[/quote]
,
be a random point on the smaller arc 
of the circumcircle of square 
,
be the intersection point of segments 
and 
.
to the circumcircle of the triangle 
are 
and 
, where 
is the center of the square. Prove that points 
,
,
and 
such that
such that: 
such that for every 
is satisfied? Explain the answer.
be an integer, and let 
be positive real numbers such that 
.![\[(1 + a_2)^2 (1 + a_3)^3 \dotsm (1 + a_n)^n > n^n.\]](http://latex.artofproblemsolving.com/7/7/f/77fb719e975225b59c64f1f3aa102c98a6c25f50.png)
colors, there exists a monochromatic regular hexagon.
is the smallest number such that if 
,
with two colors there exists a monochromatic arithmetic progression of length 
.
.
and 
.
points in plane. Prove that the number of isosceles triangles having their vertices among these 
.
.
and 
,
denotes the sum of all positive divisors of 
denotes the largest prime divisor of 
be a polynomial with real coefficients.
.
cannot have 2024 real roots.
and 
.
for all real numbers 
.
grid, one can find 
points such that no four of them are vertices of a square with sides parallel to lines of the grid. Imagine yourself as Erdos (!) and guess what is the best exponent instead of 
!
denote the set of positive integers. Find all functions 
satisfying the condition
for all 
be a scalene triangle, 
its medial triangle, and 
)
).
be the circumcevian triangle of 
be the pedal triangle of 
be the reflection in 
of 
.
,
cyclically.
be the reflection in 
of 
,
cyclically.
be the reflection in 
of 
,
cyclically.
,
,
and the orthocentroidal circle of 
Y by Amir Hossein, dave_mathsguru, 799786, itslumi, Adventure10, Mango247, fuzzball2023, and 1 other user
Let be given a triangle 
and its internal angle bisector 
. The line intersects the circumcircle 
of triangle at 
and 
. Circle 
with diameter 
cuts again at 
. Prove that 
is the symmedian line of triangle .
and its internal angle bisector 
. The line intersects the circumcircle 
of triangle at 
and 
. Circle 
with diameter 
cuts again at 
. Prove that 
is the symmedian line of triangle .
be midpoint of 
.
,
,
.
then:
,
,
,
(a.a), 
,
.
(thanks to 
,
(Combine 
)
is a harmonic quadrilateral. Therefore, 
,
is the circumcenter of 
.
the midpoint of 
.
is cyclic (let 
its circumcenter) . Then 
are collinear . So if the circumcircle of 
, then 
. So by polars 
is tangent to 
is the 
meets the circumcircle of 
, then 
. But also because 
we have that 
is the incenter of 
so 
and because 
we also have 
. This shows that 
is cyclic and so 
and we are done .
cuts 
at 
are incenter and B-excenter of 
is a circle with center 
The cross ratio 
is harmonic and 
is radical axis of 
Pairwise radical axes 
of 
meet at their radical center 
and 
is cyclic. Its circumcircle is the B-Apollonius circle of 
at 
. Denote the following points : the middlepoint ![$ [BC]$](http://latex.artofproblemsolving.com/3/5/5/3550468aa97af843ef34b8868728963dec043efe.png)
;
for which 
; the point 
; the 
-
,
; the point 
. Then 
, i.e. the quadrilateral 
is cyclically.
- the tangent in 
to 
. Observe that 
and 
.
and the line 
is the 
.
is harmonically. Since 
results 
.
(a.a.) . Thus, 
, i.e. 
.
,
, i.e. 
.
obtain 
and in consequence, 
. Since 
obtain 
, i.e. the quadrilateral 
.
.
the midpoint of segment 
meets 
,we have 
are collinear.
is cyclic and 
is cylic ,we have :
.
.
.
is harmonically,so 
are collinear.
and denote by 
the intersection of two tangent to 
,respectively.
lie on the polar of 
wrt 
,we obtain 
are collinear.
(where 
at a point 
such that 
.![\[ \angle{MBC} = \angle{MTA} = \angle{ABF}.
\]](http://latex.artofproblemsolving.com/2/3/9/239532c7002a5d2f2d9ff6e8e73e2630d0ff7ad5.png)
.
be 
.
must lie on line 
,
;
;
is the midpoint of arc 
is nothing but the midpoint of 
is the reflection of the median about the angle bisector, i.e. the symmedian. 
be the antipode of 
in 
.
,
.
lies on 
,
,
.
denote the circle with diameter 
.
at 
and 
.
followed by a reflection about 
,
.
and 
are isogonal, as desired. 
will be swapped, hence circle with diameter 
fixed, 
with center 
passes through 
Take 
so 
and 
collinear. DDIT on 
means that 
and 
are isogonal from existing isogonal pairs 
and 
.
concur clearly. This is as 
is projected to 
.
is harmonic conj of 
are collinear and 
is cyclic. Note that 
so 
are also collinear and we get:![\[
\angle FAD=\angle FAE = \angle FNE = \angle DNM=\angle DAM
\]](http://latex.artofproblemsolving.com/0/a/b/0ab0704c1156f3fab3189527e5b838287d79b6d2.png)
and we're done since 
,
is fixed, therefore 
2024 AIME I P10 
,
and 
is cyclic, it follows, that 
,
,
be 
is co line.
by angle bisector, so square 
is harmonic, so 
