[/quote]
,
of positive integers, with 
prime, such that 
is a perfect square.
be an integer. Two players, Alice and Bob, play the following game on the complete graph 
:
,![\[
\left( \frac{1}{4} - \varepsilon \right) n^2 + n.
\]](http://latex.artofproblemsolving.com/a/9/c/a9cca00e87c643af32eb75a2501b59060f3504fb.png)
,
(where 
)
,
with 
vertices that contains a Hamiltonian cycle 
and has more than 
edges. For which lengths 
must the graph 
?
be a given integer, 
.
vertices satisfying the condition: for any two non-adjacent vertices 
and 
in graph 
.
satisfying the inequality 
,
?
and graph 
?
,
(i.e., a Hamiltonian path)?
be a set of 2025 positive real numbers. For a subset 
,
as the median of 
when all elements of 
.![\[
P(A) = \sum_{\substack{T \subseteq A \\ |T| \text{ odd}}} M_T, \quad Q(A) = \sum_{\substack{T \subseteq A \\ |T| \text{ even}}} M_T.
\]](http://latex.artofproblemsolving.com/1/d/f/1df2ea8487289111939a53dcc49576fbe6fdef37.png)
Find the smallest real number 
such that for any set ![\[
P(A) - Q(A) \leq C \cdot \max(A),
\]](http://latex.artofproblemsolving.com/c/d/e/cdea343d11c722a423e857aeb01202863405773e.png)
where 
denotes the largest element in 
,
,
number. If a positive integer leaves a remainder that belongs to 
number.
be the Euler's totient function.
,
such that for any 
,
and 
,
is a non-negative power of 
?
,
and 
satisfying:![\[
\varphi(k_i) \in \left ( \frac{x}{a} ,x \right ], i = 1,2; \quad \varphi(k_1) \neq \varphi(k_2).
\]](http://latex.artofproblemsolving.com/6/c/b/6cbe59f15b51439d1202267f521b573a52f1fe67.png)
And these two different 
correspond to the same 
and 
as described in (1), yet 
.
as the number of elements in set 
.
and 
.
with 
.
such that for all 
,
.
such that, for all integers 
that satisfy 
,![\[f(a)^2+f(b)^2+f(c)^2=2f(a)f(b)+2f(b)f(c)+2f(c)f(a).\]](http://latex.artofproblemsolving.com/a/9/e/a9edf1a3207d29facc970b6ee288d4fabdd43a4b.png)
denotes the set of integers.)
be a 
board. Some of its cells are colored black in such a way that every 
board of 
black cells. Find the maximum amount of black cells that the board may have.
such that 
is unbounded and![\[2f(m)f(n)-f(n-m)-1\]](http://latex.artofproblemsolving.com/0/f/7/0f77b5addbf8f5e603eb1b6923cdac795cd132e9.png)
is a perfect square for all integer 
Y by Davi-8191, dangerousliri, itslumi, centslordm, Adventure10, Mango247, ItsBesi, Rounak_iitr, and 2 other users
Let 
be a cyclic quadrilateral whose diagonals 
and 
meet at 
. The extensions of the sides 
and 
beyond 
and 
meet at 
. Let 
be the point such that 
is a parallelogram, and let 
be the image of under reflection in . Prove that 
are concyclic.
be a cyclic quadrilateral whose diagonals 
and 
meet at 
. The extensions of the sides 
and 
beyond 
and 
meet at 
. Let 
be the point such that 
is a parallelogram, and let 
be the image of under reflection in . Prove that 
are concyclic.This post has been edited 2 times. Last edited by v_Enhance, Sep 7, 2013, 2:19 AM
Reason: pint -> point
Reason: pint -> point
and 
are isogonal wrt angle 
.
,![\[\angle FHD=\pi - \angle HFD -\angle HDF=\pi -\angle ZFC -\angle ZCF=\pi - \angle DGF.\]](http://latex.artofproblemsolving.com/f/b/1/fb1a624a8676daae33f946d2957525c4f5f721c0.png)
![\[\angle FHD=\pi - \angle HFD -\angle HDF=\pi - \angle HFD -\angle ZFC -\angle ZCF=\pi - \angle DGF.\]](http://latex.artofproblemsolving.com/2/a/a/2aacf2d7dc47d04af498b8f9ce92de218888ed47.png)
,
by the properties of cyclic quads and reflections?
