AoPS Academy
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, 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.
such that ![\[f(xy-1)+f(x)f(y)=2xy-1,\]](http://latex.artofproblemsolving.com/8/8/8/888ca39f2b7f8cec6d6426bee28d40eade40a66e.png)
for any real numbers 
and 
![\[
f : \mathbb{R}^+ \to \mathbb{R}
\]](http://latex.artofproblemsolving.com/e/7/b/e7bfc5b236fb6f00ef328f850b1b14632cbf8416.png)
satisfying the condition that for all 
,![\[
f(x + y^2) \geq f(x) + y.
\]](http://latex.artofproblemsolving.com/3/a/a/3aa20083835682dbd81be692eba65cab19e923e5.png)
of positive integers with 
prime, 
, and 
, such that 
is a multiple of 
.
kawaii if it satisfies 
(
is the number of positive factors of , while 
is the number of integers not more than that are relatively prime with ). Find all that is kawaii.
a point in the space and 
the centroid of a tetrahedron 
. Prove that:
be a logarithmic spiral centered at the origin (ie curve satisfying for any point 
on it, line 
makes a fixed angle with the tangent to at ). Let 
be a rectangular hyperbola centered at the origin, scaled such that it is tangent to the logarithmic spiral at some point.
on the spiral, the polar of wrt. is tangent to the spiral.
be a monic polynomial so that there exists another real coefficients 
that satisfy![\[P(x^2-2)=P(x)Q(x)\]](http://latex.artofproblemsolving.com/6/9/7/697faa929e4fda7a6e9b1cd97849bd42ffc14306.png)
Determine all complex roots that are possible from that satisfies (a), determine whether should have real coefficients or not.
with exactly 9 elements that is subset of 
. Prove that there exist two subsets 
and 
that satisfy the following: and are non-empty subsets from , and are equal, and
is an empty subset.
-excenter of 
is 
and 
. The foot from 
to 
is 
and 
. The intersection of 
and 
is . The circumcenter of 
is .
is tangent to the -excircle if is on the incircle of .
with 
and 
. Lines and intersect at , and lines 
and 
intersect at 
. Let 
be the midpoints of sides 
, respectively. Let and 
be points on segment 
and 
, respectively, so that 
is the angle bisector of 
and 
is the angle bisector of 
. Prove that 
is parallel to 
if and only if 
divides into two triangles with equal area.
for which, for all 
, the following identity holds:![\[
f(x) f(yf(x)) + y f(xy) = \frac{f\left(\frac{x}{y}\right)}{y} + \frac{f\left(\frac{y}{x}\right)}{x}
\]](http://latex.artofproblemsolving.com/9/0/c/90c180110402e1a32b70edb2b0a03a28727457d1.png)
is chosen. Prove that there exists a prime number that divides infinitely many terms of the sequence 
, where![\[
b_k = a^{k^k} \cdot 2^{2^k - k} + 1.
\]](http://latex.artofproblemsolving.com/1/7/5/1751a60482d729a36c71b77ac9c978e724f40da0.png)
a number of 
distinct points, so that the distance (measured on the chord) between any two points (from the considered points) is a rational number?
, and consider a chord [a,b] on the circle. Then if there is a point with rational distance to a and to b, by ptolemy's theorem it has rational distance to every other point in the structure. (since the quadrilateral is cyclic)
, and represent a second point as ![$[\cos(t),\sin(t)]$](http://latex.artofproblemsolving.com/d/7/5/d75f3abd1173f85d678788a68a1722b0fc3eb518.png)
. Then we find that 
has to be rational, so there are an infinity of points at rational distance of that one point, more specifically all points with rational y-ordinate... but then I miss the rigorous argument to complete the proof... anyone able to finish?
s.t. 
are both rational. Call such angles good. By angle-sum formulas, if 
are good, then 
are also good.
on the circle 
, let 
be the angle subtended by . Now inductively construct points on s.t. all angles formed by them are good; for 1,2 take any good angle. If there are points chosen, pick a good angle and a marked point s.t. the point on with 
is distinct from the points. Since there are infinitely many good angles but finitely many marked points, such exists. For a previously marked point we have 
for suitable choices for the two 
. Since 
are both good, it follows that 
is good, which finishes induction by adding .
work: since 
for on the circle, it follows that is rational, and so we're done.
s.t. 
s.t. 
s.t. 
and 
was arbitrary , so for any no. of points we can have that distance between any two points is rational