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.
. A convex polygon is said to be 
-good if it contains lattice points where any three of them are not collinear.-good convex polygon with area at most 
.
so that any -good convex polygon has area at least 
.
and 
be concentric circles, with in the interior of . From a point 
on one draws the tangent 
to (
). Let 
be the second point of intersection of and , and let 
be the midpoint of . A line passing through intersects at 
and 
in such a way that the perpendicular bisectors of 
and 
intersect at a point 
on . Find, with proof, the ratio 
.
with 
and let be 
it’s circumcircle. Diameters 
and 
are drawn. Circle 
interescts 
at 
. Circle 
intersects 
at 
.(
lies between 
and ). Prove that lines 
and 
intersect at circle 
.
has been partitioned into disjoint pairs 
(
) so that for all 
, 
equals 
or 
. Prove that the sum ![\[ |a_1-b_1|+|a_2-b_2|+\cdots +|a_{999}-b_{999}| \]](http://latex.artofproblemsolving.com/0/4/4/04422a7d2266b92a6b3f215b698f078cda85091f.png)
ends in the digit 
.
such that 
for all positive integers 
.
be a triangle with 
Line bisector of 
intersects at point 
. Prove that 
.
, and for all 
, define the sequence 
by the recurrence![\[
a_{n+1} = a_n^2 + 1
\]](http://latex.artofproblemsolving.com/3/1/1/3114595ded8c8961d611afaab9ad19e254973001.png)
Prove that there is no natural number 
such that![\[
\prod_{k=1}^{n} \left( a_k^2 + a_k + 1 \right)
\]](http://latex.artofproblemsolving.com/a/4/f/a4f88ee7e3d4e53e369319f74cd9b12be26c9681.png)
is a perfect square.
are randomly arranged in the cells of a 
square (
). For any pair of numbers situated on the same row or on the same column the ratio of the greater number to the smaller number is calculated. Let us call the characteristic of the arrangement the smallest of these 
fractions. What is the highest possible value of the characteristic ?
be distinct real numbers such that![\[
a + b + c + \frac{1}{abc} = \frac{19}{2}
\]](http://latex.artofproblemsolving.com/5/1/e/51eef77358db2b6c98f0fb0fb4e30db837a03642.png)
Find the maximum possible value of 
.
has centroid 
. A line parallel to 
passing through intersects the circumcircle of at . Let lines 
and intersect at . Suppose a point 
is chosen on such that the tangent of the circumcircle of 
at , the tangent of the circumcircle of at and concur. Prove that 
.
be some prime number, and write 
, 
, and 
, with 
.![\[ xyz\left(\frac{ab}{c}+ \frac{ac}{b}+ \frac{bc}{a}\right) ~=~ x^2y^2p^{\alpha+\beta-\gamma} +x^2z^2p^{\alpha-\beta+\gamma} +y^2z^2p^{-\alpha+\beta+\gamma}. \]](http://latex.artofproblemsolving.com/b/3/c/b3cfc4dc631cd7ca59e391dcc292c90a7011d0d0.png)
If 
, then the first two terms in the right hand side are integers (as the exponent of is non-negative). Hence also the third term 
must be integer. But 
is relatively prime with , and the exponent of is strictly negative; a contradiction.
, and symmetric arguments yield 
and 
.![\[ \frac{ab}{c}~=~ \frac{xy}{z}p^{\alpha+\beta-\gamma} \]](http://latex.artofproblemsolving.com/9/3/0/9301d42402fc95c4bb479ea2f8f594885ed15977.png)
the prime occurs with non-negative exponent., the number 
is integer.
and 
are integers by an analogous argument.
be the exponent of in the prime factorisation of 
. 
implies![\[abc \mid (ab)^2+(ac)^2+(bc)^2\]](http://latex.artofproblemsolving.com/5/d/0/5d0357f71a4582da8e9ecee75a76bcd9c3c42ba8.png)
which is equivalent to![\[v_p(a)+v_p(b)+v_p(c) \leq v_p((ab)^2+(ac)^2+(bc)^2)\]](http://latex.artofproblemsolving.com/d/e/5/de5e19e9e100c2362d3ee6f469323cebb376a091.png)
for every prime . We now have to proof that![\[v_p(c) \leq v_p(a)+v_p(b), \ v_p(b) \leq v_p(a)+v_p(c) \ \text{and} \ v_p(a) \leq v_p(b)+v_p(c)\]](http://latex.artofproblemsolving.com/8/0/d/80de4f99adfd0aae8133bad43d075001663bd5e4.png)
are true for every prime .
from which it immidiately follows that 
and 
. Thus it remains to proof 
. Since this is obvious when 
, we can say that 
is strictly greater than 
which implies that 
and since this must be greater or equal than 
it follows that 
for every prime meaning that 
, 
and 
must be integers.
