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.
= 2∠AMP.
Let ![$[n]$](http://latex.artofproblemsolving.com/0/d/e/0deff9086fd7840ef23860f5bc924f1d4d2d2310.png)
denote the set of natural numbers less than or equal to 
![$f(m,n) = \sum_{(x_1,x_2,x_3, \cdots, x_m) \in [n]^{m}} \frac{x_1}{x_1 + x_2 + x_3 + \cdots + x_m} \binom{n}{x_1} \binom{n}{x_2} \binom{n}{x_3} \cdots \binom{n}{x_m} 2^{\left(\sum_{i=1}^{m} x_i\right)}$](http://latex.artofproblemsolving.com/f/d/f/fdf7b2fcfb960e93d59f3f84cef4372262c69499.png)
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 -digit number divisible by 
all of whose digits are odd.
be an acute triangle with 
. Let 
be the centroid of triangle 
and let 
be the foot of the perpendicular from 
to side 
. The median 
intersects the circumcircle 
of triangle at a second point 
. Let 
be the point where intersects . The line 
intersects the circle at a point 
, such that lies between and . Prove that the points 
and lie on a circle.
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 
. (This connection is directed.) When she does so, if usamon has an electron and usamon does not, then the electron jumps from to . In any other case, nothing happens. In addition, the physicist cannot tell whether an electron jumps during any given step. The physicist's goal is to isolate two usamons that she is sure are currently in the same state. Is there any series of diode usage that makes this possible?
be 
and let the orthocenter of be 
. Then prove that the isogonal conjugate of wrt lies on the Euler line 
of .
be and let the orthocenter of be . Then prove that the isogonal conjugate of wrt lies on the Euler line of .
with circumcenter 
orthic triangle 
tangential triangle 
Let 
be the isogonal conjugate (WRT ) of the isotomic conjugate of 
WRT and 
be the medial triangle of Then is the homothetic center of 
be the midpoint of 
and 
be the intersection of 
with 
respectively, then lie on the X -midline 
of Let 
be the excentral triangle of 
then it's clear that 
and 
are homothetic, so 
are concurrent.
lie on the circle 
with center and diameter 
so 
and 
are antiparallel WRT 
and 
lies on 
Similarly, we can prove lies on 
so is the homothetic center of 
with orthocenter 
circumcenter 
Let 
be the isotomic conjugate of WRT and 
be the circum-rectangular hyperbola of passing through 
Then 
is a tangent of 
be the orthic triangle of and 
be the Nagel point of 
Since is the crosspoint of and 
WRT 
so 
is tangent to the Jerabek hyperbola 
of at On the other hand, by Lemma we get passes through the isogonal conjugate of WRT so is tangent to at 
be the isotomic conjugate of the circumcenter of WRT and let be the circum-rectangular hyperbola of passing through Notice the tangent of at passes through the isogonal conjugate 
of WRT 
so by Corollary we conclude that lies on the Euler line of 
, given a point 
, define it's isotomic conjugate 
, and isogonal conjugate of be 
be circumcenter, centroid, orthocenter respectively.
tangent to the Jerabek hyperbola 
of 
lies on , 
are collinear, 
and the image of isotomic conjugate of is 
are collinear. 
so 
is tangent to . Let the center of be 
, the midpoint of 
be , We want to proof 
are collinear, it's well-known that this line pass through 
and Tarry point (the fourth intersection of 
and Kiepert hyperbola of .)
be 
, lies on so lies on . Since both 
are isotomic conjugate, so 
pass through the isotomic conjugate of the infinitely point on , consider the image of this line after isotomic and isogonal conjugate, it will lead to the fact that are collinear, as desired., is the centroid of orthic triangle wrt 
.
-Orthocenter of . Parallel fron to 
intersect the 
at 
resp.
cyclically. 
: Inverse of respect to circle 
. Define 
cyclically.
: Circumcenter of 
lies on Euler line of . 
.
.
Let 
.
,
.
.
,
.
.
.
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)