[/quote]
,
,
on 
satisfies 
and 
.
be the point where the perpendicular from 
to line 
meets the angle bisector of 
.
holds.
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.
such that 
.
that verify the following equation for all 
:
be a positive integer. Determine all sequences 
of positive integers such that
for all positive integers 
.
and 
be two circles with respective centres 
and 
intersecting in two distinct points 
and 
such that 
is an obtuse angle. Let the circumcircle of 
intersect 
and 
.
intersect 
; let the line 
intersect 
.
are concyclic.
be an acute scalene triangle with circumcenter 
.
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 
respectively. Let 
be the intersection of 
and 
.
and 
.
be the line through 
and let 
be the reflection of 
is tangent to 
.
where 
are integers in 
are each labeled with a positive integer less than or equal to 
such that every label is used exactly 
and 
such that the following property holds: For any two lattice points 
and 
that are both labeled 
,
and 
are both divisible by 
be an odd prime. Define 
over the positive integers as follows:
where 
such that 
.
such that 
and 
is a perfect square.
that are relatively prime to 
.
reducible over the rational numbers, given that it has integer coefficients?
students, each pair of students have watched a movie together. Let the set of movies watched by a student be his movie collection. If every student has watched every movie at most once, at least how many different movie collections can these students have?
be non-negative real numbers such that 
Prove that
and 
.
Y by bcp123, megarnie, Adventure10, Mango247
In 
, 
. 
and 
are the feet of the internal and external angle bisectors of 
, respectively. 
is the midpoint of segment 
, and 
is the circumcircle of 
. Suppose 
is on the minor arc 
of such that 
is tangent to . 
meets again at 
, and the line through perpendicular to meets 
at 
. Prove 
is tangent to the circumcircle of 
.
Proposed by Ray Li
, 
. 
and 
are the feet of the internal and external angle bisectors of 
, respectively. 
is the midpoint of segment 
, and 
is the circumcircle of 
. Suppose 
is on the minor arc 
of such that 
is tangent to . 
meets again at 
, and the line through perpendicular to meets 
at 
. Prove 
is tangent to the circumcircle of 
.Proposed by Ray Li
is harmonic, then 
circle 
with diameter 
is orthogonal to 
Inversion with center 
takes 
into itself and carries 
due to conformity. 
hence 
is perpendicular bisector of 
is S-isosceles 
,
are bisectors of 
which, being onto the circle of diameter 
and 
of the circle 
and 
,
and therefore 
.
and note that it is orthogonal to 
to be the intersection of 
since 
bisects 
are collinear and so 
,
.
because 
since 
which immediately implies what we wanted.
.
.
.
,
and consequently 
.
are the internal and external angle bisectors of 
.
,
,
.
,
as reference points, the problem can be reinterpreted as follows:
for collinear points 
and suppose the circles with diameters 
,
intersect the circle with diameter 
is tangent to 
.
To be honest I found the G4 more challenging than this.![[asy]
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By the angle bisector theorem, 
so 
Since 
implying 
and quadrilateral 
is cyclic, 
so 
which implies 
is harmonic. Hence, 
where we have used the fact that 
which follows from PoP. Hence, 
which implies that 
Since 
this forces 
By our previous observations, we have that 
so 
is isoceles with 
Since 
must bisect 
Hence, 
and 
Since 
It follows that 
completing the proof. 
,
,
.
is isosceles, or 
.
being a harmonic pencil implies 
.
which implies the result.
.
and 
.
.
.
of 
.
as 
.
.
.
.
.
.
bisects 
is cyclic and 
bisects 
.
so since 
we have that 
and since using MacLaurins theorem and PoP we have 
so 
meaning that 
as 
Thus we are done 
,
,
,
bisects 
.
is a diameter of 
bisects 
and 
,
,
.
so 
and hence 
is diametrically opposite to 
is a harmonic bundle, so 
is actually the diameter perpendicular to 
