[/quote]
,
and 
on a given circle meet at point 
.
and let 
meet the circle again at 
parallel to 
intersects 
.
is parallel to 
.
,
be positive real numbers, such that 
.![\[\frac{x^3}{2 + x} + \frac{y^3}{2 + y} + \frac{z^3}{2 + z} \ge 1.\]](http://latex.artofproblemsolving.com/b/e/5/be5819a67c3cd78f2dea35fdccf48688c720ce3c.png)
When does the equality hold?
of positive integers such that ![\[ k!=(2^n-1)(2^n-2)(2^n-4)\cdots(2^n-2^{n-1}). \]](http://latex.artofproblemsolving.com/3/5/3/353bbb15bac205053c26208af8cac7c32a296f55.png)
Proposed by Gabriel Chicas Reyes, El Salvador
![\[\frac{\operatorname{lcm}(x, y) + \operatorname{lcm}(y, z)}{\operatorname{lcm}(x, z)}\]](http://latex.artofproblemsolving.com/1/e/8/1e8b3e2d9a4d0917ec622449b058702700542f22.png)
for positive integers 
,
,
be prime, let 
be a positive integer, show that ![\[ \gcd\left({p - 1 \choose n - 1}, {p + 1 \choose n}, {p \choose n + 1}\right) = \gcd\left({p \choose n - 1}, {p - 1 \choose n}, {p + 1 \choose n + 1}\right). \]](http://latex.artofproblemsolving.com/9/d/4/9d409ce186320241eb358b5fbbb8a5a66120899c.png)
.
lie in such a way that 
.
and 
be the projection of 
is half as long as 
be an isosceles triangle with 
and circumcircle 
.
.
be any point on 
with center 
intersects 
and 
,
,
,
.
such that for every 
![\[
f_n(x) = \frac{x}{1 + nx}, \quad \text{if and only if } f(x) = \frac{x}{1 + x}, \quad \forall x \geq 0.
\]](http://latex.artofproblemsolving.com/0/f/c/0fc36d9264eb7e103128c489aeae521a859c1fd4.png)
(For 
represents 
.
be the incentre of a non-equilateral triangle 
be the 
be the reflection of 
be the reflection of line 
in 
.
,
and line 
analogously. Let 
where 
.
.
such that 
is a perfect square and 
is a prime number.
,
,
,![\[ \frac{a^3}{bc} + \frac{b^3}{ca} + \frac{c^3}{ab} \geq a+b+c \]](http://latex.artofproblemsolving.com/9/f/a/9fa6cca20b3620fbb32ee782196e96f6fffa0063.png)
such that for any real value of 
we have:
Y by GeoKing, Rounak_iitr
Let 
and 
be the circumcenter and orthocenter of a scalene triangle 
, respectively. Let 
be the intersection point of the lines 
and 
. Suppose the line 
meets the side at 
. Let 
and 
be the second intersection points of the circumcircles of 
and 
with the circumcircle of 
, respectively. Show that the four points 
and lie on a circle.
and 
be the circumcenter and orthocenter of a scalene triangle 
, respectively. Let 
be the intersection point of the lines 
and 
. Suppose the line 
meets the side at 
. Let 
and 
be the second intersection points of the circumcircles of 
and 
with the circumcircle of 
, respectively. Show that the four points 
and lie on a circle.This post has been edited 1 time. Last edited by a_507_bc, May 3, 2024, 10:18 AM
and 
Also, let 
be the midpoint of 
Note that 
is the nine-point circle of 
which we will denote with 
are collinear.
be the antipode of 
are collinear, so if 
intersects 
for a second time at point 
,
However, we know that 
so 
and we are done.
are collinear.
centered at 
Note that under this inversion, we have that 
and 
,
Hence, 
and 
.
Since 
,
Note that 
and in particular 
Therefore, 
is cyclic iff 
is cyclic. However, points 
all lie on the circle with diameter 
and we may conclude.
be the orthic triangle, let 
where 
are midpoints of 
,
inversion the queue points go to 
are collinear but this follows as 
are antipodes of 
and 
.
other then 
we have 
cyclic.
inversion about 
, 
and 
.
.
are cyclic 
cyclic
Let 
be the antipodes of 
on 
be the foot of the altitude from 
to 
.
is cyclic.![\[\frac{r+a+b+c}{2}=a\iff r=a-b-c\]](http://latex.artofproblemsolving.com/c/0/5/c051a406274e309a65d0de4a89bb599563e6f8df.png)
![\[s=\frac{\overline{(a+b+c)}(a-b-c)+(a+b+c)\overline{(a-b-c)}}{2\overline{(a+b+c)}}=\frac{(ab+bc+ca)(a-b-c)+(a+b+c)(bc-ab-ac)}{2(ab+bc+ca)}\]](http://latex.artofproblemsolving.com/3/8/3/383e4c5c5a7ca65986086aba5a692fb8272bc82f.png)
![\[s=\frac{-ab^2-abc-ac^2}{(ab+bc+ca)}\]](http://latex.artofproblemsolving.com/a/f/6/af66ef05323a993327c979578b110720881e2403.png)
![\[\frac{k-r}{k-s}.\frac{l-s}{l-r}=\frac{c-a}{-b+\frac{ab^2+abc+ac^2}{(ab+bc+ca)}}.\frac{-c+\frac{ab^2+abc+ac^2}{(ab+bc+ca)}}{b-a}\]](http://latex.artofproblemsolving.com/c/1/4/c146933f20d5e61696444ac8f8da2c3de745e0c3.png)
![\[=\frac{c-a}{b-a}.\frac{ab^2-bc^2}{ac^2-b^2c}=\frac{b(c-a)(ab-c^2)}{c(b-a)(ac-b^2)}\]](http://latex.artofproblemsolving.com/4/c/b/4cb2084c042395f4ce7ba514fef4f67c30a41312.png)
![\[\frac{b(c-a)(ab-c^2)}{c(b-a)(ac-b^2)}\overset{?}{=}\overline{(\frac{b(c-a)(ab-c^2)}{c(b-a)(ac-b^2)})}=\frac{(a-c)(c^2-ab)b}{(a-b)(b^2-ac)c}\]](http://latex.artofproblemsolving.com/a/c/d/acd7c2dba9886660414589d983fadc86dd17e313.png)
Which is true as desired
be the orthic triangle of 
.
and 
are the midpoints of 
is 
,
lies on line 
.
,
:
;
.
is cyclic which means 
.
is circle with diameter 
means 
at 
and 
be the point 
.
be a point on the line 
so that 
.
.
