2024 SSMO Team Round Problems/Problem 9
Problem
Let 
be an equiangular octagon such that 
and 
. The radius of the largest circle that fits inside the octagon can be expressed as 
where \(a,b,\) and \(c\) are integers and \(c\) is squarefree. Find 
Solution
Note that 
and 
Thus, the diameter of the circle is the least distance between any pair of these opposite parallel sides. Note that a side of length 
oriented at a 
-degree angle relative to a pair of opposite parallel sides contributes 
to the distance and a side of length oriented perpendicularly contributes to the distance. Obviously, we only have to add up the contributions of one run of three sides; for example for 
and 
we only have to consider sides from 
to 
or 
to 
However, it doesn't matter which run we pick since opposite sides are congruent. Let the distances be 
and 
respectively, with the order given in the first sentence. We have \begin{align*}
a &= 10 + \frac{8}{\sqrt2} + \frac{12}{\sqrt2} = 10 + 10\sqrt2, \\
b &= 12 + \frac{10}{\sqrt2} + \frac{6}{\sqrt2} = 12 + 8\sqrt2, \\
c &= 6 + \frac{12}{\sqrt2} + \frac{8}{\sqrt2} = 6 + 10\sqrt2, \text{ and} \\
d &= 8 + \frac{6}{\sqrt2} + \frac{10}{\sqrt2} = 8 + 8\sqrt2. \end{align*} Obviously, 
and 
We also have 
since 
So, the least distance between pairs of opposite parallel sides is 
and the radius of the largest circle that fits inside the octagon is 
Thus, the answer is 
~SMO_Team
