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 
, 
.
be right-angled. Let 
be the mirror image of the vertex
in the side , let 
be the mirror image of in and 
the mirror image of in . Find the ratio of areas 
.
and is not neccesarily the midpoint of and . so you do need to use the fact that is right
. Then, 
. Complex Shoelace finishes.
and 
Prove that
Prove that
.
.
.
.
,
Similarly, we have 
and 
.![$\frac{[A'B'C']}{[ABC]}$](http://latex.artofproblemsolving.com/b/d/9/bd98ea2a52520a6496a91cd245ccc0528dc6cfd1.png)
is the absolute value of 
which is 
so ![$\frac{[ABC]}{[A'B'C]'} = \boxed{\frac{1}{4}}$](http://latex.artofproblemsolving.com/6/d/1/6d14f2ec7e907f1b53df9d378b48f62829a53c85.png)
.
are named in counterclockwise order and that 
about 
.
is the rotation of 
clockwise and 
is the rotation of 
counterclockwise.
and 
,
is a rhombus with area 4 times that of 
.![$[A''B''C]=[AB''A''B]+[ABC]+([A''BC]-[AB''C])=5[ABC]+([A''BC]-[AB''C])$](http://latex.artofproblemsolving.com/7/7/2/772e8310e4e3b77e2831c800b8e1d6bcc25b3b29.png)
have lengths 
respectively, then ![$[A''BC]=\frac{1}{2}ac\cos 3\theta$](http://latex.artofproblemsolving.com/b/5/6/b5611f7914322877f9b91ee241f46ae8ec18ee30.png)
and ![$[AB''C]=\frac{1}{2}bc\sin 3\theta$](http://latex.artofproblemsolving.com/c/0/4/c0479e688fdb739e9cfcbe6a4baaee20647b1398.png)
![$[A''BC]-[AB''C]=\frac{1}{2}c^2(\sin\theta\cos 3\theta-\cos\theta\sin 3\theta)=-\frac{1}{2}c^2\sin 2\theta$](http://latex.artofproblemsolving.com/4/2/7/42725c31e2d09f16a3c74d8f97c8c0ad97fe0f85.png)
![$[ABC]=\frac{1}{2}c^2\sin\theta\cos\theta$](http://latex.artofproblemsolving.com/5/a/b/5abc1c840a4a08c11de6be29be52c16933830bb1.png)
,![$[A''BC]-[AB''C]=-2[ABC]$](http://latex.artofproblemsolving.com/3/b/7/3b78dbef520d902405d291cebe8bea6608a1b667.png)
![$[A''B''C]=(5-2)[ABC]=3$](http://latex.artofproblemsolving.com/9/9/a/99a6f47d6c75933a67bc19b0063f2b5d37185571.png)
,