It is well known that 
.
be positive numbers such that 
.![\[T = \dfrac{bc}{\sqrt{a}+3}+\dfrac{ca}{\sqrt{b}+3}+\dfrac{ab}{\sqrt{c}+3} \]](http://latex.artofproblemsolving.com/6/d/5/6d570d06679f17452790b90171428d645e97fe9f.png)
be the incenter of a triangle 
with 
,
be the midpoint of the arc 
of the circumcircle of the triangle. The perpendicular line to 
passing through 
at point 
.
intersects the circumcircle of triangle 
at point 
.
is tangent to the circumcircle of triangle 
such that there exist infinitely many rational numbers 
satisfying![\[
\left| \frac{q}{p} - \frac{\sqrt{5} - 1}{2} \right| < \frac{\alpha}{p^2}.
\]](http://latex.artofproblemsolving.com/1/1/6/1162c6a7215bff2f9fdaa042a6952f86524e9e92.png)
,
and circumcircle 
.
be the midpoints of sides 
,
,
and let 
be the tangency points of 
be the intersections of line 
with line 
and line 
,
be the midpoint of arc 
of 
.
bisects 
.
be the set of integers, and let 
be a function. Prove that there are infinitely many integers 
such that the function 
defined by 
is not bijective.
,
such that 
.
be the foot of the altitude from 
to 
,
be the reflection of 
.
intersects line 
.
lie on a circle 
and point 
and 
are tangent to 
are collinear, and (iii) 
.
bisects 
.
Y by MathLuis, Rounak_iitr
Let 
be an acute triangle with point 
lie on the plane such that 
is a parallelogram. 
is the orthocenter of 
intersects 
at 
and 
intersects 
at 
The circle with diameter 
intersects the circumcircle of 
again at 
Prove that: The circumcircle of 
passes through the reflection point of over 
Proposed by MathLuis
be an acute triangle with point 
lie on the plane such that 
is a parallelogram. 
is the orthocenter of 
intersects 
at 
and 
intersects 
at 
The circle with diameter 
intersects the circumcircle of 
again at 
Prove that: The circumcircle of 
passes through the reflection point of over 
Proposed by MathLuis
This post has been edited 1 time. Last edited by Noob_at_math_69_level, Dec 18, 2023, 7:40 PM
then 
has equal power with respect to 
,
are collinear, where 
is the reflection of 
then by Ceva-Menelaus, 
.
passes through the midpoint of 
-
,
is the reflection of 
is harmonic, and 
are collinear as desired. 
,
is the antipode of 
are collinear. 
is cyclic. If we prove that 
is concurrent at 
gives the desired result.
is 
.
is a parallelogram so 
.
are concurrent.
(well-known Ceva + Menelaus config) and 
.
,
is concurrent.
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where 
To prove this Notice the spiral similarity at 
and 
then do some ratio chasing to get the desired conclusion.
be the reflection of 
q
.
,
q
are concurrent and 
concur thus, power of point at the concurrency point finishes
and similarly 
.
and 
,
to 
.
bisects 
must bisect 
is the reflection of 
where 
,
concur. 
is the reflection of 
