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.
(where 
are the digits and 
) we define the \textit{permutation sum} as the value 
For example the permutation sum of 20253 is 
Let 
and 
be two fivedigit integers with the same permutation sum.
.
be a triangle. Let 
be a points on the segment 
such that 
. Let 
be the mid-point of 
. Let 
intersect 
in 
and 
in 
respectively. Determine 
.
, Its incircle 
and 
escircle 
are tangent to at 
and 
respectively. 
intersects line at 
. Line intersects at , and line intersects at 
. Line 
intersects at 
. Prove that 
.
it is fixed positiv number.
, where 
are integer numbers .
that satisfy the following condition for any real numbers 
and 
such that:
be a triangle with circumcenter 
. Let 
be the midpoints of 
and 
respectively and let 
be the projection of on 
. Let be the projection of 
on 
. Let 
intersect 
at points 
and 
. Let 
intersct at points 
. Let 
intersect 
at 
and 
intersect at 
. Then show that 
are concurrent.
+1 w
and 
respectively be the incentre, circumcentre and circumcircle of triangle . Points 
are chosen on , such that 
, and point 
is the foot of the altitude from 
to 
. If 
are similarly defined, prove that lines 
and 
concurr on 
.
such that, for all real numbers 
with 
, 
other than a polynomial satisfying
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)
Prove that
and let 
and 
and 
respectively. Let the circle with centre 
and 
.
,
and 
,
.
and 
be feet of perpendiculars from 
and 
respectively. Prove that 
,
and 
are concurrent.
.
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