[/quote]
,
. Prove that: 
,
be the orthocenter of it and 
be any point on the side 
.
are on the segments 
,
and 
are cyclic. The segments 
and 
intersect at 
is a point on 
such that 
is tangent to the circumcircle of triangle 
at 
and 
intersect at 
.
and 
and 
then prove 
be the set of positive real numbers.
such that there exists a function 
with the property:![\[
f(y^2f(x) + y + c) = xf(x+y^2)
\]](http://latex.artofproblemsolving.com/e/3/0/e30ee6d9f69cc43422a5c2b9813192a632d359cb.png)
for all 
.
be non-negative real numbers such that 
Prove that
and 
.
,![\[ a^{b+c} b^{c+a} c^{a+b} \leq 1 . \]](http://latex.artofproblemsolving.com/0/c/e/0ce097b0b507403c959c2992cae1217ecdd7e581.png)
be a finite sequence of positive integers. Prove that there exist nonnegative integers 
and 
such that 
holds for all integers 
be a positive integer. Dominoes are placed on a 
board in such a way that every cell of the board is adjacent to exactly one cell covered by a domino. For each 
or 
.
of positive real numbers that satisfy the system:![\[
\begin{aligned}
11bc - 36b - 15c &= abc \\
12ca - 10c - 28a &= abc \\
13ab - 21a - 6b &= abc.
\end{aligned}
\]](http://latex.artofproblemsolving.com/1/7/0/170e1cfaee90f5c37428fa75192ac889261a8f83.png)
be positive reals. Prove that there is a cyclic permutation 
of 
such that the inequality:![\[
\frac{a}{xa + yb + zc} + \frac{b}{xb + yc + za} + \frac{c}{xc + ya + zb} \geq \frac{3}{x + y + z}
\]](http://latex.artofproblemsolving.com/5/5/3/553527fc40a1b7d29dbdfb4b5e63635308f46c85.png)
and 
.
.
,
.
with diameter 
,
at a point 
.
meets line 
at 
.
and 
intersect at a point 
on 
.
Prove that: 
is the sum of 
square numbers
Let 
and 
is the sum of 
is the sum of 
Y by Pluto1708, GammaBetaAlpha, itslumi, Adventure10, Rounak_iitr
Let 
be the orthocenter of acute-angled triangle 
and 
be the midpoint of 
. Line 
cuts 
at 
. Consider point 
such that 
is the perpendicular bisector of 
. Segments 
and 
intersect at 
. Show that 
is perpendicular to .
Proposed by Germán Puga
be the orthocenter of acute-angled triangle 
and 
be the midpoint of 
. Line 
cuts 
at 
. Consider point 
such that 
is the perpendicular bisector of 
. Segments 
and 
intersect at 
. Show that 
is perpendicular to .Proposed by Germán Puga
This post has been edited 1 time. Last edited by plagueis, May 27, 2020, 7:12 AM
be reflection of 
wrt ![\[AM\cdot AA'=\tfrac{AH}{2}\cdot 2\cdot AT=AH\cdot AT=AD\cdot AC\]](http://latex.artofproblemsolving.com/3/4/7/347badecb147910ca0cef9dd27c61db31bc942d8.png)
.Hence 
is cylic.Also clearly 
is a cylic trapezium.Thus ![\[\angle{FCD}=\angle{MCD}=\angle{MA'D}=\angle{AED}=\angle{FED}\]](http://latex.artofproblemsolving.com/4/f/7/4f77c35aebbe5fa56846a5461590fd83aa2d2fbb.png)
.Hence 
is cylic.Hence 
.
as desired
![[asy]
import olympiad;
import geometry;
size(10cm);
pair O=origin;
pair A,B,C,D,E,T,AT,H,M,K,F;
A=dir(110);
B=dir(210);
C=dir(330);
D=foot(B,A,C);
T=foot(A,B,C);
AT=2*T-A;
H=orthocenter(A,B,C);
M=(A+H)/2;
K=foot(D,B,C);
E=2*K-D;
dot(A^^B^^C^^D^^E^^H^^AT^^T^^M);
draw("$A$",A,N);
draw("$B$",B,W);
draw("$C$",C,NE);
draw("$D$",D,NE);
draw("$E$",E,SW);
draw("$T$",T,NE);
draw("$H$",H,NE);
draw("$M$",M,NE);
draw("$A'$",AT,SW);
draw(segment(A,AT));
draw(circumcircle(A,B,C));
draw(circumcircle(B,D,C));
draw(circumcircle(M,D,C),dashed);
draw(circumcircle(A,D,E),dashed);
draw(A--B--C--A);
[/asy]](http://latex.artofproblemsolving.com/6/7/7/6771422aa5b241037dfb185b9fa9d7d4bdfa5521.png)
onto 
be 
respectively. Let 
. It suffices to prove that 
collinear. Its well known that 
are tangents in 
.
where 
is harmonic quaderilateral 
and we are done
are concurrent.
.
gives the desired result
collinear.
along with some trivial angle chase
intersect 
.
is cyclic
.
.
be the orthogonal projection of 
and 
.
.
.
too and they both belong on 
.
.
and 
and 
,
.
and 
.
to 
.
where 
.
so 
are colinear as desired.
is the harmonic.
.
and 
which leads to: ![\[DAM \sim DEC \implies DAE \sim DMC \implies \angle AED = \angle MCD \implies FDCE \text{ is cyclic} \implies BF \perp CM. \]](http://latex.artofproblemsolving.com/6/e/c/6ec999e53d6f74e3e7b64d81db720b78157708f5.png)
as the orthic triangle. Let 
be orthocenter WRT 
,
be foot from 
to 
and 
be reflection of 
and 
are cyclic, By POP, 
is also cyclic and Using the fact that, 
and 
are cyclic, By Radical Axes Theorem, 
the inversion centered at 
.
is cyclic with diameter 
,
be the reflection of 
we have 
,
.
.
coincides with its image, so 
,
be a point on 
is cyclic,we intend to show that 
therefore 
we definitely have 
collinear thus 
.
is bisected by 
.
.
and 
are parallel, we know that 
.
,
.
Therefore, we have that 
,
,
.
.
and 
be the perpendicular line to 
be the intersections of 
and 
and 
are orthogonal to 
.
is a parallelogram. Thus, 
.
.
is cyclic and 
.
,
is cyclic.
belong to a Simson line (perpendicular feet from 
)
.
.
is tangent to 
,![\[ \angle MAF=\angle MAE=\angle AED=\angle F'ED=\angle F'CD=\angle F'DM \]](http://latex.artofproblemsolving.com/5/e/c/5ec0bff00c2f0525e30eb4aa6c6d846695a9c062.png)
which means 
is cyclic. So 
.
are collinear. So 
.
,
,
.
so 
is cyclic and 
,
is cyclic and 
.
.
,
is cyclic. Then,
since 
as well, i.e. 
.
,
