AoPS Academy
boxes 
, 
, 
,..,
. The 
-th box 
contains balls.
greater than 
and choose 
a multiple of .
, 
, 
and move the 
balls to the box 
.?
be nonnegative real numbers such that 
. Prove that![\[
\frac{a}{4 - b} + \frac{b}{4 - c} + \frac{c}{4 - a} + \frac{1}{16}(1 - a)^2(1 - b)^2(1 - c)^2 \leq 1,
\]](http://latex.artofproblemsolving.com/5/b/d/5bd3349071e075519bd986c845c500125b7d46f8.png)
and determine all such triples 
where the equality holds.
in this order. Prove that![\[a_1a_3+a_3a_5+a_5a_7+a_7a_9+a_9a_1=a_2a_4+a_4a_6+a_6a_8+a_8a_{10}+a_{10}a_2.\]](http://latex.artofproblemsolving.com/9/e/e/9ee73bbdc4b4f2cad2eb3fcfb3dbdf76b6200b4d.png)
is the orthocenter of an acute triangle 
, and let 
be the midpoint of 
. Suppose 
meets 
and 
at 
respectively. 
meets 
at 
, and the line through perpendicular to meets 
at 
. Prove that 
are collinear.
![$f:[1,8]\to R$](http://latex.artofproblemsolving.com/c/e/c/cec97ea0372e37977e8a77e8e77518a2344c1c43.png)
that:
and the distribution function is 
then
and find the expression of 
![$\{x_n\} \subseteq [0,1]$](http://latex.artofproblemsolving.com/8/e/4/8e4206325387485d73765c9d7070ecb1ce18ff60.png)
, there exists some constant 
, for any 
, among the sequence 
and 
could be chosen to satisfy 
and 
.
be a continuous function . Suppose![\[f(x)=\frac{1}{t} \int^t_0 (f(x+y)-f(y))\,dy\]](http://latex.artofproblemsolving.com/c/0/c/c0c48c4259d4e98562c21a23cf4a6841704b8e82.png)
and all 
. Then show that there exists a constant 
such that 
, 
, 
, if f ' (0)=p and f ' (5)=q, then prove f ' (-5) = q
is differentiable and 
for all 
. Show that for some 
, 
.
on the real line such that for all 
, ![\[ \left( f(x) \right)^2 = \displaystyle\int_0^x \left[ \left( f(t) \right)^2 + \left( f'(t) \right)^2 \right] \, \mathrm{d}t + 1990. \]](http://latex.artofproblemsolving.com/e/7/9/e79b1d8b17010c15a0f6397c8775e4e268a66a2e.png)
be the foot of the internal bisector of the angle 
of the triangle . The straight line which joins the incenters of the triangles 
and 
cut and at and 
, respectively.
and 
meet on the bisector 
.
be the incenter of triangle , 
and 
the incenters of triangles and respectively.
and 
.
with bisector 
we get ![\[AI = \frac{AB \cdot IO_1}{BO_1}\]](http://latex.artofproblemsolving.com/c/b/1/cb1aa7b2e0dabc5fa7d3babcfcb951c1c3fd446a.png)
we get ![\[AI = \frac{AC \cdot IO_2}{CO_2}\]](http://latex.artofproblemsolving.com/2/d/3/2d3bd2ad46de3b5b4003d02896b93cceb81ff63c.png)
![\[ \frac{AB \cdot IO_1}{BO_1} = \frac{AC \cdot IO_2}{CO_2} => \frac{BO_1 \cdot IO_2}{IO_1 \cdot CO_2} = \frac{AB}{AC}= \frac{BD}{DC}\]](http://latex.artofproblemsolving.com/f/3/c/f3c63222b150ab79ad08b409a10346b7b899ddd4.png)
![\[\frac{AM\cdot BD \cdot CN}{BM\cdot CD \cdot AN} = 1 \leftrightarrow \frac{BD}{DC} = \frac{BM\cdot AN}{AM\cdot CN} \]](http://latex.artofproblemsolving.com/6/f/b/6fb2e5ed67b0ce94bbbac0eb19a23108e373eda1.png)
![\[\frac{AN}{AM} = \frac{\sin \angle AMN}{\sin \angle ANM}\]](http://latex.artofproblemsolving.com/7/5/1/75125503154b8e2cf14ce182748717d3a4d9f193.png)
and 
and therefore by sine law on triangles 
and 
we get ![\[BM =\frac{\sin \angle IO_1O_2 \cdot BO_1}{\sin \angle AMN}\]](http://latex.artofproblemsolving.com/8/3/0/8301d5596fdcebd0cbea285d8897581ebb685a68.png)
and ![\[CN =\frac{\sin \angle IO_2O_1 \cdot CO_2}{\sin \angle ANM} \]](http://latex.artofproblemsolving.com/0/1/a/01a70d086f349131aef7d9d24569fd8d369f4b44.png)
and therefore we obtain ![\[\frac{BM}{CN} = \frac{\sin \angle IO_1O_2 \cdot BO_1 \cdot \sin \angle ANM}{\sin \angle IO_2O_1 \cdot CO_2 \cdot \sin \angle AMN}= \frac{IO_2 \cdot BO_1 \cdot \sin \angle ANM}{IO_1 \cdot CO_2 \cdot \sin \angle AMN} \]](http://latex.artofproblemsolving.com/8/6/c/86c8f769370192957857b53b8bb07a81b7837c4e.png)
![\[\frac{BM\cdot AN}{AM\cdot CN} = \frac{IO_2 \cdot BO_1}{IO_1 \cdot CO_2 } = \frac{BD}{DC} \]](http://latex.artofproblemsolving.com/7/1/1/711c2cc5e7462cd78d7896cad4e782e288820f2a.png)
and we are done.
be the incenters of 
Internal bisectors of 
cut at 
and external bisector of 
cuts at 
Let 
cut 
at 
Since 
also bisect 
it follows that 
But, 
thus 
and 
Therefore, if 
we have 
the incenters of the triangles 
respectively and let be the point 
bisects the angle 
and 
we conclude that the points 
are in Harmonic Division.
is also Harmonic and then, we have that the points 
are in Harmonic Division.
where 
we conclude that 
lies on and the proof is completed.
