AoPS Academy
[/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.
+1 w
be a triangle with circumcircle 
and incenter 
and let 
be the midpoint of 
. The points 
, 
, 
are selected on sides , 
, 
such that 
, 
, and 
. Suppose that the circumcircle of 
intersects at a point 
other than 
. Prove that lines 
and 
meet on .
, 
, 
, 
, 
, 
initially contains one coin. The following operations are allowed
, 
, remove one coin from and add two coins to 
;
, 
, remove one coin from and swap the contents (maybe empty) of the boxes 
and 
., , , , become empty, while box contains exactly 
coins. be a triangle, 
its orthocenter, 
its circumcenter, and 
its circumradius. Let be the reflection of the point across the line 
, let be the reflection of the point 
across the line 
, and let be the reflection of the point 
across the line 
. Prove that the points , and are collinear if and only if 
.
be a real number and let 
be arbitrary real numbers in 
. The sequences 
and 
are then defined recursively by
. Prove that 
for all sufficiently large 
.
is a prime number
is inscribed in circle 
. Let denote its incenter and 
its -excenter. Let 
denote the midpoint of arc 
. Line 
meets a second time at 
. The perpendicular to 
at meets sides 
and at and respectively. The circumcircle of 
meets 
a second time at 
, and the circumcircle of 
meets 
a second time at 
. Prove that 
passes through the antipodal to on .
be a triangle with circumcircle and incentre . A line 
intersects the lines , 
, and 
at points , , and , respectively, distinct from the points , , , and . The perpendicular bisectors 
, 
, and 
of the segments 
, 
, and 
, respectively determine a triangle 
. Show that the circumcircle of the triangle is tangent to .
be a positive integer. Prove that there are infinitely many primes, each of which divides at least one term of the sequence 
, 
.
and 
such that the intersection points of the parabola 
with the axis 
and the point 
form an obtuse triangle.
, 
, 
.
or 
will be obtuse when 
is a double circular arc. The first half of the double arc arc is the part of the circle that passes through 
and a particular point has endpoints on and and passes through . The second half of the double arc is the reflection of the first half about .-axis at different locations than another locus, we can arbitrarily restrict to the -axis to find which loci satisfy 
. We can see trigonometrically that his occurs when 
. Since 
corresponds to the locus of the circle with radius 
and centered on 
, we also have an obtuse triangle when 
Prove that
