and 
be the midpoint of 
. Let the perpendicular bisector of 
,
intersect the lines 
and 
, respectively. We call the pair 
,
is cyclic.
and 
are interesting. Prove that 
for which 
and 
are both perfect squares.
denotes the sum of the positive integer divisors of 
Let 
be the least positive integer such that 
is divisible by 
for all positive integers 
be the set of integers. Find all functions 
such that ![\[xf(2f(y)-x)+y^2f(2x-f(y))=\frac{f(x)^2}{x}+f(yf(y))\]](http://latex.artofproblemsolving.com/9/f/d/9fdb11dbf102dfe0783b1d31466e276420e3913d.png)
for all 
with 
.
be the second intersection of the circumcircle of triangle 
with the internal bisector of the angle 
.
be the circumcenter of triangle 
and 
the orthocenter of triangle 
.
is equal to the circumradius of triangle 
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 
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 
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 
.
,
.
we obtain that 
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 
is the circumradius.
and 
are isogonal lines wrt 
.
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
,
.
similarly, now the first condition is satisfied so we only need to show that they are concurrent.
,![\[
\frac{BD}{CD}=\frac{\sin \angle B \cdot DH_A \cdot \sin \angle BDH_A}{\sin \angle C \cdot DH_A \cdot \sin 180-\angle BDH_A}=\frac{\sin \angle B}{\sin \angle C}
\]](http://latex.artofproblemsolving.com/a/c/7/ac7fb93f5ba0024c38e73325da9a88c92a95d5a6.png)
deriving the other ratios similarly we get that![\[
\frac{BC \cdot CE \cdot AF}{DC \cdot EA \cdot FB}=\frac{\sin \angle A \cdot \sin \angle B \cdot \sin \angle C}{\sin \angle A \cdot \sin \angle B \cdot \sin \angle C}=1
\]](http://latex.artofproblemsolving.com/5/e/0/5e0ef1700787c93d224b3df4983054163a51e0fc.png)
so we're done by Ceva.
