,
be the result.
,
is the Euler's totient theorem, which calculates the number of integers less than 
.
for 
.
be the remainder 
leaves when you divide it by 
.
be your predicted USA(J)MO score.![$\lfloor \frac{x + \sqrt[r]{r^2} + r \ln(r)}{r} \rfloor$](http://latex.artofproblemsolving.com/c/d/3/cd3f7d917b0ec944ec3017e1f737c538e1edd10e.png)
.
unless of your age can be written as 
.
,
,
away. What is the least number of moves that Carina can make in order for triangle 
to have area 2021?
in the coordinate plane where 
are both integers, not necessarily positive.)
be an integer. We say that an arrangement of the numbers 
,
,
in a 
table is row-valid if the numbers in each row can be permuted to form an arithmetic progression, and column-valid if the numbers in each column can be permuted to form an arithmetic progression. For what values of 
be collinear points (in order) and 
a point in plane. Consider the disc 
of center 
,
.![$\mathcal{D}\cap [AC]$](http://latex.artofproblemsolving.com/0/5/6/0569d4130943000380a87b72be211965b5ff02ad.png)
is either the empty set or a segment of length at most 
.![$P(X)\in\mathbb{C}[X]$](http://latex.artofproblemsolving.com/9/9/e/99e8cde28b0b9bc9414425ebc30fa2dc0386fb37.png)
be a polynomial of degree ![\[\sup_{x\in[0,1]}|P(x)|\le(2n+1)^{n+1}\int\limits_{0}^{1}|P(x)|\mathrm{d}x.\]](http://latex.artofproblemsolving.com/9/1/6/91647cd75888c5e9d291280c8ae3f199cc6d1bfd.png)
![$f \colon [0,1] \to \mathbb{R} $](http://latex.artofproblemsolving.com/2/1/8/21859d37ed035abd07d0be112f49d52598fa477e.png)
be a differentiable function such that its derivative is an integrable function on ![$[0,1]$](http://latex.artofproblemsolving.com/e/8/6/e861e10e1c19918756b9c8b7717684593c63aeb8.png)
,
.![\[ \int_0^1 (xf'(x))^2 dx \geq 12 \cdot \left( \int_0^1 xf(x) dx\right)^2 \]](http://latex.artofproblemsolving.com/2/3/e/23e33eed39fc1e20486bbb86261d3566e8fd2017.png)
be an odd prime number, and 
be an odd number not divisible by 
be a field with 
elements, and 
be the set of elements of 
,
.
has at least 
and 
be a continuous function for which there exists an antiderivative 
,
,
,
.
for all 
be a positive integer, ![$g \in \mathbb{R}[X]$](http://latex.artofproblemsolving.com/0/1/e/01e7ebcebcb9f16935716ee859f7acfaa66c29d4.png)
,
be a polynomial with all of its roots being real, and 
for any 
for all 
be two matrices such that 
.
;
,
is differentiable, with continuous first derivative.
and for any two sequences 
,
,
for any positive integer 
is convergent.
and 
,
:![\[(f(x)-g(y))(f'(x)-g'(y))(f''(x)-g''(y))=0\]](http://latex.artofproblemsolving.com/0/2/9/029a935e28ca0a9b36e65702689cca91eeb8f614.png)
![$f\colon[0,1]\rightarrow \mathbb{R}$](http://latex.artofproblemsolving.com/f/d/b/fdb1fb0216ab9d17007b1b7b03d20d1060b88531.png)
be a continuous function. Suppose that for each 
,![\[f_t\colon[0,1-t]\rightarrow\mathbb{R}, f_t(x)=f(x+t)-f(x)\]](http://latex.artofproblemsolving.com/4/f/c/4fc6a66d0600a999c000136a26ea23dfcc269cc3.png)
has an unique global minimum point, which we will denote by 
.
,
is constant zero.
Y by AlastorMoody, rart-of-problem-solving, mathematicsy, megarnie, Jc426, HWenslawski, jhu08, Adventure10, Mango247, GeoKing
The isosceles triangle 
, with 
, is inscribed in the circle 
. Let 
be a variable point on the arc 
that does not contain 
, and let 
and 
denote the incenters of triangles 
and 
, respectively.
Prove that as varies, the circumcircle of triangle 
passes through a fixed point.
, with 
, is inscribed in the circle 
. Let 
be a variable point on the arc 
that does not contain 
, and let 
and 
denote the incenters of triangles 
and 
, respectively.Prove that as
varies, the circumcircle of triangle 
passes through a fixed point.
be the midpoint of arc 
not containing 
and ray 
bisects 
,
(by external version of Fact 5). Let 
,
be the second intersections of 
and 
with circumcircle. Now 
by Fact 5 again twice, also 
,
,
,
and 
gives a nice complex solution. That's what I did during the test after spending a half hour trying to bash it with messy 20-term expressions.
?
is right so then I went back to midpoint of arc.
intersects 
(say at point X) is at a point that is not the incenter. Note that 
and 
where I is the incenter of ABC are supplementary, so I cannot intersect the circle
.
denote the midpoints of minor 
,
respectively.
and 
are congruent.
and 
pass through 
.
and 
.
and 
by Fact 5, and the claim is proven by SAS. 
bisects the external 
,
So 
,
was cyclic but ended up with opposite angles adding to more than 180). Just curious if there's some insight on how someone can draw figures accurately enough to deduce that the midpoint of the arc is the correct locus.
,
be the midpoints of minor arcs 
,
on 
,
are collinear, so we have 
.
and 
and by symmetry, we have 
,
,
,
,
,
,![\[M_BB = M_BI_B = M_CC = M_CI_C.\]](http://latex.artofproblemsolving.com/5/c/7/5c7d8cc28225e4920ffe0ba702625ec7b9ca36e4.png)
from SAS, so![\[\angle I_BMI_C = \angle M_BMM_C = \angle I_BPI_C \implies I_BI_CPM \text{ cyclic}. \quad \blacksquare\]](http://latex.artofproblemsolving.com/6/1/7/6170ee5ddae58918654a949057d972927a83c1a4.png)
,
and 
.
and paired with the fact that 
we know that 
with ratio 
.
and hence 
and so we're done.
,
,
,
,
,
,
and 
are given by 
and 
respectively, and the arc midpoint of minor arc 
.
appear in counterclockwise order. Then for 
and 
,
and 
.
,
,
,
and 
yields the arc midpoints 
for triangle 
are concyclic, or the expression 
is real. This is clear as the expression equals its conjugate.
,
,
Also, note that 
so the claim follows by SAS.
and
so 
containing 
,
,
,
be the minor arc midpoints of 
.
and clearly 
and 
.
to 
as desired.
.
,
and that ![\[I_B M_C = M_C A = A M_B = I_C M_B.\]](http://latex.artofproblemsolving.com/a/2/a/a2ab1a8b9189723a0a006bb2ec64ac277f145ede.png)
,
passes through 
are the 
,
so it suffices to show 
.
and 
have degree 
have degree 
has degree 
cases:
it is clear since 
and 
works. Finally, if 
then 
.
,
,
,
,
,
.
Which is real bc 
,
be the point on 
, so we want to show 
.
so there is a spiral similarity with center 
to 
.
,
is on 
and similarly 
is on 
.
and 
are similar. Angle chase 
and the symmetrical equality implies 
and 
and we are done.
