[/quote]
,
, where 
always occurs at 
non-zero digits such that deleting consecutively the digit of the left, in each step, we obtain a divisor of the previous number.
there exists an 
all of whose digits are odd.
be an acute triangle with 
.
be the centroid of triangle 
and let 
be the foot of the perpendicular from 
to side 
.
intersects the circumcircle 
of triangle 
.
be the point where 
intersects the circle 
,
and 
atoms called usamons. Each usamon either has one electron or zero electrons, and the physicist can't tell the difference. The physicist's only tool is a diode. The physicist may connect the diode from any usamon 
to any other usamon 
.
.
Let 
.![$$ \frac{1}{2}\geq \frac{a}{a^2+3 }+ \frac{b}{b^2+3} \geq \frac{1}{4}(\frac{1}{\sqrt[3]{3}}+\sqrt[3]{3}-1)$$](http://latex.artofproblemsolving.com/f/9/1/f9119e071c12778037616c624c7b9ca75356e8e3.png)
![$$ \frac{1}{2}\geq \frac{a}{a^3+3 }+ \frac{b}{b^3+3} \geq \frac{1}{2\sqrt[3]{9}}$$](http://latex.artofproblemsolving.com/2/b/2/2b2d8a011229fe88a81cb75f9b018ad3a6245faa.png)
![$$ \frac{1}{2}\geq \frac{a}{a^2+ab+2}+ \frac{b}{b^2+ ab+2} \geq \frac{4\sqrt[3]{3}+3\sqrt[3]{9}-6}{17}$$](http://latex.artofproblemsolving.com/5/f/6/5f673cd494b1909ca237db63e96f949c24b6f2ad.png)
![$$ \frac{1}{2}\geq \frac{a}{a^3+ab+2}+ \frac{b}{b^3+ ab+2} \geq \frac{\sqrt[3]{3}}{5}$$](http://latex.artofproblemsolving.com/b/c/3/bc343a07c88dbd98ebffdfb78f311b7b82683d65.png)
-
be a unit square. Points 
are chosen outside 
. Let 
,
.
is at most 
.
is divisible by 
.
and 
Prove that
Y by Rounak_iitr
Is there a way to do this without drawing obscure auxiliary lines? (the auxiliary lines might not be obscure I might just be calling them obscure)
For example I tried rotating triangle MBC 80 degrees around point C (so the BC line segment would now lie on segment AC) but I couldn't get any results. Any help would be appreciated!
For example I tried rotating triangle MBC 80 degrees around point C (so the BC line segment would now lie on segment AC) but I couldn't get any results. Any help would be appreciated!
Attachments:
and 
.
,
,
,
,
,
,
,
.
,
makes 
,
!
!
