AoPS Academy
. Prove that: 
, 
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.
be the set of positive real numbers.
such that there exists a function 
with the property:![\[
f(y^2f(x) + y + c) = xf(x+y^2)
\]](http://latex.artofproblemsolving.com/e/3/0/e30ee6d9f69cc43422a5c2b9813192a632d359cb.png)
for all 
.
![\[\lfloor x \rfloor\]](http://latex.artofproblemsolving.com/d/3/f/d3f8facf62413e1f17c89b9c183a8735c2defe8b.png)
denote the greatest integer less than or equal to x . What is the sum of the squares of the real numbers x for which ![\[ x^2 - 20\lfloor x \rfloor + 19 = 0 \]](http://latex.artofproblemsolving.com/f/a/9/fa96f2d6217a16029e5f03499e0a83deffdfa28a.png)
of the pizza, where 
is a positive integer. Find .
Is there an elegant way to prove this?
is called stylish if the set 
can be partitioned into 
pairs such that the sum of the elements in each pair is a power of 
. For example, 
is stylish because the set 
can be partitioned as 
, with sums , 
, and respectively. Determine the number of stylish numbers less than 
.
to the cross product 
reveals that a negative determinant [of 
] corresponds to a matrix which reverses the direction of the cross product of two vectors.[/quote]
and 
, but am confused on how to prove this statement for general vectors 
and 
although its supposed to be a quick comparison.
chance of landing heads when flipped. Krithik is playing a game where he starts with 
point. Every turn, he flips the coin, and if it lands heads, he gains point, and if it lands tails, he loses point. However, after the turn, if he has a negative number of points, his point counter resets to . Krithik wins when he earns 
points. Find the expected number of turns until Krithik wins.
be an isosceles triangle with 
Let 
be a point on the circumcircle of on minor arc 
Let 
intersect the extension of 
at 
Let 
be the midpoint of segment 
and let 
be the intersection of 
and 
Let the extension of 
intersect 
and the circumcircle of at 
and 
respectively. Given that 
and 
compute the length of 
are on circle 
. 
is incenter of 
, is midpoint of arc 
. 
is a circle that is tangent to 
and 
and tangent to at 
. ( is in ) and and are on a line.
.
with i from 1 to 4 circles internally tangent to C, arranged around C in the increasing order of the indexes (1->2->3->4). Take 
be the length of the external common tangent of and 
. Then the analogous of Ptolemy's thm takes place: 
.
based on the proof of the fact (1), mentioned by grobber in the above post #2#, which appeared in the topic Mixtilinear incircle of Leon.
We denote the points 
where 
is the circumcircle of the given triangle 
and it is easy to show that 
from 
where 
from the trapezium 
we have that the point 
is the midpoint of the segment 
and then, we conclude that 
where 
is the circumcenter of 
It is easy to show that 
where 
because of the incenter 
of 
is the orthocenter of the triangle 
as well.
and because of 
we conclude that the line segment 
passes through the point 
as the symmetric point of 
with respect to and then ( from 
), as the midpoint of the arc 
of 
are collinear and the proof is completed.
-invert, which sends 
to the excentral touchpoint , to the 
-excenter 
, and to the intersection point of the -external bisector and . It suffices to show that 
are concyclic now, but this is obvious.
, , and 
as 
, 
, and , and the midpoints of arcs and as 
and 
. Homothety tells us 
and 
collinear. Pascal on 
then tells us that 
collinear, and noting 
implies 
.
and 
are a median and symmedian, respectively, of 
, so 
. As a result, it's clear that will pass through the top point of . 
and 
then prove 
and 
Prove that
Prove that
![\[\mathcal{P}(x) = x^{108}+x^{102}+x^{96}+2x^{54}+3x^{36}+4x^{24}+5x^{18}+6\]](http://latex.artofproblemsolving.com/e/b/2/eb28918359848dd07784a3c2c6a34b934743c016.png)
be 
.![\[\dfrac{r_1^6+r_2^6+\dots+r_{108}^6}{r_1^6r_2^6+r_1^6r_3^6+\dots+r_{107}^6r_{108}^6}\]](http://latex.artofproblemsolving.com/3/1/5/315341a7adf35bc04e40ee0685a011c9f7b76115.png)
without Newton Sums.
