[/quote]
,
and 
. Prove that
Where 
be defined by![\[
a_1 = 1, \quad a_{n+1} = a_n + \frac{1}{\sqrt[2024]{a_n}} \quad \text{for } n \geq 1, \, n \in \mathbb{N}
\]](http://latex.artofproblemsolving.com/8/8/f/88f423d87c08b20dc552703fcf30f2a8e9585902.png)
Prove that![\[
a_n^{2025} >n^{2024}
\]](http://latex.artofproblemsolving.com/a/6/8/a6806c817edb7ce29a399981137c1a3e93f63c63.png)
for all positive integers 
.
and 
Prove that
Where 
Find the range of values that can be taken by :
are positive reals.
![\[ a = b\cos C + c\cos B. \]](http://latex.artofproblemsolving.com/7/e/d/7ed699f18ede191e5b89bc79d7f0592294347818.png)
be an integer. In a configuration of an 
board, each of the 
cells contains an arrow, either pointing up, down, left, or right. Given a starting configuration, Turbo the snail starts in one of the cells of the board and travels from cell to cell. In each move, Turbo moves one square unit in the direction indicated by the arrow in her cell (possibly leaving the board). After each move, the arrows in all of the cells rotate 
counterclockwise. We call a cell good if, starting from that cell, Turbo visits each cell of the board exactly once, without leaving the board, and returns to her initial cell at the end. Determine, in terms of 
,
such that
for all reals 
satisfying 
is a connected graph with 
edges. Prove that it is possible to label the edges 
in such a way that at each vertex which belongs to two or more edges, the greatest common divisor of the integers labeling those edges is equal to 1.
belongs to at most one edge. The graph 
is connected if for each pair of distinct vertices 
there is some sequence of vertices 
such that each pair 
is joined by an edge of 
.
and 
. Prove that
Where 
be the set of real numbers. Determine all functions 
that satisfy the equation![\[f(x^2f(x+y))=f(xyf(x))+xf(x)^2\]](http://latex.artofproblemsolving.com/b/4/b/b4b1e22e937394550f13cfdab142ca5bf4a26aea.png)
for all real numbers 
and 
.
of positive integers is called central if for every positive integer 
terms of the sequence is equal to 
of positive integers such that for every central sequence 
there are infinitely many positive integers 
.
be reals such that 
. Prove that
Let 
. Prove that
and in 
steps and has to reach 
is valid but 
is not valid.
such that
holds for all 
.
Y by Davi-8191, tenplusten, anantmudgal09, OlympusHero, Adventure10, HWenslawski, Jc426, math_comb01, Mango247, Rounak_iitr, and 3 other users
In a convex quadrilateral 
, the diagonal 
bisects neither the angle 
nor the angle 
. The point 
lies inside and satisfies ![\[\angle PBC=\angle DBA\quad\text{and}\quad \angle PDC=\angle BDA.\]](//latex.artofproblemsolving.com/c/2/0/c20761f3eadd054958f40259f3d1c05f26279783.png)
Prove that is a cyclic quadrilateral if and only if 
.
, the diagonal 
bisects neither the angle 
nor the angle 
. The point 
lies inside and satisfies ![\[\angle PBC=\angle DBA\quad\text{and}\quad \angle PDC=\angle BDA.\]](http://latex.artofproblemsolving.com/c/2/0/c20761f3eadd054958f40259f3d1c05f26279783.png)
Prove that is a cyclic quadrilateral if and only if 
.Attachments:
This post has been edited 2 times. Last edited by djmathman, Aug 1, 2015, 2:53 AM
Reason: formatting
Reason: formatting
,
,
,
be the orthogonal projections of the point P on the lines AB, BC, CD, DA, and let 
,
,
,
be the reflections of P in these lines, or, equivalently, the reflections of P in the points 
is the line BD. Hence, the points 
, we have 
.
and 
,
.
.
.
and 
,
has the segment AP as diameter. Therefore, by the Extended Law of Sines, we have
,
.
.
along line 
has remained the same, so 
,
is the perpendicular bisector of 
are the symmetrics of 
respectively wrt 
is an isosceles trapezoid which has 
to 
and 
respectively on the circumcircle of 
and 
since they subtend equal arcs. The circle also gives us 
by equal arcs 
and 
.
since they are vertical, and 
by subtraction. The equal angles and two adjacent equal sides force 
and 
at 
and 
.
and triangles 
and 
are similar in ratio 
.
by subtraction, and triangles 
and 
are also similar in ratio 
and 
are similar in ratio 
.
is 
by hypothesis, so 
.
is cyclic, its reflection 
is also cyclic and the original quadrilateral 
. ]
to 
,
.
,
=
and by an analogous argument that 
.
so that by definition, 
.
,
be the associated intersection point for quadrilateral 
.
is cyclic, so you agree that 
.
,
,
while 
.
.
Let 
and 
Notice 
so 
and 
Similarly, 
Since 
it lies on the perpendicular bisector of 
which is equivalent to the perpendicular bisector of 
Construct 
on 
such that 
Similarly construct 
on 
such that 
Notice 
as 
are isogonal conjugates wrt 
so 
and 
Hence, 
and 
is cyclic isosceles trapezoid. Also, 
and 
so 
Similarly, 
is cyclic isosceles trapezoid and so 
is cyclic. 
intersect the circumcircle of 
Let 
Note that 
so 
This implies 
Similarly, 
is a cyclic trapezoid so 
Thus it must also lie on the perpendicular bisector of 
be reflections of 
Since 
is a reflection of 
over 
so 
is cyclic. Also, 
Thus, 
is cyclic. This implies 
so 
is isosceles trapezoid. Thus, 
so the reflection of 
is also on that circle. Thus, 
and 
and 
are both isosceles trapezoids and therefore 
(ii) Assume 
be a point such that 
,
which implies 
,
bisects 
.
be the circumcenter of 
is cyclic. Since 
,
and 
or 
is cyclic. However, the latter is absurd because 
and 
,
on 
to the corresponding point 
.
is cyclic, then define 
.
such that 
and 
,
so one direction is completed
.
