AoPS Academy
+2 w
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
and 
be positive integers with 
. Let 
be a polynomial of degree with real coefficients, nonzero constant term, and no repeated roots. Suppose that for any real numbers 
such that the polynomial 
divides , the product 
is zero. Prove that has a nonreal root.
x square board has the row and column numbered from 
to , square in 
row and 
column get symbolized by square 
. Subset 
of squares on the board is called "good" subset if two random squares 
belong to satisfy that the squares 
with 
or 
are not belong to . Find the minimum number of "good" distinct subsets such that each square on the board belongs to only one subset
be a triangle and let 
and 
denote its incentre and circumcentre respectively. Let 
be the circle through 
and 
which is tangent to the incircle of the triangle ; the circles 
and 
are defined similarly. The circles and meet at a point 
distinct from ; the points 
and 
are defined similarly. Prove that the lines 
and 
are concurrent at a point on the line 
.
defined as follows:![\[
a_1 = \frac{\sqrt{6}}{3},
\quad
a_{n+1} = a_n + \frac{1}{3a_n},
\quad
\forall n \in \mathbb{N}.
\]](http://latex.artofproblemsolving.com/8/9/2/892a2676214a3c837d6b18a10a4f6f83bc099f74.png)
![\[
0 \le a_n \sqrt{6} - 2\sqrt{n}
\le
\frac{1}{4\sqrt{n}}
\Bigl(
1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n}
\Bigr),
\quad
\forall n \in \mathbb{N}.
\]](http://latex.artofproblemsolving.com/f/0/8/f0824edd96ddfeaa107d729466134812ad7ea777.png)
b,For each 
, define![\[
b_n = \frac{3a_n^2 - 2n - 1}{\ln(n+1)}.
\]](http://latex.artofproblemsolving.com/a/9/3/a937b379140df5aa51d7ede1a685d263c3229e96.png)
Compute the limit 
(i know 
well). i want to study so if you guys have some easy or normal problems please send me
has a right angle at . Let be the point on line 
such that 
and 
lies between and . Point 
is chosen so that 
and 
is the bisector of 
. Point 
is chosen so that 
and 
is the bisector of 
. Let 
be the midpoint of . Let 
be the point such that 
is a parallelogram. Prove that 
and 
are concurrent.
satisfying the following conditions:
for all 
Define the set ![\[S = \left\{ \left( \sum_{i=1}^{10} a_i x_i, \sum_{i=1}^{10} a_i y_i \right) : a_1, a_2, \cdots, a_{10} \in \{0, 1\} \right\},\]](http://latex.artofproblemsolving.com/d/e/b/deb50830b9fe10c86ea169c044c1b26f515b2514.png)
then 
，and any rectangular strip of width 1 covers at most two points of S.
is such that 
for any positive integers 
. Assume there exists a positive integer such that 
for all positive integers 
. Determine all possible values of 
.
, 
, 
, 
, 
in some order.
be an acute triangle with orthocenter 
. Let 
be the point such that the quadrilateral 
is a parallelogram. Let be the point on the line 
such that bisects 
. Suppose that the line intersects the circumcircle of the triangle 
at and 
. Prove that 
.
times as many students as Lara’s high school. The two high schools have a total of
students. How many students does Portia’s high school have?
Solving with Cramer's Rule gives 
or .
people. Then portia's has 
. Thus 
. Thus 
, and 
. Then Portia's high school has students.
From the above result, the number of students in Lara's school is 
. Hence the answer is (C) 
.
satisfying: 
--> 
--> 
so (C)
to get 
.
Then, 
,
.
and 
students, respectively. We have the following systems of equations: 
Plugging the first equation into the second, we have 
Finally, we have 
students and Portia's school have 
students. So 
.
.
so 
so the answer is 
.