AoPS Academy
are positive real numbers such that 
, Prove that ![\[ a^{b+c} b^{c+a} c^{a+b} \leq 1 . \]](http://latex.artofproblemsolving.com/0/c/e/0ce097b0b507403c959c2992cae1217ecdd7e581.png)
+1 w
be a finite sequence of positive integers. Prove that there exist nonnegative integers 
and 
such that 
holds for all integers 
be a positive integer. Dominoes are placed on a 
board in such a way that every cell of the board is adjacent to exactly one cell covered by a domino. For each , determine the largest number of dominoes that can be placed in this way.
or 
. Dominoes are placed on the board in such a way that each domino covers exactly two cells of the board, and dominoes do not overlap. Two cells are said to be adjacent if they are different and share a common side.)
![\[\lfloor x \rfloor\]](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 
we havev 
. Construct 
and on continue of 
and 
respectively such that 
. Suppose that 
meets circumcircle of 
in 
. (
). Prove that is on the bisector of 
.
, 
, 
. By cosine law applied in triangle , 
. (1) By cosine law in triangle 
, ![\[EF^2=(x+z)^2-(x+z)(y+z)+(y+z)^2\]](http://latex.artofproblemsolving.com/7/6/a/76a0979da0da7465b15dad60512b42003194468f.png)
. Using (1), ![\[EF=\sqrt{z(x+y+2z)}\]](http://latex.artofproblemsolving.com/b/2/c/b2c7c13e1cccbbdabcedff3912fed471f9c4b0d3.png)
. As 
lie on one circle, 
, from which 
. So, 
. It means that 
. Finally, 
, which means that lies on the bisector of 
, as desired.
, if we have 
and 
, then is concyclic.
:
so we have 
. because we have 
and so is concyclic. is not on the internal bisector of 
, so internal bisector of meets circumcircle of 
at 
. 
is bisector, so we have 
.
so 
is concyclic, so we have 
.
.
, 
, 
so by lemma we have 
is concyclic. so 
is concyclic, so we have 
so 
so 
and hence 
so we have 
so 
is not , and so is not on .
where 
is center of A_excircle.but 
and form that lies on and 
is cyclic and we are done.
with 
, 
and 
intersect 
and respectively at 
and 
, so that 
, where 
is the circumcenter of , hence 
, i.e. 
, or intersects the circle 
at , the midpoint of the arc 
, done. from cosine theorem, then from 
finding 
and, finally, 
and proving 
.
and by cosine formula we get 
, where 
are the sizes of 
. Taking 
on 
produced so that 
we get 
, and is tangent to 
, with 
.
be the center of the rotation: 
are concyclic and 
are concyclic and 
, hence 
are perpendicular bisectors of and 
, respectively.
are colinear, so 
. And it easy to prove 
be the bisector of
midpoint of arc 
of the circle 
; then 
is equilateral, 
cyclic, 
(s.a.s), hence a 
rotation around maps to and to , or 
equilateral, i.e. 
; consequently 
is a kite, i.e. is angle bisector of 
, or 
, hence 
is angle bisector.
are cyclic.
we get 
and 
and using 
with letting 
and expanding the above expression using 
or 
and hence 
bisect 
.
and 
and 
so 
is cyclic.
and 
and 
so 
and 
are congruent so 
so 
which implies is angle bisector of .
and 
.
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.
be on 
.
be feet of perpendiculars from 
and 
respectively. Notice
so 
so 
so 
as desired