.
both belong to the 
,
.
and thus 
is also cylcic. Hence 
is cyclic.
denote the reflection of 
and 
be 
and 
respectively, and the pedal triangle of 
.
,
,
are cyclic with diameters 
,
respectively.
and 
Now, because 
.
,
.
In addition, we have 
Thus, 
is cyclic if and only if 
.
It follows that 
is cyclic, as this is the harder portion.
and 
and 
.
and 
to be isogonal in this context.
and 
,
and 
excenters, respectively. Now comes the key claim:
and 
are both cyclic.
is cyclic, as the others are analogous. Observe that 
is the angle bisector of 
,
,
!
on 
,
.
bisects 
.
is the external angle bisector of 
,
.
,
.
and 
.
,
is an isoceles trapezoid, so 
.
.
,
is an isoceles trapezoid as well. This means that 
and 
share a perpendicular bisector.
and 
.
to meet the circumcircle of 
respectively.
so it follows that 
and 
which implies that 
is an isosceles trapezoid with legs 
we have 
so 
It follows that 
is an isosceles trapezoid with legs 
so 
is parallel with 
we have 
so 
Since 
so 
as desired. 
such that the fourth point is not outside the circumcircle of those three points.
then 
and let 
be the second intersections of the lines 
with 
(a property we proved in the necessity that it must be isosceles), but since 
intersects that circle at only two points, either 
is an isosceles triangle. We proceed when 
because the other case is analogous.
Since P lies on the perpendicular bisector of 
,
,
which implies that 
It follows that P lies on the perpendicular bisector of 
,
is an isosceles trapezoid (perpendicular bisectors are parallel). Therefore, since 
by SSS congruence, we have 
Hence, 
,
.
share a perpendicular bisector, and the result follows.
.
and 
similarly. The triangles 
and 
are congruent as 
,
.
,
is cyclic, and similarly 
is cyclic. It follows that all the points 
are concyclic, as needed.
and 
,
and 
.
and 
are isosceles trapezoids (with 
)
are collinear, as are 
,
which lies on the perpendicular bisector of 
by symmetry.
be points and 
such that 
are isogonal in 
,
etc. The isogonality condition is equivalent to
Upon expansion and simplification this means that any 
)
,
,
and 
.
,
.
is placed at 
,
so 
but 
.
are valid positions for 
, then construct 
is cyclic. Let 
intersect again a
.
be the midpoint of 
and 
the midpoint of 
are collinear.
.
which finishes.
.
are isogonaly conjugate with respect to three of the four angle, so 
means orange, 
means green and 
,![\[ \measuredangle PAC = \measuredangle ACP \iff \measuredangle BAC + \measuredangle CBD + \measuredangle DCA + \measuredangle ADB = 0. \qquad (\spadesuit). \]](http://latex.artofproblemsolving.com/7/e/5/7e57e142c8f1fa0c11c8588dfc18cc3f5af78fa0.png)
A careless reader would be forgiven for thinking that 
implies the problem or at least one direction, but it turns out the situation is more subtle. The second part analyzes the angle conditions more carefully and provides a complete proof.
.
.
,![[asy]
import geometry;
size(9cm); pair A = (-0.50099,1.69422); pair B = (-1.8,-0.6); pair C = (0.09182,-1.79600); pair D = (1.55802,-0.60384); pair P = (-1.04359,-0.19363);
draw(circumcircle(A,B,C), gray+dashed); filldraw(A--B--C--D--cycle, invisible, black); draw(B--D); draw(A--C);
draw(C--P--A, brown); draw(B--P--D, gray);
markangle(radius=15, n=2, P, A, C, red, StickIntervalMarker(1,1, red)); markangle(radius=15, n=2, A, C, P, red, StickIntervalMarker(1,1, red)); markangle(radius=10, n=3, A, P, B, deepgreen, StickIntervalMarker(1,2, deepgreen)); markangle(radius=10, n=3, C, P, D, deepgreen, StickIntervalMarker(1,1, deepgreen)); markangle(radius=12, n=1, B, A, C, blue); markangle(radius=12, n=1, C, B, D, blue); markangle(radius=12, n=1, D, C, A, blue); markangle(radius=12, n=1, A, D, B, blue);
dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$P$", P, dir(45));
[/asy]](http://latex.artofproblemsolving.com/0/e/c/0eca105f1f7fddffcbb253a96d5434767c875c18.png)
and 
in the claim (colored green with three rings) so that their only dependence on 
and 
(colored red with two rings), as follows: 
Since the claim says 
,
in terms of only 
These four latter angles are colored blue with one ring in the figure. This proves 
.![\[ \measuredangle BAC + \measuredangle CBD + \measuredangle DCA + \measuredangle ADB = 0 \]](http://latex.artofproblemsolving.com/b/8/6/b86d4b46e9417c6e66aa186a00663048a68a1057.png)
is equivalent to 
.
meets line 
.
meets line 
,
.
(that is, 
)
if and only if 
is not degenerate) and the second lemma turns our angle condition into 
and hence 
be the point for which 
is cyclic and harmonic; then 
.
(the corresponding Apollonian circle only meets line 