.
be their isogonal conjugates in this triangle 
implies that 
and similarly 
lie on the perpendicular bisector of 
.
,
be the midpoint of arc 
and 
the midpoint of the other arc ![\[
2\cdot \angle DCX=180-\angle DFC=\angle BDC+\angle BCA+\angle BCD=\angle FCE+\angle FCG=\angle DCE^{*}+\angle DCG^{*}
\]](http://latex.artofproblemsolving.com/0/2/4/024016691685ca3294c69893a12386d2bd273273.png)
and 
are the bisectors of 
.
is harmonic, so 
are isogonal as desired. 
.
,![\[
180-\angle FED=\angle DFE+\angle ADB=\angle PFE+\angle FCP=\angle CPG=\angle FGD
\]](http://latex.artofproblemsolving.com/a/2/4/a24197fc684b82d435cae0bb2a5e1e7bdba63477.png)
and 
,
by AA similarity. Thus, 
.
and 
,
by AA similarity. Therefore, 
.
,![\[ \angle{FAE}=180-\angle{DAC}=180-\angle{DBC}=\angle{DBF}=\angle{GCF}\]](http://latex.artofproblemsolving.com/b/d/a/bda990979cae501d6867eb416ce25e7daba59804.png)
by SAS similarity.
and 
.
by SSS congruency, so 
.
and 
.
is cyclic, and because 
,![\[\angle{GHF}=180-\angle{XHF}=180-\angle{XAF}=\angle{DAX}=\angle{GDF},\]](http://latex.artofproblemsolving.com/c/8/5/c858b5c152daf195f74fb2090aefbd6b008ce527.png)
is cyclic. //
.
.
.
.
so there exists a spiral similarity centered at 
.
.
is cyclic.
by parallel lines. So we are done.
and 
intersect at 
.
to be cyclic because 
.
.
cyclic and 
.
is on the circumcircle of 
and 
.
.
,
is cyclic.
and 
lies on the line 
such that 
.
cyclic and 
,
are parallelogram.
are concyclic. 
![[asy]
import graph; size(10cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -13.56, xmax = 25.44, ymin = -5.49, ymax = 13.49; /* image dimensions */
pen ffxfqq = rgb(1,0.5,0); pen qqwuqq = rgb(0,0.39,0); pen xdxdff = rgb(0.49,0.49,1);
/* draw figures */
draw((-3.236895072206104,1.716522681043244)--(2.195238943204361,2.358939626736120));
draw((2.055826808152264,0.02425299063591571)--(-3.236895072206104,1.716522681043244));
draw((-0.4580563678168402,2.045154682386091)--(-0.6131947965902200,0.8776332488183485));
draw((2.195238943204361,2.358939626736120)--(2.055826808152264,0.02425299063591571));
draw((-0.4580563678168402,2.045154682386091)--(2.055826808152264,0.02425299063591571));
draw((-0.6131947965902200,0.8776332488183485)--(2.195238943204361,2.358939626736120));
draw((2.195238943204361,2.358939626736120)--(3.893636125792478,0.9936042076656184));
draw((3.893636125792478,0.9936042076656184)--(2.055826808152264,0.02425299063591571));
draw(circle((0.9800000000000012,1.260000000000001), 1.638436447437424));
/* dots and labels */
dot((2.195238943204361,2.358939626736120),dotstyle);
label("$D$", (2.256494727886761,2.450823303759719), NE * labelscalefactor);
dot((-0.4580563678168402,2.045154682386091),dotstyle);
label("$A$", (-0.3928179596270162,2.144544380347722), NE * labelscalefactor);
dot((-0.6131947965902200,0.8776332488183485),dotstyle);
label("$B$", (-0.5459574213330147,0.9653705252115314), NE * labelscalefactor);
dot((2.055826808152264,0.02425299063591571),dotstyle);
label("$C$", (2.118669212351362,0.1231034858285383), SE * labelscalefactor);
dot((-3.236895072206104,1.716522681043244),dotstyle);
label("$F$", (-3.179956162676191,1.807637564594524), NE * labelscalefactor);
dot((0.3574296255641460,1.389588409706417),dotstyle);
label("$E$", (0.4188211874147766,1.486044695011927), N * labelscalefactor);
dot((3.893636125792478,0.9936042076656184),dotstyle);
label("$G$", (3.956342752823345,1.087882094576330), NE * labelscalefactor);
dot((0.1820146103966589,2.872858546171026),dotstyle);
label("$H$", (0.2503677795381781,2.971497473560114), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]](http://latex.artofproblemsolving.com/e/2/d/e2d19445eaa35b21da904bc6a0bef2a2d7b15b55.png)
Similarly, 
Hence, 
Also, since 
and 
(we could also get this by power of a point, but this is basically how PoP is proved). Combining them, 
Now, note that 
so 
Similarly, 
Finally, note that 
so we're done. 
