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.
be positive real numbers such that 
. Show that![\[ \frac{x^2-yz}{x^2+x}+\frac{y^2-zx}{y^2+y}+\frac{z^2-xy}{z^2+z}\leq 0. \]](http://latex.artofproblemsolving.com/e/a/1/ea1f56cc594862673040411c3ad322cb00cfe904.png)
formed by using the digits 
once each such that the number 
is divisible by 
for 
.
be an odd prime and 
is a bijection and 
is an integer such that 
Show that 
is a multiple of .
be a set of 
positive integers with the property that the product of any two distinct members of will not exceed 1994. Show that there are numbers 
and 
in the set such that the gcd of and is greater than 1.
and 
: 
, knowing that the median 
makes an angle of 
with side 
and an angle of 
with side 
.
such that
;
and 
are relatively prime;
.
are positive real numbers such that 
, prove that ![\[ (1+a)(1+b)(1+c) \geq 8 (1-a)(1-b)(1-c) . \]](http://latex.artofproblemsolving.com/0/b/e/0bedd9ed8362ac5202845f8a4d0d24519423e9b4.png)
on the board. and write instead the numbers 
and 
.
, such that 
be an acute triangle with 
. Let 
be its circumcircle, 
its orthocenter, and 
the foot of the altitude from . Let 
be the midpoint of 
. Let 
be the point on 
such that 
and let 
be the point on such that 
. Assume that the points , , 
, and are all different and lie on in this order.
and 
are tangent to each other. with real coefficients is the average of two monic polynomials of degree with real roots.
be defined by![\[
f(n) = \left\lfloor \frac{n}{3} \right\rfloor + \left\lfloor \frac{n+1}{5} \right\rfloor + \left\lfloor \frac{n+2}{7} \right\rfloor,
\]](http://latex.artofproblemsolving.com/e/a/f/eafc3b0e7225da09bb62c906572dbc1b84a275a0.png)
for any 
.
is neither injective nor surjective.
:
with the points lying in this order and that any side doesn't cross any other. is concave, without loss of generality let it be concave at 
. Then since the incircle is the largest circle that is completely inside a triangle, and since lies strictly inside 
, the inradius of is strictly greater than that of 
, which is contrary to our assumption. Thus the quadrilateral is convex.![$[F]$](http://latex.artofproblemsolving.com/9/4/3/943a895f28610b01bafc5382abba8c4ced6234fc.png)
we denote the area of a figure , by 
we denote the semiperimeter of the triangle 
, by 
we denote the inradius of the triangle .![$[ABC]/s_{ABC}=[ABD]/s_{ABD}=[DBC]/s_{DBC}=[ADC]/s_{ADC}=[ABCD]/(s_{ABC}+s_{ACD})=[ABCD]/(s_{ABD}+s_{BCD})$](http://latex.artofproblemsolving.com/1/4/8/14897920572603a667fcceda583a3a70e0c8e0c4.png)
, which implies that the sum of the semiperimeters of triangles and 
is the same as that of 
and 
.
.
, from which the others follow similarly. about the perpendicular bisector of 
to 
. Draw the semicircle 
with center as and radius 
, with lying on one of the diameters. Clearly should be on because 
. Suppose 
.
on the . The inradius of 
is given by
where 
, 
, both of which are constant, and 
.
intersect at 
. We claim that as a function of there is atmost one X such that 
where the function is 
as varies from ray 
to , excluding 
, (noting that the function is continuous and differentiable in the interval), for then, either the inradius increases from to the maximum and then decreases till or the inradius increases till . In either case, the inradius of 
is not the inradius of any 
, because the inradius strictly decreases in the second case, and in the first case because the radius can't decrease as much to equal the inradius of because the inradius of 
is more than that of as is inside , and due to the condition we prove, because otherwise it would increase after decreasing, which would violate the condition, for the slope of the tangent of the graph of the function will be 0 again after becoming 0 once. This analysis would imply that would not lie in the interval 
, and thus that and 
intersect outside the quadrilateral, which would be a contradiction, because it would violate the ordering we gave to the quadrilateral, and thus leading to a contradiction to our assumption that in the given interval such that . is equivalent to
Note that can't be in the interval ![$[\pi/2,\pi]$](http://latex.artofproblemsolving.com/e/5/4/e54d4ed06f8d2742d8c85c2920a0cbcdb2ecad2c.png)
because the LHS would otherwise be negative or 0, while the RHS is not at the same time. So the only such possibility is when 
.
, the equation becomes 
. Define 
. This is clearly continuous. When 
, 
, and 
but when 
, 
because then the triangle degenerates. When 
, 
, and 
So there exists an odd number of solutions for acute , because in the graph of 
, the places where it cuts the axis are precisely the points of change of sign. Since the equation on separating the expression with the radical on the other side and squaring yields a quartic expression which can have at most 4 real roots, there can be either 1 or 3 roots for which is acute. In particular, the expression is 
where the 
stand for some coefficients we don't have to work with. From here, we can see that 3 roots don't exist for which is acute, because when we expand the quartic, we observe that the sum of roots is -4, and the product of the roots is 1, from which if there were 3 real roots, then there would have been 4 real roots, as complex roots occur in conjugates, and at least 2 roots must be negative since product is 1 and sum is negative, which gives us 
2 roots for which is in the interval 
, which is a contradiction, which leads to 
and thus to the fact that 
due to the reflection. Similarly we show that the other pairs of triangles having the same base (as one of the sides of the quadrilateral) are congruent, and due to the transitivity of congruences, we conclude that all triangles are congruent. \ \ \ 
We can also prove the fact that is a rectangle using the conclusion of the solution.
here, we get that the diagonals are equal,lets call them x.
moves.
,
for all 
