AoPS Academy
, angle 
. Point 
lies on 
such that 
. Line parallel to 
passing trough intersects 
at 
. Prove that 
.
be a parallelogram of center 
. Prove that for any point 
, there exist unique points 
and 
such that is the center of mass of 
.
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) 
of positive integers such that
, then it must not contain edge 
.
with circumcenter and orthocenter 
, let 
be an arbitrary point on the circumcircle of triangle such that does not lie on line 
and that line 
is not parallel to line 
. Let 
be the point on the circumcircle of triangle such that 
is perpendicular to , and let 
be the point on line such that 
. Let and 
be the points on the circumcircle of triangle such that 
is a diameter, and 
and 
are parallel. Let 
be the midpoint of 
.
and 
are perpendicular.
and 
are perpendicular.
be nonnegative real numbers such that 
. Find min
and 
. Prove that
.
, then![\[
a+b+a^2+b^2 \geq 1 \geq 28-6\sqrt{21}.
\]](http://latex.artofproblemsolving.com/6/7/7/677ab9d63defffddb7a8c986c2e2b4ddb4e8267e.png)
, then![\[
\frac{1}{b^2-b+1} \leq \frac{1}{a^2-a+1}, a^2 + a + 1 \geq b^2 + b + 1
\]](http://latex.artofproblemsolving.com/e/8/f/e8fdd29170a984431f76f172d52feb9724f7edfc.png)
which implies that![\[
\frac{a^2+a+1}{a^2-a+1}+\frac{b^2+b+1}{b^2-b+1} \geq \frac{a^2+a+1}{b^2-b+1}+\frac{b^2+b+1}{a^2-a+1} \geq 3.
\]](http://latex.artofproblemsolving.com/8/a/3/8a3d11ba00dc8246113ff3c5881989a12be3f4c9.png)
![\[
3 \geq 6-\left(\frac{a^2+a+1}{a^2-a+1}+\frac{b^2+b+1}{b^2-b+1}\right)
=\frac{2(a-1)^2}{a^2-a+1}+\frac{2(b-1)^2}{b^2-b+1}
\geq \frac{2(1-a+1-b)^2}{a^2-a+1+b^2-b+1}.
\]](http://latex.artofproblemsolving.com/2/2/2/222d49414291c3fd180471f92d561d4d0f69b171.png)
and 
. Then,![\[
3\left(u^2-u+1-2v\right) \geq 2\,(2-u)^2.
\]](http://latex.artofproblemsolving.com/2/a/4/2a4022474a7acca1c57c06b05a98c40e269e5c80.png)
![\[
u^2-4v \geq \frac{u^2-10u+4}{3}.
\]](http://latex.artofproblemsolving.com/c/2/4/c24df229451c119fb2ac4cac1352030302fa6f9a.png)
![\[
a+b+a^2+b^2 = u^2-2v+u = \frac{u^2-4v}{2}+\frac{u^2}{2}+u,
\]](http://latex.artofproblemsolving.com/4/9/4/49466166b20f3199205c898f5de4e53481919827.png)
if 
, then![\[
\frac{u^2-4v}{2}+\frac{u^2}{2}+u \geq \frac{u^2-10u+4}{6}+\frac{u^2}{2}+u
=\frac{2}{3}\,(u^2-u+1).
\]](http://latex.artofproblemsolving.com/d/7/2/d72ef57efa4ba4f14b344a02649dd4c961867fd7.png)
is monotonically decreasing on ![\( u\in \left(0,5-\sqrt{21}\right] \)](http://latex.artofproblemsolving.com/2/b/d/2bde58e2769d4081e0740823e379e1ed883a7ca5.png)
, it follows that![\[
\frac{2}{3}\,(u^2-u+1) \geq \frac{2}{3}\Bigl[(5-\sqrt{21})^2-(5-\sqrt{21})+1\Bigr]
= 28-6\sqrt{21}.
\]](http://latex.artofproblemsolving.com/d/b/c/dbc2c7413768ebff082b089c4c28ffa2197d6a03.png)
, then since 
, we have![\[
\frac{u^2-4v}{2}+\frac{u^2}{2}+u \geq \frac{u^2}{2}+u \geq \frac{(5-\sqrt{21})^2}{2}+(5-\sqrt{21})
= 28-6\sqrt{21}.
\]](http://latex.artofproblemsolving.com/b/4/7/b4785f87dac778351daf902c356f8c2409418183.png)
![\[
a+b+a^2+b^2 \geq 28-6\sqrt{21}.
\]](http://latex.artofproblemsolving.com/2/8/1/28126ac3dc744ba5841d2065f3965fb53db64d91.png)
and 
Prove that
then prove 
.
Let 
.
Let 
.
.
Let 
.
then it suffices to prove 
and 
hence 
with equality at 
and 
Show that
Let 
Show that
Let 
Show that
Let 
Show that
Let 
be reals such that 
Show that
Let 
Show that