[/quote]
,
be a circle inscribed inside a rhombus 
.
and 
be variable points on 
and 
respectively, such as 
is always the tangent line to 
is the same.
,
such that any graph with 
vertices and at least 
passing through point 
is an oblique strophoid.
.
and 
externally on the sides 
and 
respectively.
meets 
at 
,
meets 
at 
and 
denotes circumcenter of 
and 
meets again on 
.
be a positive integer. Show that any polynomial ![$Q\in\mathbb{Z}[X]$](http://latex.artofproblemsolving.com/e/8/2/e8286b6f2b9790f39c88849a1415d86f85c0b92c.png)
with all positive coefficients such that
is a perfect square for any 
must be a constant polynomial.
be an acute triangle and let 
be the foot of the 
be the midpoint of 
with diameter 
meets 
at 
,
with diameter 
at 
.
,
are concyclic.
![\[
\text{Find all prime numbers } (p, q) \text{ such that } p^2 + 3pq + q^2 \text{ is a fifth power of an integer.}
\]](http://latex.artofproblemsolving.com/c/f/8/cf855cb087102024a50d41f49c67e7fd1ef22767.png)
having 
distinct real roots 
)
: 
.
defined on all real numbers and taking real values such that ![\[f(f(y)) + f(x - y) = f(xf(y) - x),\]](http://latex.artofproblemsolving.com/f/5/2/f528fc4047e21247663d4993913eadf346315be7.png)
for all real numbers 
.
such that 
.
![$P\in \mathbb{R}[x]$](http://latex.artofproblemsolving.com/7/4/c/74cb2247069abb8a22666e87fcbd69a55a015004.png)
for which there exists 
such that for all 
we have: ![\[P\left(x+\frac1n\right)+P\left(x-\frac1n\right)=2P(x).\]](http://latex.artofproblemsolving.com/9/0/9/90954cdeade9d13149e1aaa03df2c87dff157041.png)
are taken such that the point 
.
and 
,
and 
.
.
Y by Adventure10, Mango247, and 1 other user
Look at these fractions. At firs step we have 
and 
, and at each step we write 
between 
and 
, and we do this forever
![\[ \begin{array}{ccccccccccccccccccccccccc}\frac{0}{1}&&&&&&&&\frac{1}{0}\\ \frac{0}{1}&&&&\frac{1}{1}&&&&\frac{1}{0}\\ \frac{0}{1}&&\frac{1}{2}&&\frac{1}{1}&&\frac{2}{1}&&\frac{1}{0}\\ \frac{0}{1}&\frac{1}{3}&\frac{1}{2}&\frac{2}{3}&\frac{1}{1}&\frac{3}{2}&\frac{2}{1}&\frac{3}{1}&\frac{1}{0}\\ &&&&\dots\end{array}\]](//latex.artofproblemsolving.com/b/c/9/bc90d41735c7fd23b56ff37cbe71c4bfdecf70ba.png)
a) Prove that each of these fractions is irreducible.
b) In the plane we have put infinitely many circles of diameter 1, over each integer on the real line, one circle. The inductively we put circles that each circle is tangent to two adjacent circles and real line, and we do this forever. Prove that points of tangency of these circles are exactly all the numbers in part a(except).
c) Prove that in these two parts all of positive rational numbers appear.
If you don't understand the numbers, look at here.
and 
, and at each step we write 
between 
and 
, and we do this forever![\[ \begin{array}{ccccccccccccccccccccccccc}\frac{0}{1}&&&&&&&&\frac{1}{0}\\ \frac{0}{1}&&&&\frac{1}{1}&&&&\frac{1}{0}\\ \frac{0}{1}&&\frac{1}{2}&&\frac{1}{1}&&\frac{2}{1}&&\frac{1}{0}\\ \frac{0}{1}&\frac{1}{3}&\frac{1}{2}&\frac{2}{3}&\frac{1}{1}&\frac{3}{2}&\frac{2}{1}&\frac{3}{1}&\frac{1}{0}\\ &&&&\dots\end{array}\]](http://latex.artofproblemsolving.com/b/c/9/bc90d41735c7fd23b56ff37cbe71c4bfdecf70ba.png)
a) Prove that each of these fractions is irreducible.
b) In the plane we have put infinitely many circles of diameter 1, over each integer on the real line, one circle. The inductively we put circles that each circle is tangent to two adjacent circles and real line, and we do this forever. Prove that points of tangency of these circles are exactly all the numbers in part a(except
).
c) Prove that in these two parts all of positive rational numbers appear.
If you don't understand the numbers, look at here.
t
this figure doesn't change.now use induction...
appear in 
row...
and 
which proves the induction step. It is obvious that it is true at the beginning of the process.
.
is a number generated in one of the processes, then 
are also produced.
number, counting from the beginning of the sequence, is 
,
.
)
as the first and the last fraction is exactly the sequence generated by 
as the first and the last fraction, plus 
.
are generated somewhere in the second sequence, then 
are generated in the first sequence. The next fraction appearing between them is 
which is one unit more than 
(the one appearing in the second sequence)
step of the process, the number appearing just before 
is 
.
are exactly one unit more than the numbers appearing between 
.
does, and if 
of the x axis, maps circles to circles, and tangent ones to tangent ones. The circle above 
.
and 
.
.
.
is the same as replacing 
and 
,
.
,
appeared in the process and we know that their tangency points with the x axis are rational numbers.
to the circle above 
'
which has a rational tangency point. Since our operations map rational numbers to rational numbers, the original tangency point should have been a rational number; which completes the proof.