, 
, and 
. Consider the polynomial![\[f(x)=(x-q)(x-r)(x-s),\]](http://latex.artofproblemsolving.com/4/0/5/40510fe4c2758733bdc7b88861a1c2020bbb01a6.png)
and note that by Vieta's, since![\[q+r+s=\frac{ab}{c}+ \frac{ac}{b}+ \frac{bc}{a} \text{ (by problem conditions)},\]](http://latex.artofproblemsolving.com/e/2/6/e269dfd2b367168b6caff346bf1a09e849368fce.png)
![\[qr+qs+rs=\left(\frac{ab}{c}\right)\left(\frac{ac}{b}\right)+\left(\frac{ab}{c}\right)\left(\frac{bc}{a}\right)+\left(\frac{ac}{b}\right)\left(\frac{bc}{a}\right)=a^2+b^2+c^2,\]](http://latex.artofproblemsolving.com/3/b/0/3b07f53692c131f58c213cbbde79963db6826637.png)
and![\[qrs=\left(\frac{ab}{c}\right)\left(\frac{ac}{b}\right)\left(\frac{bc}{a}\right)=abc,\]](http://latex.artofproblemsolving.com/e/3/9/e39cc2cb4efeedfc10cd8b788c98a293f8f7f274.png)
are all integers (because 
, 
, are integers), we know that the coefficients of 
are all integers. Additionally, we also know that the roots of , which are 
, 
, and 
, are all rational, and since has leading coefficient , by Rational Root Theorem, we know that the roots of must be integers. Therefore , , and are all integers, finishing the problem.
It is 
which has integer coefficients. The solutions to this are rational numbers, but by Rational Root Theorem, they are integers.
is an integer polynomial. Thus, by the rational root theorem, since the roots are rational, we have that the roots also must be integers.
be three integers for which the sum![\[ \frac{ab}{c}+ \frac{ac}{b}+ \frac{bc}{a}\]](http://latex.artofproblemsolving.com/6/c/6/6c60a7b9763c00e8f3a2a6a5da3d6c43bc0b06b3.png)
is integer.![\[ \frac{ab}{c}, \quad \frac{ac}{b},\quad \frac{bc}{a}\]](http://latex.artofproblemsolving.com/0/3/8/038176eab9c9cc9d94076805ea36bddcfe28da27.png)
is integer.
.![$f\in\mathbb Z[x]$](http://latex.artofproblemsolving.com/d/3/c/d3c2eb0df94d0b288dba41c852ac4431e46af9a1.png)
.
are roots of 
,
.
.
which implies that 
,
is the largest power of 
.![\[\nu_p(a)+\nu_p(b)+\nu_p(c) \leq \nu_p((ab)^2+(ac)^2+(bc)^2)\]](http://latex.artofproblemsolving.com/3/c/8/3c88e09072145aa2c5ce71e7276126ba99c8d9f3.png)
.![\[\nu_p(a) \leq \nu_p(b)+\nu_p(c)\]](http://latex.artofproblemsolving.com/0/6/2/062c461ebc539bcd493a400dddd4765f94282ad3.png)
![\[ \nu_p(a) \leq \nu_p(b)+\nu_p(c). \ \square\]](http://latex.artofproblemsolving.com/9/8/b/98ba16d8d24820ead89538b1cf0a7aa64940389e.png)
.![$P(x)\in Z[x]$](http://latex.artofproblemsolving.com/a/2/9/a2971f8b530421becd05d4808c7e242c6406c8d9.png)
,
Then, the polynomial has coefficients
Now, the roots to this are rational, and since all of the coefficients are integers, and it is monic, we have that all of the roots must be integers. Thus, we are done.
The LHS expands to 
RRT finishes from here, as by the given condition and that 
are integers, the polynomial is monic with integer coefficnets. Thus all the roots (for which we need to prove are integers) must be integers by RRT and we are done.