[/quote]
,
such that 
and 
.
be a trapezoid with parallel sides 
.
and 
lie on the line segments 
and 
,
.
and 
on the line segment 
satisfying ![\[\angle{APB} = \angle{BCD}\qquad\text{and}\qquad \angle{CQD} = \angle{ABC}.\]](http://latex.artofproblemsolving.com/b/1/a/b1ae208955104081bafe221832509d9dd20eeb83.png)
Prove that the points 
and 
are concyclic.
,
,
with center 
and radius 
.
intersects the circle 
.
and 
.
.
.
to 
and 
as the reflections of 
be the circumcircle of 
.
with 
with 
can be inscribed within a circle.
![\[\sqrt{a+h_B}+\sqrt{b+h_C}+\sqrt{c+h_A}=\sqrt{a+h_C}+\sqrt{b+h_A}+\sqrt{c+h_B}\]](http://latex.artofproblemsolving.com/5/d/6/5d6881be29327bcdc3e7bd76c221968578c7f50b.png)
is isosceles.
be all the divisors of 
.
are all divisors for some 
(
)
be a sequence of positive real numbers that satisfies 
and 
for every natural number 
for every natural number 
?
be positive real numbers such that : 
. Prove that : 
be the set of positive integers. Find all functions 
such that![\[ m^2 + f(n) \mid mf(m) +n \]](http://latex.artofproblemsolving.com/2/f/4/2f409d1de993f1af8fd839bb8e9f87a57e1b8608.png)
and 
(ans- 
)
satisfying 
Prove that 
When does equality hold?
Y by samrocksnature, Adventure10, Mango247, Rounak_iitr
Cyclic quadrilateral 
has circumcircle 
. Points 
and 
are the midpoints of 
and 
, and 
and 
lie on 
and 
respectively such that 
passes through 
and 
. Let 
meet 
at 
. Denote 
as the circumcenter of 
. Line 
intersects and 
at 
and 
respectively. Suppose 
is a parallelogram. Prove that 
.
has circumcircle 
. Points 
and 
are the midpoints of 
and 
, and 
and 
lie on 
and 
respectively such that 
passes through 
and 
. Let 
meet 
at 
. Denote 
as the circumcenter of 
. Line 
intersects and 
at 
and 
respectively. Suppose 
is a parallelogram. Prove that 
.This post has been edited 6 times. Last edited by djmathman, Jul 1, 2019, 2:22 PM
Reason: formatting
Reason: formatting
lie on 
,
,
,respectively. 
with center 
first and then 
with diameters 
respectively, so the intersection of 
cannot lie on the circle with center 
,
)
are 
respectively. 
.
at 
respectively. Suppose 
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[/asy]](http://latex.artofproblemsolving.com/9/c/5/9c5135472eb846741f1230ee9e7f13283157e423.png)
be the orthocenter of 
lies on 
.
are equal, combine 
is a diameter of the 
p
.
lies on 
as well as 
extende
.
,
is the circumcenter of 
.
is a parallelogra
.
cyclic, combine 
bisects 
.
bisects 
,
collinear 
,
,
.
.
thus the circumradii of 
?
we have 
,
and 
are similar; triangles 
and 
are congruent. Efforts to prove these led me in a long chain of circular reasoning though. I even tried a homothetic shift from 
to 
and marked the images of 
of 
forms an isosceles trapezium; so we are done, if 
,
is a parallelogram. So, instead I tried marking 
!
.
at all.
;
first, where the angle formed with these two lines should be equal to 
as they are 
rotations. Then I drew 
.
!
line (which is 
in 
,
,
,
are reflections of 
intersecting on 
.
and 
are reflections across 
.
;
!
cyclic produces a simpler solution.
should be 
with reference triangle 
.
and 
where 
.
is a diameter of the nine-point circle ![\[\angle CMO = \angle CNO = 90^{\circ}\]](http://latex.artofproblemsolving.com/9/1/3/913582937f559c3c273045e5790d0f9fe6cb099f.png)
we have that 
.
and 
so 
to 
,
.
with diameter ![\[CM = CP = CN = CH.\]](http://latex.artofproblemsolving.com/8/b/1/8b158ff90f4cea01ec0c62fecc76dc75801aa6c3.png)
Most importantly, 
so 
.
since that would imply![\[\angle ASR = \angle BAC = \angle DAC = \angle ARS\]](http://latex.artofproblemsolving.com/1/2/b/12bc5251a0b6e6d55b034fb2246941a081883622.png)
which would finish.
be the reflection of 
are reflections of each other over ![\[\angle ES'F = 2\angle EHF = 2\angle MHN = \angle MCN = \angle BCD\]](http://latex.artofproblemsolving.com/7/2/2/722badf8c7ca427b4c6ae19797cdcbcf4a06c38f.png)
hence 
is actually cyclic and since 
by circumradius it actually holds that the line 
bisects 
.
is clearly a parallelogram so 
and we are done. 
