[/quote]
,
and 
Prove that
Let 
Prove that
and 
Prove that
is given, find the largest real number 
such that there exists a geometric sequence 
with common ratio 
for all positive integers 
.
denotes the distance from the real number 
to the nearest integer.
,
.
.
is your favourite function, 
is your mother's favourite function, and 
is your father's favourite function. It was discovered that for any reals 
,
Prove that you are boring.
is 
,
,
,
are positive integers, and 
are odd primes.
to be 
,![$[x]$](http://latex.artofproblemsolving.com/b/c/e/bceb7b14e55d33a8bca29b7863ad3cdae95afce4.png)
is the largest integer no greater than 
,![$\{x\}=x-[x]$](http://latex.artofproblemsolving.com/6/1/0/61036a03415081af535bd57c408b7be79aae9458.png)
.
,
such as there exists an infinite geometric sequence with common ratio 
and prove 
,
.
such that 
.
corresponds to 
,
be the 
-
and 
are placed on line 
such that 
is between 
and 
,
be the 
,
be the reflection of the orthocenter of triangle 
be positive real variables. Prove that
Edit: Thank you, Nguyenhuyen-AG
![$$\iff \sum_{\mathrm{cyc}}\frac{(a-b)^2}{b}\ge \frac{1}{a^2+b^2+c^2}\cdot\left[a(b-c)^2+b(c-a)^2+c(a-b)^2+\frac{1}{2}(a+b+c)\big((a-b)^2+(b-c)^2+(c-a)^2\big)\right]$$](http://latex.artofproblemsolving.com/b/6/b/b6b4e84a550f4308cb4b9a5afe17ace9b808bc36.png)
![$$\iff \sum_{\mathrm{cyc}}(a-b)^2\left[\frac{1}{b}-\frac{a+b+3c}{2(a^2+b^2+c^2)}\right]\ge 0.$$](http://latex.artofproblemsolving.com/a/9/c/a9ca83aa0efd15cc70cd608655eec335ad303166.png)
I think someone will complete it by 
method.
has no solutions in integers 
,
.
Y by Davi-8191, mathematicsy, Adventure10, chessgocube, HWenslawski, ImSh95, Mango247, Amir Hossein
Let 
be the circumcenter and 
the orthocenter of an acute triangle 
. Show that there exist points 
, 
, and 
on sides 
, 
, and 
respectively such that ![\[ OD + DH = OE + EH = OF + FH\]](//latex.artofproblemsolving.com/a/8/7/a874af4cb802b9b341b055140a08e096aaf2112e.png)
and the lines 
, 
, and 
are concurrent.
be the circumcenter and 
the orthocenter of an acute triangle 
. Show that there exist points 
, 
, and 
on sides 
, 
, and 
respectively such that ![\[ OD + DH = OE + EH = OF + FH\]](http://latex.artofproblemsolving.com/a/8/7/a874af4cb802b9b341b055140a08e096aaf2112e.png)
and the lines 
, 
, and 
are concurrent.
.
equals 
.
are the intersection points between 
the reflections of 
across 
respectively. All we have to prove is that 
,
and 
are concurrent.
,
and 
,
is acute, so 
,
.
.
and 
,
and 
.
and 
,![\[ \dfrac {BD}{DC} = \dfrac { \sin B }{\sin C }.\]](http://latex.artofproblemsolving.com/4/1/b/41b575c3a62f879fe19923244df8edcf4426ed38.png)
and 
,
,
,
concur at the Gergonne point of 
.
be any line, and let 
and 
be any points on the same side of 
be the reflection of 
be the intersection of 
and 
.
by the triangle inequality, since 
,
,
,
,
,
hit 
hit 
hit 
lie on the circumcircle of 
,
is the circumradius of 
(since 
are the reflections of 
I chose to scale because I felt it was the most elementary solution; anyone who knows anything about ellipses should be able to understand the solution I gave.
![\[ \frac { BD } { \sin B } = \frac { A'D } { \sin \angle{DBA'} } \quad \wedge \quad \frac { DC } { \sin C } = \frac { A'D } { \sin \angle{DCA'} } \]](http://latex.artofproblemsolving.com/d/c/a/dca38f07efd63ab71a099e4490506565de8f17eb.png)
![\[ \sin \angle{DBA'} = \sin \angle{DCA'} \]](http://latex.artofproblemsolving.com/6/0/5/6059a6e389664ba1602680ca40a1637f7d039913.png)
I unfortunately can't read the entire problems. Only half of the picture is displayed. Do you know why? 
be the reflections of 
respectively. Let 
and 
.
,
.
,
are the points we are looking for.
,
.
,
.
is an isosceles trapezoid, implying that 
and 
.
and 
.
is cyclic, so 
and 
.
and 
,![\[ \frac{BD}{DC}=\frac{\frac{DD'\sin\angle{DD'B}}{\sin\angle{DBD'}}}{\frac{DD'\sin\angle{DD'C}}{\sin\angle{DCD'}}}=\frac{\sin B\cos B}{\sin C\cos C}=\frac{\sin2B}{\sin2C}\]](http://latex.artofproblemsolving.com/c/b/c/cbc066a21c7fe1bb5180e2af1a97a2527bc45033.png)
and 
.![\[\frac{BD}{DC}\cdot\frac{CE}{EA}\cdot\frac{AF}{FB}=\frac{\sin2B}{\sin2C}\cdot\frac{\sin2C}{\sin2A}\cdot\frac{\sin2A}{\sin2B}=1,\]](http://latex.artofproblemsolving.com/9/f/c/9fc4f0c16807ee5e2f21e7ab5398ea5367a2667e.png)
are concurrent. //
lie on an ellipse with focus 
.
is 
are points of tangency) or use a projective transformation taking the conic to a circle. Note that tangency is preserved for the latter since it is one-to-one). Now, since 
be the mirror images of 
and 
.
