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.
such that for any positive integer 
, satisfies 
.
and 
lie respectively on sides 
and 
of a triangle 
and 
. Segments 
and 
cross at 
. Circumscribed circles of triangles 
and 
cross again at point 
different from . Prove that point lies on the bisector of angle 
.
satisfying 
and their sum is an even number, consider the quadratic polynomial:![\[
P(x) = x^2 - (m^2 - m + 1)x + (m^2 - n^2 - m)(n^2 + 1).
\]](http://latex.artofproblemsolving.com/d/5/1/d51aaddfddfe06cc91abfdd3476a155770a8a5fd.png)
are positive integers but are not perfect squares.
such that![\[
ad = bc, \quad a + d = 2^k, \quad b + c = 2^m
\]](http://latex.artofproblemsolving.com/1/1/d/11d16732e3a77af46ad17adecf104a442a5d7f67.png)
for some positive integers 
and 
.
be a triangle with incenter 
and intouch triangle 
. Let be the foot of the perpendicular from 
onto 
. Assume that 
, intersect the sides , in 
respectively. Finally, let the rays 
, 
meet the circumcircle of 
in , 
respectively. Prove that the tangent from to the incircle and the line 
meet on the circumcircle of .
be a triangle with circum-circle 
. Let 
be a point in the interior of triangle which is also on the bisector of 
. Let 
meet in 
respectively. Suppose 
is the point of intersection of 
with 
; and 
is the point of intersection of 
with 
. Prove that 
is parallel to 
.
such that 
is a perfect square but none of the integers 
are perfect squares.
be an acute-angled triangle with 
. The altitudes 
and 
intersect at 
. Let 
be the midpoint of 
. Point 
is chosen on the extension of 
beyond and point 
is chosen on the segment 
such that 
. Prove that points 
and are collinear. be an acute-angled triangle with 
. Let be the incentre of triangle , and let 
be the points where the incircle touches the sides 
respectively. Let 
meet the line at 
respectively. Further assume that both and 
are outside the triangle . Prove that
are concyclic.
is also the incentre of triangle 
.
and 
, prove
Compute -![\begin{align*}
\frac{\mathrm{d}}{\mathrm{d}x} \bigg[ \int_{0}^{e^x} \log ( t ) \cos^4 ( t ) \mathrm{d}t \bigg]
\end{align*}](http://latex.artofproblemsolving.com/8/6/f/86fb48db976012106bd30f74cb6ada8c6e6d7800.png)
For 
define 
Determine the open interval(s) (if any) where 
is decreasing and all the open interval(s) (if any) where is increasing.
Determine all the local minima of (if any) and all the local maxima of (if any) 
![$\frac{d}{dx} [\int_0^{e^x} \log (t) \cos ^4(t) dt]$](http://latex.artofproblemsolving.com/c/c/5/cc57ad163c659a09cb0cf4c3dda9198ac398cfd3.png)
.
.
.
.
.
if we put 
get 
.
is decreasing in
and increasing in
.
has local minimum at .and no local Maxima but has global maximum.
Compute - For define Determine the open interval(s) (if any) where is decreasing and all the open interval(s) (if any) where is increasing. Determine all the local minima of (if any) and all the local maxima of (if any) 
by newton leibntiz (which is nothing but an application of FTC) we get, ![$\frac{d}{dx}\left[\int_0^{e^x}\log(t)\cos^4(t)dt\right]=e^x(x\cos^4(e^{x}))$](http://latex.artofproblemsolving.com/b/5/0/b50f25dff4ee175490c62f92c8b7a8ea5a9a6413.png)
so decreasing in the interval 
and increasing in 
vanishes only at and we observe that 
hence it has a local minima at 
such that![\[
f(xy) = f(x)f(y) \;-\; f(x + y) \;+\; 1,
\quad \forall x, y \in \mathbb{R}.
\]](http://latex.artofproblemsolving.com/7/a/5/7a51a7be2306d18f4e80db1f5aac6e9bbf25c25c.png)
.
bisects 
.
dec on 
and inc on 
is a local minima (easy to see from above)