,
,
,
And, let 
So, 
forms an orthocentric system
,
,
,
Next, as we know that 
have the same radius, say 
So, its clear that in 
we have 
as the diameter of it's NPC.
hence, 
;
,
as the foot of altitudes from 
respectively, ergo 
in 
We also observe that 
so, 
So, 
,
.
to be the center of 
As 
both are perpendicular to 
,
,
,
is a parallelogram, using 
,
so 
are concyclic. And as 
due to 
,
as the bisector of 
which is 
,
are collinear, whence 
Now, as 
So, 
.
,
(draw in median and use euler line), we have 
,
is the euler circle. Furthermore, since 
,
.
.
and let 
,
,
,
,
lies on a circle, this means 
,
,
lie on a circle, and 
bisects 
.
,
results in 
,
.
bisects ![\[\angle AQR = \angle CAQ = \angle CAB = \angle ARQ\]](http://latex.artofproblemsolving.com/7/1/a/71a1efb28cb2585ff1970cec299fe8cf7ad55625.png)
This implies 
,
be the point on the circle with diameter 
.
and 
we find that 
are collinear. Thus 
,
,
.
.
.
is the nine-point circle of 
,
.
,
.
,
implies that 
are collinear. Now note that the conclusion is equivalent to 
to 
where 
is diametrically opposite 
are collinear, which means that 
,
Let 
be the point on 
such that 
It is well known that 
is the reflection of 
Therefore, since 
circles 
and 
we know 
as needed. 
Then, 
goes to 
so we are done. 
respectively in 
and 
.
so quadrilateral 
is cyclic. In particular, 
bisects 
and thus passes through 
Since 
is parallel to the bisector of 
implying the desired statement.
and more importantly that 
immediately after drawing the sketch and well L567 carried me for the rest of the solution.
is the perpendicular bisector of 
is also perpendicular to 
is the nine-point circle of 
is cyclic and is also tangent to 
and 
are perpendicular to 
and consequently, 
and therefore 
be the center of 
is a parallelogram.
meaning that 
is cyclic and 
meaning that 
which coincides with the angle bisector of 
which implies 
which implies 
as 
as desired.
is collinear.
)
cyclic, and as 
bisect 
which implies 
is cyclic.
is a parallelogram.
as desired.
or 
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unitsize(0.8cm);
pair A, B, C, D, E, F, H, M, N, O, P, S, O1, Q, R;
O=(0,0);
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F=(4,0);
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R=extension(P,O,A,B);
O1=E+F-S;
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label("$S$", S, dir(90));
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label("$Q$", Q, SW);
label("$R$", R, dir(90));
[/asy]](http://latex.artofproblemsolving.com/d/2/a/d2af58d9ee5a77cb35b826eb8893fc211e2622e9.png)
.
and 
,
,
,
.
is a parallelogram, so it suffices to show 
,
implies 
,![[asy]size(12cm);
pair E = dir(142.15); pair F = dir(37.85); pair H = dir(275); pair P = E+H+F; pair M = extension(F, P, E, H); pair N = extension(E, P, F, H); pair C = midpoint(H--P); pair D = 2*N-C; pair B = 2*M-C; pair A = extension(B, E, D, F);
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draw(M--F, lightcyan); draw(N--E, lightcyan);
dot("$E$", E, dir(E)); dot("$F$", F, dir(F)); dot("$H$", H, dir(H)); dot("$P$", P, dir(P)); dot("$M$", M, dir(M)); dot("$N$", N, dir(N)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$B$", B, dir(B)); dot("$A$", A, dir(A)); dot("$O$", O, dir(O)); dot("$S$", S, dir(210));
/* TSQ Source:
E = dir 142.15 F = dir 37.85 H = dir 275 P = E+H+F M = extension F P E H N = extension E P F H C = midpoint H--P D = 2*N-C B = 2*M-C A = extension B E D F
E--F--H--cycle 0.1 lightgreen / deepgreen O = midpoint E--F circumcircle C M N 0.1 lightred / red CP O C blue circumcircle P E F blue A--B--C--D--cycle 0.1 lightcyan / deepcyan S = circumcenter P E F R210
M--F lightcyan N--E lightcyan
*/
[/asy]](http://latex.artofproblemsolving.com/1/9/d/19d6c327ad5a6f03387d9647f8d7f94207f6b039.png)
,
,
,
and 
)
bisects 
,
.![\[ \measuredangle EAF = \measuredangle BAD = \measuredangle BCD = \measuredangle MCN = \measuredangle MON = 2 \measuredangle FHE \]](http://latex.artofproblemsolving.com/9/9/4/994aaf1cd9c44381df364d92e6ef5165fc930af6.png)
so 
.
,
bisects 
are collinear. Since 
is a parallelogram, this completes the proof. 
and 
,
be the reflection of 
and 
,
is the nine-point circle of triangle 
.
,
is cyclic.
.
,
bisects 
by homothety at 
and desired result follows.
