[/quote]
,
that verify the following equation for all 
:
is a prime number
.
be the feet of the altitudes from 
to 
respectively. Let 
be the circle with diameter 
.
be the intersection of 
and 
.
be the orthocenter of 
be the intersection of 
and 
be the lines through 
tangent to 
respectively. Let 
be the intersection of 
and 
.
be the intersection of 
and 
.
be the line through 
and let 
be the reflection of 
is tangent to 
.
be a positive integer, and let its positive divisors be![\[
d_1 < d_2 < \cdots < d_k.
\]](http://latex.artofproblemsolving.com/6/3/e/63ec82aba0d6c4eece89af6a1247a89498616281.png)
Define 
to be the number of ordered pairs 
with 
such that 
.
.![\[
n = p_1^{e_1} p_2^{e_2} \cdots p_k^{e_k},
\]](http://latex.artofproblemsolving.com/2/7/2/272dc2cc33ebbf695311cb61567fa0297e1299b6.png)
where the 
are distinct primes and the 
are positive integers.
.
meets the side 
and 
are 
,
and 
beyond 
,
and 
and 
.
is a rectangle if and only if 
.
satisfying
Prove
is a positive integer. Call all positive divisors of 
and 
beautiful divisors and difference of any 
that verify the following equation for all 
:
.
passing through 
passing through 
,
.
and 
meet at 
.
be a point such that 
is tangent to 
is tangent to 
are collinear.
.
When does equality hold?
denote the set of all positive integers less than 
that are relatively prime to 
.
reducible over the rational numbers, given that it has integer coefficients?
blocks, each weighing at least 
.
with 
you can choose a subset of the blocks whose total weight is at least 
.
Y by wizixez, Rounak_iitr
Let 
be an acute triangle. Tangents on the circumscribed circle of triangle at points 
and 
intersect at point 
. Let 
and 
be a foot of the altitudes from onto 
and 
and let 
be the midpoint of 
. Prove:
A) Prove that is the orthocenter of the triangle 
.
B) Prove that
cuts 
in half.
be an acute triangle. Tangents on the circumscribed circle of triangle at points 
and 
intersect at point 
. Let 
and 
be a foot of the altitudes from onto 
and 
and let 
be the midpoint of 
. Prove:A) Prove that
is the orthocenter of the triangle 
.B) Prove that
cuts 
in half.
and it follows that
and 
and 
comes out to be 
So we want to show that this midpoint is collinear with 
is equal to its own conjugate. We will in fact show that this value is 
which proves the desired result.
First we take a pause and evalute 
We have that 
And therefore 
Going back to the original equation:
as desired. Since conjugates of real numbers are equal to themselves we have proved that 
and 
,
,
.
is cyclic. Let 
meet 
.
.
meet 
.
and 
are altitudes and their intersection, 
.
.
and 
are cyclic, 
and 
.
is a parallelogram and 
therefore 
is a parallelogram
is the orchocenter
