AoPS Academy
be prime, let 
be a positive integer, show that ![\[ \gcd\left({p - 1 \choose n - 1}, {p + 1 \choose n}, {p \choose n + 1}\right) = \gcd\left({p \choose n - 1}, {p - 1 \choose n}, {p + 1 \choose n + 1}\right). \]](http://latex.artofproblemsolving.com/9/d/4/9d409ce186320241eb358b5fbbb8a5a66120899c.png)
be a right triangle with a right angle in 
and a circumcenter 
. On the sides 
and 
, the points 
and 
lie in such a way that 
. Let 
and 
be the projection of and on 
, respectively. Prove that 
is half as long as .
be an isosceles triangle with 
and circumcircle 
. The line through 
perpendicular to is denoted by 
. Furthermore, let 
be any point on . The circle 
with center and radius 
intersects once more at point 
and the circumcircle once more at point 
. Prove that the points 
and lie on a straight line.
?
, and the sum of the roots is 
. So we need to find a combination of integers that multiply to and add up to . Let's call the roots of the equation p, q, and r. From Vieta's, we get that 
and 
. Looking at the factors of , which is 
, we try to split the numbers in a way that gives us the correct sum and product. Trying 3, -2, and -8, we see that they add up to 
and multiply to 
, so they work. That means the roots of the polynomial are -3, -2, and -8, and the factorization is 
. Multiplying it out, we get 
, so we find that a = 46.
with 
, inscribed in circle 
.
be the orthocenter of triangle 
, and 
be the midpoint of 
., parallel to 
, intersects lines 
and 
at points 
and 
, respectively.
be the circumcenter of triangle 
. Prove that: Points 
are collinear.
and 
If each person remains there for 
minutes, what's the probability that all three people meet each other?, all the roots added will give 
. I also learned that if every single root, multiplies every single root, it will give 
. I also learned that if all the roots are multiplied, it will give 
. Is this right? And is there any purpose for these equations?
and 
Then the number of elements in the set ${C⊆A:C∩B=ϕ}$ is
The altitudes 
, intersect
. The tangents at and 
of intersect at 
. Let 
be the midpoint of . 
intersects 
, 
intersects 
at 
lie on the same circle.
intersects the circle at the second point 
The line intersects at the second point 
. Prove that 
are collinear.
is such that 
for any positive integers 
. Assume there exists a positive integer such that 
for all positive integers 
. Determine all possible values of 
.
is such that for any positive integers . Assume there exists a positive integer such that for all positive integers . Determine all possible values of .
and 
are complex numbers such that
and solve for the area of a hexagon in that 
square.
satisfies the condition and 
.
.
and 
,
for all 
.
,
for all non-negative integers 
and 
,![\[
g(a, b) + g(c, b) = g(ac, b^2).
\]](http://latex.artofproblemsolving.com/9/1/d/91dd66cfb45616ccfaa1691e942e816060decf3d.png)
denote the assumption on 
in this equation. For 
,
depends only on the value of 
for all prime numbers 
.![\[
h(x) = \sum_{p \text{ prime}} h(p)\nu_p(x). \tag{2}
\]](http://latex.artofproblemsolving.com/6/8/b/68b83a00528564c6a54d5e2ac13d438b4266babf.png)
or 
.![\[
h(45) = g(45, 1) = 2g(3, 1) + g(5, 1) = 0.
\]](http://latex.artofproblemsolving.com/8/e/0/8e02c396ee210441ab72d69160d7f57de8ab2805.png)
and 
follows that 
.
we get that 
.
from 
,
,
,
.
,![\[
g(n, x) = g(n, x^n) + (x^2 - 1)h(x) + (x^4 - 1)h(x) + \ldots.
\]](http://latex.artofproblemsolving.com/4/4/b/44b899e553765ac801138cdc87c7900ea3c368da.png)
for some 
,![\[
g(n, x) + (x^2 - 1)h(x) = g(n, x^2) \le g(n, (x + 1)^2) = g(n, x + 1) + (x^2 - 1)h(x + 1).
\]](http://latex.artofproblemsolving.com/9/d/0/9d07a67f85a3666f1bb39273106e7079e9f29a88.png)
,
are constants. Therefore 
for all 
from the beginning of the problem to case 
.
or 
,
for every prime 
because 
is monotonically increasing. Let 
.
,
(the case 
is analogous). Then we have![\[
h(a^k) = kh(5^k)(7) = kh(5)(7) = h(b^k).
\]](http://latex.artofproblemsolving.com/0/1/7/017694b47f0bd9755bf3e2458aaea70e3f458104.png)
.
,![\[
q + 1 = \frac{a^{q + 1} - 1}{a - 1} > \frac{b^{q + 1} - 1}{b - 1} > b^q + 1.
\]](http://latex.artofproblemsolving.com/f/5/f/f5fdfd3799129cc139d96f18b8d683e47f815745.png)
.
,
and 
.