be a triangle with 
let 
be the circumcircle of 
and let 
be its radius. Point 
is chosen on 
such taht 
and point 
is the foot of the perpendicular from 
.
mets 
.
is chosen on line 
such that 
and 
lie on a line in that order. Finally, let 
be a point satisfying 
and 
.
,
has two real roots, so that the one is the square of the other.
,
be all positive integers smaller than 
such that 
for all 
is the largest positive integer that divides both 
.
.
such that: ![\[ x = \sqrt{ x - \frac{1}{x} } + \sqrt{ 1 - \frac{1}{x} } \]](http://latex.artofproblemsolving.com/e/c/f/ecf7fb3c2805290f2cb07b7fdd7c73e1c7a55b99.png)
Y by TheDarkPrince, MNJ2357, pablock, MATH1945, Darghy, Hamel, ltf0501, Supercali, parmenides51, amar_04, megarnie, tiendung2006, Adventure10
As promised, here is the problems used for 
IMOC. Hope that there isn't any typo ... 
https://drive.google.com/file/d/1Pa1JzsfDb6OVuuDHvov7yYpxVdnHPmEv/view?usp=sharing
IMOC. Hope that there isn't any typo ... https://drive.google.com/file/d/1Pa1JzsfDb6OVuuDHvov7yYpxVdnHPmEv/view?usp=sharing
.
and 
,
and blah blah cauchy
,
,
and its root 
,
,
is non-constant,thus 
.
,
.
Now we will show that 
and will finish the problem thanks to Cauchy.
we have that 
or 
.
or 
.
for any 
,
only. Let 
be an arbitrarily small number, now it is easy to see that 
finishes the problem.
Now, by C-S we have that
)
.
and 
,
also, hence 
.
Can anyone help me how to continue after showing injectivity ...
,
and 
are roots of 
.
is equivalent to 
,
(Try 
)
are 
can be expressed as the sum of 
,
is valid expression, of course.
,
is valid expression.
,
,
is valid expression.
,
,
is valid expression.
,
,
,
such that 
and 
cannot be expressed as the sum of 
positive divisors of 
can be expressed as the sum of 
such that 
and 
.
be the smallest positive integer such that 
.
,
,
.
,
,
and 
cannot be expressed as the sum of 
then 
or 
,
.
,
for sufficiently large 
for all 
.
such that 
implies 
.
there exists 
such that 
.![\[f^{(n+1)}(y) = f^{(n)}(f(y)) = f^{f(t_n)}(f(y)) = t_n+2f(y),\]](http://latex.artofproblemsolving.com/3/4/6/3461215efe906d8e0cb573c276e11ee979c7ae71.png)
![\[f^{(n+1)}(y) = f^{f(t_{n+1})}(y) = t_{n+1}+2y,\]](http://latex.artofproblemsolving.com/4/3/2/43233a7b2f6ec381cc35ae6c0f15c9858d96bdb2.png)
and so 
for all 
should be 
in order to have 
such that 
.
is periodic and thus bounded. Therefore for sufficiently large 
for all 
is such that 
implies 
.
for all ![\[f^{(n)}(y) = f^{f(t_n)}(y) = y,\]](http://latex.artofproblemsolving.com/f/1/2/f12be52b60560d6eac552db6d5c725845584c742.png)
![\[f^{(n+1)}(y) = f^{f(t_{n+1})}(y) = y,\]](http://latex.artofproblemsolving.com/7/a/9/7a9dd3043e18dc92af30c4ea7647a229b9dc70c7.png)
and so![\[f(y) = f(f^{(n)}(y)) = f^{(n+1)}(y) = y,\]](http://latex.artofproblemsolving.com/1/7/0/170c4300e7ab7b5cb5d127cbfbb0b8025bb7a7c0.png)
as desired.
.
and 
,
.
After that it is an easy infinite descent.
,
components (with ugly method), but I haven’t proved the opposite direction.
.
.
).It can be found that any two squares have Descartes distance at least 
is the reflection of 
over 
.
and let 
be the foot from 
to 
are collinear, and now we have that 
by Spiral Sim so if 
is the midpoint of ![\[
\angle PIA = \angle PNM = \angle D_0NM = \angle D_1NM
\]](http://latex.artofproblemsolving.com/c/c/a/ccad96702781c56fff3970c14ed05b2ced3afad2.png)
where 
is the extouch point. Let 
meet 
at ![\[
\angle ND_1Q = \angle MND_1 = \angle PIA = \angle QIN,
\]](http://latex.artofproblemsolving.com/e/7/e/e7ef1d6eb0cc9be283936c6e99945928dde79e72.png)
so 
is cyclic. In particular, ![\[
\angle IQT = \angle IND_1 = \angle AIT = \angle AXT,
\]](http://latex.artofproblemsolving.com/5/b/2/5b2893c206e31ac21cffc85a84ad37eebf46c05d.png)
so 
.
by homothety centered at 
.
doesn’t require accurate construction. It can be generalised that for a 
chessboard, the final situation will be 
connected components. But I somehow found that the way that 8-mino tiling the plane is not optimal for special 
and I don't like to calculate the 
term :p
chessboard can be put 
.
,
is used). For Bob’s strategy, I am quite sure that the tiling with 
can be constructed by letting any 
that 
.
problem
is missing.
