AoPS Academy
be a triangle with 
. Points 
and 
lie on sides 
and 
, respectively, such that 
and 
. Segments 
and 
meet at 
. Determine whether or not it is possible for segments , , 
, 
, 
, 
to all have integer lengths.
+2 w
be the orthocenter of acute triangle , let 
be the foot of the altitude from 
to , and let 
be the reflection of across 
. Suppose that the circumcircle of triangle 
intersects line at two distinct points 
and 
. Prove that is the midpoint of 
.
for Girls, PRISMS Online Math Meet for Girls, is happening this spring! https://www.prismsus.org/events/prom/home/index
be a convex polygon with 
sides, 
. Any set of 
diagonals of that do not intersect in the interior of the polygon determine a triangulation of into 
triangles. If is regular and there is a triangulation of consisting of only isosceles triangles, find all the possible values of .
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 
is it possible to transform any row-valid arrangement into a column-valid arrangement by permuting the numbers in each row?
, and , respectively, located at the origin of the coordinate plane. In a move, Carina may move a pin to an adjacent lattice point at distance 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 
and 
are both integers, not necessarily positive.)
be a quadratic polynomial with real coefficients satisfying ![\[x^2 - 2x + 2 \le P(x) \le 2x^2 - 4x + 3\]](http://latex.artofproblemsolving.com/e/8/a/e8a2e0fe543bc5b638d0c26e7fd227358e57449d.png)
for all real numbers 
, and suppose 
. Find 
.
and 
. Thus, 
and 
. Since 
is intuitively of the form 
, using the fact that and , we get:
Subtracting, 
.Thus, 
.
be a quadratic polynomial with real coefficients satisfying for all real numbers , and suppose . Find .
and 
..
.
has a vertex at 
.
since 
is its minimum value.
. Then, subtracting 
from the given inequality, we have 
. Then we must have 
. Note that 
. Hence, 
, so 
. Direct computation yields 
,
,
is a constant between 1 and 2.
and set the expression equal to 
to get 
into 
gives 
that it is some multiple of 
,
and 
so 
and 
.
.
t
to 
.
,
and would make this a clean number. So 
to 
is 
.
.
and 
.
You found 
does not define a quadratic equation, and there may be an infinite number of other quadratics that fit between the other two quadratics.
and 
,
That means your quadratic, 
.
,
.
.
we have 
.
and 
and 
So that 
and 
So equality holds and 
.
.
we get 
and so, 
.
is always nonnegative and equal to 
.
gives 
,
gives 
.
.
for some real 
.
we have that 
so 
and 
.
.
.
for some constant 
,
.
.
.
,
for some ![$a\in[1,2]$](http://latex.artofproblemsolving.com/5/3/3/533c5dc8056f6007980976464cb6c6a106d1a1fa.png)
.
,
,
and ![$a\in [1,2]$](http://latex.artofproblemsolving.com/b/4/b/b4b99478edad44af3c2c4f9551f312c79c52143b.png)
,
,
,
satisfy 
,
,
.
Subtracting 
Therefore,
and
So, the largest value of 
and the smallest is 
.
We can assume that 
,
.
.
.
, vertex is 
, so 
this gives 
so 
comes down as 
,
, so this comes out as 
, so 
and 
, so 
, solve for these two equations to get 
, 
.
,
.
and 
we can probably say that 
.
,
.
.
for some 
.
.
means that 
.
,
.
.
.
.
.
.
and we get 
.
.