and that 
is the perpendicular bisector of 
.
and we are done.
it's clear that 
.
be the intersection of 
,
.
so 
is a parallelogram 
and we're done. 
and 
.
and 
the two intersections of line 
are the solutions of the equation 
,
and 
.
is its own projection on chord 
.
is the projection of 
.
and 
,
,
