[/quote]
,
satisfying
Prove
be a right angled triangle with 
.
be the midpoint of 
,
be an arbitrary point on 
.
over 
intersects lines 
and 
and 
,
and 
.
lies on 
.
and 
.
be the points on 
and 
,
.
and 
is on the circumference of triangle 
and 
,
.
past 
and point 
such that 
.
be on the opposite side of 
from 
such that 
and 
.
intersect the circumcircle of 
again at 
,
intersect the circumcircle of 
again at 
.
,
are collinear.
with the following property: the 
positive divisors of 
such that for 
,
is a perfect square.
such that there exists a sequence of positive real numbers 
,
,![\[
a_1 + \cdots + a_{n+1} < \alpha \cdot a_n.
\]](http://latex.artofproblemsolving.com/9/b/7/9b73deb46032844b324288b02470ff0623ac0ff1.png)
Proposed by Pavle Martinović
and 
such that![\[
\frac{1}{\gcd(x, y)} + \frac{3}{xy} + \frac{y}{\operatorname{lcm}(x, y)} = y,
\]](http://latex.artofproblemsolving.com/c/8/2/c828f866a97dbb38fe5b6d7b8d190b3684365938.png)
where 
denotes the greatest common divisor of 
denotes their least common multiple.
and satisfy ![\[ a^2+b^2+c^2 +abc = 4 . \]](http://latex.artofproblemsolving.com/d/3/a/d3ac01295b077425b5eaae556faa4cd678bce32a.png)
Show that ![\[ 0 \le ab + bc + ca - abc \leq 2. \]](http://latex.artofproblemsolving.com/2/0/9/209b3fcb56c01c335bbcb8d5fb781e28e49a0798.png)
that satisfies: for all different 
,
?
with the common external tangent lines of the circumcircles of triangles 
and 
,![\[\left(\frac{PA}{XY}\right)^2+\frac{PB\cdot PC}{AB\cdot AC}=1.\]](http://latex.artofproblemsolving.com/0/8/0/080a689aea002f1b997294cb6aa48d3334881be7.png)
Y by KevinYang2.71, OronSH, aliz, Pengu14, vincentwant, EpicBird08, llbaobao
Can someone explain to me how this is a zero and not a 5? I wrote the Vieta's equivalent of "two consec zero coefficients", which was worth 5 points
I messed up the numbering and I believe that is the underlying cause of the misgrade but if someone sees any other error, please let me know so I don't wrongly email MAA.
Update: I posted this while flipping out upon seeing a zero on my P2 wanting to find a way to somehow appeal - it genuinely felt like 24JMO4 all over again. Thankfully, this -5 did not game-end my score this year
I messed up the numbering and I believe that is the underlying cause of the misgrade but if someone sees any other error, please let me know so I don't wrongly email MAA.
Update: I posted this while flipping out upon seeing a zero on my P2 wanting to find a way to somehow appeal - it genuinely felt like 24JMO4 all over again. Thankfully, this -5 did not game-end my score this year
Attachments:
This post has been edited 6 times. Last edited by Mathandski, Apr 23, 2025, 12:01 AM
