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.
be a triangle with circumcircle 
. The tangents at 
and 
intersect at 
. The circumcircle of triangle 
intersects the line 
at 
and 
is the midpoint of 
. Prove that the lines 
and 
intersect on .
![\[x^4-(2\cdot10^{10}+1)x^2-x+10^{20}+10^{10}-1=0\]](http://latex.artofproblemsolving.com/b/9/d/b9d355433e896e8b97f28c2e4235140fd2348e77.png)
correct to four decimal places.
be an integer, 
a real number, and 
be positive real numbers such that 
. Prove that:![\[
\frac{1 + x_1^k}{1 + x_2} + \frac{1 + x_2^k}{1 + x_3} + \cdots + \frac{1 + x_n^k}{1 + x_1} \geq n.
\]](http://latex.artofproblemsolving.com/0/8/d/08de0c74a4e36b50a64d17875d3fd93eeb5b52de.png)
be positive real numbers. Prove that there exists a cyclic permutation 
of 
such that for all positive real numbers 
the following holds:![\[
\frac{a}{x\,a + y\,b + z\,c}
\;+\;
\frac{b}{x\,b + y\,c + z\,a}
\;+\;
\frac{c}{x\,c + y\,a + z\,b}
\;\ge\;
\frac{3}{x + y + z}.
\]](http://latex.artofproblemsolving.com/1/5/2/15261b3f712c2cc8ffbeb92a9711e06ad0992b5d.png)
be positive integers such that 
, 
, and 
are perfect squares. Prove that 
is also a perfect square.
be distinct primes and let 
be nonnegative integers. Define![\[
m \;=\;
\frac12
\Bigl(\prod_{i=2}^k p_i^{a_i}\Bigr)
\Bigl(\prod_{i=1}^k(p_i+1)\;+\;\sum_{i=1}^k(p_i-1)\Bigr),
\]](http://latex.artofproblemsolving.com/7/9/a/79af85375957986298d37c00d288b6d402f40106.png)
![\[
n \;=\;
\frac12
\Bigl(\prod_{i=2}^k p_i^{a_i}\Bigr)
\Bigl(\prod_{i=1}^k(p_i+1)\;-\;\sum_{i=1}^k(p_i-1)\Bigr).
\]](http://latex.artofproblemsolving.com/e/2/3/e23799ff03400d5eade3731bb83456fcd702a5e1.png)
Prove that![\[
p^2-1 \;\bigm|\; p\,m \;-\; n.
\]](http://latex.artofproblemsolving.com/3/7/e/37e4535b59ef183908730946213429e9ebc05dd7.png)
, and 
be positive real numbers. Prove that![\[\frac{a^4 + 3}{b} + \frac{b^4 + 3}{c} + \frac{c^4 + 3}{a} \ge 12.\]](http://latex.artofproblemsolving.com/1/f/d/1fdfd64c9aafa8741ad1594db46cc16a403b1a99.png)
When does equality hold?
with 
. Let 
be the midpoint of side 
, and let 
and 
be points on segments 
and 
, respectively, such that 
. Let 
be the circumcircle of 
, and 
the circumcircle of 
. The common tangent 
to and , which lies closer to point 
, touches and at points 
and 
, respectively. Let the line 
intersect again at 
, and the line 
intersect again at 
. Prove that the circumcircle of triangle 
is tangent to both and .
islands of the Arctic Ocean. Every bear sometimes swims from one island to another. It turned out that every bear made at least one swim in a year, but no two bears made equal swams. At the same time, exactly one swim was made between each two islands and 
: either from to or from to . Prove that there were no bears on some island at the beginning and at the end of the year.
and 
such that 
.
be a convex polyhedron with the following properties: has exactly 
edges. differ by at most 
. with 
colors such that for each color 
and any two distinct vertices 
, there exists a path from 
to 
all of whose edges have color . for which such a polyhedron exists.
such that
divides 
for every 
, and
we have![\[
f\bigl(f(a)+kb\bigr)\;=\;f\bigl(a + k\,f(b)\bigr).
\]](http://latex.artofproblemsolving.com/c/f/f/cff41352422c8765978b8fe6dc966ba934a9df49.png)
be an acute scalene triangle. Denote by 
the circle with diameter 
, and let 
be the contact points of the tangents from to , chosen so that and 
lie on opposite sides of 
and and 
lie on opposite sides of 
. Similarly, let 
be the circle with diameter 
, with tangents from touching at 
, and 
the circle with diameter , with tangents from touching at 
.
are concurrent.
![\[a^2+b^2+16c^2 = 9k^2 + 1.\]](http://latex.artofproblemsolving.com/5/4/5/545f8edbdebe4e759c5a70bd3d489e5d53c9147c.png)
yields 
.
,
must be divisible by 
,
,
![\[18+16c^2=9k^2+1\implies 17=9k^2-16c^2=(3k-4c)(3k+4c).\]](http://latex.artofproblemsolving.com/7/1/b/71bef806084778dee85166f201f023c8f79e611a.png)
We must then have 
and 
- solving yields 
and 
.
is a solution.
as well. Then ![\[3^2+b^2+16\cdot 3^2=9k^2+1\implies 152=9k^2-b^2=(3k-b)(3k+b).\]](http://latex.artofproblemsolving.com/5/6/0/5600ae0d937f7d9c246cc8748b9dd36d9371b59f.png)
The prime factorization of 
is 
.
and 
must have the same parity yields only two cases to check.
and 
,
and 
.
is prime, the triples 
work.
and 
,
and 
.
is prime, the triples 
both work.
that work subject to 
