2024 SSMO Team Round Problems/Problem 10
Problem
The side-lengths of a convex cyclic quadrilateral 
are integers and 
. Find all possible values of the perimeter of .
Solution
Let 
From Ptolemy's Theorem, we have 
So, the equation is equivalent to
\begin{align*}
(ad+bc)^2 &= (ac+bd)^2-72\implies\\
72&=a^2c^2+b^2d^2-a^2d^2-b^2c^2\implies\\
72&=(a^2-b^2)(c^2-d^2).
\end{align*}
WLOG, assume that 
Since the difference of two perfect squares cannot be 
we have 
Clearly, 
has no solutions, as 
For 
we have 
and 
This means 
Finally, for 
we have 
and 
This gives 
It is easy to verify that all three of these possibilities work, giving an answer of ![\[(7+5+2+1)+(5+1+2+1)+(3+1+5+4) = 15+9+13 = \boxed{37}.\]](http://latex.artofproblemsolving.com/3/7/3/37315c670a3766287d7eb4bc8e80bd25e28f351b.png)
~SMO_Team