,
and 
.
.
is minimal, so the points 
,
,
at 
,
be the reflections of 
respectively. Let 
intersect 
.
,
.
,
.
and 
have the same circumradius, so 
Then, by Ratio Lemma, 
Taking the cyclic product of this over 
yields 1 as everything cancels, so by Ceva's, 
be the reflection of 
.
analogously. Now observe that we obviously have 
,
concur. For now, let 
be the ellipse with focii at 
.
,
.
in place of 
;
we used is that 
,
,
,
concur. By ellipse properties 
are the reflections of 
satisfy the length condition with a constant sum of 
.
So 
and analogously 
,
as desired. 
.
,![\[OD+DH=OD+DA'=OA'=R,\]](http://latex.artofproblemsolving.com/5/0/7/507540daa70c9da023b47267610d8e6f8cbf8116.png)
and similarly, we also find that ![\[\angle DHA'=\angle OAH=\angle BAC-\angle BAH-\angle OAC=a-(90-b)-(90-b)=180-(2b+a)=b-c,\]](http://latex.artofproblemsolving.com/c/c/8/cc8aa10ee4a5fafa8bf2c54ca9818c6d5964c06c.png)
by orthocenter properties. Additionally, we also get that 
,![\[\angle DHC=90-\angle DHA'-\angle DCH=c.\]](http://latex.artofproblemsolving.com/b/9/f/b9f77bee0cfc4facc018e89287231187be5c2aff.png)
Now using Law of Sines, we get that![\[BD=DH*\frac{\sin b}{\cos c},\]](http://latex.artofproblemsolving.com/c/0/2/c02fc3bd09ca6db8726f0e954cfc23d27013cc07.png)
and![\[CD=DH*\frac{\sin c}{\cos b},\]](http://latex.artofproblemsolving.com/8/c/a/8ca29ef1ef6f5bc8a1bad527ed2efb8b4bbaa561.png)
which gives that![\[\frac{BD}{CD}=\frac{\sin b\cos b}{\sin c\cos c}=\frac{2\sin b\cos b}{2\sin c\cos c}=\frac{\sin 2b}{\sin 2c}.\]](http://latex.artofproblemsolving.com/7/0/4/70486eb1aeeb0e297419c6c4a32a596fc9989ebd.png)
Similarly, we get that 
and 
.![\[\frac{BD}{CD}*\frac{CE}{EA}*\frac{AF}{FB}=1,\]](http://latex.artofproblemsolving.com/5/0/a/50ab301460b1a2a9751d4b2b441e0f6620938ef8.png)
which means that 
,![\[OD+DH = OD+DH_A = R,\]](http://latex.artofproblemsolving.com/d/c/5/dc5a0f4ba038e96d88062cd1c71ef9bce39085a3.png)
![\[\prod \frac{BD}{DC} = \prod \frac{BH_A \cdot \sin \angle BH_AD}{CH_A \cdot \sin \angle CH_AD} = \prod \frac{\sin
B \cos B}{\sin C \cos C} = 1. \quad \blacksquare\]](http://latex.artofproblemsolving.com/f/e/d/fed9f9800c1bc63b3ba77a34cf6e4f4ad03d73f0.png)
.
,
and 
.
is a reflection of H over BC, because of famous orthocenter property, similarly for 
and 
.
,
and 
.
,
and 
DO + DH = EO + EH = FO + FH = R. So it is only left to show that AD, BE and CF are concurrent. Now using the ratio lemma 
,
and 
it follows that AD, BE and CF are concurrent, which was what we needed to show 
in 
, 
and 
the reflections of the orthocenter 
, 
and 
with sides 
.![\[OD+DH = OD + DH_a = R \]](http://latex.artofproblemsolving.com/1/e/f/1ef77f7c366de6a1405a75cf9c09447d36d366cc.png)
where ![\[OD+DH = OE + EH = OF + FH = R\]](http://latex.artofproblemsolving.com/c/8/f/c8f9318f90bc43991b4947ccba0339261902fddc.png)
which implies that ![\[\measuredangle ODC = \measuredangle H_a DB = \measuredangle BDH\]](http://latex.artofproblemsolving.com/4/b/5/4b52dc1040a3ddb8eb943efd89ddbf6324b66d86.png)
which implies that side 
and 
altitudes respectively intersect the sides and the circumcircle at 
and 
.
.
which is constant, where 
and 
are isogonal lines wrt 
.
be the intersection of 
with the circumcircle we have,
which implies the isogonality. Now we can use Steiner's Ratio theorem and the law of sines to compute 
Applying a similar argument to the other points we obtain
and we're done.
and define 
and a similar expression holds for the other ratios 
But, 
so 
and we are done by Ceva's Theorem. QED
