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.
satisfy the following two conditions:
exist such that 
.
, exist such that 
.
+1 w
and 
.
satisfies the following three conditions:
All elements of are a subset of 
.
.
of the set 
, let 
. Find the maximum number of elements in the set 
that satisfies all of the three following conditions: are subsets of .
of , 
is not empty.
of , 
.
+1 w
, then
. Half chord 
be a cyclic 
-gon and let 
for all 
. Define 
as the intersection of diagonals 
and 
for all integers . satisfying 
for all integers . Prove that the points 
are concyclic.
be a triangle and let 
and 
be variable points on sides 
and 
, respectively, such that 
. Let 
denote the point on 
such that 
is parallel to 
, and let 
denote the point on 
such that 
is parallel to 
. Prove that as 
vary, the circumcircle of 
passes through a fixed point, other than . satisfies 
. Let the incenter of triangle be 
, which touches 
at 
, respectively. Let 
be the midpoint of 
. Let the circle centered at passing through 
intersect 
at 
, respecively. Let line 
meet at 
, line 
meet 
at 
. Prove that the three lines 
are concurrent.
and 
Prove that
Let and 
Prove that
is given, find the largest real number 
such that there exists a geometric sequence 
with common ratio satisfying
for all positive integers 
. Here, 
denotes the distance from the real number 
to the nearest integer.
, where the common ratio 
. Given that
.
is your favourite function, 
is your mother's favourite function, and 
is your father's favourite function. It was discovered that for any reals 
, 
Prove that you are boring. is 
, then is called a perfect number. For example, the sum of the positive divisors of 6 is 
, hence 6 is a perfect number.
, where 
are positive integers, and 
are odd primes.
be an acute triangle with circumcircle 
. Let 
and be respectively the midpoints of the arcs 
and 
of . Show that the inradius of triangle 
is not less than the inradius of triangle .
where the notations are what they usually mean. Note also that the incenter of is the DeLongChamps point of triangle . We also see that triangle is acute angled, being similar to the intouch triangle of . Now the result which we seek to prove is the following: In acute angled triangle 
with orthocenter 
, incenter 
and circumcenter 
, we have 
.
is the median of the set 
. Since 
are isogonal conjugates, we see that 
bisects angle 
and 
bisects angle 
. Let 
and 
. We claim that 
is a convex quadrilateral and lies in its interior. We can see that 
is a convex quadrilateral by our angle condition on 
. Therefore, 
lies on opposite sides of 
and hence is convex. Now, lies in its interior immediately follows. Observe that 
and 
. Let 
extended meet the boundary of the quadrilateral 
at point . We clearly have 
. Therefore, 
. The result holds.
(due to some technical issue) got a special prize.
and 
, so we just need that 
.
), we know that the area of 
is greater than or equal to that of . It's quite easy to see that the semi-perimeters of the two triangles is the same. Since 
, the conclusion follows immediately.
where the notations are what they usually mean. Note also that the incenter of is the DeLongChamps point of triangle . We also see that triangle is acute angled, being similar to the intouch triangle of . Now the result which we seek to prove is the following: In acute angled triangle with orthocenter , incenter and circumcenter , we have . is the median of the set . Since are isogonal conjugates, we see that bisects angle and bisects angle . Let and . We claim that is a convex quadrilateral and lies in its interior. We can see that is a convex quadrilateral by our angle condition on . Therefore, lies on opposite sides of and hence is convex. Now, lies in its interior immediately follows. Observe that and . Let extended meet the boundary of the quadrilateral at point . We clearly have . Therefore, . The result holds. (due to some technical issue) got a special prize.
are 
. It is well known that 
in 
. Thus,
where 
and 
are inradii of and respectively.
. Thus,
Claim: 
is concave in the interval ![$[0,\pi]$](http://latex.artofproblemsolving.com/e/c/e/ece27ba559bd93cdf2416aa0f88fc8d7d85e5791.png)
. Thus by Jensen's Inequality we have
Multiplying all, we get
Also, using 
where 
we have
Combining the last two inequalities we get the desired claim. Thus, the inradius of triangle 
is not less than the inradius of triangle . 
(
)
and 
Prove that
