2022 AMC 12A Problems/Problem 25
Contents
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 1
Suppose that with a pair the circle is an excircle. Then notice that the hypotenuse must be , so it must be the case that Similarly, if with a pair the circle is an incircle, the hypotenuse must be , leading to the same equation.
Notice that this equation can be simplified through SFFT to Thus, we want the smallest such that this equation has at least distinct pairs for which this holds. The obvious choice to check is . In this case, since has positive factors, we get pairs, and we get another two if the factors are or vice versa. One can check that for smaller values of , we do not even get close to possible pairs.
When , the smallest possible -value is clearly when the factors are negative. When this occurs, (or vice versa), so the mimimal is . The largest possible -value occurs when the largest of and are maximized. This occurs when the factors are and , leading to (or vice-versa), leading to a maximal of .
Hence the answer is .
~ bluelinfish
Solution 2
Case 1: The tangent and the origin are on the opposite sides of the circle.
In this case, .
We can easily prove that
Recall that .
Taking square of (1) and reorganizing all terms, (1) is converted as
Case 2: The tangent and the origin are on the same sides of the circle.
In this case, .
We can easily prove that
Recall that .
Taking square of (2) and reorganizing all terms, (2) is converted as
Putting both cases together, for given , we look for solutions of and satisfying with either or .
Now, we need to find the smallest , such that the number of feasible solutions of is at least 14.
For equation 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:
, .
, .
, .
, .
, .
, .
, .
Therefore, .
Solution 3
As is the length of the segment and ， . The equation for the line that passes through and is .
By the point-line distance formula from point to line
To find and for the smallest , we will list Pythagorean triples with relatively prime elements from the smallest. Notice that we only need to find triples with , the other triples will simply be . will not equal because then , meaning that , , and cannot all be integers.
The smallest triples are: , , , , , ,
Therefore, .
Video Solution by Math-X (Let's understand the question first)
https://youtu.be/7yAh4MtJ8a8?si=2dTtBDEZLIjN3YWS&t=9728 ~Math-X
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution by MOP 2024
~r00tsOfUnity
Video Solution by Steven Chen
Video Solution (5 Minutes)
~MathProblemSolvingSkills.com
See also
2022 AMC 12A (Problems • Answer Key • Resources) | |
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