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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
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Revision as of 21:58, 11 November 2022
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
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 opposite 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, .
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
See Also
2022 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 24 |
Followed by Last Problem |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.