which is true since 
.
.
.
.
.
so 
are concyclic indeed.
Thus, we have 
Then 
done.
be the reflection of 
.
is a parallelogram. Now, notice that we have
and thus 
is in fact also a parallelogram. Now, this means
and thus 
must be cyclic. But now, note that 
is in fact congruent to quadrilateral 
as necessary). Thus, in turn 
,
.
So opposite angles in quadrilateral 
implying the points are concyclic 
from 
and 
.![\[\measuredangle FGD = \measuredangle FEB = \measuredangle FED = \measuredangle FHD. \quad \blacksquare\]](http://latex.artofproblemsolving.com/6/b/d/6bd34ac05f9bd7057f00410d707f3bdbf5955b46.png)
![\[\angle FDG = 180^{\circ} - \angle DAC = 180^{\circ} - \angle DBC = \angle FBE.\]](http://latex.artofproblemsolving.com/5/c/5/5c5fcb1d1a47cef176ac4a166478facd21e7cad1.png)
Furthermore,
,![\[\angle FGD = \angle FEB = 180^{\circ} - \angle FED = \angle FHD, \]](http://latex.artofproblemsolving.com/3/f/4/3f4c79f553def0458adc0fa51fc0f9d00ae2ac86.png)
hence 
H
F
G
and 
so 
.
,
is cyclic.
and
the conclusion follows by SAS.
.
.![\[ \angle FAE = \angle DAE = \angle DBC = \angle GCF, \]](http://latex.artofproblemsolving.com/a/0/1/a015ae45050c60578608a4a5b1b6b5fd4001d4aa.png)
and furthermore,![\[ \frac{FA}{FC} = \frac{AB}{CD} = \frac{AE}{ED} = \frac{AE}{CG}, \]](http://latex.artofproblemsolving.com/2/f/d/2fd3d21b029c19787eb4cf7620daa00587d2dac0.png)
so the claim is proven.
,
we have that 
(by a well known lemma). So we get
.
is a point such as 
is parallelogram. Then 
and 
so 
is parallelogram. Hence 
or 
lie on a circle. From this, we have 
or 
lie on a circle
and 
are similar and since 
and 
are similar too with literally the same exact common ratio
and 
so 
and 
are similar so 
is collinear so 
and since 
,
![\[180^{\circ}=\angle FHD+\angle FGD=\angle FED+\angle FGD=360^{\circ}-\angle DFE-\angle FDE-\angle GFD-\angle FDG.\]](http://latex.artofproblemsolving.com/d/5/8/d58637ccdc49aad614a79e4da824035c2c1cdeb8.png)
But note that![\[\angle FDE+\angle FDG=\angle FDE+\angle CDG+\angle EDC+\angle FCE=180^{\circ}-\angle F,\]](http://latex.artofproblemsolving.com/4/1/e/41e25fbe81234046c5928a9f731f2dad590c8759.png)
and 
by the First Isogonality Lemma, so we are done. 
.
and 
.
and 
are isogonal with respect to 
