2022 AMC 12A Problems/Problem 25
Problem
A circle with integer radius is centered at . Distinct line segments of length connect points to for and are tangent to the circle, where , , and are all positive integers and . What is the ratio for the least possible value of ?
Solution
Case 1: The tangent and the origin are on the opposite sides of the circle.
In this case, .
We can easily prove that \[ a + b - 2 r = c . \hspace{1cm} (1) \]
Recall that .
Taking square of (1) and reorganizing all terms, (1) is converted as \[ \left( a - 2 r \right) \left( b - 2 r \right) = 2 r^2 . \]
Case 2: The tangent and the origin are on the opposite sides of the circle.
In this case, .
We can easily prove that \[ 2 r - a - b = c . \hspace{1cm} (2) \]
Recall that .
Taking square of (2) and reorganizing all terms, (2) is converted as \[ \left( a - 2 r \right) \left( b - 2 r \right) = 2 r^2 . \]
Putting both cases together, for given , we look for solutions of and satisfying \[ \left( a - 2 r \right) \left( b - 2 r \right) = 2 r^2 , \] with either or .
Now, we need to find the smallest , such that the number of feasible solutions of is at least 14.
For equation \[ uv = 2 r^2 , \] we observe that the R.H.S. is a not a perfect square. Thus, the number of positive is equal to the number of positive divisors of .
Second, for each feasible positive solution , its opposite is also a solution. However, corresponds to a feasible solution if with and , but may not lead to a feasible solution if with and .
Recall that we are looking for that leads to at least 14 solutions. Therefore, the above observations imply that we must have , such that has least 7 positive divisors.
Following this guidance, we find the smallest is 6. This leads to the following solutions:
\begin{enumerate} \item , . \item , . \item , . \item , . \item , . \item , . \item , . \end{enumerate}
Therefore, .
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)